Any Math wizards?

[1 Bad F N Z]

Lets say the Landcruiser weighs 4000 lbs., and each axle carries 2000 lbs. each. Wheel base is 90 inches center to center. Now I've put my spare tire rack on with full cans and spare tire. I've just added 225 lbs. 30 inches farther back from the rear axle. How much weight did I take from the front axle? And the formula? Kevin.

[inVERt'D]

I'm no math wizard, especially after 12am, but I think the problem as you describe it could be modeled statically as a class 1 lever and fulcrum with the lever having weight. I'll leave it at that for tonight, since I'm dead tired.

[6spdYJ]

This has been bugging me all day. It will depend on your front and rear spring rates. It won't actually take any weight from the front axle as much as just move it rearward. If the springs have a rate of infinity (solid) - which they won't, you have to know the weight on the frame (body, t-case, etc...) and how it's dispersed front and rear. Now measure from the front of the frame to the back of the frame, taking note of the rear axle placement. Divide the length forward by the length rearward. Now divide your extra 250 pounds by this figure. This should get you close.
The difficulty is in the spring rates. If the front rates are high, it will push more weight rearward. Same if your rear springs are soft. If it's opposite, soft front springs may sit no different but the back may stay at the same height and possibly ride smoother. Maybe play with a center of gravity calculator, or make a few calls to custom spring shops with a strong reputation.

[aw12345]

Lets say the Landcruiser weighs 4000 lbs., and each axle carries 2000 lbs. each. Wheel base is 90 inches center to center. Now I've put my spare tire rack on with full cans and spare tire. I've just added 225 lbs. 30 inches farther back from the rear axle. How much weight did I take from the front axle? And the formula? Kevin.

Simple people would go to a scale, weigh the front axle before and after you mount the jungle gym on the rear, Easy peasy

[inVERt'D]

This has been bugging me all day. It will depend on your front and rear spring rates. It won't actually take any weight from the front axle as much as just move it rearward.

I meant to get to this today, but I'll shoot for tomorrow or the next. It's a leverage problem. The only real question is, where is the fulcrum? Is it over the rear axle? Start by trying to solve the problem as he described it, because you don't have any other information. No model is perfect, but some are useful. If each axle has 2000# on it (not true, unless perhaps it's on an incline) I think you can consider the COG to be centered between them.

Another way to approach the problem would be to ask how much the front rises over the front axle centerline in inches x. Then you can solve the problem by figuring the lbs torque required to lift 2000# x inches.

[aw12345]

I meant to get to this today, but I'll shoot for tomorrow or the next. It's a leverage problem. The only real question is, where is the fulcrum? Is it over the rear axle? Start by trying to solve the problem as he described it, because you don't have any other information. No model is perfect, but some are useful. If each axle has 2000# on it (not true, unless perhaps it's on an incline) I think you can consider the COG to be centered between them.

Another way to approach the problem would be to ask how much the front rises over the front axle centerline in inches x. Then you can solve the problem by figuring the lbs torque required to lift 2000# x inches.

The latter is usable. measure ride height in rear and front then after the jungle gym is on remeasure and with those numbers and some math you could arrive and a fairly accurate number. Still using a scale is a heck of a lot easier. However the problem when it counts is when you go on steep inclines that is when the weight transfer matters most and is somewhat harder to calculate