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

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mep wrote:It is a function of load.

And so is a contact patch.
timbo

Joined: 22 Oct 2007

autogyro, I find it fascinating that someone can produce 367 postings, all completely void of any engineering substance?
"Bernoulli is a nine-letter name"
xpensive

Joined: 22 Nov 2008

xpensive wrote: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?

My concern is that using the energy-method is it even possible to calculate wheel-slip? The only time I've ever seen this worked out was using the method I've been using. If there's another way, fine. I was under the impression that the energy-method was useful when the vehicle was not traction limited, but at all other times you should use the force-based method (what I've been doing)?

'10-'11 Head of Powertrain - Glasgow University Formula Student
Scotracer

Joined: 22 Apr 2008
Location: Edinburgh, Scotland, UK

My comment was with regards to wheel-torque, when if you know the wheel-radius, you also get the thrust from that.

What wheel-radius did you use anyway?
"Bernoulli is a nine-letter name"
xpensive

Joined: 22 Nov 2008

xpensive wrote:My comment was with regards to wheel-torque, when if you know the wheel-radius, you also get the thrust from that.

What wheel-radius did you use anyway?

Current F1 slick diameter, 660mm.

'10-'11 Head of Powertrain - Glasgow University Formula Student
Scotracer

Joined: 22 Apr 2008
Location: Edinburgh, Scotland, UK

OK. Your graph explains the reason for wheel-spin from zero speed as wheel-torque from 480kW becomes theoretically enormous at very low speeds. If we would toy with the idea of a mu of 2.5 and a rear wheel contact-load of 5 kN, maximum thrust is then 12.5 kN, which equals a wheel-torque of 4125 Nm with the 660 mm wheels.

Still with 480 kW of power, said thrust would be at 38.4 m/s, or 138 km/h.

In other words, under the above conditions with 480 kW applied all the way, the F1 car would have wheel-spin up to the speed of 138 km/h, is that reasonable?
"Bernoulli is a nine-letter name"
xpensive

Joined: 22 Nov 2008

xpensive wrote:autogyro, I find it fascinating that someone can produce 367 postings, all completely void of any engineering substance?

I think you will find that the comment is more accurately applied to you, when everyone realizes that your question cannot be answered.
autogyro

Joined: 4 Oct 2009

Not by you anyway, that much is for certain.
"Bernoulli is a nine-letter name"
xpensive

Joined: 22 Nov 2008

xpensive wrote:Not by you anyway, that much is for certain.

I am way out of my depth in the company of brilliance such as yours.
I will not post again on anything.
I will let you ol boys carry on playing with your Math
Vrrm Vrrm
autogyro

Joined: 4 Oct 2009

xpensive wrote:OK. Your graph explains the reason for wheel-spin from zero speed as wheel-torque from 480kW becomes theoretically enormous at very low speeds. If we would toy with the idea of a mu of 2.5 and a rear wheel contact-load of 5 kN, maximum thrust is then 12.5 kN, which equals a wheel-torque of 4125 Nm with the 660 mm wheels.

Still with 480 kW of power, said thrust would be at 38.4 m/s, or 138 km/h.

In other words, under the above conditions with 480 kW applied all the way, the F1 car would have wheel-spin up to the speed of 138 km/h, is that reasonable?

Is the mu of 2.5 that of pure sliding friction or rolling?
I have a feeling this coefficient is this high because it may represent static wheels scraping along a surface, (braking coefficient like in a lock up) instead of rolling with localized slippage and deformation at the contact patch.

The wheel torque has to be a limiting value, it must not be worked back from another equation.
As you know, torque is limited to the gearing at what ever road speed and engine speed. So it should be like a given parameter.

From dyno chart, you have power and torque at what ever speed you want.
you know your torque to the wheel axle; only the engine can change this, anything more will decelerate or just stall the engine. ie if mu is increased to the point where the engine cannot deliver it will simply stall, kind of like a big load on an electric circuit.

this is how i suppose could work,

Power = T x omega (angular velocity), (1)
omega = rpm/60, = V wheel (V road x slip) / wheel radius

Net Thrust = Force at radius (using 2)- rolling resistance at wheels ( reaction force x coef of rolling resistance)

Net Thrust x wheel radius = I wheel x angular accel, (I = second moment of inertia , alpha = r x accel at road.)

at top speed, drag force = thrust, no acceleration so alpha = 0

from this: thrust x wheel radius = I x alpha = 0

so it follows, 0 = (force at radius - rolling resistance) x radius

the torque to the wheels balances the rolling resistance. The torque component of the engine is still there working to stay up to speed, just within the wheel system.
For Sure!!
ringo

Joined: 29 Mar 2009

Those relationships seems both reasonable as well as a good start ringo, try aome numerical values perhaps?

The issue of static vs dynamic friction, or stick-slip if you wish, is always interesting, any ideas there?
"Bernoulli is a nine-letter name"
xpensive

Joined: 22 Nov 2008

xpensive wrote:Still with 480 kW of power, said thrust would be at 38.4 m/s, or 138 km/h.

In other words, under the above conditions with 480 kW applied all the way, the F1 car would have wheel-spin up to the speed of 138 km/h, is that reasonable?

I'm getting close to that same figure plotting what I imagine is thrust available from the engine vs. thrust able to go down on the ground. They intersect @ ~39 m/s.

Conclusion? Some combination of:
1) your favorite driver is pussyfooting it up to this speed
2) he's flat before then cause he's brave or a little stupid
3) the squat helps
4) downforce is more than we think (or I'm guesstimating)
Loud idiot in red since 2010
United States Grand Prix Club, because there's more to racing than NASCAR
jon-mullen

Joined: 10 Sep 2008
Location: Big Blue Nation

That's what I thought the graph would look like!

'10-'11 Head of Powertrain - Glasgow University Formula Student
Scotracer

Joined: 22 Apr 2008
Location: Edinburgh, Scotland, UK

Sure.

SUM(Fy) = 4N - mg - (1/2)*d*A*L*v^2

So the normal N at one wheel is, N = (mg + (1/2)*d*A*L*v^2)/4

So the maximum the thrust at one wheel can be (red line on graph) is given by mu * N.

That's with AL = 4 m^2, d = 1.2 kg/m^3, m = 605 kg, mu = 2.5

If anyone doesn't like those assumptions, here's the Excel so you can play with it yourself.

And I'll go bump autogyro's Transmission Concept thread since he's obviously dying to tell you about it.
Loud idiot in red since 2010
United States Grand Prix Club, because there's more to racing than NASCAR
jon-mullen

Joined: 10 Sep 2008
Location: Big Blue Nation

When I originally asked I didn't realise, from my own doofusity, that you'd already put them in the graph legend

Those look fine to me. Excellent work, gang!

'10-'11 Head of Powertrain - Glasgow University Formula Student
Scotracer

Joined: 22 Apr 2008
Location: Edinburgh, Scotland, UK

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