Vehicle weight distribution

Question

I've been creating a 2D (above view) vehicle simulation as a hobby project, and (as is often the case) a problem which I assumed would be simple has turned into the sticking point of the project.

Essentially, I want to be able to calculate the total load on each tire given arbitrary vehicle configurations.

Assuming we just had an 200kg engine, and it was placed exactly in the centerpoint between 4 tires, suspended there by magical zero-mass connections, there would be 500N evenly distributed across each tire. If you put it directly over the front axle between the two tires, there'd (presumably) be 1000N on each front tire and nothing on the rear tires.

But what if it were overhanging past the front tires? Would there be less than 1000N on each front tire, and negative load on the rear tires due to the front axle acting as a lever?

Then on top of that, what about adding other components, like a fuel tank offset to one side?

In googling for this I've found all sorts of confusing things, and I have a feeling I'm overcomplicating it as a result.

Is there a general solution for such a system?

EDIT: I have mocked up a testing app, which is available here: http://www.fastswf.com/H4o9u_c You can move the tires or blue weights by clicking and dragging. Hold A while clicking to add a weight. Each weight is 1000N.

My method was to project the weight on the line between the middle of the front tires and the middle of the rear tires, use this to find the front/back weight distribution, then do the same projection for the middle point between the left tires and right tires. The amount of force seems to add up (barring rounding errors it's always the same as the net weight), but I don't know if this is correct or even ballpark correct. One thing that concerns me is that the black circle (intersection point between the 4 tires) and the intersection between the mid-way centerpoints is quite different, and I'm not sure which one is more relevant to calculate from.

For the moment I am not including the tires in the distribution since they're infinitely stiff and grounded, but for arguments sake they can be 20kg. I have just added a mockup to the original post which shows my proposed solution, but I'm not sure if it's correct at all. — Elliott Smith, Oct 11 '16 at 05:17
Possible duplicate of Estimate the reaction force on each leg of a 4-legged table — John Alexiou, Oct 14 '16 at 19:35
This problem is identical to resolving the weight on the table legs given arbitrary center of mass location. I hope the link above helps. — John Alexiou, Oct 14 '16 at 19:36
Also know that the traction circle for FWD or RWD cars is a circle for braking, but an ellipse for acceleration, limited to the traction of the wheels with power (and the weight on them). — John Alexiou, Oct 14 '16 at 19:58

score 0 · Answer 1 · answered Oct 14 '16 at 17:30

In general, the center of mass has to be inside a triangle formed by a subset of supporting points. You are assuming that there is some elasticity in the suspension of the wheels, so all four are carrying the load. In that case, you can write down the equations for balancing torque. You can compute torque about any point, but it is easiest to calculate it around the load.

What you end up with is three equations - one that says the sum of the forces on the tires must equal the total load, and two that balances the torque to zero about the X and Y axis. So you have a total of three independent equations, and a unique solution for exactly three tires (showing that three tires is necessary and sufficient for a stable solution; if the forces all have to point in the same direction, the load has to be inside the triangle of the three wheels). But since you have four wheels, you have an additional degree of freedom - and in principle infinitely many possible solutions.

The only way to get around that is to make certain assumptions about the load distribution to the fourth wheel. This could be done by using symmetry, or by assuming a certain elastic behavior of the suspension (so that the load can be carried by the wheel that is "outside the triangle").

score 0 · Accepted Answer · answered Oct 14 '16 at 18:24

Let's assume the coordinates of the tires as (0.0);(4,0);(0,1.5);(4,1,5) meters. and the engine mass =m, placed at coordinates (x,y).

The share of front axel and rear axel bearing m is given by this equation.
Tributary load to front = m.(1-x/4).
The rear axel load is = m.x/4.
The right side tires load = m.y/1.5.
The left side tires load = m(1-y/1.5).

and if your engine x is greater than 4 or less than zero you have negative load on the tire that you can counter with the ballast weight or weight of the transmission, etc.
The above equations will apply to all other parts of the car and will determine its CG.

Vehicle weight distribution

2 Answers2