## Sunday, August 3, 2014

### error in bike rack force calculation?

I was looking at an excellent web page in how to make your own wall-mounted bike rack (link here). But one aspect of it was bugging me... the calculation of the force acting on the hook in the wall.

I encourage you to look at that site -- it's a bit annoying in that it requires you to click through 5-separate ad-laden pages to see it all, but I suppose that pays the bills.

Here's my diagram of how the rack would look on my wall... either in the recommended arrangement of hanging the bikes wheel up, or in a way a bike shop friend of mine recommended, wheel down. Either or a mix would be compatible with the 2-tier arrangement of hooks. With only a single tier, you'd need to alternate up-down.

Here's a separate example of such a rack with bikes hanging from the front wheel:

Going back to the original web page where the bikes are also hung from the front wheel, the part I'm most interested in here is on page 3, which described the physics. Here's the diagram from the web page, although the physics are essentially the same if the bike is hung from the rear wheel:

The argument is the net torque on the wheel about the point of contact with the wall must be zero. The wheel is free to rotate about the front hub. The frame is pulling down on the front hub with force (M - Mfw)g, where M is the mass of the bike, Mfw is the mass of the front wheel, and g is the acceleration of gravity. Additionally, gravity is working on the front wheel with force Mfw g, operating through the center of mass. Therefore the total torque operating about the point of contact with the wall, in the absence of the hook, is M g Rw, where Rw is the rolling radius of the wheel (L1 in the diagram). To prevent the system from accelerating about that point of contact, a counterbalancing torque must be applied at the hook. The argument is since torque must be force times distance, the magnitude of this force equals the weight of the bike multiplied by the ratio of the wheel radius to the distance from the wall contact point to the hook contact point. So if the distances are equal, the hook is pulling on the rim with the bike weight.

There's an obvious problem with this analysis. The full weight of the bike needs to be supported in the vertical dimension, all at the point of contact from the hook.

So where did the argument go wrong? The problem is that the hook and the hub are not the only components of torque operating on the front wheel. The bottom wheel is also in contact with the wall (or the bike would rotate). This point of contact also exerts a torque about the point of contact between the front wheel and the wall.

Going back to the point that the full weight of the bicycle must be supported by the front wheel, I analyze the torque on the front wheel about the hub, rather than about the point of contact with the wall. The combination of the frame and gravity are pulling on the hub with force M g, vertically. This exerts zero torque about the hub. This force must must be balanced by forces applied at the other two points of contact to the front wheel: the wall contact point and the hook contact point.

Assume there's negligible friction, and therefore the contact point between the front tire and the wall supports essentially zero vertical force: the point of contact between the wheel and the wall supports purely horizontal force. This point thus contributes nothing to supporting the bike vertically. If the hook came loose, the bike would fall. I think this is a fair assumption.

Then the vertical force supported by the hook-rim contact is M g, the full weight of the bike.

We now need to balance the rotational torque on the front wheel. The forces acting on the hub (the bike weight) exert no torque: the forces are acting at zero distance from the wheel's spin axis. The forces acting on the wall contact point also exert zero torque: the assumption is these are perpendicular to the tangent of the wheel, or in other words that they operate along the radius of the wheel, and radial forces exert zero torque. So all torque must be exerted by the hook, and since total torque must be zero, the hook must exert a purely radial force.

Since we know the perpendicular component of the hook's force is M g, and we know the net force must be radial, then it's easy to calculate that the horizontal component of the hook force must be M g / sine θ, where θ is the angle of the contact point relative to the horizontal.

This result can be checked trivially in two limits. In one case the bike is hanging from a hook in the ceiling. Then the hook supports the bike weight, no more. The sine of a right angle is 1, so the formula works. In the other limit, the hook is nailed into the wall with infinite force, forcing the rim to be contacted at the same point the tire contacts the wall, which is an angle of zero. In this case the supporting force is infinite. Obviously something would fail (the hook would pull out of the wall, or the wheel would deform to move the contact point).

The net magnitude of the force operating on the hook is the quadrature sum of the horizontal and vertical components:

F = M g sqrt[ 1 + 1 / sin2 θ ],

where θ is the angle of the hook relative to horizontal, and M g is the weight of the bike.

So, for the case in the web page, where the support is an equilateral triangle, sin θ = sqrt[3] / 2, and therefore instead of a net force M g, which the web page predicts, the actual net force would be M g sqrt [ 2/3 + 1 ] = M g [ 5 / 3 ] ≈ 1.29 M g. So the rim needs to support 29% more than the bike weight instead of just the bike weight.

But what about the original argument about torque operating about the point of contact between the front wheel and the wall? The answer is that is balanced by the point of contact between the bottom wheel and the wall. The excess is supported there as a force operating over the length of the wheelbase.