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 |
|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|
0.95 | cropmath-v1 | {
"new_value": 1.6,
"old_value": 1.8,
"parameter": "Qt"
} | [
"CMF029",
"CMF030"
] | {
"PDT": 35.6,
"Qt": 1.6,
"Tmean": 18.5
} | 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_sensitivity_t1_0000 | 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: 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: 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... | 3 | Piecewise power penalty | sensitivity | {
"PDT": 35.6,
"Qt": 1.8,
"Tmean": 18.5
} | Given current measurements PDT=35.6 physiological days, Tmean=18.5 C, Qt=1.8 C day in temperature stress, what would the Low-temperature fertility factor be if Qt were adjusted to 1.6? 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 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=35.6 physiological days, Tmean=18.5 ... | dev | Temperature stress | 1 | CMF029 |
0.166 | cropmath-v1 | {
"new_value": 2200,
"old_value": 2121.5,
"parameter": "Pool"
} | [
"CMF001",
"CMF014"
] | {
"Coef_K_Decom": 0.001,
"Coef_K_Trans": 0.264,
"F_notill": 0.9,
"F_pH": 0.40900000000000003,
"F_temp": 0.838,
"F_texture": 0.77,
"F_water": 0.432,
"Pool": 2200
} | 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_sensitivity_t1_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 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... | 3 | Chained multiplicative decomposition | sensitivity | {
"Coef_K_Decom": 0.001,
"Coef_K_Trans": 0.264,
"F_notill": 0.9,
"F_pH": 0.40900000000000003,
"F_temp": 0.838,
"F_texture": 0.77,
"F_water": 0.432,
"Pool": 2121.5
} | Given current measurements Pool=2121.5 g C/m2, Coef_K_Decom=0.001 d-1, F_temp=0.838 fraction, F_water=0.432 fraction, F_texture=0.77 fraction, F_pH=0.409 fraction, F_notill=0.9 fraction, Coef_K_Trans=0.264 fraction in carbon-nitrogen cycling soil carbon, what would the carbon-nitrogen cycling carbon pool respiration lo... | {
"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 | 1 | CMF001 |
27.35 | cropmath-v1 | null | [
"CMF025",
"CMF020"
] | {
"Tair": 30.2
} | methane_emission | {
"expression": "4.4 + 0.76 * Tair",
"id": "CMF024",
"latex": "T_{soil}=4.4+0.76T_{air}",
"name": "土壤温度估算",
"name_en": "Soil temperature from air temperature",
"precision": 2,
"unit": "C"
} | CMF024_direct_t1_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: 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 ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.
Output un... | 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 ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.
Output un... | Formula: Tsoil = 4.4 + 0.76 * Tair
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 | Relevant formula name: Soil temperature from air temperature (土壤温度估算).
Formula: Tsoil = 4.4 + 0.76 * Tair
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 | 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... | 1 | Linear | direct | {
"Tair": 30.2
} | Given Tair=30.2 C, calculate the Soil temperature from air temperature for a methane. The result should be in C. | {
"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 temperature from air temperature.
Use the verified formula:
\[T_{soil}=4.4+0.76T_{air}\]
In calculation-friendly notation, this is:
Tsoil = 4.4 + 0.76 * Tair.
Substitute the relevant values:
Tair=30.2 C.
The computed value is 27.35 C.
Therefore, the final numeric answer is \boxed{27... | dev | Methane | 1 | CMF025 |
12.55 | cropmath-v1 | null | [
"CMF032",
"CMF035"
] | {
"PS": 0.23700000000000002,
"Po": 10.5
} | growth_yield | {
"expression": "Po + sqrt(1 / PS)",
"id": "CMF040",
"latex": "P_c=P_o+\\sqrt{1/PS}",
"name": "临界日长",
"name_en": "Critical daylength",
"precision": 2,
"unit": "h"
} | CMF040_chain_2_t2_0000 | 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: 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) means natural logarithm, and min/max select the smaller/larger value.
Output unit: h
Parameters:
- Po: optim... | 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) means natural logarithm, and min/max select the smaller/larger value.
Output unit: h
Parameters:
- Po: optim... | Formula: Pc = Po + sqrt(1 / PS)
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: h | Relevant formula name: Critical daylength (临界日长).
Formula: Pc = Po + sqrt(1 / PS)
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: h | A retrieved formula is listed below. It may look relevant, but use it only if it truly matches the problem.
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-... | 3 | Power response | chain_2 | {
"PS": 0.23700000000000002,
"Po": 10.5
} | In phenology, the Critical daylength is obtained by chaining intermediate calculations. With inputs Po=10.5 h, PS=0.237 dimensionless, determine the numeric value in 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 Critical daylength.
Use the verified formula:
\[P_c=P_o+\sqrt{1/PS}\]
In calculation-friendly notation, this is:
Pc = Po + sqrt(1 / PS).
Substitute the relevant values:
Po=10.5 h, PS=0.237 dimensionless.
The computed value is 12.55 h.
Therefore, the final numeric answer is \boxed{12.55}.... | dev | Phenology | 2 | CMF032 |
275.26 | cropmath-v1 | null | [
"CMF039",
"CMF034"
] | {
"DTGA": 407.9,
"RG": 50.3,
"RM": 96.1,
"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_direct_t1_0000 | 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: 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: 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... | 1 | Subtraction and conversion | direct | {
"DTGA": 407.9,
"RG": 50.3,
"RM": 96.1,
"b": 0.05
} | Given DTGA=407.9 kg/ha/day, RM=96.1 kg/ha/day, RG=50.3 kg/ha/day, b=0.05 fraction, calculate the Daily dry matter increment 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 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=407.9 kg/ha/day, RM=96.1 kg/ha/day, RG=50.3 kg/ha/day, b=0.05 fraction.
The computed valu... | dev | Respiration | 1 | CMF039 |
1.518 | cropmath-v1 | null | [
"CMF010",
"CMF011"
] | {
"Coef_T_M": 0.76,
"Td": 33.8,
"Tmax_active": 42,
"Tmin_active": 3,
"Topt_active": 25
} | carbon_nitrogen_cycle | {
"expression": "if Td<=Tmin_active or Td>=Tmax_active: 0; else: ((Td-Tmin_active)/(Topt_active-Tmin_active))^(1+Coef_T_M) * ((Tmax_active-Td)/(Tmax_active-Topt_active))^(1-Coef_T_M)",
"id": "CMF013",
"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}... | CMF013_branch_t2_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 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}
Formula: F_T = if Td<=Tmin_active or Td>=Tmax_active: 0; else: ((Td-Tmin_active)/(Topt_active-Tmin_active))^... | 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}
Formula: F_T = if Td<=Tmin_active or Td>=Tmax_active: 0; else: ((Td-Tmin_active)/(Topt_active-Tmin_active))^... | Formula: F_T = if Td<=Tmin_active or Td>=Tmax_active: 0; else: ((Td-Tmin_active)/(Topt_active-Tmin_active))^(1+Coef_T_M) * ((Tmax_active-Td)/(Tmax_active-Topt_active))^(1-Coef_T_M)
Operator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max... | Relevant formula name: carbon-nitrogen cycling crop temperature activity factor (作物温度活性因子).
Formula: F_T = if Td<=Tmin_active or Td>=Tmax_active: 0; else: ((Td-Tmin_active)/(Topt_active-Tmin_active))^(1+Coef_T_M) * ((Tmax_active-Td)/(Tmax_active-Topt_active))^(1-Coef_T_M)
Operator notes: * means multiplication, / 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 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... | 4 | Beta temperature response | branch | {
"Coef_T_M": 0.76,
"Td": 33.8,
"Tmax_active": 42,
"Tmin_active": 3,
"Topt_active": 25
} | Calculate carbon-nitrogen cycling crop temperature activity factor in carbon-nitrogen cycling stress with these inputs: Td=33.8 C, Tmin_active=3.0 C, Topt_active=25.0 C, Tmax_active=42.0 C, Coef_T_M=0.76 dimensionless. The computation may follow different paths based on piecewise thresholds. Give the numeric result in ... | {
"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 temperature activity factor.
Use the verified formula:
\[F_T=\left(\frac{T_d-T_{min}}{T_{opt}-T_{min}}\right)^{1+M}\left(\frac{T_{max}-T_d}{T_{max}-T_{opt}}\right)^{1-M}\]
In calculation-friendly notation, this is:
F_T = if Td<=Tmin_active or Td>=Tmax_active:... | dev | carbon-nitrogen cycling stress | 2 | CMF010 |
9.179 | cropmath-v1 | null | [
"CMF022",
"CMF023"
] | {
"CH4RiceEfC": 0.386,
"Ebl": 4.01,
"P": 15.119,
"W": 4936.9,
"Wmax": 12837.1
} | 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_chain_2_t1_0000 | false | Candidate formulas are listed below. Select the physiologically appropriate formula before calculating.
Candidate 1:
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 ... | 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 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 ^ ... | 4 | Chained sum | chain_2 | {
"CH4RiceEfC": 0.386,
"Ebl": 4.01,
"P": 15.119,
"W": 4936.9,
"Wmax": 12837.1
} | Given Ebl=4.01 g CH4/m2/day, P=15.119 g CH4/m2/day, CH4RiceEfC=0.386 fraction, W=4936.9 kg/ha, Wmax=12837.1 kg/ha, compute Total methane emission in methane. This requires sequential calculation through intermediate variables. Report the final 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=4.01 g CH4/m2/day, P=15.119 g CH4/m2/day, CH4RiceEfC=0.386 fraction, W=... | dev | Methane | 1 | CMF022 |
0.629 | cropmath-v1 | null | [
"CMF009",
"CMF004"
] | {
"Biom_plant_yesterday": 221,
"Coef_Rm": 0.0016,
"DVI": 1.15,
"Tmn": 25.3
} | 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_direct_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 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 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 | Q10 with DVI correction | direct | {
"Biom_plant_yesterday": 221,
"Coef_Rm": 0.0016,
"DVI": 1.15,
"Tmn": 25.3
} | In a carbon-nitrogen cycling respiration, the carbon-nitrogen cycling plant maintenance respiration is computed as part of carbon-nitrogen cycling respiration modeling. With Biom_plant_yesterday=221.0 g C/m2, Coef_Rm=0.0016 g C/g C/s, Tmn=25.3 C, DVI=1.15 dimensionless, 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 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 | CMF009 |
12.27 | cropmath-v1 | null | [
"CMF035",
"CMF052"
] | {
"PS": 0.318,
"Po": 10.5
} | growth_yield | {
"expression": "Po + sqrt(1 / PS)",
"id": "CMF040",
"latex": "P_c=P_o+\\sqrt{1/PS}",
"name": "临界日长",
"name_en": "Critical daylength",
"precision": 2,
"unit": "h"
} | CMF040_chain_2_t0_0001 | 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: 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) means natural logarithm, and min/max select the smaller/larger value.
Output unit: h
Parameters:
- Po: optim... | 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) means natural logarithm, and min/max select the smaller/larger value.
Output unit: h
Parameters:
- Po: optim... | Formula: Pc = Po + sqrt(1 / PS)
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: h | Relevant formula name: Critical daylength (临界日长).
Formula: Pc = Po + sqrt(1 / PS)
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: h | A retrieved formula is listed below. It may look relevant, but use it only if it truly matches the problem.
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
Operator no... | 3 | Power response | chain_2 | {
"PS": 0.318,
"Po": 10.5
} | A phenology module estimates Critical daylength using intermediate drivers. Use the provided inputs in sequence. The measurements are Po=10.5 h, PS=0.318 dimensionless. Compute the critical daylength and give the final numeric value in 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 Critical daylength.
Use the verified formula:
\[P_c=P_o+\sqrt{1/PS}\]
In calculation-friendly notation, this is:
Pc = Po + sqrt(1 / PS).
Substitute the relevant values:
Po=10.5 h, PS=0.318 dimensionless.
The computed value is 12.27 h.
Therefore, the final numeric answer is \boxed{12.27}.... | dev | Phenology | 0 | CMF035 |
0.825 | cropmath-v1 | null | [
"CMF016",
"CMF020"
] | {
"Eh": -133
} | methane_emission | {
"expression": "if Eh<-150: 1; else: exp(-1.7*(1+Eh/150))",
"id": "CMF018",
"latex": "F_{Eh}=1\\;(Eh<-150),\\quad e^{-1.7(1+Eh/150)}",
"name": "氧化还原电位因子",
"name_en": "Redox potential factor",
"precision": 3,
"unit": "fraction"
} | CMF018_branch_t0_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: 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 multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the sm... | 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 multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the sm... | Formula: FEh = if Eh<-150: 1; else: exp(-1.7*(1+Eh/150))
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: Redox potential factor (氧化还原电位因子).
Formula: FEh = if Eh<-150: 1; else: exp(-1.7*(1+Eh/150))
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: 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... | 4 | Piecewise exponential | branch | {
"Eh": -133
} | A methane involves a stage that may fall on either side of a piecewise threshold. The measured inputs are Eh=-133.0 mV. Determine the redox potential 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 Redox potential factor.
Use the verified formula:
\[F_{Eh}=1\;(Eh<-150),\quad e^{-1.7(1+Eh/150)}\]
In calculation-friendly notation, this is:
FEh = if Eh<-150: 1; else: exp(-1.7*(1+Eh/150)).
Substitute the relevant values:
Eh=-133.0 mV.
The computed value is 0.825 fraction.
Therefore, th... | dev | Methane | 0 | CMF016 |
1.000 | cropmath-v1 | null | [
"CMF059",
"CMF057"
] | {
"ExtremeT": 42.5,
"MaxT": 35.1,
"Tmax": 33.9
} | environmental_response | {
"expression": "if Tmax<=MaxT: 1; if Tmax>ExtremeT: 0; else: max(1-(Tmax-MaxT)/(ExtremeT-MaxT), 0)",
"id": "CMF058",
"latex": "F_H=1,\\;0,\\;\\text{or}\\;1-\\frac{T_{max}-MaxT}{ExtremeT-MaxT}",
"name": "环境响应高温响应",
"name_en": "environmental-response heat stress response",
"precision": 3,
"unit": "fraction... | CMF058_direct_t0_0001 | 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 heat stress response (环境响应高温响应)
Formula (LaTeX): F_H=1,\;0,\;\text{or}\;1-\frac{T_{max}-MaxT}{ExtremeT-MaxT}
Formula: F_Heat = if Tmax<=MaxT: 1; if Tmax>ExtremeT: 0; else: max(1-(Tmax-MaxT)/(ExtremeT-MaxT), 0)
Operator notes: * means multiplication, / means division, ** or ^ means p... | Formula name: environmental-response heat stress response (环境响应高温响应)
Formula (LaTeX): F_H=1,\;0,\;\text{or}\;1-\frac{T_{max}-MaxT}{ExtremeT-MaxT}
Formula: F_Heat = if Tmax<=MaxT: 1; if Tmax>ExtremeT: 0; else: max(1-(Tmax-MaxT)/(ExtremeT-MaxT), 0)
Operator notes: * means multiplication, / means division, ** or ^ means p... | Formula: F_Heat = if Tmax<=MaxT: 1; if Tmax>ExtremeT: 0; else: max(1-(Tmax-MaxT)/(ExtremeT-MaxT), 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 heat stress response (环境响应高温响应).
Formula: F_Heat = if Tmax<=MaxT: 1; if Tmax>ExtremeT: 0; else: max(1-(Tmax-MaxT)/(ExtremeT-MaxT), 0)
Operator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max ... | A retrieved formula is listed below. It may look relevant, but use it only if it truly matches the problem.
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+exp(0.01*(T... | 2 | Piecewise linear stress | direct | {
"ExtremeT": 42.5,
"MaxT": 35.1,
"Tmax": 33.9
} | A environmental-response formula system calculation estimates environmental-response heat stress response. The measured inputs are Tmax=33.9 C, MaxT=35.1 C, ExtremeT=42.5 C. What is the simple heat stress response? 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 environmental-response heat stress response.
Use the verified formula:
\[F_H=1,\;0,\;\text{or}\;1-\frac{T_{max}-MaxT}{ExtremeT-MaxT}\]
In calculation-friendly notation, this is:
F_Heat = if Tmax<=MaxT: 1; if Tmax>ExtremeT: 0; else: max(1-(Tmax-MaxT)/(ExtremeT-MaxT), 0).
Substitute the re... | dev | environmental-response formula system | 0 | CMF059 |
77.039 | cropmath-v1 | {
"new_value": 0.9,
"old_value": 0.78,
"parameter": "SI"
} | [
"CMF021",
"CMF022"
] | {
"SI": 0.9,
"VI": 1,
"W": 5517.7
} | methane_emission | {
"expression": "0.0018 * VI * SI * W ** 1.25",
"id": "CMF015",
"latex": "C_r=0.0018VI\\cdot SI\\cdot W^{1.25}",
"name": "根系分泌碳底物",
"name_en": "Root-exudate carbon substrate",
"precision": 3,
"unit": "g C/m2/day"
} | CMF015_sensitivity_t1_0001 | 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: Root-exudate carbon substrate (根系分泌碳底物)
Formula (LaTeX): C_r=0.0018VI\cdot SI\cdot W^{1.25}
Formula: Cr = 0.0018 * VI * SI * W ** 1.25
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... | Formula name: Root-exudate carbon substrate (根系分泌碳底物)
Formula (LaTeX): C_r=0.0018VI\cdot SI\cdot W^{1.25}
Formula: Cr = 0.0018 * VI * SI * W ** 1.25
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... | Formula: Cr = 0.0018 * VI * SI * W ** 1.25
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 C/m2/day | Relevant formula name: Root-exudate carbon substrate (根系分泌碳底物).
Formula: Cr = 0.0018 * VI * SI * W ** 1.25
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 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: 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 | Power product | sensitivity | {
"SI": 0.78,
"VI": 1,
"W": 5517.7
} | Given current measurements VI=1.0 fraction, SI=0.78 fraction, W=5517.7 kg/ha in methane, what would the Root-exudate carbon substrate be if SI were adjusted to 0.9? 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 Root-exudate carbon substrate.
Use the verified formula:
\[C_r=0.0018VI\cdot SI\cdot W^{1.25}\]
In calculation-friendly notation, this is:
Cr = 0.0018 * VI * SI * W ** 1.25.
Substitute the relevant values:
VI=1.0 fraction, SI=0.9 fraction, W=5517.7 kg/ha.
The computed value is 77.039 g C... | dev | Methane | 1 | CMF021 |
0.09 | cropmath-v1 | {
"new_value": 32.7,
"old_value": 29.7,
"parameter": "PDT"
} | [
"CMF054",
"CMF031"
] | {
"B_above": 12138,
"PDT": 32.7,
"Y_grain": 6390
} | growth_yield | {
"expression": "(Y_grain / B_above / 0.87) / (1 + exp(-0.2804 * (PDT - 39)))",
"id": "CMF042",
"latex": "PIP=\\frac{HI/0.87}{1+e^{-0.2804(PDT-39)}}",
"name": "穗分配指数",
"name_en": "Panicle partitioning index",
"precision": 2,
"unit": "fraction"
} | CMF042_sensitivity_t1_0000 | false | Candidate formulas are listed below. Select the physiologically appropriate formula before calculating.
Candidate 1:
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: * ... | 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 multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/... | 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 multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/... | Formula: PIP = (Y_grain / B_above / 0.87) / (1 + exp(-0.2804 * (PDT - 39)))
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: Panicle partitioning index (穗分配指数).
Formula: PIP = (Y_grain / B_above / 0.87) / (1 + exp(-0.2804 * (PDT - 39)))
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: 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 | Multi-step sigmoid | sensitivity | {
"B_above": 12138,
"PDT": 29.7,
"Y_grain": 6390
} | Given current measurements Y_grain=6390.0 kg/ha, B_above=12138.0 kg/ha, PDT=29.7 physiological days in dry matter partitioning, what would the Panicle partitioning index be if PDT were adjusted to 32.7? 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 Panicle partitioning index.
Use the verified formula:
\[PIP=\frac{HI/0.87}{1+e^{-0.2804(PDT-39)}}\]
In calculation-friendly notation, this is:
PIP = (Y_grain / B_above / 0.87) / (1 + exp(-0.2804 * (PDT - 39))).
Substitute the relevant values:
Y_grain=6390.0 kg/ha, B_above=12138.0 kg/ha, ... | dev | Dry matter partitioning | 1 | CMF054 |
0.06 | cropmath-v1 | null | [
"CMF053",
"CMF038"
] | {
"PDT": 38.5,
"Tmax_day": 39.8
} | growth_yield | {
"expression": "if 32<=PDT<=39 and Tmax_day>36.6: 1/(1+exp(0.853*(Tmax_day-36.6))); else: 1",
"id": "CMF035",
"latex": "HTF=\\frac{1}{1+e^{0.853(T_m-36.6)}}",
"name": "高温结实因子",
"name_en": "High-temperature fertility factor",
"precision": 2,
"unit": "fraction"
} | CMF035_direct_t1_0000 | false | Candidate formulas are listed below. Select the physiologically appropriate formula before calculating.
Candidate 1:
Formula name: Leaf area index from SLA (比叶面积法叶面积指数)
Formula (LaTeX): LAI=SLA\cdot AWL
Formula: LAI = SLA * AWL
Operator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means... | 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
Operator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural lo... | 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
Operator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural lo... | Formula: HTF = if 32<=PDT<=39 and Tmax_day>36.6: 1/(1+exp(0.853*(Tmax_day-36.6))); 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: High-temperature fertility factor (高温结实因子).
Formula: HTF = if 32<=PDT<=39 and Tmax_day>36.6: 1/(1+exp(0.853*(Tmax_day-36.6))); 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/lar... | A retrieved formula is listed below. It may look relevant, but use it only if it truly matches the problem.
Formula name: Specific leaf area (比叶面积)
Formula (LaTeX): SLA=(-0.0002GDD^2-0.5604GDD+581.04)(SLA_c/200)
Formula: SLA = if GDD<=1200: (-0.0002*GDD^2-0.5604*GDD+581.04)*(SLAc/200); else: 200*(SLAc/200)
Operator not... | 2 | Piecewise logistic | direct | {
"PDT": 38.5,
"Tmax_day": 39.8
} | Given PDT=38.5 physiological days, Tmax_day=39.8 C, calculate the High-temperature fertility factor for a temperature stress. 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 High-temperature fertility factor.
Use the verified formula:
\[HTF=\frac{1}{1+e^{0.853(T_m-36.6)}}\]
In calculation-friendly notation, this is:
HTF = if 32<=PDT<=39 and Tmax_day>36.6: 1/(1+exp(0.853*(Tmax_day-36.6))); else: 1.
Substitute the relevant values:
PDT=38.5 physiological days, ... | dev | Temperature stress | 1 | CMF053 |
0.92 | cropmath-v1 | null | [
"CMF028",
"CMF027"
] | {
"AWL": 7061,
"SLA": 0.00013000000000000002
} | growth_yield | {
"expression": "SLA * AWL",
"id": "CMF038",
"latex": "LAI=SLA\\cdot AWL",
"name": "比叶面积法叶面积指数",
"name_en": "Leaf area index from SLA",
"precision": 2,
"unit": "m2/m2"
} | CMF038_direct_t1_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: Leaf area index from SLA (比叶面积法叶面积指数)
Formula (LaTeX): LAI=SLA\cdot AWL
Formula: LAI = SLA * AWL
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
Parameters:
- SLA... | Formula name: Leaf area index from SLA (比叶面积法叶面积指数)
Formula (LaTeX): LAI=SLA\cdot AWL
Formula: LAI = SLA * AWL
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
Parameters:
- SLA... | Formula: LAI = SLA * AWL
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: Leaf area index from SLA (比叶面积法叶面积指数).
Formula: LAI = SLA * AWL
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: 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 ... | 1 | Product | direct | {
"AWL": 7061,
"SLA": 0.00013000000000000002
} | Given SLA=0.00013 ha/kg, AWL=7061.0 kg/ha, calculate the Leaf area index from SLA for a leaf area growth. The result should be 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 Leaf area index from SLA.
Use the verified formula:
\[LAI=SLA\cdot AWL\]
In calculation-friendly notation, this is:
LAI = SLA * AWL.
Substitute the relevant values:
SLA=0.00013 ha/kg, AWL=7061.0 kg/ha.
The computed value is 0.92 m2/m2.
Therefore, the final numeric answer is \boxed{0.92}.... | dev | Leaf area growth | 1 | CMF028 |
18.864 | cropmath-v1 | null | [
"CMF006",
"CMF009"
] | {
"Coef_alfa": 16.16,
"Coef_beta": 129.6,
"Leaf_N": 1.3599999999999999,
"PAR": 785.1
} | 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_t2_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: 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 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*L... | 1 | Michaelis-Menten response | direct | {
"Coef_alfa": 16.16,
"Coef_beta": 129.6,
"Leaf_N": 1.3599999999999999,
"PAR": 785.1
} | In a carbon-nitrogen cycling photosynthesis, the carbon-nitrogen cycling optimal photosynthetic rate is computed as part of carbon-nitrogen cycling photosynthesis modeling. With Coef_alfa=16.16 umol CO2/gN/s, Coef_beta=129.6 umol/m2/s, PAR=785.1 umol/m2/s, Leaf_N=1.36 gN/m2, determine the numeric value in umol CO2/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 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=16.16 umol... | dev | carbon-nitrogen cycling photosynthesis | 2 | CMF006 |
5.811 | cropmath-v1 | null | [
"CMF002",
"CMF014"
] | {
"Temp": 30.1,
"is_soil_temp": 1
} | 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_t2_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 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 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
Oper... | 4 | Q10 temperature response | branch | {
"Temp": 30.1,
"is_soil_temp": 1
} | Calculate carbon-nitrogen cycling soil carbon temperature factor in carbon-nitrogen cycling soil carbon with these inputs: Temp=30.1 C, is_soil_temp=1.0 0/1. The computation may follow different paths based on piecewise thresholds. Give the numeric result 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 | 2 | CMF002 |
0.290 | cropmath-v1 | null | [
"CMF030",
"CMF048"
] | {
"BFF": 0.86,
"DTE": 0.337,
"IE": 0.73,
"PDT": 51.3,
"RPE": 0.26
} | 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_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: 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, / ... | 2 | Stage-dependent product | direct | {
"BFF": 0.86,
"DTE": 0.337,
"IE": 0.73,
"PDT": 51.3,
"RPE": 0.26
} | A phenology calculation estimates Daily physiological effectiveness. The measured inputs are PDT=51.3 physiological days, DTE=0.337 fraction, IE=0.73 dimensionless, RPE=0.26 fraction, BFF=0.86 dimensionless. What is the daily physiological effectiveness? Give the final numeric value 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=51.... | dev | Phenology | 0 | CMF030 |
0.758 | cropmath-v1 | {
"new_value": 0.579,
"old_value": 0.643,
"parameter": "WaterC"
} | [
"CMF023",
"CMF015"
] | {
"WaterC": 0.579
} | 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_t1_0000 | false | Candidate formulas are listed below. Select the physiologically appropriate formula before calculating.
Candidate 1:
Formula name: Root-exudate carbon substrate (根系分泌碳底物)
Formula (LaTeX): C_r=0.0018VI\cdot SI\cdot W^{1.25}
Formula: Cr = 0.0018 * VI * SI * W ** 1.25
Operator notes: * means multiplication, / means divis... | 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: 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... | 2 | Exponential quadratic | sensitivity | {
"WaterC": 0.643
} | Given current measurements WaterC=0.643 fraction in methane, what would the Water-content decomposition factor be if WaterC were adjusted to 0.579? 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 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=0.579 fraction.
The computed value is 0.758 fraction.
Therefore... | dev | Methane | 1 | CMF023 |
0.502 | cropmath-v1 | {
"new_value": 32.2,
"old_value": 29.3,
"parameter": "PDT"
} | [
"CMF026",
"CMF048"
] | {
"PDT": 32.2
} | 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_t2_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: 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: 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 ... | 2 | Linear | sensitivity | {
"PDT": 29.3
} | In photosynthesis, we analyze how Canopy extinction coefficient responds to parameter changes. Baseline: PDT=29.3 physiological days. After changing PDT from 29.3 to 32.2, compute the new Canopy extinction coefficient. Give the 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=32.2 physiological days.
The computed value is 0.502 dimensionless.
Therefore, the final numeric answe... | dev | Photosynthesis | 2 | CMF026 |
0.654 | cropmath-v1 | null | [
"CMF037",
"CMF043"
] | {
"BFF": 0.89,
"DTE": 0.735,
"IE": 0.8,
"PDT": 46.1,
"RPE": 0.7000000000000001
} | 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_t2_0001 | 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: 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, ex... | 4 | Stage-dependent product | branch | {
"BFF": 0.89,
"DTE": 0.735,
"IE": 0.8,
"PDT": 46.1,
"RPE": 0.7000000000000001
} | Calculate Daily physiological effectiveness in phenology with these inputs: PDT=46.1 physiological days, DTE=0.735 fraction, IE=0.8 dimensionless, RPE=0.7 fraction, BFF=0.89 dimensionless. The computation may follow different paths based on piecewise thresholds. Give the numeric result 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=46.... | dev | Phenology | 2 | CMF037 |
19.87 | cropmath-v1 | null | [
"CMF029",
"CMF050"
] | {
"GDD": 774.4,
"SLAc": 146.5
} | growth_yield | {
"expression": "if GDD<=1200: (-0.0002*GDD^2-0.5604*GDD+581.04)*(SLAc/200); else: 200*(SLAc/200)",
"id": "CMF053",
"latex": "SLA=(-0.0002GDD^2-0.5604GDD+581.04)(SLA_c/200)",
"name": "比叶面积",
"name_en": "Specific leaf area",
"precision": 2,
"unit": "cm2/g"
} | CMF053_branch_t2_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: Specific leaf area (比叶面积)
Formula (LaTeX): SLA=(-0.0002GDD^2-0.5604GDD+581.04)(SLA_c/200)
Formula: SLA = if GDD<=1200: (-0.0002*GDD^2-0.5604*GDD+581.04)*(SLAc/200); else: 200*(SLAc/200)
Operator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural log... | Formula name: Specific leaf area (比叶面积)
Formula (LaTeX): SLA=(-0.0002GDD^2-0.5604GDD+581.04)(SLA_c/200)
Formula: SLA = if GDD<=1200: (-0.0002*GDD^2-0.5604*GDD+581.04)*(SLAc/200); else: 200*(SLAc/200)
Operator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural log... | Formula: SLA = if GDD<=1200: (-0.0002*GDD^2-0.5604*GDD+581.04)*(SLAc/200); else: 200*(SLAc/200)
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: cm2/g | Relevant formula name: Specific leaf area (比叶面积).
Formula: SLA = if GDD<=1200: (-0.0002*GDD^2-0.5604*GDD+581.04)*(SLAc/200); else: 200*(SLAc/200)
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... | 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... | 4 | Piecewise polynomial | branch | {
"GDD": 774.4,
"SLAc": 146.5
} | Calculate Specific leaf area in leaf area growth with these inputs: GDD=774.4 C day, SLAc=146.5 cm2/g. The computation may follow different paths based on piecewise thresholds. Give the numeric result in cm2/g. | {
"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 Specific leaf area.
Use the verified formula:
\[SLA=(-0.0002GDD^2-0.5604GDD+581.04)(SLA_c/200)\]
In calculation-friendly notation, this is:
SLA = if GDD<=1200: (-0.0002*GDD^2-0.5604*GDD+581.04)*(SLAc/200); else: 200*(SLAc/200).
Substitute the relevant values:
GDD=774.4 C day, SLAc=146.5 ... | dev | Leaf area growth | 2 | CMF029 |
0.976 | cropmath-v1 | null | [
"CMF017",
"CMF022"
] | {
"WaterC": 0.321
} | 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_direct_t2_0001 | 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: 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: 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, ... | 1 | Exponential quadratic | direct | {
"WaterC": 0.321
} | In a methane, the Water-content decomposition factor is computed as part of methane modeling. With WaterC=0.321 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 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=0.321 fraction.
The computed value is 0.976 fraction.
Therefore... | dev | Methane | 2 | CMF017 |
75.74 | cropmath-v1 | null | [
"CMF061",
"CMF058"
] | {
"F_CO2": 1.19,
"F_Heat": 0.45,
"F_Temp": 0.75,
"F_Water": 0.8,
"RUE": 1.99,
"SRAD": 24.3,
"fSolar": 0.39
} | 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_t2_0000 | false | Candidate formulas are listed below. Select the physiologically appropriate formula before calculating.
Candidate 1:
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... | 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": 1.19,
"F_Heat": 0.45,
"F_Temp": 0.75,
"F_Water": 0.8,
"RUE": 1.99,
"SRAD": 24.3,
"fSolar": 0.39
} | In an environmental-response formula system, the environmental-response daily biomass increment is computed as part of environmental-response formula calculation. With RUE=1.99 g/MJ, fSolar=0.39 fraction, SRAD=24.3 MJ/m2/day, F_CO2=1.19 fraction, F_Temp=0.75 fraction, F_Water=0.8 fraction, F_Heat=0.45 fraction, determi... | {
"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=1.99 g/MJ, f... | dev | environmental-response formula system | 2 | CMF061 |
14.89 | cropmath-v1 | {
"new_value": 13.8,
"old_value": 12,
"parameter": "Tair"
} | [
"CMF015",
"CMF018"
] | {
"Tair": 13.8
} | methane_emission | {
"expression": "4.4 + 0.76 * Tair",
"id": "CMF024",
"latex": "T_{soil}=4.4+0.76T_{air}",
"name": "土壤温度估算",
"name_en": "Soil temperature from air temperature",
"precision": 2,
"unit": "C"
} | CMF024_sensitivity_t2_0001 | false | Candidate formulas are listed below. Select the physiologically appropriate formula before calculating.
Candidate 1:
Formula name: Root-exudate carbon substrate (根系分泌碳底物)
Formula (LaTeX): C_r=0.0018VI\cdot SI\cdot W^{1.25}
Formula: Cr = 0.0018 * VI * SI * W ** 1.25
Operator notes: * means multiplication, / means divis... | 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 ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.
Output un... | 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 ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.
Output un... | Formula: Tsoil = 4.4 + 0.76 * Tair
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 | Relevant formula name: Soil temperature from air temperature (土壤温度估算).
Formula: Tsoil = 4.4 + 0.76 * Tair
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 | A retrieved formula is listed below. It may look relevant, but use it only if it truly matches the problem.
Formula name: Root-exudate carbon substrate (根系分泌碳底物)
Formula (LaTeX): C_r=0.0018VI\cdot SI\cdot W^{1.25}
Formula: Cr = 0.0018 * VI * SI * W ** 1.25
Operator notes: * means multiplication, / means division, ** or... | 2 | Linear | sensitivity | {
"Tair": 12
} | In methane, we analyze how Soil temperature from air temperature responds to parameter changes. Baseline: Tair=12.0 C. After changing Tair from 12.0 to 13.8, compute the new Soil temperature from air temperature. Give the numeric value in C. | {
"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 temperature from air temperature.
Use the verified formula:
\[T_{soil}=4.4+0.76T_{air}\]
In calculation-friendly notation, this is:
Tsoil = 4.4 + 0.76 * Tair.
Substitute the relevant values:
Tair=13.8 C.
The computed value is 14.89 C.
Therefore, the final numeric answer is \boxed{14... | dev | Methane | 2 | CMF015 |
254.39 | cropmath-v1 | {
"new_value": 5.23,
"old_value": 5.81,
"parameter": "LAI"
} | [
"CMF005",
"CMF011"
] | {
"Coef_K": 0.71,
"Coef_rou": 0.18,
"LAI": 5.23,
"Radi": 583.4
} | 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_sensitivity_t1_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: 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 | Exponential ratio | sensitivity | {
"Coef_K": 0.71,
"Coef_rou": 0.18,
"LAI": 5.8100000000000005,
"Radi": 583.4
} | Given current measurements Radi=583.4 W/m2, LAI=5.81 m2/m2, Coef_rou=0.18 fraction, Coef_K=0.71 dimensionless in carbon-nitrogen cycling photosynthesis, what would the carbon-nitrogen cycling canopy absorbed PAR be if LAI were adjusted to 5.23? Provide 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 | 1 | CMF005 |
5.576 | cropmath-v1 | null | [
"CMF010",
"CMF011"
] | {
"Biom_plant_yesterday": 641.5,
"Coef_Rm": 0.0033,
"DVI": 1.12,
"Tmn": 30.2
} | 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_branch_t0_0001 | false | Candidate formulas are listed below. Select the physiologically appropriate formula before calculating.
Candidate 1:
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 mult... | 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 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... | 4 | Q10 with DVI correction | branch | {
"Biom_plant_yesterday": 641.5,
"Coef_Rm": 0.0033,
"DVI": 1.12,
"Tmn": 30.2
} | A carbon-nitrogen cycling respiration involves a stage that may fall on either side of a piecewise threshold. The measured inputs are Biom_plant_yesterday=641.5 g C/m2, Coef_Rm=0.0033 g C/g C/s, Tmn=30.2 C, DVI=1.12 dimensionless. Determine the carbon-nitrogen cycling plant maintenance respiration and report the final ... | {
"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 | 0 | CMF010 |
0.433 | cropmath-v1 | null | [
"CMF015",
"CMF020"
] | {
"Com": 4.036,
"Cr": 10.96,
"FEh": 0.128,
"TI": 0.774
} | 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_chain_2_t0_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: Root-exudate carbon substrate (根系分泌碳底物)
Formula (LaTeX): C_r=0.0018VI\cdot SI\cdot W^{1.25}
Formula: Cr = 0.0018 * VI * SI * W ** 1.25
Operator notes: * means multiplication, / means division, ** or... | 3 | Maximum with chained drivers | chain_2 | {
"Com": 4.036,
"Cr": 10.96,
"FEh": 0.128,
"TI": 0.774
} | A methane module estimates Methane production using intermediate drivers. Use the provided inputs in sequence. The measurements are FEh=0.128 fraction, TI=0.774 fraction, Cr=10.96 g C/m2/day, Com=4.036 g C/m2/day. Compute the methane production and give the final 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=0.128 fraction, TI=0.774 fraction, Cr=10.96 g C/m2/day, Com=4.036 g C/m2/day.
T... | dev | Methane | 0 | CMF015 |
0.90 | cropmath-v1 | {
"new_value": 42.4,
"old_value": 47.1,
"parameter": "PDT"
} | [
"CMF038",
"CMF040"
] | {
"PDT": 42.4
} | 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_t0_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: 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: Leaf area index from SLA (比叶面积法叶面积指数)
Formula (LaTeX): LAI=SLA\cdot AWL
Formula: LAI = SLA * AWL
Operator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x... | 2 | Quadratic polynomial | sensitivity | {
"PDT": 47.1
} | A dry matter partitioning calculation involves sensitivity analysis. The current measurements are PDT=47.1 physiological days. A field technician asks what the shoot partitioning index would become if PDT changed to 42.4. 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 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=42.4 physiological days.
The computed value is 0.90 fraction.
Th... | dev | Dry matter partitioning | 0 | CMF038 |
4.656 | cropmath-v1 | null | [
"CMF012",
"CMF014"
] | {
"Biom_plant_yesterday": 757.9,
"Coef_Rm": 0.003,
"DVI": 0.76,
"Tmn": 23.3
} | 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_branch_t1_0000 | false | Candidate formulas are listed below. Select the physiologically appropriate formula before calculating.
Candidate 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_Wa... | 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.... | 4 | Q10 with DVI correction | branch | {
"Biom_plant_yesterday": 757.9,
"Coef_Rm": 0.003,
"DVI": 0.76,
"Tmn": 23.3
} | For carbon-nitrogen cycling respiration, compute the carbon-nitrogen cycling plant maintenance respiration. The inputs are Biom_plant_yesterday=757.9 g C/m2, Coef_Rm=0.003 g C/g C/s, Tmn=23.3 C, DVI=0.76 dimensionless. Note that this formula uses piecewise logic depending on parameter values. Report the final numeric v... | {
"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 | 1 | CMF012 |
2.90 | cropmath-v1 | null | [
"CMF061",
"CMF059"
] | {
"Tbase": 8.2,
"Tmean": 11.1
} | 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_t1_0001 | false | Candidate formulas are listed below. Select the physiologically appropriate formula before calculating.
Candidate 1:
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... | 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": 8.2,
"Tmean": 11.1
} | Given Tmean=11.1 C, Tbase=8.2 C, calculate the environmental-response daily thermal time for an environmental-response formula system. The result should be 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=11.1 C, Tbase=8.2 C.
The computed value is 2.90 C day.
Therefore, the final... | dev | environmental-response formula system | 1 | CMF061 |
3.15 | cropmath-v1 | {
"new_value": 8500,
"old_value": 8345,
"parameter": "AWL"
} | [
"CMF041",
"CMF053"
] | {
"AWL": 8500,
"SLA": 0.00037000000000000005
} | growth_yield | {
"expression": "SLA * AWL",
"id": "CMF038",
"latex": "LAI=SLA\\cdot AWL",
"name": "比叶面积法叶面积指数",
"name_en": "Leaf area index from SLA",
"precision": 2,
"unit": "m2/m2"
} | CMF038_sensitivity_t2_0000 | false | Candidate formulas are listed below. Select the physiologically appropriate formula before calculating.
Candidate 1:
Formula name: Leaf area index from SLA (比叶面积法叶面积指数)
Formula (LaTeX): LAI=SLA\cdot AWL
Formula: LAI = SLA * AWL
Operator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means... | Formula name: Leaf area index from SLA (比叶面积法叶面积指数)
Formula (LaTeX): LAI=SLA\cdot AWL
Formula: LAI = SLA * AWL
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
Parameters:
- SLA... | Formula name: Leaf area index from SLA (比叶面积法叶面积指数)
Formula (LaTeX): LAI=SLA\cdot AWL
Formula: LAI = SLA * AWL
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
Parameters:
- SLA... | Formula: LAI = SLA * AWL
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: Leaf area index from SLA (比叶面积法叶面积指数).
Formula: LAI = SLA * AWL
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: 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)
... | 2 | Product | sensitivity | {
"AWL": 8345,
"SLA": 0.00037000000000000005
} | In leaf area growth, we analyze how Leaf area index from SLA responds to parameter changes. Baseline: SLA=0.00037 ha/kg, AWL=8345.0 kg/ha. After changing AWL from 8345.0 to 8500.0, compute the new Leaf area index from SLA. Give the 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 Leaf area index from SLA.
Use the verified formula:
\[LAI=SLA\cdot AWL\]
In calculation-friendly notation, this is:
LAI = SLA * AWL.
Substitute the relevant values:
SLA=0.00037 ha/kg, AWL=8500.0 kg/ha.
The computed value is 3.15 m2/m2.
Therefore, the final numeric answer is \boxed{3.15}.... | dev | Leaf area growth | 2 | CMF041 |
0.0065 | cropmath-v1 | {
"new_value": 25.9,
"old_value": 28.8,
"parameter": "PDT"
} | [
"CMF030",
"CMF027"
] | {
"PDT": 25.9
} | 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_sensitivity_t0_0001 | 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: 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: 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, / ... | 2 | Linear decline | sensitivity | {
"PDT": 28.8
} | A respiration calculation involves sensitivity analysis. The current measurements are PDT=28.8 physiological days. A field technician asks what the maintenance respiration coefficient at optimum temperature would become if PDT changed to 25.9. Report the new final 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=25.9 physiological days.
The computed value is 0.0065 d-1.
There... | dev | Respiration | 0 | CMF030 |
0.652 | cropmath-v1 | null | [
"CMF016",
"CMF025"
] | {
"OMN": 195.9,
"SI": 0.65,
"TI": 0.68,
"WI": 0.279
} | 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_chain_2_t2_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: 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: 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... | 3 | Multiplicative | chain_2 | {
"OMN": 195.9,
"SI": 0.65,
"TI": 0.68,
"WI": 0.279
} | In methane, the Fresh organic matter carbon decomposition is obtained by chaining intermediate calculations. With inputs WI=0.279 fraction, SI=0.65 fraction, TI=0.68 fraction, OMN=195.9 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 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=0.279 fraction, SI=0.65 fraction, TI=0.68 fraction, OMN=195... | dev | Methane | 2 | CMF016 |
0.20 | cropmath-v1 | null | [
"CMF028",
"CMF030"
] | {
"B_above": 13156,
"PDT": 39.3,
"Y_grain": 4369
} | growth_yield | {
"expression": "(Y_grain / B_above / 0.87) / (1 + exp(-0.2804 * (PDT - 39)))",
"id": "CMF042",
"latex": "PIP=\\frac{HI/0.87}{1+e^{-0.2804(PDT-39)}}",
"name": "穗分配指数",
"name_en": "Panicle partitioning index",
"precision": 2,
"unit": "fraction"
} | CMF042_branch_t1_0001 | false | Candidate formulas are listed below. Select the physiologically appropriate formula before calculating.
Candidate 1:
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: * ... | 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 multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/... | 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 multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/... | Formula: PIP = (Y_grain / B_above / 0.87) / (1 + exp(-0.2804 * (PDT - 39)))
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: Panicle partitioning index (穗分配指数).
Formula: PIP = (Y_grain / B_above / 0.87) / (1 + exp(-0.2804 * (PDT - 39)))
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: 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 | Multi-step sigmoid | branch | {
"B_above": 13156,
"PDT": 39.3,
"Y_grain": 4369
} | For dry matter partitioning, compute the Panicle partitioning index. The inputs are Y_grain=4369.0 kg/ha, B_above=13156.0 kg/ha, PDT=39.3 physiological days. 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 Panicle partitioning index.
Use the verified formula:
\[PIP=\frac{HI/0.87}{1+e^{-0.2804(PDT-39)}}\]
In calculation-friendly notation, this is:
PIP = (Y_grain / B_above / 0.87) / (1 + exp(-0.2804 * (PDT - 39))).
Substitute the relevant values:
Y_grain=4369.0 kg/ha, B_above=13156.0 kg/ha, ... | dev | Dry matter partitioning | 1 | CMF028 |
40.80 | cropmath-v1 | {
"new_value": 5430,
"old_value": 4936,
"parameter": "GrainYield"
} | [
"CMF012",
"CMF011"
] | {
"GrainYield": 5430,
"NUptake": 133.1
} | 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_t1_0000 | false | Candidate formulas are listed below. Select the physiologically appropriate formula before calculating.
Candidate 1:
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_Wa... | 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 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.... | 2 | Ratio | sensitivity | {
"GrainYield": 4936,
"NUptake": 133.1
} | Given current measurements GrainYield=4936.0 kg/ha, NUptake=133.1 kg N/ha in nitrogen, what would the Nitrogen use efficiency be if GrainYield were adjusted to 5430.0? Provide 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=5430.0 kg/ha, NUptake=133.1 kg N/ha.
The computed value is 40.80 kg grain/kg N.
Therefor... | dev | Nitrogen | 1 | CMF012 |
68.1 | cropmath-v1 | null | [
"CMF036",
"CMF035"
] | {
"SDEPTH": 3.3
} | growth_yield | {
"expression": "45 + 7 * SDEPTH",
"id": "CMF028",
"latex": "EMGDD=45+7SDEPTH",
"name": "出苗热量需求",
"name_en": "Emergence thermal requirement",
"precision": 1,
"unit": "C day"
} | CMF028_direct_t0_0000 | false | Candidate formulas are listed below. Select the physiologically appropriate formula before calculating.
Candidate 1:
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... | 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 e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.
Output unit: C day
Paramete... | 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 e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.
Output unit: C day
Paramete... | Formula: EMGDD = 45 + 7 * SDEPTH
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: Emergence thermal requirement (出苗热量需求).
Formula: EMGDD = 45 + 7 * SDEPTH
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: 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... | 1 | Linear | direct | {
"SDEPTH": 3.3
} | A phenology calculation estimates Emergence thermal requirement. The measured inputs are SDEPTH=3.3 cm. What is the emergence thermal requirement? 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 Emergence thermal requirement.
Use the verified formula:
\[EMGDD=45+7SDEPTH\]
In calculation-friendly notation, this is:
EMGDD = 45 + 7 * SDEPTH.
Substitute the relevant values:
SDEPTH=3.3 cm.
The computed value is 68.1 C day.
Therefore, the final numeric answer is \boxed{68.1}.
#### 68.... | dev | Phenology | 0 | CMF036 |
0.000 | cropmath-v1 | null | [
"CMF055",
"CMF058"
] | {
"Tbase": 8.4,
"Tmean": 3.3,
"Topt": 22.7
} | 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_branch_t0_0001 | 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 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_... | 4 | Piecewise linear | branch | {
"Tbase": 8.4,
"Tmean": 3.3,
"Topt": 22.7
} | A environmental-response formula system involves a stage that may fall on either side of a piecewise threshold. The measured inputs are Tmean=3.3 C, Tbase=8.4 C, Topt=22.7 C. Determine the simple temperature response 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 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=3.3 C, Tbase=8.4 C... | dev | environmental-response formula system | 0 | CMF055 |
1.000 | cropmath-v1 | {
"new_value": 42,
"old_value": 41.5,
"parameter": "t_soil"
} | [
"CMF018",
"CMF016"
] | {
"Q10": 3.41,
"t_soil": 42
} | methane_emission | {
"expression": "Q10^((min(t_soil,30)-30)/10)",
"id": "CMF023",
"latex": "TI=Q_{10}^{(t_{sl}-30)/10}",
"name": "甲烷温度指数",
"name_en": "Methane temperature index",
"precision": 3,
"unit": "multiplier"
} | CMF023_sensitivity_t2_0001 | false | Candidate formulas are listed below. Select the physiologically appropriate formula before calculating.
Candidate 1:
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, / me... | 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 power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.
Output unit... | 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 power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.
Output unit... | Formula: TI = Q10^((min(t_soil,30)-30)/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: fraction | Relevant formula name: Methane temperature index (甲烷温度指数).
Formula: TI = Q10^((min(t_soil,30)-30)/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: fraction | A retrieved formula is listed below. It may look relevant, but use it only if it truly matches the problem.
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 multiplication, / mea... | 2 | Q10 with temperature clipping | sensitivity | {
"Q10": 3.41,
"t_soil": 41.5
} | In methane, we analyze how Methane temperature index responds to parameter changes. Baseline: Q10=3.41 dimensionless, t_soil=41.5 C. After changing t_soil from 41.5 to 42.0, compute the new Methane temperature 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 Methane temperature index.
Use the verified formula:
\[TI=Q_{10}^{(t_{sl}-30)/10}\]
In calculation-friendly notation, this is:
TI = Q10^((min(t_soil,30)-30)/10).
Substitute the relevant values:
Q10=3.41 dimensionless, t_soil=42.0 C.
The computed value is 1.000 fraction.
Therefore, the fi... | dev | Methane | 2 | CMF018 |
0.66 | cropmath-v1 | {
"new_value": 8095,
"old_value": 7039,
"parameter": "Y_grain"
} | [
"CMF028",
"CMF047"
] | {
"B_above": 12285,
"Y_grain": 8095
} | growth_yield | {
"expression": "Y_grain / B_above",
"id": "CMF034",
"latex": "HI=\\frac{Y_{grain}}{B_{above}}",
"name": "收获指数",
"name_en": "Harvest index",
"precision": 2,
"unit": "fraction"
} | CMF034_sensitivity_t1_0000 | false | Candidate formulas are listed below. Select the physiologically appropriate formula before calculating.
Candidate 1:
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... | 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, ln(x) means natural logarithm, and min/max select the smaller/larger value.
Output unit: fraction
Parameters... | 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, ln(x) means natural logarithm, and min/max select the smaller/larger value.
Output unit: fraction
Parameters... | Formula: HI = Y_grain / B_above
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: Harvest index (收获指数).
Formula: HI = Y_grain / B_above
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: 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 ... | 2 | Ratio | sensitivity | {
"B_above": 12285,
"Y_grain": 7039
} | Given current measurements Y_grain=7039.0 kg/ha, B_above=12285.0 kg/ha in dry matter partitioning, what would the Harvest index be if Y_grain were adjusted to 8095.0? 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 Harvest index.
Use the verified formula:
\[HI=\frac{Y_{grain}}{B_{above}}\]
In calculation-friendly notation, this is:
HI = Y_grain / B_above.
Substitute the relevant values:
Y_grain=8095.0 kg/ha, B_above=12285.0 kg/ha.
The computed value is 0.66 fraction.
Therefore, the final numeric an... | dev | Dry matter partitioning | 1 | CMF028 |
1.000 | cropmath-v1 | null | [
"CMF062",
"CMF060"
] | {
"ExtremeT": 43.3,
"MaxT": 34.5,
"Tmax": 32.7
} | environmental_response | {
"expression": "if Tmax<=MaxT: 1; if Tmax>ExtremeT: 0; else: max(1-(Tmax-MaxT)/(ExtremeT-MaxT), 0)",
"id": "CMF058",
"latex": "F_H=1,\\;0,\\;\\text{or}\\;1-\\frac{T_{max}-MaxT}{ExtremeT-MaxT}",
"name": "环境响应高温响应",
"name_en": "environmental-response heat stress response",
"precision": 3,
"unit": "fraction... | CMF058_branch_t2_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 heat stress response (环境响应高温响应)
Formula (LaTeX): F_H=1,\;0,\;\text{or}\;1-\frac{T_{max}-MaxT}{ExtremeT-MaxT}
Formula: F_Heat = if Tmax<=MaxT: 1; if Tmax>ExtremeT: 0; else: max(1-(Tmax-MaxT)/(ExtremeT-MaxT), 0)
Operator notes: * means multiplication, / means division, ** or ^ means p... | Formula name: environmental-response heat stress response (环境响应高温响应)
Formula (LaTeX): F_H=1,\;0,\;\text{or}\;1-\frac{T_{max}-MaxT}{ExtremeT-MaxT}
Formula: F_Heat = if Tmax<=MaxT: 1; if Tmax>ExtremeT: 0; else: max(1-(Tmax-MaxT)/(ExtremeT-MaxT), 0)
Operator notes: * means multiplication, / means division, ** or ^ means p... | Formula: F_Heat = if Tmax<=MaxT: 1; if Tmax>ExtremeT: 0; else: max(1-(Tmax-MaxT)/(ExtremeT-MaxT), 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 heat stress response (环境响应高温响应).
Formula: F_Heat = if Tmax<=MaxT: 1; if Tmax>ExtremeT: 0; else: max(1-(Tmax-MaxT)/(ExtremeT-MaxT), 0)
Operator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max ... | A retrieved formula is listed below. It may look relevant, but use it only if it truly matches the problem.
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, exp(x) means... | 4 | Piecewise linear stress | branch | {
"ExtremeT": 43.3,
"MaxT": 34.5,
"Tmax": 32.7
} | Calculate environmental-response heat stress response in environmental-response formula system with these inputs: Tmax=32.7 C, MaxT=34.5 C, ExtremeT=43.3 C. The computation may follow different paths based on piecewise thresholds. Give the numeric result 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 heat stress response.
Use the verified formula:
\[F_H=1,\;0,\;\text{or}\;1-\frac{T_{max}-MaxT}{ExtremeT-MaxT}\]
In calculation-friendly notation, this is:
F_Heat = if Tmax<=MaxT: 1; if Tmax>ExtremeT: 0; else: max(1-(Tmax-MaxT)/(ExtremeT-MaxT), 0).
Substitute the re... | dev | environmental-response formula system | 2 | CMF062 |
0.648 | cropmath-v1 | {
"new_value": 22.2,
"old_value": 20.2,
"parameter": "Tmean"
} | [
"CMF059",
"CMF056"
] | {
"Tbase": 9.5,
"Tmean": 22.2,
"Topt": 29.1
} | 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_0001 | 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 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+exp(0.01*(T... | 3 | Piecewise linear | sensitivity | {
"Tbase": 9.5,
"Tmean": 20.2,
"Topt": 29.1
} | Given current measurements Tmean=20.2 C, Tbase=9.5 C, Topt=29.1 C in environmental-response formula system, what would the environmental-response temperature response be if Tmean were adjusted to 22.2? 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=22.2 C, Tbase=9.5 ... | dev | environmental-response formula system | 1 | CMF059 |
0.14 | cropmath-v1 | null | [
"CMF033",
"CMF046"
] | {
"PDT": 30.5
} | 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_chain_2_t2_0000 | 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: 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: 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... | 3 | Residual from shoot partitioning | chain_2 | {
"PDT": 30.5
} | In dry matter partitioning, the Root partitioning index is obtained by chaining intermediate calculations. With inputs PDT=30.5 physiological days, 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 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=30.5 physiological days.
The computed value is 0.14 fraction.
Therefore, the final numeric an... | dev | Dry matter partitioning | 2 | CMF033 |
0.157 | cropmath-v1 | null | [
"CMF002",
"CMF011"
] | {
"Crop_Water_max": 0.42,
"Crop_Water_min": 0.22,
"Soil_Water": 0.10200000000000001,
"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_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 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 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
Oper... | 2 | Piecewise water response | direct | {
"Crop_Water_max": 0.42,
"Crop_Water_min": 0.22,
"Soil_Water": 0.10200000000000001,
"Wilt_point": 0.08
} | A carbon-nitrogen cycling water calculation estimates carbon-nitrogen cycling crop water stress factor. The measured inputs are Soil_Water=0.102 cm3/cm3, Crop_Water_max=0.42 cm3/cm3, Crop_Water_min=0.22 cm3/cm3, Wilt_point=0.08 cm3/cm3. What is the carbon-nitrogen cycling crop water stress factor? Give the final numeri... | {
"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 | 0 | CMF002 |
38.61 | cropmath-v1 | null | [
"CMF052",
"CMF028"
] | {
"DTGA": 148.5,
"Rg": 0.26
} | growth_yield | {
"expression": "Rg * DTGA",
"id": "CMF046",
"latex": "RG=R_g\\cdot DTGA",
"name": "生长呼吸",
"name_en": "Growth respiration",
"precision": 2,
"unit": "kg/ha/day"
} | CMF046_direct_t1_0000 | false | Candidate formulas are listed below. Select the physiologically appropriate formula before calculating.
Candidate 1:
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) m... | 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 natural logarithm, and min/max select the smaller/larger value.
Output unit: kg/ha/day
Parameters:
- Rg: growth r... | 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 natural logarithm, and min/max select the smaller/larger value.
Output unit: kg/ha/day
Parameters:
- Rg: growth r... | Formula: RG = Rg * DTGA
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: Growth respiration (生长呼吸).
Formula: RG = Rg * DTGA
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 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... | 1 | Product | direct | {
"DTGA": 148.5,
"Rg": 0.26
} | Given Rg=0.26 fraction, DTGA=148.5 kg/ha/day, calculate the Growth 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 Growth respiration.
Use the verified formula:
\[RG=R_g\cdot DTGA\]
In calculation-friendly notation, this is:
RG = Rg * DTGA.
Substitute the relevant values:
Rg=0.26 fraction, DTGA=148.5 kg/ha/day.
The computed value is 38.61 kg/ha/day.
Therefore, the final numeric answer is \boxed{38.61... | dev | Respiration | 1 | CMF052 |
0.736 | cropmath-v1 | null | [
"CMF062",
"CMF061"
] | {
"I50A": 388.4,
"I50B": 398.6,
"TT": 1058.6,
"Tsum": 1779.5,
"fSolarMax": 0.89,
"fSolar_water": 0.86
} | environmental_response | {
"expression": "min(fSolarMax/(1+exp(-0.01*(TT-I50A))), fSolarMax/(1+exp(0.01*(TT-(Tsum-I50B))))) * min(fSolar_water, 1)",
"id": "CMF059",
"latex": "fSolar=\\min(f_1,f_2)F_{solar,water}",
"name": "环境响应光截获比例",
"name_en": "environmental-response solar interception fraction",
"precision": 3,
"unit": "fracti... | CMF059_direct_t2_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 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+exp(0.01*(TT-(Tsum-I50B))))) * min(fSolar_water, 1)
Operator notes: * means multiplication, / means division, ** or ^ m... | 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+exp(0.01*(TT-(Tsum-I50B))))) * min(fSolar_water, 1)
Operator notes: * means multiplication, / means division, ** or ^ m... | Formula: fSolar = min(fSolarMax/(1+exp(-0.01*(TT-I50A))), fSolarMax/(1+exp(0.01*(TT-(Tsum-I50B))))) * min(fSolar_water, 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 solar interception fraction (环境响应光截获比例).
Formula: fSolar = min(fSolarMax/(1+exp(-0.01*(TT-I50A))), fSolarMax/(1+exp(0.01*(TT-(Tsum-I50B))))) * min(fSolar_water, 1)
Operator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means n... | A retrieved formula is listed below. It may look relevant, but use it only if it truly matches the problem.
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, exp(x) means... | 2 | Double logistic | direct | {
"I50A": 388.4,
"I50B": 398.6,
"TT": 1058.6,
"Tsum": 1779.5,
"fSolarMax": 0.89,
"fSolar_water": 0.86
} | In an environmental-response formula system, the environmental-response solar interception fraction is computed as part of environmental-response formula calculation. With fSolarMax=0.89 fraction, TT=1058.6 C day, I50A=388.4 C day, I50B=398.6 C day, Tsum=1779.5 C day, fSolar_water=0.86 fraction, determine the numeric v... | {
"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 solar interception fraction.
Use the verified formula:
\[fSolar=\min(f_1,f_2)F_{solar,water}\]
In calculation-friendly notation, this is:
fSolar = min(fSolarMax/(1+exp(-0.01*(TT-I50A))), fSolarMax/(1+exp(0.01*(TT-(Tsum-I50B))))) * min(fSolar_water, 1).
Substitute t... | dev | environmental-response formula system | 2 | CMF062 |
12.42 | cropmath-v1 | {
"new_value": 0.27,
"old_value": 0.235,
"parameter": "PS"
} | [
"CMF027",
"CMF051"
] | {
"PS": 0.27,
"Po": 10.5
} | growth_yield | {
"expression": "Po + sqrt(1 / PS)",
"id": "CMF040",
"latex": "P_c=P_o+\\sqrt{1/PS}",
"name": "临界日长",
"name_en": "Critical daylength",
"precision": 2,
"unit": "h"
} | CMF040_sensitivity_t1_0000 | false | Candidate formulas are listed below. Select the physiologically appropriate formula before calculating.
Candidate 1:
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 - ... | 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) means natural logarithm, and min/max select the smaller/larger value.
Output unit: h
Parameters:
- Po: optim... | 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) means natural logarithm, and min/max select the smaller/larger value.
Output unit: h
Parameters:
- Po: optim... | Formula: Pc = Po + sqrt(1 / PS)
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: h | Relevant formula name: Critical daylength (临界日长).
Formula: Pc = Po + sqrt(1 / PS)
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: h | A retrieved formula is listed below. It may look relevant, but use it only if it truly matches the problem.
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*BF... | 2 | Power response | sensitivity | {
"PS": 0.23500000000000001,
"Po": 10.5
} | Given current measurements Po=10.5 h, PS=0.235 dimensionless in phenology, what would the Critical daylength be if PS were adjusted to 0.27? Provide the numeric value in 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 Critical daylength.
Use the verified formula:
\[P_c=P_o+\sqrt{1/PS}\]
In calculation-friendly notation, this is:
Pc = Po + sqrt(1 / PS).
Substitute the relevant values:
Po=10.5 h, PS=0.27 dimensionless.
The computed value is 12.42 h.
Therefore, the final numeric answer is \boxed{12.42}.
... | dev | Phenology | 1 | CMF027 |
101.31 | cropmath-v1 | {
"new_value": 0.23,
"old_value": 0.2,
"parameter": "Rg"
} | [
"CMF050",
"CMF039"
] | {
"DTGA": 440.5,
"Rg": 0.23
} | growth_yield | {
"expression": "Rg * DTGA",
"id": "CMF046",
"latex": "RG=R_g\\cdot DTGA",
"name": "生长呼吸",
"name_en": "Growth respiration",
"precision": 2,
"unit": "kg/ha/day"
} | CMF046_sensitivity_t2_0001 | false | Candidate formulas are listed below. Select the physiologically appropriate formula before calculating.
Candidate 1:
Formula name: Relative photoperiod effectiveness (相对光周期效应)
Formula (LaTeX): RPE=1,\;1-PS(P-P_o)^2,\;0
Formula: RPE = if P<=Po: 1; if P<=Po+sqrt(1/PS): 1-PS*(P-Po)^2; else: 0
Operator notes: * means mult... | 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 natural logarithm, and min/max select the smaller/larger value.
Output unit: kg/ha/day
Parameters:
- Rg: growth r... | 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 natural logarithm, and min/max select the smaller/larger value.
Output unit: kg/ha/day
Parameters:
- Rg: growth r... | Formula: RG = Rg * DTGA
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: Growth respiration (生长呼吸).
Formula: RG = Rg * DTGA
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: Relative photoperiod effectiveness (相对光周期效应)
Formula (LaTeX): RPE=1,\;1-PS(P-P_o)^2,\;0
Formula: RPE = if P<=Po: 1; if P<=Po+sqrt(1/PS): 1-PS*(P-Po)^2; else: 0
Operator notes: * means multiplication... | 2 | Product | sensitivity | {
"DTGA": 440.5,
"Rg": 0.2
} | In respiration, we analyze how Growth respiration responds to parameter changes. Baseline: Rg=0.2 fraction, DTGA=440.5 kg/ha/day. After changing Rg from 0.2 to 0.23, compute the new Growth respiration. 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 Growth respiration.
Use the verified formula:
\[RG=R_g\cdot DTGA\]
In calculation-friendly notation, this is:
RG = Rg * DTGA.
Substitute the relevant values:
Rg=0.23 fraction, DTGA=440.5 kg/ha/day.
The computed value is 101.31 kg/ha/day.
Therefore, the final numeric answer is \boxed{101.... | dev | Respiration | 2 | CMF050 |
0.024 | cropmath-v1 | null | [
"CMF017",
"CMF024"
] | {
"P": 16.848,
"Wr": 1725.2,
"t_soil": 33
} | methane_emission | {
"expression": "min(0.7 * ln(t_soil) / Wr, 0.9) * P",
"id": "CMF016",
"latex": "E_{bl}=\\min(0.7\\ln(t_{soil})/W_r,0.9)P",
"name": "甲烷冒泡排放",
"name_en": "Methane ebullition emission",
"precision": 3,
"unit": "g CH4/m2/day"
} | CMF016_chain_2_t1_0000 | false | Candidate formulas are listed below. Select the physiologically appropriate formula before calculating.
Candidate 1:
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, / me... | 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 division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/lar... | 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 division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/lar... | Formula: Ebl = min(0.7 * ln(t_soil) / Wr, 0.9) * P
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 ebullition emission (甲烷冒泡排放).
Formula: Ebl = min(0.7 * ln(t_soil) / Wr, 0.9) * P
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: 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, ... | 3 | Logarithmic minimum | chain_2 | {
"P": 16.848,
"Wr": 1725.2,
"t_soil": 33
} | Given P=16.848 g CH4/m2/day, t_soil=33.0 C, Wr=1725.2 kg/ha, compute Methane ebullition emission in methane. This requires sequential calculation through intermediate variables. Report the final 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 ebullition emission.
Use the verified formula:
\[E_{bl}=\min(0.7\ln(t_{soil})/W_r,0.9)P\]
In calculation-friendly notation, this is:
Ebl = min(0.7 * ln(t_soil) / Wr, 0.9) * P.
Substitute the relevant values:
P=16.848 g CH4/m2/day, t_soil=33.0 C, Wr=1725.2 kg/ha.
The computed valu... | dev | Methane | 1 | CMF017 |
448.60 | cropmath-v1 | {
"new_value": 2.74,
"old_value": 2.49,
"parameter": "LAI"
} | [
"CMF011",
"CMF004"
] | {
"Coef_K": 0.74,
"Coef_rou": 0.12,
"LAI": 2.74,
"Radi": 588.1
} | 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_sensitivity_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 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: 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 | Exponential ratio | sensitivity | {
"Coef_K": 0.74,
"Coef_rou": 0.12,
"LAI": 2.49,
"Radi": 588.1
} | A carbon-nitrogen cycling photosynthesis calculation involves sensitivity analysis. The current measurements are Radi=588.1 W/m2, LAI=2.49 m2/m2, Coef_rou=0.12 fraction, Coef_K=0.74 dimensionless. A field technician asks what the carbon-nitrogen cycling canopy absorbed par would become if LAI changed to 2.74. Report th... | {
"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 | 0 | CMF011 |
0.65 | cropmath-v1 | null | [
"CMF032",
"CMF044"
] | {
"B_above": 16042,
"PDT": 35.1,
"Y_grain": 5337
} | growth_yield | {
"expression": "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)))",
"id": "CMF045",
"latex": "PIS=1-PIGL-PIP",
"name": "茎鞘分配指数",
"name_en": "Stem partitioning index",
"precision": 2,
"unit": "fraction"
} | CMF045_direct_t1_0000 | 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: 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)))
Operator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means na... | 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)))
Operator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means na... | 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)))
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... | Relevant formula name: Stem partitioning index (茎鞘分配指数).
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)))
Operator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and m... | A retrieved formula is listed below. It may look relevant, but use it only if it truly matches the problem.
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-... | 2 | Residual with piecewise components | direct | {
"B_above": 16042,
"PDT": 35.1,
"Y_grain": 5337
} | Given PDT=35.1 physiological days, Y_grain=5337.0 kg/ha, B_above=16042.0 kg/ha, calculate the Stem 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 Stem partitioning index.
Use the verified formula:
\[PIS=1-PIGL-PIP\]
In calculation-friendly notation, this is:
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))).
Substitute the relevant values:
PDT=35.1 physiologica... | dev | Dry matter partitioning | 1 | CMF032 |
0.015 | cropmath-v1 | null | [
"CMF020",
"CMF023"
] | {
"P": 8.889,
"Wr": 1500.3,
"t_soil": 34.8
} | methane_emission | {
"expression": "min(0.7 * ln(t_soil) / Wr, 0.9) * P",
"id": "CMF016",
"latex": "E_{bl}=\\min(0.7\\ln(t_{soil})/W_r,0.9)P",
"name": "甲烷冒泡排放",
"name_en": "Methane ebullition emission",
"precision": 3,
"unit": "g CH4/m2/day"
} | CMF016_direct_t2_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 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 division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/lar... | 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 division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/lar... | Formula: Ebl = min(0.7 * ln(t_soil) / Wr, 0.9) * P
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 ebullition emission (甲烷冒泡排放).
Formula: Ebl = min(0.7 * ln(t_soil) / Wr, 0.9) * P
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 | Logarithmic minimum | direct | {
"P": 8.889,
"Wr": 1500.3,
"t_soil": 34.8
} | In a methane, the Methane ebullition emission is computed as part of methane modeling. With P=8.889 g CH4/m2/day, t_soil=34.8 C, Wr=1500.3 kg/ha, 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 ebullition emission.
Use the verified formula:
\[E_{bl}=\min(0.7\ln(t_{soil})/W_r,0.9)P\]
In calculation-friendly notation, this is:
Ebl = min(0.7 * ln(t_soil) / Wr, 0.9) * P.
Substitute the relevant values:
P=8.889 g CH4/m2/day, t_soil=34.8 C, Wr=1500.3 kg/ha.
The computed value... | dev | Methane | 2 | CMF020 |
6.134 | cropmath-v1 | null | [
"CMF022",
"CMF024"
] | {
"CH4RiceEfC": 0.244,
"Ebl": 5.626,
"P": 2.569,
"W": 6214.2,
"Wmax": 10951.2
} | 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_chain_2_t2_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: 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 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 ^ ... | 4 | Chained sum | chain_2 | {
"CH4RiceEfC": 0.244,
"Ebl": 5.626,
"P": 2.569,
"W": 6214.2,
"Wmax": 10951.2
} | In methane, the Total methane emission is obtained by chaining intermediate calculations. With inputs Ebl=5.626 g CH4/m2/day, P=2.569 g CH4/m2/day, CH4RiceEfC=0.244 fraction, W=6214.2 kg/ha, Wmax=10951.2 kg/ha, 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 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=5.626 g CH4/m2/day, P=2.569 g CH4/m2/day, CH4RiceEfC=0.244 fraction, W=... | dev | Methane | 2 | CMF022 |
35.555 | cropmath-v1 | null | [
"CMF006",
"CMF013"
] | {
"Coef_alfa": 16.85,
"Coef_beta": 158.8,
"Leaf_N": 3.39,
"PAR": 261.8
} | 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_t2_0001 | 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: 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 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*L... | 1 | Michaelis-Menten response | direct | {
"Coef_alfa": 16.85,
"Coef_beta": 158.8,
"Leaf_N": 3.39,
"PAR": 261.8
} | In a carbon-nitrogen cycling photosynthesis, the carbon-nitrogen cycling optimal photosynthetic rate is computed as part of carbon-nitrogen cycling photosynthesis modeling. With Coef_alfa=16.85 umol CO2/gN/s, Coef_beta=158.8 umol/m2/s, PAR=261.8 umol/m2/s, Leaf_N=3.39 gN/m2, determine the numeric value in umol CO2/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 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=16.85 umol... | dev | carbon-nitrogen cycling photosynthesis | 2 | CMF006 |
2.161 | cropmath-v1 | null | [
"CMF014",
"CMF008"
] | {
"Temp": 18.8,
"is_soil_temp": 1
} | 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_direct_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 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 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_W... | 2 | Q10 temperature response | direct | {
"Temp": 18.8,
"is_soil_temp": 1
} | Given Temp=18.8 C, is_soil_temp=1.0 0/1, calculate the carbon-nitrogen cycling soil carbon temperature factor for a carbon-nitrogen cycling soil carbon. 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 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 | 1 | CMF014 |
0.698 | cropmath-v1 | null | [
"CMF010",
"CMF009"
] | {
"Temp": 5.9,
"is_soil_temp": 1
} | 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_0001 | false | Candidate formulas are listed below. Select the physiologically appropriate formula before calculating.
Candidate 1:
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 mult... | 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 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... | 4 | Q10 temperature response | branch | {
"Temp": 5.9,
"is_soil_temp": 1
} | A carbon-nitrogen cycling soil carbon involves a stage that may fall on either side of a piecewise threshold. The measured inputs are Temp=5.9 C, is_soil_temp=1.0 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 | CMF010 |
0.814 | cropmath-v1 | {
"new_value": 5.59,
"old_value": 4.86,
"parameter": "pH"
} | [
"CMF014",
"CMF003"
] | {
"pH": 5.59
} | carbon_nitrogen_cycle | {
"expression": "1 / (1 + exp(2.5 * (5 - pH)))",
"id": "CMF010",
"latex": "F_{pH}=\\frac{1}{1+e^{2.5(5-pH)}}",
"name": "土壤pH分解因子",
"name_en": "carbon-nitrogen cycling soil pH decomposition factor",
"precision": 3,
"unit": "fraction"
} | CMF010_sensitivity_t2_0000 | false | Candidate formulas are listed below. Select the physiologically appropriate formula before calculating.
Candidate 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_Wa... | 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, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select... | 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, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select... | Formula: F_pH = 1 / (1 + exp(2.5 * (5 - pH)))
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 pH decomposition factor (土壤pH分解因子).
Formula: F_pH = 1 / (1 + exp(2.5 * (5 - pH)))
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 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_W... | 2 | Logistic | sensitivity | {
"pH": 4.86
} | In carbon-nitrogen cycling soil carbon, we analyze how carbon-nitrogen cycling soil pH decomposition factor responds to parameter changes. Baseline: pH=4.86 pH. After changing pH from 4.86 to 5.59, compute the new carbon-nitrogen cycling soil pH decomposition 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 carbon-nitrogen cycling soil pH decomposition factor.
Use the verified formula:
\[F_{pH}=\frac{1}{1+e^{2.5(5-pH)}}\]
In calculation-friendly notation, this is:
F_pH = 1 / (1 + exp(2.5 * (5 - pH))).
Substitute the relevant values:
pH=5.59 pH.
The computed value is 0.814 fraction.
Therefor... | dev | carbon-nitrogen cycling soil carbon | 2 | CMF014 |
1.00 | cropmath-v1 | null | [
"CMF037",
"CMF053"
] | {
"PDT": 15.7,
"a": 0.023
} | 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_direct_t1_0000 | false | Candidate formulas are listed below. Select the physiologically appropriate formula before calculating.
Candidate 1:
Formula name: Specific leaf area (比叶面积)
Formula (LaTeX): SLA=(-0.0002GDD^2-0.5604GDD+581.04)(SLA_c/200)
Formula: SLA = if GDD<=1200: (-0.0002*GDD^2-0.5604*GDD+581.04)*(SLAc/200); else: 200*(SLAc/200)
Op... | 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: 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, ex... | 2 | Piecewise exponential decay | direct | {
"PDT": 15.7,
"a": 0.023
} | Given PDT=15.7 physiological days, a=0.023 dimensionless, calculate the Physiological age factor for a 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 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=15.7 physiological days, a=0.023 dimensionless.
The comp... | dev | Photosynthesis | 1 | CMF037 |
25.45 | cropmath-v1 | null | [
"CMF023",
"CMF022"
] | {
"Tair": 27.7
} | methane_emission | {
"expression": "4.4 + 0.76 * Tair",
"id": "CMF024",
"latex": "T_{soil}=4.4+0.76T_{air}",
"name": "土壤温度估算",
"name_en": "Soil temperature from air temperature",
"precision": 2,
"unit": "C"
} | CMF024_direct_t2_0001 | false | Candidate formulas are listed below. Select the physiologically appropriate formula before calculating.
Candidate 1:
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 ^ ... | 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 ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.
Output un... | 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 ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.
Output un... | Formula: Tsoil = 4.4 + 0.76 * Tair
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 | Relevant formula name: Soil temperature from air temperature (土壤温度估算).
Formula: Tsoil = 4.4 + 0.76 * Tair
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 | 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... | 1 | Linear | direct | {
"Tair": 27.7
} | In a methane, the Soil temperature from air temperature is computed as part of methane modeling. With Tair=27.7 C, determine the numeric value in C. | {
"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 temperature from air temperature.
Use the verified formula:
\[T_{soil}=4.4+0.76T_{air}\]
In calculation-friendly notation, this is:
Tsoil = 4.4 + 0.76 * Tair.
Substitute the relevant values:
Tair=27.7 C.
The computed value is 25.45 C.
Therefore, the final numeric answer is \boxed{25... | dev | Methane | 2 | CMF023 |
0.374 | cropmath-v1 | null | [
"CMF036",
"CMF050"
] | {
"BFF": 0.6900000000000001,
"DTE": 0.331,
"IE": 1.13,
"PDT": 2.4,
"RPE": 0.64
} | 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_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: 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: 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 | Stage-dependent product | direct | {
"BFF": 0.6900000000000001,
"DTE": 0.331,
"IE": 1.13,
"PDT": 2.4,
"RPE": 0.64
} | A phenology calculation estimates Daily physiological effectiveness. The measured inputs are PDT=2.4 physiological days, DTE=0.331 fraction, IE=1.13 dimensionless, RPE=0.64 fraction, BFF=0.69 dimensionless. What is the daily physiological effectiveness? Give the final numeric value 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=2.4... | dev | Phenology | 0 | CMF036 |
2.086 | cropmath-v1 | null | [
"CMF003",
"CMF001"
] | {
"Day_Length_hour": 11.13,
"F_CO2": 1.21,
"F_T": 0.42,
"F_Water": 0.36,
"Photo_Rate_opt": 23.71
} | 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_t2_0000 | 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: 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: 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... | 2 | Multiplicative conversion | direct | {
"Day_Length_hour": 11.13,
"F_CO2": 1.21,
"F_T": 0.42,
"F_Water": 0.36,
"Photo_Rate_opt": 23.71
} | In a carbon-nitrogen cycling photosynthesis, the carbon-nitrogen cycling gross primary production is computed as part of carbon-nitrogen cycling photosynthesis modeling. With Photo_Rate_opt=23.71 umol CO2/m2/s, F_T=0.42 fraction, F_Water=0.36 fraction, F_CO2=1.21 fraction, Day_Length_hour=11.13 h, determine the 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 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=23.71 u... | dev | carbon-nitrogen cycling photosynthesis | 2 | CMF003 |
0.972 | cropmath-v1 | null | [
"CMF025",
"CMF015"
] | {
"OMS": 1220.1,
"SI": 1.33,
"TI": 0.435,
"WI": 0.459
} | 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_0001 | false | Candidate formulas are listed below. Select the physiologically appropriate formula before calculating.
Candidate 1:
Formula name: Root-exudate carbon substrate (根系分泌碳底物)
Formula (LaTeX): C_r=0.0018VI\cdot SI\cdot W^{1.25}
Formula: Cr = 0.0018 * VI * SI * W ** 1.25
Operator notes: * means multiplication, / means divis... | 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: 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... | 1 | Multiplicative | direct | {
"OMS": 1220.1,
"SI": 1.33,
"TI": 0.435,
"WI": 0.459
} | In a methane, the Soil organic matter carbon decomposition is computed as part of methane modeling. With WI=0.459 fraction, SI=1.33 fraction, TI=0.435 fraction, OMS=1220.1 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=0.459 fraction, SI=1.33 fraction, TI=0.435 fraction, OMS=122... | dev | Methane | 2 | CMF025 |
135.57 | cropmath-v1 | null | [
"CMF047",
"CMF031"
] | {
"GDD": 597.9,
"SLAc": 155.4
} | growth_yield | {
"expression": "if GDD<=1200: (-0.0002*GDD^2-0.5604*GDD+581.04)*(SLAc/200); else: 200*(SLAc/200)",
"id": "CMF053",
"latex": "SLA=(-0.0002GDD^2-0.5604GDD+581.04)(SLA_c/200)",
"name": "比叶面积",
"name_en": "Specific leaf area",
"precision": 2,
"unit": "cm2/g"
} | CMF053_branch_t1_0000 | false | Candidate formulas are listed below. Select the physiologically appropriate formula before calculating.
Candidate 1:
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 multip... | Formula name: Specific leaf area (比叶面积)
Formula (LaTeX): SLA=(-0.0002GDD^2-0.5604GDD+581.04)(SLA_c/200)
Formula: SLA = if GDD<=1200: (-0.0002*GDD^2-0.5604*GDD+581.04)*(SLAc/200); else: 200*(SLAc/200)
Operator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural log... | Formula name: Specific leaf area (比叶面积)
Formula (LaTeX): SLA=(-0.0002GDD^2-0.5604GDD+581.04)(SLA_c/200)
Formula: SLA = if GDD<=1200: (-0.0002*GDD^2-0.5604*GDD+581.04)*(SLAc/200); else: 200*(SLAc/200)
Operator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural log... | Formula: SLA = if GDD<=1200: (-0.0002*GDD^2-0.5604*GDD+581.04)*(SLAc/200); else: 200*(SLAc/200)
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: cm2/g | Relevant formula name: Specific leaf area (比叶面积).
Formula: SLA = if GDD<=1200: (-0.0002*GDD^2-0.5604*GDD+581.04)*(SLAc/200); else: 200*(SLAc/200)
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... | 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, ... | 4 | Piecewise polynomial | branch | {
"GDD": 597.9,
"SLAc": 155.4
} | For leaf area growth, compute the Specific leaf area. The inputs are GDD=597.9 C day, SLAc=155.4 cm2/g. Note that this formula uses piecewise logic depending on parameter values. Report the final numeric value in cm2/g. | {
"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 Specific leaf area.
Use the verified formula:
\[SLA=(-0.0002GDD^2-0.5604GDD+581.04)(SLA_c/200)\]
In calculation-friendly notation, this is:
SLA = if GDD<=1200: (-0.0002*GDD^2-0.5604*GDD+581.04)*(SLAc/200); else: 200*(SLAc/200).
Substitute the relevant values:
GDD=597.9 C day, SLAc=155.4 ... | dev | Leaf area growth | 1 | CMF047 |
1.519 | cropmath-v1 | null | [
"CMF023",
"CMF015"
] | {
"OMS": 1545.9,
"SI": 0.92,
"TI": 0.658,
"WI": 0.541
} | 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_chain_2_t0_0001 | false | Candidate formulas are listed below. Select the physiologically appropriate formula before calculating.
Candidate 1:
Formula name: Root-exudate carbon substrate (根系分泌碳底物)
Formula (LaTeX): C_r=0.0018VI\cdot SI\cdot W^{1.25}
Formula: Cr = 0.0018 * VI * SI * W ** 1.25
Operator notes: * means multiplication, / means divis... | 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 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 | Multiplicative | chain_2 | {
"OMS": 1545.9,
"SI": 0.92,
"TI": 0.658,
"WI": 0.541
} | A methane module estimates Soil organic matter carbon decomposition using intermediate drivers. Use the provided inputs in sequence. The measurements are WI=0.541 fraction, SI=0.92 fraction, TI=0.658 fraction, OMS=1545.9 g C/m2. Compute the soil organic matter carbon decomposition and give the final numeric value in g ... | {
"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=0.541 fraction, SI=0.92 fraction, TI=0.658 fraction, OMS=154... | dev | Methane | 0 | CMF023 |
28.104 | cropmath-v1 | null | [
"CMF009",
"CMF014"
] | {
"Day_Length_hour": 14.04,
"F_CO2": 1.31,
"F_T": 0.9500000000000001,
"F_Water": 0.6900000000000001,
"Photo_Rate_opt": 53.96
} | 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_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 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... | 4 | Multiplicative conversion | chain_2 | {
"Day_Length_hour": 14.04,
"F_CO2": 1.31,
"F_T": 0.9500000000000001,
"F_Water": 0.6900000000000001,
"Photo_Rate_opt": 53.96
} | In carbon-nitrogen cycling photosynthesis, the carbon-nitrogen cycling gross primary production is obtained by chaining intermediate calculations. With inputs Photo_Rate_opt=53.96 umol CO2/m2/s, F_T=0.95 fraction, F_Water=0.69 fraction, F_CO2=1.31 fraction, Day_Length_hour=14.04 h, determine the numeric value in g C/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 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=53.96 u... | dev | carbon-nitrogen cycling photosynthesis | 2 | CMF009 |
3.42 | cropmath-v1 | {
"new_value": 8153,
"old_value": 7090,
"parameter": "AWL"
} | [
"CMF053",
"CMF030"
] | {
"AWL": 8153,
"SLA": 0.00042
} | growth_yield | {
"expression": "SLA * AWL",
"id": "CMF038",
"latex": "LAI=SLA\\cdot AWL",
"name": "比叶面积法叶面积指数",
"name_en": "Leaf area index from SLA",
"precision": 2,
"unit": "m2/m2"
} | CMF038_sensitivity_t0_0000 | false | Candidate formulas are listed below. Select the physiologically appropriate formula before calculating.
Candidate 1:
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 multipli... | Formula name: Leaf area index from SLA (比叶面积法叶面积指数)
Formula (LaTeX): LAI=SLA\cdot AWL
Formula: LAI = SLA * AWL
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
Parameters:
- SLA... | Formula name: Leaf area index from SLA (比叶面积法叶面积指数)
Formula (LaTeX): LAI=SLA\cdot AWL
Formula: LAI = SLA * AWL
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
Parameters:
- SLA... | Formula: LAI = SLA * AWL
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: Leaf area index from SLA (比叶面积法叶面积指数).
Formula: LAI = SLA * AWL
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: Specific leaf area (比叶面积)
Formula (LaTeX): SLA=(-0.0002GDD^2-0.5604GDD+581.04)(SLA_c/200)
Formula: SLA = if GDD<=1200: (-0.0002*GDD^2-0.5604*GDD+581.04)*(SLAc/200); else: 200*(SLAc/200)
Operator not... | 2 | Product | sensitivity | {
"AWL": 7090,
"SLA": 0.00042
} | A leaf area growth calculation involves sensitivity analysis. The current measurements are SLA=0.00042 ha/kg, AWL=7090.0 kg/ha. A field technician asks what the leaf area index from sla would become if AWL changed to 8153.0. 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 Leaf area index from SLA.
Use the verified formula:
\[LAI=SLA\cdot AWL\]
In calculation-friendly notation, this is:
LAI = SLA * AWL.
Substitute the relevant values:
SLA=0.00042 ha/kg, AWL=8153.0 kg/ha.
The computed value is 3.42 m2/m2.
Therefore, the final numeric answer is \boxed{3.42}.... | dev | Leaf area growth | 0 | CMF053 |
11.71 | cropmath-v1 | null | [
"CMF044",
"CMF028"
] | {
"LAI": 2.8200000000000003,
"PAR": 14.18,
"k": 0.62
} | 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_direct_t0_0001 | false | Candidate formulas are listed below. Select the physiologically appropriate formula before calculating.
Candidate 1:
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 multip... | 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: 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... | 1 | Product with exponential saturation | direct | {
"LAI": 2.8200000000000003,
"PAR": 14.18,
"k": 0.62
} | A photosynthesis calculation estimates Intercepted photosynthetically active radiation. The measured inputs are PAR=14.18 MJ/m2/day, k=0.62 dimensionless, LAI=2.82 m2/m2. What is the intercepted photosynthetically active radiation? Give the 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=14.18 MJ/m2/day, k=0.62 dimensionless, LAI=2.82 m2/m2.
T... | dev | Photosynthesis | 0 | CMF044 |
1.07 | cropmath-v1 | {
"new_value": 0.64,
"old_value": 0.56,
"parameter": "beta"
} | [
"CMF046",
"CMF026"
] | {
"Cx": 377,
"beta": 0.64
} | 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_0000 | 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: 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: 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 | Logarithmic | sensitivity | {
"Cx": 377,
"beta": 0.56
} | A photosynthesis calculation involves sensitivity analysis. The current measurements are Cx=377.0 ppm, beta=0.56 dimensionless. A field technician asks what the co2 response factor would become if beta changed to 0.64. 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=377.0 ppm, beta=0.64 dimensionless.
The computed value is 1.07 fraction.
Therefore, the final numer... | dev | Photosynthesis | 0 | CMF046 |
26.59 | cropmath-v1 | null | [
"CMF023",
"CMF016"
] | {
"Tair": 29.2
} | methane_emission | {
"expression": "4.4 + 0.76 * Tair",
"id": "CMF024",
"latex": "T_{soil}=4.4+0.76T_{air}",
"name": "土壤温度估算",
"name_en": "Soil temperature from air temperature",
"precision": 2,
"unit": "C"
} | CMF024_direct_t0_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: 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 ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.
Output un... | 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 ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.
Output un... | Formula: Tsoil = 4.4 + 0.76 * Tair
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 | Relevant formula name: Soil temperature from air temperature (土壤温度估算).
Formula: Tsoil = 4.4 + 0.76 * Tair
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 | 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... | 1 | Linear | direct | {
"Tair": 29.2
} | A methane calculation estimates Soil temperature from air temperature. The measured inputs are Tair=29.2 C. What is the soil temperature from air temperature? Give the final numeric value in C. | {
"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 temperature from air temperature.
Use the verified formula:
\[T_{soil}=4.4+0.76T_{air}\]
In calculation-friendly notation, this is:
Tsoil = 4.4 + 0.76 * Tair.
Substitute the relevant values:
Tair=29.2 C.
The computed value is 26.59 C.
Therefore, the final numeric answer is \boxed{26... | dev | Methane | 0 | CMF023 |
0.173 | cropmath-v1 | {
"new_value": 0.41,
"old_value": 0.45,
"parameter": "RPE"
} | [
"CMF044",
"CMF029"
] | {
"BFF": 1.07,
"DTE": 0.422,
"IE": 0.67,
"PDT": 8.3,
"RPE": 0.41000000000000003
} | 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_sensitivity_t1_0000 | 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: 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: 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 | Stage-dependent product | sensitivity | {
"BFF": 1.07,
"DTE": 0.422,
"IE": 0.67,
"PDT": 8.3,
"RPE": 0.45
} | Given current measurements PDT=8.3 physiological days, DTE=0.422 fraction, IE=0.67 dimensionless, RPE=0.45 fraction, BFF=1.07 dimensionless in phenology, what would the Daily physiological effectiveness be if RPE were adjusted to 0.41? Provide the numeric value 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=8.3... | dev | Phenology | 1 | CMF044 |
1.454 | cropmath-v1 | null | [
"CMF014",
"CMF005"
] | {
"CO2": 613,
"Coef_B_CO2": 0.77
} | 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_t0_0001 | false | Candidate formulas are listed below. Select the physiologically appropriate formula before calculating.
Candidate 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_Wa... | 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 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_W... | 1 | Logarithmic | direct | {
"CO2": 613,
"Coef_B_CO2": 0.77
} | A carbon-nitrogen cycling photosynthesis calculation estimates carbon-nitrogen cycling CO2 fertilization factor. The measured inputs are CO2=613.0 ppm, Coef_B_CO2=0.77 dimensionless. What is the carbon-nitrogen cycling co2 fertilization 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 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=613.0 ppm, Coef_B_CO2=0.77 dimensionless.
The computed value is 1... | dev | carbon-nitrogen cycling photosynthesis | 0 | CMF014 |
17.20 | cropmath-v1 | null | [
"CMF027",
"CMF043"
] | {
"Am": 28,
"FCO2": 1.16,
"FPA": 0.85,
"FT": 0.89,
"NNI": 0.84,
"WDF": 0.7000000000000001
} | 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_chain_2_t2_0000 | 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: 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: 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*BF... | 4 | Multiplicative with minimum stress | chain_2 | {
"Am": 28,
"FCO2": 1.16,
"FPA": 0.85,
"FT": 0.89,
"NNI": 0.84,
"WDF": 0.7000000000000001
} | In photosynthesis, the Actual maximum assimilation rate is obtained by chaining intermediate calculations. With inputs Am=28.0 kg CO2/ha/h, FCO2=1.16 fraction, FPA=0.85 fraction, FT=0.89 fraction, NNI=0.84 fraction, WDF=0.7 fraction, determine the numeric value 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=28.0 kg CO2/ha/h, FCO2=1.16 fraction, FPA... | dev | Photosynthesis | 2 | CMF027 |
1.00 | cropmath-v1 | null | [
"CMF028",
"CMF039"
] | {
"P": 10.46,
"PS": 0.095,
"Po": 10.5
} | growth_yield | {
"expression": "if P<=Po: 1; if P<=Po+sqrt(1/PS): 1-PS*(P-Po)^2; else: 0",
"id": "CMF050",
"latex": "RPE=1,\\;1-PS(P-P_o)^2,\\;0",
"name": "相对光周期效应",
"name_en": "Relative photoperiod effectiveness",
"precision": 2,
"unit": "fraction"
} | CMF050_branch_t1_0000 | false | Candidate formulas are listed below. Select the physiologically appropriate formula before calculating.
Candidate 1:
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... | Formula name: Relative photoperiod effectiveness (相对光周期效应)
Formula (LaTeX): RPE=1,\;1-PS(P-P_o)^2,\;0
Formula: RPE = if P<=Po: 1; if P<=Po+sqrt(1/PS): 1-PS*(P-Po)^2; 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... | Formula name: Relative photoperiod effectiveness (相对光周期效应)
Formula (LaTeX): RPE=1,\;1-PS(P-P_o)^2,\;0
Formula: RPE = if P<=Po: 1; if P<=Po+sqrt(1/PS): 1-PS*(P-Po)^2; 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... | Formula: RPE = if P<=Po: 1; if P<=Po+sqrt(1/PS): 1-PS*(P-Po)^2; 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: fraction | Relevant formula name: Relative photoperiod effectiveness (相对光周期效应).
Formula: RPE = if P<=Po: 1; if P<=Po+sqrt(1/PS): 1-PS*(P-Po)^2; 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... | 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 ... | 4 | Piecewise quadratic | branch | {
"P": 10.46,
"PS": 0.095,
"Po": 10.5
} | For phenology, compute the Relative photoperiod effectiveness. The inputs are P=10.46 h, Po=10.5 h, PS=0.095 dimensionless. 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 Relative photoperiod effectiveness.
Use the verified formula:
\[RPE=1,\;1-PS(P-P_o)^2,\;0\]
In calculation-friendly notation, this is:
RPE = if P<=Po: 1; if P<=Po+sqrt(1/PS): 1-PS*(P-Po)^2; else: 0.
Substitute the relevant values:
P=10.46 h, Po=10.5 h, PS=0.095 dimensionless.
The compute... | dev | Phenology | 1 | CMF028 |
0.182 | cropmath-v1 | null | [
"CMF023",
"CMF025"
] | {
"CH4RiceEfC": 0.217,
"W": 6048.7,
"Wmax": 11870.1
} | methane_emission | {
"expression": "CH4RiceEfC * (1 - W / Wmax) ** 0.25",
"id": "CMF019",
"latex": "F_p=C(1-W/W_{max})^{0.25}",
"name": "植株传输因子",
"name_en": "Plant methane transport factor",
"precision": 3,
"unit": "fraction"
} | CMF019_direct_t2_0000 | false | Candidate formulas are listed below. Select the physiologically appropriate formula before calculating.
Candidate 1:
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 ^ ... | 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 division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.
... | 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 division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.
... | Formula: Fp = 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: fraction | Relevant formula name: Plant methane transport factor (植株传输因子).
Formula: Fp = 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: fraction | 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... | 1 | Power attenuation | direct | {
"CH4RiceEfC": 0.217,
"W": 6048.7,
"Wmax": 11870.1
} | In a methane, the Plant methane transport factor is computed as part of methane modeling. With CH4RiceEfC=0.217 fraction, W=6048.7 kg/ha, Wmax=11870.1 kg/ha, 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 Plant methane transport factor.
Use the verified formula:
\[F_p=C(1-W/W_{max})^{0.25}\]
In calculation-friendly notation, this is:
Fp = CH4RiceEfC * (1 - W / Wmax) ** 0.25.
Substitute the relevant values:
CH4RiceEfC=0.217 fraction, W=6048.7 kg/ha, Wmax=11870.1 kg/ha.
The computed value i... | dev | Methane | 2 | CMF023 |
0.0022 | cropmath-v1 | null | [
"CMF041",
"CMF037"
] | {
"NNI": 0.58,
"RP": 0.0038,
"WDF": 0.73
} | 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_chain_2_t1_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: 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: 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)
... | 3 | Minimum stress multiplier | chain_2 | {
"NNI": 0.58,
"RP": 0.0038,
"WDF": 0.73
} | Given RP=0.0038 d-1, NNI=0.58 fraction, WDF=0.73 fraction, compute Actual LAI relative growth rate in leaf area growth. This requires sequential calculation through intermediate variables. Report the final 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=0.0038 d-1, NNI=0.58 fraction, WDF=0.73 fraction.
The computed value is 0.0022 d-1.
Therefore, the f... | dev | Leaf area growth | 1 | CMF041 |
24.14 | cropmath-v1 | null | [
"CMF049",
"CMF037"
] | {
"ABIOMASS": 5647,
"PDT": 55.4,
"Q10": 1.88,
"T": 32.4,
"To": 29.5
} | 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_chain_2_t1_0001 | false | Candidate formulas are listed below. Select the physiologically appropriate formula before calculating.
Candidate 1:
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 multiplicati... | 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: 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... | 4 | Product with Q10 temperature response | chain_2 | {
"ABIOMASS": 5647,
"PDT": 55.4,
"Q10": 1.88,
"T": 32.4,
"To": 29.5
} | Given PDT=55.4 physiological days, ABIOMASS=5647.0 kg/ha, Q10=1.88 dimensionless, T=32.4 C, To=29.5 C, compute Maintenance respiration in respiration. This requires sequential calculation through intermediate variables. Report the final 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 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=55.4 physiological days, AB... | dev | Respiration | 1 | CMF049 |
243.95 | cropmath-v1 | {
"new_value": 187.7,
"old_value": 170.6,
"parameter": "SLAc"
} | [
"CMF031",
"CMF027"
] | {
"GDD": 488,
"SLAc": 187.7
} | growth_yield | {
"expression": "if GDD<=1200: (-0.0002*GDD^2-0.5604*GDD+581.04)*(SLAc/200); else: 200*(SLAc/200)",
"id": "CMF053",
"latex": "SLA=(-0.0002GDD^2-0.5604GDD+581.04)(SLA_c/200)",
"name": "比叶面积",
"name_en": "Specific leaf area",
"precision": 2,
"unit": "cm2/g"
} | CMF053_sensitivity_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: Specific leaf area (比叶面积)
Formula (LaTeX): SLA=(-0.0002GDD^2-0.5604GDD+581.04)(SLA_c/200)
Formula: SLA = if GDD<=1200: (-0.0002*GDD^2-0.5604*GDD+581.04)*(SLAc/200); else: 200*(SLAc/200)
Operator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural log... | Formula name: Specific leaf area (比叶面积)
Formula (LaTeX): SLA=(-0.0002GDD^2-0.5604GDD+581.04)(SLA_c/200)
Formula: SLA = if GDD<=1200: (-0.0002*GDD^2-0.5604*GDD+581.04)*(SLAc/200); else: 200*(SLAc/200)
Operator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural log... | Formula: SLA = if GDD<=1200: (-0.0002*GDD^2-0.5604*GDD+581.04)*(SLAc/200); else: 200*(SLAc/200)
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: cm2/g | Relevant formula name: Specific leaf area (比叶面积).
Formula: SLA = if GDD<=1200: (-0.0002*GDD^2-0.5604*GDD+581.04)*(SLAc/200); else: 200*(SLAc/200)
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... | 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(... | 3 | Piecewise polynomial | sensitivity | {
"GDD": 488,
"SLAc": 170.6
} | A leaf area growth calculation involves sensitivity analysis. The current measurements are GDD=488.0 C day, SLAc=170.6 cm2/g. A field technician asks what the specific leaf area would become if SLAc changed to 187.7. Report the new final numeric value in cm2/g. | {
"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 Specific leaf area.
Use the verified formula:
\[SLA=(-0.0002GDD^2-0.5604GDD+581.04)(SLA_c/200)\]
In calculation-friendly notation, this is:
SLA = if GDD<=1200: (-0.0002*GDD^2-0.5604*GDD+581.04)*(SLAc/200); else: 200*(SLAc/200).
Substitute the relevant values:
GDD=488.0 C day, SLAc=187.7 ... | dev | Leaf area growth | 0 | CMF031 |
0.025 | cropmath-v1 | {
"new_value": 0.53,
"old_value": 0.48,
"parameter": "S_Water"
} | [
"CMF057",
"CMF059"
] | {
"ARID": 1.8399999999999999,
"S_Water": 0.53
} | environmental_response | {
"expression": "max(0, 1 - S_Water * ARID)",
"id": "CMF061",
"latex": "F_W=\\max(0,1-S_WARID)",
"name": "环境响应水分响应",
"name_en": "environmental-response water stress response",
"precision": 3,
"unit": "response factor"
} | CMF061_sensitivity_t0_0001 | 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 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, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger... | 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, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger... | Formula: F_Water = max(0, 1 - S_Water * ARID)
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 water stress response (环境响应水分响应).
Formula: F_Water = max(0, 1 - S_Water * ARID)
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: 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... | 2 | Maximum linear | sensitivity | {
"ARID": 1.8399999999999999,
"S_Water": 0.48
} | A environmental-response formula system calculation involves sensitivity analysis. The current measurements are ARID=1.84 dimensionless, S_Water=0.48 dimensionless. A field technician asks what the simple water stress response would become if S_Water changed to 0.53. 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 environmental-response water stress response.
Use the verified formula:
\[F_W=\max(0,1-S_WARID)\]
In calculation-friendly notation, this is:
F_Water = max(0, 1 - S_Water * ARID).
Substitute the relevant values:
ARID=1.84 dimensionless, S_Water=0.53 dimensionless.
The computed value is 0.... | dev | environmental-response formula system | 0 | CMF057 |
0.936 | cropmath-v1 | null | [
"CMF024",
"CMF020"
] | {
"WaterC": 0.455
} | 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_direct_t1_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 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 ^ means po... | 1 | Exponential quadratic | direct | {
"WaterC": 0.455
} | Given WaterC=0.455 fraction, calculate the Water-content decomposition factor for a methane. 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 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=0.455 fraction.
The computed value is 0.936 fraction.
Therefore... | dev | Methane | 1 | CMF024 |
0.094 | cropmath-v1 | null | [
"CMF013",
"CMF002"
] | {
"Coef_Plant_TN_uptake": 0.20400000000000001,
"DVI": 1.28,
"NPP": 8.58,
"N_requirement_max_day": 2.99,
"N_shoot": 5,
"Ratio_N_C_photo_product": 0.011
} | 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_branch_t1_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 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 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}
... | 4 | Piecewise minimum | branch | {
"Coef_Plant_TN_uptake": 0.20400000000000001,
"DVI": 1.28,
"NPP": 8.58,
"N_requirement_max_day": 2.99,
"N_shoot": 5,
"Ratio_N_C_photo_product": 0.011
} | For carbon-nitrogen cycling nitrogen, compute the carbon-nitrogen cycling crop nitrogen demand. The inputs are N_shoot=5.0 g N/m2, Coef_Plant_TN_uptake=0.204 d-1, N_requirement_max_day=2.99 g N/m2/day, NPP=8.58 g C/m2/day, Ratio_N_C_photo_product=0.011 g N/g C, DVI=1.28 dimensionless. Note that this formula uses piecew... | {
"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 | CMF013 |
1.286 | cropmath-v1 | null | [
"CMF062",
"CMF055"
] | {
"CO2": 557,
"CO2_RUE": 0.138
} | 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_t0_0001 | 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 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, exp(x) means... | 2 | Piecewise linear cap | direct | {
"CO2": 557,
"CO2_RUE": 0.138
} | A environmental-response formula system calculation estimates environmental-response CO2 response. The measured inputs are CO2=557.0 ppm, CO2_RUE=0.138 fraction. What is the simple co2 response? 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 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=557.0 ppm, CO2_RUE=0.138 f... | dev | environmental-response formula system | 0 | CMF062 |
0.0082 | cropmath-v1 | null | [
"CMF035",
"CMF030"
] | {
"PDT": 9.3
} | 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_t1_0001 | false | Candidate formulas are listed below. Select the physiologically appropriate formula before calculating.
Candidate 1:
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 multipli... | 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: 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
Operator no... | 1 | Linear decline | direct | {
"PDT": 9.3
} | Given PDT=9.3 physiological days, calculate the Maintenance respiration coefficient at optimum temperature for a respiration. The result should be 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=9.3 physiological days.
The computed value is 0.0082 d-1.
Theref... | dev | Respiration | 1 | CMF035 |
1.02 | cropmath-v1 | null | [
"CMF044",
"CMF047"
] | {
"AWL": 2747,
"SLA": 0.00037000000000000005
} | growth_yield | {
"expression": "SLA * AWL",
"id": "CMF038",
"latex": "LAI=SLA\\cdot AWL",
"name": "比叶面积法叶面积指数",
"name_en": "Leaf area index from SLA",
"precision": 2,
"unit": "m2/m2"
} | CMF038_direct_t1_0001 | false | Candidate formulas are listed below. Select the physiologically appropriate formula before calculating.
Candidate 1:
Formula name: Leaf area index from SLA (比叶面积法叶面积指数)
Formula (LaTeX): LAI=SLA\cdot AWL
Formula: LAI = SLA * AWL
Operator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means... | Formula name: Leaf area index from SLA (比叶面积法叶面积指数)
Formula (LaTeX): LAI=SLA\cdot AWL
Formula: LAI = SLA * AWL
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
Parameters:
- SLA... | Formula name: Leaf area index from SLA (比叶面积法叶面积指数)
Formula (LaTeX): LAI=SLA\cdot AWL
Formula: LAI = SLA * AWL
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
Parameters:
- SLA... | Formula: LAI = SLA * AWL
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: Leaf area index from SLA (比叶面积法叶面积指数).
Formula: LAI = SLA * AWL
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: 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... | 1 | Product | direct | {
"AWL": 2747,
"SLA": 0.00037000000000000005
} | Given SLA=0.00037 ha/kg, AWL=2747.0 kg/ha, calculate the Leaf area index from SLA for a leaf area growth. The result should be 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 Leaf area index from SLA.
Use the verified formula:
\[LAI=SLA\cdot AWL\]
In calculation-friendly notation, this is:
LAI = SLA * AWL.
Substitute the relevant values:
SLA=0.00037 ha/kg, AWL=2747.0 kg/ha.
The computed value is 1.02 m2/m2.
Therefore, the final numeric answer is \boxed{1.02}.... | dev | Leaf area growth | 1 | CMF044 |
0.592 | cropmath-v1 | {
"new_value": -103.8,
"old_value": -90.3,
"parameter": "Eh"
} | [
"CMF016",
"CMF024"
] | {
"Eh": -103.8
} | methane_emission | {
"expression": "if Eh<-150: 1; else: exp(-1.7*(1+Eh/150))",
"id": "CMF018",
"latex": "F_{Eh}=1\\;(Eh<-150),\\quad e^{-1.7(1+Eh/150)}",
"name": "氧化还原电位因子",
"name_en": "Redox potential factor",
"precision": 3,
"unit": "fraction"
} | CMF018_sensitivity_t1_0001 | 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: 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 multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the sm... | 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 multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the sm... | Formula: FEh = if Eh<-150: 1; else: exp(-1.7*(1+Eh/150))
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: Redox potential factor (氧化还原电位因子).
Formula: FEh = if Eh<-150: 1; else: exp(-1.7*(1+Eh/150))
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: 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... | 3 | Piecewise exponential | sensitivity | {
"Eh": -90.3
} | Given current measurements Eh=-90.3 mV in methane, what would the Redox potential factor be if Eh were adjusted to -103.8? 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 Redox potential factor.
Use the verified formula:
\[F_{Eh}=1\;(Eh<-150),\quad e^{-1.7(1+Eh/150)}\]
In calculation-friendly notation, this is:
FEh = if Eh<-150: 1; else: exp(-1.7*(1+Eh/150)).
Substitute the relevant values:
Eh=-103.8 mV.
The computed value is 0.592 fraction.
Therefore, th... | dev | Methane | 1 | CMF016 |
2.409 | cropmath-v1 | null | [
"CMF024",
"CMF016"
] | {
"Com": 6.896,
"Cr": 63.36,
"FEh": 0.354,
"TI": 0.289
} | 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_chain_2_t2_0001 | false | Candidate formulas are listed below. Select the physiologically appropriate formula before calculating.
Candidate 1:
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, / me... | 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: 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 ^ means po... | 3 | Maximum with chained drivers | chain_2 | {
"Com": 6.896,
"Cr": 63.36,
"FEh": 0.354,
"TI": 0.289
} | In methane, the Methane production is obtained by chaining intermediate calculations. With inputs FEh=0.354 fraction, TI=0.289 fraction, Cr=63.36 g C/m2/day, Com=6.896 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=0.354 fraction, TI=0.289 fraction, Cr=63.36 g C/m2/day, Com=6.896 g C/m2/day.
T... | dev | Methane | 2 | CMF024 |
0.58 | cropmath-v1 | {
"new_value": 9000,
"old_value": 9991,
"parameter": "B_above"
} | [
"CMF043",
"CMF026"
] | {
"B_above": 9000,
"Y_grain": 5253
} | growth_yield | {
"expression": "Y_grain / B_above",
"id": "CMF034",
"latex": "HI=\\frac{Y_{grain}}{B_{above}}",
"name": "收获指数",
"name_en": "Harvest index",
"precision": 2,
"unit": "fraction"
} | CMF034_sensitivity_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: 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, ln(x) means natural logarithm, and min/max select the smaller/larger value.
Output unit: fraction
Parameters... | 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, ln(x) means natural logarithm, and min/max select the smaller/larger value.
Output unit: fraction
Parameters... | Formula: HI = Y_grain / B_above
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: Harvest index (收获指数).
Formula: HI = Y_grain / B_above
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: 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 | Ratio | sensitivity | {
"B_above": 9991,
"Y_grain": 5253
} | In dry matter partitioning, we analyze how Harvest index responds to parameter changes. Baseline: Y_grain=5253.0 kg/ha, B_above=9991.0 kg/ha. After changing B_above from 9991.0 to 9000.0, compute the new Harvest 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 Harvest index.
Use the verified formula:
\[HI=\frac{Y_{grain}}{B_{above}}\]
In calculation-friendly notation, this is:
HI = Y_grain / B_above.
Substitute the relevant values:
Y_grain=5253.0 kg/ha, B_above=9000.0 kg/ha.
The computed value is 0.58 fraction.
Therefore, the final numeric ans... | dev | Dry matter partitioning | 2 | CMF043 |
0.961 | cropmath-v1 | {
"new_value": 1.4,
"old_value": 1.27,
"parameter": "BD"
} | [
"CMF005",
"CMF003"
] | {
"BD": 1.4,
"Saturate_Water": 0.48,
"Soil_Water": 0.41100000000000003
} | carbon_nitrogen_cycle | {
"expression": "if Soil_Water>=Saturate_Water: 0.65; else: 0.49*exp(3.88*(Soil_Water/BD)-5.4*(Soil_Water/BD)^2)",
"id": "CMF012",
"latex": "F_W=0.49e^{3.88W-5.4W^2}",
"name": "土壤碳分解水分因子",
"name_en": "carbon-nitrogen cycling soil carbon water factor",
"precision": 3,
"unit": "fraction"
} | CMF012_sensitivity_t0_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: 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)-5.4*(Soil_Water/BD)^2)
Operator notes: * means multiplication, / means division, ** or ^ means power, exp(x) m... | 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)-5.4*(Soil_Water/BD)^2)
Operator notes: * means multiplication, / means division, ** or ^ means power, exp(x) m... | Formula: F_W_soil = if Soil_Water>=Saturate_Water: 0.65; else: 0.49*exp(3.88*(Soil_Water/BD)-5.4*(Soil_Water/BD)^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: carbon-nitrogen cycling soil carbon water factor (土壤碳分解水分因子).
Formula: F_W_soil = if Soil_Water>=Saturate_Water: 0.65; else: 0.49*exp(3.88*(Soil_Water/BD)-5.4*(Soil_Water/BD)^2)
Operator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural lo... | 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 + ... | 3 | Piecewise exponential | sensitivity | {
"BD": 1.27,
"Saturate_Water": 0.48,
"Soil_Water": 0.41100000000000003
} | A carbon-nitrogen cycling soil carbon calculation involves sensitivity analysis. The current measurements are Soil_Water=0.411 cm3/cm3, Saturate_Water=0.48 cm3/cm3, BD=1.27 g/cm3. A field technician asks what the carbon-nitrogen cycling soil carbon water factor would become if BD changed to 1.4. Report the new 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 soil carbon water factor.
Use the verified formula:
\[F_W=0.49e^{3.88W-5.4W^2}\]
In calculation-friendly notation, this is:
F_W_soil = if Soil_Water>=Saturate_Water: 0.65; else: 0.49*exp(3.88*(Soil_Water/BD)-5.4*(Soil_Water/BD)^2).
Substitute the relevant values:
... | dev | carbon-nitrogen cycling soil carbon | 0 | CMF005 |
18.30 | cropmath-v1 | {
"new_value": 30,
"old_value": 33.3,
"parameter": "Tmean"
} | [
"CMF062",
"CMF058"
] | {
"Tbase": 11.7,
"Tmean": 30
} | 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_sensitivity_t1_0001 | false | Candidate formulas are listed below. Select the physiologically appropriate formula before calculating.
Candidate 1:
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 divis... | 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 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, exp(x) means... | 2 | Maximum linear | sensitivity | {
"Tbase": 11.7,
"Tmean": 33.3
} | Given current measurements Tmean=33.3 C, Tbase=11.7 C in environmental-response formula system, what would the environmental-response daily thermal time be if Tmean were adjusted to 30.0? Provide the 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=30.0 C, Tbase=11.7 C.
The computed value is 18.30 C day.
Therefore, the fin... | dev | environmental-response formula system | 1 | CMF062 |
487.96 | cropmath-v1 | null | [
"CMF001",
"CMF008"
] | {
"Coef_K": 0.38,
"Coef_rou": 0.23,
"LAI": 2.51,
"Radi": 485.8
} | 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_t0_0000 | 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 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: 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... | 3 | Exponential ratio | chain_2 | {
"Coef_K": 0.38,
"Coef_rou": 0.23,
"LAI": 2.51,
"Radi": 485.8
} | A carbon-nitrogen cycling photosynthesis module estimates carbon-nitrogen cycling canopy absorbed PAR using intermediate drivers. Use the provided inputs in sequence. The measurements are Radi=485.8 W/m2, LAI=2.51 m2/m2, Coef_rou=0.23 fraction, Coef_K=0.38 dimensionless. Compute the carbon-nitrogen cycling canopy absor... | {
"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 | 0 | CMF001 |
0.094 | cropmath-v1 | {
"new_value": 42,
"old_value": 43.8,
"parameter": "ExtremeT"
} | [
"CMF056",
"CMF062"
] | {
"ExtremeT": 42,
"MaxT": 33.5,
"Tmax": 41.2
} | environmental_response | {
"expression": "if Tmax<=MaxT: 1; if Tmax>ExtremeT: 0; else: max(1-(Tmax-MaxT)/(ExtremeT-MaxT), 0)",
"id": "CMF058",
"latex": "F_H=1,\\;0,\\;\\text{or}\\;1-\\frac{T_{max}-MaxT}{ExtremeT-MaxT}",
"name": "环境响应高温响应",
"name_en": "environmental-response heat stress response",
"precision": 3,
"unit": "fraction... | CMF058_sensitivity_t2_0001 | false | Candidate formulas are listed below. Select the physiologically appropriate formula before calculating.
Candidate 1:
Formula name: environmental-response heat stress response (环境响应高温响应)
Formula (LaTeX): F_H=1,\;0,\;\text{or}\;1-\frac{T_{max}-MaxT}{ExtremeT-MaxT}
Formula: F_Heat = if Tmax<=MaxT: 1; if Tmax>ExtremeT: 0;... | Formula name: environmental-response heat stress response (环境响应高温响应)
Formula (LaTeX): F_H=1,\;0,\;\text{or}\;1-\frac{T_{max}-MaxT}{ExtremeT-MaxT}
Formula: F_Heat = if Tmax<=MaxT: 1; if Tmax>ExtremeT: 0; else: max(1-(Tmax-MaxT)/(ExtremeT-MaxT), 0)
Operator notes: * means multiplication, / means division, ** or ^ means p... | Formula name: environmental-response heat stress response (环境响应高温响应)
Formula (LaTeX): F_H=1,\;0,\;\text{or}\;1-\frac{T_{max}-MaxT}{ExtremeT-MaxT}
Formula: F_Heat = if Tmax<=MaxT: 1; if Tmax>ExtremeT: 0; else: max(1-(Tmax-MaxT)/(ExtremeT-MaxT), 0)
Operator notes: * means multiplication, / means division, ** or ^ means p... | Formula: F_Heat = if Tmax<=MaxT: 1; if Tmax>ExtremeT: 0; else: max(1-(Tmax-MaxT)/(ExtremeT-MaxT), 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 heat stress response (环境响应高温响应).
Formula: F_Heat = if Tmax<=MaxT: 1; if Tmax>ExtremeT: 0; else: max(1-(Tmax-MaxT)/(ExtremeT-MaxT), 0)
Operator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max ... | 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 stress | sensitivity | {
"ExtremeT": 43.8,
"MaxT": 33.5,
"Tmax": 41.2
} | In environmental-response formula system, we analyze how environmental-response heat stress response responds to parameter changes. Baseline: Tmax=41.2 C, MaxT=33.5 C, ExtremeT=43.8 C. After changing ExtremeT from 43.8 to 42.0, compute the new environmental-response heat stress response. Give the numeric value in fract... | {
"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 heat stress response.
Use the verified formula:
\[F_H=1,\;0,\;\text{or}\;1-\frac{T_{max}-MaxT}{ExtremeT-MaxT}\]
In calculation-friendly notation, this is:
F_Heat = if Tmax<=MaxT: 1; if Tmax>ExtremeT: 0; else: max(1-(Tmax-MaxT)/(ExtremeT-MaxT), 0).
Substitute the re... | dev | environmental-response formula system | 2 | CMF056 |
1.00 | cropmath-v1 | {
"new_value": 22.9,
"old_value": 19.9,
"parameter": "Tmean"
} | [
"CMF054",
"CMF047"
] | {
"PDT": 37.4,
"Qt": 17.4,
"Tmean": 22.9
} | 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_sensitivity_t0_0000 | false | Candidate formulas are listed below. Select the physiologically appropriate formula before calculating.
Candidate 1:
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 multip... | 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: 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... | 3 | Piecewise power penalty | sensitivity | {
"PDT": 37.4,
"Qt": 17.4,
"Tmean": 19.9
} | A temperature stress calculation involves sensitivity analysis. The current measurements are PDT=37.4 physiological days, Tmean=19.9 C, Qt=17.4 C day. A field technician asks what the low-temperature fertility factor would become if Tmean changed to 22.9. 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 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=37.4 physiological days, Tmean=22.9 ... | dev | Temperature stress | 0 | CMF054 |
1.00 | cropmath-v1 | {
"new_value": 23,
"old_value": 22,
"parameter": "Tmean"
} | [
"CMF026",
"CMF047"
] | {
"PDT": 33.6,
"Qt": 3.6,
"Tmean": 23
} | 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_sensitivity_t2_0000 | 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: 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 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 | Piecewise power penalty | sensitivity | {
"PDT": 33.6,
"Qt": 3.6,
"Tmean": 22
} | In temperature stress, we analyze how Low-temperature fertility factor responds to parameter changes. Baseline: PDT=33.6 physiological days, Tmean=22.0 C, Qt=3.6 C day. After changing Tmean from 22.0 to 23.0, compute the new Low-temperature fertility 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 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=33.6 physiological days, Tmean=23.0 ... | dev | Temperature stress | 2 | CMF026 |
68.00 | cropmath-v1 | null | [
"CMF037",
"CMF035"
] | {
"ABIOMASS": 8842,
"PDT": 20.6,
"Q10": 1.88,
"T": 30.6,
"To": 29.2
} | 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_t2_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: 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: 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, ex... | 2 | Product with Q10 temperature response | direct | {
"ABIOMASS": 8842,
"PDT": 20.6,
"Q10": 1.88,
"T": 30.6,
"To": 29.2
} | In a respiration, the Maintenance respiration is computed as part of respiration modeling. With PDT=20.6 physiological days, ABIOMASS=8842.0 kg/ha, Q10=1.88 dimensionless, T=30.6 C, To=29.2 C, 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 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=20.6 physiological days, AB... | dev | Respiration | 2 | CMF037 |
0.576 | cropmath-v1 | null | [
"CMF038",
"CMF036"
] | {
"BFF": 0.67,
"DTE": 0.86,
"IE": 1.09,
"PDT": 33.7,
"RPE": 0.59
} | 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_chain_2_t1_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: 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: Leaf area index from SLA (比叶面积法叶面积指数)
Formula (LaTeX): LAI=SLA\cdot AWL
Formula: LAI = SLA * AWL
Operator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x... | 4 | Stage-dependent product | chain_2 | {
"BFF": 0.67,
"DTE": 0.86,
"IE": 1.09,
"PDT": 33.7,
"RPE": 0.59
} | Given PDT=33.7 physiological days, DTE=0.86 fraction, IE=1.09 dimensionless, RPE=0.59 fraction, BFF=0.67 dimensionless, compute Daily physiological effectiveness in phenology. This requires sequential calculation through intermediate variables. Report the final numeric value 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=33.... | dev | Phenology | 1 | CMF038 |
1.240 | cropmath-v1 | null | [
"CMF009",
"CMF005"
] | {
"CO2": 550,
"Coef_B_CO2": 0.5
} | 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_0000 | false | Candidate formulas are listed below. Select the physiologically appropriate formula before calculating.
Candidate 1:
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 / (Bio... | 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 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... | 1 | Logarithmic | direct | {
"CO2": 550,
"Coef_B_CO2": 0.5
} | Given CO2=550.0 ppm, Coef_B_CO2=0.5 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=550.0 ppm, Coef_B_CO2=0.5 dimensionless.
The computed value is 1.... | dev | carbon-nitrogen cycling photosynthesis | 1 | CMF009 |
1.000 | cropmath-v1 | null | [
"CMF009",
"CMF010"
] | {
"Crop_Water_max": 0.42,
"Crop_Water_min": 0.22,
"Soil_Water": 0.339,
"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_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 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 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... | 4 | Piecewise water response | branch | {
"Crop_Water_max": 0.42,
"Crop_Water_min": 0.22,
"Soil_Water": 0.339,
"Wilt_point": 0.08
} | For carbon-nitrogen cycling water, compute the carbon-nitrogen cycling crop water stress factor. The inputs are Soil_Water=0.339 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 | CMF009 |
7.802 | cropmath-v1 | null | [
"CMF005",
"CMF014"
] | {
"Day_Length_hour": 9.05,
"F_CO2": 1.02,
"F_T": 0.84,
"F_Water": 0.5,
"Photo_Rate_opt": 46.58
} | 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_chain_2_t0_0000 | false | Candidate formulas are listed below. Select the physiologically appropriate formula before calculating.
Candidate 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_Wa... | 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 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 + ... | 4 | Multiplicative conversion | chain_2 | {
"Day_Length_hour": 9.05,
"F_CO2": 1.02,
"F_T": 0.84,
"F_Water": 0.5,
"Photo_Rate_opt": 46.58
} | A carbon-nitrogen cycling photosynthesis module estimates carbon-nitrogen cycling gross primary production using intermediate drivers. Use the provided inputs in sequence. The measurements are Photo_Rate_opt=46.58 umol CO2/m2/s, F_T=0.84 fraction, F_Water=0.5 fraction, F_CO2=1.02 fraction, Day_Length_hour=9.05 h. Compu... | {
"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=46.58 u... | dev | carbon-nitrogen cycling photosynthesis | 0 | CMF005 |
0.678 | cropmath-v1 | null | [
"CMF013",
"CMF014"
] | {
"Biom_plant_yesterday": 691.1,
"Coef_Rm": 0.0013000000000000002,
"DVI": 1.63,
"Tmn": 23
} | 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_direct_t1_0000 | 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 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}
... | 2 | Q10 with DVI correction | direct | {
"Biom_plant_yesterday": 691.1,
"Coef_Rm": 0.0013000000000000002,
"DVI": 1.63,
"Tmn": 23
} | Given Biom_plant_yesterday=691.1 g C/m2, Coef_Rm=0.0013 g C/g C/s, Tmn=23.0 C, DVI=1.63 dimensionless, calculate the carbon-nitrogen cycling plant maintenance respiration for a carbon-nitrogen cycling respiration. 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 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 | 1 | CMF013 |
4.020 | cropmath-v1 | {
"new_value": 39.5,
"old_value": 35.91,
"parameter": "Photo_Rate_opt"
} | [
"CMF003",
"CMF006"
] | {
"Day_Length_hour": 12.71,
"F_CO2": 0.97,
"F_T": 0.49,
"F_Water": 0.39,
"Photo_Rate_opt": 39.5
} | 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_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: 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: 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 | Multiplicative conversion | sensitivity | {
"Day_Length_hour": 12.71,
"F_CO2": 0.97,
"F_T": 0.49,
"F_Water": 0.39,
"Photo_Rate_opt": 35.91
} | In carbon-nitrogen cycling photosynthesis, we analyze how carbon-nitrogen cycling gross primary production responds to parameter changes. Baseline: Photo_Rate_opt=35.91 umol CO2/m2/s, F_T=0.49 fraction, F_Water=0.39 fraction, F_CO2=0.97 fraction, Day_Length_hour=12.71 h. After changing Photo_Rate_opt from 35.91 to 39.5... | {
"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=39.5 um... | dev | carbon-nitrogen cycling photosynthesis | 2 | CMF003 |
6.640 | cropmath-v1 | null | [
"CMF013",
"CMF006"
] | {
"Day_Length_hour": 13.35,
"F_CO2": 1.23,
"F_T": 0.54,
"F_Water": 0.54,
"Photo_Rate_opt": 32.1
} | 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_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 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 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}
... | 4 | Multiplicative conversion | chain_2 | {
"Day_Length_hour": 13.35,
"F_CO2": 1.23,
"F_T": 0.54,
"F_Water": 0.54,
"Photo_Rate_opt": 32.1
} | In carbon-nitrogen cycling photosynthesis, the carbon-nitrogen cycling gross primary production is obtained by chaining intermediate calculations. With inputs Photo_Rate_opt=32.1 umol CO2/m2/s, F_T=0.54 fraction, F_Water=0.54 fraction, F_CO2=1.23 fraction, Day_Length_hour=13.35 h, determine the numeric value in g C/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 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=32.1 um... | dev | carbon-nitrogen cycling photosynthesis | 2 | CMF013 |
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