From the yield check to the manure rate

Nutrient planning · yield component method to spreader setting

Walking the corn in early October: the thousandth-acre counts average 7.5 ears per square metre, a representative ear carries about 520 kernels, and this hybrid runs near 90,000 kernels to the bushel at a 25.4 kg test weight. Grain corn removes about 12.5 kg of nitrogen per tonne harvested. The dairy's solid manure tests 5.5 kg N per tonne, of which about half is available in the first season. Estimate the yield, the nitrogen a matching crop removes, and the manure rate that would replace it.

EkYR_mR_n

Every number in this problem is editable — change any value below and the whole chain recalculates.

Given
  • E = 7.5 ears/m²Ear count
  • k = 520 kernelsKernels per ear
  • K = 90,000 kernels/buKernels per bushel, hybrid
  • w_b = 25.4 kgTest weight per bushel
  • k_N = 12.5 kg/tNitrogen removal per tonne of grain
  • c = 5.5 kg/tManure nitrogen content
  • a = 50 %First-season availability
Determine
  1. (a)the estimated yield
  2. (b)the nitrogen the crop removes at that yield
  3. (c)the manure rate that replaces it
Step 1 of 3(a) · solve for Estimated yield

The yield component method is just multiplication with field-sized error bars: ears times kernels is kernels per square metre, and the hybrid's kernels-per-bushel turns count into mass. The counts here say 11.0 t/ha — carry it as an estimate wearing ±10 %, because the kernel count is a guess about grain fill that October hasn't finished making.

EYkKwb
Rearranged for Y
Y=EkKwbY = \frac{E \, k}{K} \, w_b
Your values, in your units
Y=(7.5 per m2)(520 kernels/ear)(90,000 kernels/bu)(25.4 kg)Y = \frac{\left( 7.5\ \text{per m}^{2} \right) \cdot \left( 520\ \text{kernels/ear} \right)}{\left( 90{,}000\ \text{kernels/bu} \right)} \cdot \left( 25.4\ \text{kg} \right)
Answer
Y=11.007 t/haY = 11.007\ \text{t/ha}

Carried onward at full precision, not this rounded figure.

Open the Corn Yield Estimate (Yield Component Method) solver →

Step 2 of 3(b) · solve for Nutrient removed

Removal is what leaves on the truck, not what the crop touched: 11.0 t/ha at 12.5 kg N per tonne is 137.6 kg/ha of nitrogen gone with the grain. Stover returned to the field is why this figure is smaller than uptake — and why baling the stalks changes the arithmetic.

YkR
Rearranged for R
R=YkR = Y \, k
11.007 t/hacarried from step 1
Your values, in your units
R=(1.10067 kg/m2)(12.5 kg/t)R = \left( 1.10067\ \text{kg/m}^{2} \right) \cdot \left( 12.5\ \text{kg/t} \right)
Converted to base units
R=(11.0067 t/ha)(12.5 kg/t)R = \left( 11.0067\ \text{t/ha} \right) \cdot \left( 12.5\ \text{kg/t} \right)
Answer
R=137.58 kg/haR = 137.58\ \text{kg/ha}

Carried onward at full precision, not this rounded figure.

Open the Nutrient Removal by a Crop solver →

Step 3 of 3(c) · solve for Manure rate

Manure nitrogen is discounted twice — once for what the load contains, once for what this season can actually use. At 5.5 kg/t and 50 % availability, replacing 137.6 kg/ha takes just over 50 t/ha of manure: a heavy but real spreader setting, and the moment to ask whether phosphorus is about to be over-applied.

cRmRna
Rearranged for R_m
Rm=RncaR_m = \frac{R_n}{c \, a}
137.58 kg/hacarried from step 2
Your values, in your units
Rm=(0.0137583 kg/m2)(5.5 kg/t)(50 %)R_m = \frac{\left( 0.0137583\ \text{kg/m}^{2} \right)}{\left( 5.5\ \text{kg/t} \right) \cdot \left( 50\ \text{\%} \right)}
Converted to base units
Rm=(137.583 kg/ha)(5.5 kg/t)(50 %)R_m = \frac{\left( 137.583\ \text{kg/ha} \right)}{\left( 5.5\ \text{kg/t} \right) \cdot \left( 50\ \text{\%} \right)}
Answer
Rm=50.03 t/haR_m = 50.03\ \text{t/ha}

Carried onward at full precision, not this rounded figure.

Open the Manure Rate for a Nitrogen Target solver →

Answer

Therefore the counts estimate 11.0 t/ha, a crop like it removes about 137.6 kg/ha of nitrogen, and replacing that with 5.5 kg/t manure at half availability takes 50.0 t/ha — at which point the limiting number is usually no longer nitrogen but the phosphorus riding along with it.

Why this order

The chain runs the direction nutrient planning actually runs: yield first, because every removal table is written per tonne of harvest; removal second, because that — not uptake — is what has left the field for good; the spreader setting last, because manure is a fertiliser whose analysis you inherit rather than choose. The two discounts in step (c) are the part worth internalising. The content c is a lab number that varies load to load; the availability a is an agronomic judgement — roughly half of solid manure's nitrogen is organic and mineralises over years, not months. Skip the availability discount and the rate looks halved and the crop runs yellow by July.

The trap at the end is the one the final statement names: manure carries its nutrients in fixed proportion, and a rate set to satisfy nitrogen usually over-supplies phosphorus two- to three-fold. Nitrogen decides the rate; phosphorus decides how many years in a row you may use it.

Carried values move at full precision, not the rounded figure shown — chaining rounded numbers compounds error.