Consider a vehicle with four wheels and suspension (coil and dampers) rolling over an arbitrary surface. What would be the magnitude (not direction) of the normal force on each wheel? Is there a method in dynamics to handle such a predicament? I am familiar with methods in dealing with static indeterminacy, but I am not sure if these methods hold for this situation.
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$\begingroup$ I think you need to better identify your question, whether the indeterminacy is regarding the frame-wheel assembly or the suspension-wheel assembly (global system vs local system),; and, to spell out what are the methods that you considered not applicable. $\endgroup$– r13Dec 1, 2021 at 16:54
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$\begingroup$ Well you could express the dynamics of the system as a larangian with holonomic constraints. Or just use a multibody simulator because deriving the whole damn thing took a entire 500 page book to do. $\endgroup$– joojaaDec 1, 2021 at 22:00
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2 Answers
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There are many ways to address this, depending on the level of accuracy that is required. IMHO the simplest approach would assume a infinitely rigid chassis and spring of the same stiffness. In that case you would be able to work with the symmetry of either the longitudinal (along the motion).
Alternatively the transverse axis can be used. If you use the transverse, effectively what you would be doing is you use the bicycle model. I.e. you assume that both front wheel are one -- essentially calculating the forces on each axis.
With the bicycle model, you'd only use the longitudinal acceleration and inlination. Once you have the the forces of each axis you can use the centrifugal acceleration and/or inclination to distribute the forces on each axis.
Closed forms equation can be derived for this, but largely depend on the frame of reference you are working.
if you wanted a more detailed model, you'd need to account for the deformation of the chassis and the suspension system. This is way more complicated, and in most cases the results are --usually -- in the same order as the simpler case mentioned above.
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One reasonable way is to calculate the CG of the car and assign its centripetal forces to different tires.
$$F_{centripetal}=m\frac{V^2}{R}$$
- R-radius of turn
- $\frac{V^2}{R}$ is centripetal acceleration.
Let's set the origin at the geographic center of the car and x-axis and y-axis along the length and width f the car. Say your CG is at the x of the car near its middle and its y is at the center axis, at the height of 25cm above the road.
So we apply the X and Y components of the moment of $F_{centripetal}*25$ plus any forces due to car acceleration or breaking to the car at its CG knowing that this overturning momentum of $F_{c_x}$ adds or subtracts to the tributary loads of each tire or in extreme cases can even lift and roll the car.
For example, let's say the weight of the car is distributed 30% to each rear wheel and 20% to each front wheel and we have the sum of moments component resolved to $$F_x=-0.2m\ and \ F_y=0.60m$$
Then the lateral contribution of momentum is,
$$0.6M*25CM / \text{width of the car}= F_{turn}$$
This $F_{turn}$ will add to the left side and subtract from the right side, inversely proportional to the distances of the rear and front tires from the CG of the car. Then we do the same for the x component of momentum and add/subtract to the front side and rear. The Sum of all these on each tire is its normal force.
Of course, as we know things are more complicated than this as we get to suspension and changing of the footprint of the tires and their angle to corner better.
Here is a figure of a car in a turn and the force on it.
