'Nother post for emphasis..



Pay careful attention. I'll speak slowly so TJ can understand.

Q. What happens to the leaf spring of this flexed Revolver suspension when the tire first gets pushed up into the wheel well, from this position?

A. For the first couple inches, there is very little resistance as the revolver is folding up. The spring is not starting to compress until after the revolver is folded back neatly into place.

Q. If there were a normal shackle in that photo, what would happen as soon as the tire started being pushed up into place?

A. The leaf spring would begin compressing instantaneously, providing a resistant force.

Q. How does a leaf spring work?

A. When a force is applied, either upwards or downwards, a spring flexes based on it's spring constant, to resist the force.

Q. If the leaf spring does not compress, as the tire was raised, was any force being resisted?

A. No. It was dangling in it's natural state.

Q. How is traction generated?

A. It's the normal force (perpendicular) to the ground, multiplied by the friction coefficient of the two surfaces.

Q. What would the traction (aka, friction) be as soon as the tire began to be compressed in the revolver situation as in that picture?

A. The weight of the tire, hub, axle, etc. multiplied by the friction coefficient between the tire & the surface. Until the revolver is folded back onto itself, there is no additional force acting downward on the spring.

Q. What would the normal force be in the normal shackle situation as soon as the tire began to be compressed?

A. It would be the weight of the tire, hub, axle, etc., PLUS the force provided by the spring,. all of that multiplied by the friction coefficient.

Q. Wouldn't the force of the spring be acting negatively (if down is positive) on the assemblies, therefore negating the regular spring advantage?

A. Yes, but no. Yes, the math of it would show the spring would technically be pulling the axle assembly upwards, therefore a reduction in traction from the total. However, since the spring is still connected to the revolver, the revolver can't actually droop w/o taking the spring properties into effect. So at the point the revolver starts opening, it has already reached the maximum up-pull of the leaf spring. Therefore, it's already at it's lowest amount of traction available in the equation. It does not change at any point in the revolver travel.

So for example, if the spring was pulling -5 lbs of force (remember, up is negative), it reaches that -5 lbs at the exact instant the revolver begins to open up. Whether the revolver is open 0" to 3", the same 5 lbs are pulling up. But on the regular leaf suspension that flexes to the same point, the full -5 lbs of pull is only at the lowest point, aka, the equivalent of the 3" open revolver. As soon as the regular shackle leaf pack begins compressing, this -5 lbs of pull begins to reduce to 0.

So no, the spring will not negate the advantage of the regular spring. Both a revolver and the leaf pack have the most amount of traction at the point the spring force is 0. However, the revolver will not even start to begin to reach that point until after it has already closed together.

Conclusion:

Through the handy-dandy method of linearity, you can move the origin of force for the spring pack down to the most fully open position (fully flexed out) of the wheel, and begin from there. This makes it easier to visualize (except for TJ, which will probably argue this point, though he would be wrong to do so. This is simple statics/dynamics at work, here, and the calculations/rules for doing this are taught in 100 level engineering classes; in other words, it ain't rocket science.)

Once you do this, you will find the revolver doesn't start adding traction from the spring until it's fully closed, but the regular leaf spring starts adding traction from the moment it starts compressing.

In other words, and one more time for those that have fallen asleep:

IF YOU WANT FLEX WITH TRACTION, GET A REAL LEAF PACK. IF YOU WANT FLEX FOR AN RTI RAMP, USE REVOLVERS.