While I prefer the sliding beer indicator, I can't pass an opportunity to pull out my engineering mechanics statics book.
My calcs were done for a 2WD base and 4-wheelers will be better off due to a wider base.
The problem statement is "Find the maximum hieght where P is the force that may be exerted without tipping the van over."
h = b\2u = 37.2"
b = base = 67"
u = friction coefficient for rubber on dry smooth pavement = 0.9 static (or 0.8 kinetic)
You want the van's center of gravity (cg) to be below h. P is going to be determined by the van's gross wieght (assume GVWR if the van is not overloaded), plus the force of acceleration if you are moving.
The 2010 Ford E250 RB center of gravity is 32" (floor hieght) and 31" for the E350 SD. Each conversion is going to be unique and you will have to calculate your new cg based on the weight and location of each added item in relation to the original cg for summation of your new cg.
Interestingly, Ford sets the maximum allowable cg for converters at 45.8" for an E250 with 8,000 GVWR. However, they reduce this to 39.2" when their Roll Stability Control (RSC) system is installed. Due to their liability, I figure these are conservative figures. Results from the academic formula appear to be even more conservative (not sure why, but it most likely has to do with the model used; i.e., a solid rectangle versus an object with wheels).
Being at an angle will raise the van's center of gravity. My trigonometry is rusty. Can anybody lend a hand with the math and tell us how many additional inches you lift the van when it is tilted at 30 degrees?