Acceleration and braking figures at different speeds?

[g-force_addict]

I know at medium speeds acceleration can actually be faster than at low speeds because highforce provides increased traction.

So does anyone has a speed vs Gs or time vs Gs chart or table for both acceleration and braking.
If not, please help me do the math from time vs speed tables.


tnx

[Tim.Wright]

You have twitter? Ask Lewis, he should be able to give you the information you need.

Tim

[Greg Locock]

Lapsim is your friend. It is free.

Alternatively, buy Ferrari Formula 1 by Peter Wright if old numbers are good enough.

[mzivtins]

I love that a Nitro RC car is probably the quickest thing to accelerate from 0 - 20mph with four wheels... well and two!

[raymondu999]

Lateral and rearwards acceleration (cornering and braking) would benefit from the downforce, but I think forward acceleration is hampered by drag, and is probably more drag vs power limited.

[xxChrisxx]

If you want to do this yourself and build a model to learn the principles, stat with the basics. You can build yourself a crude model in excel. It's also probably easier to do this as a x=speed y=force/acceleration.

The first thing to do would be to define the maximum acceleration limited by engine power. This will asymptote @ zero speed. And slowly drop off. You can also add in drag to this curve.

The next thing to do would be to overlay a maximum acceleration limited by traction curve, based on tyre data and load.

The area below he curves then indicates the maximum possible linear acceleration envelope.


When you get a sensible looking model, you can then add complexity to more accurately approximate reality.

[Stradivarius]

I have tried to calculate the accelerations, but I have not been able to get accurate figures of the grip level at different speeds. As a first guess, I assumed that the friction coefficient of an f1-tyre is around 1. This means that the maximum acceleration through a corner without any aerodynamic downforce, assuming a perfectly ballanced car, would be 1 g. If a car is doing 2 g through a corner at constant speed, this means that the aerodynamic downforce is equal to the wheight of the car. I have seen som numbers, several years ago, that an f1 car produces aerodynamic downforce equal to its own wheight at around 120 km/h. So at 120 km/h an f1 car would be able to do 2 g around a corner. It would, however, not be able to do 2 g straight line acceleration, since the front wheels are not contributing. I would think that about 60% of the total downforce (gravity + aerodynamic) is acting on the rear wheels, so in this case, the maximum acceleration would be 1.2 g, when ignoring air drag.

I tried to look at the information on f1.com for different circuits to find data to enable me to determine the car's grip level at different speeds. I realize of course that this will depend on the circuit, as the downforce levels are different, but even when just looking at one track, it was difficult to find data that are consistent with the theory. Probably, the reason is simply lack of data. For example, I couldn't find any information of the inclination of the surface. At Monza, according to the circuit layout on f1.com, the speed through the second Lesmo turn is 178 km/h. And the centripetal acceleration is 3.8 g. But when comparing these number to those of the first Lesmo turn and Parabolica, it doesn't add up, unless there is a difference in the camber or something like that. Both Lesmo 1 and Parabolica are taken at higher speed, which should mean more downforce, but the centripetal acceleration is smaller. Another reason for this difference could be harder forward acceleration, which would put greater demands on the rear tyres. But in this case I fail to see how this could be the case.

Another approach I made was to look at the end of the start/finish straight at Monza, before braking. There the speed is 340 km/h and the acceleration is 0.13 g. This information should allow us to determine the drag coefficient as a function of the engine power. If we assume that the engine power is P = 800 hp (590 kW), and the car's mass is m = 650 kg, the drag force at v = 340 km/h (94.4 m/s) is equal to P/v - m*0.13*g. The drag force is proportional to the square of the velocity, so this should give us some idea of the drag force at velocities. Unfortunately, this doesn't add up when looking at different parts of the track, and again I expect that the inclination of the surface is to blame. Just before Parabolica, the speed is 335 km/h and the acceleration is 0.29 g. Just before Ascari, the speed is 332 and the acceleration is only 0.14 g. One way to find an estimate could be to look at the average of all the points at the circuit where this information is provided and assume that the average inclination is 0. It would of course be necessary to exclude all points where the speed is so low that it can be expected that it is the grip and not the engine power which limits the acceleration. But I am afraid that such calculation will give less accurate figures than those one can easily find on the web. I just checked on wikipedia and found some information on the topic, which is probably not accurate, but nevertheless gives some idea: http://en.wikipedia.org/wiki/Formula_One_car

[Tommy Cookers]

surely the frictional coeff of the tyre is about 1.7 ?

most high speed circuits (incl Monza ?) have some fast corners that are now (with only 750 bhp in F1) power-limited ?

[Greg Locock]

Stradivarius, welcome to the world of reverse engineering. So far as I can tell you are on the right track, and the frustrations you are finding are typical of not having enough data about one car. I may have some spare time tomorrow night to dig out my rather old figures for vehicle parameters. They were entirely based on what I could glean from the book I mentioned, which has quite a lot of useful stuff in it.

[tathan]

I guess Tim Wright was being facetious (although I did lol) but he might be onto something... The gradient of the speed plot on the telemetry Hamilton tweeted should be everything you need? dv/dt is the acceleration remember. You have both the axis labelled so can reproduce in excel by putting a selection of points in for bits you know are straight line accels or decels.

[Stradivarius]

surely the frictional coeff of the tyre is about 1.7 ?

most high speed circuits (incl Monza ?) have some fast corners that are now (with only 750 bhp in F1) power-limited ?

How can you be sure that the frictional coefficient is 1.7? I would then expect that we would not see accelerations much lower than 1.7 g, even in slow corners. Based on observation, I would think that the friction coefficient is close to 1, although it changes slightly depending on different factors, like the normal force the track surface and the tyre compound. Super soft should have the highest friction coefficient.

There is not many turns at Monza which are power-limited. The only turns I can imagine are power-limited are turns, 9 and 10 (Variante Ascari) in addition to turn 3. At least there, the speed increases. The Lesmo turns and Parabolica, however, are definitely not power-limited, since there is a braking zone going into them. There wouldn't be any reason for breaking before a turn if the speed was power-limited and not grip-limited. Coming down towards Parabolica, the speed exceeds 330 km/h, while the speed at the apex is only 215 km/h.

[xxChrisxx]

How can you be sure that the frictional coefficient is 1.7? I would then expect that we would not see accelerations much lower than 1.7 g, even in slow corners.

A 0-100 kph time of 2 seconds is 1.4g avg. assuming no slipping and 650kg load distributed equally. And that has minimal aero input. So you are talking at least an equivilant friction coeffient of 1.2.
1.7 seems a bit high, but 1.3-1.4 doesn't seem too outrageous.


I actually don't agree that looking at a specific track and working backwards is the best way to go. There are too many variables to extract any useful information with such a cursory look. It's too difficult to tell if your assumptions are wrong, or the measured data includes something you haven't accounted for.

Then again I've always been a fan of the bottom up approach to modelling. As you can get 90% of the way there with back of the fag packet calculations.

[Tommy Cookers]

@Stradivarius
eg in the Aero, Chassis & Tyres section the thread 'Define Tyre Grip' shows 1.7 ideal (old data ex Carroll Smith?)
how else does eg MotoGP corner around 60deg lean ?
friction coeff of 1 is about right for a 1960 race tyre, sorry to be blunt
if you have centripetal accels of 1 presumably there is simultaneous high longitudinal acceleration

in straight line acceleration a surprisingly large factor is the necessary rotational acceleration of wheels'tyres transmission etc
(especially at low speeds, you can't just convert engine power into thrust and treat the vehicle as just a mass with translational inertia only)

[Tim.Wright]

I guess Tim Wright was being facetious (although I did lol) but he might be onto something... The gradient of the speed plot on the telemetry Hamilton tweeted should be everything you need? dv/dt is the acceleration remember. You have both the axis labelled so can reproduce in excel by putting a selection of points in for bits you know are straight line accels or decels.

I was only half joking with my suggestion. Reverse engineering the sheet from Lewis will be a lot more exact than using hypothetical accelerations and speeds from a track map.

I dint know if acceleration was on the sheet, but if not you could calculate it from the velocity and a track map.

Then you've speed, ax, and ay. Everything you need.

Tim

[Stradivarius]

@Stradivarius
eg in the Aero, Chassis & Tyres section the thread 'Define Tyre Grip' shows 1.7 ideal (old data ex Carroll Smith?)
how else does eg MotoGP corner around 60deg lean ?
friction coeff of 1 is about right for a 1960 race tyre, sorry to be blunt
if you have centripetal accels of 1 presumably there is simultaneous high longitudinal acceleration

You are right that if there is tangential acceleration, it will affect (reduce) the centripetal accelereation. But since the data at f1.com is usually taken from the apex, where the speed is lowest, I don't think there is any significant tangential acceleration. If you are braking all the time until you start accelerating again, your driving is not optimal, as you could have held a higher speed through the corner by lifting the brake earlier and accelerating later.

I still find it hard to explain how the sideways acceleration is smaller than 1 through any turn on any track, if the friction coefficient was significantly higher than 1, except of course, for the earlier mentioned camber of the track. However, I don't think there is any camber at turn 1 at Monza, so why is the "g-force" as they call it on f1.com only 0.73 g? By the way, g-force generally means total acceleration, not only centripetal component, isn't that so?

in straight line acceleration a surprisingly large factor is the necessary rotational acceleration of wheels'tyres transmission etc
(especially at low speeds, you can't just convert engine power into thrust and treat the vehicle as just a mass with translational inertia only)

I actually thought about this, but didn't expect rotational energy to account for a significant part of the total energy, so I didn't bother to mention it. Now I have tried to look at some numbers and check it for the wheels.

I don't know the exact weight of the wheels, but I have seen someone suggesting the rear wheels might be around 30 pounds, while the front wheels are about 22 pounds. To be conservative I assume that all wheels have a mass m = 15 kg. To be even more conservative, I assume that all the mass is placed at the maximum radius, so that the moment of inertia of each wheel is equal to I = m*r^2. This means that the rotational energy of the wheel is equal to the translational energy of the wheel. If we compare this to the translational energy of the whole f1-car, that the ratio is equal to the ratio between the mass of the wheels and the total mass of the car: 4*15 kg/650 kg = 9.2%, which is worth mentioning, but doesn't really change the situation much. Keep in mind that this is an over-estimate, since most of the mass is located at a smaller radius, and also since the total mass of the 4 wheels is probably smaller than 60 kg.

There is other rotating parts in the drive-train, but I believe they have an even smaller mass and certainly significantly smaller moments of inertia. The rotational speeds, however, is higher, but I find it hard to believe that rotational energy accounts for more than a few percent of the total energy, at the most. Does anyone know the specifications of the flywheel used in f1? I would assum this would account for most of the rotational energy apart from the wheels.

I don't really know this, but I don't expect the total rotational energy to be much more than 5% of the total kinetic energy.

And of course, I am well aware (and in fact think I demonstrated) that the data from circuit maps won't give any accurate figures. I was simply trying, in lack of more accurate data, to establish some figures, but I realize that they are probably less accurate than most figures available as trivia around the web.