1. Count the number of ears in 1/1000th of an acre.
The row length equal to 1/1000th of an acre can be determine using the following equation:
Row length equal to 1/1000th of acre (in feet) = 43,560 / (row spacing/12) / 1000
where there are 43,560 ft2 per acre, row spacing is in inches, and there are 12 inches per foot.
The following table contains row lengths equal to 1/1000th of an acre for several row widths common to row-crop production.
Table 1. The length of row in 1/1000th of an acre for several crop row widths.
|
Row Spacing (inches)
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|||
|
20
|
30
|
38
|
|
| Row length = 1/1000th of an acre |
26 ft 2 in
|
17 ft 5 in
|
13 ft 9 in
|
2. Count the number of kernel rows per ear on every fifth ear in the 1/1000th of an acre and average.
3. Count the number of kernels per row on the five ears and average. Do not count kernels that are less than half the size of the majority of the kernels on the ear.
4. Estimate yield by multiplying the number of ears by the number of rows per ear by the number of kernels per row and divide by 90* (ear number x row number per ear x kernel number per row/ 90*).
5. Adjust for conditions during kernel filling. The equation above works best under growth conditions that result in full-sized kernels. Stress during kernel filling can reduce kernel size. Under these conditions, yield can be more accurately estimated by reducing the yield estimate from the equation by the amount kernel size was reduced by the stress.
6. Several counts should be made randomly throughout a field. Five measurements per twenty acres is a minimum for getting a sample that will be representative of the field.
1. What is the corn yield (Bu/A) estimate from a 20-acre area with the following data?
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Count
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Number of ears in 1/1000th of an acre
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Average kernel rows per ear
|
Average kernels per row
|
|
1
|
30
|
16
|
30
|
|
2
|
28
|
15
|
32
|
|
3
|
29
|
17
|
28
|
|
4
|
31
|
16
|
31
|
|
5
|
28
|
15
|
33
|
Count 1 = 30 x 16 x 30 / 90 = 160 Bu/A
Count 2 = 28 x 15 x 32 / 90 = 149 Bu/A
Count 3 = 29 x 17 x 28 / 90 = 153 Bu/A
Count 4 = 31 x 16 x 31 / 90 = 171 Bu/A
Count 5 = 28 x 15 x 33 / 90 = 154 Bu/AAverage = 160 + 149 + 153 + 171 + 154 / 5 = 157 Bu/A
2. A late-summer drought has reduced the expected kernel size for the area in problem 1 by 10%. What is the yield estimate for the area taking the smaller kernel size into account?
Because the kernel size is 10% smaller than normal the estimated yield should be reduced by 10%. Therefore, the yield estimate for the drought-stressed crop will be 90% of the yield estimate for a normal crop.
157 Bu/A x 0.90 = 141 Bu/A