guido@twitch.UUCP ( G.Bertocci) (05/30/85)
> First, the relative front-to-rear roll stiffness (which certainly IS > affected by sway bars etc.) has a great effect on how much of the > total weight trasnfer is borne by the front vs. rear wheels. > -- > Gordon V. Cormack CS Department, University of Waterloo The transfer of weight from rear to front is also a fallacy. There are only two ways to transfer load from rear to front or vice versa. One is to move the center of mass fore or aft, which it does not do. The second is to introduce a force with a fore or aft component, ie. accelerating or braking. When a car is cornering at a constant velocity there is no such force since centripetal force is perpendicular the direction of travel. In other words, the sum of the load on the rear tires is the same when the car is parked as it is when the car is cornering at a constant velocity. (neglecting aerodynamics). This can be shown with the following diagram. This is a sideways view of the car with RT= rear tires and FT = front tires. Mt Mt = total downward force (mass*grav) | Mrt = downward force on rear tires. v Mft = downward force on front tires. CM A = distance from RT to point P Mrt . Mft B = distance from FT to point P | . | v A .P B v RT..........................FT Mt = Mrt + Mft Mrt = Mt*(B/A+B) Mft = Mt*(A/A+B) These are all the forces when the car is parked. When the car is cornering at a constant velocity you have a force at CM (centripetal force) however, it is perpendicular to the axis shown, (going in or out of screen) so there is no component that can change Mrt and Mft. One might make the argument that a car might not be aimed directly in the direction it is traveling (ie. the nose might be inside the tail or vice versa) however, that seems to be a second order problem and not part of the current discussion. Most of the pictures you see with the inside rear wheel off the ground are taken as the car enters the turn while braking, which is transfering the load to the front. > Finally, who said weight transfer had anything to do with increased > or decreased cornering power? I do not understand the argument that > begins: assuming tires have a constant coefficient of friction... > Assuming that, the weight distribution makes absolutely no difference. > Of course, tires are not perfect frictional devices, and that is why > one tries to transfer roll stiffness to the lighter end of the car. > But there is no simple formula for what that does. > -- > Gordon V. Cormack CS Department, University of Waterloo The who are articles in Car & Driver, Road & Track, Korman Autoworks catalog, BMW Roundel, and others including postings on netnews. Frictional force is calculated by a coefficient times an orthogonal force. FF = K*Fk FF= frictional force, K=coefficient of friction Fm = orthogonal force, ie. weight of car. My statement that if tires had a constant coefficient of friction weight distribution is irrelevant, is due to the above equation. If K for tires is truly constant then doubling Fk doubles FF. Which means that you could load all the weight of a car on the outside edge of one tire and generate the same cornering force. It would also mean that controlling camber would also be irrelevant. For tires, K is a function of Fk, decreasing as Fk increases. -- Guido Bertocci AT&T Bell Labs Holmdel, NJ ...!ihnp4!houxm!twitch!guido
gvcormack@watdaisy.UUCP (Gordon V. Cormack) (05/31/85)
> > First, the relative front-to-rear roll stiffness (which certainly IS > > affected by sway bars etc.) has a great effect on how much of the > > total weight transfer is borne by the front vs. rear wheels. > > -- > > Gordon V. Cormack CS Department, University of Waterloo > > The transfer of weight from rear to front is also a fallacy. > In other words, the sum of the load on the rear tires > is the same when the car is parked as it is when > the car is cornering at a constant velocity. > Guido Bertocci > AT&T Bell Labs > Holmdel, NJ > > ...!ihnp4!houxm!twitch!guido The latter is true, but in no way contradicts what I said. Weight transfer is not a fallacy; it is real. What is fallacy is that stiff suspension can alter (significantly) the amount of weight transfer for a given lateral acceleration. What I said is that the LATERAL weight transfer (that is caused by cornering) can be distributed to the front and rear wheels using anti-roll bars. Let me give an example. Suppose a car weighs 2000 lb. and is perfectly balanced front/rear and side/side. Each wheel bears 500 lb. static weight. Suppose further that the car is cornering left at such a rate as to incur 100 lb. lateral weight transfer. That means that the right side gets 100 lb. "heavier" and the left side gets 100 lb. "lighter". If roll stiffness rates of the front and rear are equal, the weight distribution is: front left: 450 front right: 550 rear left: 450 rear right: 550 On the other hand, if the rear roll stiffness is made infinite, the weight distribution is: front left: 500 front right: 500 rear left: 400 rear right: 600 The total weight of the car is still 2000 lb. and the lateral weight transfer is still 100 lb. But the weight transfer is borne by the rear wheels. It is easy to see that if, in the second case, the weight transfer were 500 lb., the left rear wheel would lift from the ground. Even with a wheel off the ground, the car is not in particular danger of rolling over (assuming that there is, in fact, some roll stiffness in the front). It would take 1000 lb. weight transfer (twice the current cornering force) to roll the car. -- Gordon V. Cormack CS Department, University of Waterloo gvcormack@watdaisy.uucp gvcormack%watdaisy@waterloo.csnet