# Jamming & linear guiding

For a linear guiding I need to know the concept and understand the jamming limits. See picture. If I choose concept 1 or concept 2, what will be the value of X? How can I calculate this?

• Because of the extra clearance at C, this is essentially a single cylindrical fit, with offset force. A rough practical guideline is to compare "X" to the distance between the axis of the shaft near A, to how far out the offset load is applied. I.e. if X < 50, very roughly, be wary. Depends much on friction, of course – Pete W Feb 22 at 18:22
• more specifically, you can calculate the normal force at A and B, as a function of F, multiply each of those by friction coefficient. If the sum of those products at A,B is greater than F, it jams. The effect of the 10deg angle is usually negligible – Pete W Feb 22 at 18:25
• If I use concept 1 then I think jamming will appear very quickly. Assume the distance F to B = 50 and the angel is 0 degree => Normal Force on point B = F /x The friction force on the contact point B must be < 0,74 x the Normal force. X = 50/0,74 = ~ 67,5! If I use concept 2 the contact on C will result in tilting left turn. Then no jamming will appear. Is my reasoning correct? – laurens van Lieshout Feb 24 at 8:11
• concept 1- almost, except I think friction from normal force has to be counted in two places. If the R = axis offset distance, N = normal force, and neglecting the angle, then N=FR/X as you have, but F = 2μN because there is normal force in two places, both producing friction in same direction. Thus X=2μR. – Pete W Feb 24 at 11:18
• for concept 2, I agree, but the normal force is primarily from Fsinθ, so the jamming condition is μFsinθ > Fcosθ, which is never for small angles. If the force is passing through a point other than the center, it would become interesting – Pete W Feb 24 at 11:38