Combines Math problem for group

[Old_Pokey]

Keep in mind that the tank must run empty at the same time the cart gets full. To fill the 1000 bushel cart at 180 bu. per minute, will take 5.55 minutes.IJIJIJIJIJIJMaybeIJ

 

[Ed_Boysun]

Not as many Deeres here as there used to be. More silver around, all the time ;-) Unless you're talking about the other deersIJ Kind of puzzling, as guys I've talked to haven't seen as many as last year. We had deep snow cover here last winter and I think many of them moved further south to get away from it. Further south has been pretty dry and don't think they had a lot of forage to overwinter with. Maybe the populations will be down, which really doesn't hurt my feelings, as I don't need their help with harvesting my grain. No Deeres or deers needed in my wheat fields. Ed in Montana

 

[Al_gorithm]

Have we drawn a conclusion yetIJ

 

[catmand]

one thing you could do is figure the distance needed to gather one thousnad bushels and cross check it from there.

 

[Mav]

I was perusing through the posts and noticed the IH boys were the first to answer this question CORRECTlY without the muddle. Congrats to theredgreenshow and Farm Kid2 for the answer. Also, I took the liberty to calculate a formula that gives the minimum required grain cart size (bu) while unloading on the go by inputting the following: (1) grain tank size of combine (bu),(2) speed of the combine (mi_hr),(3) header size (ft),(4) yield of crop (bu_ac),and (5) unloader rate (bu_s). Size of Grain Cart = (1) _ [1 - 0.00003367 x (2) x (3) x (4) _ (5)] For Example typing the stated problem into a TI-xx calculator would look like this: 495 _ (1 0.00003367 x 5 x 45 x 200 _ 3) = 1000.1 bushel grain cart

 

[preacherman]

Since your call name is Al gorithm, I thought you might be interested in the real solution to this problem. Simple algebra would represent it like this: S= C+(S_U*H) where S= size of Grain Cart in bu, C= combine tank in bu., U= unloading rate in bu_sec, and H=harvest rate in bu_sec. Solving for the combine tank variable we get, C=S-(SH_U). The exact answer to your question is a combine tank size of 494.9 bushels. However, if the hired man gets in that big Gleaner and decides to push it to 6.5 mph, then you would only need a 343 bu grain tank. If grandpa takes it for a spin and is afraid to go over 3.5 mph because he knows what a plugged combine looks like, the grain tank needs to hold 646 bu! The real problem is not with the size of the combine grain tank. It is how are you going to find enough semis to haul the stuff off. At the harvest rate you suggested, you would be filling a 1000 bu semi every 11.1 minutes which amounts to 5400 bushels_hour. I hope your elevator has a bigger leg than ours does! This is all in good fun. Happy harvesting.

 

[Rolf]

Sorry but I have to say this "Who cares, Show me the money"!!!!!! with a big enough paddock (Field) I look to the dollars these day's!!!! and as long as the fuel pump on the this monstrosity is not pumping in the dollar value of fuel faster that what this thing can pump out in dollar value of grain, Im happy camper!!!! (grin) Rolf

 

[theredgreenshow]

Nice work. My college advisor (Doctor of Mechanical Engineering) used to say to us: "Don't let the higher math get to you."

 

[jacco]

spell out the 30" rows and your theory on the defintion of a 'run' before you pose this question to your ninth grade FFA class next year....

 

[Old_Pokey]

Personally I'd like to thank you to Al gorithm for posting this quiz. It has brought a breath of fresh freindly air to the message board which it has been needing lately. It got a lot of people thinking and participating in finding the answer and if you check out tbran's answer on the gleaner page, you'll see it brought in some humor too. Thanks Al gorithm, and keep'em coming.