Traction-force as a function of Power and Speed.

[xpensive]

There was an interesting discussion on a now long since forgotten thread about how much traction-force the F1 tyres can handle and at what downforce, ie what is the "coefficient of friction". I've given the issue some thought and came up with the following simplified model, where rolling friction is set to nil:

Dynamics tells us that Power (Watt) is Force (Newton) times Speed (Meters per second). If we imagine that the car has a continous wheel-power of 480 kW (650 Hp), through a magic-CVT, the traction-force will theoretically come down with speed going up according to this:

At 20 m/s (72 km/h), traction-force is 24 kN, at 30 m/s-16 kN, at 40 m/s-12 kN and so on to the point where traction-force balances the air-resistance, at some 300 km/h.

As the speed goes up, downfoce will increase of course, but at what speed does the wheel-spin stop and how much is the vertical contact-force between the rear wheels and the surface at that speed?

[autogyro]

There was an interesting discussion on a now long since forgotten thread about how much traction-force the F1 tyres can handle and at what downforce, ie what is the "coefficient of friction". I've given the issue some thought and came up with the following simplified model, where rolling friction is set to nil:

Dynamics tells us that Power (Watt) is Force (Newton) times Speed (Meters per second). If we imagine that the car has a continous wheel-power of 480 kW (650 Hp), though a magic-CVT, the traction-force will theoretically come down with speed going up according to this:

At 20 m/s (72 km/h), traction-force is 24 kN, at 30 m/s-16 kN, at 40 m/s-12 kN and so on to the point where traction-force balances the air-resistance, at some 300 km/h.

As the speed goes up, downfoce will increase of course, but at what speed does the wheel-spin stop and how much is the vertical contact-force between the rear wheels and the surface at that speed?

Give the geartrain the right range of gear ratio's either CVT, TVT or stepped and there will not be any wheel spin.

[Ciro Pabón]

Well, for the car to accelerate you have to have wheel-spin (sort of ).

I suppose you refer to this thread, X: viewtopic.php?f=1&t=6451

There it was shown this graph, analyzing the sped vs time for Fisichella at Albert Park in 2007 (data extracted by Reca using a Fourier transformation of the sound of the engine, taken from a video, to get the rpm and using gearbox ratios already known):

Speed vs time for Fisichella


The rolling force is neglectable, or a very approximate calculation shows this (sorry, the red box means nothing, check the "Rolling power" cell at position B20: a mere 9400 watts or miserable 13 hp).

Power to the wheels simple calculation


From that, it was relatively simple to get this, by iterating the same procedure between different points in that "peak":

By "Fourth acceleration" I mean the fourth peak from left to right in the previous picture.


Nice, I've forgotten that one. However, I'd say this:

Wheel spin stops at top speed (acceleration equals zero).

The total vertical force is more or less 5 kilonewtons at 250 kph (cell B13). I know I have the worksheet interred somewhere, I'll check as soon as I have time.

You can reconstruct the same worksheet, the equations are in the C column: those are approximations, but the figures must be pretty good, they are taken from the first graph, that should be more or less accurate.

The main uncertainty is in the drag coefficient, but nobody protested back then about that number... perhaps someone will find an error (thanks in advance! I'd appreciate that) or maybe someone can provide a better figure for drag coefficient now.

The number is in the range of downforce I've read elsewhere. Anyway, from the speed/time graph you can extract whatever you want about the force, the power, etc. Enjoy.

[WhiteBlue]

The amount of traction that tyres can handle is be determined by their geometry, construction and compound. I think it is also interesting to discuss how those factors will change in 2010 and what it will do to the traction.

First of all the traction is dependent of the temperature and the state of the tread. If the tyres are already run down the capability of the tread to absorb heat and dispose of it will quickly disappear. The traction will break down very quickly from there.

Next years front tyres will be stiffer in construction and will have harder compounds according to Bridgestone. The contact patch will be smaller. The two aspects are meant to compensate each other but as always there will probably not be equilibrium and the traction will either go up or down.

My guess at this point is that tyres will have initially less traction (tyre blankets will be banned) and that the traction loss over the distance will be reduced compared to todays tyres. It would make for a more durable tyre which necessitates fewer pit stops than the 2009 front tyres. Having said that, one has to consider that front tyre wear hasn't been that critical for a long time. I think that the last time that was an issue was Istanbul 2007 with Hamilton's car.

So the real question might be how the stiffer front tyres will impact on the rear tyres . Those are the wheels which suffer from degradation and blistering earlier than the front tyres. I guess that they will also profit from a harder compound and rear tyre blistering will be less of an issue next year.

[jon-mullen]

Give the geartrain the right range of gear ratio's either CVT, TVT or stepped and there will not be any wheel spin.

CVT? Is that a Cat-Valued Thread? I thought you'd seen enough of those.

There was an interesting discussion on a now long since forgotten thread about how much traction-force the F1 tyres can handle and at what downforce, ie what is the "coefficient of friction". I've given the issue some thought and came up with the following simplified model, where rolling friction is set to nil:

Dynamics tells us that Power (Watt) is Force (Newton) times Speed (Meters per second). If we imagine that the car has a continous wheel-power of 480 kW (650 Hp), though a magic-CVT, the traction-force will theoretically come down with speed going up according to this:

At 20 m/s (72 km/h), traction-force is 24 kN, at 30 m/s-16 kN, at 40 m/s-12 kN and so on to the point where traction-force balances the air-resistance, at some 300 km/h.

As the speed goes up, downfoce will increase of course, but at what speed does the wheel-spin stop and how much is the vertical contact-force between the rear wheels and the surface at that speed?

Thanks, x, finally I can review for my Physics final on Monday and be on F1Tech at the same time! This is what my sad mix of kinematics, Newton's Laws and Work-Energy came up with: (WB will note it is in metric and dimensionally consistent to boot)



Suppose a Formula One car accelerates from a standing start to 100 km/h in 1.7 seconds (1) and is subject to drag and negative lift forces. Find the coefficient of kinetic friction mu.



which doesn't strike me as completely UNreasonable, but I'm not betting anyone's farm on it. The bit I'm not totally sold on is rewriting the drag and lift equations in terms of a and x using vf^2 = v0^2 + 2a(xf-xi). I know it works if you use it to get the work because you're integrating anyway, but I'm not sure about using it in the sum of Fys. If someone thinks it's wrong, let me know, you're not going to hurt my feelings...I'll take being wrong on the interwebs over being wrong on my final ANY day!

[xpensive]

Holy crap, an engineer on F1T?! Send young Mullen all the well-wishings for his xam!

- autogyro, you have once and for all proven to me you don't have the foggiest of what you are talking about.

- Ciro, Impressive as usual, but I still don't understand the yellow curve, care to xplain?

- WB...er, perhaps you should read my initial post again?

- Jon, damnit, finally an engineer.

Let's see now, wheelspin should have stopped at 30 m/s (108 km/h or 67 mph), when traction-force from 480 kW would be 16 kN. A CoF of 2.5 would mean a contact force on the rear wheels of 6400 N (640 kg). Considering most of the gravitational load will be in the rear due to forward accelleration, which even if aerodynamical downforce is rather low at that speed, seems most reasonable

[RH1300S]

Actually for me, I found Autogyro's comment though provoking. He is right - think about it. How the dickens you make it happen I don't know - you would surely need some for of traction control to tell the gear box which ratio gives you all the thrust the car can use.

Nonetheless - Ciro's answer was clear and dealt well with the question. Ciro - thanks for putting the forumlas on show - they help a novice like me begin to understand

[xpensive]

RH, can you please study Ciro's yellow curve and xplain to me what it says?

But also plot this curve on your PC; Traction force on the Y- and car-speed on the X-axis, when Traction-force (Newton) is the constant Power of 480 kW, divided by car-speed (meters per second.

Power is still Force times Speed.

[RH1300S]

I think you are politely telling me I'm talking out of my ar** - Quite probably I am, I can't explain what the yellow curve says.

I need it in basic language to understand.

However - what I THINK Autogyro is saying is that the gear ratios limit to amount of THRUST available at the driven wheels, so with an infinite choice of ratios you could give the back wheels just as much THRUST as they can use.

To me that was a thought provoking comment.

I would agree that it wasn't very helpful in answering your question.

[Richard]

Actually for me, I found Autogyro's comment though provoking

Me too. A steady incremental application of power would enable the vehicle to maintain optimum balance of grip/spin throughout the whole period of acceleration.

[xpensive]

The condition for the xample given Richard, was a continous application of 480 kW all the way from zero-speed up to 85 meters per second. Plot the curve and I'm certain you will see what I mean. You too Ciro.

[Richard]

The condition for the xample given Richard, was a continous application of 480 kW all the way from zero-speed up to 85 meters per second. Plot the curve and I'm certain you will see what I mean. You too Ciro.

aha - you are talking of an engine and transmission that delivers a constant 480kW to the wheels at all speeds?

[xpensive]

You are very perceptive Richard, this way my simplified xample above should try to give an intellectual understanding to why a vehicle has wheel-spin from zero-speed, question being when it stops and at what relation between traction-force and contact such.

[RH1300S]

I keep re-reading your first post xpensive - I thought I understood what you were saying.

But re-reading it - are you using the expression 'traction force' to mean rate of acceleration?

It seems to me that a constant application of 480kw at the wheels would give constant accleration in the absence of anything to change that rate. Aero drag being one thing that changes the rate - increasing as speed increases - therefore slowing the rate of acceleration until top speed is reached (acceleration ceases).

As the speed goes up, downfoce will increase of course, but at what speed does the wheel-spin stop and how much is the vertical contact-force between the rear wheels and the surface at that speed?

But your last paragraph doesn't seem to need to relate constant power at the wheels.

Are you simply asking at what speed could an F1 car stop being traction limited and be able to apply all it's available power to the road?

Obviously the answer will vary from car to car, track to track and depending on weather conditions, tyres etc. But in principle I'm sure you could choose an approximation of 'normal' dry conditions and car setup for soemone to give you an answer - which I don't think was answered by jon-mullen, although he might have provided some of the tools to work it out with (honestly I don't know - as I'm sure you have guessed )

[Jersey Tom]

I'd be surprised if mu was anywhere near 2.5