four wheel weight transfer

I'm attempting to derive equations for the load on each wheel of a car given the longtitudinal and lateral forces on each tyre. I'm ignoring the effects of suspension, or in other words this thing has no suspension. I think this is a safe assumption - kind of like a street kart. It seems easy enough - you calculate the torque each wheel is exerting on the chassis by taking the cross product of its location with its force. The force contains an unknown z component. You add the torques to get total torque. The pitching torque must be zero and the rolling torque must be zero. And, the sum of all four vertical forces must equal mg - the weight of the car. That gives us three equations with four forces to solve for, and I can't see how to tighten up the problem by including more information. Intuitively, I can see that in a static situation like a four legged table resting on flat ground, my three conditions above are insufficient since they allow for pairs of opposite legs to vary against each other - I hope this description makes sense. The question is - in the above example, a four legged table on flat ground, how do we know that all legs carry equal weight? That is, why does each leg carry 1/4 of total weight, instead of it being: 1/3 1/6 1/6 1/3 or something? And how can we generalise this fact to a more complex situation like a the street kart undergoing forces? I appreciate any help or comments. Thanks, Ruben

"Ruben" <[email hidden]> wrote in message news:[email hidden]...

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> I'm attempting to derive equations for the load on each wheel of a car > given the longtitudinal and lateral forces on each tyre. > > I'm ignoring the effects of suspension, or in other words this thing > has no suspension. I think this is a safe assumption - kind of like a > street kart. > > It seems easy enough - you calculate the torque each wheel is exerting > on the chassis by taking the cross product of its location with its > force. The force contains an unknown z component. > > You add the torques to get total torque. The pitching torque must be > zero and the rolling torque must be zero. And, the sum of all four > vertical forces must equal mg - the weight of the car. > > That gives us three equations with four forces to solve for, and I > can't see how to tighten up the problem by including more information. > > Intuitively, I can see that in a static situation like a four legged > table resting on flat ground, my three conditions above are > insufficient since they allow for pairs of opposite legs to vary > against each other - I hope this description makes sense. > > The question is - in the above example, a four legged table on flat > ground, how do we know that all legs carry equal weight? That is, why > does each leg carry 1/4 of total weight, instead of it being: > > 1/3 1/6 > 1/6 1/3

Because we assume the center of gravity is in the middle of the table top. If the center of gravity changes, then the weight distribution will be assymetrical.

On Thu, 14 Oct 2004 07:04:42 GMT, "Saint Cad" <[email hidden]> wrote: >Because we assume the center of gravity is in the middle of the table top. >If the center of gravity changes, then the weight distribution will be >assymetrical. > But when you look away is the table still there?

"Ruben" <[email hidden]> wrote in message news:[email hidden]...

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> I'm attempting to derive equations for the load on each wheel of a car > given the longtitudinal and lateral forces on each tyre. > > I'm ignoring the effects of suspension, or in other words this thing > has no suspension. I think this is a safe assumption - kind of like a > street kart. > > It seems easy enough - you calculate the torque each wheel is exerting > on the chassis by taking the cross product of its location with its > force. The force contains an unknown z component. > > You add the torques to get total torque. The pitching torque must be > zero and the rolling torque must be zero. And, the sum of all four > vertical forces must equal mg - the weight of the car. > > That gives us three equations with four forces to solve for, and I > can't see how to tighten up the problem by including more information. > > Intuitively, I can see that in a static situation like a four legged > table resting on flat ground, my three conditions above are > insufficient since they allow for pairs of opposite legs to vary > against each other - I hope this description makes sense. > > The question is - in the above example, a four legged table on flat > ground, how do we know that all legs carry equal weight? That is, why > does each leg carry 1/4 of total weight, instead of it being:

I think it is only true for a perfect table on perfectly flat ground. In the real world the distribution would be unequal. I guess one reason for having suspension is to balance things up, (by effectively making some table legs longer than others). F1 cars, with next to no suspension, can happily drive on 3 wheels, well, at least up to the next corner.

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> > 1/3 1/6 > 1/6 1/3 > > or something? And how can we generalise this fact to a more complex > situation like a the street kart undergoing forces? > > I appreciate any help or comments. Thanks, > > Ruben

Hello! Ruben wrote: > I'm attempting to derive equations for the load on each wheel of a car > given the longtitudinal and lateral forces on each tyre. > While this may certainly be done, if you are trying to build a realtime car simulator you should never, ever even try to calculate such things :). The weight transfers in a simulation should be a result of suspension and rigid body physics, not something to be directly simulated. Work out the rather more basic physics properly and the weight transfers will come out right. If, howerer, this is an analysis of car behaviour you'd like to be doing, read on...

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> The question is - in the above example, a four legged table on flat > ground, how do we know that all legs carry equal weight? That is, why > does each leg carry 1/4 of total weight, instead of it being: > > 1/3 1/6 > 1/6 1/3 > > or something? And how can we generalise this fact to a more complex > situation like a the street kart undergoing forces? >

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 ;) The uniform weight distribution can only be achieved with a suspension, and even then only for a symmetric weight distribution and no external forces. In a situation where the car is accelerating, it is precisely this underdetermination of loads that is used to ballance out understeer/oversteer effects. Stiffening the front springs of the car will mean that that the outside front wheel will take more load than it would with softer springs in a corner, leading to more understeer due to load sensitivity of the tyres, for example. You must therefore include suspension modelling into your model if you really want to determine load transfers of a car. The question again is, do you really want to do it or rather make it a result of the (realatively simpler) rigid body/springs computation. Cheers, -Gregor

In article <eypbd.349$_%5.262@trnddc07>, "Saint Cad" <[email hidden]> writes:

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> > "Ruben" <[email hidden]> wrote in message > news:[email hidden]... >> I'm attempting to derive equations for the load on each wheel of a car >> given the longtitudinal and lateral forces on each tyre. >> >> I'm ignoring the effects of suspension, or in other words this thing >> has no suspension. I think this is a safe assumption - kind of like a >> street kart. >> >> It seems easy enough - you calculate the torque each wheel is exerting >> on the chassis by taking the cross product of its location with its >> force. The force contains an unknown z component. >> >> You add the torques to get total torque. The pitching torque must be >> zero and the rolling torque must be zero. And, the sum of all four >> vertical forces must equal mg - the weight of the car. >> >> That gives us three equations with four forces to solve for, and I >> can't see how to tighten up the problem by including more information. >> >> Intuitively, I can see that in a static situation like a four legged >> table resting on flat ground, my three conditions above are >> insufficient since they allow for pairs of opposite legs to vary >> against each other - I hope this description makes sense. >> >> The question is - in the above example, a four legged table on flat >> ground, how do we know that all legs carry equal weight? That is, why >> does each leg carry 1/4 of total weight, instead of it being: >> >> 1/3 1/6 >> 1/6 1/3 > > Because we assume the center of gravity is in the middle of the table top. > If the center of gravity changes, then the weight distribution will be > assymetrical.

Nope. You've misunderstood. Take a table that is 1/4 1/4 1/4 1/4 Now slip a piece of cardboard under the upper left leg 1/2 0 0 1/2 You haven't moved the center of gravity. You've made the ground uneven. If you have a perfectly rigid table (or car) and a perfectly rigid floor (or race track) and no suspension then the 1/4 1/4 1/4 1/4 solution is unreasonable. The floor won't be perfectly flat and at least one of the four legs won't be touching the floor. Try solving the system of equations with each leg, in turn, lifted from the floor. At least one solution should be viable. The table will tilt that way and stay on three legs. The fourth leg (or wheel) is at least momentarily irrelevant. It is only in an environment where the give in the suspension is greater than the unevenness in the road where you approach the ideal 1/4 1/4 1/4 1/4 distribution that you would naively expect. John Briggs

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. "Ruben" <[email hidden]> wrote in message news:[email hidden]...

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> I'm attempting to derive equations for the load on each wheel of a car > given the longtitudinal and lateral forces on each tyre. > > I'm ignoring the effects of suspension, or in other words this thing > has no suspension. I think this is a safe assumption - kind of like a > street kart. > > It seems easy enough - you calculate the torque each wheel is exerting > on the chassis by taking the cross product of its location with its > force. The force contains an unknown z component. > > You add the torques to get total torque. The pitching torque must be > zero and the rolling torque must be zero. And, the sum of all four > vertical forces must equal mg - the weight of the car. > > That gives us three equations with four forces to solve for, and I > can't see how to tighten up the problem by including more information. > > Intuitively, I can see that in a static situation like a four legged > table resting on flat ground, my three conditions above are > insufficient since they allow for pairs of opposite legs to vary > against each other - I hope this description makes sense. > > The question is - in the above example, a four legged table on flat > ground, how do we know that all legs carry equal weight? That is, why > does each leg carry 1/4 of total weight, instead of it being: > > 1/3 1/6 > 1/6 1/3 > > or something? And how can we generalise this fact to a more complex > situation like a the street kart undergoing forces? > > I appreciate any help or comments. Thanks, > > Ruben

D'oh, that was of course the weight transfers, not the total load as you asked for. But still it should be possible to figure out the static weights based on the lateral and longitudinal cg location and then add these to the weight transfers to get the total load. "Haqsau" <[email hidden]> wrote in message news:[email hidden]...

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> 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. > > > > "Ruben" <[email hidden]> wrote in message > news:[email hidden]... >> I'm attempting to derive equations for the load on each wheel of a car >> given the longtitudinal and lateral forces on each tyre. >>

Ruben wrote:

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> I'm attempting to derive equations for the load on each wheel of a car > given the longtitudinal and lateral forces on each tyre. > > I'm ignoring the effects of suspension, or in other words this thing > has no suspension. I think this is a safe assumption - kind of like a > street kart. > > It seems easy enough - you calculate the torque each wheel is exerting > on the chassis by taking the cross product of its location with its > force. The force contains an unknown z component. > > You add the torques to get total torque. The pitching torque must be > zero and the rolling torque must be zero. And, the sum of all four > vertical forces must equal mg - the weight of the car. > > That gives us three equations with four forces to solve for, and I > can't see how to tighten up the problem by including more information. > > Intuitively, I can see that in a static situation like a four legged > table resting on flat ground, my three conditions above are > insufficient since they allow for pairs of opposite legs to vary > against each other - I hope this description makes sense.

You seem to have forgotten to include the position of the weight force mg in your calculations, or at least I think you have... -- Phil http://www.usefilm.com/photographer/31307.html

Gregor Veble <[email hidden]> wrote in message news:<ckld00$2f9$[email hidden]>... > 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

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.

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> > 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

"Ruben" <[email hidden]> wrote in message news:[email hidden]...

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> Gregor Veble <[email hidden]> wrote in message > news:<ckld00$2f9$[email hidden]>... > >> 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

"Haqsau" <[email hidden]> wrote in message news:<[email hidden]>...

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> 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

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 non-rectangular wheel locations when you aren't accounting for suspension deflection. I thought you were trying to keep it simple. :o) 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" <[email hidden]> wrote in message news:[email hidden]...

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> > 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