Power vs Torque
Discussion
I felt this may be an interesting topic for conversation.
I have asked a few pro's in the game of mapping/dyno's and an ecu manufacturer but not found a precise the answer to;
"how much power can we put down before we lose traction"
Obviously it depends how much load is on the tyres, the tyre width, weight transfer, road conditions etc.
If you take an average 7 type car as an example. Call it 600kgs with fuel and driver.
25% weight distribution to make calcs easier so 150kg on each tyre.
Call it a dry road with road tyres that are say 200mm wide.
To make it easier lets assume there is no loss in the drive due to transmission/bearings etc.
If you dial in weight transfer this may confuse things.
Ratio wise, lets call it 5-1 via first gear and 4-1 via the open (not LSD) diff.
If we assume we have just 100 ft lbs at 6000 rpm, x by 5 in first gear =500, x4 for the diff and that gives 2000 ft lbs of torque at the tyre contact point assuming the wheel is 2 feet dia. That would be a one foot lever from hub centre to the road.
We then need to divide this by the 2 wheels.
I select LAUNCH control or just hit the gas up to 6000 rpm, flick my foot off the clutch and hold tight.
Question is, how much torque can we apply before we lose traction. Opinions very welcome.
Clutch slip will soften the hit on the tyres and rpm/torque will not be constant, but say at the maximum, how many ft lbs will those tyres transmit? Please let me know if I have missed something.
Maybe modern traction control cars have a memory showing how much torque in which gear was available at the point of slip.
I have asked a few pro's in the game of mapping/dyno's and an ecu manufacturer but not found a precise the answer to;
"how much power can we put down before we lose traction"
Obviously it depends how much load is on the tyres, the tyre width, weight transfer, road conditions etc.
If you take an average 7 type car as an example. Call it 600kgs with fuel and driver.
25% weight distribution to make calcs easier so 150kg on each tyre.
Call it a dry road with road tyres that are say 200mm wide.
To make it easier lets assume there is no loss in the drive due to transmission/bearings etc.
If you dial in weight transfer this may confuse things.
Ratio wise, lets call it 5-1 via first gear and 4-1 via the open (not LSD) diff.
If we assume we have just 100 ft lbs at 6000 rpm, x by 5 in first gear =500, x4 for the diff and that gives 2000 ft lbs of torque at the tyre contact point assuming the wheel is 2 feet dia. That would be a one foot lever from hub centre to the road.
We then need to divide this by the 2 wheels.
I select LAUNCH control or just hit the gas up to 6000 rpm, flick my foot off the clutch and hold tight.
Question is, how much torque can we apply before we lose traction. Opinions very welcome.
Clutch slip will soften the hit on the tyres and rpm/torque will not be constant, but say at the maximum, how many ft lbs will those tyres transmit? Please let me know if I have missed something.
Maybe modern traction control cars have a memory showing how much torque in which gear was available at the point of slip.
Edited by Stuart Mills on Thursday 18th September 14:21
It's easiest to work out the maximum linear force at the contact patch and work backwards towards the engine from that if you want to work out an accurate figure in isolation.
I think what launch control does, is have a hard-coded rough figure and then use a very fast acting traction control type system to keep the correct slip.
I think what launch control does, is have a hard-coded rough figure and then use a very fast acting traction control type system to keep the correct slip.
Edited by kambites on Thursday 18th September 14:13
What a good Thursday afternoon question!
It all starts with a very simple formula:
F<u.R
F is friction, u is the coefficient of friction and R is reaction
At the limit, ie when the wheel is just about to start slipping the maximum friction force becomes;
F = u. R
R, the reaction, is the weight on the wheel, so more weight (rear engined 911) means more Reaction which leads to a greater Force before the wheel will spin. Your caterham 7’s Reaction per wheel you state as being 250 kg. 250 kg is 2500 Newtons
Working out u is the tricky bit. Although there is no mention of surface area (ie width of tyre) in the basic F = u.R formula it comes into play when determining a value for u. Very hard rubber has a low coefficient of friction, but the softer (and stickier) the rubber then the greater the coefficient of friction. At some point the grip will be lost not by slipping the tyre over the road, but by shearing off small layers of tyre rubber or small layers of road (black stripes down the road or piles of gravel being thrown backwards). This is when width of tyre starts being important as more width means more shear force. The balance point for tyre design u value therefore depends not only on the tyre rubber but also on the road material, the temperature and the weight on the tyre, plus if the road is slippery (wet) or not.
Anyway, for an upper bound calculation I’m pretty sure you can use a value for u of 1.0 (perfect friction), which is the best you’ll ever get with perfectly balanced tyre design on a decent dry tarmac.
So F = u. R can now be written as F = 1.0 x 2500 N
Which is a nice easy calculation to give F = 2500 N
To get from F to torque is simply; Torque = F. Radius of Wheel
Radius you state as 1 foot, which is 0.3 metres
Torque = 2500 x 0.3 = 750 N.m
Converting Newton.metres into lbs.feet is done by google to give 750N.m = 550 lbs.feet of torque. This is the torque at the wheel hub.
2 wheels are driven so total torque to be supplied to the wheels to make them loose traction is 2 x 550 = 1100 lbs.feet
The engine you say delivers 100 lbs.feet of torque at 6000 rpm and the gearbox and final drive reduce engine rotation into wheel hub rotation by a combined factor of 200. Ie in first gear the wheels turn 200 times slower than the engine. This factors up the torque at the hubs by 200, so as you say there is 2000 lbs.feet of torque available.
2000 lbs.feet of available torque > 1100 lbs.feet, so I suspect you can spin the wheels on your caterham 7 by revving to 6000 rpm and dropping the clutch!
The amount of acceleration this will provide is another simple formula; F = m. a
F is the same F as above = 2500 N per driven wheel, or 5000 N in total, and M is the mass of the car plus driver; M = 600 kg
2500 = 600. a
Or jiggling around; a = 2500/600 = 4.2 m/s, which is nearly half a g in acceleration.
It all starts with a very simple formula:
F<u.R
F is friction, u is the coefficient of friction and R is reaction
At the limit, ie when the wheel is just about to start slipping the maximum friction force becomes;
F = u. R
R, the reaction, is the weight on the wheel, so more weight (rear engined 911) means more Reaction which leads to a greater Force before the wheel will spin. Your caterham 7’s Reaction per wheel you state as being 250 kg. 250 kg is 2500 Newtons
Working out u is the tricky bit. Although there is no mention of surface area (ie width of tyre) in the basic F = u.R formula it comes into play when determining a value for u. Very hard rubber has a low coefficient of friction, but the softer (and stickier) the rubber then the greater the coefficient of friction. At some point the grip will be lost not by slipping the tyre over the road, but by shearing off small layers of tyre rubber or small layers of road (black stripes down the road or piles of gravel being thrown backwards). This is when width of tyre starts being important as more width means more shear force. The balance point for tyre design u value therefore depends not only on the tyre rubber but also on the road material, the temperature and the weight on the tyre, plus if the road is slippery (wet) or not.
Anyway, for an upper bound calculation I’m pretty sure you can use a value for u of 1.0 (perfect friction), which is the best you’ll ever get with perfectly balanced tyre design on a decent dry tarmac.
So F = u. R can now be written as F = 1.0 x 2500 N
Which is a nice easy calculation to give F = 2500 N
To get from F to torque is simply; Torque = F. Radius of Wheel
Radius you state as 1 foot, which is 0.3 metres
Torque = 2500 x 0.3 = 750 N.m
Converting Newton.metres into lbs.feet is done by google to give 750N.m = 550 lbs.feet of torque. This is the torque at the wheel hub.
2 wheels are driven so total torque to be supplied to the wheels to make them loose traction is 2 x 550 = 1100 lbs.feet
The engine you say delivers 100 lbs.feet of torque at 6000 rpm and the gearbox and final drive reduce engine rotation into wheel hub rotation by a combined factor of 200. Ie in first gear the wheels turn 200 times slower than the engine. This factors up the torque at the hubs by 200, so as you say there is 2000 lbs.feet of torque available.
2000 lbs.feet of available torque > 1100 lbs.feet, so I suspect you can spin the wheels on your caterham 7 by revving to 6000 rpm and dropping the clutch!
The amount of acceleration this will provide is another simple formula; F = m. a
F is the same F as above = 2500 N per driven wheel, or 5000 N in total, and M is the mass of the car plus driver; M = 600 kg
2500 = 600. a
Or jiggling around; a = 2500/600 = 4.2 m/s, which is nearly half a g in acceleration.
Dynamically a typical tyre can achieve a Mu of well over 1.
And, you need to include the inertia of the engine, clutch, drivetrain and crucially, the wheel/tyre package.
Then we get to the thorny issue of dynamic weight transfer under acceleration (which is makes it easier to continue to accelerate at an increase rate than start accelerating in the first place!)
And that's before we consider drivetrain torsional stiffness, and suspension geometry (anti squat, dynamic camber change etc)
And, you need to include the inertia of the engine, clutch, drivetrain and crucially, the wheel/tyre package.
Then we get to the thorny issue of dynamic weight transfer under acceleration (which is makes it easier to continue to accelerate at an increase rate than start accelerating in the first place!)
And that's before we consider drivetrain torsional stiffness, and suspension geometry (anti squat, dynamic camber change etc)
Nice walkthrough Lorneg, though I did notice you said the overall drivetrain reduction is a factor of 200 - 20 seems more reasonable, to avoid confusion 
Max Torque is of course right about all the complexities involved in simulating this accurately, but for discussion here it's probably best to simplify, dropping weight transfer and rotational inertia effects at least.
Here's something that might interest you, it's called a road load graph:
Unfortunately this example lacks a scale, but what it shows is the available tractive effort in each gear against road speed. This is effectively the engine's torque curve scaled to equate the available rev range to the roadspeed in each gear, and then multiplied by the reduction that gear gives.
Also shown is a "road resistance" curve (which I would call running resistance, but Google beggars can't be choosers) - this is the sum of all drag forces on the car, namely aerodynamic drag and tyre rolling resistance and windage. This one is assuming level ground and no headwind but of course you could add terms for those too. At the point where the running resistance curve crosses the tractive effort curve, your resistance equals your available motive force, so you're at your top speed.
For this topic, I've thrown on a red line which is the maximum tractive effort you can transmit for a given tyre on a given road surface. For most road cars this will be more or less constant with increasing speed, though of course if you generated significant downforce it would curve upwards as more vertical load became available as the speed rises. Notice that this arbitrary example is capable of overcoming tyre grip in first gear, but the other gears are nowhere close.
You can produce curves like this for your own car, all you need is an extremely basic torque map (a dozen points across the rev range would do it) and the gear ratios
Max Torque is of course right about all the complexities involved in simulating this accurately, but for discussion here it's probably best to simplify, dropping weight transfer and rotational inertia effects at least.
Here's something that might interest you, it's called a road load graph:
Unfortunately this example lacks a scale, but what it shows is the available tractive effort in each gear against road speed. This is effectively the engine's torque curve scaled to equate the available rev range to the roadspeed in each gear, and then multiplied by the reduction that gear gives.
Also shown is a "road resistance" curve (which I would call running resistance, but Google beggars can't be choosers) - this is the sum of all drag forces on the car, namely aerodynamic drag and tyre rolling resistance and windage. This one is assuming level ground and no headwind but of course you could add terms for those too. At the point where the running resistance curve crosses the tractive effort curve, your resistance equals your available motive force, so you're at your top speed.
For this topic, I've thrown on a red line which is the maximum tractive effort you can transmit for a given tyre on a given road surface. For most road cars this will be more or less constant with increasing speed, though of course if you generated significant downforce it would curve upwards as more vertical load became available as the speed rises. Notice that this arbitrary example is capable of overcoming tyre grip in first gear, but the other gears are nowhere close.
You can produce curves like this for your own car, all you need is an extremely basic torque map (a dozen points across the rev range would do it) and the gear ratios
Max_Torque said:
Dynamically a typical tyre can achieve a Mu of well over 1.
And, you need to include the inertia of the engine, clutch, drivetrain and crucially, the wheel/tyre package.
Then we get to the thorny issue of dynamic weight transfer under acceleration (which is makes it easier to continue to accelerate at an increase rate than start accelerating in the first place!)
And that's before we consider drivetrain torsional stiffness, and suspension geometry (anti squat, dynamic camber change etc)
ahh, Mu can be greater than 1. I was thinking the calculated acceleration was a bit wimpy, but it was a simplified attempt at doing it and the answer came out in a reasonable ball-park so I thought I'd go with it. Normally when I try this type of calculation I end up with an answer of zero or infinity!And, you need to include the inertia of the engine, clutch, drivetrain and crucially, the wheel/tyre package.
Then we get to the thorny issue of dynamic weight transfer under acceleration (which is makes it easier to continue to accelerate at an increase rate than start accelerating in the first place!)
And that's before we consider drivetrain torsional stiffness, and suspension geometry (anti squat, dynamic camber change etc)
The complications surround the fact that a tyre does not follow "classical friction", but a strange combination of multiple effects, that occur at different speeds. Google "Magic tyre model" for more info.
Generally the "static friction" is extremely high, as the relatively soft stationary tyre has plenty of time to physically deform around the rough road surface and "mechanically" grip that surface. As soon as it starts to revolve, the effective friction co-efficient falls dramatically. The as vehicle speed builds, dynamic effects, such as windage (or "waterage" if it's raining) start to occur, and in fact, the tyre contact patch changes shape as the different loadings on it wax and wane (ie centripedal acceleration of the tyre carcass vs normal load from the vehicles mass minus any lift forces (most road cars have positive lift co-efficients btw)
Generally the "static friction" is extremely high, as the relatively soft stationary tyre has plenty of time to physically deform around the rough road surface and "mechanically" grip that surface. As soon as it starts to revolve, the effective friction co-efficient falls dramatically. The as vehicle speed builds, dynamic effects, such as windage (or "waterage" if it's raining) start to occur, and in fact, the tyre contact patch changes shape as the different loadings on it wax and wane (ie centripedal acceleration of the tyre carcass vs normal load from the vehicles mass minus any lift forces (most road cars have positive lift co-efficients btw)
McSam said:
You can produce curves like this for your own car, all you need is an extremely basic torque map (a dozen points across the rev range would do it) and the gear ratios
yeah, but I'm a middle aged bloke with an old SL, so I'm not interested in acceleration and top speed anymore. Way back when though I would be able to put a scale to the diagram:Maximum possible car speed = 168 mph (911 carrera with a starchip engine management upgrade). Knackers the gearbox though.
Stuart Mills said:
"how much power can we put down before we lose traction"
Let's be clear here. At launch (i.e. zero vehicle speed) you can put down zero power - that is, unless you count the relatively useless power that is warming the tyres and pavement.Of course, if you could travel at a million miles an hour, then a 155 section tyre could put down Megawatts, at least for a short time before it melted or exploded from the centripetal accelerations...
So, there's your upper and lower bounds, approximately speaking.
That's why it's important we don't consider power, but instead tractive effort, the force at the wheels. Power is force times velocity, hence it being zero at rest and massive at high speeds, but force is an absolute and is simply a function of the engine's torque and the drivetrain reduction. Much easier to deal with.
If you want a dead quick and easy way to determine how much engine torque can be applied before you lose grip, this will do it:
Tractive effort = engine torque (Nm) * overall gear reduction / tyre rolling radius (m)
Maximum tractive effort to be supported before the tyres are overwhelmed = vertical load on driven wheels (N) * adhesion coefficient of tyre to road (typically around 1 for road tyres on tarmac)
If the first figure is larger, you'll spin up. If not, sufficient grip (or insufficient power, in my opinion )
If you want a dead quick and easy way to determine how much engine torque can be applied before you lose grip, this will do it:
Tractive effort = engine torque (Nm) * overall gear reduction / tyre rolling radius (m)
Maximum tractive effort to be supported before the tyres are overwhelmed = vertical load on driven wheels (N) * adhesion coefficient of tyre to road (typically around 1 for road tyres on tarmac)
If the first figure is larger, you'll spin up. If not, sufficient grip (or insufficient power, in my opinion
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