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.RARS computes just onequantity, the vector , and accounts for all forces and torques on the tyre through that one variable.The advantages of the approach, when it comes to computer simulation, are• very simple math, easy to code and debug• very fast conversion from velocity to force• one table lookup and one interpolationThe limitations, of course, are that RARS cannot account for detailed tyre physics and important effects like suspension geometry and dynamics, so the whole scheme trades off accuracy for simplicity.However, as a quick-and-dirty approximation, it'sremarkably effective.91The Physics of Racing,Part 20 : Four-Point StaticsBrian Beckman, PhD©Copyright January 2001In this instalment, we analyse the stability of a single wheel, a bicycle, tricycle, and, finally, of a four-wheeled vehicle.In the offing, we introduce force moments, vector cross products, matrices and linear algebra, and some interesting facts about how the number of wheels on a vehicle relate to the number of dimensions of space and to the practice of weight jacking on a race car.Consider a single bicycle wheel and tyre combination, all by itself, just standing on the ground.Call it a unicycle wheel.It almost immediately falls over.The reason is that its centre of gravity (CG) is above the ground, but its contact patch (CP) is ON the ground.Assuming that the CP doesn't slide, then the ground will resist any force put on it with an equal and opposite force.If the wheel begins to tip ever so slightly sideways, Earth's gravitation, pulling on the CG, and the reaction force, pushing mostly up and a little sideways on the CP, conspire to twist the wheel even more sideways down toward the ground.In other words, if it tips over just a little, it will have an overwhelming tendency to tip over ALL the way.The following figure shows the wheel precariously balanced on its CP:The next figure shows the wheel just starting to tip over.One can easily see that the weight, pulling down on the CG, and the reaction force, pushing up on the CP, will quickly knock the wheel down to the ground.At any instance of time, the tendency for the wheel to fall over is measured by the moment of the forces about some arbitrary point.The moment of a force about a point is the magnitude of the force times the perpendicular distance of the force line from the point.We suggest this perpendicular distance in the following diagram with a small right triangle.Since the CP is not sliding, by assumption, it's fixed in inertial space and is an ideal candidate for a moment centre: the point about which to compute force moments.There is also a small, sidewayscomponent to the ground's force on the CP, but we ignore that in the diagram.92More generally, the moment of a force vector can be thought of as a vector in its own right (at least in three dimensions, it can; the story is more complicated in four or more dimensions).This vector is the cross product of the moment arm and the force vector.The moment arm is a vector drawn from the moment centre to the point of application of a force.In the diagram above, the moment arm of the gravitational force is along the hypotenuse of the little triangle and points upwards.The cross product of a moment arm, r, and a force, F, is defined to be a certain vector that is perpendicular to both r and F.There are lots of vectors perpendicular to both r and F if r and F are not collinear, and there are NO vectors perpendicular to both of them if they are collinear.In any event, we're looking for the particular one that satisfies some properties.Suppose r has components ( rx, ry, rz) and F has components ( Fx, Fy, Fz).Let the vector we're seeking be T.The conditions that T be perpendicular to r and to F can be written as follows (assuming you understand the much-simpler inner product or dot product of vectors-ifnot, take a search for "inner products" at http://www.britannica.com, for one of many Internet sources):T r = 0 = Txrx + Tyry + TzrzT F = 0 = TxFx + TyFy + TzFzIt's easy to check that the following vector satisfies these two equations in three unknowns:T r F = ( ryFz - rzFy, rzFx - rxFz, rxFy - ryFx) It's a little more subtle to check that the magnitude of T, written, is the magnitude ofr times the magnitude of F times the sin of the angle between them in the plane they form
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