Is a chair with N+1 wheels ever less stable than a chair with N wheels?

Question

My workplace has a policy to provide standing desks upon request, but no policy to provide chairs of matching height. (I work for the government...) We can buy our own, or build our own. However they added a stipulation that the chairs must be bought or built with five or more wheels.

Since five is an awkward symmetry to measure and to sit in, I'd probably build my chair with six wheels. Or perhaps eight because I can start with a square and lop off the corners to make an octagon base. Or maybe make the base a perfect circle and line the bottom with innumerable tiny wheels?

This set me to wondering: is there any situation in which a chair with N+1 wheels is strictly less stable than a chair with N wheels?

To make the question interesting and rule out trivial answers, assume that the wheels are more-or-less at the chair's perimeter and more-or-less evenly distributed. Assume the floor is flat.

Answer 1

Assuming all the wheels are evenly spaced on the same circle, then more wheels is always more stable than less wheels. However, there is diminishing return as the number of wheels gets large.

The metric of stability is how far from the center of the circle the center of mass can be before the chair tips over. The chair is stable whenever the center of mass is inside the polygon formed by all the wheel points. The worst case is in the center of one of the edges, since these are the closest points on the polygon to the center of the circle. In the limit, with infinite number of support points, the minimum distance to instability is the radius.

We can therefore quantify stability as the minimum distance to instability relative to the radius. A value of 1 is the maximum, with infinite support points. After a little trig, it's easy to see that this stability metric is:

  S = cos(Π/N)

where N is the number of support points. The stability metrics for values of N up to 20 are:

   N      S
----   ----
   2   0.00
   3   0.50
   4   0.71
   5   0.81
   6   0.87
   7   0.90
   8   0.92
   9   0.94
  10   0.95
  11   0.96
  12   0.97
  13   0.97
  14   0.97
  15   0.98
  16   0.98
  17   0.98
  18   0.98
  19   0.99
  20   0.99

Office chairs commonly use N=5, which is a tradeoff between being good enough but not too expensive. The extra 7% stability from adding a 6th wheel isn't worth the cost. Or, put another way, you can achieve the same stability as 6 wheels by using 5 wheels but growing the base another 7% outward.

Answer 2

I think the "health and safety" regulation about 5 wheels is a compromise between stability and cost.

If your weight is on the edge of the chair seat and the chair has only 3 wheels, it is much less stable if you are in line with one of the wheels than if you rotate through 60 degrees to be mid-way between two wheels. That might happen (1) because the occupant turns on the seat, (2) if the chair is moving and one wheel hits an obstruction which rotates the legs, or (3) the occupant is sitting centrally on the chair, but moves from "leaning forward" to "leaning backward".

The result could be that a stable sitting position suddenly becomes unstable. For a 3-legged chair, the minimum stable distance for a load offset from the center is only half the maximum stable distance.

A bigger number of wheels reduces this issue, but increases the friction that has to be overcome to point all the wheels in the correct direction to move the chair. It is also more expensive to manufacture. In the worst case situation of tipping over, all the load on the chair is carried on just one wheel independent of how many wheels the chair has, so increasing the number of wheels doesn't allow the size of each leg and wheel to be reduced!

The "min:max stability ratio" of 0.5 for a 3-wheel chair increases to about 0.7 for 4 wheels, 0.8 for 5, and 0.9 for 7 wheels. IIRC, the safety regulations in the UK changed from 4 to 5 wheels some time during the 1970s.