Engine torque multiplied when at wheels?

[NCpaj]

Gday

well i was flick through a couple of old 4wdm mags and there was an article on the 3.0 Gu V's the 4.2Gu---blah blah blah. But what struck me in the article was the dyno readouts. They say that you lose power from the time it goes from the flywheel to the wheels, i understand that-the whole mechanical inefficeny of the gearbox/heat noise thing.
But it also said that torque is mulitplied by the time it gets to the wheels.

How does that work? is it only multiplied by the lenght of one side of the axels? (ie perpendicular lever arm?)

its one of those things that has just been bugging me all week

cheers-

[1MadEngineer]

i think they call them diff gears

[11_evl]

X2
im guessing that the two trucks have different diff or box or transfer gears or even tyre size. all make a difference to final output.

[thehanko]

as far as my understanding of torque goes the only change will be through in efficiency.

torque is a constant - however the force and distance from the point of rotation can change as long as its in relation to one another.

eg.

torque = force x distance from point of rotation.

so if you have 100Nm on a 1m diameter tyre the force at the point of contact is 100N.

but on a 2m tyre the force is 1/2 that at 50N, or on a 50cm tyre twice that at 200N.

So from engine to wheels the torque is constant however by gearing it down it gives more force at the wheels.

when they say torque is multiplied, what they really mean is that the drive line effectively gears down (reduces the size of the wheel). thus increasing the force at the point of contact.

[macca81]

its just lower gearing. you have more torque at the wheels simply by putting smaller wheels on

[Dirty]

Basically hp = torque x rpm (divided by a constant 5252)

So if you change the gearing the rpm value changes for the wheel and thus the torque also needs to change for the fixed hp from the engine.

Spin the wheels faster (higher gearing) => less torque
Spin the wheels slower (lower gearing) => more torque

This is all at the axle, you then need to take into consideration the diameter of the tyre as larger tyres will deliver less torque.

- David.

[lay80n]

This also relates to the gear used during the dyno work. Technically you should use a 1:1 gear (most 4th gears in 5 speed box's) to remove as much reduction as possible. It is possible to fudge dyno results by running in other gears. Your HP is also altered, as HP is just a function of torque and RPM combined with a factor.

Its HP = (torque in lbft * Speed in rpm) / 5252



Layto....

[Jacked]

Torque convertors in autos also increase torque

[RED60]

so if you have 100Nm on a 1m diameter tyre the force at the point of contact is 100N.

but on a 2m tyre the force is 1/2 that at 50N, or on a 50cm tyre twice that at 200N.

So from engine to wheels the torque is constant however by gearing it down it gives more force at the wheels.

when they say torque is multiplied, what they really mean is that the drive line effectively gears down (reduces the size of the wheel). thus increasing the force at the point of contact.

Your diameters of tyres are wrong for your answers to be correct. eg at 100Nm the tyre diam is 2m.....

[Gwagensteve]

Torque convertors in autos also increase torque

only when they are slipping AFAIK. Less slip= less torque multiplication.

Obviously when you're cruising in a modern car with the converter locked there is no torque multiplication.

Steve.

[NCpaj]

as far as my understanding of torque goes the only change will be through in efficiency.

torque is a constant - however the force and distance from the point of rotation can change as long as its in relation to one another.

eg.

torque = force x distance from point of rotation.

so if you have 100Nm on a 1m diameter tyre the force at the point of contact is 100N.

but on a 2m tyre the force is 1/2 that at 50N, or on a 50cm tyre twice that at 200N.

So from engine to wheels the torque is constant however by gearing it down it gives more force at the wheels.

when they say torque is multiplied, what they really mean is that the drive line effectively gears down (reduces the size of the wheel). thus increasing the force at the point of contact.

ok cool, it all makes sense now.
thanks fellas.

[RED60]

as far as my understanding of torque goes the only change will be through in efficiency.

torque is a constant - however the force and distance from the point of rotation can change as long as its in relation to one another.

eg.

torque = force x distance from point of rotation.

so if you have 100Nm on a 1m diameter tyre the force at the point of contact is 100N.

but on a 2m tyre the force is 1/2 that at 50N, or on a 50cm tyre twice that at 200N.

So from engine to wheels the torque is constant however by gearing it down it gives more force at the wheels.

when they say torque is multiplied, what they really mean is that the drive line effectively gears down (reduces the size of the wheel). thus increasing the force at the point of contact.

ok cool, it all makes sense now.
thanks fellas.

ncpaj, read my post 3 up..... the principal is correct but the numbers are wrong

[dogbreath_48]

as far as my understanding of torque goes the only change will be through in efficiency.

torque is a constant - however the force and distance from the point of rotation can change as long as its in relation to one another.

eg.

torque = force x distance from point of rotation.

so if you have 100Nm on a 1m diameter tyre the force at the point of contact is 100N.

but on a 2m tyre the force is 1/2 that at 50N, or on a 50cm tyre twice that at 200N.

So from engine to wheels the torque is constant however by gearing it down it gives more force at the wheels.

when they say torque is multiplied, what they really mean is that the drive line effectively gears down (reduces the size of the wheel). thus increasing the force at the point of contact.

ok cool, it all makes sense now.
thanks fellas.

ncpaj, read my post 3 up..... the principal is correct but the numbers are wrong

radius vs. diameter

[-Scott-]

its just lower gearing. you have more torque at the wheels simply by putting smaller wheels on

Not really.

Smaller wheels has, in many respects, the same effect as shorter gears, but strictly speaking the torque at the wheel (the axis of rotation) hasn't changed.

As thehanko wrote, Torque = Force x Radius, so Force = Torque / Radius. Smaller wheels (well, tyres, actually) reduces the radius component, which increases the force on the road. This means more push, to accelerate, or to climb the hill.

On a rolling dynamometer, the dyno isn't measuring the torque at the driving wheel, it's measuring the torque at its roller, which is driven by the wheel (tyre...) Fitting a smaller tyre is effectively reducing the gear ratio driving the dyno roller, which will "measure" more torque. If the dyno's torque measurement is corrected for tyre size, there shouldn't be a noticeable difference between the two.

[PGS 4WD]

This also relates to the gear used during the dyno work. Technically you should use a 1:1 gear (most 4th gears in 5 speed box's) to remove as much reduction as possible. It is possible to fudge dyno results by running in other gears. Your HP is also altered, as HP is just a function of torque and RPM combined with a factor.

Its HP = (torque in lbft * Speed in rpm) / 5252



Layto....

Wouldn't necessarily agree with using 1:1 all the time, most dyno's are speed limited between 220 and 250 kph. 36 inch tyres, 4.11:1 diffs and 6000 rpm make for some high speeds. It's not such fun at that speed, also with the higher wind drag on the tyre at high wheel speed the resultant measured power is significantly less, especially with agressive tyres. I use the gear that gives constant traction with the lowest top speed, often 3rd in manuals and sometimes second on three speed autos depending on the power output.
Forget about torque, if you log RPM and make all comparison relative to the same RPM then an increase in power must mean an increase in torque.

Joel

[KiwiBacon]

Basically hp = torque x rpm (divided by a constant 5252)

Only for those using obselete US measurements (ft-lbs).

In SI it's
power (watts) = torque (Nm) * rotating speed (radians/second).

radians/second = rpm*pi/30

So the whole thing becomes.
Power = Torque * rpm * pi/30

[lay80n]

This also relates to the gear used during the dyno work. Technically you should use a 1:1 gear (most 4th gears in 5 speed box's) to remove as much reduction as possible. It is possible to fudge dyno results by running in other gears. Your HP is also altered, as HP is just a function of torque and RPM combined with a factor.

Its HP = (torque in lbft * Speed in rpm) / 5252



Layto....

Wouldn't necessarily agree with using 1:1 all the time, most dyno's are speed limited between 220 and 250 kph. 36 inch tyres, 4.11:1 diffs and 6000 rpm make for some high speeds. It's not such fun at that speed, also with the higher wind drag on the tyre at high wheel speed the resultant measured power is significantly less, especially with agressive tyres. I use the gear that gives constant traction with the lowest top speed, often 3rd in manuals and sometimes second on three speed autos depending on the power output.
Forget about torque, if you log RPM and make all comparison relative to the same RPM then an increase in power must mean an increase in torque.

Joel

Cool. I have never been around big tyres on Dyno's, only "normal" size tyres

Layto....

[RockyF75]

So would it be safe to say, that when in low range, there's more torque going to the wheels?

Cause I tried to tell my mate this when he was trying to do 2wd low burnouts , telling him he has a better chance of breaking stuff cause low range will increase the torque and strain on the drivertrain. But I couldn't really explain it.

[KiwiBacon]

So would it be safe to say, that when in low range, there's more torque going to the wheels?

Cause I tried to tell my mate this when he was trying to do 2wd low burnouts , telling him he has a better chance of breaking stuff cause low range will increase the torque and strain on the drivertrain. But I couldn't really explain it.

The torque you can put on the wheels is limited by the traction they've got.
So burnouts in high or low range will put similar torque through the axles if the ground conditions are the same.

But in high range you've got more chance of shock-loading them enough to break something (like catching a rock when you're spinning through mud).

[thehanko]

as far as my understanding of torque goes the only change will be through in efficiency.

torque is a constant - however the force and distance from the point of rotation can change as long as its in relation to one another.

eg.

torque = force x distance from point of rotation.

so if you have 100Nm on a 1m diameter tyre the force at the point of contact is 100N.

but on a 2m tyre the force is 1/2 that at 50N, or on a 50cm tyre twice that at 200N.

So from engine to wheels the torque is constant however by gearing it down it gives more force at the wheels.

when they say torque is multiplied, what they really mean is that the drive line effectively gears down (reduces the size of the wheel). thus increasing the force at the point of contact.

ok cool, it all makes sense now.
thanks fellas.

ncpaj, read my post 3 up..... the principal is correct but the numbers are wrong

radius vs. diameter

i was even thinking radius as i wrote it... meh you get the idea.

[NCpaj]

its just lower gearing. you have more torque at the wheels simply by putting smaller wheels on

Not really.

Smaller wheels has, in many respects, the same effect as shorter gears, but strictly speaking the torque at the wheel (the axis of rotation) hasn't changed.

As thehanko wrote, Torque = Force x Radius, so Force = Torque / Radius. Smaller wheels (well, tyres, actually) reduces the radius component, which increases the force on the road. This means more push, to accelerate, or to climb the hill.

On a rolling dynamometer, the dyno isn't measuring the torque at the driving wheel, it's measuring the torque at its roller, which is driven by the wheel (tyre...) Fitting a smaller tyre is effectively reducing the gear ratio driving the dyno roller, which will "measure" more torque. If the dyno's torque measurement is corrected for tyre size, there shouldn't be a noticeable difference between the two.

ok, well i think ive got it all worked out now, good post scott---so i guess thats why some dynos are the type that bolt onto the wheel studs to eliminate the wheel size factor?

[RED60]

its just lower gearing. you have more torque at the wheels simply by putting smaller wheels on

Not really.

Smaller wheels has, in many respects, the same effect as shorter gears, but strictly speaking the torque at the wheel (the axis of rotation) hasn't changed.

As thehanko wrote, Torque = Force x Radius, so Force = Torque / Radius. Smaller wheels (well, tyres, actually) reduces the radius component, which increases the force on the road. This means more push, to accelerate, or to climb the hill.

On a rolling dynamometer, the dyno isn't measuring the torque at the driving wheel, it's measuring the torque at its roller, which is driven by the wheel (tyre...) Fitting a smaller tyre is effectively reducing the gear ratio driving the dyno roller, which will "measure" more torque. If the dyno's torque measurement is corrected for tyre size, there shouldn't be a noticeable difference between the two.

ok, well i think ive got it all worked out now, good post scott---so i guess thats why some dynos are the type that bolt onto the wheel studs to eliminate the wheel size factor?

Bolt on dynos are better I think... everything else being equal, they remove another variable...

[croozy]

Here is what happens when you have a little too much torque at the wrong time.
a result I realy didn't need

[KiwiBacon]

Here is what happens when you have a little too much torque at the wrong time.
a result I realy didn't need

That's a great photo, it's not often you see something that's twisted that far without breaking.

What vehicle is it off? I presume rear axle shaft into the diff?

[macca81]

Here is what happens when you have a little too much torque at the wrong time.
a result I realy didn't need

crickey... impresive.

[croozy]

That's a great photo, it's not often you see something that's twisted that far without breaking.

What vehicle is it off? I presume rear axle shaft into the diff?

That is actually the output shaft that I just took out of my t700.
It is behind a small block chev in an fj40.
It's getting a special place in the shed to remind me how much it is costing me to fix everything.


Funny you should say crikey macca81 'cause thats the name of the car, it was given by my young bloke.
http://www.4wdmonthly.com.au/shed/index ... =1120&im=1