1. Is there a difference in value of torque messured at the wheels and at the crankshaft?
2. If yes what does the value of Torque mean (in specs for example - my Skoda Octavia has 120 Nm)? Is it torque messured between the clutch and engine or is it torque messured at the wheels?
3. I was searching for a link to the site where all aspects of "why the car is moving" was described. I know there was one link posted on the forum but I couln't find it. Anyone knows one?
4. How can you calculate the increase in RPM per unit of time at full throttle?

Quote from himself :

1. Is there a difference in value of torque messured at the wheels and at the crankshaft?
2. If yes what does the value of Torque mean (in specs for example - my Skoda Octavia has 120 Nm)? Is it torque messured between the clutch and engine or is it torque messured at the wheels?

1. Yes, there is a difference. You have less torque at the wheels due to drivetrain losses.

2. Most cars give these value measured at the flywheel.

4) you mean that how fast the rpms of the engine rise when you stab the throttle on neutral? If so it's all about the engine performance and weight of the parts that are rotating plus bearing friction + other things like other moving parts attached to the engine etc...

But why would you like to calculate that? Or did you mean the acceleration of the on gear?

1. You have less torque at the wheels due to drivetrain losses. However torque multiplication may mean you have more torque at the wheels (low gears, say 1st) or less at the wheels (high gears, say 6th)

2. The torque will be measured at the flywheel.

3. Sorry I don't know what you're talking about. Cars move because the wheels transmit a force to the road, either longitudinally, laterally or both. The forces are generated either by turning (slip angles), driveshaft torque (from the transmission), or braking torque.

4. Using the cars gearing (including wheel size) you can work out the tractive effort (force at the wheels) at any speed/engine revs from the engine torque. This gives you your instantaneous force on the road (for more accurate analysis include slip ratios and transmission losses), from which you can work out the rate of change. Countering that force is aero drag and rolling resistance. Thus you now have the sum of all forces. If you know the mass of the car (and ideally the rate at which it varies), you can make a fairly complex 'simulation' of how fast the car will accelerate. And if you know the acceleration and you know all the previous data you can easily find the rate of change of RPM.

I hope thats okay, and neither too complex nor too simple.

Thanks for help.

Quote :

Sorry I don't know what you're talking about. Cars move because the wheels transmit a force to the road, either longitudinally, laterally or both. The forces are generated either by turning (slip angles), driveshaft torque (from the transmission), or braking torque.

I ment there was once a web site where all the equations and calculations of forces acting during car moving, accelerating, turning were completely described. The link was on the forum and I cant find it. I know the theory but the exect eqations are not known to me.

Quote :

you mean that how fast the rpms of the engine rise when you stab the throttle on neutral?

Yes. I thought when I can calculate that on neutral I would be able to calculate the wheel rotation speed after including the power losses and gear/diff ratios.

5. Another question. I am modelling the engine at 2000 rpm. I use the torque generated by engine - for example 60Nm. Then I multiply it by ratios at gearbox at 1st gear. [2.97:1] - result is aproximately 180Nm at wheels. Then I have a wheel with radius of 25 cm. Does it mean I have 720 N force at both wheels?

6. Do I need the information about power generated by engine then?

5. No, you have 720N force in total.

6. Nope, totally useless for performance predictions.

Quote from tristancliffe :

6. Nope, totally useless for performance predictions.

Not entirely. For guesstimates and predictions, power is a useful quantity. For doing a simulation like you described, you need the whole torque curve. The peak torque figure itself is never particularly useful.

I never mentioned peak torque, and we're not talking guesstimates here

Quote :

4) you mean that how fast the rpms of the engine rise when you stab the throttle on neutral?

Quote from himself :

Yes. I thought when I can calculate that on neutral I would be able to calculate the wheel rotation speed after including the power losses and gear/diff ratios.

That "value" is plain useless. And only way to get the perfect values would be to test it in real life. But I wouldn't suggest anyone to do so, because revving an engine without any load isn't exactly a smart move

Quote from himself :

1. Is there a difference in value of torque messured at the wheels and at the crankshaft?

sounds like a simple question at first but gets somewhat complicated
first of all as long as theres no significant amount of torsion on the axes that connect the crankshaft to the wheels or slip in the clutch etc the amount of power (torque*rpm) is pretty much constant everywhere in the transmission
the gears however change the rpm so the torque at the wheels will be slightly/significantly in lower gears different from the torque at the crankshaft

the problem with this logic however is that the torque numbers you see when you buy a car are the maximum available torque without drivetrain losses
now lets get back to the power figures as its a lot easier to understand drivetrain losses in terms of power ... a significant par of the engines power output is lost in the transmission line and end up as heat in the oils/axes/cogwheels etc
the point of all this is that the effctive available torque when you measure it at the wheels will be lower than what youd expect when you just take the engine torque figures and the gearing but the torque you would measure at the crankshaft and at the wheel would be the same (considering the gearing of course)

Quote :

2. If yes what does the value of Torque mean (in specs for example - my Skoda Octavia has 120 Nm)? Is it torque messured between the clutch and engine or is it torque messured at the wheels?

measured at the crankshaft without the tranmussion installed

Quote :

4. How can you calculate the increase in RPM per unit of time at full throttle?

well youll have to search for graphs that show torque over rpm and throttle-plate-angle from those you can work out the power reduce that by the amout of power you lose in the transmission calculate the rpm of the wheels at the current rpm of the engine divide those two and youll have the torque at the wheels
with that figue you can calculate the force acting on the wheels reduce that number by drag and some more or less random number from tyre physics and calculate the acceleration through F = ma
for there on out its a tedious task of iterating those steps a few hundred tmes to get a smooth curve

Quote from himself :

6. Do I need the information about power generated by engine then?

yes as drivetrain losses make a lot more sense in terms of power

Would it be a good aproximation if:

I calculated 720N at both wheels (few posts above) (so 360 at each). Then using mass of a car 1300 kg I have acceleration of a = 0.55 m / s*s. Would it be accurate if having 25% of power loss at drivetrain I multiply 0.55 by 75%?

Depends on exactly what you want, but it'll be a reasonable guess.

Remember that if you use the maximum peak torque value in that calculation you get the maximum acceleration value, ie. you can't use it to calculate 0-100kph values etc.

Other thing would be air resistance as it builds as the speed increases, expotentially. One way of getting some estimates would be to create curves and/or "lines" that would be close to the torque curve. And just integrate from there...

Questions phase 2:

About Locked Diff. Lets say that maximal torque at left wheel without spinning it is 200 Nm and 800 Nm on the right. Reaching 400 Nm both wheels are proppeled with 200 Nm. Then the right wheel is being "filled" with torque untill it reaches its peak value. So we have 1000 Nm at diff -> 200 @ left and 800 @ right one. But what next? Increasing torque at diff to 1200 Nm will it be distributed equally to both wheels then? (300 / 900) or unequally proportional to maximal non-sliping value? (240 / 1060) ?

The main physics programmer of RBR (Richard Burns Rally) gave a lecture on how differentials work. He said that in a differential, both wheels get the same torque - "it's just the way it works mechanically". It's the power that is different (the effect, the work done etc - the twisting energy is the same).

Quote from Mikkomattic :

in a differential, both wheels get the same torque - "it's just the way it works mechanically".

Isn't that true for open diffs only?

Quote from himself :

About Locked Diff. Lets say that maximal torque at left wheel without spinning it is 200 Nm and 800 Nm on the right. Reaching 400 Nm both wheels are proppeled with 200 Nm. Then the right wheel is being "filled" with torque untill it reaches its peak value. So we have 1000 Nm at diff -> 200 @ left and 800 @ right one. But what next? Increasing torque at diff to 1200 Nm will it be distributed equally to both wheels then? (300 / 900) or unequally proportional to maximal non-sliping value? (240 / 1060) ?

Additional torque (over 1000 Nm) will cause angular accelaration of both wheels. Because they are locked together, they will have equal acceleration. They also have the same moment of inertia, so I think this additional torque will be split equally, 300 / 900 in this case. In fact 280 / 880 might be more correct, because some torque will be needed to accelerate the axle.

Quote from Mikkomattic :

The main physics programmer of RBR (Richard Burns Rally) gave a lecture on how differentials work. He said that in a differential, both wheels get the same torque - "it's just the way it works mechanically". It's the power that is different (the effect, the work done etc - the twisting energy is the same).

If both wheels have the same traction and equal weight distribution on them then ok - Torque is equal. But I can not agree that EVERY time the torque is the same. Just imagine one wheel hanging in the air like moment after hitting a curb. If the torque is distributed on both wheels equally what happens with the torque on the wheel hanging in the air? Disappears?

Do you have a link to the lecture?

Thanks for answers guys

With a wheel in the air the torque acting on the wheel through the axle will be used to increase the rotational speed of the wheel, which has non-zero moment of inertia.

Quote from w126 :

With a wheel in the air the torque acting on the wheel through the axle will be used to increase the rotational speed of the wheel, which has non-zero moment of inertia.

However it is locked differential If it makes one wheel go faster so the other one too. So it actually goes to the other wheel because that one is on the ground and have more friction than one in the air. So the whole torque goes to one wheel <=> it is not distributed equally.

Yes, for locked differential the torque is not distributed equally most of the time. But for open differential it is distributed equally all the time.

Anyone could be so kind to tell me what is the function of Slipratio to Friction of the tyre to the asphalt? For example when the wheel is not the propelling one and the car starts to move the friction cause the wheel to rotate.

... and the function of Slipratio to lateral friction.

I know that pacejka formula tells something about but i cant find just acuurately these functions.

Thanks for help.