## Is tyre slip caused by power or torque?

Here are our CFD links and discussions about aerodynamics, suspension, driver safety and tyres. Please stick to F1 on this forum.
Mystery Steve wrote:
Jersey Tom wrote:
autogyro wrote:I think going through all the complex Math is exactly what the question was posted for. It remains impossible to answer.

...I'm still goin with plain and simple 'torque.'

Yeah, I thought your blog entry did an excellent job of answering this question, but I guess I was wrong...

Plain and simple: power is the measure of the time rate at which work is performed. Torque is the result of a force acting on a moment arm. A tire is only capable of providing some finite level of longitudinal force. This critical force is multiplied by the tire radius to get a critical torque that the tire can handle. If the torque applied to the wheel by the engine is greater than this critical torque, you've got slip. Power is related to the torque by the gearing. You could have a one-million horsepower engine, but for some gearing ratio (granted it would be ridiculous, but just being theoretical..) you won't get enough torque applied to the wheel.

Is tyre slip caused by power or torque?
No
autogyro

Joined: 4 Oct 2009

You're right autogyro. Tire slip is created as a byproduct of unicorn procreation.
Grip is a four letter word.

2 is the new #1.
Jersey Tom

Joined: 29 May 2006
Location: Huntersville, NC

autogyro wrote:As you add weight the drive wheels will grip to higher rpm until you reach 5.5 thousand rpm but that is only because you reach 5.5 before you reach 8.5.
If you carry on adding weight you will reach a point where 8.5 has to be reached before traction is broken.
The point being?

No that's not what I meant, of course if the tyres start to slip at 5.5 i don't expect them to stop at 8.5, that would not make any sense.
This is what i meant:

fizzer wrote:Well this is purely anecdotal but...

I have an RX-8. I autocross it on very sticky AutoX specific compound tires (Hoosier A6 285/30/R-18s).

Clutch drop at 5.5k - engine bogs, tires hop, ugly launch.

Clutch drop at 8k - immediate spin up, nice predictable launch.

I'm sure there's plenty of other factors, but the tires break traction way easier near peak hp. Same thing in turns, I can slam the throttle at 5k without upsetting the chassis nearly as much if I was steady state and went full throttle at 7k.

And that really answers my question and does match with my real world experience. The second part even addresses RJSA very valid point of stored kinetic energy.

I was wondering though if numbers could be put in place to get some final theoretical closure on the issue.

engineguru00 wrote:Figure I would actually register to answer this, I have been lurking around for a while. I did a good amount of work with wheel slip when gearing an FSAE car last year around a CVT transmission.

The amount of longitudinal force relies on a few things. Tires produce their maximum force at a certain slip ratio, which is not to be confused with runaway slip. Tires also vary their longitudinal thrust with load, although it isn't a linear relationship. Temperature and inclination angle both play a big role as well and neither are constant. Most SR vs LF vs Load plots you see are on a belt setup which doesn't accurately predict tire capacity in real life during operation, so you normally have to add a correction factor to take this into account and scale down all the values. A good driver can modulate the throttle to keep you at this peak of force, so you take that multiplied by your correction factor to get your maximum longitudinal force.

We all know torque is force times the distance it is being applied. With tires this distance is the loaded tire radius, which is not a constant value as it changes with load transfer. You now know how much torque you need to provide. You then work back through the half-shafts, differential, and transmission and take into account all the gear ratios and transmission losses to get the amount of engine torque required to get this longitudinal acceleration.

Now that all that ground work is laid out, to the question. The answer is "Yes". Your SR vs LF vs Load plot looks like a mountain on most tires, so once you exceed the peak force you actually have less tractive capacity and the wheel will spin faster and easier while actually providing less force. This is because you have no transitioned from static friction to kinetic friction at the contact patch. If you notice through all of this I have said nothing about HP and that is because it has nothing to do with wheel spin. You could spin the tires at peak torque or peak horsepower if the car has enough torque to pass the peak on your tire plot. Sorry if its a bit of a long winded answer, but thats the engineering behind it.

"Horsepower may get the car off the showroom floor, but torque is what gets it off the line" -C.S.

EDIT: I didn't go into the whole inertial side in this, but thats another story for another day. I just got out of my class on system modeling and don't really want to go through that again.

Thank you very much, that explanation does make sense. That being said real life experience in my case is very close to engineguru00's description of what happens in mid corner and/or launch acceleration. Thus I was hoping someone here with engineering background (which i certainly don't have) could put some numbers behind either answer.
Alejandro L.
alelanza

Joined: 16 Jun 2008
Location: San José, Costa Rica

I'm sure there's plenty of other factors, but the tires break traction way easier near peak hp. Same thing in turns, I can slam the throttle at 5k without upsetting the chassis nearly as much if I was steady state and went full throttle at 7k.

And that really answers my question and does match with my real world experience. The second part even addresses RJSA very valid point of stored kinetic energy.

Not it doesn't. Two absolutelly different things. From 5K rmp to 7K rpm on the same gear you are going 40% faster on the same turn, so it's much easier to upset the car when you are faster.

If the speed is the same, then when you are at 7K rmp you should be 1 or 2 gears down from when you are at 5K.

I'm impressed how people fail to grasp relationship between force, speed and power.
rjsa

Joined: 2 Mar 2007

alelanza wrote:
Thank you very much, that explanation does make sense. That being said real life experience in my case is very close to engineguru00's description of what happens in mid corner and/or launch acceleration. Thus I was hoping someone here with engineering background (which i certainly don't have) could put some numbers behind either answer.

Welcome. I cannot give practical numbers because so many different factors affect it. I did a spreadsheet for one of our cars to help calculate gearing more effectively, but it was rather project specific and cannot be easily transfered to other projects.

Mid corner is another matter though. Most tires develop their peak longitudinal force at zero inclination angle, where as peak lateral capacity is attained at a few degrees of inclination. Depending on how your car is setup, your force values can change a good amount between these two points. At the same time you are now using some of the car's tractive effort for lateral force, so you have less longitudinal grip available. As rjsa said, this higher speed now means you are using more of your tires grip to turn and you now have less grip available for acceleration. In addition to this, you now have lateral weight transfer (which increases with lateral acceleration) affecting how much longitudinal thrust each tire can provide. Your differential then also now comes into play with if/how it biases torque, which can have a big impact on how your car will want to put power down mid corner.
Chris
UB Motorsports: Formula SAE '08-'10
Captain '09-'10
engineguru00

Joined: 18 Mar 2010
Location: Buffalo,NY

Oh, my... Don't you have mercy? Well, I don't, so another long answer.

First, simple: what twists a tyre is torque. Stay there, don't let trick questions confound you. You slip the tyre at max torque more easily. End. No "ifs", "buts" or "I guess".

Second, my daily nag:

For the love of Pete, in the name of James Watt's sacred slide rule, please, people. This is a technical forum. No steaming piles, please.

a. There is no shame in answering: "I don't know" or "I was mistaken". It's called scientific method.

b. Also, there is no shame in saying: "Look, these are my numbers. What do you think?" instead of implying people is stupid, ehem, even if we (and I'm included there) are. Concentrate on the original question.

So, autogyro, JerseyTom stop it.

Third, allow me to complicate my life. Imagine, for sake of simplicity, that:

a. you have a small engine, so its power it's not enough to make the tyre slip.
b. you don't shift gears
c. you slam the throttle and keep it there (as God intended)

When do you have more force on the wheels? (twisting them, of course) Read ahead if you want a few ideas (and a recommendation to win your drag races).

Your car has a power-rpm "curve" like this one:

232 hp @ 8500 rpm
159 lb-ft @ 5500 rpm

Damn, we also have to deal with gringo units... (I'm starting to become racist ).

So, 8500 rpm and 232 hp is like (wait a minute)... All right, these are my numbers

Alelanza's car numbers (in bold, the power curve numbers, in italics my calculations, in colour red and blue, the important numbers)

Now, even if you don't read this (and maybe you won't, I understand), your car engine axle gives you 194 newtons at one meter distance when you're at max power (last number in red, lower left) or it gives you 216 newtons at the same one meter fulcrum when you're at max torque (red number, row 9).

Like this (for dummies!):

Your car has a gearbox and wheels (I think).

Imagine your gearbox and rear/front transmission have a gear relationship of, I don't know, 9 to 1.

That is, when your car is in first gear, for every turn of the engine, the axle at the end of the gearbox turns 3 times and your differential box also turns 3 times per each revolution of the gearbox, for a final relationship of 9 to 1.

In English: every time your engine turns 9 times, your wheels turn 1 time. That would be more or less typical for the kind of cars I drive.

So, you multiply your torque per nine. Power stays the same, because the rotational speed (how fast your wheels turn) is 9 times slower.

In the end your car receives a force of 595 kilograms when at max power, or 660 kilograms at max torque (blue numbers, bottom of the worksheet, assuming 13 inches radius wheels) that impulses it forward.

Amazingly, when you have 25% or so less power (232 hp at max power vs 167 hp at max torque) your force on the ground, the force that really counts, is 10% larger (as I said, 595 kg of force at max power vs 660 kg of force at max torque)

So, your tires slip easier with max torque. The force on them is larger. QED

Of course, your wheels can develop, more or less, the same force as the weight of your car on them. You have two wheels that transmit power (unless you drive a 4x4), so each wheel receives 300 (max power) or 330 kg of force (max torque).

If your car, I don't know, weighs 1.000 kg (a number I picked because I can divide it by 4 easily) then each wheel has 250 kg of weight on it, assuming perfect 50/50 weight distribution (more like 25/25/25/25).

So, tyres will slip anyway, either at max torque or max power. The force you're trying to put on the tyres is larger than its weight (I'm also assuming you have tires with a friction factor of 1).

However, seriously: if you want to win your drag race, learn at what rpm point your tires slip (hear them screeching). Now, put the pedal to the metal and when you're reaching that rpm (which means you reached max torque), go easy on the throttle. Past that point, squeeeeeeeze again the throttle (gentle there, remember I said 10% difference: can you move your throttle forward in ten steps? I can ).

Why (I think) engineguru00 gives a different answer, in his experience? Easy: when you hold the clutch at mid-point, what you're doing is changing the final gear relationship. The clutch is slipping, so your 9 to 1 engine turns per wheel turn doesn't apply. What does it means? That his car has not enough power to slip the wheels, so, he's trying to apply more force (using more torque, burning some millimeters out of its clutch). Actually, I bet the car weighs more than 1.000 kg and/or tyres give him more than 1.0 coefficient of friction (in real life, high spec tyres give you something like 1.5 cf).

Finally, if I'm wrong (autogyro, are you listening?), thanks, people. I want to learn and answer the original question, instead of starting a row, no matter if you're right (and you are, JTom).
Ciro
Ciro Pabón

Joined: 10 May 2005

Figures are great Ciro, at least for working out how much torque or HP for different variables.
However if you applied 1000000 ft Ibs of torque or even Mexican jumping beans to a low enough gear ratio the tyre would not slip.
If the tyre had high enough traction though it would break the drive train.
autogyro

Joined: 4 Oct 2009

Oh, c'mon, autogyro, I said stop it. Please, man.

I gave the force at the wheels, first gear, regular car, regular wheels. For alelanza and his rx.

Show me what's wrong or where I am speculating, I'm not judging your answer, ehem (second bell). I'm asking to change the attitude of controversy for its own sake.

Torque. Got it? You don't agree, give a different answer, give us numbers instead of opinions. Change friction factor, change gear relations or whatever you want (preferably for real cars of this planet), and give us your tips to drag, engineer.

Check the practical, real life answer engineguru00 gave to us, or the dry, to the point one of JTom, if I may recommend something. We're all ears, but I don't like a bit useless controversies: this is about (or at least it tries to be) race engineering, not a contest of egos. If we wanted to be in that mood, we would be at Pitpass forum...

We don't recognize authority.
Ciro
Ciro Pabón

Joined: 10 May 2005

Ciro Pabón wrote:Oh, c'mon, autogyro, I said stop it. Please, man.

I gave the force at the wheels, first gear, regular car, regular wheels. For alelanza and his rx.

Show me what's wrong or where I am speculating, I'm not judging your answer, ehem (second bell). I'm asking to change the attitude of controversy for its own sake.

Torque. Got it? You don't agree, give a different answer, give us numbers instead of opinions. Change friction factor, change gear relations or whatever you want (preferably for real cars of this planet), and give us your tips to drag, engineer.

Check the practical, real life answer engineguru00 gave to us, or the dry, to the point one of JTom, if I may recommend something. We're all ears, but I don't like a bit useless controversies: this is about (or at least it tries to be) race engineering, not a contest of egos. If we wanted to be in that mood, we would be at Pitpass forum...

We don't recognize authority.

Sorry Ciro, I think I understand what you are trying to motivate and I fully support the engineering and the math.
I for one am not trying to cause controversy just for its own sake.
I am well aware of my habit of doing this, I do it hopeing to stimulate debate.
However, in this case, the question and thread heading do not give enough information to allow a meaningful answer IMO.
Please carry on the debate though, I would certainly like to be proven wrong.
I promise no more contradictory posts on this one Ciro.
autogyro

Joined: 4 Oct 2009

autogyro wrote:However if you applied 1000000 ft Ibs of torque or even Mexican jumping beans to a low enough gear ratio the tyre would not slip.

Yes it would.
Grip is a four letter word.

2 is the new #1.
Jersey Tom

Joined: 29 May 2006
Location: Huntersville, NC

Ciro Pabón wrote:Oh, my... Don't you have mercy? Well, I don't, so another long answer...

Long rant

Muchísimas gracias Ciro, that does put to rest the whole thing in my head, though I'll have to check the graph from home as imageshack is blocked here
And if you ever come back to San José let me know, we'll have a beer and i'll make sure to bring all 3 girls along for the ride, we'll also do a few burnouts just to impress them with the power of our torques
Alejandro L.
alelanza

Joined: 16 Jun 2008
Location: San José, Costa Rica

Ciro Pabón wrote:Why (I think) engineguru00 gives a different answer, in his experience? Easy: when you hold the clutch at mid-point, what you're doing is changing the final gear relationship. The clutch is slipping, so your 9 to 1 engine turns per wheel turn doesn't apply. What does it means? That his car has not enough power to slip the wheels, so, he's trying to apply more force (using more torque, burning some millimeters out of its clutch). Actually, I bet the car weighs more than 1.000 kg and/or tyres give him more than 1.0 coefficient of friction (in real life, high spec tyres give you something like 1.5 cf).

Car I do work on is a little under 200kg without the driver in race prep. It uses a belt driven Continuously Variable Transmission, so I have gear ratios rangine from 3.375:1 to 0.8:1 at the transmission. There is then a 428 chain drive to the rear end. Only slipping in the system is at the belt if I overcome its grip, but I haven't done this in a while. Tire data shows they have a 1.6 coefficient of friction, but my experience on the track is that they actually have a hot frictional coefficient of 1.4. Engine I currently run only has 50HP, but 45 ft-lbs of torque. Due to the light weight of the car, its actually fairly easy to slip the tires on the car with a fixed ratio gearbox (what most teams run) off the line. This doesn't happen with a CVT as the secondary is torque sensing and allows the clutches to simply shift to a higher ratio if it slips automatically. 0-60 time is in the high 3s to low 4s.

I will try to put up some of the calculations and plots that I have done to show what my post said earlier in more than tire dynamics fundamentals. I just have to blank out some names and numbers first on the tire plots. I am pretty sure we are both agreeing on the same point for the same reasons, but just explaining it in slightly different ways.
Chris
UB Motorsports: Formula SAE '08-'10
Captain '09-'10
engineguru00

Joined: 18 Mar 2010
Location: Buffalo,NY

What beats me is this: If tire slip is really bigger at maximum tourqe than at maximum power, why does keeping the revs as close as possible to maximum power while going through the gears, give faster acceleration than keeping them as close as possible to maximum tourque? I can't get my head around it, I guess I am just stupid.
Roland Ehnström

Joined: 10 Jan 2008
Location: Norrtälje, Sweden

Roland Ehnström wrote:What beats me is this: If tire slip is really bigger at maximum tourqe than at maximum power, why does keeping the revs as close as possible to maximum power while going through the gears, give faster acceleration than keeping them as close as possible to maximum tourque? I can't get my head around it, I guess I am just stupid.

Gears. Even though the torque rolls off after its peak... if you were to switch gears at that point, the torque at the wheels would be less than if you were to go slightly higher on the revs in a higher gear.
Grip is a four letter word.

2 is the new #1.
Jersey Tom

Joined: 29 May 2006
Location: Huntersville, NC

Well, I'm not sure, Tom. Power means you have a combination of torque and rotational speed. I'm not mechanical engineer, but when you have more power, the rotational speed increases more rapidly than when you have max torque. The force on the ground can be larger at max torque, but the rate at which the tyres increase the rpm is larger at max power. I'm not thinking about real life "details", like gears.

So, another answer to Roland clever point could be that you can have large torque and produce a large force but if your wheels doesn't increase the rotation rate more quickly you will lose. The original question is about slip, not acceleration (although I know the former brings the later, so...).

engineguru, thanks, very interesting, I was wrong, thinking on what I read about keeping the engine at 8.000 rpm before hitting the clutch, we wait for your acclarations. Thanks.

Alejandro, I'm changing the host of the image and editing your quote of my post. My host wasn't working when I posted and the quote was too long. Thanks for your kind words, I might be wrong, I'm not a professional, let's wait for the "big guys" judgement, it's just an opinion.
Ciro
Ciro Pabón

Joined: 10 May 2005

PreviousNext