I've been avoiding writing this up because a) I'm lazy and b) I'm probably going to come across as sounding preachy. Still, I think it is good to get this all in one place so that the discussions about lightness and acceleration/cornering/braking can be straightened out.
Some equations that describe what we are talking about (remember to use consistent units when using these equations):
F = M * A: Force equals mass times acceleration
F_f = mu * N: Friction force available equals coefficient of friction times normal force
F_r = M * v^2 / r: Force necessary to make you go in a circle of radius 'r' at velocity 'v' equals mass times velocity squared divided by the radius of the circle.
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Case 1: Speeding up
Your engine provides a given force to the ground. You can figure out this force by measuring your torque on a dyno. Torque equals force times moment_arm, where moment_arm is going to have to include things like your gear ratios and the radius of your wheels/tires. Rearranging the first equation gives:
A = F / M
So, you can increase acceleration either with increasing the force put out by your engine or decreasing the mass. Since you have a redline that you can't rev past, you have to shift gears. Everytime you upshift you choose a gear ratio that decreases the amount of torque delivered to the wheels. This is why a low torque engine that revs high (i.e. makes good horsepower) can accelerate a car faster than an engine with more torque but a lower redline.
You can also increase acceleration by decreasing the mass you have to move. This is typically considered to be the way Lotus looks at things. Less is more...
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Case 2: Traction limited acceleration
This can apply to accelerating the car but more typically applies to braking and cornering. Let's look at braking first.
Case 2a: Braking
In most cars braking is limited by the amount of friction force provided by the tires. From the equations above we have:
F = M * A and
F = mu * N
In the case of braking, A is the amount of deceleration you get. Let's first think about this in terms of a car that has one wheel. In that case, N is the mass of the car and setting the two equations equal we get:
M * A = mu * M
Of course, the Ms cancel and we get:
A = mu
So, the deceleration you get is basically equal to the friction coefficient of the tires against the pavement. If you cut the friction coefficient in half, the amount of deceleration you get is half as much.
If you want to translate acceleration into braking distance from a given speed down to zero speed:
distance = (velocity^2) / (2 * A)
So, if you cut the acceleration in half, you double your stopping distance. Note: This all assumes constant braking deceleration.
Of course, you actually have four tires on a car that each carry some fraction of the mass of the car (let's say each tire supports 1/4 of the mass). Now we would have:
M * A = (mu * 1/4 * M) + (mu * 1/4 * M) + (mu * 1/4 * M) + (mu * 1/4 * M)
Clearly, if you add all four terms up, you end up with what we had before...A = mu.
The world isn't quite this simple of course. As some people have pointed out, mu is not constant with respect to the normal force (the amount of downforce on a tire). Typically, as the normal force increases, mu decreases. For the sake of argument (that is to say, I have no reason to believe this is exactly how tires interact with pavement), let's say that:
mu = 10 m/s^2 * (1.0 - (M/5000 kg))
FWIW, 10 m/s^2 is about one 'g'. So, if we load a tire that has this kind of friction coefficient with 1000 kg, we get a friction coefficient of:
mu = 8 m/s^2
If we lighten the car and make the tire bear just 500 kg,
mu = 9 m/s^2
Since A=mu, this means the car will stop a little more than 10% faster when the weight is cut in half.
There is more to this though. The tire is in contact with the ground over an area. A given square centimeter of contact patch only supports a fraction of the car's weight. If you increase the size of the contact patch, you decrease the amount of weight that each square centimeter of contact patch has to support. Since mu increases when the amount of weight being supported decreases, you can expect that increasing the size of the contact patch will increase the amount of friction available to you.
Tire dynamics are tricky though and I'm a bit out of my depth to discuss them in any detail. However, you roughly know the amount of weight per square centimeter that your tires support. Let's say you inflate your tires to 30 lbs/in^2. If you discount the amount of weight supported by the sidewalls of your tires you can say that each square inch of contact patch is supporting 30 lbs of car (I recommend switching to metric units if you actually play with these numbers though). You can use this information to weigh your car...drive each tire over a piece of paper and trace the outline of the contact patch. Measure the area of the contact patches, add the areas together and multiply by the tire pressure. This should give you the weight of the car (assuming the sidewalls don't support any of the weight of the car...which they actually do, so this won't be 100% right)
If you decrease the air pressure in your tires you will increase the size of your contact patch. This decreases the amount of weight supported by each square centimeter and therefore increases mu. This MAY mean that you can brake harder. Remember that the sidewalls of any performance tire are pretty stiff. If you drop your pressure too much you may actually REDUCE the amount of tire that is actually working for you...you may make it so that only the edge of the tire (the part under the sidewall) is the only part working...this would mean that you've made your mu go down and you'll loose grip.
You also have to take into account weight transfer from the rear to the front of the car under braking as well if you want to accurately calculate braking distances.
This is pretty much the extent of my tire knowledge...my impression is that there are many more details and variables that effect things (e.g. deformation of the tire).
All of the above applies to acceleration (if it is traction limited) as well since acceleration is just like deceleration, just in the opposite direction. You'll note that dragsters quite often have very low tire pressures on the driven tires in order to maximize grip.
Case 2b: Cornering
All of the above applies to cornering as well, but radial acceleration (physics speak for cornering) is slightly different.
F_r = M * v^2 / r
F_f = mu * M
So,
M * v^2 / r = mu * M
yielding,
v = sqrt(mu * r)
So, the velocity you can go around a corner is equal to the square root of the friction coefficient times the radius of the corner. All the same arguments about the friction coefficient stated in the braking example apply here. Deformation of the tire can be quite obvious under cornering...most of us have seen pictures of sidewalls rolling over. Clearly you have to have enough inflation to avoid that since you aren't really even using the tire if its sidewall is sufficiently rolled over.
-----------------------------------
Again, I'm sure there are many more details concerning tire dynamics (contact patch shapes, camber, etc), however I believe the above points are useful for understanding the basics.
I'll get down off my soapbox now...
Some equations that describe what we are talking about (remember to use consistent units when using these equations):
F = M * A: Force equals mass times acceleration
F_f = mu * N: Friction force available equals coefficient of friction times normal force
F_r = M * v^2 / r: Force necessary to make you go in a circle of radius 'r' at velocity 'v' equals mass times velocity squared divided by the radius of the circle.
--------------------------------------
Case 1: Speeding up
Your engine provides a given force to the ground. You can figure out this force by measuring your torque on a dyno. Torque equals force times moment_arm, where moment_arm is going to have to include things like your gear ratios and the radius of your wheels/tires. Rearranging the first equation gives:
A = F / M
So, you can increase acceleration either with increasing the force put out by your engine or decreasing the mass. Since you have a redline that you can't rev past, you have to shift gears. Everytime you upshift you choose a gear ratio that decreases the amount of torque delivered to the wheels. This is why a low torque engine that revs high (i.e. makes good horsepower) can accelerate a car faster than an engine with more torque but a lower redline.
You can also increase acceleration by decreasing the mass you have to move. This is typically considered to be the way Lotus looks at things. Less is more...
--------------------------------------------
Case 2: Traction limited acceleration
This can apply to accelerating the car but more typically applies to braking and cornering. Let's look at braking first.
Case 2a: Braking
In most cars braking is limited by the amount of friction force provided by the tires. From the equations above we have:
F = M * A and
F = mu * N
In the case of braking, A is the amount of deceleration you get. Let's first think about this in terms of a car that has one wheel. In that case, N is the mass of the car and setting the two equations equal we get:
M * A = mu * M
Of course, the Ms cancel and we get:
A = mu
So, the deceleration you get is basically equal to the friction coefficient of the tires against the pavement. If you cut the friction coefficient in half, the amount of deceleration you get is half as much.
If you want to translate acceleration into braking distance from a given speed down to zero speed:
distance = (velocity^2) / (2 * A)
So, if you cut the acceleration in half, you double your stopping distance. Note: This all assumes constant braking deceleration.
Of course, you actually have four tires on a car that each carry some fraction of the mass of the car (let's say each tire supports 1/4 of the mass). Now we would have:
M * A = (mu * 1/4 * M) + (mu * 1/4 * M) + (mu * 1/4 * M) + (mu * 1/4 * M)
Clearly, if you add all four terms up, you end up with what we had before...A = mu.
The world isn't quite this simple of course. As some people have pointed out, mu is not constant with respect to the normal force (the amount of downforce on a tire). Typically, as the normal force increases, mu decreases. For the sake of argument (that is to say, I have no reason to believe this is exactly how tires interact with pavement), let's say that:
mu = 10 m/s^2 * (1.0 - (M/5000 kg))
FWIW, 10 m/s^2 is about one 'g'. So, if we load a tire that has this kind of friction coefficient with 1000 kg, we get a friction coefficient of:
mu = 8 m/s^2
If we lighten the car and make the tire bear just 500 kg,
mu = 9 m/s^2
Since A=mu, this means the car will stop a little more than 10% faster when the weight is cut in half.
There is more to this though. The tire is in contact with the ground over an area. A given square centimeter of contact patch only supports a fraction of the car's weight. If you increase the size of the contact patch, you decrease the amount of weight that each square centimeter of contact patch has to support. Since mu increases when the amount of weight being supported decreases, you can expect that increasing the size of the contact patch will increase the amount of friction available to you.
Tire dynamics are tricky though and I'm a bit out of my depth to discuss them in any detail. However, you roughly know the amount of weight per square centimeter that your tires support. Let's say you inflate your tires to 30 lbs/in^2. If you discount the amount of weight supported by the sidewalls of your tires you can say that each square inch of contact patch is supporting 30 lbs of car (I recommend switching to metric units if you actually play with these numbers though). You can use this information to weigh your car...drive each tire over a piece of paper and trace the outline of the contact patch. Measure the area of the contact patches, add the areas together and multiply by the tire pressure. This should give you the weight of the car (assuming the sidewalls don't support any of the weight of the car...which they actually do, so this won't be 100% right)
If you decrease the air pressure in your tires you will increase the size of your contact patch. This decreases the amount of weight supported by each square centimeter and therefore increases mu. This MAY mean that you can brake harder. Remember that the sidewalls of any performance tire are pretty stiff. If you drop your pressure too much you may actually REDUCE the amount of tire that is actually working for you...you may make it so that only the edge of the tire (the part under the sidewall) is the only part working...this would mean that you've made your mu go down and you'll loose grip.
You also have to take into account weight transfer from the rear to the front of the car under braking as well if you want to accurately calculate braking distances.
This is pretty much the extent of my tire knowledge...my impression is that there are many more details and variables that effect things (e.g. deformation of the tire).
All of the above applies to acceleration (if it is traction limited) as well since acceleration is just like deceleration, just in the opposite direction. You'll note that dragsters quite often have very low tire pressures on the driven tires in order to maximize grip.
Case 2b: Cornering
All of the above applies to cornering as well, but radial acceleration (physics speak for cornering) is slightly different.
F_r = M * v^2 / r
F_f = mu * M
So,
M * v^2 / r = mu * M
yielding,
v = sqrt(mu * r)
So, the velocity you can go around a corner is equal to the square root of the friction coefficient times the radius of the corner. All the same arguments about the friction coefficient stated in the braking example apply here. Deformation of the tire can be quite obvious under cornering...most of us have seen pictures of sidewalls rolling over. Clearly you have to have enough inflation to avoid that since you aren't really even using the tire if its sidewall is sufficiently rolled over.
-----------------------------------
Again, I'm sure there are many more details concerning tire dynamics (contact patch shapes, camber, etc), however I believe the above points are useful for understanding the basics.
I'll get down off my soapbox now...
