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#1




four wheel weight transfer
Gregor Veble > wrote in message >...
> If you ever tried to ballance any four legged table you should know that > its legs NEVER carry the same ammount of weight, no matter what stuff > you try to squeeze underneath the leg hanging in the air Unless the table is on carpet! Mine is on carpet, which makes it look as though all legs are carrying equal weight. But, I suppose this is a bit like having suspension. Yes, I see now  a proper model of suspension is what is needed to model weight transfer  only, this is a fair bit more complicated than I anticipated. It seems to me that the chassis must actually be able to pitch and roll which means that the suspension is not fixed at 90 degrees to the chassis. Is this right? In any case, I am not going to try to model this for now. It's for a 2d game  so all that detail would be wasted. But isn't there an aproximate way of working out weight transfer from the tyre forces? On the other hand, perhaps it's not worth doing at all if not done properly... Ruben 
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#2




Hello!
Ruben wrote: > > Unless the table is on carpet! Mine is on carpet, which makes it look > as though all legs are carrying equal weight. But, I suppose this is a > bit like having suspension. Yup, as others have said, only with a suspension which has much more give than the irregularities in the ground can you expect something close to a uniform weight distribution. > > Yes, I see now  a proper model of suspension is what is needed to > model weight transfer  only, this is a fair bit more complicated than > I anticipated. It seems to me that the chassis must actually be able > to pitch and roll which means that the suspension is not fixed at 90 > degrees to the chassis. Is this right? > Yeah, pitch and roll is needed in order to compress the springs/dampers, which in turn provide the load on the wheels (and vice versa, as Sir Isaac tells you). But, with stronger springs the pitching and rolling is vastly reduced. It is the relative spring stifnesses that matter in individual load transfer, not their actual magnitude as long as the road is sufficiently flat. The way the suspension works is then really dependant on what sort of suspension you're dealing with. You can simplify the system a lot, though, by thinking about the contact point of the tyre moving up and down relative to the chassis, and from the position and velocity of that contact point relative to the chassis you get the reaction force for both the spring and the damper, and that will already give really nice, plausible results. > In any case, I am not going to try to model this for now. It's for a > 2d game  so all that detail would be wasted. But isn't there an > aproximate way of working out weight transfer from the tyre forces? On > the other hand, perhaps it's not worth doing at all if not done > properly... > Since you mention you will be dealing with a 2D game, what one can say something about is the transfer between left and right side of the car and the same for front and back transfer, but only for the total transfer in each direction, not the individual wheel loads. For the forward/backward total transfer, you have to assume that the car is quasi stationary with respect to pitching (no rotating accelerations). Then the sum of the torques originating from the loads on the front and the rear wheels as well as the torques originating from the longitudinal forces must be 0 with respect to the c.o.g. One must also make the wheel loads match the force of gravity and the mass*vertical acceleration, while the longitudinal forces should provide longitudinal acceleration. From these equations you can determine the total front and rear loads. The same can be done for the left/right transfer, but again only for the total one, not load on individual wheels. But still, IMHO the code that calculates things from first principle (which is rigid body rotations and springs/dampers) will be much simpler and will look much more natural, too. Cheers, Gregor 
#3




"Ruben" > wrote in message om... > Gregor Veble > wrote in message > >... > >> If you ever tried to ballance any four legged table you should know >> that >> its legs NEVER carry the same ammount of weight, no matter what stuff >> you try to squeeze underneath the leg hanging in the air > > Unless the table is on carpet! Mine is on carpet, which makes it look > as though all legs are carrying equal weight. But, I suppose this is a > bit like having suspension. > > Yes, I see now  a proper model of suspension is what is needed to > model weight transfer  only, this is a fair bit more complicated than > I anticipated. It seems to me that the chassis must actually be able > to pitch and roll which means that the suspension is not fixed at 90 > degrees to the chassis. Is this right? > > In any case, I am not going to try to model this for now. It's for a > 2d game  so all that detail would be wasted. But isn't there an > aproximate way of working out weight transfer from the tyre forces? On > the other hand, perhaps it's not worth doing at all if not done > properly.. In cases of rigid structure, you have to model the stiffnesses and geometry of the 4 contact points to determine the loads in the 4 members. For car tires with suspension, you can ignore this and calculate the load distribution from a simple freebody if you know the location of the center of mass or better still, the mass distribution. For table legs , there will be a certain amount of loading and deflection in 3 of the members before the 4th is loaded, if at all. Plastic deformation (the carpet) will reduce the problem to a simple freebody similar to car tires. PH 
#4




"Haqsau" > wrote in message >...
> Way too complicated. With no suspension deflection and two orthogonal > forces (longitudinal and lateral) wouldn't it be possible to solve each > plane individually and then add the results? For example: > > cg height = Zcg > longitudinal force = Fx > lateral force = Fy > wheelbase = WB > track = T > > longitudinal weight transfer = (Fx * Zcg) / WB > lateral weight transfer = (Fy * Zcg) / T > > Now just add the weight transfers with the correct sign at each wheel. I > believe this will give you the correct solution given the conditions of the > problem as you state it. I don't think I understand how you arrive at this. Doesn't it matter where the centre of mass is located  apart from its height? Also, this is assuming the wheels are in a rectangle, which may not be the case. If you are being accurate about steering, then the wheels rotate about a point which is external to the wheel. Also, the front and rear track may be different. Ruben 
#5




For the static weight distribution yes, the location matters, but for
dynamic weight *transfer* only the height of the cg matters. This is under the assumption that you have already resolved the lateral and longitudinal forces so that they are truly orthogonal to each other and to the vertical direction. It is this relationship of the forces that allows you to assume that the car's opposing force is acting through the cg and parallel to the ground. I'm not sure why you are trying to account for nonrectangular wheel locations when you aren't accounting for suspension deflection. I thought you were trying to keep it simple. ) If you for example resolve the roll moments and lateral forces into the yz plane, i.e. looking at the vehicle from the rear, unequal track width means you now have 2 moments opposing the roll moment. In order to solve this you need an equation relating the two moments to each other. The normal way of doing this is to include the deflection characteristics of the suspension and chassis, as others have mentioned. So if you really are just trying to do a simple 2D problem and ignore deflections you should probably also keep the wheels in a rectangular arrangement so that the problem is solvable under the constraints you are using. "Ruben" > wrote in message m... > > I don't think I understand how you arrive at this. Doesn't it matter > where the centre of mass is located  apart from its height? > > Also, this is assuming the wheels are in a rectangle, which may not be > the case. If you are being accurate about steering, then the wheels > rotate about a point which is external to the wheel. Also, the front > and rear track may be different. > > Ruben 
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