The "general formula of thumb" used by the ag machinery industry to estimate rates of productivity at 82.5% efficiency, which is the generally accepted standard rate of agricultural machinery operations, including combines, is:
Ground speed x Width / 10 =
Ex. 1: 3.5 mph x 36 ft / 10 = 12.6 ac/hr
80 bu x 12.6 ac/hr = 1008 bph
Ex. 2: 4.0 mph x 36 ft / 10 = 14.4 ac/hr
80 bu x 14.4 ac/hr = 1152 bph
1008 - 1152 bushels per hour is a safe estimate of what you'll likely be able to cover with your new 8010 and 36 foot draper.
I don't know what else could be used, there isn't really another formula that's going to give you any better or any significant different (advantageous) out come. You can adjust the efficiency rate in the long version to adjust to different applications and conditions. Combine efficiency rates take into account any down or idle time, opening a field, turning on the headlands, unloading, stopping to pick up the land lord, neighbor or kids to ride around, when Nature calls, the auger cart is full and all trucks are gone, etc. That's why I think 75% is more realistic rate of measure for harvesting. 82-85% is fine for tillage, etc. because you aren't relying on the support crew like harvesting. In harvesting, everyone has to run at 82.5%, not just the combine. Might take some of the fun out of it if so.
There is a good way of figurering your efficiancy rate out. Just look at your engine hours and seperator hours. Log them as you start on a new field and log them when you finish a field.
That will give you a pretty accurate idea. And honestly i think 75% as an average on the whole season is way too high. but i dont know to tell you the truth. Just like you pointed out Muddy, the combine itself is just a very little brick in the puzzle
Muddy your going to give up 21% productivity?
5280 X 3.5 X 36 / 43560 (sq ft / ac) = 15.2 not 12.6
By the way your simplified area calculation actually works metrically.
5.6 km X 11m = 61.6 / 10 = 6.16 ha
I'm sure I'll have all North American farmers converted to metric by sunset.
Or maybe when heck freezes over.
1 of the 2.
Muddy, I really agree wth you, given the overal variabes. After all, this was more about overall efficiency, rather than a test of just acres/bu per hour, etc.
Yes, I use that formula to calculate just the above, but it won't tell me or anyone here, exactly what's done over a 10-16 hour day.
Simpler calculation is just header width x ground speed >acre for approx acres/hour. However, even that's not exact, counting for overlap, speed variances and so on.
Now you just gotta love acre counters, yield monitors and GPS!
Muddy your going to give up 21% productivity?
5280 X 3.5 X 36 / 43560 (sq ft / ac) = 15.2 not 12.6
By the way your simplified area calculation actually works metrically.
5.6 km X 11m = 61.6 / 10 = 6.16 ha
I'm sure I'll have all North American farmers converted to metric by sunset.
Or maybe when heck freezes over.
1 of the 2.
Don
Don, your equation is correct, if you are not accounting for efficiency. Ground speed x Working Width / 10 is the productivity formula at 82.5% efficiency. I probably do give up 15-20% efficiency, just like you and everyone else does. Hence, the reason why 82.5% has been settled on by every mfg as the published / quoted generally accepted rate of efficiency.
I understand, interesting that an efficiency rating just happens to fit a simplified formula.
I'm impressed you think I'm 80% efficient.
You haven't seen our haywire harvest crew have you?
This machine is going to Alabama we no-till all of our grain crops, so we didn't need the mud hog. Other than that it has everything else. We also got the small tube rotor. No corn. Just soybeans and wheat.
It is supposed to be in this coming week. I'll try to get some pics of the rotor and post them. We should be in wheat about the 10th of June.
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