## Traction-force as a function of Power and Speed.

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Quite obviously Scot, with a continous power of 480 kW, thrust at zero speed becomes infinite, hence wheel-spin.
But what happens to the power?

Power will never be anything but force times speed, why power now will be thrust times peripheral wheel-speed.

Thoughts on static frcition vs dynamic such, JT?

"Bernoulli is a nine-letter name"
xpensive
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Joined: 22 Nov 2008

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xpensive wrote:Thoughts on static frcition vs dynamic such, JT?

Not quite as simple as static and dynamic as Coulomb would describe. Can't disclose much here, but there is public domain information on this subject, in the field of tribology.

Hint: a Google search of 'rubber friction velocity' has some interesting results.
Grip is a four letter word.

2 is the new #1.
Jersey Tom
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Joined: 29 May 2006
Location: Huntersville, NC

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Just a few graphs of this seemingly perfect CVT. I don't know if this is correct so anyone with experience feel free to tear apart...

So, if anyone can at least give me the equation to work out maximum torque a wheel can take before spinning I can work out the minimum speed needed to run full throttle. I hope!

'10-'11 Head of Powertrain - Glasgow University Formula Student
Scotracer
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Joined: 22 Apr 2008
Location: Edinburgh, Scotland, UK

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Power also= Torque x angular velocity. Which is useful if considering gear ratio and engine speed.
In a perfectly mechanical case like a sprocket and chain, it could account for the wheel forces.
However with a slipping wheel, the difference of the measured power to the ground and that of the measured power to the wheel axle should give wasted power to slipping. Energy wasted on deformation can also be considered.
Should be able to work back from there and get the frictional force and coefficient.

Concerning zero slip on the wheels at terminal speed, i don't think that is the case, though i could be wrong. It only makes sense if the wheel forces and the drag forces are in the same system.
The tyre has to deform to generate wheel forces, and with the engine putting torque to deform the tyre, and with the friction with the road still present, slip is also still present.
The aerodynamic force balances the net force of the tyre system, but within the tyre system the slip is still present.
If my reasoning is wrong correct me, I am not really knowledgeable on tyres.
For Sure!!
ringo
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Joined: 29 Mar 2009

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Torque, engine or wheel-wise, is insignificant here, Scot.Power is force times speed, nothing else.

Jon?
"Bernoulli is a nine-letter name"
xpensive
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Joined: 22 Nov 2008

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Torque, engine or wheel-wise, is insignificant here, Scot. Power is force times speed, nothing else.

Jon?
"Bernoulli is a nine-letter name"
xpensive
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Joined: 22 Nov 2008

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xpensive wrote:Torque, engine or wheel-wise, is insignificant here, Scot. Power is force times speed, nothing else.

Jon?

Torque is what defines wheel-spin! Power = Torque x Engine speed. Power and torque are different measures of the same thing.

To calculate the speed at which you can run full throttle the best way is to work out the point at which the Critical Torque* crosses the line in this graph:

*Here I am defining Critical Torque as the torque required to slip the rear wheels beyond their working slip region.

'10-'11 Head of Powertrain - Glasgow University Formula Student
Scotracer
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Joined: 22 Apr 2008
Location: Edinburgh, Scotland, UK

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xpensive wrote:Torque, engine or wheel-wise, is insignificant here, Scot.Power is force times speed, nothing else.

Jon?

why is it irrelevant? Maybe i misread what is being investigated?

Engine power was mentioned. It makes sense to me if you are talking about slip.
For Sure!!
ringo
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Joined: 29 Mar 2009

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Scotracer wrote:
xpensive wrote:Thanx Scot, most eduacative graphs. One of my points is that applying 480 kW at zero speed would lead to infinite thrust.

So, what happens to all that power at zero-speed wheel-spin? Rather simple actually, power is still force times speed, but the speed is under such circumstances is the peripheral speed of the wheel itself.

"Friction" for dicsussion, dynamic such vs static, anyone?

Btw, thank you for giving space to old-fart thinking, Ciro.

I'm doing a little piece for this thread at the moment. With a CVT you really shouldn't get any wheelspin as a correctly designed one will be preset for the peak torque the back wheels can take under acceleration. I am instead doing a scenario where it's a conventional gearbox. I'm having to take a few liberties on the assumptions so bear with me. I should have it up later.

I think it's safe to assume that to keep the idea of 480kW @ 0m/s possible we have a clutch involved. I don't think even tyres of mu = 2.5 are quite ready for infinite thrust

Replace the clutch with electric traction from 0m/s for the first few yards, using a dual ratio from two ratios used together in the gearbox of around 5.5:1.
Then when inertia is established, use a slightly taller first gear ratio for stepped first (IC) followed by six more finishing with direct top.
Do not bother with a CVT, it would need to much power applied to the toroidal traction fluid to ever be efficient.
The Math should be interesting.
autogyro
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Joined: 4 Oct 2009

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I typically find it wise to try and minimize parameters involved. When "Torque at wheel", is simply "Thrust" times "Wheel-radius", can we stick to "Thrust" to avoid confusion?
"Bernoulli is a nine-letter name"
xpensive
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Joined: 22 Nov 2008

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Xpensive, you participated in the previous thread where we posted a crude calculation of what I called the "power to the wheels".

The yellow curve you ask for, comes from this calculations, using essentially the same equations Scotracer used.

I pasted a section of the graph of speed vs time in AutoCAD. Then I traced vertical black lines every second. I callibrated the scale and with the coordinates of the points on the graph every second, I got the speed, like this:

From that figures I estimated the power to the wheels, using a frontal area and a drag coefficient. This were the results posted in that thread:

The yellow line is a graph of that estimation of total power to the wheels, that is, row 22 in the previous picture.

This gives an answer to your question. You can see that, when the car is at low speeds, most of the power is used to overcome inertia (that is, to move the damn car). At high speeds most of the power is used to overcome drag. The force at low speeds is around 700 kilograms. If you repeat the calculations at lower speeds (one of the peaks go down to 70 kph or so), you might get 900 kg, tops.

Autogyro: the tyres slip to move the car, in the sense explained in this thread, something you probably know: Must a tire slip to generate force?

I think this limits the acceleration you can give to the car, CVT or not. That 2.5 mu coefficient Jersey Tom is talking about is only to move the car laterally.

This means that even if you cannot see it the tyre is sliding "under the patch" to propel the car ahead. The ultimate power you can deliver to move ahead the car, the thrust that xpensive asks for, is limited by that "inner friction", invisible: anyway you won't get over 1.5 Gs, I think.

The 2.5 mu that JTom mentions only works to move laterally the car. You can easily see from Reca's graph, from the top of your head, that when the car is accelerating, it's not limited by the gearbox, but by the rubber, if you follow my drift.
Ciro
Ciro Pabón
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Joined: 10 May 2005

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I'm sorry Ciro, but your yellow curve still makes very little sense to me, when it claims the car to have only 150 hp at the wheels at 150 km/h? Equally, 400 hp would not be anywhere near enough to propel an F1 car to 300 km/h.

Anyway, in order to take a 700 kg object from zero to 100 km/h, or 27.8 m/s, in 1.50 seconds, you would need a constant accelleration of 18.5 m/s^2, or almost 2g, which in turn should ask for a continous thrust of 13.0 kN

Even if you have all the 700 kg on the rear wheels, that would still call for a mu of 1.89.
"Bernoulli is a nine-letter name"
xpensive
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Joined: 22 Nov 2008

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xpensive wrote:Even if you have all the 700 kg on the rear wheels, that would still call for a mu of 1.89.

And what's the problem? Rubber sticks to asphalt and it does not follow Amontons' 2nd law. In wiki it is quoted that rubber on concrete has mu of 1.7 and F1 tyre heated to 100degC can easily produce higher value.
timbo
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Joined: 22 Oct 2007

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timbo wrote:
xpensive wrote:Even if you have all the 700 kg on the rear wheels, that would still call for a mu of 1.89.

And what's the problem? Rubber sticks to asphalt and it does not follow Amontons' 2nd law. In wiki it is quoted that rubber on concrete has mu of 1.7 and F1 tyre heated to 100degC can easily produce higher value.

mu 1.82?

Torque at the wheels is the key.
With a lower gear ratio the torque is higher for less engine power.
Maximum tyre limited acceleration is then possible at lower speeds with less power applied.
Now design a gearbox that is not a torque sapping CVT to do the job.

Oh and electric vehicles can also benefit hugely from gearbox technology, it is just that the electronic engineers and designers do not know it yet.
autogyro
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Joined: 4 Oct 2009

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One important thing everybody seems to forget here is that the friction coefficient (µ) of a tire is not a constant figure.
It is a function of load.
Increasing the load on the tire will decrease the friction coefficient (µ).
A value of 2,5 might be possible when you don’t put loads on the tire.
Try to imagine a chewing gum sticking to the road. It has almost zero load but you need a lot of force to get it removed. So the calculated friction coefficient for this condition would be pretty high, but that is not a condition we can use for a racecar. Putting the slightest load on your tire means that the friction coefficient will dramatically degrease. You will easily end up with values around 1. In fact that is the reason why a light and well balanced racecar performs better than a heavy one.
mep
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Joined: 11 Oct 2003
Location: Germany

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