answer stringlengths 4 6 | benchmark_version stringclasses 1
value | change dict | distractor_formula_ids listlengths 2 2 | effective_parameters unknown | family stringclasses 4
values | formula dict | id stringlengths 21 26 | is_gold bool 1
class | knowledge_distractor stringlengths 1.95k 3.54k | knowledge_domain stringlengths 530 1.23k | knowledge_formula stringlengths 381 1.03k | knowledge_formula_expr stringlengths 216 407 | knowledge_name stringlengths 269 491 | knowledge_wrong stringlengths 638 1.34k | level int64 1 4 | math_type stringclasses 50
values | mode stringclasses 4
values | parameters unknown | problem stringlengths 105 438 | prompts dict | solution stringlengths 321 682 | split stringclasses 1
value | subject stringclasses 17
values | template_id int64 0 2 | wrong_formula_id stringlengths 6 6 |
|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|
0.028 | cropmath-v1 | {
"new_value": 1.8,
"old_value": 1.65,
"parameter": "DVI"
} | [
"CMF012",
"CMF013"
] | {
"Biom_plant_yesterday": 131,
"Coef_Rm": 0.0005,
"DVI": 1.8,
"Tmn": 17.4
} | carbon_nitrogen_cycle | {
"expression": "0.0432*24*(Biom_plant_yesterday/0.45)*Coef_Rm*2^((Tmn-25)/10) / (if DVI>1: DVI^2; else: 1)",
"id": "CMF008",
"latex": "R_m=0.0432\\cdot24(Biom/0.45)C_m2^{(T-25)/10}",
"name": "植株维持呼吸",
"name_en": "carbon-nitrogen cycling plant maintenance respiration",
"precision": 3,
"unit": "g C/m2/day"... | CMF008_sensitivity_t2_0001 | false | Candidate formulas are listed below. Select the physiologically appropriate formula before calculating.
Candidate 1:
Formula name: carbon-nitrogen cycling plant maintenance respiration (植株维持呼吸)
Formula (LaTeX): R_m=0.0432\cdot24(Biom/0.45)C_m2^{(T-25)/10}
Formula: Rm = 0.0432*24*(Biom_plant_yesterday/0.45)*Coef_Rm*2^(... | Formula name: carbon-nitrogen cycling plant maintenance respiration (植株维持呼吸)
Formula (LaTeX): R_m=0.0432\cdot24(Biom/0.45)C_m2^{(T-25)/10}
Formula: Rm = 0.0432*24*(Biom_plant_yesterday/0.45)*Coef_Rm*2^((Tmn-25)/10) / (if DVI>1: DVI^2; else: 1)
Operator notes: * means multiplication, / means division, ** or ^ means powe... | Formula name: carbon-nitrogen cycling plant maintenance respiration (植株维持呼吸)
Formula (LaTeX): R_m=0.0432\cdot24(Biom/0.45)C_m2^{(T-25)/10}
Formula: Rm = 0.0432*24*(Biom_plant_yesterday/0.45)*Coef_Rm*2^((Tmn-25)/10) / (if DVI>1: DVI^2; else: 1)
Operator notes: * means multiplication, / means division, ** or ^ means powe... | Formula: Rm = 0.0432*24*(Biom_plant_yesterday/0.45)*Coef_Rm*2^((Tmn-25)/10) / (if DVI>1: DVI^2; else: 1)
Operator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.
Output unit: g C/m2/day | Relevant formula name: carbon-nitrogen cycling plant maintenance respiration (植株维持呼吸).
Formula: Rm = 0.0432*24*(Biom_plant_yesterday/0.45)*Coef_Rm*2^((Tmn-25)/10) / (if DVI>1: DVI^2; else: 1)
Operator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, ... | A retrieved formula is listed below. It may look relevant, but use it only if it truly matches the problem.
Formula name: carbon-nitrogen cycling soil carbon water factor (土壤碳分解水分因子)
Formula (LaTeX): F_W=0.49e^{3.88W-5.4W^2}
Formula: F_W_soil = if Soil_Water>=Saturate_Water: 0.65; else: 0.49*exp(3.88*(Soil_Water/BD)-5.... | 3 | Q10 with DVI correction | sensitivity | {
"Biom_plant_yesterday": 131,
"Coef_Rm": 0.0005,
"DVI": 1.65,
"Tmn": 17.4
} | In carbon-nitrogen cycling respiration, we analyze how carbon-nitrogen cycling plant maintenance respiration responds to parameter changes. Baseline: Biom_plant_yesterday=131.0 g C/m2, Coef_Rm=0.0005 g C/g C/s, Tmn=17.4 C, DVI=1.65 dimensionless. After changing DVI from 1.65 to 1.8, compute the new carbon-nitrogen cycl... | {
"C": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ..... | Identify the target quantity as carbon-nitrogen cycling plant maintenance respiration.
Use the verified formula:
\[R_m=0.0432\cdot24(Biom/0.45)C_m2^{(T-25)/10}\]
In calculation-friendly notation, this is:
Rm = 0.0432*24*(Biom_plant_yesterday/0.45)*Coef_Rm*2^((Tmn-25)/10) / (if DVI>1: DVI^2; else: 1).
Substitute the rel... | dev | carbon-nitrogen cycling respiration | 2 | CMF012 |
10.44 | cropmath-v1 | null | [
"CMF029",
"CMF035"
] | {
"Am": 53.7,
"FCO2": 1.27,
"FPA": 0.52,
"FT": 0.64,
"NNI": 0.46,
"WDF": 0.89
} | growth_yield | {
"expression": "Am * FCO2 * FPA * FT * min(NNI, WDF)",
"id": "CMF026",
"latex": "AMAX=Am\\cdot F_{CO2}\\cdot F_{PA}\\cdot F_T\\cdot\\min(NNI,WDF)",
"name": "实际最大同化速率",
"name_en": "Actual maximum assimilation rate",
"precision": 2,
"unit": "kg CO2/ha/h"
} | CMF026_direct_t1_0000 | false | Candidate formulas are listed below. Select the physiologically appropriate formula before calculating.
Candidate 1:
Formula name: High-temperature fertility factor (高温结实因子)
Formula (LaTeX): HTF=\frac{1}{1+e^{0.853(T_m-36.6)}}
Formula: HTF = if 32<=PDT<=39 and Tmax_day>36.6: 1/(1+exp(0.853*(Tmax_day-36.6))); else: 1
O... | Formula name: Actual maximum assimilation rate (实际最大同化速率)
Formula (LaTeX): AMAX=Am\cdot F_{CO2}\cdot F_{PA}\cdot F_T\cdot\min(NNI,WDF)
Formula: AMAX = Am * FCO2 * FPA * FT * min(NNI, WDF)
Operator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and ... | Formula name: Actual maximum assimilation rate (实际最大同化速率)
Formula (LaTeX): AMAX=Am\cdot F_{CO2}\cdot F_{PA}\cdot F_T\cdot\min(NNI,WDF)
Formula: AMAX = Am * FCO2 * FPA * FT * min(NNI, WDF)
Operator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and ... | Formula: AMAX = Am * FCO2 * FPA * FT * min(NNI, WDF)
Operator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.
Output unit: kg CO2/ha/h | Relevant formula name: Actual maximum assimilation rate (实际最大同化速率).
Formula: AMAX = Am * FCO2 * FPA * FT * min(NNI, WDF)
Operator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.
Output unit: kg CO2/ha/h | A retrieved formula is listed below. It may look relevant, but use it only if it truly matches the problem.
Formula name: CO2 response factor (二氧化碳影响因子)
Formula (LaTeX): F_{CO2}=1+\beta\ln(C_x/340)
Formula: FCO2 = 1 + beta * ln(Cx / 340)
Operator notes: * means multiplication, / means division, ** or ^ means power, exp... | 2 | Multiplicative with minimum stress | direct | {
"Am": 53.7,
"FCO2": 1.27,
"FPA": 0.52,
"FT": 0.64,
"NNI": 0.46,
"WDF": 0.89
} | Given Am=53.7 kg CO2/ha/h, FCO2=1.27 fraction, FPA=0.52 fraction, FT=0.64 fraction, NNI=0.46 fraction, WDF=0.89 fraction, calculate the Actual maximum assimilation rate for a photosynthesis. The result should be in kg CO2/ha/h. | {
"C": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ..... | Identify the target quantity as Actual maximum assimilation rate.
Use the verified formula:
\[AMAX=Am\cdot F_{CO2}\cdot F_{PA}\cdot F_T\cdot\min(NNI,WDF)\]
In calculation-friendly notation, this is:
AMAX = Am * FCO2 * FPA * FT * min(NNI, WDF).
Substitute the relevant values:
Am=53.7 kg CO2/ha/h, FCO2=1.27 fraction, FPA... | dev | Photosynthesis | 1 | CMF029 |
18.10 | cropmath-v1 | null | [
"CMF061",
"CMF062"
] | {
"Tbase": 11.5,
"Tmean": 29.6
} | environmental_response | {
"expression": "max(Tmean - Tbase, 0)",
"id": "CMF056",
"latex": "dTT=\\max(T_{mean}-T_{base},0)",
"name": "每日有效积温",
"name_en": "environmental-response daily thermal time",
"precision": 2,
"unit": "C day"
} | CMF056_direct_t0_0000 | false | Candidate formulas are listed below. Select the physiologically appropriate formula before calculating.
Candidate 1:
Formula name: environmental-response yield (环境响应产量)
Formula (LaTeX): Yield=Biomass\cdot HI
Formula: Yield = Biomass * HI
Operator notes: * means multiplication, / means division, ** or ^ means power, ex... | Formula name: environmental-response daily thermal time (每日有效积温)
Formula (LaTeX): dTT=\max(T_{mean}-T_{base},0)
Formula: dTT = max(Tmean - Tbase, 0)
Operator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value... | Formula name: environmental-response daily thermal time (每日有效积温)
Formula (LaTeX): dTT=\max(T_{mean}-T_{base},0)
Formula: dTT = max(Tmean - Tbase, 0)
Operator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value... | Formula: dTT = max(Tmean - Tbase, 0)
Operator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.
Output unit: C day | Relevant formula name: environmental-response daily thermal time (每日有效积温).
Formula: dTT = max(Tmean - Tbase, 0)
Operator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.
Output unit: C day | A retrieved formula is listed below. It may look relevant, but use it only if it truly matches the problem.
Formula name: environmental-response water stress response (环境响应水分响应)
Formula (LaTeX): F_W=\max(0,1-S_WARID)
Formula: F_Water = max(0, 1 - S_Water * ARID)
Operator notes: * means multiplication, / means division,... | 1 | Maximum linear | direct | {
"Tbase": 11.5,
"Tmean": 29.6
} | A environmental-response formula system calculation estimates environmental-response daily thermal time. The measured inputs are Tmean=29.6 C, Tbase=11.5 C. What is the simple daily thermal time? Give the final numeric value in C day. | {
"C": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ..... | Identify the target quantity as environmental-response daily thermal time.
Use the verified formula:
\[dTT=\max(T_{mean}-T_{base},0)\]
In calculation-friendly notation, this is:
dTT = max(Tmean - Tbase, 0).
Substitute the relevant values:
Tmean=29.6 C, Tbase=11.5 C.
The computed value is 18.10 C day.
Therefore, the fin... | dev | environmental-response formula system | 0 | CMF061 |
8.46 | cropmath-v1 | {
"new_value": 0.57,
"old_value": 0.52,
"parameter": "k"
} | [
"CMF034",
"CMF043"
] | {
"LAI": 1.98,
"PAR": 12.51,
"k": 0.5700000000000001
} | growth_yield | {
"expression": "PAR * (1 - exp(-k * LAI))",
"id": "CMF047",
"latex": "IPAR = PAR \\cdot (1 - e^{-k \\cdot LAI})",
"name": "截获光合有效辐射",
"name_en": "Intercepted photosynthetically active radiation",
"precision": 2,
"unit": "MJ/m2/day"
} | CMF047_sensitivity_t0_0001 | false | Candidate formulas are listed below. Select the physiologically appropriate formula before calculating.
Candidate 1:
Formula name: Harvest index (收获指数)
Formula (LaTeX): HI=\frac{Y_{grain}}{B_{above}}
Formula: HI = Y_grain / B_above
Operator notes: * means multiplication, / means division, ** or ^ means power, exp(x) m... | Formula name: Intercepted photosynthetically active radiation (截获光合有效辐射)
Formula (LaTeX): IPAR = PAR \cdot (1 - e^{-k \cdot LAI})
Formula: IPAR = PAR * (1 - exp(-k * LAI))
Operator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select t... | Formula name: Intercepted photosynthetically active radiation (截获光合有效辐射)
Formula (LaTeX): IPAR = PAR \cdot (1 - e^{-k \cdot LAI})
Formula: IPAR = PAR * (1 - exp(-k * LAI))
Operator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select t... | Formula: IPAR = PAR * (1 - exp(-k * LAI))
Operator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.
Output unit: MJ/m2/day | Relevant formula name: Intercepted photosynthetically active radiation (截获光合有效辐射).
Formula: IPAR = PAR * (1 - exp(-k * LAI))
Operator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.
Output unit: MJ/m2/day | A retrieved formula is listed below. It may look relevant, but use it only if it truly matches the problem.
Formula name: Harvest index (收获指数)
Formula (LaTeX): HI=\frac{Y_{grain}}{B_{above}}
Formula: HI = Y_grain / B_above
Operator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ... | 2 | Product with exponential saturation | sensitivity | {
"LAI": 1.98,
"PAR": 12.51,
"k": 0.52
} | A photosynthesis calculation involves sensitivity analysis. The current measurements are PAR=12.51 MJ/m2/day, k=0.52 dimensionless, LAI=1.98 m2/m2. A field technician asks what the intercepted photosynthetically active radiation would become if k changed to 0.57. Report the new final numeric value in MJ/m2/day. | {
"C": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ..... | Identify the target quantity as Intercepted photosynthetically active radiation.
Use the verified formula:
\[IPAR = PAR \cdot (1 - e^{-k \cdot LAI})\]
In calculation-friendly notation, this is:
IPAR = PAR * (1 - exp(-k * LAI)).
Substitute the relevant values:
PAR=12.51 MJ/m2/day, k=0.57 dimensionless, LAI=1.98 m2/m2.
T... | dev | Photosynthesis | 0 | CMF034 |
20.32 | cropmath-v1 | {
"new_value": 120,
"old_value": 113.3,
"parameter": "RM"
} | [
"CMF031",
"CMF050"
] | {
"DTGA": 194.1,
"RG": 54.8,
"RM": 120,
"b": 0.05
} | growth_yield | {
"expression": "(DTGA - RM - RG) / (1 - b)",
"id": "CMF033",
"latex": "GCR=\\frac{DTGA-RM-RG}{1-b}",
"name": "日干物质增量",
"name_en": "Daily dry matter increment",
"precision": 2,
"unit": "kg/ha/day"
} | CMF033_sensitivity_t2_0001 | false | Candidate formulas are listed below. Select the physiologically appropriate formula before calculating.
Candidate 1:
Formula name: Light interception fraction (光能截获率)
Formula (LaTeX): fPAR = 1 - e^{-k \cdot LAI}
Formula: fPAR = 1 - exp(-k * LAI)
Operator notes: * means multiplication, / means division, ** or ^ means p... | Formula name: Daily dry matter increment (日干物质增量)
Formula (LaTeX): GCR=\frac{DTGA-RM-RG}{1-b}
Formula: GCR = (DTGA - RM - RG) / (1 - b)
Operator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.
Output unit... | Formula name: Daily dry matter increment (日干物质增量)
Formula (LaTeX): GCR=\frac{DTGA-RM-RG}{1-b}
Formula: GCR = (DTGA - RM - RG) / (1 - b)
Operator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.
Output unit... | Formula: GCR = (DTGA - RM - RG) / (1 - b)
Operator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.
Output unit: kg/ha/day | Relevant formula name: Daily dry matter increment (日干物质增量).
Formula: GCR = (DTGA - RM - RG) / (1 - b)
Operator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.
Output unit: kg/ha/day | A retrieved formula is listed below. It may look relevant, but use it only if it truly matches the problem.
Formula name: Light interception fraction (光能截获率)
Formula (LaTeX): fPAR = 1 - e^{-k \cdot LAI}
Formula: fPAR = 1 - exp(-k * LAI)
Operator notes: * means multiplication, / means division, ** or ^ means power, exp(... | 2 | Subtraction and conversion | sensitivity | {
"DTGA": 194.1,
"RG": 54.8,
"RM": 113.3,
"b": 0.05
} | In respiration, we analyze how Daily dry matter increment responds to parameter changes. Baseline: DTGA=194.1 kg/ha/day, RM=113.3 kg/ha/day, RG=54.8 kg/ha/day, b=0.05 fraction. After changing RM from 113.3 to 120.0, compute the new Daily dry matter increment. Give the numeric value in kg/ha/day. | {
"C": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ..... | Identify the target quantity as Daily dry matter increment.
Use the verified formula:
\[GCR=\frac{DTGA-RM-RG}{1-b}\]
In calculation-friendly notation, this is:
GCR = (DTGA - RM - RG) / (1 - b).
Substitute the relevant values:
DTGA=194.1 kg/ha/day, RM=120.0 kg/ha/day, RG=54.8 kg/ha/day, b=0.05 fraction.
The computed val... | dev | Respiration | 2 | CMF031 |
0.040 | cropmath-v1 | {
"new_value": 0.257,
"old_value": 0.285,
"parameter": "WI"
} | [
"CMF020",
"CMF018"
] | {
"OMS": 298.4,
"SI": 0.6900000000000001,
"TI": 0.25,
"WI": 0.257
} | methane_emission | {
"expression": "WI * SI * TI * 0.003 * OMS",
"id": "CMF021",
"latex": "OMSC=WI\\cdot SI\\cdot TI\\cdot0.003\\cdot OMS",
"name": "土壤有机质碳分解量",
"name_en": "Soil organic matter carbon decomposition",
"precision": 3,
"unit": "g C/m2/day"
} | CMF021_sensitivity_t2_0000 | false | Candidate formulas are listed below. Select the physiologically appropriate formula before calculating.
Candidate 1:
Formula name: Redox potential factor (氧化还原电位因子)
Formula (LaTeX): F_{Eh}=1\;(Eh<-150),\quad e^{-1.7(1+Eh/150)}
Formula: FEh = if Eh<-150: 1; else: exp(-1.7*(1+Eh/150))
Operator notes: * means multiplicat... | Formula name: Soil organic matter carbon decomposition (土壤有机质碳分解量)
Formula (LaTeX): OMSC=WI\cdot SI\cdot TI\cdot0.003\cdot OMS
Formula: OMSC = WI * SI * TI * 0.003 * OMS
Operator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the... | Formula name: Soil organic matter carbon decomposition (土壤有机质碳分解量)
Formula (LaTeX): OMSC=WI\cdot SI\cdot TI\cdot0.003\cdot OMS
Formula: OMSC = WI * SI * TI * 0.003 * OMS
Operator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the... | Formula: OMSC = WI * SI * TI * 0.003 * OMS
Operator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.
Output unit: g C/m2/day | Relevant formula name: Soil organic matter carbon decomposition (土壤有机质碳分解量).
Formula: OMSC = WI * SI * TI * 0.003 * OMS
Operator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.
Output unit: g C/m2/day | A retrieved formula is listed below. It may look relevant, but use it only if it truly matches the problem.
Formula name: Fresh organic matter carbon decomposition (新鲜有机质碳分解量)
Formula (LaTeX): OMNC=WI\cdot SI\cdot TI\cdot0.027\cdot OMN
Formula: OMNC = WI * SI * TI * 0.027 * OMN
Operator notes: * means multiplication, /... | 2 | Multiplicative | sensitivity | {
"OMS": 298.4,
"SI": 0.6900000000000001,
"TI": 0.25,
"WI": 0.28500000000000003
} | In methane, we analyze how Soil organic matter carbon decomposition responds to parameter changes. Baseline: WI=0.285 fraction, SI=0.69 fraction, TI=0.25 fraction, OMS=298.4 g C/m2. After changing WI from 0.285 to 0.257, compute the new Soil organic matter carbon decomposition. Give the numeric value in g C/m2/day. | {
"C": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ..... | Identify the target quantity as Soil organic matter carbon decomposition.
Use the verified formula:
\[OMSC=WI\cdot SI\cdot TI\cdot0.003\cdot OMS\]
In calculation-friendly notation, this is:
OMSC = WI * SI * TI * 0.003 * OMS.
Substitute the relevant values:
WI=0.257 fraction, SI=0.69 fraction, TI=0.25 fraction, OMS=298.... | dev | Methane | 2 | CMF020 |
1.00 | cropmath-v1 | {
"new_value": 32.6,
"old_value": 29.6,
"parameter": "T"
} | [
"CMF028",
"CMF044"
] | {
"T": 32.6,
"Tb": 10,
"Tmax": 45,
"Tol": 24,
"Tou": 34
} | growth_yield | {
"expression": "if Tb<=T<Tol: sin(pi/2*(T-Tb)/(Tol-Tb)); if Tol<=T<Tou: 1; if Tou<=T<Tmax: cos(pi/2*(T-Tou)/(Tmax-Tou)); else: 0",
"id": "CMF032",
"latex": "F_T=\\sin(\\frac{T-T_b}{T_{ol}-T_b}\\frac{\\pi}{2}); 1; \\cos(\\frac{T-T_{ou}}{T_{max}-T_{ou}}\\frac{\\pi}{2})",
"name": "温度响应因子",
"name_en": "Temperatu... | CMF032_sensitivity_t1_0001 | false | Candidate formulas are listed below. Select the physiologically appropriate formula before calculating.
Candidate 1:
Formula name: Shoot partitioning index (地上部分配指数)
Formula (LaTeX): PISH=-8.42\times10^{-5}PDT^2+0.01PDT+0.63
Formula: PISH = -0.0000842 * PDT^2 + 0.01 * PDT + 0.63
Operator notes: * means multiplication,... | Formula name: Temperature response factor (温度响应因子)
Formula (LaTeX): F_T=\sin(\frac{T-T_b}{T_{ol}-T_b}\frac{\pi}{2}); 1; \cos(\frac{T-T_{ou}}{T_{max}-T_{ou}}\frac{\pi}{2})
Formula: F_T = if Tb<=T<Tol: sin(pi/2*(T-Tb)/(Tol-Tb)); if Tol<=T<Tou: 1; if Tou<=T<Tmax: cos(pi/2*(T-Tou)/(Tmax-Tou)); else: 0
Operator notes: * mea... | Formula name: Temperature response factor (温度响应因子)
Formula (LaTeX): F_T=\sin(\frac{T-T_b}{T_{ol}-T_b}\frac{\pi}{2}); 1; \cos(\frac{T-T_{ou}}{T_{max}-T_{ou}}\frac{\pi}{2})
Formula: F_T = if Tb<=T<Tol: sin(pi/2*(T-Tb)/(Tol-Tb)); if Tol<=T<Tou: 1; if Tou<=T<Tmax: cos(pi/2*(T-Tou)/(Tmax-Tou)); else: 0
Operator notes: * mea... | Formula: F_T = if Tb<=T<Tol: sin(pi/2*(T-Tb)/(Tol-Tb)); if Tol<=T<Tou: 1; if Tou<=T<Tmax: cos(pi/2*(T-Tou)/(Tmax-Tou)); else: 0
Operator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.
Output unit: fracti... | Relevant formula name: Temperature response factor (温度响应因子).
Formula: F_T = if Tb<=T<Tol: sin(pi/2*(T-Tb)/(Tol-Tb)); if Tol<=T<Tou: 1; if Tou<=T<Tmax: cos(pi/2*(T-Tou)/(Tmax-Tou)); else: 0
Operator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and... | A retrieved formula is listed below. It may look relevant, but use it only if it truly matches the problem.
Formula name: Emergence thermal requirement (出苗热量需求)
Formula (LaTeX): EMGDD=45+7SDEPTH
Formula: EMGDD = 45 + 7 * SDEPTH
Operator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means ... | 3 | Piecewise trigonometric | sensitivity | {
"T": 29.6,
"Tb": 10,
"Tmax": 45,
"Tol": 24,
"Tou": 34
} | Given current measurements T=29.6 C, Tb=10.0 C, Tol=24.0 C, Tou=34.0 C, Tmax=45.0 C in temperature response, what would the Temperature response factor be if T were adjusted to 32.6? Provide the numeric value in fraction. | {
"C": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ..... | Identify the target quantity as Temperature response factor.
Use the verified formula:
\[F_T=\sin(\frac{T-T_b}{T_{ol}-T_b}\frac{\pi}{2}); 1; \cos(\frac{T-T_{ou}}{T_{max}-T_{ou}}\frac{\pi}{2})\]
In calculation-friendly notation, this is:
F_T = if Tb<=T<Tol: sin(pi/2*(T-Tb)/(Tol-Tb)); if Tol<=T<Tou: 1; if Tou<=T<Tmax: co... | dev | Temperature response | 1 | CMF028 |
1.15 | cropmath-v1 | {
"new_value": 0.0047,
"old_value": 0.0041,
"parameter": "RP"
} | [
"CMF043",
"CMF045"
] | {
"GDD": 275.6,
"LAI0": 0.315,
"RP": 0.0047
} | growth_yield | {
"expression": "LAI0 * exp(RP * GDD)",
"id": "CMF037",
"latex": "LAI_P=LAI_0e^{RP\\cdot GDD}",
"name": "潜在叶面积指数",
"name_en": "Potential leaf area index",
"precision": 2,
"unit": "m2/m2"
} | CMF037_sensitivity_t0_0000 | false | Candidate formulas are listed below. Select the physiologically appropriate formula before calculating.
Candidate 1:
Formula name: Stem partitioning index (茎鞘分配指数)
Formula (LaTeX): PIS=1-PIGL-PIP
Formula: PIS = 1 - (if PDT<26: 0.54-0.0046*PDT; else: 1.4532*exp(-0.0492*PDT)) - (Y_grain/B_above/0.87)/(1+exp(-0.2804*(PDT... | Formula name: Potential leaf area index (潜在叶面积指数)
Formula (LaTeX): LAI_P=LAI_0e^{RP\cdot GDD}
Formula: LAIP = LAI0 * exp(RP * GDD)
Operator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.
Output unit: m2/... | Formula name: Potential leaf area index (潜在叶面积指数)
Formula (LaTeX): LAI_P=LAI_0e^{RP\cdot GDD}
Formula: LAIP = LAI0 * exp(RP * GDD)
Operator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.
Output unit: m2/... | Formula: LAIP = LAI0 * exp(RP * GDD)
Operator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.
Output unit: m2/m2 | Relevant formula name: Potential leaf area index (潜在叶面积指数).
Formula: LAIP = LAI0 * exp(RP * GDD)
Operator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.
Output unit: m2/m2 | A retrieved formula is listed below. It may look relevant, but use it only if it truly matches the problem.
Formula name: Root partitioning index (地下部分配指数)
Formula (LaTeX): PIRO=1-PISH
Formula: PIRO = 1 - (-0.0000842*PDT^2 + 0.01*PDT + 0.63)
Operator notes: * means multiplication, / means division, ** or ^ means power,... | 2 | Exponential growth | sensitivity | {
"GDD": 275.6,
"LAI0": 0.315,
"RP": 0.0041
} | A leaf area growth calculation involves sensitivity analysis. The current measurements are LAI0=0.315 m2/m2, RP=0.0041 d-1, GDD=275.6 C day. A field technician asks what the potential leaf area index would become if RP changed to 0.0047. Report the new final numeric value in m2/m2. | {
"C": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ..... | Identify the target quantity as Potential leaf area index.
Use the verified formula:
\[LAI_P=LAI_0e^{RP\cdot GDD}\]
In calculation-friendly notation, this is:
LAIP = LAI0 * exp(RP * GDD).
Substitute the relevant values:
LAI0=0.315 m2/m2, RP=0.0047 d-1, GDD=275.6 C day.
The computed value is 1.15 m2/m2.
Therefore, the f... | dev | Leaf area growth | 0 | CMF043 |
10.60 | cropmath-v1 | null | [
"CMF057",
"CMF059"
] | {
"Tbase": 5.1,
"Tmean": 15.7
} | environmental_response | {
"expression": "max(Tmean - Tbase, 0)",
"id": "CMF056",
"latex": "dTT=\\max(T_{mean}-T_{base},0)",
"name": "每日有效积温",
"name_en": "environmental-response daily thermal time",
"precision": 2,
"unit": "C day"
} | CMF056_direct_t0_0001 | false | Candidate formulas are listed below. Select the physiologically appropriate formula before calculating.
Candidate 1:
Formula name: environmental-response solar interception fraction (环境响应光截获比例)
Formula (LaTeX): fSolar=\min(f_1,f_2)F_{solar,water}
Formula: fSolar = min(fSolarMax/(1+exp(-0.01*(TT-I50A))), fSolarMax/(1+e... | Formula name: environmental-response daily thermal time (每日有效积温)
Formula (LaTeX): dTT=\max(T_{mean}-T_{base},0)
Formula: dTT = max(Tmean - Tbase, 0)
Operator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value... | Formula name: environmental-response daily thermal time (每日有效积温)
Formula (LaTeX): dTT=\max(T_{mean}-T_{base},0)
Formula: dTT = max(Tmean - Tbase, 0)
Operator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value... | Formula: dTT = max(Tmean - Tbase, 0)
Operator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.
Output unit: C day | Relevant formula name: environmental-response daily thermal time (每日有效积温).
Formula: dTT = max(Tmean - Tbase, 0)
Operator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.
Output unit: C day | A retrieved formula is listed below. It may look relevant, but use it only if it truly matches the problem.
Formula name: environmental-response CO2 response (环境响应二氧化碳响应)
Formula (LaTeX): F_{CO2}=f(CO2,CO2_{RUE})
Formula: F_CO2 = if CO2>=700: 1+CO2_RUE*350/100; else: max(CO2_RUE*CO2*0.01+1-0.01*350*CO2_RUE, 1)
Operator... | 1 | Maximum linear | direct | {
"Tbase": 5.1,
"Tmean": 15.7
} | A environmental-response formula system calculation estimates environmental-response daily thermal time. The measured inputs are Tmean=15.7 C, Tbase=5.1 C. What is the simple daily thermal time? Give the final numeric value in C day. | {
"C": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ..... | Identify the target quantity as environmental-response daily thermal time.
Use the verified formula:
\[dTT=\max(T_{mean}-T_{base},0)\]
In calculation-friendly notation, this is:
dTT = max(Tmean - Tbase, 0).
Substitute the relevant values:
Tmean=15.7 C, Tbase=5.1 C.
The computed value is 10.60 C day.
Therefore, the fina... | dev | environmental-response formula system | 0 | CMF057 |
0.44 | cropmath-v1 | {
"new_value": 22.4,
"old_value": 20.4,
"parameter": "PDT"
} | [
"CMF036",
"CMF031"
] | {
"PDT": 22.4
} | growth_yield | {
"expression": "if PDT<26: 0.54-0.0046*PDT; else: 1.4532*exp(-0.0492*PDT)",
"id": "CMF041",
"latex": "PIGL=0.54-0.0046PDT\\;(PDT<26),\\quad1.4532e^{-0.0492PDT}\\;(PDT\\ge26)",
"name": "绿叶分配指数",
"name_en": "Green leaf partitioning index",
"precision": 2,
"unit": "fraction"
} | CMF041_sensitivity_t2_0000 | false | Candidate formulas are listed below. Select the physiologically appropriate formula before calculating.
Candidate 1:
Formula name: Green leaf partitioning index (绿叶分配指数)
Formula (LaTeX): PIGL=0.54-0.0046PDT\;(PDT<26),\quad1.4532e^{-0.0492PDT}\;(PDT\ge26)
Formula: PIGL = if PDT<26: 0.54-0.0046*PDT; else: 1.4532*exp(-0.... | Formula name: Green leaf partitioning index (绿叶分配指数)
Formula (LaTeX): PIGL=0.54-0.0046PDT\;(PDT<26),\quad1.4532e^{-0.0492PDT}\;(PDT\ge26)
Formula: PIGL = if PDT<26: 0.54-0.0046*PDT; else: 1.4532*exp(-0.0492*PDT)
Operator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means... | Formula name: Green leaf partitioning index (绿叶分配指数)
Formula (LaTeX): PIGL=0.54-0.0046PDT\;(PDT<26),\quad1.4532e^{-0.0492PDT}\;(PDT\ge26)
Formula: PIGL = if PDT<26: 0.54-0.0046*PDT; else: 1.4532*exp(-0.0492*PDT)
Operator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means... | Formula: PIGL = if PDT<26: 0.54-0.0046*PDT; else: 1.4532*exp(-0.0492*PDT)
Operator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.
Output unit: fraction | Relevant formula name: Green leaf partitioning index (绿叶分配指数).
Formula: PIGL = if PDT<26: 0.54-0.0046*PDT; else: 1.4532*exp(-0.0492*PDT)
Operator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.
Output uni... | A retrieved formula is listed below. It may look relevant, but use it only if it truly matches the problem.
Formula name: Canopy extinction coefficient (冠层消光系数)
Formula (LaTeX): K=0.0087PDT+0.2222
Formula: K = 0.0087 * PDT + 0.2222
Operator notes: * means multiplication, / means division, ** or ^ means power, exp(x) me... | 3 | Piecewise linear/exponential | sensitivity | {
"PDT": 20.4
} | In dry matter partitioning, we analyze how Green leaf partitioning index responds to parameter changes. Baseline: PDT=20.4 physiological days. After changing PDT from 20.4 to 22.4, compute the new Green leaf partitioning index. Give the numeric value in fraction. | {
"C": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ..... | Identify the target quantity as Green leaf partitioning index.
Use the verified formula:
\[PIGL=0.54-0.0046PDT\;(PDT<26),\quad1.4532e^{-0.0492PDT}\;(PDT\ge26)\]
In calculation-friendly notation, this is:
PIGL = if PDT<26: 0.54-0.0046*PDT; else: 1.4532*exp(-0.0492*PDT).
Substitute the relevant values:
PDT=22.4 physiolog... | dev | Dry matter partitioning | 2 | CMF036 |
0.563 | cropmath-v1 | null | [
"CMF007",
"CMF006"
] | {
"Coef_K_Decom": 0.0102,
"Coef_K_Trans": 0.152,
"F_notill": 1.1400000000000001,
"F_pH": 0.18,
"F_temp": 1.209,
"F_texture": 0.9590000000000001,
"F_water": 0.532,
"Pool": 514.7
} | carbon_nitrogen_cycle | {
"expression": "Pool*Coef_K_Decom*F_temp*F_water*F_texture*F_pH*F_notill*(1-Coef_K_Trans)",
"id": "CMF009",
"latex": "R=PoolK F_TF_WF_{texture}F_{pH}F_{notill}(1-K_{trans})",
"name": "碳库呼吸损失",
"name_en": "carbon-nitrogen cycling carbon pool respiration loss",
"precision": 3,
"unit": "g C/m2/day"
} | CMF009_chain_2_t2_0000 | false | Candidate formulas are listed below. Select the physiologically appropriate formula before calculating.
Candidate 1:
Formula name: carbon-nitrogen cycling carbon pool respiration loss (碳库呼吸损失)
Formula (LaTeX): R=PoolK F_TF_WF_{texture}F_{pH}F_{notill}(1-K_{trans})
Formula: Respiration = Pool*Coef_K_Decom*F_temp*F_wate... | Formula name: carbon-nitrogen cycling carbon pool respiration loss (碳库呼吸损失)
Formula (LaTeX): R=PoolK F_TF_WF_{texture}F_{pH}F_{notill}(1-K_{trans})
Formula: Respiration = Pool*Coef_K_Decom*F_temp*F_water*F_texture*F_pH*F_notill*(1-Coef_K_Trans)
Operator notes: * means multiplication, / means division, ** or ^ means pow... | Formula name: carbon-nitrogen cycling carbon pool respiration loss (碳库呼吸损失)
Formula (LaTeX): R=PoolK F_TF_WF_{texture}F_{pH}F_{notill}(1-K_{trans})
Formula: Respiration = Pool*Coef_K_Decom*F_temp*F_water*F_texture*F_pH*F_notill*(1-Coef_K_Trans)
Operator notes: * means multiplication, / means division, ** or ^ means pow... | Formula: Respiration = Pool*Coef_K_Decom*F_temp*F_water*F_texture*F_pH*F_notill*(1-Coef_K_Trans)
Operator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.
Output unit: g C/m2/day | Relevant formula name: carbon-nitrogen cycling carbon pool respiration loss (碳库呼吸损失).
Formula: Respiration = Pool*Coef_K_Decom*F_temp*F_water*F_texture*F_pH*F_notill*(1-Coef_K_Trans)
Operator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/m... | A retrieved formula is listed below. It may look relevant, but use it only if it truly matches the problem.
Formula name: carbon-nitrogen cycling optimal photosynthetic rate (最适光合速率)
Formula (LaTeX): P_{opt}=\alpha N_{leaf}\frac{PAR}{PAR+\beta}
Formula: Photo_Rate_opt = (Coef_alfa * Leaf_N) * PAR / (PAR + Coef_beta)
Op... | 4 | Chained multiplicative decomposition | chain_2 | {
"Coef_K_Decom": 0.0102,
"Coef_K_Trans": 0.152,
"F_notill": 1.1400000000000001,
"F_pH": 0.18,
"F_temp": 1.209,
"F_texture": 0.9590000000000001,
"F_water": 0.532,
"Pool": 514.7
} | In carbon-nitrogen cycling soil carbon, the carbon-nitrogen cycling carbon pool respiration loss is obtained by chaining intermediate calculations. With inputs Pool=514.7 g C/m2, Coef_K_Decom=0.0102 d-1, F_temp=1.209 fraction, F_water=0.532 fraction, F_texture=0.959 fraction, F_pH=0.18 fraction, F_notill=1.14 fraction,... | {
"C": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ..... | Identify the target quantity as carbon-nitrogen cycling carbon pool respiration loss.
Use the verified formula:
\[R=PoolK F_TF_WF_{texture}F_{pH}F_{notill}(1-K_{trans})\]
In calculation-friendly notation, this is:
Respiration = Pool*Coef_K_Decom*F_temp*F_water*F_texture*F_pH*F_notill*(1-Coef_K_Trans).
Substitute the re... | dev | carbon-nitrogen cycling soil carbon | 2 | CMF007 |
1.00 | cropmath-v1 | {
"new_value": 24,
"old_value": 21.8,
"parameter": "PDT"
} | [
"CMF049",
"CMF029"
] | {
"PDT": 24,
"a": 0.039
} | growth_yield | {
"expression": "if PDT<28: 1; else: exp(-a*(PDT-28))",
"id": "CMF030",
"latex": "F_{PA}=1\\;(PDT<28),\\quad e^{-a(PDT-28)}\\;(PDT\\ge 28)",
"name": "生理年龄因子",
"name_en": "Physiological age factor",
"precision": 2,
"unit": "fraction"
} | CMF030_sensitivity_t1_0001 | false | Candidate formulas are listed below. Select the physiologically appropriate formula before calculating.
Candidate 1:
Formula name: CO2 response factor (二氧化碳影响因子)
Formula (LaTeX): F_{CO2}=1+\beta\ln(C_x/340)
Formula: FCO2 = 1 + beta * ln(Cx / 340)
Operator notes: * means multiplication, / means division, ** or ^ means ... | Formula name: Physiological age factor (生理年龄因子)
Formula (LaTeX): F_{PA}=1\;(PDT<28),\quad e^{-a(PDT-28)}\;(PDT\ge 28)
Formula: FPA = if PDT<28: 1; else: exp(-a*(PDT-28))
Operator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the... | Formula name: Physiological age factor (生理年龄因子)
Formula (LaTeX): F_{PA}=1\;(PDT<28),\quad e^{-a(PDT-28)}\;(PDT\ge 28)
Formula: FPA = if PDT<28: 1; else: exp(-a*(PDT-28))
Operator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the... | Formula: FPA = if PDT<28: 1; else: exp(-a*(PDT-28))
Operator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.
Output unit: fraction | Relevant formula name: Physiological age factor (生理年龄因子).
Formula: FPA = if PDT<28: 1; else: exp(-a*(PDT-28))
Operator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.
Output unit: fraction | A retrieved formula is listed below. It may look relevant, but use it only if it truly matches the problem.
Formula name: Maintenance respiration coefficient at optimum temperature (最适温度维持呼吸系数)
Formula (LaTeX): RM(T_o)=0.0091-0.0001PDT
Formula: RMTo = 0.0091 - 0.0001 * PDT
Operator notes: * means multiplication, / mean... | 3 | Piecewise exponential decay | sensitivity | {
"PDT": 21.8,
"a": 0.039
} | Given current measurements PDT=21.8 physiological days, a=0.039 dimensionless in photosynthesis, what would the Physiological age factor be if PDT were adjusted to 24.0? Provide the numeric value in fraction. | {
"C": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ..... | Identify the target quantity as Physiological age factor.
Use the verified formula:
\[F_{PA}=1\;(PDT<28),\quad e^{-a(PDT-28)}\;(PDT\ge 28)\]
In calculation-friendly notation, this is:
FPA = if PDT<28: 1; else: exp(-a*(PDT-28)).
Substitute the relevant values:
PDT=24.0 physiological days, a=0.039 dimensionless.
The comp... | dev | Photosynthesis | 1 | CMF049 |
0.16 | cropmath-v1 | null | [
"CMF047",
"CMF027"
] | {
"PDT": 27.6
} | growth_yield | {
"expression": "1 - (-0.0000842*PDT^2 + 0.01*PDT + 0.63)",
"id": "CMF043",
"latex": "PIRO=1-PISH",
"name": "地下部分配指数",
"name_en": "Root partitioning index",
"precision": 2,
"unit": "fraction"
} | CMF043_direct_t1_0001 | false | Candidate formulas are listed below. Select the physiologically appropriate formula before calculating.
Candidate 1:
Formula name: Root partitioning index (地下部分配指数)
Formula (LaTeX): PIRO=1-PISH
Formula: PIRO = 1 - (-0.0000842*PDT^2 + 0.01*PDT + 0.63)
Operator notes: * means multiplication, / means division, ** or ^ me... | Formula name: Root partitioning index (地下部分配指数)
Formula (LaTeX): PIRO=1-PISH
Formula: PIRO = 1 - (-0.0000842*PDT^2 + 0.01*PDT + 0.63)
Operator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.
Output unit: ... | Formula name: Root partitioning index (地下部分配指数)
Formula (LaTeX): PIRO=1-PISH
Formula: PIRO = 1 - (-0.0000842*PDT^2 + 0.01*PDT + 0.63)
Operator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.
Output unit: ... | Formula: PIRO = 1 - (-0.0000842*PDT^2 + 0.01*PDT + 0.63)
Operator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.
Output unit: fraction | Relevant formula name: Root partitioning index (地下部分配指数).
Formula: PIRO = 1 - (-0.0000842*PDT^2 + 0.01*PDT + 0.63)
Operator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.
Output unit: fraction | A retrieved formula is listed below. It may look relevant, but use it only if it truly matches the problem.
Formula name: Intercepted photosynthetically active radiation (截获光合有效辐射)
Formula (LaTeX): IPAR = PAR \cdot (1 - e^{-k \cdot LAI})
Formula: IPAR = PAR * (1 - exp(-k * LAI))
Operator notes: * means multiplication, ... | 1 | Residual from shoot partitioning | direct | {
"PDT": 27.6
} | Given PDT=27.6 physiological days, calculate the Root partitioning index for a dry matter partitioning. The result should be in fraction. | {
"C": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ..... | Identify the target quantity as Root partitioning index.
Use the verified formula:
\[PIRO=1-PISH\]
In calculation-friendly notation, this is:
PIRO = 1 - (-0.0000842*PDT^2 + 0.01*PDT + 0.63).
Substitute the relevant values:
PDT=27.6 physiological days.
The computed value is 0.16 fraction.
Therefore, the final numeric an... | dev | Dry matter partitioning | 1 | CMF047 |
0.343 | cropmath-v1 | null | [
"CMF040",
"CMF042"
] | {
"T": 15,
"TS": 1.19,
"Tb": 10,
"Tm": 42,
"To": 30
} | growth_yield | {
"expression": "(((T - Tb)/(To - Tb)) * (((Tm - T)/(Tm - To)) ** ((Tm - To)/(To - Tb)))) ** TS",
"id": "CMF051",
"latex": "RTE=\\left[\\frac{T-T_b}{T_o-T_b}\\left(\\frac{T_m-T}{T_m-T_o}\\right)^{(T_m-T_o)/(T_o-T_b)}\\right]^{TS}",
"name": "相对热效应",
"name_en": "Relative thermal effectiveness",
"precision": 3... | CMF051_direct_t0_0000 | false | Candidate formulas are listed below. Select the physiologically appropriate formula before calculating.
Candidate 1:
Formula name: Critical daylength (临界日长)
Formula (LaTeX): P_c=P_o+\sqrt{1/PS}
Formula: Pc = Po + sqrt(1 / PS)
Operator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e... | Formula name: Relative thermal effectiveness (相对热效应)
Formula (LaTeX): RTE=\left[\frac{T-T_b}{T_o-T_b}\left(\frac{T_m-T}{T_m-T_o}\right)^{(T_m-T_o)/(T_o-T_b)}\right]^{TS}
Formula: RTE = (((T - Tb)/(To - Tb)) * (((Tm - T)/(Tm - To)) ** ((Tm - To)/(To - Tb)))) ** TS
Operator notes: * means multiplication, / means division... | Formula name: Relative thermal effectiveness (相对热效应)
Formula (LaTeX): RTE=\left[\frac{T-T_b}{T_o-T_b}\left(\frac{T_m-T}{T_m-T_o}\right)^{(T_m-T_o)/(T_o-T_b)}\right]^{TS}
Formula: RTE = (((T - Tb)/(To - Tb)) * (((Tm - T)/(Tm - To)) ** ((Tm - To)/(To - Tb)))) ** TS
Operator notes: * means multiplication, / means division... | Formula: RTE = (((T - Tb)/(To - Tb)) * (((Tm - T)/(Tm - To)) ** ((Tm - To)/(To - Tb)))) ** TS
Operator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.
Output unit: fraction | Relevant formula name: Relative thermal effectiveness (相对热效应).
Formula: RTE = (((T - Tb)/(To - Tb)) * (((Tm - T)/(Tm - To)) ** ((Tm - To)/(To - Tb)))) ** TS
Operator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larg... | A retrieved formula is listed below. It may look relevant, but use it only if it truly matches the problem.
Formula name: Critical daylength (临界日长)
Formula (LaTeX): P_c=P_o+\sqrt{1/PS}
Formula: Pc = Po + sqrt(1 / PS)
Operator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) ... | 2 | Beta temperature response | direct | {
"T": 15,
"TS": 1.19,
"Tb": 10,
"Tm": 42,
"To": 30
} | A phenology calculation estimates Relative thermal effectiveness. The measured inputs are T=15.0 C, Tb=10.0 C, To=30.0 C, Tm=42.0 C, TS=1.19 dimensionless. What is the relative thermal effectiveness? Give the final numeric value in fraction. | {
"C": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ..... | Identify the target quantity as Relative thermal effectiveness.
Use the verified formula:
\[RTE=\left[\frac{T-T_b}{T_o-T_b}\left(\frac{T_m-T}{T_m-T_o}\right)^{(T_m-T_o)/(T_o-T_b)}\right]^{TS}\]
In calculation-friendly notation, this is:
RTE = (((T - Tb)/(To - Tb)) * (((Tm - T)/(Tm - To)) ** ((Tm - To)/(To - Tb)))) ** T... | dev | Phenology | 0 | CMF040 |
204.19 | cropmath-v1 | null | [
"CMF003",
"CMF005"
] | {
"Coef_K": 0.6000000000000001,
"Coef_rou": 0.15,
"LAI": 5.57,
"Radi": 411.2
} | carbon_nitrogen_cycle | {
"expression": "Radi * 0.44 * 4.6 * (1-Coef_rou) * (1-exp(-Coef_K*LAI)) / (Coef_K*LAI)",
"id": "CMF006",
"latex": "PAR=Radi\\cdot0.44\\cdot4.6(1-\\rho)\\frac{1-e^{-KLAI}}{KLAI}",
"name": "冠层吸收光合有效辐射",
"name_en": "carbon-nitrogen cycling canopy absorbed PAR",
"precision": 2,
"unit": "umol/m2/s"
} | CMF006_chain_2_t2_0001 | false | Candidate formulas are listed below. Select the physiologically appropriate formula before calculating.
Candidate 1:
Formula name: carbon-nitrogen cycling canopy absorbed PAR (冠层吸收光合有效辐射)
Formula (LaTeX): PAR=Radi\cdot0.44\cdot4.6(1-\rho)\frac{1-e^{-KLAI}}{KLAI}
Formula: PAR = Radi * 0.44 * 4.6 * (1-Coef_rou) * (1-exp... | Formula name: carbon-nitrogen cycling canopy absorbed PAR (冠层吸收光合有效辐射)
Formula (LaTeX): PAR=Radi\cdot0.44\cdot4.6(1-\rho)\frac{1-e^{-KLAI}}{KLAI}
Formula: PAR = Radi * 0.44 * 4.6 * (1-Coef_rou) * (1-exp(-Coef_K*LAI)) / (Coef_K*LAI)
Operator notes: * means multiplication, / means division, ** or ^ means power, exp(x) me... | Formula name: carbon-nitrogen cycling canopy absorbed PAR (冠层吸收光合有效辐射)
Formula (LaTeX): PAR=Radi\cdot0.44\cdot4.6(1-\rho)\frac{1-e^{-KLAI}}{KLAI}
Formula: PAR = Radi * 0.44 * 4.6 * (1-Coef_rou) * (1-exp(-Coef_K*LAI)) / (Coef_K*LAI)
Operator notes: * means multiplication, / means division, ** or ^ means power, exp(x) me... | Formula: PAR = Radi * 0.44 * 4.6 * (1-Coef_rou) * (1-exp(-Coef_K*LAI)) / (Coef_K*LAI)
Operator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.
Output unit: umol/m2/s | Relevant formula name: carbon-nitrogen cycling canopy absorbed PAR (冠层吸收光合有效辐射).
Formula: PAR = Radi * 0.44 * 4.6 * (1-Coef_rou) * (1-exp(-Coef_K*LAI)) / (Coef_K*LAI)
Operator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the sm... | A retrieved formula is listed below. It may look relevant, but use it only if it truly matches the problem.
Formula name: Nitrogen use efficiency (氮利用效率)
Formula (LaTeX): NUE=\frac{GrainYield}{NUptake}
Formula: NUE = GrainYield / NUptake
Operator notes: * means multiplication, / means division, ** or ^ means power, exp... | 3 | Exponential ratio | chain_2 | {
"Coef_K": 0.6000000000000001,
"Coef_rou": 0.15,
"LAI": 5.57,
"Radi": 411.2
} | In carbon-nitrogen cycling photosynthesis, the carbon-nitrogen cycling canopy absorbed PAR is obtained by chaining intermediate calculations. With inputs Radi=411.2 W/m2, LAI=5.57 m2/m2, Coef_rou=0.15 fraction, Coef_K=0.6 dimensionless, determine the numeric value in umol/m2/s. | {
"C": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ..... | Identify the target quantity as carbon-nitrogen cycling canopy absorbed PAR.
Use the verified formula:
\[PAR=Radi\cdot0.44\cdot4.6(1-\rho)\frac{1-e^{-KLAI}}{KLAI}\]
In calculation-friendly notation, this is:
PAR = Radi * 0.44 * 4.6 * (1-Coef_rou) * (1-exp(-Coef_K*LAI)) / (Coef_K*LAI).
Substitute the relevant values:
Ra... | dev | carbon-nitrogen cycling photosynthesis | 2 | CMF003 |
0.901 | cropmath-v1 | {
"new_value": 0.566,
"old_value": 0.629,
"parameter": "TI"
} | [
"CMF025",
"CMF024"
] | {
"OMN": 249.2,
"SI": 0.91,
"TI": 0.5660000000000001,
"WI": 0.26
} | methane_emission | {
"expression": "WI * SI * TI * 0.027 * OMN",
"id": "CMF020",
"latex": "OMNC=WI\\cdot SI\\cdot TI\\cdot0.027\\cdot OMN",
"name": "新鲜有机质碳分解量",
"name_en": "Fresh organic matter carbon decomposition",
"precision": 3,
"unit": "g C/m2/day"
} | CMF020_sensitivity_t1_0001 | false | Candidate formulas are listed below. Select the physiologically appropriate formula before calculating.
Candidate 1:
Formula name: Soil temperature from air temperature (土壤温度估算)
Formula (LaTeX): T_{soil}=4.4+0.76T_{air}
Formula: Tsoil = 4.4 + 0.76 * Tair
Operator notes: * means multiplication, / means division, ** or ... | Formula name: Fresh organic matter carbon decomposition (新鲜有机质碳分解量)
Formula (LaTeX): OMNC=WI\cdot SI\cdot TI\cdot0.027\cdot OMN
Formula: OMNC = WI * SI * TI * 0.027 * OMN
Operator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select th... | Formula name: Fresh organic matter carbon decomposition (新鲜有机质碳分解量)
Formula (LaTeX): OMNC=WI\cdot SI\cdot TI\cdot0.027\cdot OMN
Formula: OMNC = WI * SI * TI * 0.027 * OMN
Operator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select th... | Formula: OMNC = WI * SI * TI * 0.027 * OMN
Operator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.
Output unit: g C/m2/day | Relevant formula name: Fresh organic matter carbon decomposition (新鲜有机质碳分解量).
Formula: OMNC = WI * SI * TI * 0.027 * OMN
Operator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.
Output unit: g C/m2/day | A retrieved formula is listed below. It may look relevant, but use it only if it truly matches the problem.
Formula name: Water-content decomposition factor (水分分解因子)
Formula (LaTeX): WI=0.49e^{3.88W_c-5.4W_c^2}
Formula: WI = 0.49 * exp(3.88 * WaterC - 5.4 * WaterC^2)
Operator notes: * means multiplication, / means divi... | 2 | Multiplicative | sensitivity | {
"OMN": 249.2,
"SI": 0.91,
"TI": 0.629,
"WI": 0.26
} | Given current measurements WI=0.26 fraction, SI=0.91 fraction, TI=0.629 fraction, OMN=249.2 g C/m2 in methane, what would the Fresh organic matter carbon decomposition be if TI were adjusted to 0.566? Provide the numeric value in g C/m2/day. | {
"C": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ..... | Identify the target quantity as Fresh organic matter carbon decomposition.
Use the verified formula:
\[OMNC=WI\cdot SI\cdot TI\cdot0.027\cdot OMN\]
In calculation-friendly notation, this is:
OMNC = WI * SI * TI * 0.027 * OMN.
Substitute the relevant values:
WI=0.26 fraction, SI=0.91 fraction, TI=0.566 fraction, OMN=249... | dev | Methane | 1 | CMF025 |
1.605 | cropmath-v1 | null | [
"CMF055",
"CMF056"
] | {
"CO2": 739,
"CO2_RUE": 0.17300000000000001
} | environmental_response | {
"expression": "if CO2>=700: 1+CO2_RUE*350/100; else: max(CO2_RUE*CO2*0.01+1-0.01*350*CO2_RUE, 1)",
"id": "CMF057",
"latex": "F_{CO2}=f(CO2,CO2_{RUE})",
"name": "环境响应二氧化碳响应",
"name_en": "environmental-response CO2 response",
"precision": 3,
"unit": "multiplier"
} | CMF057_direct_t2_0000 | false | Candidate formulas are listed below. Select the physiologically appropriate formula before calculating.
Candidate 1:
Formula name: environmental-response CO2 response (环境响应二氧化碳响应)
Formula (LaTeX): F_{CO2}=f(CO2,CO2_{RUE})
Formula: F_CO2 = if CO2>=700: 1+CO2_RUE*350/100; else: max(CO2_RUE*CO2*0.01+1-0.01*350*CO2_RUE, 1... | Formula name: environmental-response CO2 response (环境响应二氧化碳响应)
Formula (LaTeX): F_{CO2}=f(CO2,CO2_{RUE})
Formula: F_CO2 = if CO2>=700: 1+CO2_RUE*350/100; else: max(CO2_RUE*CO2*0.01+1-0.01*350*CO2_RUE, 1)
Operator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural... | Formula name: environmental-response CO2 response (环境响应二氧化碳响应)
Formula (LaTeX): F_{CO2}=f(CO2,CO2_{RUE})
Formula: F_CO2 = if CO2>=700: 1+CO2_RUE*350/100; else: max(CO2_RUE*CO2*0.01+1-0.01*350*CO2_RUE, 1)
Operator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural... | Formula: F_CO2 = if CO2>=700: 1+CO2_RUE*350/100; else: max(CO2_RUE*CO2*0.01+1-0.01*350*CO2_RUE, 1)
Operator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.
Output unit: fraction | Relevant formula name: environmental-response CO2 response (环境响应二氧化碳响应).
Formula: F_CO2 = if CO2>=700: 1+CO2_RUE*350/100; else: max(CO2_RUE*CO2*0.01+1-0.01*350*CO2_RUE, 1)
Operator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select t... | A retrieved formula is listed below. It may look relevant, but use it only if it truly matches the problem.
Formula name: environmental-response daily biomass increment (环境响应日生物量增量)
Formula (LaTeX): dB=10RUEfSolarSRADF_{CO2}F_T\min(F_W,F_H)
Formula: dBiomass = 10 * RUE * fSolar * SRAD * F_CO2 * F_Temp * min(F_Water, F_... | 2 | Piecewise linear cap | direct | {
"CO2": 739,
"CO2_RUE": 0.17300000000000001
} | In an environmental-response formula system, the environmental-response CO2 response is computed as part of environmental-response formula calculation. With CO2=739.0 ppm, CO2_RUE=0.173 fraction, determine the numeric value in fraction. | {
"C": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ..... | Identify the target quantity as environmental-response CO2 response.
Use the verified formula:
\[F_{CO2}=f(CO2,CO2_{RUE})\]
In calculation-friendly notation, this is:
F_CO2 = if CO2>=700: 1+CO2_RUE*350/100; else: max(CO2_RUE*CO2*0.01+1-0.01*350*CO2_RUE, 1).
Substitute the relevant values:
CO2=739.0 ppm, CO2_RUE=0.173 f... | dev | environmental-response formula system | 2 | CMF055 |
6.456 | cropmath-v1 | null | [
"CMF009",
"CMF001"
] | {
"Day_Length_hour": 11.99,
"F_CO2": 1.1400000000000001,
"F_T": 0.49,
"F_Water": 0.43,
"Photo_Rate_opt": 51.89
} | carbon_nitrogen_cycle | {
"expression": "Photo_Rate_opt * F_T * F_Water * F_CO2 * Day_Length_hour * 0.0432",
"id": "CMF002",
"latex": "GPP=P_{opt}F_TF_WF_{CO2}DL\\cdot0.0432",
"name": "总初级生产力",
"name_en": "carbon-nitrogen cycling gross primary production",
"precision": 3,
"unit": "g C/m2/day"
} | CMF002_direct_t1_0001 | false | Candidate formulas are listed below. Select the physiologically appropriate formula before calculating.
Candidate 1:
Formula name: carbon-nitrogen cycling carbon pool respiration loss (碳库呼吸损失)
Formula (LaTeX): R=PoolK F_TF_WF_{texture}F_{pH}F_{notill}(1-K_{trans})
Formula: Respiration = Pool*Coef_K_Decom*F_temp*F_wate... | Formula name: carbon-nitrogen cycling gross primary production (总初级生产力)
Formula (LaTeX): GPP=P_{opt}F_TF_WF_{CO2}DL\cdot0.0432
Formula: GPP = Photo_Rate_opt * F_T * F_Water * F_CO2 * Day_Length_hour * 0.0432
Operator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means nat... | Formula name: carbon-nitrogen cycling gross primary production (总初级生产力)
Formula (LaTeX): GPP=P_{opt}F_TF_WF_{CO2}DL\cdot0.0432
Formula: GPP = Photo_Rate_opt * F_T * F_Water * F_CO2 * Day_Length_hour * 0.0432
Operator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means nat... | Formula: GPP = Photo_Rate_opt * F_T * F_Water * F_CO2 * Day_Length_hour * 0.0432
Operator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.
Output unit: g C/m2/day | Relevant formula name: carbon-nitrogen cycling gross primary production (总初级生产力).
Formula: GPP = Photo_Rate_opt * F_T * F_Water * F_CO2 * Day_Length_hour * 0.0432
Operator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smalle... | A retrieved formula is listed below. It may look relevant, but use it only if it truly matches the problem.
Formula name: carbon-nitrogen cycling carbon pool respiration loss (碳库呼吸损失)
Formula (LaTeX): R=PoolK F_TF_WF_{texture}F_{pH}F_{notill}(1-K_{trans})
Formula: Respiration = Pool*Coef_K_Decom*F_temp*F_water*F_textur... | 2 | Multiplicative conversion | direct | {
"Day_Length_hour": 11.99,
"F_CO2": 1.1400000000000001,
"F_T": 0.49,
"F_Water": 0.43,
"Photo_Rate_opt": 51.89
} | Given Photo_Rate_opt=51.89 umol CO2/m2/s, F_T=0.49 fraction, F_Water=0.43 fraction, F_CO2=1.14 fraction, Day_Length_hour=11.99 h, calculate the carbon-nitrogen cycling gross primary production for a carbon-nitrogen cycling photosynthesis. The result should be in g C/m2/day. | {
"C": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ..... | Identify the target quantity as carbon-nitrogen cycling gross primary production.
Use the verified formula:
\[GPP=P_{opt}F_TF_WF_{CO2}DL\cdot0.0432\]
In calculation-friendly notation, this is:
GPP = Photo_Rate_opt * F_T * F_Water * F_CO2 * Day_Length_hour * 0.0432.
Substitute the relevant values:
Photo_Rate_opt=51.89 u... | dev | carbon-nitrogen cycling photosynthesis | 1 | CMF009 |
113.00 | cropmath-v1 | null | [
"CMF061",
"CMF062"
] | {
"F_CO2": 1.26,
"F_Heat": 1,
"F_Temp": 0.7000000000000001,
"F_Water": 0.65,
"RUE": 2.5,
"SRAD": 12.13,
"fSolar": 0.65
} | environmental_response | {
"expression": "10 * RUE * fSolar * SRAD * F_CO2 * F_Temp * min(F_Water, F_Heat)",
"id": "CMF055",
"latex": "dB=10RUEfSolarSRADF_{CO2}F_T\\min(F_W,F_H)",
"name": "环境响应日生物量增量",
"name_en": "environmental-response daily biomass increment",
"precision": 2,
"unit": "kg/ha/day"
} | CMF055_direct_t0_0001 | false | Candidate formulas are listed below. Select the physiologically appropriate formula before calculating.
Candidate 1:
Formula name: environmental-response yield (环境响应产量)
Formula (LaTeX): Yield=Biomass\cdot HI
Formula: Yield = Biomass * HI
Operator notes: * means multiplication, / means division, ** or ^ means power, ex... | Formula name: environmental-response daily biomass increment (环境响应日生物量增量)
Formula (LaTeX): dB=10RUEfSolarSRADF_{CO2}F_T\min(F_W,F_H)
Formula: dBiomass = 10 * RUE * fSolar * SRAD * F_CO2 * F_Temp * min(F_Water, F_Heat)
Operator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x)... | Formula name: environmental-response daily biomass increment (环境响应日生物量增量)
Formula (LaTeX): dB=10RUEfSolarSRADF_{CO2}F_T\min(F_W,F_H)
Formula: dBiomass = 10 * RUE * fSolar * SRAD * F_CO2 * F_Temp * min(F_Water, F_Heat)
Operator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x)... | Formula: dBiomass = 10 * RUE * fSolar * SRAD * F_CO2 * F_Temp * min(F_Water, F_Heat)
Operator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.
Output unit: kg/ha/day | Relevant formula name: environmental-response daily biomass increment (环境响应日生物量增量).
Formula: dBiomass = 10 * RUE * fSolar * SRAD * F_CO2 * F_Temp * min(F_Water, F_Heat)
Operator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the ... | A retrieved formula is listed below. It may look relevant, but use it only if it truly matches the problem.
Formula name: environmental-response water stress response (环境响应水分响应)
Formula (LaTeX): F_W=\max(0,1-S_WARID)
Formula: F_Water = max(0, 1 - S_Water * ARID)
Operator notes: * means multiplication, / means division,... | 2 | Multiplicative minimum stress | direct | {
"F_CO2": 1.26,
"F_Heat": 1,
"F_Temp": 0.7000000000000001,
"F_Water": 0.65,
"RUE": 2.5,
"SRAD": 12.13,
"fSolar": 0.65
} | A environmental-response formula system calculation estimates environmental-response daily biomass increment. The measured inputs are RUE=2.5 g/MJ, fSolar=0.65 fraction, SRAD=12.13 MJ/m2/day, F_CO2=1.26 fraction, F_Temp=0.7 fraction, F_Water=0.65 fraction, F_Heat=1.0 fraction. What is the simple daily biomass increment... | {
"C": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ..... | Identify the target quantity as environmental-response daily biomass increment.
Use the verified formula:
\[dB=10RUEfSolarSRADF_{CO2}F_T\min(F_W,F_H)\]
In calculation-friendly notation, this is:
dBiomass = 10 * RUE * fSolar * SRAD * F_CO2 * F_Temp * min(F_Water, F_Heat).
Substitute the relevant values:
RUE=2.5 g/MJ, fS... | dev | environmental-response formula system | 0 | CMF061 |
1.00 | cropmath-v1 | {
"new_value": 17.8,
"old_value": 16.2,
"parameter": "PDT"
} | [
"CMF042",
"CMF041"
] | {
"PDT": 17.8,
"a": 0.035
} | growth_yield | {
"expression": "if PDT<28: 1; else: exp(-a*(PDT-28))",
"id": "CMF030",
"latex": "F_{PA}=1\\;(PDT<28),\\quad e^{-a(PDT-28)}\\;(PDT\\ge 28)",
"name": "生理年龄因子",
"name_en": "Physiological age factor",
"precision": 2,
"unit": "fraction"
} | CMF030_sensitivity_t2_0000 | false | Candidate formulas are listed below. Select the physiologically appropriate formula before calculating.
Candidate 1:
Formula name: Green leaf partitioning index (绿叶分配指数)
Formula (LaTeX): PIGL=0.54-0.0046PDT\;(PDT<26),\quad1.4532e^{-0.0492PDT}\;(PDT\ge26)
Formula: PIGL = if PDT<26: 0.54-0.0046*PDT; else: 1.4532*exp(-0.... | Formula name: Physiological age factor (生理年龄因子)
Formula (LaTeX): F_{PA}=1\;(PDT<28),\quad e^{-a(PDT-28)}\;(PDT\ge 28)
Formula: FPA = if PDT<28: 1; else: exp(-a*(PDT-28))
Operator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the... | Formula name: Physiological age factor (生理年龄因子)
Formula (LaTeX): F_{PA}=1\;(PDT<28),\quad e^{-a(PDT-28)}\;(PDT\ge 28)
Formula: FPA = if PDT<28: 1; else: exp(-a*(PDT-28))
Operator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the... | Formula: FPA = if PDT<28: 1; else: exp(-a*(PDT-28))
Operator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.
Output unit: fraction | Relevant formula name: Physiological age factor (生理年龄因子).
Formula: FPA = if PDT<28: 1; else: exp(-a*(PDT-28))
Operator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.
Output unit: fraction | A retrieved formula is listed below. It may look relevant, but use it only if it truly matches the problem.
Formula name: Panicle partitioning index (穗分配指数)
Formula (LaTeX): PIP=\frac{HI/0.87}{1+e^{-0.2804(PDT-39)}}
Formula: PIP = (Y_grain / B_above / 0.87) / (1 + exp(-0.2804 * (PDT - 39)))
Operator notes: * means mult... | 3 | Piecewise exponential decay | sensitivity | {
"PDT": 16.2,
"a": 0.035
} | In photosynthesis, we analyze how Physiological age factor responds to parameter changes. Baseline: PDT=16.2 physiological days, a=0.035 dimensionless. After changing PDT from 16.2 to 17.8, compute the new Physiological age factor. Give the numeric value in fraction. | {
"C": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ..... | Identify the target quantity as Physiological age factor.
Use the verified formula:
\[F_{PA}=1\;(PDT<28),\quad e^{-a(PDT-28)}\;(PDT\ge 28)\]
In calculation-friendly notation, this is:
FPA = if PDT<28: 1; else: exp(-a*(PDT-28)).
Substitute the relevant values:
PDT=17.8 physiological days, a=0.035 dimensionless.
The comp... | dev | Photosynthesis | 2 | CMF042 |
0.94 | cropmath-v1 | null | [
"CMF052",
"CMF032"
] | {
"PDT": 29.3,
"Qt": 7.8,
"Tmean": 18.4
} | growth_yield | {
"expression": "if 26<=PDT<=39 and Tmean<22: 1-(4.6+0.054*Qt^1.56)/100; else: 1",
"id": "CMF039",
"latex": "LTF=1-\\frac{4.6+0.054Q_t^{1.56}}{100}",
"name": "低温结实因子",
"name_en": "Low-temperature fertility factor",
"precision": 2,
"unit": "fraction"
} | CMF039_direct_t0_0001 | false | Candidate formulas are listed below. Select the physiologically appropriate formula before calculating.
Candidate 1:
Formula name: Low-temperature fertility factor (低温结实因子)
Formula (LaTeX): LTF=1-\frac{4.6+0.054Q_t^{1.56}}{100}
Formula: LTF = if 26<=PDT<=39 and Tmean<22: 1-(4.6+0.054*Qt^1.56)/100; else: 1
Operator not... | Formula name: Low-temperature fertility factor (低温结实因子)
Formula (LaTeX): LTF=1-\frac{4.6+0.054Q_t^{1.56}}{100}
Formula: LTF = if 26<=PDT<=39 and Tmean<22: 1-(4.6+0.054*Qt^1.56)/100; else: 1
Operator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, an... | Formula name: Low-temperature fertility factor (低温结实因子)
Formula (LaTeX): LTF=1-\frac{4.6+0.054Q_t^{1.56}}{100}
Formula: LTF = if 26<=PDT<=39 and Tmean<22: 1-(4.6+0.054*Qt^1.56)/100; else: 1
Operator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, an... | Formula: LTF = if 26<=PDT<=39 and Tmean<22: 1-(4.6+0.054*Qt^1.56)/100; else: 1
Operator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.
Output unit: fraction | Relevant formula name: Low-temperature fertility factor (低温结实因子).
Formula: LTF = if 26<=PDT<=39 and Tmean<22: 1-(4.6+0.054*Qt^1.56)/100; else: 1
Operator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.
Ou... | A retrieved formula is listed below. It may look relevant, but use it only if it truly matches the problem.
Formula name: Actual LAI relative growth rate (实际叶面积相对增长率)
Formula (LaTeX): R=RP\cdot\min(NNI,WDF)
Formula: R = RP * min(NNI, WDF)
Operator notes: * means multiplication, / means division, ** or ^ means power, ex... | 2 | Piecewise power penalty | direct | {
"PDT": 29.3,
"Qt": 7.8,
"Tmean": 18.4
} | A temperature stress calculation estimates Low-temperature fertility factor. The measured inputs are PDT=29.3 physiological days, Tmean=18.4 C, Qt=7.8 C day. What is the low-temperature fertility factor? Give the final numeric value in fraction. | {
"C": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ..... | Identify the target quantity as Low-temperature fertility factor.
Use the verified formula:
\[LTF=1-\frac{4.6+0.054Q_t^{1.56}}{100}\]
In calculation-friendly notation, this is:
LTF = if 26<=PDT<=39 and Tmean<22: 1-(4.6+0.054*Qt^1.56)/100; else: 1.
Substitute the relevant values:
PDT=29.3 physiological days, Tmean=18.4 ... | dev | Temperature stress | 0 | CMF052 |
0.922 | cropmath-v1 | null | [
"CMF011",
"CMF002"
] | {
"Crop_Water_max": 0.42,
"Crop_Water_min": 0.22,
"Soil_Water": 0.511,
"Wilt_point": 0.08
} | carbon_nitrogen_cycle | {
"expression": "if Soil_Water<=Wilt_point: 0; if Soil_Water<Crop_Water_min: (Soil_Water-Wilt_point)/(Crop_Water_min-Wilt_point); if Soil_Water<Crop_Water_max: 1; else: 0.5*(1+(1-Soil_Water)/(1-Crop_Water_max))",
"id": "CMF014",
"latex": "F_W=f(\\theta,\\theta_{wp},\\theta_{min},\\theta_{max})",
"name": "作物水分胁迫... | CMF014_branch_t1_0001 | false | Candidate formulas are listed below. Select the physiologically appropriate formula before calculating.
Candidate 1:
Formula name: carbon-nitrogen cycling soil carbon temperature factor (土壤碳分解温度因子)
Formula (LaTeX): F_T=2.4^{(T_{soil}-10)/10}\;\text{where}\;T_{soil}=T\;\text{or}\;4.4+0.76T
Formula: F_T_soil = if is_soi... | Formula name: carbon-nitrogen cycling crop water stress factor (作物水分胁迫因子)
Formula (LaTeX): F_W=f(\theta,\theta_{wp},\theta_{min},\theta_{max})
Formula: F_Water = if Soil_Water<=Wilt_point: 0; if Soil_Water<Crop_Water_min: (Soil_Water-Wilt_point)/(Crop_Water_min-Wilt_point); if Soil_Water<Crop_Water_max: 1; else: 0.5*(1... | Formula name: carbon-nitrogen cycling crop water stress factor (作物水分胁迫因子)
Formula (LaTeX): F_W=f(\theta,\theta_{wp},\theta_{min},\theta_{max})
Formula: F_Water = if Soil_Water<=Wilt_point: 0; if Soil_Water<Crop_Water_min: (Soil_Water-Wilt_point)/(Crop_Water_min-Wilt_point); if Soil_Water<Crop_Water_max: 1; else: 0.5*(1... | Formula: F_Water = if Soil_Water<=Wilt_point: 0; if Soil_Water<Crop_Water_min: (Soil_Water-Wilt_point)/(Crop_Water_min-Wilt_point); if Soil_Water<Crop_Water_max: 1; else: 0.5*(1+(1-Soil_Water)/(1-Crop_Water_max))
Operator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) mean... | Relevant formula name: carbon-nitrogen cycling crop water stress factor (作物水分胁迫因子).
Formula: F_Water = if Soil_Water<=Wilt_point: 0; if Soil_Water<Crop_Water_min: (Soil_Water-Wilt_point)/(Crop_Water_min-Wilt_point); if Soil_Water<Crop_Water_max: 1; else: 0.5*(1+(1-Soil_Water)/(1-Crop_Water_max))
Operator notes: * means... | A retrieved formula is listed below. It may look relevant, but use it only if it truly matches the problem.
Formula name: carbon-nitrogen cycling soil carbon temperature factor (土壤碳分解温度因子)
Formula (LaTeX): F_T=2.4^{(T_{soil}-10)/10}\;\text{where}\;T_{soil}=T\;\text{or}\;4.4+0.76T
Formula: F_T_soil = if is_soil_temp>=0.... | 4 | Piecewise water response | branch | {
"Crop_Water_max": 0.42,
"Crop_Water_min": 0.22,
"Soil_Water": 0.511,
"Wilt_point": 0.08
} | For carbon-nitrogen cycling water, compute the carbon-nitrogen cycling crop water stress factor. The inputs are Soil_Water=0.511 cm3/cm3, Crop_Water_max=0.42 cm3/cm3, Crop_Water_min=0.22 cm3/cm3, Wilt_point=0.08 cm3/cm3. Note that this formula uses piecewise logic depending on parameter values. Report the final numeric... | {
"C": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ..... | Identify the target quantity as carbon-nitrogen cycling crop water stress factor.
Use the verified formula:
\[F_W=f(\theta,\theta_{wp},\theta_{min},\theta_{max})\]
In calculation-friendly notation, this is:
F_Water = if Soil_Water<=Wilt_point: 0; if Soil_Water<Crop_Water_min: (Soil_Water-Wilt_point)/(Crop_Water_min-Wil... | dev | carbon-nitrogen cycling water | 1 | CMF011 |
52.19 | cropmath-v1 | null | [
"CMF039",
"CMF053"
] | {
"ABIOMASS": 7608,
"PDT": 39.5,
"Q10": 2.17,
"T": 31.8,
"To": 28.1
} | growth_yield | {
"expression": "(0.0091 - 0.0001 * PDT) * ABIOMASS * Q10**((T-To)/10)",
"id": "CMF048",
"latex": "RM=(0.0091-0.0001\\cdot PDT)\\cdot ABIOMASS\\cdot Q_{10}^{(T-T_o)/10}",
"name": "维持呼吸",
"name_en": "Maintenance respiration",
"precision": 2,
"unit": "kg/ha/day"
} | CMF048_direct_t1_0001 | false | Candidate formulas are listed below. Select the physiologically appropriate formula before calculating.
Candidate 1:
Formula name: Low-temperature fertility factor (低温结实因子)
Formula (LaTeX): LTF=1-\frac{4.6+0.054Q_t^{1.56}}{100}
Formula: LTF = if 26<=PDT<=39 and Tmean<22: 1-(4.6+0.054*Qt^1.56)/100; else: 1
Operator not... | Formula name: Maintenance respiration (维持呼吸)
Formula (LaTeX): RM=(0.0091-0.0001\cdot PDT)\cdot ABIOMASS\cdot Q_{10}^{(T-T_o)/10}
Formula: RM = (0.0091 - 0.0001 * PDT) * ABIOMASS * Q10**((T-To)/10)
Operator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logari... | Formula name: Maintenance respiration (维持呼吸)
Formula (LaTeX): RM=(0.0091-0.0001\cdot PDT)\cdot ABIOMASS\cdot Q_{10}^{(T-T_o)/10}
Formula: RM = (0.0091 - 0.0001 * PDT) * ABIOMASS * Q10**((T-To)/10)
Operator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logari... | Formula: RM = (0.0091 - 0.0001 * PDT) * ABIOMASS * Q10**((T-To)/10)
Operator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.
Output unit: kg/ha/day | Relevant formula name: Maintenance respiration (维持呼吸).
Formula: RM = (0.0091 - 0.0001 * PDT) * ABIOMASS * Q10**((T-To)/10)
Operator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.
Output unit: kg/ha/day | A retrieved formula is listed below. It may look relevant, but use it only if it truly matches the problem.
Formula name: Low-temperature fertility factor (低温结实因子)
Formula (LaTeX): LTF=1-\frac{4.6+0.054Q_t^{1.56}}{100}
Formula: LTF = if 26<=PDT<=39 and Tmean<22: 1-(4.6+0.054*Qt^1.56)/100; else: 1
Operator notes: * mean... | 2 | Product with Q10 temperature response | direct | {
"ABIOMASS": 7608,
"PDT": 39.5,
"Q10": 2.17,
"T": 31.8,
"To": 28.1
} | Given PDT=39.5 physiological days, ABIOMASS=7608.0 kg/ha, Q10=2.17 dimensionless, T=31.8 C, To=28.1 C, calculate the Maintenance respiration for a respiration. The result should be in kg/ha/day. | {
"C": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ..... | Identify the target quantity as Maintenance respiration.
Use the verified formula:
\[RM=(0.0091-0.0001\cdot PDT)\cdot ABIOMASS\cdot Q_{10}^{(T-T_o)/10}\]
In calculation-friendly notation, this is:
RM = (0.0091 - 0.0001 * PDT) * ABIOMASS * Q10**((T-To)/10).
Substitute the relevant values:
PDT=39.5 physiological days, AB... | dev | Respiration | 1 | CMF039 |
0.32 | cropmath-v1 | {
"new_value": 5.6,
"old_value": 4.9,
"parameter": "PDT"
} | [
"CMF036",
"CMF040"
] | {
"PDT": 5.6
} | growth_yield | {
"expression": "1 - (-0.0000842*PDT^2 + 0.01*PDT + 0.63)",
"id": "CMF043",
"latex": "PIRO=1-PISH",
"name": "地下部分配指数",
"name_en": "Root partitioning index",
"precision": 2,
"unit": "fraction"
} | CMF043_sensitivity_t2_0001 | false | Candidate formulas are listed below. Select the physiologically appropriate formula before calculating.
Candidate 1:
Formula name: Canopy extinction coefficient (冠层消光系数)
Formula (LaTeX): K=0.0087PDT+0.2222
Formula: K = 0.0087 * PDT + 0.2222
Operator notes: * means multiplication, / means division, ** or ^ means power,... | Formula name: Root partitioning index (地下部分配指数)
Formula (LaTeX): PIRO=1-PISH
Formula: PIRO = 1 - (-0.0000842*PDT^2 + 0.01*PDT + 0.63)
Operator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.
Output unit: ... | Formula name: Root partitioning index (地下部分配指数)
Formula (LaTeX): PIRO=1-PISH
Formula: PIRO = 1 - (-0.0000842*PDT^2 + 0.01*PDT + 0.63)
Operator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.
Output unit: ... | Formula: PIRO = 1 - (-0.0000842*PDT^2 + 0.01*PDT + 0.63)
Operator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.
Output unit: fraction | Relevant formula name: Root partitioning index (地下部分配指数).
Formula: PIRO = 1 - (-0.0000842*PDT^2 + 0.01*PDT + 0.63)
Operator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.
Output unit: fraction | A retrieved formula is listed below. It may look relevant, but use it only if it truly matches the problem.
Formula name: Canopy extinction coefficient (冠层消光系数)
Formula (LaTeX): K=0.0087PDT+0.2222
Formula: K = 0.0087 * PDT + 0.2222
Operator notes: * means multiplication, / means division, ** or ^ means power, exp(x) me... | 2 | Residual from shoot partitioning | sensitivity | {
"PDT": 4.9
} | In dry matter partitioning, we analyze how Root partitioning index responds to parameter changes. Baseline: PDT=4.9 physiological days. After changing PDT from 4.9 to 5.6, compute the new Root partitioning index. Give the numeric value in fraction. | {
"C": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ..... | Identify the target quantity as Root partitioning index.
Use the verified formula:
\[PIRO=1-PISH\]
In calculation-friendly notation, this is:
PIRO = 1 - (-0.0000842*PDT^2 + 0.01*PDT + 0.63).
Substitute the relevant values:
PDT=5.6 physiological days.
The computed value is 0.32 fraction.
Therefore, the final numeric ans... | dev | Dry matter partitioning | 2 | CMF036 |
0.000 | cropmath-v1 | {
"new_value": 25.4,
"old_value": 23.1,
"parameter": "Topt"
} | [
"CMF056",
"CMF059"
] | {
"Tbase": 4.3,
"Tmean": 0.2,
"Topt": 25.4
} | environmental_response | {
"expression": "if Tmean>=Topt: 1; else: max((Tmean-Tbase)/(Topt-Tbase), 0)",
"id": "CMF060",
"latex": "F_T=1\\;\\text{or}\\;\\max((T-T_b)/(T_o-T_b),0)",
"name": "环境响应温度响应",
"name_en": "environmental-response temperature response",
"precision": 3,
"unit": "fraction"
} | CMF060_sensitivity_t1_0000 | false | Candidate formulas are listed below. Select the physiologically appropriate formula before calculating.
Candidate 1:
Formula name: environmental-response temperature response (环境响应温度响应)
Formula (LaTeX): F_T=1\;\text{or}\;\max((T-T_b)/(T_o-T_b),0)
Formula: F_Temp = if Tmean>=Topt: 1; else: max((Tmean-Tbase)/(Topt-Tbase... | Formula name: environmental-response temperature response (环境响应温度响应)
Formula (LaTeX): F_T=1\;\text{or}\;\max((T-T_b)/(T_o-T_b),0)
Formula: F_Temp = if Tmean>=Topt: 1; else: max((Tmean-Tbase)/(Topt-Tbase), 0)
Operator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means nat... | Formula name: environmental-response temperature response (环境响应温度响应)
Formula (LaTeX): F_T=1\;\text{or}\;\max((T-T_b)/(T_o-T_b),0)
Formula: F_Temp = if Tmean>=Topt: 1; else: max((Tmean-Tbase)/(Topt-Tbase), 0)
Operator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means nat... | Formula: F_Temp = if Tmean>=Topt: 1; else: max((Tmean-Tbase)/(Topt-Tbase), 0)
Operator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.
Output unit: fraction | Relevant formula name: environmental-response temperature response (环境响应温度响应).
Formula: F_Temp = if Tmean>=Topt: 1; else: max((Tmean-Tbase)/(Topt-Tbase), 0)
Operator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larg... | A retrieved formula is listed below. It may look relevant, but use it only if it truly matches the problem.
Formula name: environmental-response daily thermal time (每日有效积温)
Formula (LaTeX): dTT=\max(T_{mean}-T_{base},0)
Formula: dTT = max(Tmean - Tbase, 0)
Operator notes: * means multiplication, / means division, ** or... | 3 | Piecewise linear | sensitivity | {
"Tbase": 4.3,
"Tmean": 0.2,
"Topt": 23.1
} | Given current measurements Tmean=0.2 C, Tbase=4.3 C, Topt=23.1 C in environmental-response formula system, what would the environmental-response temperature response be if Topt were adjusted to 25.4? Provide the numeric value in fraction. | {
"C": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ..... | Identify the target quantity as environmental-response temperature response.
Use the verified formula:
\[F_T=1\;\text{or}\;\max((T-T_b)/(T_o-T_b),0)\]
In calculation-friendly notation, this is:
F_Temp = if Tmean>=Topt: 1; else: max((Tmean-Tbase)/(Topt-Tbase), 0).
Substitute the relevant values:
Tmean=0.2 C, Tbase=4.3 C... | dev | environmental-response formula system | 1 | CMF056 |
1.100 | cropmath-v1 | null | [
"CMF008",
"CMF003"
] | {
"Temp": 8.8,
"is_soil_temp": 0
} | carbon_nitrogen_cycle | {
"expression": "if is_soil_temp>=0.5: 2.4**((Temp-10)*0.1); else: 2.4**((4.4+0.76*Temp-10)*0.1)",
"id": "CMF011",
"latex": "F_T=2.4^{(T_{soil}-10)/10}\\;\\text{where}\\;T_{soil}=T\\;\\text{or}\\;4.4+0.76T",
"name": "土壤碳分解温度因子",
"name_en": "carbon-nitrogen cycling soil carbon temperature factor",
"precision... | CMF011_branch_t0_0000 | false | Candidate formulas are listed below. Select the physiologically appropriate formula before calculating.
Candidate 1:
Formula name: carbon-nitrogen cycling soil carbon temperature factor (土壤碳分解温度因子)
Formula (LaTeX): F_T=2.4^{(T_{soil}-10)/10}\;\text{where}\;T_{soil}=T\;\text{or}\;4.4+0.76T
Formula: F_T_soil = if is_soi... | Formula name: carbon-nitrogen cycling soil carbon temperature factor (土壤碳分解温度因子)
Formula (LaTeX): F_T=2.4^{(T_{soil}-10)/10}\;\text{where}\;T_{soil}=T\;\text{or}\;4.4+0.76T
Formula: F_T_soil = if is_soil_temp>=0.5: 2.4**((Temp-10)*0.1); else: 2.4**((4.4+0.76*Temp-10)*0.1)
Operator notes: * means multiplication, / means... | Formula name: carbon-nitrogen cycling soil carbon temperature factor (土壤碳分解温度因子)
Formula (LaTeX): F_T=2.4^{(T_{soil}-10)/10}\;\text{where}\;T_{soil}=T\;\text{or}\;4.4+0.76T
Formula: F_T_soil = if is_soil_temp>=0.5: 2.4**((Temp-10)*0.1); else: 2.4**((4.4+0.76*Temp-10)*0.1)
Operator notes: * means multiplication, / means... | Formula: F_T_soil = if is_soil_temp>=0.5: 2.4**((Temp-10)*0.1); else: 2.4**((4.4+0.76*Temp-10)*0.1)
Operator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.
Output unit: fraction | Relevant formula name: carbon-nitrogen cycling soil carbon temperature factor (土壤碳分解温度因子).
Formula: F_T_soil = if is_soil_temp>=0.5: 2.4**((Temp-10)*0.1); else: 2.4**((4.4+0.76*Temp-10)*0.1)
Operator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, a... | A retrieved formula is listed below. It may look relevant, but use it only if it truly matches the problem.
Formula name: carbon-nitrogen cycling plant maintenance respiration (植株维持呼吸)
Formula (LaTeX): R_m=0.0432\cdot24(Biom/0.45)C_m2^{(T-25)/10}
Formula: Rm = 0.0432*24*(Biom_plant_yesterday/0.45)*Coef_Rm*2^((Tmn-25)/1... | 4 | Q10 temperature response | branch | {
"Temp": 8.8,
"is_soil_temp": 0
} | A carbon-nitrogen cycling soil carbon involves a stage that may fall on either side of a piecewise threshold. The measured inputs are Temp=8.8 C, is_soil_temp=0.0 0/1. Determine the carbon-nitrogen cycling soil carbon temperature factor and report the final numeric value in fraction. | {
"C": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ..... | Identify the target quantity as carbon-nitrogen cycling soil carbon temperature factor.
Use the verified formula:
\[F_T=2.4^{(T_{soil}-10)/10}\;\text{where}\;T_{soil}=T\;\text{or}\;4.4+0.76T\]
In calculation-friendly notation, this is:
F_T_soil = if is_soil_temp>=0.5: 2.4**((Temp-10)*0.1); else: 2.4**((4.4+0.76*Temp-10... | dev | carbon-nitrogen cycling soil carbon | 0 | CMF008 |
1.000 | cropmath-v1 | null | [
"CMF001",
"CMF006"
] | {
"Crop_Water_max": 0.42,
"Crop_Water_min": 0.22,
"Soil_Water": 0.28800000000000003,
"Wilt_point": 0.08
} | carbon_nitrogen_cycle | {
"expression": "if Soil_Water<=Wilt_point: 0; if Soil_Water<Crop_Water_min: (Soil_Water-Wilt_point)/(Crop_Water_min-Wilt_point); if Soil_Water<Crop_Water_max: 1; else: 0.5*(1+(1-Soil_Water)/(1-Crop_Water_max))",
"id": "CMF014",
"latex": "F_W=f(\\theta,\\theta_{wp},\\theta_{min},\\theta_{max})",
"name": "作物水分胁迫... | CMF014_direct_t2_0000 | false | Candidate formulas are listed below. Select the physiologically appropriate formula before calculating.
Candidate 1:
Formula name: carbon-nitrogen cycling canopy absorbed PAR (冠层吸收光合有效辐射)
Formula (LaTeX): PAR=Radi\cdot0.44\cdot4.6(1-\rho)\frac{1-e^{-KLAI}}{KLAI}
Formula: PAR = Radi * 0.44 * 4.6 * (1-Coef_rou) * (1-exp... | Formula name: carbon-nitrogen cycling crop water stress factor (作物水分胁迫因子)
Formula (LaTeX): F_W=f(\theta,\theta_{wp},\theta_{min},\theta_{max})
Formula: F_Water = if Soil_Water<=Wilt_point: 0; if Soil_Water<Crop_Water_min: (Soil_Water-Wilt_point)/(Crop_Water_min-Wilt_point); if Soil_Water<Crop_Water_max: 1; else: 0.5*(1... | Formula name: carbon-nitrogen cycling crop water stress factor (作物水分胁迫因子)
Formula (LaTeX): F_W=f(\theta,\theta_{wp},\theta_{min},\theta_{max})
Formula: F_Water = if Soil_Water<=Wilt_point: 0; if Soil_Water<Crop_Water_min: (Soil_Water-Wilt_point)/(Crop_Water_min-Wilt_point); if Soil_Water<Crop_Water_max: 1; else: 0.5*(1... | Formula: F_Water = if Soil_Water<=Wilt_point: 0; if Soil_Water<Crop_Water_min: (Soil_Water-Wilt_point)/(Crop_Water_min-Wilt_point); if Soil_Water<Crop_Water_max: 1; else: 0.5*(1+(1-Soil_Water)/(1-Crop_Water_max))
Operator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) mean... | Relevant formula name: carbon-nitrogen cycling crop water stress factor (作物水分胁迫因子).
Formula: F_Water = if Soil_Water<=Wilt_point: 0; if Soil_Water<Crop_Water_min: (Soil_Water-Wilt_point)/(Crop_Water_min-Wilt_point); if Soil_Water<Crop_Water_max: 1; else: 0.5*(1+(1-Soil_Water)/(1-Crop_Water_max))
Operator notes: * means... | A retrieved formula is listed below. It may look relevant, but use it only if it truly matches the problem.
Formula name: carbon-nitrogen cycling CO2 fertilization factor (作物二氧化碳施肥因子)
Formula (LaTeX): F_{CO2}=1+B\ln(CO2/340)
Formula: F_CO2 = 1 + Coef_B_CO2 * ln(CO2 / 340)
Operator notes: * means multiplication, / means... | 2 | Piecewise water response | direct | {
"Crop_Water_max": 0.42,
"Crop_Water_min": 0.22,
"Soil_Water": 0.28800000000000003,
"Wilt_point": 0.08
} | In a carbon-nitrogen cycling water, the carbon-nitrogen cycling crop water stress factor is computed as part of carbon-nitrogen cycling water modeling. With Soil_Water=0.288 cm3/cm3, Crop_Water_max=0.42 cm3/cm3, Crop_Water_min=0.22 cm3/cm3, Wilt_point=0.08 cm3/cm3, determine the numeric value in fraction. | {
"C": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ..... | Identify the target quantity as carbon-nitrogen cycling crop water stress factor.
Use the verified formula:
\[F_W=f(\theta,\theta_{wp},\theta_{min},\theta_{max})\]
In calculation-friendly notation, this is:
F_Water = if Soil_Water<=Wilt_point: 0; if Soil_Water<Crop_Water_min: (Soil_Water-Wilt_point)/(Crop_Water_min-Wil... | dev | carbon-nitrogen cycling water | 2 | CMF001 |
1.266 | cropmath-v1 | null | [
"CMF056",
"CMF062"
] | {
"CO2": 630,
"CO2_RUE": 0.095
} | environmental_response | {
"expression": "if CO2>=700: 1+CO2_RUE*350/100; else: max(CO2_RUE*CO2*0.01+1-0.01*350*CO2_RUE, 1)",
"id": "CMF057",
"latex": "F_{CO2}=f(CO2,CO2_{RUE})",
"name": "环境响应二氧化碳响应",
"name_en": "environmental-response CO2 response",
"precision": 3,
"unit": "multiplier"
} | CMF057_direct_t1_0000 | false | Candidate formulas are listed below. Select the physiologically appropriate formula before calculating.
Candidate 1:
Formula name: environmental-response CO2 response (环境响应二氧化碳响应)
Formula (LaTeX): F_{CO2}=f(CO2,CO2_{RUE})
Formula: F_CO2 = if CO2>=700: 1+CO2_RUE*350/100; else: max(CO2_RUE*CO2*0.01+1-0.01*350*CO2_RUE, 1... | Formula name: environmental-response CO2 response (环境响应二氧化碳响应)
Formula (LaTeX): F_{CO2}=f(CO2,CO2_{RUE})
Formula: F_CO2 = if CO2>=700: 1+CO2_RUE*350/100; else: max(CO2_RUE*CO2*0.01+1-0.01*350*CO2_RUE, 1)
Operator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural... | Formula name: environmental-response CO2 response (环境响应二氧化碳响应)
Formula (LaTeX): F_{CO2}=f(CO2,CO2_{RUE})
Formula: F_CO2 = if CO2>=700: 1+CO2_RUE*350/100; else: max(CO2_RUE*CO2*0.01+1-0.01*350*CO2_RUE, 1)
Operator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural... | Formula: F_CO2 = if CO2>=700: 1+CO2_RUE*350/100; else: max(CO2_RUE*CO2*0.01+1-0.01*350*CO2_RUE, 1)
Operator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.
Output unit: fraction | Relevant formula name: environmental-response CO2 response (环境响应二氧化碳响应).
Formula: F_CO2 = if CO2>=700: 1+CO2_RUE*350/100; else: max(CO2_RUE*CO2*0.01+1-0.01*350*CO2_RUE, 1)
Operator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select t... | A retrieved formula is listed below. It may look relevant, but use it only if it truly matches the problem.
Formula name: environmental-response daily thermal time (每日有效积温)
Formula (LaTeX): dTT=\max(T_{mean}-T_{base},0)
Formula: dTT = max(Tmean - Tbase, 0)
Operator notes: * means multiplication, / means division, ** or... | 2 | Piecewise linear cap | direct | {
"CO2": 630,
"CO2_RUE": 0.095
} | Given CO2=630.0 ppm, CO2_RUE=0.095 fraction, calculate the environmental-response CO2 response for an environmental-response formula system. The result should be in fraction. | {
"C": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ..... | Identify the target quantity as environmental-response CO2 response.
Use the verified formula:
\[F_{CO2}=f(CO2,CO2_{RUE})\]
In calculation-friendly notation, this is:
F_CO2 = if CO2>=700: 1+CO2_RUE*350/100; else: max(CO2_RUE*CO2*0.01+1-0.01*350*CO2_RUE, 1).
Substitute the relevant values:
CO2=630.0 ppm, CO2_RUE=0.095 f... | dev | environmental-response formula system | 1 | CMF056 |
0.308 | cropmath-v1 | {
"new_value": 40,
"old_value": 39.5,
"parameter": "T"
} | [
"CMF036",
"CMF052"
] | {
"T": 40,
"TS": 1.76,
"Tb": 10,
"Tm": 42,
"To": 30
} | growth_yield | {
"expression": "(((T - Tb)/(To - Tb)) * (((Tm - T)/(Tm - To)) ** ((Tm - To)/(To - Tb)))) ** TS",
"id": "CMF051",
"latex": "RTE=\\left[\\frac{T-T_b}{T_o-T_b}\\left(\\frac{T_m-T}{T_m-T_o}\\right)^{(T_m-T_o)/(T_o-T_b)}\\right]^{TS}",
"name": "相对热效应",
"name_en": "Relative thermal effectiveness",
"precision": 3... | CMF051_sensitivity_t0_0001 | false | Candidate formulas are listed below. Select the physiologically appropriate formula before calculating.
Candidate 1:
Formula name: Actual LAI relative growth rate (实际叶面积相对增长率)
Formula (LaTeX): R=RP\cdot\min(NNI,WDF)
Formula: R = RP * min(NNI, WDF)
Operator notes: * means multiplication, / means division, ** or ^ means... | Formula name: Relative thermal effectiveness (相对热效应)
Formula (LaTeX): RTE=\left[\frac{T-T_b}{T_o-T_b}\left(\frac{T_m-T}{T_m-T_o}\right)^{(T_m-T_o)/(T_o-T_b)}\right]^{TS}
Formula: RTE = (((T - Tb)/(To - Tb)) * (((Tm - T)/(Tm - To)) ** ((Tm - To)/(To - Tb)))) ** TS
Operator notes: * means multiplication, / means division... | Formula name: Relative thermal effectiveness (相对热效应)
Formula (LaTeX): RTE=\left[\frac{T-T_b}{T_o-T_b}\left(\frac{T_m-T}{T_m-T_o}\right)^{(T_m-T_o)/(T_o-T_b)}\right]^{TS}
Formula: RTE = (((T - Tb)/(To - Tb)) * (((Tm - T)/(Tm - To)) ** ((Tm - To)/(To - Tb)))) ** TS
Operator notes: * means multiplication, / means division... | Formula: RTE = (((T - Tb)/(To - Tb)) * (((Tm - T)/(Tm - To)) ** ((Tm - To)/(To - Tb)))) ** TS
Operator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.
Output unit: fraction | Relevant formula name: Relative thermal effectiveness (相对热效应).
Formula: RTE = (((T - Tb)/(To - Tb)) * (((Tm - T)/(Tm - To)) ** ((Tm - To)/(To - Tb)))) ** TS
Operator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larg... | A retrieved formula is listed below. It may look relevant, but use it only if it truly matches the problem.
Formula name: Canopy extinction coefficient (冠层消光系数)
Formula (LaTeX): K=0.0087PDT+0.2222
Formula: K = 0.0087 * PDT + 0.2222
Operator notes: * means multiplication, / means division, ** or ^ means power, exp(x) me... | 3 | Beta temperature response | sensitivity | {
"T": 39.5,
"TS": 1.76,
"Tb": 10,
"Tm": 42,
"To": 30
} | A phenology calculation involves sensitivity analysis. The current measurements are T=39.5 C, Tb=10.0 C, To=30.0 C, Tm=42.0 C, TS=1.76 dimensionless. A field technician asks what the relative thermal effectiveness would become if T changed to 40.0. Report the new final numeric value in fraction. | {
"C": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ..... | Identify the target quantity as Relative thermal effectiveness.
Use the verified formula:
\[RTE=\left[\frac{T-T_b}{T_o-T_b}\left(\frac{T_m-T}{T_m-T_o}\right)^{(T_m-T_o)/(T_o-T_b)}\right]^{TS}\]
In calculation-friendly notation, this is:
RTE = (((T - Tb)/(To - Tb)) * (((Tm - T)/(Tm - To)) ** ((Tm - To)/(To - Tb)))) ** T... | dev | Phenology | 0 | CMF036 |
0.294 | cropmath-v1 | {
"new_value": 8.2,
"old_value": 7.1,
"parameter": "PDT"
} | [
"CMF046",
"CMF053"
] | {
"PDT": 8.2
} | growth_yield | {
"expression": "0.0087 * PDT + 0.2222",
"id": "CMF036",
"latex": "K=0.0087PDT+0.2222",
"name": "冠层消光系数",
"name_en": "Canopy extinction coefficient",
"precision": 3,
"unit": "dimensionless"
} | CMF036_sensitivity_t0_0001 | false | Candidate formulas are listed below. Select the physiologically appropriate formula before calculating.
Candidate 1:
Formula name: Canopy extinction coefficient (冠层消光系数)
Formula (LaTeX): K=0.0087PDT+0.2222
Formula: K = 0.0087 * PDT + 0.2222
Operator notes: * means multiplication, / means division, ** or ^ means power,... | Formula name: Canopy extinction coefficient (冠层消光系数)
Formula (LaTeX): K=0.0087PDT+0.2222
Formula: K = 0.0087 * PDT + 0.2222
Operator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.
Output unit: dimensionl... | Formula name: Canopy extinction coefficient (冠层消光系数)
Formula (LaTeX): K=0.0087PDT+0.2222
Formula: K = 0.0087 * PDT + 0.2222
Operator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.
Output unit: dimensionl... | Formula: K = 0.0087 * PDT + 0.2222
Operator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.
Output unit: dimensionless | Relevant formula name: Canopy extinction coefficient (冠层消光系数).
Formula: K = 0.0087 * PDT + 0.2222
Operator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.
Output unit: dimensionless | A retrieved formula is listed below. It may look relevant, but use it only if it truly matches the problem.
Formula name: Growth respiration (生长呼吸)
Formula (LaTeX): RG=R_g\cdot DTGA
Formula: RG = Rg * DTGA
Operator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natur... | 2 | Linear | sensitivity | {
"PDT": 7.1
} | A photosynthesis calculation involves sensitivity analysis. The current measurements are PDT=7.1 physiological days. A field technician asks what the canopy extinction coefficient would become if PDT changed to 8.2. Report the new final numeric value in dimensionless. | {
"C": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ..... | Identify the target quantity as Canopy extinction coefficient.
Use the verified formula:
\[K=0.0087PDT+0.2222\]
In calculation-friendly notation, this is:
K = 0.0087 * PDT + 0.2222.
Substitute the relevant values:
PDT=8.2 physiological days.
The computed value is 0.294 dimensionless.
Therefore, the final numeric answer... | dev | Photosynthesis | 0 | CMF046 |
1.00 | cropmath-v1 | {
"new_value": 33.8,
"old_value": 37.6,
"parameter": "T"
} | [
"CMF044",
"CMF038"
] | {
"T": 33.8,
"Tb": 10,
"Tmax": 45,
"Tol": 24,
"Tou": 34
} | growth_yield | {
"expression": "if Tb<=T<Tol: sin(pi/2*(T-Tb)/(Tol-Tb)); if Tol<=T<Tou: 1; if Tou<=T<Tmax: cos(pi/2*(T-Tou)/(Tmax-Tou)); else: 0",
"id": "CMF032",
"latex": "F_T=\\sin(\\frac{T-T_b}{T_{ol}-T_b}\\frac{\\pi}{2}); 1; \\cos(\\frac{T-T_{ou}}{T_{max}-T_{ou}}\\frac{\\pi}{2})",
"name": "温度响应因子",
"name_en": "Temperatu... | CMF032_sensitivity_t2_0001 | false | Candidate formulas are listed below. Select the physiologically appropriate formula before calculating.
Candidate 1:
Formula name: Temperature response factor (温度响应因子)
Formula (LaTeX): F_T=\sin(\frac{T-T_b}{T_{ol}-T_b}\frac{\pi}{2}); 1; \cos(\frac{T-T_{ou}}{T_{max}-T_{ou}}\frac{\pi}{2})
Formula: F_T = if Tb<=T<Tol: si... | Formula name: Temperature response factor (温度响应因子)
Formula (LaTeX): F_T=\sin(\frac{T-T_b}{T_{ol}-T_b}\frac{\pi}{2}); 1; \cos(\frac{T-T_{ou}}{T_{max}-T_{ou}}\frac{\pi}{2})
Formula: F_T = if Tb<=T<Tol: sin(pi/2*(T-Tb)/(Tol-Tb)); if Tol<=T<Tou: 1; if Tou<=T<Tmax: cos(pi/2*(T-Tou)/(Tmax-Tou)); else: 0
Operator notes: * mea... | Formula name: Temperature response factor (温度响应因子)
Formula (LaTeX): F_T=\sin(\frac{T-T_b}{T_{ol}-T_b}\frac{\pi}{2}); 1; \cos(\frac{T-T_{ou}}{T_{max}-T_{ou}}\frac{\pi}{2})
Formula: F_T = if Tb<=T<Tol: sin(pi/2*(T-Tb)/(Tol-Tb)); if Tol<=T<Tou: 1; if Tou<=T<Tmax: cos(pi/2*(T-Tou)/(Tmax-Tou)); else: 0
Operator notes: * mea... | Formula: F_T = if Tb<=T<Tol: sin(pi/2*(T-Tb)/(Tol-Tb)); if Tol<=T<Tou: 1; if Tou<=T<Tmax: cos(pi/2*(T-Tou)/(Tmax-Tou)); else: 0
Operator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.
Output unit: fracti... | Relevant formula name: Temperature response factor (温度响应因子).
Formula: F_T = if Tb<=T<Tol: sin(pi/2*(T-Tb)/(Tol-Tb)); if Tol<=T<Tou: 1; if Tou<=T<Tmax: cos(pi/2*(T-Tou)/(Tmax-Tou)); else: 0
Operator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and... | A retrieved formula is listed below. It may look relevant, but use it only if it truly matches the problem.
Formula name: Shoot partitioning index (地上部分配指数)
Formula (LaTeX): PISH=-8.42\times10^{-5}PDT^2+0.01PDT+0.63
Formula: PISH = -0.0000842 * PDT^2 + 0.01 * PDT + 0.63
Operator notes: * means multiplication, / means d... | 3 | Piecewise trigonometric | sensitivity | {
"T": 37.6,
"Tb": 10,
"Tmax": 45,
"Tol": 24,
"Tou": 34
} | In temperature response, we analyze how Temperature response factor responds to parameter changes. Baseline: T=37.6 C, Tb=10.0 C, Tol=24.0 C, Tou=34.0 C, Tmax=45.0 C. After changing T from 37.6 to 33.8, compute the new Temperature response factor. Give the numeric value in fraction. | {
"C": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ..... | Identify the target quantity as Temperature response factor.
Use the verified formula:
\[F_T=\sin(\frac{T-T_b}{T_{ol}-T_b}\frac{\pi}{2}); 1; \cos(\frac{T-T_{ou}}{T_{max}-T_{ou}}\frac{\pi}{2})\]
In calculation-friendly notation, this is:
F_T = if Tb<=T<Tol: sin(pi/2*(T-Tb)/(Tol-Tb)); if Tol<=T<Tou: 1; if Tou<=T<Tmax: co... | dev | Temperature response | 2 | CMF044 |
0.0046 | cropmath-v1 | null | [
"CMF049",
"CMF037"
] | {
"NNI": 0.79,
"RP": 0.0058000000000000005,
"WDF": 0.96
} | growth_yield | {
"expression": "RP * min(NNI, WDF)",
"id": "CMF052",
"latex": "R=RP\\cdot\\min(NNI,WDF)",
"name": "实际叶面积相对增长率",
"name_en": "Actual LAI relative growth rate",
"precision": 4,
"unit": "d-1"
} | CMF052_direct_t2_0001 | false | Candidate formulas are listed below. Select the physiologically appropriate formula before calculating.
Candidate 1:
Formula name: Actual LAI relative growth rate (实际叶面积相对增长率)
Formula (LaTeX): R=RP\cdot\min(NNI,WDF)
Formula: R = RP * min(NNI, WDF)
Operator notes: * means multiplication, / means division, ** or ^ means... | Formula name: Actual LAI relative growth rate (实际叶面积相对增长率)
Formula (LaTeX): R=RP\cdot\min(NNI,WDF)
Formula: R = RP * min(NNI, WDF)
Operator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.
Output unit: d-1... | Formula name: Actual LAI relative growth rate (实际叶面积相对增长率)
Formula (LaTeX): R=RP\cdot\min(NNI,WDF)
Formula: R = RP * min(NNI, WDF)
Operator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.
Output unit: d-1... | Formula: R = RP * min(NNI, WDF)
Operator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.
Output unit: d-1 | Relevant formula name: Actual LAI relative growth rate (实际叶面积相对增长率).
Formula: R = RP * min(NNI, WDF)
Operator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.
Output unit: d-1 | A retrieved formula is listed below. It may look relevant, but use it only if it truly matches the problem.
Formula name: Maintenance respiration coefficient at optimum temperature (最适温度维持呼吸系数)
Formula (LaTeX): RM(T_o)=0.0091-0.0001PDT
Formula: RMTo = 0.0091 - 0.0001 * PDT
Operator notes: * means multiplication, / mean... | 1 | Minimum stress multiplier | direct | {
"NNI": 0.79,
"RP": 0.0058000000000000005,
"WDF": 0.96
} | In a leaf area growth, the Actual LAI relative growth rate is computed as part of leaf area growth modeling. With RP=0.0058 d-1, NNI=0.79 fraction, WDF=0.96 fraction, determine the numeric value in d-1. | {
"C": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ..... | Identify the target quantity as Actual LAI relative growth rate.
Use the verified formula:
\[R=RP\cdot\min(NNI,WDF)\]
In calculation-friendly notation, this is:
R = RP * min(NNI, WDF).
Substitute the relevant values:
RP=0.0058 d-1, NNI=0.79 fraction, WDF=0.96 fraction.
The computed value is 0.0046 d-1.
Therefore, the f... | dev | Leaf area growth | 2 | CMF049 |
398.53 | cropmath-v1 | null | [
"CMF026",
"CMF030"
] | {
"DTGA": 502.1,
"RG": 61.2,
"RM": 62.3,
"b": 0.05
} | growth_yield | {
"expression": "(DTGA - RM - RG) / (1 - b)",
"id": "CMF033",
"latex": "GCR=\\frac{DTGA-RM-RG}{1-b}",
"name": "日干物质增量",
"name_en": "Daily dry matter increment",
"precision": 2,
"unit": "kg/ha/day"
} | CMF033_chain_2_t2_0001 | false | Candidate formulas are listed below. Select the physiologically appropriate formula before calculating.
Candidate 1:
Formula name: Actual maximum assimilation rate (实际最大同化速率)
Formula (LaTeX): AMAX=Am\cdot F_{CO2}\cdot F_{PA}\cdot F_T\cdot\min(NNI,WDF)
Formula: AMAX = Am * FCO2 * FPA * FT * min(NNI, WDF)
Operator notes... | Formula name: Daily dry matter increment (日干物质增量)
Formula (LaTeX): GCR=\frac{DTGA-RM-RG}{1-b}
Formula: GCR = (DTGA - RM - RG) / (1 - b)
Operator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.
Output unit... | Formula name: Daily dry matter increment (日干物质增量)
Formula (LaTeX): GCR=\frac{DTGA-RM-RG}{1-b}
Formula: GCR = (DTGA - RM - RG) / (1 - b)
Operator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.
Output unit... | Formula: GCR = (DTGA - RM - RG) / (1 - b)
Operator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.
Output unit: kg/ha/day | Relevant formula name: Daily dry matter increment (日干物质增量).
Formula: GCR = (DTGA - RM - RG) / (1 - b)
Operator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.
Output unit: kg/ha/day | A retrieved formula is listed below. It may look relevant, but use it only if it truly matches the problem.
Formula name: Actual maximum assimilation rate (实际最大同化速率)
Formula (LaTeX): AMAX=Am\cdot F_{CO2}\cdot F_{PA}\cdot F_T\cdot\min(NNI,WDF)
Formula: AMAX = Am * FCO2 * FPA * FT * min(NNI, WDF)
Operator notes: * means ... | 3 | Subtraction and conversion | chain_2 | {
"DTGA": 502.1,
"RG": 61.2,
"RM": 62.3,
"b": 0.05
} | In respiration, the Daily dry matter increment is obtained by chaining intermediate calculations. With inputs DTGA=502.1 kg/ha/day, RM=62.3 kg/ha/day, RG=61.2 kg/ha/day, b=0.05 fraction, determine the numeric value in kg/ha/day. | {
"C": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ..... | Identify the target quantity as Daily dry matter increment.
Use the verified formula:
\[GCR=\frac{DTGA-RM-RG}{1-b}\]
In calculation-friendly notation, this is:
GCR = (DTGA - RM - RG) / (1 - b).
Substitute the relevant values:
DTGA=502.1 kg/ha/day, RM=62.3 kg/ha/day, RG=61.2 kg/ha/day, b=0.05 fraction.
The computed valu... | dev | Respiration | 2 | CMF026 |
0.792 | cropmath-v1 | {
"new_value": 0.05,
"old_value": 0.045,
"parameter": "Ratio_N_C_photo_product"
} | [
"CMF011",
"CMF006"
] | {
"Coef_Plant_TN_uptake": 0.117,
"DVI": 0.49,
"NPP": 15.84,
"N_requirement_max_day": 2.09,
"N_shoot": 5.24,
"Ratio_N_C_photo_product": 0.05
} | carbon_nitrogen_cycle | {
"expression": "if NPP<0: 0; if N_shoot<3.5 and DVI<0.3: Coef_Plant_TN_uptake*N_shoot; else: min(N_requirement_max_day, Ratio_N_C_photo_product*NPP)",
"id": "CMF004",
"latex": "N_d=C_uN_{shoot}\\;\\text{or}\\;\\min(N_{max},r_NNPP)",
"name": "作物需氮量",
"name_en": "carbon-nitrogen cycling crop nitrogen demand",
... | CMF004_sensitivity_t1_0000 | false | Candidate formulas are listed below. Select the physiologically appropriate formula before calculating.
Candidate 1:
Formula name: carbon-nitrogen cycling crop nitrogen demand (作物需氮量)
Formula (LaTeX): N_d=C_uN_{shoot}\;\text{or}\;\min(N_{max},r_NNPP)
Formula: N_demand = if NPP<0: 0; if N_shoot<3.5 and DVI<0.3: Coef_Pl... | Formula name: carbon-nitrogen cycling crop nitrogen demand (作物需氮量)
Formula (LaTeX): N_d=C_uN_{shoot}\;\text{or}\;\min(N_{max},r_NNPP)
Formula: N_demand = if NPP<0: 0; if N_shoot<3.5 and DVI<0.3: Coef_Plant_TN_uptake*N_shoot; else: min(N_requirement_max_day, Ratio_N_C_photo_product*NPP)
Operator notes: * means multiplic... | Formula name: carbon-nitrogen cycling crop nitrogen demand (作物需氮量)
Formula (LaTeX): N_d=C_uN_{shoot}\;\text{or}\;\min(N_{max},r_NNPP)
Formula: N_demand = if NPP<0: 0; if N_shoot<3.5 and DVI<0.3: Coef_Plant_TN_uptake*N_shoot; else: min(N_requirement_max_day, Ratio_N_C_photo_product*NPP)
Operator notes: * means multiplic... | Formula: N_demand = if NPP<0: 0; if N_shoot<3.5 and DVI<0.3: Coef_Plant_TN_uptake*N_shoot; else: min(N_requirement_max_day, Ratio_N_C_photo_product*NPP)
Operator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger v... | Relevant formula name: carbon-nitrogen cycling crop nitrogen demand (作物需氮量).
Formula: N_demand = if NPP<0: 0; if N_shoot<3.5 and DVI<0.3: Coef_Plant_TN_uptake*N_shoot; else: min(N_requirement_max_day, Ratio_N_C_photo_product*NPP)
Operator notes: * means multiplication, / means division, ** or ^ means power, exp(x) mean... | A retrieved formula is listed below. It may look relevant, but use it only if it truly matches the problem.
Formula name: carbon-nitrogen cycling soil carbon temperature factor (土壤碳分解温度因子)
Formula (LaTeX): F_T=2.4^{(T_{soil}-10)/10}\;\text{where}\;T_{soil}=T\;\text{or}\;4.4+0.76T
Formula: F_T_soil = if is_soil_temp>=0.... | 4 | Piecewise minimum | sensitivity | {
"Coef_Plant_TN_uptake": 0.117,
"DVI": 0.49,
"NPP": 15.84,
"N_requirement_max_day": 2.09,
"N_shoot": 5.24,
"Ratio_N_C_photo_product": 0.045
} | Given current measurements N_shoot=5.24 g N/m2, Coef_Plant_TN_uptake=0.117 d-1, N_requirement_max_day=2.09 g N/m2/day, NPP=15.84 g C/m2/day, Ratio_N_C_photo_product=0.045 g N/g C, DVI=0.49 dimensionless in carbon-nitrogen cycling nitrogen, what would the carbon-nitrogen cycling crop nitrogen demand be if Ratio_N_C_phot... | {
"C": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ..... | Identify the target quantity as carbon-nitrogen cycling crop nitrogen demand.
Use the verified formula:
\[N_d=C_uN_{shoot}\;\text{or}\;\min(N_{max},r_NNPP)\]
In calculation-friendly notation, this is:
N_demand = if NPP<0: 0; if N_shoot<3.5 and DVI<0.3: Coef_Plant_TN_uptake*N_shoot; else: min(N_requirement_max_day, Rati... | dev | carbon-nitrogen cycling nitrogen | 1 | CMF011 |
0.481 | cropmath-v1 | null | [
"CMF048",
"CMF042"
] | {
"BFF": 0.72,
"DTE": 0.668,
"IE": 0.67,
"PDT": 39.5,
"RPE": 0.5
} | growth_yield | {
"expression": "if PDT<8: DTE*IE; if PDT<18: DTE*RPE; if PDT<32: DTE; if PDT<57: DTE*BFF; else: 0",
"id": "CMF027",
"latex": "DPE=DTE\\cdot IE,\\;DTE\\cdot RPE,\\;DTE,\\;DTE\\cdot BFF",
"name": "每日生理效应",
"name_en": "Daily physiological effectiveness",
"precision": 3,
"unit": "physiological days/day"
} | CMF027_direct_t1_0001 | false | Candidate formulas are listed below. Select the physiologically appropriate formula before calculating.
Candidate 1:
Formula name: Maintenance respiration (维持呼吸)
Formula (LaTeX): RM=(0.0091-0.0001\cdot PDT)\cdot ABIOMASS\cdot Q_{10}^{(T-T_o)/10}
Formula: RM = (0.0091 - 0.0001 * PDT) * ABIOMASS * Q10**((T-To)/10)
Opera... | Formula name: Daily physiological effectiveness (每日生理效应)
Formula (LaTeX): DPE=DTE\cdot IE,\;DTE\cdot RPE,\;DTE,\;DTE\cdot BFF
Formula: DPE = if PDT<8: DTE*IE; if PDT<18: DTE*RPE; if PDT<32: DTE; if PDT<57: DTE*BFF; else: 0
Operator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ... | Formula name: Daily physiological effectiveness (每日生理效应)
Formula (LaTeX): DPE=DTE\cdot IE,\;DTE\cdot RPE,\;DTE,\;DTE\cdot BFF
Formula: DPE = if PDT<8: DTE*IE; if PDT<18: DTE*RPE; if PDT<32: DTE; if PDT<57: DTE*BFF; else: 0
Operator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ... | Formula: DPE = if PDT<8: DTE*IE; if PDT<18: DTE*RPE; if PDT<32: DTE; if PDT<57: DTE*BFF; else: 0
Operator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.
Output unit: physiological days/day | Relevant formula name: Daily physiological effectiveness (每日生理效应).
Formula: DPE = if PDT<8: DTE*IE; if PDT<18: DTE*RPE; if PDT<32: DTE; if PDT<57: DTE*BFF; else: 0
Operator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the small... | A retrieved formula is listed below. It may look relevant, but use it only if it truly matches the problem.
Formula name: Maintenance respiration (维持呼吸)
Formula (LaTeX): RM=(0.0091-0.0001\cdot PDT)\cdot ABIOMASS\cdot Q_{10}^{(T-T_o)/10}
Formula: RM = (0.0091 - 0.0001 * PDT) * ABIOMASS * Q10**((T-To)/10)
Operator notes:... | 2 | Stage-dependent product | direct | {
"BFF": 0.72,
"DTE": 0.668,
"IE": 0.67,
"PDT": 39.5,
"RPE": 0.5
} | Given PDT=39.5 physiological days, DTE=0.668 fraction, IE=0.67 dimensionless, RPE=0.5 fraction, BFF=0.72 dimensionless, calculate the Daily physiological effectiveness for a phenology. The result should be in physiological days/day. | {
"C": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ..... | Identify the target quantity as Daily physiological effectiveness.
Use the verified formula:
\[DPE=DTE\cdot IE,\;DTE\cdot RPE,\;DTE,\;DTE\cdot BFF\]
In calculation-friendly notation, this is:
DPE = if PDT<8: DTE*IE; if PDT<18: DTE*RPE; if PDT<32: DTE; if PDT<57: DTE*BFF; else: 0.
Substitute the relevant values:
PDT=39.... | dev | Phenology | 1 | CMF048 |
11.269 | cropmath-v1 | null | [
"CMF020",
"CMF023"
] | {
"Com": 6.15,
"Cr": 88.53,
"FEh": 0.601,
"TI": 0.715
} | methane_emission | {
"expression": "max(0, 0.27 * FEh * (TI * Cr + Com))",
"id": "CMF022",
"latex": "P=\\max(0,0.27F_{Eh}(TI\\cdot C_r+Com))",
"name": "甲烷产生量",
"name_en": "Methane production",
"precision": 3,
"unit": "g CH4/m2/day"
} | CMF022_direct_t2_0000 | false | Candidate formulas are listed below. Select the physiologically appropriate formula before calculating.
Candidate 1:
Formula name: Methane production (甲烷产生量)
Formula (LaTeX): P=\max(0,0.27F_{Eh}(TI\cdot C_r+Com))
Formula: P = max(0, 0.27 * FEh * (TI * Cr + Com))
Operator notes: * means multiplication, / means division... | Formula name: Methane production (甲烷产生量)
Formula (LaTeX): P=\max(0,0.27F_{Eh}(TI\cdot C_r+Com))
Formula: P = max(0, 0.27 * FEh * (TI * Cr + Com))
Operator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.
O... | Formula name: Methane production (甲烷产生量)
Formula (LaTeX): P=\max(0,0.27F_{Eh}(TI\cdot C_r+Com))
Formula: P = max(0, 0.27 * FEh * (TI * Cr + Com))
Operator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.
O... | Formula: P = max(0, 0.27 * FEh * (TI * Cr + Com))
Operator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.
Output unit: g CH4/m2/day | Relevant formula name: Methane production (甲烷产生量).
Formula: P = max(0, 0.27 * FEh * (TI * Cr + Com))
Operator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.
Output unit: g CH4/m2/day | A retrieved formula is listed below. It may look relevant, but use it only if it truly matches the problem.
Formula name: Fresh organic matter carbon decomposition (新鲜有机质碳分解量)
Formula (LaTeX): OMNC=WI\cdot SI\cdot TI\cdot0.027\cdot OMN
Formula: OMNC = WI * SI * TI * 0.027 * OMN
Operator notes: * means multiplication, /... | 1 | Maximum with chained drivers | direct | {
"Com": 6.15,
"Cr": 88.53,
"FEh": 0.601,
"TI": 0.715
} | In a methane, the Methane production is computed as part of methane modeling. With FEh=0.601 fraction, TI=0.715 fraction, Cr=88.53 g C/m2/day, Com=6.15 g C/m2/day, determine the numeric value in g CH4/m2/day. | {
"C": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ..... | Identify the target quantity as Methane production.
Use the verified formula:
\[P=\max(0,0.27F_{Eh}(TI\cdot C_r+Com))\]
In calculation-friendly notation, this is:
P = max(0, 0.27 * FEh * (TI * Cr + Com)).
Substitute the relevant values:
FEh=0.601 fraction, TI=0.715 fraction, Cr=88.53 g C/m2/day, Com=6.15 g C/m2/day.
Th... | dev | Methane | 2 | CMF020 |
0.0084 | cropmath-v1 | null | [
"CMF033",
"CMF051"
] | {
"PDT": 6.8
} | growth_yield | {
"expression": "0.0091 - 0.0001 * PDT",
"id": "CMF049",
"latex": "RM(T_o)=0.0091-0.0001PDT",
"name": "最适温度维持呼吸系数",
"name_en": "Maintenance respiration coefficient at optimum temperature",
"precision": 4,
"unit": "d-1"
} | CMF049_direct_t2_0001 | false | Candidate formulas are listed below. Select the physiologically appropriate formula before calculating.
Candidate 1:
Formula name: Daily dry matter increment (日干物质增量)
Formula (LaTeX): GCR=\frac{DTGA-RM-RG}{1-b}
Formula: GCR = (DTGA - RM - RG) / (1 - b)
Operator notes: * means multiplication, / means division, ** or ^ ... | Formula name: Maintenance respiration coefficient at optimum temperature (最适温度维持呼吸系数)
Formula (LaTeX): RM(T_o)=0.0091-0.0001PDT
Formula: RMTo = 0.0091 - 0.0001 * PDT
Operator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the sma... | Formula name: Maintenance respiration coefficient at optimum temperature (最适温度维持呼吸系数)
Formula (LaTeX): RM(T_o)=0.0091-0.0001PDT
Formula: RMTo = 0.0091 - 0.0001 * PDT
Operator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the sma... | Formula: RMTo = 0.0091 - 0.0001 * PDT
Operator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.
Output unit: d-1 | Relevant formula name: Maintenance respiration coefficient at optimum temperature (最适温度维持呼吸系数).
Formula: RMTo = 0.0091 - 0.0001 * PDT
Operator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.
Output unit: ... | A retrieved formula is listed below. It may look relevant, but use it only if it truly matches the problem.
Formula name: Daily dry matter increment (日干物质增量)
Formula (LaTeX): GCR=\frac{DTGA-RM-RG}{1-b}
Formula: GCR = (DTGA - RM - RG) / (1 - b)
Operator notes: * means multiplication, / means division, ** or ^ means powe... | 1 | Linear decline | direct | {
"PDT": 6.8
} | In a respiration, the Maintenance respiration coefficient at optimum temperature is computed as part of respiration modeling. With PDT=6.8 physiological days, determine the numeric value in d-1. | {
"C": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ..... | Identify the target quantity as Maintenance respiration coefficient at optimum temperature.
Use the verified formula:
\[RM(T_o)=0.0091-0.0001PDT\]
In calculation-friendly notation, this is:
RMTo = 0.0091 - 0.0001 * PDT.
Substitute the relevant values:
PDT=6.8 physiological days.
The computed value is 0.0084 d-1.
Theref... | dev | Respiration | 2 | CMF033 |
0.680 | cropmath-v1 | null | [
"CMF016",
"CMF020"
] | {
"OMS": 832.8,
"SI": 1,
"TI": 0.56,
"WI": 0.486
} | methane_emission | {
"expression": "WI * SI * TI * 0.003 * OMS",
"id": "CMF021",
"latex": "OMSC=WI\\cdot SI\\cdot TI\\cdot0.003\\cdot OMS",
"name": "土壤有机质碳分解量",
"name_en": "Soil organic matter carbon decomposition",
"precision": 3,
"unit": "g C/m2/day"
} | CMF021_direct_t2_0000 | false | Candidate formulas are listed below. Select the physiologically appropriate formula before calculating.
Candidate 1:
Formula name: Fresh organic matter carbon decomposition (新鲜有机质碳分解量)
Formula (LaTeX): OMNC=WI\cdot SI\cdot TI\cdot0.027\cdot OMN
Formula: OMNC = WI * SI * TI * 0.027 * OMN
Operator notes: * means multipl... | Formula name: Soil organic matter carbon decomposition (土壤有机质碳分解量)
Formula (LaTeX): OMSC=WI\cdot SI\cdot TI\cdot0.003\cdot OMS
Formula: OMSC = WI * SI * TI * 0.003 * OMS
Operator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the... | Formula name: Soil organic matter carbon decomposition (土壤有机质碳分解量)
Formula (LaTeX): OMSC=WI\cdot SI\cdot TI\cdot0.003\cdot OMS
Formula: OMSC = WI * SI * TI * 0.003 * OMS
Operator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the... | Formula: OMSC = WI * SI * TI * 0.003 * OMS
Operator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.
Output unit: g C/m2/day | Relevant formula name: Soil organic matter carbon decomposition (土壤有机质碳分解量).
Formula: OMSC = WI * SI * TI * 0.003 * OMS
Operator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.
Output unit: g C/m2/day | A retrieved formula is listed below. It may look relevant, but use it only if it truly matches the problem.
Formula name: Methane ebullition emission (甲烷冒泡排放)
Formula (LaTeX): E_{bl}=\min(0.7\ln(t_{soil})/W_r,0.9)P
Formula: Ebl = min(0.7 * ln(t_soil) / Wr, 0.9) * P
Operator notes: * means multiplication, / means divisi... | 1 | Multiplicative | direct | {
"OMS": 832.8,
"SI": 1,
"TI": 0.56,
"WI": 0.486
} | In a methane, the Soil organic matter carbon decomposition is computed as part of methane modeling. With WI=0.486 fraction, SI=1.0 fraction, TI=0.56 fraction, OMS=832.8 g C/m2, determine the numeric value in g C/m2/day. | {
"C": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ..... | Identify the target quantity as Soil organic matter carbon decomposition.
Use the verified formula:
\[OMSC=WI\cdot SI\cdot TI\cdot0.003\cdot OMS\]
In calculation-friendly notation, this is:
OMSC = WI * SI * TI * 0.003 * OMS.
Substitute the relevant values:
WI=0.486 fraction, SI=1.0 fraction, TI=0.56 fraction, OMS=832.8... | dev | Methane | 2 | CMF016 |
2.690 | cropmath-v1 | {
"new_value": 1.315,
"old_value": 1.461,
"parameter": "TI"
} | [
"CMF016",
"CMF020"
] | {
"Com": 3.883,
"Cr": 7.84,
"FEh": 0.7020000000000001,
"TI": 1.315
} | methane_emission | {
"expression": "max(0, 0.27 * FEh * (TI * Cr + Com))",
"id": "CMF022",
"latex": "P=\\max(0,0.27F_{Eh}(TI\\cdot C_r+Com))",
"name": "甲烷产生量",
"name_en": "Methane production",
"precision": 3,
"unit": "g CH4/m2/day"
} | CMF022_sensitivity_t1_0001 | false | Candidate formulas are listed below. Select the physiologically appropriate formula before calculating.
Candidate 1:
Formula name: Fresh organic matter carbon decomposition (新鲜有机质碳分解量)
Formula (LaTeX): OMNC=WI\cdot SI\cdot TI\cdot0.027\cdot OMN
Formula: OMNC = WI * SI * TI * 0.027 * OMN
Operator notes: * means multipl... | Formula name: Methane production (甲烷产生量)
Formula (LaTeX): P=\max(0,0.27F_{Eh}(TI\cdot C_r+Com))
Formula: P = max(0, 0.27 * FEh * (TI * Cr + Com))
Operator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.
O... | Formula name: Methane production (甲烷产生量)
Formula (LaTeX): P=\max(0,0.27F_{Eh}(TI\cdot C_r+Com))
Formula: P = max(0, 0.27 * FEh * (TI * Cr + Com))
Operator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.
O... | Formula: P = max(0, 0.27 * FEh * (TI * Cr + Com))
Operator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.
Output unit: g CH4/m2/day | Relevant formula name: Methane production (甲烷产生量).
Formula: P = max(0, 0.27 * FEh * (TI * Cr + Com))
Operator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.
Output unit: g CH4/m2/day | A retrieved formula is listed below. It may look relevant, but use it only if it truly matches the problem.
Formula name: Methane ebullition emission (甲烷冒泡排放)
Formula (LaTeX): E_{bl}=\min(0.7\ln(t_{soil})/W_r,0.9)P
Formula: Ebl = min(0.7 * ln(t_soil) / Wr, 0.9) * P
Operator notes: * means multiplication, / means divisi... | 2 | Maximum with chained drivers | sensitivity | {
"Com": 3.883,
"Cr": 7.84,
"FEh": 0.7020000000000001,
"TI": 1.461
} | Given current measurements FEh=0.702 fraction, TI=1.461 fraction, Cr=7.84 g C/m2/day, Com=3.883 g C/m2/day in methane, what would the Methane production be if TI were adjusted to 1.315? Provide the numeric value in g CH4/m2/day. | {
"C": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ..... | Identify the target quantity as Methane production.
Use the verified formula:
\[P=\max(0,0.27F_{Eh}(TI\cdot C_r+Com))\]
In calculation-friendly notation, this is:
P = max(0, 0.27 * FEh * (TI * Cr + Com)).
Substitute the relevant values:
FEh=0.702 fraction, TI=1.315 fraction, Cr=7.84 g C/m2/day, Com=3.883 g C/m2/day.
Th... | dev | Methane | 1 | CMF016 |
0.091 | cropmath-v1 | null | [
"CMF010",
"CMF005"
] | {
"Coef_Plant_TN_uptake": 0.166,
"DVI": 0.73,
"NPP": 1.55,
"N_requirement_max_day": 1.6400000000000001,
"N_shoot": 6.78,
"Ratio_N_C_photo_product": 0.059000000000000004
} | carbon_nitrogen_cycle | {
"expression": "if NPP<0: 0; if N_shoot<3.5 and DVI<0.3: Coef_Plant_TN_uptake*N_shoot; else: min(N_requirement_max_day, Ratio_N_C_photo_product*NPP)",
"id": "CMF004",
"latex": "N_d=C_uN_{shoot}\\;\\text{or}\\;\\min(N_{max},r_NNPP)",
"name": "作物需氮量",
"name_en": "carbon-nitrogen cycling crop nitrogen demand",
... | CMF004_direct_t2_0001 | false | Candidate formulas are listed below. Select the physiologically appropriate formula before calculating.
Candidate 1:
Formula name: carbon-nitrogen cycling crop nitrogen demand (作物需氮量)
Formula (LaTeX): N_d=C_uN_{shoot}\;\text{or}\;\min(N_{max},r_NNPP)
Formula: N_demand = if NPP<0: 0; if N_shoot<3.5 and DVI<0.3: Coef_Pl... | Formula name: carbon-nitrogen cycling crop nitrogen demand (作物需氮量)
Formula (LaTeX): N_d=C_uN_{shoot}\;\text{or}\;\min(N_{max},r_NNPP)
Formula: N_demand = if NPP<0: 0; if N_shoot<3.5 and DVI<0.3: Coef_Plant_TN_uptake*N_shoot; else: min(N_requirement_max_day, Ratio_N_C_photo_product*NPP)
Operator notes: * means multiplic... | Formula name: carbon-nitrogen cycling crop nitrogen demand (作物需氮量)
Formula (LaTeX): N_d=C_uN_{shoot}\;\text{or}\;\min(N_{max},r_NNPP)
Formula: N_demand = if NPP<0: 0; if N_shoot<3.5 and DVI<0.3: Coef_Plant_TN_uptake*N_shoot; else: min(N_requirement_max_day, Ratio_N_C_photo_product*NPP)
Operator notes: * means multiplic... | Formula: N_demand = if NPP<0: 0; if N_shoot<3.5 and DVI<0.3: Coef_Plant_TN_uptake*N_shoot; else: min(N_requirement_max_day, Ratio_N_C_photo_product*NPP)
Operator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger v... | Relevant formula name: carbon-nitrogen cycling crop nitrogen demand (作物需氮量).
Formula: N_demand = if NPP<0: 0; if N_shoot<3.5 and DVI<0.3: Coef_Plant_TN_uptake*N_shoot; else: min(N_requirement_max_day, Ratio_N_C_photo_product*NPP)
Operator notes: * means multiplication, / means division, ** or ^ means power, exp(x) mean... | A retrieved formula is listed below. It may look relevant, but use it only if it truly matches the problem.
Formula name: carbon-nitrogen cycling soil pH decomposition factor (土壤pH分解因子)
Formula (LaTeX): F_{pH}=\frac{1}{1+e^{2.5(5-pH)}}
Formula: F_pH = 1 / (1 + exp(2.5 * (5 - pH)))
Operator notes: * means multiplication... | 3 | Piecewise minimum | direct | {
"Coef_Plant_TN_uptake": 0.166,
"DVI": 0.73,
"NPP": 1.55,
"N_requirement_max_day": 1.6400000000000001,
"N_shoot": 6.78,
"Ratio_N_C_photo_product": 0.059000000000000004
} | In a carbon-nitrogen cycling nitrogen, the carbon-nitrogen cycling crop nitrogen demand is computed as part of carbon-nitrogen cycling nitrogen modeling. With N_shoot=6.78 g N/m2, Coef_Plant_TN_uptake=0.166 d-1, N_requirement_max_day=1.64 g N/m2/day, NPP=1.55 g C/m2/day, Ratio_N_C_photo_product=0.059 g N/g C, DVI=0.73 ... | {
"C": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ..... | Identify the target quantity as carbon-nitrogen cycling crop nitrogen demand.
Use the verified formula:
\[N_d=C_uN_{shoot}\;\text{or}\;\min(N_{max},r_NNPP)\]
In calculation-friendly notation, this is:
N_demand = if NPP<0: 0; if N_shoot<3.5 and DVI<0.3: Coef_Plant_TN_uptake*N_shoot; else: min(N_requirement_max_day, Rati... | dev | carbon-nitrogen cycling nitrogen | 2 | CMF010 |
1.392 | cropmath-v1 | null | [
"CMF056",
"CMF055"
] | {
"CO2": 703,
"CO2_RUE": 0.112
} | environmental_response | {
"expression": "if CO2>=700: 1+CO2_RUE*350/100; else: max(CO2_RUE*CO2*0.01+1-0.01*350*CO2_RUE, 1)",
"id": "CMF057",
"latex": "F_{CO2}=f(CO2,CO2_{RUE})",
"name": "环境响应二氧化碳响应",
"name_en": "environmental-response CO2 response",
"precision": 3,
"unit": "multiplier"
} | CMF057_branch_t1_0000 | false | Candidate formulas are listed below. Select the physiologically appropriate formula before calculating.
Candidate 1:
Formula name: environmental-response CO2 response (环境响应二氧化碳响应)
Formula (LaTeX): F_{CO2}=f(CO2,CO2_{RUE})
Formula: F_CO2 = if CO2>=700: 1+CO2_RUE*350/100; else: max(CO2_RUE*CO2*0.01+1-0.01*350*CO2_RUE, 1... | Formula name: environmental-response CO2 response (环境响应二氧化碳响应)
Formula (LaTeX): F_{CO2}=f(CO2,CO2_{RUE})
Formula: F_CO2 = if CO2>=700: 1+CO2_RUE*350/100; else: max(CO2_RUE*CO2*0.01+1-0.01*350*CO2_RUE, 1)
Operator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural... | Formula name: environmental-response CO2 response (环境响应二氧化碳响应)
Formula (LaTeX): F_{CO2}=f(CO2,CO2_{RUE})
Formula: F_CO2 = if CO2>=700: 1+CO2_RUE*350/100; else: max(CO2_RUE*CO2*0.01+1-0.01*350*CO2_RUE, 1)
Operator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural... | Formula: F_CO2 = if CO2>=700: 1+CO2_RUE*350/100; else: max(CO2_RUE*CO2*0.01+1-0.01*350*CO2_RUE, 1)
Operator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.
Output unit: fraction | Relevant formula name: environmental-response CO2 response (环境响应二氧化碳响应).
Formula: F_CO2 = if CO2>=700: 1+CO2_RUE*350/100; else: max(CO2_RUE*CO2*0.01+1-0.01*350*CO2_RUE, 1)
Operator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select t... | A retrieved formula is listed below. It may look relevant, but use it only if it truly matches the problem.
Formula name: environmental-response daily thermal time (每日有效积温)
Formula (LaTeX): dTT=\max(T_{mean}-T_{base},0)
Formula: dTT = max(Tmean - Tbase, 0)
Operator notes: * means multiplication, / means division, ** or... | 4 | Piecewise linear cap | branch | {
"CO2": 703,
"CO2_RUE": 0.112
} | For environmental-response formula system, compute the environmental-response CO2 response. The inputs are CO2=703.0 ppm, CO2_RUE=0.112 fraction. Note that this formula uses piecewise logic depending on parameter values. Report the final numeric value in fraction. | {
"C": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ..... | Identify the target quantity as environmental-response CO2 response.
Use the verified formula:
\[F_{CO2}=f(CO2,CO2_{RUE})\]
In calculation-friendly notation, this is:
F_CO2 = if CO2>=700: 1+CO2_RUE*350/100; else: max(CO2_RUE*CO2*0.01+1-0.01*350*CO2_RUE, 1).
Substitute the relevant values:
CO2=703.0 ppm, CO2_RUE=0.112 f... | dev | environmental-response formula system | 1 | CMF056 |
0.58 | cropmath-v1 | {
"new_value": 40.7,
"old_value": 37,
"parameter": "T"
} | [
"CMF044",
"CMF043"
] | {
"T": 40.7,
"Tb": 10,
"Tmax": 45,
"Tol": 24,
"Tou": 34
} | growth_yield | {
"expression": "if Tb<=T<Tol: sin(pi/2*(T-Tb)/(Tol-Tb)); if Tol<=T<Tou: 1; if Tou<=T<Tmax: cos(pi/2*(T-Tou)/(Tmax-Tou)); else: 0",
"id": "CMF032",
"latex": "F_T=\\sin(\\frac{T-T_b}{T_{ol}-T_b}\\frac{\\pi}{2}); 1; \\cos(\\frac{T-T_{ou}}{T_{max}-T_{ou}}\\frac{\\pi}{2})",
"name": "温度响应因子",
"name_en": "Temperatu... | CMF032_sensitivity_t1_0000 | false | Candidate formulas are listed below. Select the physiologically appropriate formula before calculating.
Candidate 1:
Formula name: Shoot partitioning index (地上部分配指数)
Formula (LaTeX): PISH=-8.42\times10^{-5}PDT^2+0.01PDT+0.63
Formula: PISH = -0.0000842 * PDT^2 + 0.01 * PDT + 0.63
Operator notes: * means multiplication,... | Formula name: Temperature response factor (温度响应因子)
Formula (LaTeX): F_T=\sin(\frac{T-T_b}{T_{ol}-T_b}\frac{\pi}{2}); 1; \cos(\frac{T-T_{ou}}{T_{max}-T_{ou}}\frac{\pi}{2})
Formula: F_T = if Tb<=T<Tol: sin(pi/2*(T-Tb)/(Tol-Tb)); if Tol<=T<Tou: 1; if Tou<=T<Tmax: cos(pi/2*(T-Tou)/(Tmax-Tou)); else: 0
Operator notes: * mea... | Formula name: Temperature response factor (温度响应因子)
Formula (LaTeX): F_T=\sin(\frac{T-T_b}{T_{ol}-T_b}\frac{\pi}{2}); 1; \cos(\frac{T-T_{ou}}{T_{max}-T_{ou}}\frac{\pi}{2})
Formula: F_T = if Tb<=T<Tol: sin(pi/2*(T-Tb)/(Tol-Tb)); if Tol<=T<Tou: 1; if Tou<=T<Tmax: cos(pi/2*(T-Tou)/(Tmax-Tou)); else: 0
Operator notes: * mea... | Formula: F_T = if Tb<=T<Tol: sin(pi/2*(T-Tb)/(Tol-Tb)); if Tol<=T<Tou: 1; if Tou<=T<Tmax: cos(pi/2*(T-Tou)/(Tmax-Tou)); else: 0
Operator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.
Output unit: fracti... | Relevant formula name: Temperature response factor (温度响应因子).
Formula: F_T = if Tb<=T<Tol: sin(pi/2*(T-Tb)/(Tol-Tb)); if Tol<=T<Tou: 1; if Tou<=T<Tmax: cos(pi/2*(T-Tou)/(Tmax-Tou)); else: 0
Operator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and... | A retrieved formula is listed below. It may look relevant, but use it only if it truly matches the problem.
Formula name: Shoot partitioning index (地上部分配指数)
Formula (LaTeX): PISH=-8.42\times10^{-5}PDT^2+0.01PDT+0.63
Formula: PISH = -0.0000842 * PDT^2 + 0.01 * PDT + 0.63
Operator notes: * means multiplication, / means d... | 3 | Piecewise trigonometric | sensitivity | {
"T": 37,
"Tb": 10,
"Tmax": 45,
"Tol": 24,
"Tou": 34
} | Given current measurements T=37.0 C, Tb=10.0 C, Tol=24.0 C, Tou=34.0 C, Tmax=45.0 C in temperature response, what would the Temperature response factor be if T were adjusted to 40.7? Provide the numeric value in fraction. | {
"C": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ..... | Identify the target quantity as Temperature response factor.
Use the verified formula:
\[F_T=\sin(\frac{T-T_b}{T_{ol}-T_b}\frac{\pi}{2}); 1; \cos(\frac{T-T_{ou}}{T_{max}-T_{ou}}\frac{\pi}{2})\]
In calculation-friendly notation, this is:
F_T = if Tb<=T<Tol: sin(pi/2*(T-Tb)/(Tol-Tb)); if Tol<=T<Tou: 1; if Tou<=T<Tmax: co... | dev | Temperature response | 1 | CMF044 |
90.97 | cropmath-v1 | null | [
"CMF010",
"CMF007"
] | {
"GrainYield": 8269,
"NUptake": 90.9
} | carbon_nitrogen_cycle | {
"expression": "GrainYield / NUptake",
"id": "CMF003",
"latex": "NUE=\\frac{GrainYield}{NUptake}",
"name": "氮利用效率",
"name_en": "Nitrogen use efficiency",
"precision": 2,
"unit": "kg grain/kg N"
} | CMF003_direct_t0_0000 | false | Candidate formulas are listed below. Select the physiologically appropriate formula before calculating.
Candidate 1:
Formula name: carbon-nitrogen cycling optimal photosynthetic rate (最适光合速率)
Formula (LaTeX): P_{opt}=\alpha N_{leaf}\frac{PAR}{PAR+\beta}
Formula: Photo_Rate_opt = (Coef_alfa * Leaf_N) * PAR / (PAR + Coe... | Formula name: Nitrogen use efficiency (氮利用效率)
Formula (LaTeX): NUE=\frac{GrainYield}{NUptake}
Formula: NUE = GrainYield / NUptake
Operator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.
Output unit: kg g... | Formula name: Nitrogen use efficiency (氮利用效率)
Formula (LaTeX): NUE=\frac{GrainYield}{NUptake}
Formula: NUE = GrainYield / NUptake
Operator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.
Output unit: kg g... | Formula: NUE = GrainYield / NUptake
Operator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.
Output unit: kg grain/kg N | Relevant formula name: Nitrogen use efficiency (氮利用效率).
Formula: NUE = GrainYield / NUptake
Operator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.
Output unit: kg grain/kg N | A retrieved formula is listed below. It may look relevant, but use it only if it truly matches the problem.
Formula name: carbon-nitrogen cycling soil pH decomposition factor (土壤pH分解因子)
Formula (LaTeX): F_{pH}=\frac{1}{1+e^{2.5(5-pH)}}
Formula: F_pH = 1 / (1 + exp(2.5 * (5 - pH)))
Operator notes: * means multiplication... | 1 | Ratio | direct | {
"GrainYield": 8269,
"NUptake": 90.9
} | A nitrogen calculation estimates Nitrogen use efficiency. The measured inputs are GrainYield=8269.0 kg/ha, NUptake=90.9 kg N/ha. What is the nitrogen use efficiency? Give the final numeric value in kg grain/kg N. | {
"C": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ..... | Identify the target quantity as Nitrogen use efficiency.
Use the verified formula:
\[NUE=\frac{GrainYield}{NUptake}\]
In calculation-friendly notation, this is:
NUE = GrainYield / NUptake.
Substitute the relevant values:
GrainYield=8269.0 kg/ha, NUptake=90.9 kg N/ha.
The computed value is 90.97 kg grain/kg N.
Therefore... | dev | Nitrogen | 0 | CMF010 |
22.164 | cropmath-v1 | null | [
"CMF013",
"CMF001"
] | {
"Coef_alfa": 19.43,
"Coef_beta": 354.5,
"Leaf_N": 1.58,
"PAR": 920.5
} | carbon_nitrogen_cycle | {
"expression": "(Coef_alfa * Leaf_N) * PAR / (PAR + Coef_beta)",
"id": "CMF007",
"latex": "P_{opt}=\\alpha N_{leaf}\\frac{PAR}{PAR+\\beta}",
"name": "最适光合速率",
"name_en": "carbon-nitrogen cycling optimal photosynthetic rate",
"precision": 3,
"unit": "umol CO2/m2/s"
} | CMF007_direct_t0_0001 | false | Candidate formulas are listed below. Select the physiologically appropriate formula before calculating.
Candidate 1:
Formula name: carbon-nitrogen cycling crop temperature activity factor (作物温度活性因子)
Formula (LaTeX): F_T=\left(\frac{T_d-T_{min}}{T_{opt}-T_{min}}\right)^{1+M}\left(\frac{T_{max}-T_d}{T_{max}-T_{opt}}\rig... | Formula name: carbon-nitrogen cycling optimal photosynthetic rate (最适光合速率)
Formula (LaTeX): P_{opt}=\alpha N_{leaf}\frac{PAR}{PAR+\beta}
Formula: Photo_Rate_opt = (Coef_alfa * Leaf_N) * PAR / (PAR + Coef_beta)
Operator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means n... | Formula name: carbon-nitrogen cycling optimal photosynthetic rate (最适光合速率)
Formula (LaTeX): P_{opt}=\alpha N_{leaf}\frac{PAR}{PAR+\beta}
Formula: Photo_Rate_opt = (Coef_alfa * Leaf_N) * PAR / (PAR + Coef_beta)
Operator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means n... | Formula: Photo_Rate_opt = (Coef_alfa * Leaf_N) * PAR / (PAR + Coef_beta)
Operator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.
Output unit: umol CO2/m2/s | Relevant formula name: carbon-nitrogen cycling optimal photosynthetic rate (最适光合速率).
Formula: Photo_Rate_opt = (Coef_alfa * Leaf_N) * PAR / (PAR + Coef_beta)
Operator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/lar... | A retrieved formula is listed below. It may look relevant, but use it only if it truly matches the problem.
Formula name: carbon-nitrogen cycling crop temperature activity factor (作物温度活性因子)
Formula (LaTeX): F_T=\left(\frac{T_d-T_{min}}{T_{opt}-T_{min}}\right)^{1+M}\left(\frac{T_{max}-T_d}{T_{max}-T_{opt}}\right)^{1-M}
... | 1 | Michaelis-Menten response | direct | {
"Coef_alfa": 19.43,
"Coef_beta": 354.5,
"Leaf_N": 1.58,
"PAR": 920.5
} | A carbon-nitrogen cycling photosynthesis calculation estimates carbon-nitrogen cycling optimal photosynthetic rate. The measured inputs are Coef_alfa=19.43 umol CO2/gN/s, Coef_beta=354.5 umol/m2/s, PAR=920.5 umol/m2/s, Leaf_N=1.58 gN/m2. What is the carbon-nitrogen cycling optimal photosynthetic rate? Give the final nu... | {
"C": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ..... | Identify the target quantity as carbon-nitrogen cycling optimal photosynthetic rate.
Use the verified formula:
\[P_{opt}=\alpha N_{leaf}\frac{PAR}{PAR+\beta}\]
In calculation-friendly notation, this is:
Photo_Rate_opt = (Coef_alfa * Leaf_N) * PAR / (PAR + Coef_beta).
Substitute the relevant values:
Coef_alfa=19.43 umol... | dev | carbon-nitrogen cycling photosynthesis | 0 | CMF013 |
1.45 | cropmath-v1 | {
"new_value": 0.81,
"old_value": 0.9,
"parameter": "beta"
} | [
"CMF054",
"CMF041"
] | {
"Cx": 593,
"beta": 0.81
} | growth_yield | {
"expression": "1 + beta * ln(Cx / 340)",
"id": "CMF029",
"latex": "F_{CO2}=1+\\beta\\ln(C_x/340)",
"name": "二氧化碳影响因子",
"name_en": "CO2 response factor",
"precision": 2,
"unit": "multiplier"
} | CMF029_sensitivity_t0_0001 | false | Candidate formulas are listed below. Select the physiologically appropriate formula before calculating.
Candidate 1:
Formula name: CO2 response factor (二氧化碳影响因子)
Formula (LaTeX): F_{CO2}=1+\beta\ln(C_x/340)
Formula: FCO2 = 1 + beta * ln(Cx / 340)
Operator notes: * means multiplication, / means division, ** or ^ means ... | Formula name: CO2 response factor (二氧化碳影响因子)
Formula (LaTeX): F_{CO2}=1+\beta\ln(C_x/340)
Formula: FCO2 = 1 + beta * ln(Cx / 340)
Operator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.
Output unit: frac... | Formula name: CO2 response factor (二氧化碳影响因子)
Formula (LaTeX): F_{CO2}=1+\beta\ln(C_x/340)
Formula: FCO2 = 1 + beta * ln(Cx / 340)
Operator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.
Output unit: frac... | Formula: FCO2 = 1 + beta * ln(Cx / 340)
Operator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.
Output unit: fraction | Relevant formula name: CO2 response factor (二氧化碳影响因子).
Formula: FCO2 = 1 + beta * ln(Cx / 340)
Operator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.
Output unit: fraction | A retrieved formula is listed below. It may look relevant, but use it only if it truly matches the problem.
Formula name: Dry grain yield (稻谷干重产量)
Formula (LaTeX): Y=AWSP\cdot f_{grain}
Formula: Y = AWSP * grain_fraction
Operator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln... | 2 | Logarithmic | sensitivity | {
"Cx": 593,
"beta": 0.9
} | A photosynthesis calculation involves sensitivity analysis. The current measurements are Cx=593.0 ppm, beta=0.9 dimensionless. A field technician asks what the co2 response factor would become if beta changed to 0.81. Report the new final numeric value in fraction. | {
"C": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ..... | Identify the target quantity as CO2 response factor.
Use the verified formula:
\[F_{CO2}=1+\beta\ln(C_x/340)\]
In calculation-friendly notation, this is:
FCO2 = 1 + beta * ln(Cx / 340).
Substitute the relevant values:
Cx=593.0 ppm, beta=0.81 dimensionless.
The computed value is 1.45 fraction.
Therefore, the final numer... | dev | Photosynthesis | 0 | CMF054 |
8.457 | cropmath-v1 | {
"new_value": 15.106,
"old_value": 16.784,
"parameter": "P"
} | [
"CMF023",
"CMF019"
] | {
"CH4RiceEfC": 0.468,
"Ebl": 2.5140000000000002,
"P": 15.106,
"W": 6006.3,
"Wmax": 11997.5
} | methane_emission | {
"expression": "Ebl + P * CH4RiceEfC * (1 - W/Wmax)^0.25",
"id": "CMF017",
"latex": "E=E_{bl}+P\\cdot C(1-W/W_{max})^{0.25}",
"name": "甲烷总排放",
"name_en": "Total methane emission",
"precision": 3,
"unit": "g CH4/m2/day"
} | CMF017_sensitivity_t2_0000 | false | Candidate formulas are listed below. Select the physiologically appropriate formula before calculating.
Candidate 1:
Formula name: Plant methane transport factor (植株传输因子)
Formula (LaTeX): F_p=C(1-W/W_{max})^{0.25}
Formula: Fp = CH4RiceEfC * (1 - W / Wmax) ** 0.25
Operator notes: * means multiplication, / means divisio... | Formula name: Total methane emission (甲烷总排放)
Formula (LaTeX): E=E_{bl}+P\cdot C(1-W/W_{max})^{0.25}
Formula: E = Ebl + P * CH4RiceEfC * (1 - W/Wmax)^0.25
Operator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger ... | Formula name: Total methane emission (甲烷总排放)
Formula (LaTeX): E=E_{bl}+P\cdot C(1-W/W_{max})^{0.25}
Formula: E = Ebl + P * CH4RiceEfC * (1 - W/Wmax)^0.25
Operator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger ... | Formula: E = Ebl + P * CH4RiceEfC * (1 - W/Wmax)^0.25
Operator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.
Output unit: g CH4/m2/day | Relevant formula name: Total methane emission (甲烷总排放).
Formula: E = Ebl + P * CH4RiceEfC * (1 - W/Wmax)^0.25
Operator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.
Output unit: g CH4/m2/day | A retrieved formula is listed below. It may look relevant, but use it only if it truly matches the problem.
Formula name: Methane temperature index (甲烷温度指数)
Formula (LaTeX): TI=Q_{10}^{(t_{sl}-30)/10}
Formula: TI = Q10^((min(t_soil,30)-30)/10)
Operator notes: * means multiplication, / means division, ** or ^ means powe... | 3 | Chained sum | sensitivity | {
"CH4RiceEfC": 0.468,
"Ebl": 2.5140000000000002,
"P": 16.784,
"W": 6006.3,
"Wmax": 11997.5
} | In methane, we analyze how Total methane emission responds to parameter changes. Baseline: Ebl=2.514 g CH4/m2/day, P=16.784 g CH4/m2/day, CH4RiceEfC=0.468 fraction, W=6006.3 kg/ha, Wmax=11997.5 kg/ha. After changing P from 16.784 to 15.106, compute the new Total methane emission. Give the numeric value in g CH4/m2/day. | {
"C": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ..... | Identify the target quantity as Total methane emission.
Use the verified formula:
\[E=E_{bl}+P\cdot C(1-W/W_{max})^{0.25}\]
In calculation-friendly notation, this is:
E = Ebl + P * CH4RiceEfC * (1 - W/Wmax)^0.25.
Substitute the relevant values:
Ebl=2.514 g CH4/m2/day, P=15.106 g CH4/m2/day, CH4RiceEfC=0.468 fraction, W... | dev | Methane | 2 | CMF023 |
0.241 | cropmath-v1 | null | [
"CMF039",
"CMF032"
] | {
"BFF": 0.88,
"DTE": 0.274,
"IE": 0.91,
"PDT": 53.3,
"RPE": 0.9400000000000001
} | growth_yield | {
"expression": "if PDT<8: DTE*IE; if PDT<18: DTE*RPE; if PDT<32: DTE; if PDT<57: DTE*BFF; else: 0",
"id": "CMF027",
"latex": "DPE=DTE\\cdot IE,\\;DTE\\cdot RPE,\\;DTE,\\;DTE\\cdot BFF",
"name": "每日生理效应",
"name_en": "Daily physiological effectiveness",
"precision": 3,
"unit": "physiological days/day"
} | CMF027_branch_t0_0000 | false | Candidate formulas are listed below. Select the physiologically appropriate formula before calculating.
Candidate 1:
Formula name: Daily physiological effectiveness (每日生理效应)
Formula (LaTeX): DPE=DTE\cdot IE,\;DTE\cdot RPE,\;DTE,\;DTE\cdot BFF
Formula: DPE = if PDT<8: DTE*IE; if PDT<18: DTE*RPE; if PDT<32: DTE; if PDT<... | Formula name: Daily physiological effectiveness (每日生理效应)
Formula (LaTeX): DPE=DTE\cdot IE,\;DTE\cdot RPE,\;DTE,\;DTE\cdot BFF
Formula: DPE = if PDT<8: DTE*IE; if PDT<18: DTE*RPE; if PDT<32: DTE; if PDT<57: DTE*BFF; else: 0
Operator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ... | Formula name: Daily physiological effectiveness (每日生理效应)
Formula (LaTeX): DPE=DTE\cdot IE,\;DTE\cdot RPE,\;DTE,\;DTE\cdot BFF
Formula: DPE = if PDT<8: DTE*IE; if PDT<18: DTE*RPE; if PDT<32: DTE; if PDT<57: DTE*BFF; else: 0
Operator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ... | Formula: DPE = if PDT<8: DTE*IE; if PDT<18: DTE*RPE; if PDT<32: DTE; if PDT<57: DTE*BFF; else: 0
Operator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.
Output unit: physiological days/day | Relevant formula name: Daily physiological effectiveness (每日生理效应).
Formula: DPE = if PDT<8: DTE*IE; if PDT<18: DTE*RPE; if PDT<32: DTE; if PDT<57: DTE*BFF; else: 0
Operator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the small... | A retrieved formula is listed below. It may look relevant, but use it only if it truly matches the problem.
Formula name: Low-temperature fertility factor (低温结实因子)
Formula (LaTeX): LTF=1-\frac{4.6+0.054Q_t^{1.56}}{100}
Formula: LTF = if 26<=PDT<=39 and Tmean<22: 1-(4.6+0.054*Qt^1.56)/100; else: 1
Operator notes: * mean... | 4 | Stage-dependent product | branch | {
"BFF": 0.88,
"DTE": 0.274,
"IE": 0.91,
"PDT": 53.3,
"RPE": 0.9400000000000001
} | A phenology involves a stage that may fall on either side of a piecewise threshold. The measured inputs are PDT=53.3 physiological days, DTE=0.274 fraction, IE=0.91 dimensionless, RPE=0.94 fraction, BFF=0.88 dimensionless. Determine the daily physiological effectiveness and report the final numeric value in physiologic... | {
"C": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ..... | Identify the target quantity as Daily physiological effectiveness.
Use the verified formula:
\[DPE=DTE\cdot IE,\;DTE\cdot RPE,\;DTE,\;DTE\cdot BFF\]
In calculation-friendly notation, this is:
DPE = if PDT<8: DTE*IE; if PDT<18: DTE*RPE; if PDT<32: DTE; if PDT<57: DTE*BFF; else: 0.
Substitute the relevant values:
PDT=53.... | dev | Phenology | 0 | CMF039 |
39.02 | cropmath-v1 | {
"new_value": 100.4,
"old_value": 111.5,
"parameter": "NUptake"
} | [
"CMF005",
"CMF001"
] | {
"GrainYield": 3918,
"NUptake": 100.4
} | carbon_nitrogen_cycle | {
"expression": "GrainYield / NUptake",
"id": "CMF003",
"latex": "NUE=\\frac{GrainYield}{NUptake}",
"name": "氮利用效率",
"name_en": "Nitrogen use efficiency",
"precision": 2,
"unit": "kg grain/kg N"
} | CMF003_sensitivity_t0_0000 | false | Candidate formulas are listed below. Select the physiologically appropriate formula before calculating.
Candidate 1:
Formula name: carbon-nitrogen cycling CO2 fertilization factor (作物二氧化碳施肥因子)
Formula (LaTeX): F_{CO2}=1+B\ln(CO2/340)
Formula: F_CO2 = 1 + Coef_B_CO2 * ln(CO2 / 340)
Operator notes: * means multiplicatio... | Formula name: Nitrogen use efficiency (氮利用效率)
Formula (LaTeX): NUE=\frac{GrainYield}{NUptake}
Formula: NUE = GrainYield / NUptake
Operator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.
Output unit: kg g... | Formula name: Nitrogen use efficiency (氮利用效率)
Formula (LaTeX): NUE=\frac{GrainYield}{NUptake}
Formula: NUE = GrainYield / NUptake
Operator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.
Output unit: kg g... | Formula: NUE = GrainYield / NUptake
Operator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.
Output unit: kg grain/kg N | Relevant formula name: Nitrogen use efficiency (氮利用效率).
Formula: NUE = GrainYield / NUptake
Operator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.
Output unit: kg grain/kg N | A retrieved formula is listed below. It may look relevant, but use it only if it truly matches the problem.
Formula name: carbon-nitrogen cycling shoot nitrogen allocation (地上部氮分配量)
Formula (LaTeX): N_{shoot}=N_{uptake}\frac{B_{shoot}}{B_{shoot}+B_{root}}
Formula: N_shoot = Plant_TN_uptake * Biom_shoot / (Biom_shoot + ... | 2 | Ratio | sensitivity | {
"GrainYield": 3918,
"NUptake": 111.5
} | A nitrogen calculation involves sensitivity analysis. The current measurements are GrainYield=3918.0 kg/ha, NUptake=111.5 kg N/ha. A field technician asks what the nitrogen use efficiency would become if NUptake changed to 100.4. Report the new final numeric value in kg grain/kg N. | {
"C": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ..... | Identify the target quantity as Nitrogen use efficiency.
Use the verified formula:
\[NUE=\frac{GrainYield}{NUptake}\]
In calculation-friendly notation, this is:
NUE = GrainYield / NUptake.
Substitute the relevant values:
GrainYield=3918.0 kg/ha, NUptake=100.4 kg N/ha.
The computed value is 39.02 kg grain/kg N.
Therefor... | dev | Nitrogen | 0 | CMF005 |
0.71 | cropmath-v1 | {
"new_value": 8.3,
"old_value": 9.2,
"parameter": "PDT"
} | [
"CMF045",
"CMF037"
] | {
"PDT": 8.3
} | growth_yield | {
"expression": "-0.0000842 * PDT^2 + 0.01 * PDT + 0.63",
"id": "CMF044",
"latex": "PISH=-8.42\\times10^{-5}PDT^2+0.01PDT+0.63",
"name": "地上部分配指数",
"name_en": "Shoot partitioning index",
"precision": 2,
"unit": "fraction"
} | CMF044_sensitivity_t2_0000 | false | Candidate formulas are listed below. Select the physiologically appropriate formula before calculating.
Candidate 1:
Formula name: Stem partitioning index (茎鞘分配指数)
Formula (LaTeX): PIS=1-PIGL-PIP
Formula: PIS = 1 - (if PDT<26: 0.54-0.0046*PDT; else: 1.4532*exp(-0.0492*PDT)) - (Y_grain/B_above/0.87)/(1+exp(-0.2804*(PDT... | Formula name: Shoot partitioning index (地上部分配指数)
Formula (LaTeX): PISH=-8.42\times10^{-5}PDT^2+0.01PDT+0.63
Formula: PISH = -0.0000842 * PDT^2 + 0.01 * PDT + 0.63
Operator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smalle... | Formula name: Shoot partitioning index (地上部分配指数)
Formula (LaTeX): PISH=-8.42\times10^{-5}PDT^2+0.01PDT+0.63
Formula: PISH = -0.0000842 * PDT^2 + 0.01 * PDT + 0.63
Operator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smalle... | Formula: PISH = -0.0000842 * PDT^2 + 0.01 * PDT + 0.63
Operator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.
Output unit: fraction | Relevant formula name: Shoot partitioning index (地上部分配指数).
Formula: PISH = -0.0000842 * PDT^2 + 0.01 * PDT + 0.63
Operator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.
Output unit: fraction | A retrieved formula is listed below. It may look relevant, but use it only if it truly matches the problem.
Formula name: Stem partitioning index (茎鞘分配指数)
Formula (LaTeX): PIS=1-PIGL-PIP
Formula: PIS = 1 - (if PDT<26: 0.54-0.0046*PDT; else: 1.4532*exp(-0.0492*PDT)) - (Y_grain/B_above/0.87)/(1+exp(-0.2804*(PDT-39)))
Ope... | 2 | Quadratic polynomial | sensitivity | {
"PDT": 9.2
} | In dry matter partitioning, we analyze how Shoot partitioning index responds to parameter changes. Baseline: PDT=9.2 physiological days. After changing PDT from 9.2 to 8.3, compute the new Shoot partitioning index. Give the numeric value in fraction. | {
"C": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ..... | Identify the target quantity as Shoot partitioning index.
Use the verified formula:
\[PISH=-8.42\times10^{-5}PDT^2+0.01PDT+0.63\]
In calculation-friendly notation, this is:
PISH = -0.0000842 * PDT^2 + 0.01 * PDT + 0.63.
Substitute the relevant values:
PDT=8.3 physiological days.
The computed value is 0.71 fraction.
The... | dev | Dry matter partitioning | 2 | CMF045 |
0.853 | cropmath-v1 | {
"new_value": 0.522,
"old_value": 0.58,
"parameter": "WaterC"
} | [
"CMF021",
"CMF019"
] | {
"WaterC": 0.522
} | methane_emission | {
"expression": "0.49 * exp(3.88 * WaterC - 5.4 * WaterC^2)",
"id": "CMF025",
"latex": "WI=0.49e^{3.88W_c-5.4W_c^2}",
"name": "水分分解因子",
"name_en": "Water-content decomposition factor",
"precision": 3,
"unit": "fraction"
} | CMF025_sensitivity_t0_0000 | false | Candidate formulas are listed below. Select the physiologically appropriate formula before calculating.
Candidate 1:
Formula name: Water-content decomposition factor (水分分解因子)
Formula (LaTeX): WI=0.49e^{3.88W_c-5.4W_c^2}
Formula: WI = 0.49 * exp(3.88 * WaterC - 5.4 * WaterC^2)
Operator notes: * means multiplication, / ... | Formula name: Water-content decomposition factor (水分分解因子)
Formula (LaTeX): WI=0.49e^{3.88W_c-5.4W_c^2}
Formula: WI = 0.49 * exp(3.88 * WaterC - 5.4 * WaterC^2)
Operator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/l... | Formula name: Water-content decomposition factor (水分分解因子)
Formula (LaTeX): WI=0.49e^{3.88W_c-5.4W_c^2}
Formula: WI = 0.49 * exp(3.88 * WaterC - 5.4 * WaterC^2)
Operator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/l... | Formula: WI = 0.49 * exp(3.88 * WaterC - 5.4 * WaterC^2)
Operator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.
Output unit: fraction | Relevant formula name: Water-content decomposition factor (水分分解因子).
Formula: WI = 0.49 * exp(3.88 * WaterC - 5.4 * WaterC^2)
Operator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.
Output unit: fraction | A retrieved formula is listed below. It may look relevant, but use it only if it truly matches the problem.
Formula name: Soil organic matter carbon decomposition (土壤有机质碳分解量)
Formula (LaTeX): OMSC=WI\cdot SI\cdot TI\cdot0.003\cdot OMS
Formula: OMSC = WI * SI * TI * 0.003 * OMS
Operator notes: * means multiplication, / ... | 2 | Exponential quadratic | sensitivity | {
"WaterC": 0.58
} | A methane calculation involves sensitivity analysis. The current measurements are WaterC=0.58 fraction. A field technician asks what the water-content decomposition factor would become if WaterC changed to 0.522. Report the new final numeric value in fraction. | {
"C": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ..... | Identify the target quantity as Water-content decomposition factor.
Use the verified formula:
\[WI=0.49e^{3.88W_c-5.4W_c^2}\]
In calculation-friendly notation, this is:
WI = 0.49 * exp(3.88 * WaterC - 5.4 * WaterC^2).
Substitute the relevant values:
WaterC=0.522 fraction.
The computed value is 0.853 fraction.
Therefore... | dev | Methane | 0 | CMF021 |
21.82 | cropmath-v1 | {
"new_value": 3500,
"old_value": 3725,
"parameter": "GrainYield"
} | [
"CMF011",
"CMF012"
] | {
"GrainYield": 3500,
"NUptake": 160.4
} | carbon_nitrogen_cycle | {
"expression": "GrainYield / NUptake",
"id": "CMF003",
"latex": "NUE=\\frac{GrainYield}{NUptake}",
"name": "氮利用效率",
"name_en": "Nitrogen use efficiency",
"precision": 2,
"unit": "kg grain/kg N"
} | CMF003_sensitivity_t2_0001 | false | Candidate formulas are listed below. Select the physiologically appropriate formula before calculating.
Candidate 1:
Formula name: Nitrogen use efficiency (氮利用效率)
Formula (LaTeX): NUE=\frac{GrainYield}{NUptake}
Formula: NUE = GrainYield / NUptake
Operator notes: * means multiplication, / means division, ** or ^ means ... | Formula name: Nitrogen use efficiency (氮利用效率)
Formula (LaTeX): NUE=\frac{GrainYield}{NUptake}
Formula: NUE = GrainYield / NUptake
Operator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.
Output unit: kg g... | Formula name: Nitrogen use efficiency (氮利用效率)
Formula (LaTeX): NUE=\frac{GrainYield}{NUptake}
Formula: NUE = GrainYield / NUptake
Operator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.
Output unit: kg g... | Formula: NUE = GrainYield / NUptake
Operator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.
Output unit: kg grain/kg N | Relevant formula name: Nitrogen use efficiency (氮利用效率).
Formula: NUE = GrainYield / NUptake
Operator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.
Output unit: kg grain/kg N | A retrieved formula is listed below. It may look relevant, but use it only if it truly matches the problem.
Formula name: carbon-nitrogen cycling soil carbon temperature factor (土壤碳分解温度因子)
Formula (LaTeX): F_T=2.4^{(T_{soil}-10)/10}\;\text{where}\;T_{soil}=T\;\text{or}\;4.4+0.76T
Formula: F_T_soil = if is_soil_temp>=0.... | 2 | Ratio | sensitivity | {
"GrainYield": 3725,
"NUptake": 160.4
} | In nitrogen, we analyze how Nitrogen use efficiency responds to parameter changes. Baseline: GrainYield=3725.0 kg/ha, NUptake=160.4 kg N/ha. After changing GrainYield from 3725.0 to 3500.0, compute the new Nitrogen use efficiency. Give the numeric value in kg grain/kg N. | {
"C": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ..... | Identify the target quantity as Nitrogen use efficiency.
Use the verified formula:
\[NUE=\frac{GrainYield}{NUptake}\]
In calculation-friendly notation, this is:
NUE = GrainYield / NUptake.
Substitute the relevant values:
GrainYield=3500.0 kg/ha, NUptake=160.4 kg N/ha.
The computed value is 21.82 kg grain/kg N.
Therefor... | dev | Nitrogen | 2 | CMF011 |
1.520 | cropmath-v1 | null | [
"CMF010",
"CMF013"
] | {
"CO2": 662,
"Coef_B_CO2": 0.78
} | carbon_nitrogen_cycle | {
"expression": "1 + Coef_B_CO2 * ln(CO2 / 340)",
"id": "CMF001",
"latex": "F_{CO2}=1+B\\ln(CO2/340)",
"name": "作物二氧化碳施肥因子",
"name_en": "carbon-nitrogen cycling CO2 fertilization factor",
"precision": 3,
"unit": "multiplier"
} | CMF001_direct_t1_0001 | false | Candidate formulas are listed below. Select the physiologically appropriate formula before calculating.
Candidate 1:
Formula name: carbon-nitrogen cycling CO2 fertilization factor (作物二氧化碳施肥因子)
Formula (LaTeX): F_{CO2}=1+B\ln(CO2/340)
Formula: F_CO2 = 1 + Coef_B_CO2 * ln(CO2 / 340)
Operator notes: * means multiplicatio... | Formula name: carbon-nitrogen cycling CO2 fertilization factor (作物二氧化碳施肥因子)
Formula (LaTeX): F_{CO2}=1+B\ln(CO2/340)
Formula: F_CO2 = 1 + Coef_B_CO2 * ln(CO2 / 340)
Operator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smal... | Formula name: carbon-nitrogen cycling CO2 fertilization factor (作物二氧化碳施肥因子)
Formula (LaTeX): F_{CO2}=1+B\ln(CO2/340)
Formula: F_CO2 = 1 + Coef_B_CO2 * ln(CO2 / 340)
Operator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smal... | Formula: F_CO2 = 1 + Coef_B_CO2 * ln(CO2 / 340)
Operator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.
Output unit: fraction | Relevant formula name: carbon-nitrogen cycling CO2 fertilization factor (作物二氧化碳施肥因子).
Formula: F_CO2 = 1 + Coef_B_CO2 * ln(CO2 / 340)
Operator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.
Output unit: ... | A retrieved formula is listed below. It may look relevant, but use it only if it truly matches the problem.
Formula name: carbon-nitrogen cycling soil pH decomposition factor (土壤pH分解因子)
Formula (LaTeX): F_{pH}=\frac{1}{1+e^{2.5(5-pH)}}
Formula: F_pH = 1 / (1 + exp(2.5 * (5 - pH)))
Operator notes: * means multiplication... | 1 | Logarithmic | direct | {
"CO2": 662,
"Coef_B_CO2": 0.78
} | Given CO2=662.0 ppm, Coef_B_CO2=0.78 dimensionless, calculate the carbon-nitrogen cycling CO2 fertilization factor for a carbon-nitrogen cycling photosynthesis. The result should be in fraction. | {
"C": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ..... | Identify the target quantity as carbon-nitrogen cycling CO2 fertilization factor.
Use the verified formula:
\[F_{CO2}=1+B\ln(CO2/340)\]
In calculation-friendly notation, this is:
F_CO2 = 1 + Coef_B_CO2 * ln(CO2 / 340).
Substitute the relevant values:
CO2=662.0 ppm, Coef_B_CO2=0.78 dimensionless.
The computed value is 1... | dev | carbon-nitrogen cycling photosynthesis | 1 | CMF010 |
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