Recent content by dougk16

Also wanted to say that one constraint is that there has to be one tire in each quadrant of the chassis, where the quadrants are defined by where the center of mass is. Basically all four tires have to surround the center of mass. Which makes sense in real life, cause otherwise the chassis...

Jeez, just typed out a whole post describing my algorithm, and hit submit, only to have a timeout and lose my work...so here's an abridged version: tire weights = [(1-topY/L)*(1-leftXTop/WTop)*m; (1-topY/L)*(1-rightXTop/WTop)*m; (1-bottomY/L)*(1-leftXBottom/WBottom)*m...

I tried it with some different numbers and configurations with 2<=n<=4. With our working set in this thread, it came back [333.33..; 166.66..; 333.33..; 666.66..], which turns out to be [4500; 5500;]; With another set [1 4 1 5; 7 9 1 1;] it came back [214.29; 428.57; 428.57; 428.57], which...

Ha, I found an algorithm that satisfies the top half of the matrix, i.e. 4500, but is a little off for the 6000 part...so close!

That looks very cool. It'd still be nice to find an easy way to solve this problem, but being able to compile C to Flash is just cool in general. I hope that project gains momentum. I've tried various permutations of my original algorithm, and even though they all satisfy the first...

Good point. I was thinking that since in our case the table legs are on springs (the car's suspension and tire deformation) it would be a little more uniform, but I suppose even if springs limited the wobble, you could easily have one leg supporting more weight. I'm working in ActionScript...

I'm not sure my solution is correct, because the 'top' two tires have the same mass share, and the bottom two have the same mass share (I lost the sheet with the exact numbers). With my solution for example, if you had a 120kg bridge (center of mass in the center) supported by three pillars...

I'm a little confused...for your solution to be exact, the center of mass has to be adjusted? I don't see how different placements of tires could change the center of mass so significantly. I mean yea technically the tires have to have some mass, and therefore they might throw off the center...

Thank you again, log. With those numbers in there your reasoning is much clearer to me. However, 1145kg seems like quite a deviation from 1500kg. Also, the tire's that are furthest from the center of mass are supporting the most mass, which might be right, but doesn't make sense to me. Is...

I appreciate your reply log. I've been puzzling over it for about half an hour now, but I'm not well-versed in math notation, and I'm not sure how to apply this to my situation. Could I trouble you for a brief example? For example, let's say we plot a rectangle (width=6, height=10, lower left...

Hi all, I'm writing a car simulation for a game engine, and this problem has me going bonkers. For a car with known mass and center of mass, how can you figure out the "share" of mass that each tire is supporting? For standard tire arrangements (4 tires on 2 axles) I know it's just a ratio...