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four wheel weight transfer



 
 
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  #1  
Old October 15th 04, 02:12 PM
Ruben
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Default 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  
Old October 15th 04, 02:41 PM
Gregor Veble
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Default

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  
Old October 16th 04, 06:35 PM
Philip Holman
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"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  
Old October 17th 04, 03:18 AM
Ruben
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"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  
Old October 17th 04, 07:45 PM
Haqsau
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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. )

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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