Quote:
Originally Posted by swamp2
Still disagree. Getting way OT but...
Consider two equal masses m, a distance d apart, say equal to a wheelbase. This is an idealized two point mass "car". This results in 50:50 weight distribution and the polar moment I(about cg) = m d^2/2. Alter the weight distribution radically so 100% is on one axle. Weight distribution = 100:0, total weight unchanged. Now here I(about cg) = 0. For the more general case you can solve for the generalized I about the cg for any value of m1 and m2.
I = (m1 m2 d^2)/(m1+m2)
Again even though m1+m2 is fixed change their ratio and you can compute exactly how I changes.
Through this simple example you should be able to see that axle weight distribution and polar moment are intricately linked. You can extrapolate to the case of a continuous variation in mass along the length rather than two point masses. In both cases though change one and you change the other. In the simple cases there will be an analytical solution, in the general case certainly not.
Good book by the way!

The flaw in your example resides in the fact that you consider the masses limited to be directly over the axles.
Take your example of the two mass car but instead of combining both masses over one axle, combine them right in the middle of the two axles. You would also reduce I to zero but would maintain the 5050 weight distribution.
As another example, let's assume a car of a given weight and polar moment. If the axles are positioned at equal distance from the centre of gravity, the car has a 5050 weight distribution. Moving the front axle forward (assuming negligeable mass for the axles themselves) will shift the weight distribution towards the rear axle, but the polar moment around the CG remains unaltered.
In both examples one parameter could be changed without impacting the other.