Engine Torques

  • #1
anj
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Main Question or Discussion Point

Hello All,

Is it possible to work back from drive torque at wheel to engine torque? Can it be described with basic equations and is the calculation same between all wheel drive and front wheel drive?

Thanks Anj,
 

Answers and Replies

  • #2
384
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Yes, you just need to know all the gear ratios in between. There will be some loss from friction of course...

So if your rear wheel torque is 1000 lb-ft of torque, and your rear axle ratio is 4.10:1, you will have 1000/4.1 = 244 lb-ft toque in the driveshaft... Repeat for the transmission gear ratio and you'll have engine torque
 
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  • #3
anj
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So basically Rear Wheel Drive
Teng = Twheel/Ratio(Trans)*Ratio(RearAxle)

So for Front Wheel Drive?
Teng = Twheel/Ratio(Trans)*Ratio(FrontAxle)

How about All Wheel Drive ?

Also is Wheel torque (Left + Right)?
 
  • #4
384
172
Should be the same regardless of front/rear/all wheel drive. There may be some odd cases with computerized transfer cases, etc that have uneven drive power splits between the front and rear or different axle ratios front and rear..

Wheel torque will be the same on both wheels (because of the differential) when measured on a dynamometer, it could all be applied to one wheel and none to the other, it really doesn't make a difference.. look at the total forward force the car is applying and go from there
 
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  • #5
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Remember to include drive wheel radius in your calculations. Twheel goes down as radius increases, so axle torque must be calculated. And yes, torque will divide into two with two-wheeled drive vehicles and into four with "locked" four-wheel drive vehicles like old Jeeps. But as mentioned, modern four-wheel drive vehicles are usually computer controlled to put torque where it's needed.
 
  • #6
Yes, you just need to know all the gear ratios in between. There will be some loss from friction of course...

So if your rear wheel torque is 1000 lb-ft of torque, and your rear axle ratio is 4.10:1, you will have 1000/4.1 = 244 lb-ft toque in the driveshaft... Repeat for the transmission gear ratio and you'll have engine torque
Given the engine supplys the power...how is the wheel torque higher than the drive shaft
 
  • #7
384
172
Given the engine supplys the power...how is the wheel torque higher than the drive shaft
The torque is higher because of the gear reduction in the differential... POWER output would be slightly less because of inefficiencies in the differential though
 
  • #8
The torque is higher because of the gear reduction in the differential... POWER output would be slightly less because of inefficiencies in the differential though
So if that is true. Which i agree it is true. What would be the effect of having an engine that operates like a wheel that spins like a rotary engine where the leverage is increase by 500% at the same rpm? Leverage being 5 times longer. Does 5 times more leverage equal 5 times the power At the same rpm?
 
  • #9
384
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I don't quite follow what you're asking... Leverage being 5 times longer means you have 1/5th of the leverage.. Power would be the same, torque would be 1/5th, and speed would be 5x
 
  • #10
I don't quite follow what you're asking... If you could increase the stroke 5 times longer So the offset is 5 times the other engines main to rod journal distance. Would you have 5 times the pressure on the crankshaft?
 
  • #11
384
172
Yes, you would, but increasing the stroke by 5x usually involves a very serious redesign of an engine.. If you keep the same bore size, you're going to have 5x the displacement, but lose a lot of volumetric efficiency due to very small valves, and the peak RPM of the engine will drop by about the same ratio as your increased stroke.. you'll make more torque, but it's unlikely you'll make much more POWER.. (the valves can only flow so much air)

If you keep to a near 'square' engine design (bore ~= stroke), and increase the stroke and bore 5x, you'll have 125x (5x5x5) the displacement, at that point, yes, you'll definitely make more power and torque of course, but the thing will be MASSIVE
 
  • #12
Yes, you would, but increasing the stroke by 5x usually involves a very serious redesign of an engine.. If you keep the same bore size, you're going to have 5x the displacement, but lose a lot of volumetric efficiency due to very small valves, and the peak RPM of the engine will drop by about the same ratio as your increased stroke.. you'll make more torque, but it's unlikely you'll make much more POWER.. (the valves can only flow so much air)

If you keep to a near 'square' engine design (bore ~= stroke), and increase the stroke and bore 5x, you'll have 125x (5x5x5) the displacement, at that point, yes, you'll definitely make more power and torque of course, but the thing will be MASSIVE
If you had intake 93% the size of the piston diameter with turbos that had pressure to get 100% volumetric efficiency at 1000 rpm to introduce fresh air at top of cylinder and allow exit of combustion charge out exhaust ports at BDC area where it takes place from 120 degrees to 210 degrees. But you only compressed air from 330 degress ATDC to TDC and only compressed bore and stroke equal distance. 10 inch bore and only compressed 10 inch of stroke. it free wheeled "no compression until 330 degrees ATDC to TDC with 5 times the friction only as added drag on rotation. Would it have 5 times the torque? And what would be the effect if the engine had no negative G forces from any parts changing trajectory "as in having parts going up and down? Lets say every moving part was in an orbit? With only centripetal and centrifugal force with stored energy from inertia "mass in an orbit tied to a central point by centripetal force?
 
  • #13
384
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You're losing me again.. Intake PORT or intake VALVE? Is the exhaust a port or a valve?

If you're talking about ported engines like most 2 cycle engines, they'll pretty much always be symmetrical.. The port that opens 120* ATDC will close 120* BTDC, so it's impossible to get 120* of power stroke with 30* of compression stroke

If you're talking about a cam/valve timed engine, it's impossible to get an intake valve 93% of the piston diameter...
 
  • #14
You're losing me again.. Intake PORT or intake VALVE? Is the exhaust a port or a valve?

If you're talking about ported engines like most 2 cycle engines, they'll pretty much always be symmetrical.. The port that opens 120* ATDC will close 120* BTDC, so it's impossible to get 120* of power stroke with 30* of compression stroke

If you're talking about a cam/valve timed engine, it's impossible to get an intake valve 93% of the piston diameter...
Not talking about either one of those designs. Something very different. No valves yet it lets air in and out and compresses from 330 degrees to TDC. Combustion from 350 degrees to after TDC. it has 10 inch pistons running on diesel. Can you calculate the pressure on the rotation given 78 sq. in. on piston with very standard diesel pressure curve in the cylinder? Keep in mind the expansion is 40 inches before it hits exhaust.
 
  • #15
Not talking about either one of those designs. Something very different. No valves yet it lets air in and out and compresses from 330 degrees to TDC. Combustion from 350 degrees to after TDC. it has 10 inch pistons running on diesel. Can you calculate the pressure on the rotation given 78 sq. in. on piston with very standard diesel pressure curve in the cylinder? Keep in mind the expansion is 40 inches before it hits exhaust.
The offset is 25 inch from centerline. As compared to a 25 inch offset of rod journal to main center
 
  • #17
255
30
I really don't understand what the OP is looking for. The only comment I have is that drive wheel power always equals net engine power less frictional drive train losses, regardless of reduction ratio. Drive wheel torque is equal to net engine torque times the reduction ratio less frictional drive train losses (so if the total reduction ratio is 5:1, output torque is net torque times 5 less frictional losses).
 

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