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| I will try to explain where you could be wrong.
In your example you are assuming that both 21,000,000 kernel samples are the same kernels per pound. If they are, then you are correct and the high test weight sample will indeed weigh 8.9% more (61/56=1.089) But if we know the kernels per pound then we know the bushels (because we know the weigh) and test weight is irrelevant.
Now lets assume that 1 sample is from a high yield area with huge kernels, say 75,000 kernels per bushel or 1339 kernels per pound, and this has 56# test weight. In this case test weight is irrelevant since you know the pounds, 21,000,000/1339=15683 pounds or 280 bushels per acre. (15683/56=280)
Now lets assume that 1 sample is from a low yielding are due to poor grain fill, say 110,000 kernels per bushels or 1964 kernels per pound, and this has a 61# test weight, once again test weight is irrelevant since we know the pounds, 21,000,000/1964=10692 pounds or 190 bushels per acre. (10,692/56=190)
Here is an example where 21,000,000 kernels per acre of a low TW corn could beat a high TW corn by 90 bushels per acre, but a truckload of high TW corn will always contain more bushels (assuming same % moisture) than a truckload of low TW corn.
Hope this helps, and if I figured something wrong I'm sure someone will correct me.
Bottom line is that when we estimate yields using the kernel count method we are GUESSING on SEEDS/POUND, that guess could be way off depending on how big or small kernels are. I prefer to do yield estimates at black layer and just shell, moisture test and weigh the ears. Down side of this is you have to wait an extra month and your results will only beat your combine by a few weeks. | |
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corn yield estimate?
