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Torque ratings 06-88 series


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Just wondering what guys are getting for pound feet of torque when you dyno your tractors? Specifically 06-88 series. We always talk about hp but hardly ever torque. Nebraska put it in they're testing in the mid fifties Farmall 400 had around 400 while the Super 88 had around 360 if I remember right. But never seen it in any other eras that I've looked at

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This has never caught on and I am not sure why.  Torque is king in low rpm engines.  When you have a engine that is make 94hp at 540pto and then still 92hp at 450pto she is kicking out some torque.  I know all the dynos have the conversions from hp to torque.

Scott

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12 minutes ago, 460 said:

This has never caught on and I am not sure why.  Torque is king in low rpm engines.  When you have a engine that is make 94hp at 540pto and then still 92hp at 450pto she is kicking out some torque.  I know all the dynos have the conversions from hp to torque.

Scott

That's why I'm interested is because I would think torque is important when it comes to equipment. I really thought that was impressive when I found the torque numbers on them old tractors considering you're getting 400 pound feet from a 50-60 hp tractor 

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You should be able to use the formula to go from hp back to torque. Hp=(tq*rpm) /5252. So Tq = (hp*5252)/rpm if my algebra still works. If you want engine tq then use engine rpm and for pto torque use pto rpm

 

Edit to add. I looked at that tractor test. Those torque numbers are at the dyno, or available at the pto. The engine speed is much higher than the pto, therefore gear reduction is happening and torque multiplication is happening. So the torque output of the engine is much less than that, somewhere near 180 lb ft. If I did my math right

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24 minutes ago, 885 said:

You should be able to use the formula to go from hp back to torque. Hp=(tq*rpm) /5252. So Tq = (hp*5252)/rpm if my algebra still works

 

Edit to add. I looked at that tractor test. Those torque numbers are at the dyno, or available at the pto. The engine speed is much higher than the pto, therefore gear reduction is happening and torque multiplication is happening. So the torque output of the engine is much less than that, somewhere near 180 lb ft. If I did my math right

Ok, So pto speed on a 400 must've been at around 1100 rpm  since the dyno was spinning near 540 at that point? I get the formula and how that may put you close but you're telling me that every engine with the same hp puts out the exact same amount of torque at the same rpm? I don't know if I buy that, i think it needs to be measured

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40 minutes ago, 885 said:

You should be able to use the formula to go from hp back to torque. Hp=(tq*rpm) /5252. So Tq = (hp*5252)/rpm if my algebra still works

 

Edit to add. I looked at that tractor test. Those torque numbers are at the dyno, or available at the pto. The engine speed is much higher than the pto, therefore gear reduction is happening and torque multiplication is happening. So the torque output of the engine is much less than that, somewhere near 180 lb ft. If I did my math right

Although I'm no expert, maybe it is as simple as a formula. In that case, im probably thinking more along the lines of torque rise that really needs to be measured 

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18 minutes ago, Farmer in training said:

Ok, So pto speed on a 400 must've been at around 1100 rpm  since the dyno was spinning near 540 at that point? I get the formula and how that may put you close but you're telling me that every engine with the same hp puts out the exact same amount of torque at the same rpm? I don't know if I buy that, i think it needs to be measured

No clue about the pto speed. Or belt in this Nebraska test. I selected the max belt hp at the top of the chart and the correlating engine rpm. So the numbers were 50.78 hp and 1450 engine rpm. I plugged them in the formula and out popped around 180 lb ft of torque not accounting for losses in the driveline since we are measuring hp at the belt pulley. The losses are small and not really important at the moment. 

But as far as your question, if we have 3  50hp engines and we are on an engine dyno and they are all measured 50 hp at 2000 rpm then yes they all have the same torque output. If we have three different 50 hp engines and we are measuring at the pto rated speed of 540 and (big and) the engines are all turning the same engine speed then yes they all have the same torque output. Now in the real world those 50hp engines are spinning different engine speeds in tractors to achieve the rated pto speed of 540. In one it may be 2000, the other 2200, another would be 2300 engine rpm. The torque output of the engines would be 131 @2000, 119 @2200, and 114 @2300 rpm respectively, but the torque output on the pto shaft itself would be the same for all three since they are all the same hp and same 540 pto speed. Same hp at the pto but different torque at the engine due to gearing in the driveline

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10 minutes ago, 885 said:

No clue about the pto speed. Or belt in this Nebraska test. I selected the max belt hp at the top of the chart and the correlating engine rpm. So the numbers were 50.78 hp and 1450 engine rpm. I plugged them in the formula and out popped around 180 lb ft of torque not accounting for losses in the driveline since we are measuring hp at the belt pulley. The losses are small and not really important at the moment. 

But as far as your question, if we have 3  50hp engines and we are on an engine dyno and they are all measured 50 hp at 2000 rpm then yes they all have the same torque output. If we have three different 50 hp engines and we are measuring at the pto rated speed of 540 and (big and) the engines are all turning the same engine speed then yes they all have the same torque output. Now in the real world those 50hp engines are spinning different speeds in tractors to achieve the reted pto speed of 540.in one it may be 2000, the other 2200, another would be 2300. The torque output of the engines would be 131, 119, and 114 respectively. Same hp at the pto but different torque at the engine due to gearing in the driveline

So another question I have is take the 806 for example. They say it has a gross hp of 110 but a pto hp at 95. Wouldn't the same gear reduction that you say is increasing the torque numbers on the Nebraska test affect (increase) the HP number at the pto? Instead you're getting loss of hp like through a driveline

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Gear reduction increases torque, but not necessarily horsepower because gear reduction slows down rotation speed. Hp and torque's relationship is muddied by the third variable of rotational speed. If you look at a hp and torque graph for car engines they peak at different engine speeds, and hp usually continues to rise as torque is falling because the engine speed is increasing at a faster rate than torque is going down. And the loss in hp from gross (engine) to the pto on the 806 comes from frictional losses through the entire driveline. Every time a gear mesh happens, every bearing that adds drag, clutch slippage in the pto unit, cooling fans, water pump, hydraulic pump, etc. Causes incremental hp loss in the tractor. I know j said earlier to ignore driveline losses, but that is mainly because we can't really measure them on our end. Im sure international knew how much they were but we don't and can only make educated guesses. And at that we would be using faulty numbers from tractor data. They list claimed pto and actual, and the tractors always were more than actual, so we would be using a claimed gross hp and real pto hp numbers and would have a closeish but wrong guess of total gross hp to pto hp power loss. 

I also meant to mention earlier that the dyno doesn't actually measure hp, hp is a calculated number using the above formula. The dyno only measures torque and engine speed so that hp can be calculated. I hope this helps

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10 minutes ago, 885 said:

Gear reduction increases torque, but not necessarily horsepower because gear reduction slows down rotation speed. And the loss in hp from gross (engine) to the pto on the 806 comes from frictional losses through the entire driveline. Every time a gear mesh happens, every bearing that adds drag, clutch slippage in the pto unit, cooling fans, water pump, hydraulic pump, etc. Causes incremental hp loss in the tractor. I know j said earlier to ignore driveline losses, but that is mainly because we can't really measure them on our end. Im sure international knew how much they were but we don't and can only make educated guesses. And at that we would be using faulty numbers from tractor data. They list claimed pto and actual, and the tractors always were more than actual, so we would be using a claimed gross hp and real pto hp numbers and would have a closeish but wrong guess of total gross hp to pto hp power loss. 

I also meant to mention earlier that the dyno doesn't actually measure hp, hp is a calculated number using the above formula. The dyno only measures torque and engine speed so that hp can be calculated. I hope this helps

Ok Thank you. I'm trying not to be ignorant and I understand that gear reduction has a different effect on hp vs torque. But, going by the Nebraska test and the hp formula, (366x1450/5252)the 400 has 101 hp. Unless that dyno somehow knew exactly what the gear reduction is in that pto how does it know what the torque should be to get the right hp number? Again I'm following what you've said so far just trying to get it all through my skull

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No worries. So you are close but the numbers should be 365.9 ft lbs which you have and the rpm should be 696 rpm and not 1450 because the torque number is the torque measured at the dyno. Since we are using torque at the dyno we use the rpm at the dyno. That should give 48.48 hp. Very close to what they published as 50.79 hp. I have a feeling that they had some kind of percent loss for the dyno or weather or something that they added back in to normalize the data for comparison sake since these test were the gold standard for comparisons. 

Earlier when I was calculating backwards for engine torque I used belt hp since it was the only hp number available to me from the test data, so I "assumed" (not accounting for loss as the gross hp should be higher) it as the engine hp and used engine rpm to find a ballpark engine torque.

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Horsepower is a calculation made from the measurements of torque and rotational speed.  If you know two of the three numbers, the missing number can be figured out.  

HP = (torque X RPM)/5252

An engine making peak horsepower power at 5252 rpm will have HP and torque numbers that are equal.  If the engine is turning half of 5252 rpm at peak hp, the the torque number will be double of the hp number. 

In any case, theoretically, the only number that really matters is HP as that is the calculation of the ability to do work.  In the real world, it’s not that clear cut and I have seen that for myself where in some cases a lower hp engine with higher torque numbers can "out pull” a higher hp engine with lower torque numbers. 

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Ok I got it now. I think the biggest problem I had was I didn't wanna believe that the 1 ton Dodge at work produced more than twice the torque that my dads 5488 does. Although I'm assuming they go by peak torque on a pickup and a tractor will run their numbers all day. I'm seeing that the torque number is nothing more than part of a formula though. 

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2 hours ago, Farmer in training said:

Ok I got it now. I think the biggest problem I had was I didn't wanna believe that the 1 ton Dodge at work produced more than twice the torque that my dads 5488 does. Although I'm assuming they go by peak torque on a pickup and a tractor will run their numbers all day. I'm seeing that the torque number is nothing more than part of a formula though. 

That’s not impossible.  I have a ’99 Dodge with a 5.9 Cummins in it, and 3 tractors that also have the CDC equivalent of the 5.9 Cummins.  The pickup engines are set up to run quite a lot more power than the tractors, presumably because the duty cycle of the pickup is lighter.  The biggest of my tractors runs 170 engine horsepower; my pickup was 235 hp, and with a chip is now presumably 295 hp.  The pickup will run rings around the tractor for power. 

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51 minutes ago, Gearclash said:

That’s not impossible.  I have a ’99 Dodge with a 5.9 Cummins in it, and 3 tractors that also have the CDC equivalent of the 5.9 Cummins.  The pickup engines are set up to run quite a lot more power than the tractors, presumably because the duty cycle of the pickup is lighter.  The biggest of my tractors runs 170 engine horsepower; my pickup was 235 hp, and with a chip is now presumably 295 hp.  The pickup will run rings around the tractor for power. 

Fair enough. I think I'm confusing torque output with lugging ability which would be better explained with torque rise. Not to mention a tractor is geared much lower so torque from the engine and torque being applied to the ground or the rear axles must be a different story

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Take that Cummins and IH466 and hold it at the 2000 rpm and lug it hard you will have a clear winner.  Although that's really not fair with the 100 more CID.  My point is all tractor engines are rated at much lower rpm hence the torque curve.

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22 hours ago, 460 said:

Take that Cummins and IH466 and hold it at the 2000 rpm and lug it hard you will have a clear winner.  Although that's really not fair with the 100 more CID.  My point is all tractor engines are rated at much lower rpm hence the torque curve.

I think you would be surprised how well the Cummins would compare.  Stock, they are not a high rpm only engine.  In my experience they pull the best in the 2000-2200 rpm range and fall off fast above 2400.  They will lug just fine, never really getting to a point that they fall off a cliff, but I prefer not to do that as it is bad operating practice. 

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