Relay-Version: version B 2.10 5/3/83; site utzoo.UUCP Posting-Version: version B 2.10.2 9/17/84; site twitch.UUCP Path: utzoo!watmath!clyde!burl!ulysses!mhuxr!mhuxt!houxm!twitch!guido From: guido@twitch.UUCP ( G.Bertocci) Newsgroups: net.auto Subject: Additional weight transfer myths Message-ID: <196@twitch.UUCP> Date: Wed, 29-May-85 23:51:53 EDT Article-I.D.: twitch.196 Posted: Wed May 29 23:51:53 1985 Date-Received: Thu, 30-May-85 20:36:06 EDT References: <191@twitch.UUCP> <194@twitch.UUCP> <7275@watdaisy.UUCP> Distribution: net Organization: AT&T Bell Labs, Holmdel Lines: 83 > 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