Combines please show method for math on unloading amount

[CKinSD]

6 mi_hr X 5280 ft_mi = 31680 ft_hr 31680 ft_hr _ 3600 sec_hr = 8.8 ft_sec speed of combine 43560 sqft_ac _ 30 ft header = 1452 ft distance of combine travel to cover 1 acre 200 bu_ac _ 1452 ft_ac = .137741046832 bu_ft .137741046832 bu_ft X 8.8 ft_sec =1.2121212 bu_sec rate of corn entering hopper 3.3 bu_sec rate of corn unloading from hopper 3.3 bu_sec - 1.212121212 bu_sec=2.087878787 bu_sec net rate of corn unloading from hopper 450 bu _ 2.087878787 bu_sec = 215.52975 sec time to unload hopper completely 450 + (215.52975 sec X 1.2121212 bu_sec) =711.2 bu total corn unloaded There is some rounding done here, but this is real close. And yes, if you're wondering, I am an engineer by profession, but I do farm a little on the side. I just read this site to keep up to date, and for some entertainment from time to time. Nothing like a little math problem to exercise the mind, though...

 

[TSTAR]

This is a classical calculus problem. You are attempting to empty a hopper that is being filled at a constant rate. In theory the hopper is never emptied. However, at some point in time what is being discharged from the hopper exceeds what is actually in the hopper. To simplify the problem a twelve (12) row head on thirty (30) inch rows covers thirty feet. This is assuming from middle to middle measurements from each outside row. At 43560 sq ft per acre the machine must traverse 1452 linear feet to over one (1) acre. With this information and traveling at 6.0 miles per hour in 200 bushel per acre grain, the machine is filling it hopper at the rate of 1.21 bushels per second. In effect you are only emptying the hopper at a rate of 2.09 (3.3 minus 1.21) bushels per second. Now the problem gets easier. At this rate of discharge the machines takes 215.31 (450 divided by 2.09) additional seconds to empty. But during this time you have harvested an additional 260.53 (215.31 times 1.21) bushels of grain. Thus at the end or when what is in the hopper is less than what is being discharged you have emptied a total of 710.53 bushels on the wagon. The answer is 710 bushels.

 

[Rotorhead]

Ok, lets take this to the next harvesting level. With a 500 bushel dump pit at the grain leg, if you want to dump 1000 bushels in 6 minutes what size leg would you need. Provided the pit is empty when you start and drive away with the pit full at the end.

 

[Seen_the_Green_light]

5,000 bph How I figured this is, need to put 500 bushels up in 6 minutes 60 minutes in an hour = 10 (500 bushels) increments in one hour = 5,000 bph

 

[Seen_the_Green_light]

5,000 bph How I figured this is, need to put 500 bushels up in 6 minutes 60 minutes in an hour = 10 (500 bushels) increments in one hour = 5,000 bph