32 replies to this topic

[Lightsun]

I'm doing a research project project about aerodynamics and I'm trying to make an equation to calculate the maximum speed that an F-1 can pass a curve, but I don't have no idea about the radius of the curves in meters. Can someone gimme some examples?

[Double Apex]

Well, the 130R corner at Suzuka used to have a 130m. radius (Surprise, I know ), but not anymore...

[Lightsun]

This was the only curve I knew... but thx anyway. It's because in my ecuation, if I put a radius like 400m, the velocity seems to be negative, and I don't know why. Does anybody know the ecuation to calculate the maximum corner speed??? (depending on the weight and the downforce)

[Lightsun]

Sorry, one more thing: Does anybody knows the ecuation to know the downforce of an F-1 car??? (to obtain it in Newtons). I thought I had it but my operations don't look very realistics...

[Vrba]

I did that calculations once and draw the graph that had the radius range of 0 to 200 meters.
For example, for the car and parameters I specified, my calculations showed the cornering speed of 216 kph for a corner with a radius od 130 m.
Please tell me your results!
I have to add that my calculations were made in 1997.....

About the downforce, the equation is:

D = 0.5*rho*A*cd*v^2

D - downforce
rho - air density (approx. 1.2 kg/m^3)
A - area of the car on which the downforce is applied (approx. plan view area)
cd - lift coefficient (the biggest problem to determine, varies depending on wings, etc.....typical values 2-4)
v - car speed

Hope this helps a bit!

Hrvoje

[xflow7]

Well, there won't be a simple expression for downforce at a given speed for an F1 car. If you make the (very crude) assumption that the car just behaves like a general lifting body then you could say that the downforce is proportional to speed squared:

i.e. L=1/2*rho*A*C_l*V^2

where:

L=downforce (N)
rho=density of air (kg/m^3)
A = frontal area (m^2)
C_l = coefficient of lift
V = speed (m/s)

Not sure what a good C_l number for a modern F1 car would be, but I would guess 2.5-3.

As for calculating cornering speed, F=ma gives,

M*V^2/R=mu*(M*g + L)

M=mass of vehicle (kg)
R= radius of turn (m)
mu=effective coefficient of friction between tire and ground
g= gravitational acceleration (9.8 N/m)

Note this is also very crude because, among other things, we are assuming a simple Coulomb friction model which is bogus.

Anyway, if you do the algebra, you get:

V=(R*mu*M*g/(m-0.5*mu*R*rho*A*C_l))^1/2

Do note the rather interesting result that if the denominator goes to zero (i.e. R=2M/(mu*rho*A*C_l) ), V goes to infinity. That is, there is a radius above which there is apparently no limit to speed. Of course this isn't real due to non-linear tire behavior, non-linear downforce behavior, and a whole bunch of other things, but it's interesting anyway.

Hope that gets you started.

[Lightsun]

LOTS OF THANKS! well, in fact, I was using the same equation as you, but I had the same problem: apparently it was ok with some examples, but I put radius of 400 kph and the result was negative or something impossible like this. Can you explain me with more details whi this happens?

[xflow7]

Happy to help.

Well, first off, I assume you meant radius of 400m.

In any case, here's a couple of things to think about.

Realize that the equation there is, in fact for V^2, and one must take the square root. From a mathematical standpoint, the value of V can be either positive or negative with equal validity. However, a negative speed in this context doesn't make sense, so we are only interested in the positive value. Having said that, most tools one would use to calculate the square root (eg calculator, matlab, excel, whatever) will return the positive value by default, so I don't think that's where your problem lies.

Secondly, as I mentioned in my other post, there is a singularity when the denominator in V^2 is zero, which occurs for:

R=2M/(mu*rho*A*C_l)

Of course, if R is less than this value, V^2 is positive and finite and everyone's happy. However, if R is LARGER than this value, things get squirrely because V^2 is now less than zero. This means that only solutions are imaginary values of V. My guess is that this is the result you got that initially confused you (understandably).

What you must recognize in order to interpret this result is that we cheated a little with the friction in writing the equation the way we did. If you recall the Coulomb friction model, what it says is NOT, in fact:

F_friction=mu*F_normal

Instead what it says is:

F_friction <= mu*F_normal

This important distinction gives us an "out" because it allows the force of friction to be only as large as necessary to provide the centripetal acceleration for the car to negotiate the corner.

The passage of R through this critical value essentially means that we've entered a regime where additional speed yields more downforce (and therefore available grip) than it does lateral acceleration and therefore the potential speed is unbounded.

Does that make sense?

[Lightsun]

that's a nice explanations. thank you very much. It will be very useful for my research project!

[RX-7]

I think I popped a blood vessel reading this thread! Very interesting, even though I have no idea how you guys do it.

[mp4]

It's not exactly F1 related but I would suggest you look at IndyCars.
In their heyday, they were lapping Indy at under 40.00 seconds. This relatates to roughly 240mph per lap.
I am guessing their corner speed would be about 231. (I am no engineer so please don't be yelling at me...)
For the curve at Suzuka, on a good day, I think the boys are doing about 185.
Before the turn at Imola was made less interesting, I'd put it down to about 190...

Cheers

[Vrba]

Originally posted by Vrba
....

D = 0.5*rho*A*cd*v^2

....

Originally posted by xflow7
....

i.e. L=1/2*rho*A*C_l*V^2

....

It's interesting that we even put the variables in the same order :-)

Hrvoje

[DOHC]

Originally posted by xflow7
V=(R*mu*M*g/(m-0.5*mu*R*rho*A*C_l))^1/2

Do note the rather interesting result that if the denominator goes to zero (i.e. R=2M/(mu*rho*A*C_l) ), V goes to infinity. That is, there is a radius above which there is apparently no limit to speed.

That means that the corner is "flat out."

@ Lightsun: Try the search feature too -- these question shave been discussed in several threads. One that I remember because I took part in it is this one, but I'm sure there are more, and they will have a good amount of additional iformation on the models and equations involved.

[xflow7]

Hrvoje,

That is funny. I didn't notice that.

DOHC,

Yeah. Not sure why I didn't just state that as such.

[Yelnats]

Lightsun; Of course F1 cars never drive the true radius of a curve, the 130 R being driven at a much larger radius due to the clipping the apex effect. So if you are trying to determine "the maximum speed that an F-1 can pass a curve" you've got some further data to consider like the width of pavement and the slowing and acceleration around the apex. A very complex problem if accuracy is important to your solution.

[Lightsun]

by the way... what about the tyre behaviour? I thought that the coefficient of friction was more or less 2.1, and it always was the same. Can you tell me something about it?

[DOHC]

Peter Wright in his F1 Technology book mentions a friction coefficient of 1.7 for modern tyres.

[Ben]

A good simple analysis of this type is presented in the applied aerodynamics chapter of racecar vehicle dynamics by milliken and milliken.

They use a friction coefficient that decays linearly and then stabilises at a contant value above a certain load.

You can get tyre data for the F3000 spec tyre from www.avonracing.com. This will at least give ballpark friction coefficient values and load sensitivity.

Ben

[Yelnats]

Regarding the friction coefficients noted above (2.1 -1.7). Its quite possible that both are correct in certain contexts.

Take the following example for;

Skid pad tests of of CART cars reveal sustained values of 1.6 gs on 100 ft radii circles. (speeds too low for significant aero downloading) This would seem to correlate well with the 1.7 G maximum quoted above. But second thought reveals that if a car is to sustain 1.6 G's the tires themselves must have a co-efficient well above thia value because of factors such as irregularities in the track, unequal distribution of forces between tires due to understeer, driving forces at the rear, lateral weight transfer, ackerman angle errors etc, all of which causes less than ideal "slip" angles.

So it's quite possible (perhaps likely) that on a test bed a tire would generate 2.1 G's but be limited to 1.7 G's on a car due to the above factors. Note that we haven't considered factors like the heavily rubbered track surface typical of many racing situations.

[550spyder]

I believe that you are assuming that a car turns in a circle. I does not. The car follows a different curve: CLOTOID
You can get the CLOTOID equation in Math books

[dosco]

Originally posted by 550spyder
I believe that you are assuming that a car turns in a circle. I does not. The car follows a different curve: CLOTOID
You can get the CLOTOID equation in Math books

Clothoid.....interesting. You learn something new every day.........

"When the path of the curve is followed with an uniform velocity, the speed of rotation is linear (in time). That's why the curve with the reverse relation is called the anti-clothoid.

It was the famous Leonhard Euler (1744) who started investigating the curve.

In the early days of railways it was perfectly adequate to form the railways with series of straight lines and flat circular curves. When speeds increased the need developed for a more gradual increase in radius of curvature R concomitant with an elevation of the outer rail, so that the transition to the circular curve became smooth.
The clothoid makes a perfect transition spiral, as its curvature increases linearly with the distance along the spiral.
A first order approximation of this spiral is the cubic spiral."

From http://www.2dcurves....al/spirale.html

Very cool.

[dosco]

Originally posted by 550spyder
I believe that you are assuming that a car turns in a circle. I does not. The car follows a different curve: CLOTOID
You can get the CLOTOID equation in Math books

I've been thinking about this during my lunch break. The definition of 'clothoid' is "the spiral path on which if traveled at constant velocity, the speed of rotation is linear in time."

1. On what track is there a "constant velocity" turn? Aren't the cars (generally speaking) either accelerating or decelerating into and out of the apex of any given turn?

2. If the 'speed of rotation' is 'linear in time,' I presume that means the angular velocity is a linear function(?). If the car is accelerating in and out of the turn, doesn't that mean that it's angular velocity is a non-linear function?

3. If the lateral acceleration of the car ramps up and down to 3g, doesn't that imply the car is not following a linear relationship (of any kind) in time?

[550spyder]

Originally posted by dosco


I've been thinking about this during my lunch break. The definition of 'clothoid' is "the spiral path on which if traveled at constant velocity, the speed of rotation is linear in time."

1. On what track is there a "constant velocity" turn? Aren't the cars (generally speaking) either accelerating or decelerating into and out of the apex of any given turn?

2. If the 'speed of rotation' is 'linear in time,' I presume that means the angular velocity is a linear function(?). If the car is accelerating in and out of the turn, doesn't that mean that it's angular velocity is a non-linear function?

3. If the lateral acceleration of the car ramps up and down to 3g, doesn't that imply the car is not following a linear relationship (of any kind) in time?

I'm only saying that the path of the car in normal conditions is a Clothoid, not a circle. The car will be able to do a circle only in the presence of a rear steering system.

Another point. The normal car turns around a point that is called Intantaneus Rotation Point (maybe the english terminology is different). This point is defined by the crossing of the two perpendiculars to the front wheels as defined by Ackerman-Jeantaud. The internal wheel steers more than the outside wheel.

In the case of race cars this point is not defined since they use anti-ackerman, the internal wheel steers less than the outside wheel and the perpenduculars never cross. The point is only defined by one single front wheel. In other words: race cars have only 3 wheels, one of them in front, for calculations purposes.

[Pioneer]

A car can travel a circular path if some (or all) of its wheels are sliding in addition to rolling.

Yes?

[550spyder]

Originally posted by Pioneer
A car can travel a circular path if some (or all) of its wheels are sliding in addition to rolling.

Yes?

Who knows?

[Greg Locock]

The car will be able to do a circle only in the presence of a rear steering system.

I'm sorry, this is wrong (I think I know what you mean, but it isn't what you said). We have a standard test called the constant radius test. We drive around a circle at gently increasing speed. This allows us to measure all sorts of interesting things.

The design of banked corners is quite an interesting application of this stuff - anyone who has driven around MIRA's high speed circuit will have experienced the awful effects of a poorly calculated transition. My local banked circuit (two parallel straights, two banked turns) can be driven around at 180 kph, during the entire circuit the steering wheel is moved slightly less than the width of a finger, if there is no wind. I think if I played with the throttle a bit I could get round hands-free.

A car can travel a circular path if some (or all) of its wheels are sliding in addition to rolling.

Yes?
--------------------------------------------------------------------------------



Who knows?

Me, yes, you could drive in a circle with one wheel locked. It would smell. In fact you could probably drive in a circle with one wheel in the air in some cars (front wheel drives with short suspension travels and sporty handling).

[Schummy]

I think that as driver turns the driving wheel approximately linearly from zero angle to a final angle A, car goes in a clothoid, then driver holds the wheel approximately in the same angle A thus car goes in a circular path. Finally as drivers turns almost linearly from angle A to zero the car follows another clothoid curve.

Driver needs a certain non-zero time to turn the driving wheel, so a combination of clothoid-circular-clothoid could be a simple approx for a constant radius curve. Curvature would go linearly from 0 to C, then constantly C and then linearly from C to curvature zero.

[550spyder]

I think we are mixing theory with pratice.

I believe Lightsun wants to develop a theory. To do this, it will be necessary to gather all the existing theory, add some new and test the theory in pratice.

My point to help him is: the car follows a clothoid, not a circle ( even though we know that in some cases it is possible). Why I' m saying this. Because the clothoid curve is the best or lesser worse aproximation to the car movement in normal conditions.

Of course, banking curves will alter the car behavior. But these especial conditions should be added later.

What I' m saying is that Lightsun should use the clothoid equation in his theory instead of circles equations. Doing so, his theory will be based in the existing theories.

Only after that, the theory should be tested and modified.

[DOHC]

Very interesting discussion.

As for the question if you can drive in a circle -- yes of course you can. In real life there is no rolling without slip, even if you go straight. Why a circle would be an impossibility escapes me. From the control theoretic point of view, the car's tarjectory is controllable, and that means that it could follow any smooth enough trajectory. (And a circle is certainly smooth enough.)

It's quite another question if you can go from travelling in a straight line to following a circular arc instantanelously. This appears, more or less obviously, to be an impossibility (hence the discussion of the clothoid curve above). It would be impossible because one cannot turn the wheels instantaneously to the desired steady state steering input, and even if one could, the slip would go through a transient in the moments immediately after, thereby corrupting the circle.

The problem is the transition from straight to circle. Such a curve has a jump discontinuity in the 2nd derivative. This implies that lateral forces must also have a jump discontinuity. That doesn't happen in real life, except when you hit something (a kerb or an oval's wall).

As for the clothoid, my guess is that a real ltrajectory is far more complex than that. And then I'm not thinking of whether the driver is sawing away at the steering wheel. I'm thinking of downforce. Ideally, as dosco points out, a car is either braking or accelerating. So in negotiating a corner, the speed should drop to a minimum, somewhere near the apex. With aerodynamically generated downforce, the max lateral force diminishes with speed. Therefore, when speed is at its lowest in the corner, the car can sustain less lateral force than immediately before or after this point of minimum speed.

It seems to me that because of aero, the turn in phase can sustain greater lateral force than the apex, and the exit, when the driver is back on the gas again, could also sustain greater lateral force. If this is correct, the trajectcory is far from a clothoid. Perhaps it could be approximated by two clothoid segments, one for the trajectory turn-in to apex, and one for the trajectory apex to exit.

This is a theoretical argument only, unsubstantiated because I have no complex simulation model or race car experience to verify it with. Nevertheless, I'm pretty sure that the speed-dependency of the downforce would be a major factor in determining the optimal tajectory through a bend.

[Schummy]

Following the downforce refinement introduced in the discussion, it would seem that, as downforce is at the minimum approximately in the apex, then it perhaps would be counter productive to have the maximum curvature in the apex. Thus speed reduction and curvature increasing should be such that lateral Gs doesn't peak in the apex. A "clothoid"-like path has precisely maximum curvature at the apex. How could it be adapted without introducing undued complications? Perhaps displacing the apex to an early apex, an late apex, or "squaring" the curve?

I assume that when it is said that car follows a clothoid it means a double clothoid (one from intro to apex with increasing cutvature and other from apex to exit with decreasing curvature). A pure clothoid is basically an arc of a double spiral with curvature always (linearly respect length of path) increasing or always decreasing and it is not suitable for a corner trayectory.

[dosco]

Originally posted by Schummy
How could it be adapted without introducing undued complications?

I assume that when it is said that car follows a clothoid it means a double clothoid (one from intro to apex with increasing cutvature and other from apex to exit with decreasing curvature). A pure clothoid is basically an arc of a double spiral with curvature always (linearly respect length of path) increasing or always decreasing and it is not suitable for a corner trayectory.

From the definition of clothoid.....

"In the early days of railways it was perfectly adequate to form the railways with series of straight lines and flat circular curves. When speeds increased the need developed for a more gradual increase in radius of curvature R concomitant with an elevation of the outer rail, so that the transition to the circular curve became smooth.
The clothoid makes a perfect transition spiral, as its curvature increases linearly with the distance along the spiral."

I think the idea is that the trajectory goes from straight, to clothoid, to circular arc, back to clothoid, back to straight.....which might work well for a simulation.

To DOHCs comments, I agree that the "real" trajectory of a car is probably more complicated.....

[Schummy]

Of course with some telemetry data from simple curves (flat, constant radius, etc) we could get something to fit different possibilities... (I suppose someone would do the analisys, not me certainly )

[RDV]

From all racing data I have seen (flat, banked, cambered, uphill, downhill, over brow corners, winged, touring, ground effect ), there is always transition (IE continuously changing radius and steering input), consider corners at which you brake, turn-in, and re accelerate, the full exploitation of your G-G diagram makes it so.... only on banked flat-out corners, IE, MIS for example, do you have a steady-state circle....