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.
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.
> 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'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 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.
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
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.
> 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.
--
Phil
http://www.usefilm.com/photographer/31307.html
If, howerer, this is an analysis of car behaviour you'd like to be
doing, read on...
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*** 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
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
> 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.
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.
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
>> 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*** 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..
PH
> 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.
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
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.