Track radius and downforce

Here are our CFD links and discussions about aerodynamics, suspension, driver safety and tyres. Please stick to F1 on this forum.
User avatar
f1maniac
0
Joined: 06 Feb 2006, 11:04
Location: India

Track radius and downforce

Post

How is the track radius and design speed at a curve of a F1 track calculated..plz tell me guyz...I am doing a project n neeed it...plzzzzzzz.

aeronut
aeronut
0
Joined: 17 Mar 2006, 03:57
Location: Los Angeles, California

Post

Dear f1maniac,

If you want to continue to develop your interest in F1, you should plan to become familiar with engineering mechanics and dynamics, and the source from which they are derived, physics. The calculation of the car’s speed around a curve of a given radius is a basic application of Newton’s Second Law of Motion: Force equals mass times acceleration or F = ma. A car going around a corner is undergoing an acceleration toward the inside of the curve called “centripetal” acceleration. The force causing this acceleration is provided by the tires, acting toward the inside of the curve. This force is caused by the friction between the tire and the road, which is affected by the tire design and composition, the weight and moment (load transfer) on each of the tires and the aerodynamic downforce on the tires, among other things.

The force keeping a car on a curve of a constant radius is equal to its mass times its velocity squared, divided by the radius of the curve.
F = m v^2 / r

Now this is also equal to “ma”, from the Second Law:
mv^2 / r = ma

The mass term on each side of the equation cancels out, so this equation is good for any weight of car: v^2 / r = a

As an example, let us take Turn 5 at Sepang, Malaysia. The “interactive circuit” at the website formula1.com says this turn can be taken at 232 km/hr at a 3.4 “g” loading. Now 1 “g” is an acceleration of 9.8 meters per second squared. So the total centripetal acceleration is 3.4 times 9.8 or 33.32 m/sec^2.

The unknown in our equation is the radius of Turn 5. To make the answer come out in meters, the speed (or velocity) is converted to 64.44 meters per second. When the equation is rearranged to r = v^2 / a

This gives a turn radius of 124.6 meters, which seems reasonable.

Of course this is all a simplification because the turn is probably not of constant radius, and the driver may be detracting from the force that the tires can apply to lateral acceleration by accelerating ahead or braking. The equation may be turned around to calculate the speed of the car, knowing the turn radius and the “g” load.

ginsu
ginsu
0
Joined: 17 Jan 2006, 02:23

Post

Wouldn't you equate the Friction Force to the Centriptal force instead of 'ma' i.e.

F = mu*N , F = m*v^2/r

mu*N = m*v^2/r

where N is Normal Force, the sum of the weight and downforce

N = (m*g + Fdownforce)

then

r = m*v^2/(mu * (m*g + Fdownforce))

mu=1.3 (estimate based on R-compound DOT radial)
g=3.4g=33.2 m/s^2
v^2 = 232 km/hr; 64.44 m/s
Fdownforce = 1500lbs (rough, rough estimate); 6600N
m = 600kg

r = 72.2 meters

Now, that's incorporating the power of downforce! No wonder why
wings are so integral to the performance of the modern F1 car.

As you can see, the variable that makes the greatest difference
is mu as that is a direct inverse multiplier. Because mu is coefficient
of friction for the tire, then we know the tires are really the most
important part of cornering.

If you have alot of downforce and a low mu, then the wing really doesn't
matter at all.
I love to love Senna.

User avatar
joseff
11
Joined: 24 Sep 2002, 11:53

Post

IIRC the tires' grip on the road is more than friction. Car tires actually make little molecular bonds with the road. This is why even cars w/o aero downforce can turn at more than 1g.

EDIT: put another way, you can say that mu > 1

User avatar
f1maniac
0
Joined: 06 Feb 2006, 11:04
Location: India

Post

Thanx...guys u hav taken a great deal of weight of my mind....Thanq very much... :D

User avatar
Ciro Pabón
106
Joined: 11 May 2005, 00:31

Post

I do not want to interfere, but, guys, please, the centrifugal force does not exist. The centripetal force is not compensated by anything: you can prove it because the car turns. If it was equilibrated, the car would continue in a straight line, like Newton's first law demands. This is one of the most difficult things to teach to highway design students, BTW... :wink:

Please, please, read part IV of Brian Beckman's "The physics of racing". Every pilot and designer should read it. Scroll down and you will find the link.

Besides, the equations offered in this thread do not take in account the superelevation and the transition curves (and the transition of the superelevation), and most important of all, that 3.4 g for a tire is a little high. I would be surprised by such a high value, but if you say so...

I really do not know if you can talk of design speed for a race track, as it is defined as:

"...the maximum safe speed that can be mantained over a specified section of highway when conditions are so favorable that the design features of the highway govern."

- AASHTO's A Policy on Geometric Desing of Highways and Streets -

This definition clearly does not apply here, so I presume you are talking of maximum speed. If you want to simulate the maximum speed, you have to take in account the weight transfer caused by braking and calculate the force on every wheel individually. The equations offered are simple approximations, anyway. More important are the equations for the tire response. I will check a little and offer you the links, when I have more time.

Finally: there is no need for "little molecular bonds" to "improve" the aerodynamic downforce. I do not know how to say this as strongly as possible, so I will scream (or its equivalent) a little: ALL THE FORCES THAT THE CAR CAN EXERT TO MOVE SIDEWAYS PASS THROUGH THE WHEEL/TRACK INTERFACE. Got it? If you had zero friction wheels, not even the most powerful aero package in the world could help you to turn a car.
Ciro

Apex
Apex
0
Joined: 08 Jul 2005, 00:54

Post

I think that was was meant by the chemical reaction of the tyre and road surfaces is simply that with a chemical reaction you can get a higher mu value than with out. Like glue basically....

The F = mu*N equation is all good but is mu not a function of N also ->
F = mu(N)*N
Dont dream it, do it.

User avatar
Ciro Pabón
106
Joined: 11 May 2005, 00:31

Post

Well, Apex, I was just trying to point out that there are no misterious chemical bonds. You could argue that all forces are electrical (at least in the engineering world of low energies). When you are sitting in a chair, the forces between electrons in the chair and in your buttocks is what keeps your a** 'hovering' over the chair. But the reality of tire grip has been explained by Mr. Bo Persson in 2001 for the first time since mankind appeared on this planet... :wink:

In a rough surface there are a lot of peaks and valleys. This roughness increments the apparent surface, apparent in the sense that is measured as width times length (the nominal area of the contact patch). But the real area of contact is assumed as a self-fractal with A = L^H, where L is the nominal lateral length and H is the Hurst exponent. This area is proportional to the normal force. I cannot stress enough the importance of the precedent phrase. Until I read this article (and I paid 28 dollars to do it, all well spent), nobody had explained to me why friction forces are proportional to normal forces. But there you have it: the real area of contact varies proportionally to normal force. This variation in contact area is, of course, invisible to the naked eye. http://prola.aps.org/abstract/PRL/v87/i11/e116101

The efective area of contact can be increased: the "bonds" that are supposed to exist do not. But if you could liquify somehow the rubber, you could fill the voids that exists between the fractal surface of the road and the rubber that is envolving it. This is precisely what F1 tyres do: they exude a liquid that simply increases the effective area of contact. Actually, F1 tyres are not worn (well, not until the "one race rule"), they are sucked dry. Finally, you have to take in account that, once you know your real microscopic contact area, all you need to know is how the rubber stores and dissipates energy. Mr. Persson's theory simply states that rubber is elastic until the yield point and plastic beyond it (like they taught you in basic materials), but that in certain modes of vibration and at certain frequencies, the modulus of elasticity can be increased a thousandfold, which is another surprising result. http://scitation.aip.org/getabs/servlet ... s&gifs=yes

You can find a short article resuming Mr. Persson theory here: http://unisci.com/stories/20022/0612023.htm. I have read that the theory is so good that you do not need a whole tire to predict its behaviour. I have heard that the tire makers use it. Please, correct me if I am wrong.

And this is Mr. Persson (hey, I love this guy, I think he is the unknown genius behind Ferrari's dominance from 2001 to 2005, but all you know I am a dreamer):

Image
Last edited by Ciro Pabón on 03 Nov 2006, 03:24, edited 1 time in total.
Ciro

ginsu
ginsu
0
Joined: 17 Jan 2006, 02:23

Post

There are many types of adhesion that a deforming tyre experiences. I believe three of them in this list can be realized easily by simplying thinking about a deforming, liquefying tire. I am not a tire expert, but there must be alot at work if mu is well over 1, and I think that F1 tires are probably more at 2.
Mechanisms of Adhesion

Five mechanisms have been proposed to explain why one material sticks to another:


Mechanical Adhesion

Two materials may be mechanically interlocked. Sewing forms a large scale mechanical bond, velcro forms one on a medium scale, and some textile adhesives form one at a small scale.


Chemical Adhesion

Two materials may form a compound at the join. The strongest joins are where atoms of the two materials swap (ionic bonding) or share (covalent bonding) outer electrons. A weaker bond is formed if oxygen, nitrogen or fluorine atoms of the two materials share a hydrogen nucleus (hydrogen bonding).


Dispersive Adhesion

Also known as Adsorption. Two materials may be held together by van der Waals forces. A van der Waals force is the attraction between two molecules that have positively and negatively charged ends. This positive and negative polarity may be a permanent property of a molecule (Keesom forces) or universally occurs in molecules, as the random movement of electrons within the molecules may result in a temporary concentration of electrons at one end (London forces).


Electrostatic Adhesion

Some conducting materials may pass electrons to form a difference in electrical charge at the join. This results in a structure similar to a capacitor and creates an attractive electrostatic force between the materials.


Diffusive Adhesion

Some materials may merge at the joint by diffusion. This may occur when the molecules of both materials are mobile and soluble in each other. This would be particularly effective with polymer chains where one end of the molecule diffuses into the other material. It is also the mechanism involved in sintering. When metal or ceramic powders are pressed together and heated, atoms diffuse from one particle to the next. This joins the particles into one.
I love to love Senna.

User avatar
Ciro Pabón
106
Joined: 11 May 2005, 00:31

Post

ginsu wrote:There are many types of adhesion that a deforming tyre experiences. I believe three of them in this list can be realized easily by simplying thinking about a deforming, liquefying tire. I am not a tire expert, but there must be alot at work if mu is well over 1, and I think that F1 tires are probably more at 2.
According to Brembo, the maximum braking accelerations for Catalunya (thanks, Reca, here) are 4 G's. This translates to a mu of 4. I am not sure here, as the distances and speeds they gave for braking imply less acceleration, if you check their calculations.

You can see in the link I give that, for example, Repsol has an stated speed of 135 kph and a radius of 44 meters. This translates to:

a = V^2/ (r*g) = (135/3.6)^2/44*9.8 = 3.3 G's

This means 3.3 mu.

I guess that in a 44 meters radius curve, the 9.5 m width gives you some slack. For curves that have to be taken more or less following a circular curve, you get lateral acceleration values closer to 2. For example, the thight Seat curve has calculated lateral acceleration of 2.2 G's and La Caixa has 1.6. I am not sure about this last one, because the layout I have is old and I guess Brembo data is for the new layout).

Reca gave me some simulated data from race games, I will find the trajectories and give some data on the "real" lateral accelerations I will calculate when I have some time.

If you read Perssons theory the same way I do, mu is constant for constant conditions. What changes is the area of the surface of contact at the microscopic level. I quote: Earlier models only made use of an averaged surface roughness. "I include all length scales – from one centimetre down to the atomic level", Persson remarks.. I think this means he had taken into account all kind of mechanisms of adhesion.
Ciro