Del Voight – Penn State Extension
There is no definitive measure to accurately assess yield. Growers can get a relative estimate to determine a potential yield without a measurement of harvest loss and other environmental effects. An estimate may be calculated using some rudimentary techniques or by simply utilizing a graphic representation and conducting a seed head survey.
Under the best conditions, each head (spike) should contain 20-21 well-developed spikelets
containing 3 seeds each, giving approximately 60-63 seeds per head. Assuming a good stand has 32- 35 heads/ft2, the potential number of kernels produced per acre would be near 91,000,000.
13,000 kernels/lb (Certified seed ranged from 11,400-17,600 seeds/lb in recent year), this would equal 7,000 lb/A yield… when divided by an assumed test weight of 60lb/bu, giving 166 bu/A.
Head/Acre = (heads/ft2 x (43,560 ft/A)
Seeds/Head = (mean no. of seeds/spikelet) x (no. of spikelets/head)
Seeds/Acre = (heads/acre) x (seeds/head)
Yield (bu/A) = (seeds/acre) ÷ 60 lbs/bu
(heads/ft2, kernels/spikelet, kernel wt/lb). The number of heads/ft will be highly variable, depending on seeding rate and level of tillering.
Since this is an estimate the following tables developed by North Carolina State Extension make the job easy by estimating the number of heads and plotting them on the graph.
Figures 17-1 through 17-4 show how average wheat yields relate to the number of heads per square foot for the most common soft red winter wheat varieties grown in North Carolina.