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Sidecut Radius

post #1 of 5
Thread Starter 
I suspect this is a thread that will be of interest to Physics Man, and nobody else, but here it is anyway.

I fooled around a little bit recently with measurement of sidecut radius. My math skills are, I'm sure, on a lower plane (if you will) than Physics Man's, consisting solely of vague 30-year-old memories, combined with rusty intuitive leaps, some of which might be questionable.

In any event, there are at least three ways to determine the radius of a sidecut:

The quick way:

Look at the ski at three places: the waist (the narrowest part of the ski), where it's widest near the tip (the front end) and where it's widest toward the tail (the rear end). The engaged length (call it L) is the length from the front end to the rear end.

To simplify, assume the waist is halfway between the front end and the rear end. The width of the ski is different at the front and rear (i.e. the depth of the sidecut in front is different from in the rear). Neverthless, to simplify, treat them as if they're the same, by adjusting each so it's equal to the average.

So D = depth of sidecut = (front width / 2 + rear width / 2 - waist width) / 2.

Sketch this out on a piece of paper with the tip of the ski at the top, and the tail at the bottom; put a point way off to the right to represent the center of the circle whose arc includes the sidecut. Draw a line from the center to the front end and to the waist.

You've just drawn a right triangle, whose hypotenuse is the radius of the circle: one side (up & down) has a length equal to L/2; the other side (left & right) has a length equal to R-D.

So, applying the Pythagorean thereom:

R^2 = L^2/4 + (R-D)^2
= L^2/4 + R^2 - 2RD + D^2
R^2 - R^2 + 2RD = L^2/4 + D^2
2RD = L^2/4 + D^2
R = L^2/8D + D/2

If you want to simplify further, note that D is, at most, 30mm, so the D/2 term only affects the answer by 15mm (or .015 m). Since R is going to be something over 10 m, you don't lose any significant accuracy by omitting the D/2 term, since it's about 3 orders of magnitude less than the answer ... which is smaller than the error in your measurements.


R = L^2 / 8D

The elaborate way:

Any three distinct points (call them A, B and C) lie on one (and only one) circle.(1)(2)

By definition, the center of a circle is equidistant from all of the points on the circle.

Start with points A and B. Draw a straight line between them. The mid-point halfway along this line is equidistant from both of them. The complete set of other points equidistant between them form a line that passes through this mid-point and is perpendicular to line AB. All of the points on this "midline" are equidistant from A and B.

Do the same thing with points A and C. All of these points on the AC midline are equidistant from A and C.

The point where the AB midline and the AC midline intersect is the center of the circle. The distance from that point to A = the distance from that point to B = the distance from that point to C. (3)

So, let's apply this to the problem.

Look at the right edge of a ski. The three points we're trying to run a circle through are three points along the edge of the ski: (A) at the waist, (B) where the ski is widest near the tip and (C) where the ski is widest near the tail.

The length of the ski (along the ski's centerline) from where it's widest at the tip (next to B) to where it's widest near the tail (next to C) is the "engaged length;" call it L. Part of it is in front of the waist (Lf), and part of it is to rear of the waist (Lr).

The sidecut has a depth in front (Df) and in the rear (Dr). The depth in front is half of (the width in front minus the width at the waist); same calculation at the rear.

Go into an x-y coordinate system, with A at the origin, the Y axis extending along the length of the ski, parallel to the centerline of the ski, and the X axis (obviously) perpendicular to that.

A is at 0,0
B is at Df, Lf
C is at Dr, -Lr

The AB midpoint is at Df/2, Lf/2
the AC midpoint is at Dr/2, -Lr/2

The slope of line AB is Lf/Df
The slope of line AC is -Lr/Dr

The slope of the AB midline is -Df/Lf
The slope of the AC midline is Dr/Lr

The equation that describes the AB midline is: y - Lf/2 = -Df/Lf (x - Df/2), or:
y = -Df/Lf * x + Df^2/2Lf + Lf/2

The equation that describes the AC midline is:
y = Dr/Lr * x - Dr^2/2Lr - Lr/2

The intercept lies at Ix, Iy. To determine the x-coordinate of the intercept:

Dr/Lr * Ix - Dr^2/2Lr - Lr/2 = -Df/Lf * Ix + Df^2/2Lf + Lf/2
(Dr/Lr + Df/Lf) Ix = Df^2/2Lf + Lf/2 + Dr^2/2Lr + Lr/2

Ix = (Df^2/2Lf + Lf/2 + Dr^2/2Lr + Lr/2) / (Df/Lf + Dr/Lr)

To determine Iy, insert Ix into one of the mid-line equations above:

Iy = -Df/Lf * Ix + Df^2/2Lf + Lf/2

Apply the Pythogroean theorem to determine the radius:

R = sqrt ( Ix^2 + Iy^2).

So that's the general way to determine the sidecut radius.

As a test, apply it to the "quick-way" simplified case above. In the simplified case, the waist is at the mid-point of the engaged length, so Lf = Lr = L/2. Also, we jiggle the front and back sidecut depths and treat them as if they're equal, so Df = Dr = D. Also, it's obvious that the center of the circle will lie on the x-axis, so R = Ix.

R = Ix = (D^2/L + L/4 + D^2/L + L/4) / (2D/L + 2D/L)
= (2D^2/L + L/2) / (4D/L)
= (2D^2/L + L/2) L/4D
= 2D^2/4D + L^2/8D
= D/2 + L^2/8D

Which is the same answer obtained above.

The FIS way:

The FIS measurement procedure is a bit odd, and would seem to be susceptible to anomolous results if someone wanted to tweak a ski design to obtain a particular measurement (the way yacht designers notoriously do).

The procedures is:

"1. Measurement of W, narrowest width of the ski (or minimum width at ski middle area)."
That's straightforward enough, though I'm not sure what the parenthetical is exactly supposed to mean.

"2. Measurement of L1, the front length, and L2, back length, from the W location."
The accompanying diagram makes it clear that L1 and L2 are the length all the way to the tip and tail, not just to the widest points. That's right: all the way to the point of the tip, which includes part that isn't even on the snow, normally.

"3. Calculation of L1-20% and reading of its location, measurement of S"
The accompanying diagram shows that S is the width of the ski at "L1-20%," that is: at the point that's 80% of the way from the waist to the tip. This isn't necessarily the widest part of the ski.

"4. Calculation of L2-10% and reading of its location, measurement of H"
This is similar: H is the the width of the ski at "L2-10%," that is: at the point that's 90% of the way from the waist to the tail. Again, this isn't necessarily the widest part.

"5. Calculation of R, the ski radius."
The formula supplied is
"Radius = L^2 / 2000 x (S + H - 2 x W), with L= (L1-20%) + (L2-10%) or 0.8 x L1 + 0.9 x L2."
If you work through this (realizing that the thousands are there to adjust for using different units to measure the ski and the radius), you find it reduces to L^2 / 8D,

BUT: instead of the actual wide-points near the tip and tail, they simply move back 20% of the front length from the tip and 10% of the rear length from the tail (= 15% of the total ski length).

I suppose in most cases this is pretty close. It's interesting to think, though, what a savvy designer could do to finesse the measurement. For example: stick a long narrow tip extension on. Make it long enough, and you could get it so that "S" width is actually narrower than the waist.

I suspect they don't worry about this, because they're sufficiently autocratic that if somebody produced a ski that tried to "game" the rules, they'd just ban it.


(1) Okay, there's an exception: if the three points lie in straight line, there's NO circle on which all three lie. To put it another way, they lie on an undefined "circle" with an infinite radius.

(2) The circle has to lie in the plane defined by the three points: a circle in any other plane will only intersect the 3-points' plane at (at most) two points. In the rest of the following, we'll just assume we're staying within the plane defined by the three points ... so that (for example) the set of points that are equidistant from two of the points form a line, not a plane.

(3) In the special case referred to in (1), the two mid-lines are parallel, so there is no intercept.
post #2 of 5
I feel dizzy @#!$!#&^#@$
post #3 of 5
sjjohnston - It sure looks like you had your fun for the week with geometry! Good job. I didn't go over each line your derivations, but your conclusions certainly seem to be on the mark.

There are indeed several ways to calculate (in actuality, estimate) the sidecut radius of a ski with varying degrees of accuracy, and it looks like you derived two of them. Although I didn’t check, I believe your 2nd method is probably the same as the one here: http://www.math.chalmers.se/~olahe/Fri/skiradius.html . I have referred to this approach in the past, but since the error of method #1 is so small, the formula so simple, and since one is usually only looking for a rough but consistent estimate of the sidecut radius of various skis, method #1 is what most people use.

I need to look into it more closely, but the FIS method seems to be trying to get the best estimate of the sidecut for the just the (important) part of the edge that is in actual contact with the snow and generating all the turning forces. This seems quite reasonable.

There is also a fourth method, with which one can separately estimate the sidecut radii for the front and back parts of the ski. The derivation of this formula is not much more difficult than what you have already done, but this calculation gets quite sensitive to small errors in the widths and fore-aft position of the waist, so it’s rarely used.

Gotta run.

Tom / PM
post #4 of 5

I just stumbled across this discussion of sidecut radius calculation methods, and it got me curious.  First of all, is the curve manufacturers impart to their ski edges necessarily circular?  I'm not sure there is really any good reason why it has to be.  There was a time when it was common to refer to sidecuts as "parabolic", although I don't know if they really were parabolic, or why they should be.


It seems to me that the best way to model a circular curve, or some other near-circular curve, would be to do a least squares fit from a number of measurements to find the best fit circle, and take the radius of that.  That has the advantage of permitting an arbitrary number of measurements: less for convenience, more for increased accuracy.  All you'd need is a set of at least 3 pairs of width measurements, along with the positions along the length of the ski at which they were taken.  They wouldn't have to be at any particular locations such as the widest or narrowest points.  You'd just need a little computer program to do the least squares calculation.  It could also be made to emit the least squares residual, which would give a measure of how close the sidecut is to circular.


I haven't bothered to work out the program yet, but I wonder how that would compare to the various other methods out there.  Maybe I can find some time to try it.


P.S.  Techniques for least-squares fits to circles have been published, for example, by Gene Golub (e.g. http://www.ulb.ac.be/assoc/bms/Bulletin/sup962/gander.pdf)

Edited by renenkel - 2/20/12 at 9:20pm
post #5 of 5
Thread Starter 

No: sidecuts are not necessarily circular, and probably typically aren't exactly.


The "sidecut radius" is just a way to summarize a ski's sidecut with a single number that's meaningful, by describing the circle that approximately fits it. You could describe it more accurately, but would the description mean anything to the reader? Not really.


If you know you're describing a circle (or just act like you are), using more than three points does not make the description more accurate: there is never more than one circle that can include any three points. There's either exactly one that does, or - if the points lie in a straight line - none.


If you measure the three points at consistent and widely spaced points - ideally, the narrowest and widest parts of the ski - you maximize the extent to which using the radius of a circle as an approximatioin accurately describes the actual curve.

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