klein@ucbcad.UUCP (Mike Klein) (05/22/85)
> THE "WEIGHT TRANSFER" THEORY IS TOTALLY BOGUS. > ... > There is NO WAY one can transfer weight from left tire to > right tire or anywhere else unless one moves the CM. I think I see what you're saying. It is true that the weight of the car does not transfer unless CM moves. What does happen is that the centripetal force encountered when making a turn results in differential downward forces on the springs and gives the appearance of "shifting the weight" of the car. I am going to derive a real formula that shows the exact relationship of the downward forces on the two springs, and conclude that it is actually pretty easy to divide the body roll of a car by 3. New figure (for a right turn): <----CM FcT /|\ / | \ / V \ __ / / FmT \|\ FcR / / \ \ / / \ \ / / \ \ | | / LS-----AR------RS \ | <- angle here is alpha FcL|/__ \| I've made a few changes. Instead of looking at the left and right TIRE, you need to look at the points where the springs are attached to the body. I call these LS and RS. Exactly between them is the Axis of Rotation (AR). During cornering, CM sees two forces on it: FmT (downward) and FcT (sideways). But the real question is what the forces are on RS and LS. FmT divides itself evenly among LS and RS. FcT is divided differently; it must be divided into two forces, one from RS to CM and the other from CM to LS, shown as the ugly arrows FcR and FcL on the figure. These two must add to produce FcT. The angle between either of these forces and vertical is alpha. Now for the calculations. Both springs always have downward forces on them of FmT/2. FcT = FcR + FcL, so we get: |FcR| sin(alpha) = |FcL| sin(alpha) = 1/2 |FcT| and the downward force on the left spring is: 1/2 |FmT| + |FcL| cos(alpha) Solving we get the downward force on the left spring is 1/2 |FmT| + 1/2 |FcT| cot(alpha) and for the right spring, 1/2 |FmT| - 1/2 |FcT| cot(alpha) Since the sine of alpha is half the width of the triangle (half the track) and the cosine is the height of the triangle (distance between AR and CM), cot(alpha) is simply 2h/w. Now, |FmT| = mg and |FcT| = mG (G is how many g's you're pulling), so divide through by m/2 to normalize (we can do this because the spring rates have been multiplied by m also in the original design of the car). So the "force per unit mass" on each spring is 2 G h g +/- ------- w Now let's put a car on the skid pad. Let's say it is going a 0.8 g's, and it is the car used in the example in the original posting with h = 10" and w = 60". We get the normalized forces on the two springs being g + 0.27g and g - 0.27g Note how much smaller the roll forces are than the gravitational force! This immediately shows that to conquer excessive body roll, we use primarily the sway bars. To divide the body roll by 3 as in the posting, you do two things. First, increase the spring rate by 50% (this is a typical "high-performace" upgrade amount). That will instantly reduce our roll by 33%. The rest is done by upgrading the sway bar. We still need to halve the roll, so we must double the sway bar's spring rate. But the forces here are only 27% of those on the regular springs so this is easily possible. All we do is go from, say, a 3/4" bar to a 1" or 1-1/8" bar. That's it! > People might be wondering just why do people want stiff suspensions anyway > since weight transfer really isn't significant. > There is also an improvement during dynamics situations, (ie. a slalom) > where the the car is not in a steady state. With a stiffer > suspension, the chassis + body achieves a steady state position > quicker, with less overshoot in body roll. I think this is very important! In a boat car you overshoot all the time. The time constants involved are so long that you are always off-balance. If your suspension is underdamped, you hit a bump and the wheel keeps smacking the road surface for a while, greatly reducing your traction. Have you ever driven alongside a car that makes all these smacking noises on a freeway with expansion joints? That's what's going on. > P.S. Next week, I will give on a dissertation on the myth of > "More rubber on the road with wide tires". Yes, this is a big myth. -- -Mike Klein ...!ucbvax!ucbmerlin:klein (UUCP) klein%ucbmerlin@berkeley (ARPA)