tstop4me, short and long track speed skates do not usually have full contact with the bottom, because they do have rocker curvature. (See https://www.danielyeow.com/2011/inline-to-ice-1 for some speed skate rocker curvatures). But I'm not sure how much that affects your argument. Especially since this forum isn't devoted to speed skating.
What I wrote was that the statement "a thinner blade would cut deeper into the ice" would hold only if the blade has a flat bottom and the flat bottom is in full contact with the ice:
But note that that "the contact area is a function of the thickness of the blade" holds only for a limited scenario: for a speed skate blade ground with a flat bottom (no hollow) oriented such that the flat bottom is in full contact with the surface of the ice; i.e., the sides of the blade are orthogonal to the plane of the ice. If orthogonality is not strictly maintained, the contact area will no longer be dependent on the thickness of the blade (here I'm excluding ice so soft that the blade sinks in).
I made no comment as to how often this scenario actually occurs in practice. I used the example of a speed skate blade ground with a flat bottom because AFAIK no figure skate blades and no hockey skate blades are ground with flat bottoms. So a limited scenario for a limited blade geometry means that "a thinner blade would cut deeper into the ice" is even more of an outlier. So my analysis is not affected at all. And, in the context under discussion (dependence on the thickness of the blade), we're talking about full contact across the transverse direction of the blade (nominal inside to outside edges) not across the longitudinal direction (toe to heel). My analysis would still hold, e.g., for a figure skate blade ground with an infinite ROH (still a longitudinal rocker, but no transverse hollow), if we were comparing the same local longitudinal section of a blade with different thicknesses.
Last edited:
(of course I'm not blaming the blades).