OK. I think we all went over this before. Correct me if I'm wrong. But % slope = tan(angle) * 100

Therefore 45 dgerees = 100% slope
90 degrees is infinite slope
60 degree is 173%

angle = inv tan(% slope/100)
60% slope / 100 = .6

inv tan(.6) = 30.96375653207

60% slope = 30.9 degrees

At the risk of hyper-analysis, here goes: ;-)

The slope & side angle capibility are limited by at least two factors. One is rollover (how far on its side can you turn the beasty 'for it rolls). The other is a G (Gravity) and traction limitation. How many G's can the vehicle maintain without losing traction? These numbers are often reported for sports cars (on G pads).

G limitations apply whether on not the vehicle is in motion. Every slope applies lateral G's in the downhill direction of the slope. If you would like to know how many G's any slope will inflict, you take the SIN of the slope. For example, a 60 degree slope subjects the vehicle to .87 G's. 55 degrees is .82 G's. That seems like a lot to me, but I have no idea what if any G pad number the Hummer has. I suspect the forward G limitation is little, if any different, than the side.

So...if the 55 degrees is a G's number, it would seem like the vehicle should be able to maintain that up/down hill without losing traction. If its just a rollover number, it would not make much difference to the traction problem.

The above isn't quite true. While, yes, sin60 = .87, this figure is not representive of the G-forces applied. By this same method, one would have 1G of lateral force if the truck were at 90 degrees (vertically sitting on its butt). G-measurements around a skidpad are taken on flat surfaces and not on ramps for one reason -- you still need to have the same DOWNWARD force applied to measure the lateral traction.

The essence of what I'm saying is that although one DOES experience a lateral force of .87 x mass-of-vehicle at 60 degrees, one also experiences a reduced traction force of .5 x mass-of-vehicle (cos60). This effectively indicates a G-force of 1.73 (tan60), not .87.

Andreas

On a side slope, the amount of (gravity) force holding the wheels to the slope (i.e. friction) is also reduced.

The 55 degrees is how for you would have to tip the Hummer to the side before the Hummer's Cenger of Gravity moves outside of the tire profile. This involves several simplifications (such as weight shift from suspension and tire deformation under lateral load are ignored completely). That is why I always label this as the "theoretical" tip-over point.

Isn't the G-pad stuff for sports cars about cornering stability rather than side-slope capability? (Yes, these are very similar, but not identical.)

Percent slope is "the altitude change for a certain horizontal distance" divided by "that same horizontal distance". A 60% slope is one where it rises (or falls) 6 feet for every 10 feet traveled (horizontal, not along the slope).

A 60% slope is about 32 degrees. The Hummer will do very well at much steeper angles (more than 60 degrees), but only if there is enough traction. BTW, the calculated (i.e. not tested!) side slope limit for the hummer is about 55 degrees.

When AMG says that the Hummer will go up a 60% grade at full GVW, they mean that it will climb that grade for extended periods pulling all that weight (assuming sufficient traction, or course). For short periods, it will do much better, of course.

Dave Breggin
'95 Diesel Wagon