In an idle moment I thought to amuse myself by trying to determine the theoretical relationship between the radius of a carved turn, the sidecut radius of the ski, and the angle of inclination. The upshot is that (with a bunch of heavy-handed simplifying assumptions) the turn radius is something like the sidecut radius times the cosine of the angle of inclination.

To make the analysis more tractable for a first attempt, I'll assume that the snow is really hard and the ski doesn't penetrate into it any appreciable amount, sort of like skiing on an ice skating rink. First, imagine the case where the ski is tilted only very slightly on edge, say one degree. In this case, the shape of the edge of the ski where it contacts the ice is essentially that of the sidecut of the ski, so the radius of turn (assuming no slippage) is the same as the sidecut radius of the ski.

Now let's see what happens as you tilt the ski more on edge, say to 45 degrees. If you don't put any weight on the ski (I'm ignoring the camber of the ski here) the tip and tail will contact the ice and the middle part will be some distance above it, due to the sidecut. As you put pressure on the ski, the ski will bend until the middle contacts the ice as well. The resulting curve along which the ski contacts the ice will have a radius smaller than the sidecut radius of the ski. Again assuming no slippage, this new radius will be the radius of the carved turn made by the ski. How does this radius relate to the angle of inclination?

Let's call the angle of inclination A, and let d be the sidecut depth of the ski (the distance form the edge at the middle of the ski to a straight line joining the tip and tail edges). Let D be the "sidecut depth" of the curve along which the ski contacts the ice when pressured. Then you can show that D = d / cos A. Now how does this relate to the radius of turn?

Assuming circular sidecuts, with some geometry it is possible to show that

r = d/2 + L^2 / (8d),

where again d is the sidecut depth, L is the length of the ski, and r is the sidecut radius. Since d is very small compared to L^2, a reasonable approximation is that

r = L^2 / (8d),

so that the sidecut radius r is approximately inversely proportional to the sidecut depth d.

Now let R be the radius of the carved turn made by the ski. Using the formula above, except replacing d with D, we have

R = L^2 / (8D) = L^2 / (8 d / cos A) = r cos A.

So the radius of the carved turn should be the sidecut radius of the ski multiplied by the cosine of the angle of inclination.

I was rather surprised at this simplicity of this relationship.

In addition to the assumptions already mentioned, we are also assuming that the ski will bend enough to completely contact the ice at the given angle of inclination. Therefore this will not be valid for inclinations too close to 90 degrees.

Here are a few values for a ski with r = 27m.

A r cos A

-----------------------

0deg 27m

20 25

40 21

60 14

I wonder if these values are anywhere near realistic in practice? If you lean over at 60 degrees on hard snow, can you carve a turn something like half the radius of your sidecut?

Robert

Edited by renenkel - 9/21/12 at 8:40pm