How to find the uneven weight distribution on 4 wheels

  • #1
MStoffle
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Hello, I am working on a project where I am trying to find the weight distribution on a 4 wheeled vehicle that is traveling diagonally up an incline. You can effectively think of the vehicle as having a pitch and a yaw angle in the body frame. Using an arbitrary location of the total center of mass, I tried using equilibrium equations to find the reaction forces of all the wheels, but after taking sum of forces and moments to be zero about one of the wheels, the problem appears to be statically indeterminate. I am particularly interested in the normal component reaction forces of the wheel so I can continue a traction model for the vehicle facing any direction on the incline. I have provided imgur links to two pictures of my notes for the scenario (It may appear sloppy and I apologize) for better reference and you can see that with the end equations, I would particularly be interested in the first, second, and fourth equations for the z components, but that is still 3 equations with 4 unknowns.

https://imgur.com/3ON5rOs
https://imgur.com/j3g1zuF

I opened these images and noticed they may be blurry. The equations I have come up with taking moments about the front left wheel:
Fz,FR*(-d) + Fz,RR*(-d) + Wy,cg*(-h) + Wz,cg*(c) = 0
Fz,RL*(-b) + Fz,RR*(-b) + Wx,cg*(h) + Wz,cg*(a) = 0
Fx,FR*(-d) + Fy,RL*(-b) + Fx,RR*(-d) + Fy,RR*(-b) + Wx,cg*(c) + Wy,cg*(a) = 0
Fz,FL + Fz,FR + Fz,RL + Fz,RR - Wz,cg = 0

The other 2 remaining equations would relate sum of forces in the x and y direction.

What can I do from here to solve this problem? Am I missing something?
Any advice is greatly appreciated!
 

Answers and Replies

  • #2
jack action
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Have you tried the following:

Front view:
Fz,(F&R)R*(-d) + Wy,cg*(-h) + Wz,cg*(c) = 0
Fz,(F&R)R + Fz,(F&R)L - Wz,cg = 0

Splitting Fz,(F&R)R:
Fz,FR + Fz,RR - Fz,(F&R)R = 0
Fz,(F&R)R*(a) - Fz,RR*(b) = 0

Side view:
Fz,R(L&R)*(-b) + Wx,cg*(h) + Wz,cg*(a) = 0
Fz,R(L&R) + Fz,F(L&R) - Wz,cg = 0

Splitting Fz,F(L&R):
Fz,FL + Fz,FR - Fz,F(L&R) = 0
Fz,F(L&R)*(c) - Fz,FR*(d) = 0

You get 8 equations, 8 unknowns.
 
  • #3
MStoffle
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A few questions:

I recognize your 1st, 2nd, and 5th equations are the same as the few I mentioned as well, the moments in the x_hat and y_hat direction about the front left (FL) wheel and the full sum of all the z forces to be 0. However, would equations 3 and 7 effectively say 0 = 0 if simplified?

Fz,FR + Fz,RR - Fz,FR - Fz,RR = 0

Also, I may just be getting confused on the top view and side view perspective or how these reaction forces are being split up, but where are equations 4 and 8 taking the moments about/being derived from?

Thank you!
 
  • #4
jack action
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I don't think I made any mistake, all equations will end up to be 0 = 0 and I'm not sure how you got your equation from #3 and #7. (I tried to keep the same notation you did and it can be confusing, either on my part or yours). Although I haven't tested it with numbers.

The idea is to look at the front view and combine both axle as if there was only one. Then you know whatever you will find for the left & right sides will have to be split between the front & rear.

Similarly for the side view, you collapse both sides together as if your were working with a motorcycle. Then you know whatever you will find for the front & rear will have to be split between the left & right sides.

The way I split it, I'm assuming the vehicle frame is perfectly rigid and can be modeled as beam. Basically, if CG is perfectly in the middle, you split any combined normal force you found in two for each wheel.

In reality, if you have a suspension, the wheel rates can be different for the front and rear axles. This will affect the splitting between the front & rear axles when doing the front view. For example, if the rear axle had no suspension (perfectly rigid) and the front had a suspension, the moment reaction will be completely absorbed by the rear.

But the basic approach should be to decouple the different force and moments and add them together:
  • Do the analysis with Wz only and find Fzz for each wheel;
  • Do the analysis with Wx only and find Fzx for each wheel (side view);
  • Do the analysis with Wy only and find Fzy for each wheel (front view);
  • Add Fzz, Fzx & Fzy together for each wheel.
 
  • #5
MStoffle
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Okay, I think I was mixed up on your method and notation. I now see the terms like Fz,(F&R)R meant the both the right wheels collapsed together. Also, it seems you are splitting the problem up into a couple 2D problems rather than 3D. Continuing with this, yes, I am assuming the vehicle will be perfectly rigid and I am trying to using a generalized c.g. location, as it is not completely known yet.

I tried drawing out your method and rearranged a few equations for correct signs (I'm treating positive z as down, x -> forward, y -> right), and tried running some numbers as a system of equations in Matlab:

https://imgur.com/a/y2yEL
https://imgur.com/a/Pctna
https://imgur.com/a/YW6NJ (Matlab Picture)

Front:
(1): ΣMFL= -Fz,(F&R)R*(d) + Wz,cg*(c) - Wy,cg*(h) = 0
(2): ΣFz = Fz,(F&R)R + Fz,(F&R)L - Wz,cg = 0
(3): Fz,FR + Fz,RR - Fz,(F&R)R = 0
(4): Fz,(F&R)R*(a) - Fz,RR*(b) = 0

Side:
(5) ΣMFL = -Fz,R(R&L)*(b) + Wz,cg*(a) + Wx,cg*(h) = 0
(6) ΣFz = Fz,F(R&L) + Fz,R(R&L) - Wz,cg = 0
(7) Fz,FL + Fz,FR - Fz,F(L&R) = 0
(8) Fz,F(L&R)*(c) - Fz,FR*(d) = 0

Threw this into Matlab and got a singular matrix. I realized there was no relation for the Rear Left (RL) wheel so I thought maybe adding a relation like the following would help:

Fz,FL + Fz,RL - Fz,(F&R)L = 0

or these
Fz,RR + Fz,RL - Fz,R(L&R) = 0
Fz,FR + Fz,RR + Fz,FL + Fz,RL - Wz,cg = 0

This did allow the problem to solve and provide the correct solution for a flat surface, however, when given a pitch angle, the numbers seem to go astray. I have tried to see if there are any typos, but I am sure I have fixed them all. This lead me back to the question I was having before:

Where exactly are equations (4) and (8) derived from? I have tried drawing a picture but can't seem to understand why these equations are found/used. If this is not the issue, what else do I seem to be missing something?

Thank you!
 
  • #6
jack action
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For (4) & (8), you just assume the resultant force on one side (or axle) is acting in between the two wheels like if it was on beam. For (4) you have to go back to the side view and you put your resultant force Fz,(F&R)R at the CG and do the sum of moments about the front right wheel. Similarly for (8), you have to go back to the front view and you put your resultant force Fz,F(L&R) at the CG and do the sum of moments about the front left wheel.
 

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