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Does a lighter car have the potential to handle better???

 
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El Nagro



Joined: 22 Mar 2006
Posts: 10
Location: Lethbridge AB, CA

PostPosted: Sun Mar 26, 2006 6:19 pm    Post subject: Does a lighter car have the potential to handle better??? Reply with quote

I am having an arguement with a friend of mine, He says that with all things equal a lighter car has no advantage over a heavy car in a corner, in fact he says the heavier car actually corners harder because the load that its putting on the tires will allow it to keep traction better and corner harder. I feel that a lighter car as the advantage in not only straight line but as well in the corners, The centrifugal force of the vehicle will force it to the outside of the corner and not allow it to be as agile or be able to enter and keep its speed threw the corner. Also the lighter car as the effects of inertia on its side but thats also tied in with the centrifugal force idea.

If anyone can shed more light on this that would be great, I don't really care if I am right or wrong, I just want to know why.

Thanks,
Joe
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tErbo b00st



Joined: 24 Mar 2006
Posts: 10

PostPosted: Tue Mar 28, 2006 2:05 am    Post subject: Reply with quote

A lighter car will have EVERY advantage in cornering, except in snow and rain. I will give it a very brief explanation...this is freshman engineering stuff.

The effects of inertia and momentum are much larger than the force put on the tires.

The grip offered by the tires is simplifed to F=mN, m being the coefficient of friction (0<m<1), and N being the normal force or basically the force put downwards by the tire. This force will then be multiplied by 4 and is the force working with you to keep grip.

Momentum on the other hand is p=m*v, p being momentum, m being mass, and v being velocity. Velocity is obvisouly going to be MUCH larger than 1 (compare to F=mN), and m is going to be about a 4 times what the Normal force will be (assuming 50/50 weight distribution front/rear and left/right). This is the force that will be working against you in a turn.

You can do the math calculations on your own comparing cars, but the effect of mass will multiplied by a higher degree in the momentum equation than the friction equation. Which means as mass increases momentum will increase faster than your frictional force increase.

Again, this is very basic...but you get the idea. My gf is staring at me telling me we have to go, so if you would like I can dig into my text books and find a better answer for you.
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El Nagro



Joined: 22 Mar 2006
Posts: 10
Location: Lethbridge AB, CA

PostPosted: Tue Mar 28, 2006 6:44 pm    Post subject: Reply with quote

interesting, Thanks for the response. It was pritty basic but good enough for me, if I need more info I'll let you know.

Thanks,
Joe
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Wiisass
Head suspension nerd


Joined: 09 Nov 2005
Posts: 131
Location: Philly

PostPosted: Wed Mar 29, 2006 5:07 am    Post subject: Reply with quote

Well let's start with the advantages of a lighter car. This comparison means that all things are equal besides the overall weight of the cars. The weight distribution, suspension, engine, tires, etc are all exactly the same.

The power to weight ratio will be greater with the lighter car. The moments of inertia will be smaller for the lighter car. These moments include the polar moment, yaw moment, roll moment and pitch moment. The higher the inertia, the higher the resistance to movement. So basically, a lighter vehicle will be faster and more maneuverable meaning more responsive and easier to control.

These are the main advantages of a lower weight. Like terboboost said, a heavier car can have an advantage in the rain and snow, but in terms of racing lighter is always better.

Now if you want to talk about different driving situations. Let's start with straight line acceleration. With the greater power to weight ratio, the lighter car will accelerate faster. This is also an effect of the vehicle's inertia. You also have to consider that with a normal tire, with an increasing vertical load, the coefficient of friction of the tire decreases. This occurs both in the longitudinal direction as well as the lateral. So with a greater vehicle weight, there will be more weight on the rear axle, lowering the coefficient of friction. There will also be greater weight transfer to the rear. This means that although you will be increasing the amount of thrust able to be developed by the tires, the rate of acceleration of the heavier vehicle will be lower. This rate is determined by the amount of force produced by the tires divided by the mass of the vehicle.

Now for the lateral case. Let's just consider pure steady state cornering. The force created by the vehicle, known as the centrifugal acceleration. This force is determined by the mass of the vehicle times the velocity squared divided by the radius of the turn. So for simplicity sake, assume that the available force generated by the tires is the same. The lighter car will be able to travel through the turn faster than the heavier car. But in the real world all other factors are not equal, but the outcome will remain the same. The major factors that change are the amount of lateral load transfer and the coefficient of friction of each of the tires. The amount of lateral load transfer will be smaller with the lighter car giving all four tires a higher coefficient of friction that the heavier car. The heavier car will transfer more weight which will have the outside wheels with a lower coefficient of friction than the inside wheels. While the lighter car also has tires with a lower coefficient of friction on the outside, the difference between the inside and outside tire coefficient will be less. Now having a lower coefficient of friction doesn't necessarily mean that the tire will produce less force.

I could work out the numbers, but the lighter car will come out faster anyway.

Now that the steady state conditions have been considered, you need to consider the transient states. This is the whole time between the straights and the steady state corners. I believe that the car is always in a transient condition, but for analysis sake, steady states are easier and a good model. But that's beside the point. During the transients, the main factors contributing to the performance are the inertias. And like I have said before, the vehicle with the lower weight will have lower moments of intertia if all other things are the same.

If you don't assume that the cars are the same besides the weight, then this may not always be true. For example, let's consider the polar moment of inertia. This is one of the most frequent terms used when talking about racing. This is more general than the yaw, roll and pitch moments. Basically a car with a low polar moment of inertia has the majority of the weight closer to the center of gravity of the car. So in real life, you could have a car with a lower overall weight with most of the mass spread out to the ends of the car, and you could have a heavier car with the mass centered at the CG. So the heavier car would have a lower polar moment of inertia than the lighter car.

So it's hard to compare two different cars just based off of the weight. If they are similarly setup, you can usually assume that the lighter car will be faster in all condition, meaning racing conditions not weather.

Terboboost, a couple little problems with your explanation. Momentum probably isn't the best way to explain this, momentum isn't a force. Also, you said that the mass used in the momentum equation is four times the normal force, but it's the same as the normal force. Also, the coefficient of friction of a tire can be more than 1 and is not constant but is based on the load and slip angle.

Alright, I think that's about it for now, let me know if you have any more questions.

Tim
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tErbo b00st



Joined: 24 Mar 2006
Posts: 10

PostPosted: Wed Mar 29, 2006 7:04 pm    Post subject: Reply with quote

Good explanation.

I believe you are wrong with the Normal force. If you put the car on 4 individual scales under each tire the weight measured should be about 1/4 of the total weight of the car. Therefore the normal force will be 1/4 the total mass. Correct me if I'm wrong, as im sure you will.

The coefficient of friction I'm deffinently not sure on and you are probably correct. None of my courses have dealt w/ what exactly the coeffiction is, it is always just a given, I have always wondered what actually determines the coefficient. Moreover, I have only used numbers less than 1...so I have always assumed this was the case...apparantly I am wrong.

And yes, momentum is the wrong equation. And as I was writing it, the more I wrote the more incorrect I realized I was, but basically the same principle is there...and like I said it was very basic, and after I got done writing it I couldnt just delete it, haha.

I was going to go into centrifical forces, but as I said my gf was glaring at me. So good thing you explained it.

You deffinently know more than I do though. Are you an engineer? What kind? What racing and chassis/suspension experience do you have, you seem to know your stuff.
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Wiisass
Head suspension nerd


Joined: 09 Nov 2005
Posts: 131
Location: Philly

PostPosted: Wed Mar 29, 2006 8:27 pm    Post subject: Reply with quote

About the normal force. In the momentum equation you used, the mass is the total weight of the vehicle. The sums of the normal forces used in the friction force equation are equal to that weight. So the individual normal forces are different, but the sum of them is equal to the weight of the vehicle. I think we're both talking about the same thing, it's just the wording.

Tires and coefficients of friction is a complicated matter. Tire manufacturers do testing on their tires to determine the characteristics. This ends up as a bunch of plots comparing slip angle versus lateral force tested for a range of normal forces and slip ratio versus longitudinal force also tested over a range of normal forces. You can take these curves and determine a coefficient of friction at a given point, but this is always changing due to the nature of a tire. It gets pretty complicated. Also, the coefficient of friction of different tires can be over 1, this is how cars are able to produce more than 1g of lateral acceleration.

Thanks for the compliments. I'm a mechanical engineer finishing up my senior year. The racing experience comes from three years of Formula SAE and four years of messing with my own car. Also, lots of reading helps out a lot too. For formula, I've been doing a lot of work with dampers and supension modeling. I've also messed around with suspension design to a decent extent. And then of course fabrication of different parts, setting up the race car, rebuilding, revalving and dynoing shocks.
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El Nagro



Joined: 22 Mar 2006
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Location: Lethbridge AB, CA

PostPosted: Wed Mar 29, 2006 11:38 pm    Post subject: Reply with quote

You guys are smart, i dont know if I should even post on this board anymore, i dont think I am worthy lol Shocked
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Wiisass
Head suspension nerd


Joined: 09 Nov 2005
Posts: 131
Location: Philly

PostPosted: Wed Mar 29, 2006 11:52 pm    Post subject: Reply with quote

Definitely post here. And tell other people about the board. I'm not trying to make it just like all the other car boards, but more technical stuff like this. And of course to promote my shop and get feedback on products and ideas.

Tim
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El Nagro



Joined: 22 Mar 2006
Posts: 10
Location: Lethbridge AB, CA

PostPosted: Thu Mar 30, 2006 6:11 pm    Post subject: Reply with quote

lol I was just joking, this seams like a great forum, hope to contribe what I can to the board!! Laughing
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