Physics Of Racing...

[gixer02]

I read this thread a long time ago. It's called "Physics of Racing" http://www.pelicanparts.com/techarticles/physics_racing/index.htm It has alot of good info in it. What I was discussing with my physics teacher was weight. I told him I took like 65lbs out of my car and could feel the difference... Acceleration benifits of losing weight are obvious, but he was saying it is easier to slide with less weight. That's the logic I use when I put weight above the back tires when it snows. But when thinking about it this is what I came up with-First, this is how they explained turning a car. “When you turn the steering wheel, you are trying to get the front tires to push a little sideways on the ground, which then pushes back, by Newton's third law. When the ground pushes back, it causes a little sideways acceleration. This sideways acceleration is a change in the sideways velocity. The acceleration is proportional to the sideways force, and inversely proportional to the mass of the car, by Newton's second law. The sideways acceleration thus causes the car to veer a little sideways, which is what you wanted when you turned the wheel.”When they say “the sideways acceleration is proportional to the sideways force”- so the more force from the tires(or the more g’s they can take), then the faster you can turn-(or you can turn at a greater speed). Well that is obvious because sticky tires increase handling... Then it says “the acceleration is inversely proportional to the mass of the car” I take that as -the less the mass of the car, the greater the potential sideways acceleration is- (correct me if that is the wrong logic)In other words- lighter handles better. But that doesn’t make sense compared to the fact that more weight above the wheels means it’s harder to slide. This is what I am concluding- there is a difference between handling characteristics with x mass when you are sliding, and with that same mass when your tires are gripped. I mean if you lose control you will more quickly gain it back the more weight you have(like in the snow heavier is better) But, if you are on the track in good conditions lighter is better because it takes less force to turn the car(less mass= less inertia and so less mass= easier to change the car’s path)

Sorry that was so long but I am wondering if anyone has a better explaination of why less mass= better handling. I’m not sure if my logic is correct.

I got to thinking about this when I wondered “If you’re on a skidpad going in a circle and you keep the radius and speed the same but only reduce the mass than wouldn’t the reduction of mass enable you to go faster w/out sliding?” The equation for centrifical force is mass times velocity squared divided by the radius of the circle. So I am assuming with a lighter car you must increase speed to achieve the same g forces (than if the car were heavier) Does that sound right? I asked the physics teacher that and he said “Yes but with a lighter car it takes less to slide it” So what he is saying is when a 2000lb car that can hold up to 1G is reduced to 1500lbs it can no longer hold 1G, but it now holds less than 1G- does that sound to be correct? If anyone knows a good way to explain why lighter cars are better handling(because they are) Than that is what I am after. Thanks
Pat

[Moneyguy1]

The same principle that applies to moving an object in zero gravity. Weight is a separate issue from mass. Despite weighing "nothing" it still takes considerable effort to change the direction or speed of a massive object, even in zero gravity. The less massive the object, the less effort required to change the speed or direction. One rule of motion states that an object in motion tends to stay in motion unless operated on by a separate object. Assuming there is sufficient friction between the tire(one object) and the road (the other object), it will take less energy to redirect a lighter vehicle. Try cornering a tractor trailer at excessive speed!!!! Granted, if the friction between the road and the tire is low, such as in snow or ice, then the object acting on the vehicle (turning the wheel) is overcome by the lack of friction (the ice) and the mass of the vehicle tends to maintain the original direction.

Plus, more mass above the wheels can result in an unstable situation and if the vehicle does not begin to slide to compensate then the vehicle will tend to roll. (a 944 can outcorner an SUV) The more speed the lower the center of gravity must be to prevent this. At excessive speed (which can be calculated, the
c of g would have to be at or below surface level which is impossible.

No math, it's been too many years since college physics. Good thread.!!

Cheers!!

Bob S.

[gixer02]

You put it into words very well! What I was trying to say by saying handling is different in different situations(in snow and on the track) is what you said about the presence of friction. It would have helped my descripton had I not forgotten that little thing they call friction cheers my friend

Pat

[Z-man]

Interesting discussion here. Here are some things to consider:
** Disclaimer: I am by no means an expert, just trying to convey what I've learned myself! **

By the way, the technical term for the tires sliding is called: SLIP ANGLE. There are three different slip angles your car exhibits: front wheel slip angle, rear wheel slip angle, and whole car slip angle (read: 4 wheel drift). Read more about it in Henry Watt's book mentioned below. (General definition of SLIP ANLGE: the difference between the direction X is pointed in, vs. the direction X is actually moving, ie sliding)

1. You lost 65 lbs. So your 2800lb car (approx) now weighs 2735lbs, or 97% of it's original weight. That loss in weight is really negligable. Also: loosing weight in a well balanced (50/50) car can be tricky: if loosing the weight causes a 60/40 weight distribution, your handling will suffer. Your feeling of improved acceleration is probably a "seat of the pants" dyno: do some timed runs with and without the weight, and you'll probably notice very little, if any, change. So: weight savings can have an adverse effect on your car's handling! (Note: that's why many people stress corner balancing over loosing weight!)

2. Weight (or lack thereof) is not the only factor in handling. You already mentioned sticky tires. There's more: suspension, tire setup (pressure, camber, caster, toe-in...etc), balance, power, and don't forget about the driver! When you start to factor in all these variables, you'll see that your quest for loosing weight won't make much of a difference at all. Again.

3. With all these different factors coming into play regarding cornering and performance driving, it is important to move forward slowly: changing one thing at a time, and carefully evaluating how the change helped or hurt the capabilities of your car. It is also best to start with the cheapest stuff first too. BTW: I have a feeling that loosing more than 65 lbs in your car is going to start costing: not just money, but the 50/50 balance of your car.

4. So, what's the best thing to start with? THE DRIVER! Learn to drive your car to it's ability, and once you've done that (which could take many years), you can slowly start to modify your car's geometery. If you're not a consistent driver, you won't understand how changes alter your car, and if the changes improve or hurt your car's dynamics. Understand your car's stock setup (it's really not a bad setup, afterall), and once you can drive your car consistently to it's limits, then it's time to play.

5. The best way to improve your driving abilities is to attend autocross events, and DE track events. I suggest autocrossing first, so you can get the feel of your car before you kick up the speed a notch or two ! Track is more expensive than autox, so the more you learn before you drive on the track, the more effective your money spent there will be. I have seen too many beginners get all the goodies (sticky tires, suspension changes...etc) and get blown away by a great driver in a bone-stock setup car!

6. If you are interested in your car's handling, read some books about it. A couple I suggest: Secrets of Solo Racing by Henry Watts. Bob Bondurant's Performance Driving Handbook by Bob Bondurant. Tune to Win by Carroll Smith: gets into the geometery of suspension, tires...etc.

While the physics you talk about make sense in the classroom, things are a little different when the rubber meets the road.
-Zoltan.
Gallery updated 01-20-02:

[Lawrence Coppari]

Quoting:
Then it says “the acceleration is inversely proportional to the mass of the car” I take that as -the less the mass of the car, the greater the potential sideways acceleration is- (correct me if that is the wrong logic)In other words- lighter handles better. But that doesn’t make sense compared to the fact that more weight above the wheels means it’s harder to slide.


In the above I surmise the text is talking about Newton's second law (F=ma or a=F/m), hence the word "inverse". Here, the acceleration "a" is indeed inversely proportional to the mass.

When you corner, centrifugal force is due to a continual change in velocity towards the center of rotation; therefore, the force is directed outward. That acceleration is V*V/R where "V" is speed and "R" is radius. The force is m*V*V/R, where "m" is vehicle mass.

I feel the above quote from your post refers to straight line acceleration. Centrifugal acceleration is strictly a geometric consideration, based soley on velocity and radius as I have noted above.

[Devildog2067]

True in a classroom sense, but false. m*omega*r is only true when an object is moving at a constant speed, with a constant inward radial acceleration, in a perfect circle (or part of one).

When you add factors like acceleration out of corners, late braking in, increasing/decreasin radius turns, etc, the acceleration turns out much more complex.

As for centrifugal force: it is what we call in physics a pseudo-force or an artifactual force. It DOES NOT EXIST, ie, if you draw a free-body diagram of the car wrt the ground, there is no force radially outward exerted on the car by the ground.

The perception of centrifugal force comes from trying to use Newton's laws of motion in a noninertial reference frame (the inside of the car). If you were to draw a free-body diagram of JUST THE CAR (not the ground), you would see that there is a force radially inward to the turn. Since in YOUR REFERENCE FRAME, the car is not accelerating, the net force on it must be 0. Therefore, we add the centrifugal force as a correcting factor. Its magnitude must be the same as the radially inward CENTRIPETAL force in order to "cancel" it.

The force known as the Coriolis force works much the same way, and is a bit easier to think about. One common example: the Earth rotates at a relatively constant angular speed, and so does everything that's rigidly attached to it.

If you were to throw a ball very far straight up in the air (ok, vaccuum so we don't have to worry about wind and such), by the time it came back down the Earth would have rotated underneath it and it wouldn't come straight back down in the same spot. The Coriolis deflection is predictable and measurable; it affects weather patterns all the time.

If you want a more in-depth treatment of pseudoforces, or want to see the equation for the translation from an inertial to a rotating reference frame (you can actually see the "non-inertial terms," Coriolis and centrifugal, appear), I'd be happy to discuss it further, but I'm not going to bore anyone here. My email is swon@uiuc.edu. (And I'm a senior in applied physics, if anyone was wondering).

[gixer02]

devildog- when you say "True in a classroom sense, but false. m*omega*r is only true when an object is moving at a constant speed, with a constant inward radial acceleration, in a perfect circle (or part of one). When you add factors like acceleration out of corners, late braking in, increasing/decreasin radius turns, etc, the acceleration turns out much more complex." Yes it is indeed very complex to find it out all the forces in a real world situation. I was thinking if you had a constant speed and radius than it would be easier to explain. Would you agree only changing the mass (by reducing it) and not changing the speed or radius that it is easier to manipulate the car? Obviously looking at it that way you cannot find all the actual forces that act on the car in the real world, as there are many(i.e. downforce from wind, weight transfer etc...). My basic motivation for the post was to put reasoning behind why less massive cars handle better. My teacher said more weight is better but he was thinking along the lines of a low friction setting such as snow.

Oh and by the way, I am going to be a freshman next year at college. I want to join the Corp, as a reserve(so I can still go to school). I am planning on going with the 92 day program; bootcamp would be the first summer, training the next summer, all the time in between it's once a month and the following summers it is 2 weeks for each summer. That is the extent of information I have on that. What are you doing that you are in school too? I am interested to know. Thanks alot.

I emailed this same thing to you if by chance you do not check this thread.

Pat

[Lawrence Coppari]

The purpose of my post was to explain what I think the quote applied to without being pedantic. It's best to keep things simple when you explain. And by the way, the acceleration formula is second order in omega, not first order as you wrote.

When I was an artillery officer I learned the formulas for determining where to aim artillery take earth rotation into consideration. Howitzer shells stay aloft a long time.

[Devildog2067]

There's a fine line between not being pedantic and beng inaccurate. The idea that a centrifugal force exists that points outward through the turn is just plain wrong, it's a myth that's been perpetuated for far too long because it's simply easier to explain that way than to address the actual physics of it. But then, it's easier to tell kids that babies come from storks, too.

And yes, it's omega squared. Typo on my part, a thousand apologies.

(If you were arty, you might have heard this one (I forget the details): during one of the world wars, the British Navy was fighting down around Argentina or the Falklands somewhere, and they kept missing because the people who designed their naval gun sights forgot that Coriolis deflection goes the other way in the other hemisphere. Not sure if it's apocryphal or not).

gsxr: Pure physics formulas can't provide all the answers. For instance, formulas predict that your grip will be the same no matter what size tire you run: a skinnier tire has more weight/surface area on it and therefore sticks just as well as a fatter tire with less pressure on it. We all know that this isn't true, however. Fat rubber grips better for many reasons, but the one to keep in mind is that the racetrack is not just a physics lab.

With that in mind, the formulas say that acceleration is not mass-dependent. However, they don't take into account things like body roll and weight transfer. Also, I don't think it would be meaningful to talk about a constant-speed turn, because there's no way you'd want to keep a constant speed. You'd want to accelerate out of the turn to minimize your time.

[Lawrence Coppari]

When I was an undergraduate as you are now, my physics books talked about centrifugal force as though it were real. Guess things change.

Do they talk about negative time in your undergraduate physics books these days? They didn't when I was an undergraduate.

[Roger Hall]

I dont understand everything in this thread, here is my understanding. Please correct me if I am wrong.

Inertia wants the car to continue in a stright line turning the wheels creates friction (an object in motion stays in motion unless acted upon by an outside force). Weight transfers to the outside tires causing causing the coefficeint of friction (traction) to increase on the outside tires and decrease on the inside tires. This difference is why the car turns. The weight transfer causes stress on the tires. The amount of stress is dependant on the amount of weight. When the amount of weight is too great the stress excedes the tires ability to grip, and the car slides. Therefore the more weight the quicker the tires will lose traction in turns and slide. By adding weight you increase the stress and get better traction untill that stress exceeds the tires ability to grip.

[cruzermusic]

ice,bananas,bubble gum, don't care.....the heat of the meat is proportional to the motion

[Devildog2067]

Ok, an important point, and one that is a little hard to grasp at first:

When a car is moving, and none of the wheels are spinning/skidding, the contact patch between a tire and the ground is not moving. I say again, the point where the tire and the ground meet is STATIONARY.

Now, under most driving conditions, the contact patch remains stuck to the ground. The force of friction is given by (mu)*(Normal force), where mu is the coefficient of friction and the Normal force is the force with which the GROUND pushes on the car. (This force is equal in magnitude but opposite in direction to the weight of the car when the car is level).

The mu in this case is what's called mu[static], and it's used when two objects are not in motion wrt one another. It's LARGER than the value used when two objects are sliding against one another (mu[kinetic]).

Ok. So you try and make the car corner. Turning is a change in momentum, dp/dt. (This is the actual form that Newton stated his second law in, dp/dt is proportional to the object's mass.) dp/dt = m (dv/dt) = ma = Force which is the more common (more useful) form that we see it in today. So making the car turn is an attempt to change its momentum, which requires a force. This force comes from the car's tires. (It has to, the 4 tire contact patches are the only places where the car is connected to the ground).

As the car turns, weight shifts off the inboard tires and onto the outboard tires. Force of friction was mu*normal force (which was equal to weight on level ground) so the traction of the inboard tires gets SMALLER while the traction of the outboard tires gets bigger (you got this part backwards). This is part of why LSDs were invented; if you supply equal power to both wheels but one can grip harder than another, obviously the one with less grip will tend to spin.

Alright, now if the force of friction isn't big enough to make the car turn smoothly, something will happen. Either understeer or oversteer according to your suspension setup and weight distribution. Either way, what happens is that the contact patch unsticks itself from the ground and starts to slide. We no longer use mu(static), we use mu(kinetic) which is smaller and therefore our force of friction is smaller. Anytime any tire is sliding against the ground, the force it can apply is smaller than when it is stuck. This is why spinning the wheels in a drag start makes you slower, and why theoretically threshold braking stops you faster than 4-wheel-lockup. (Practically, your average driver will have better luck locking em up then trying to manipulate the brake).

Now, if a car starts to oversteer and you brake/come off the throttle, something potentially bad happens: weight is transferred to the front wheels. This lessens the traction of the rear wheels, and if they were sliding to begin with, they're just gonna come loose and the car will spin.

Again, keep the physics in mind but don't think about it too hard, cuz there's always loopholes between the track and the street. (For example, according to the formulas a drag launch should never spin the wheels. However, with some cars it can be advantageous to do so to let the engine run closer to its torque/rpm peak.)

Hope some of that helps you.

Mr. Coppari, they are doing some weird **** with negative time and time-like/space-like cones and intervals. We're tackling that later in the semester.

(How do you explain away the centrifugal force in Newtonian? The force equations won't balance. Were you perhaps remembering doing LaGrangian mechanics? The non-intertial terms drop out of those equations...)

[cruzermusic]

you guys are good...,,

[Lawrence Coppari]

Good explanation about the tires. I'm a mechanical engineer so we tend to stay with the more mundane stuff (positive time). My specialty is numerical modeling of heat transfer (FE). I'm work for a fairly large chemical company.

[fty]

Keep talking about this stuff and Ill be attaching a "flux capacitor" to the car....you guys are makin me nuts I wanna tear down the car tom......you guys are awsome.....-chris