15.1 What's the radius of a pure carved turn?
The sidecut radius of a ski is a design feature of a ski found by assuming the ski edge is an arc of a circle and determining the radius of that circle.
The following calculation descibes the radius of a circle if you know the parameters of a sector of the circle. A sector is found by cutting off the edge of a circle, or disc, with a straight cut that is perpendicular to the radius.
Assuming
define the variables
then the sidecut radius is
The radius of a turn is different, and it varies as the turn is being executed with the edge angle being used by the skier and with the amount of reverse camber in the ski during the turn. However, it is not too hard to calculate an approximation of the the radius of a "pure carved turn".
Geometrically, a carved turn is the composition of two curves, the sidecut of the ski and the deformed sidecut of the ski when its camber is reversed as it is pressed onto the snow. To find these curve, I assumed that the carving ski looks like a segment of a cylinder, and that the ski hill is an inclined plane. The arc the carving ski follows is the curve at the intersection of the cylinder and the inclined plane.
It will help you understand the formula if you'll do the following simple geometric demonstration. Take two identical rectangular pieces of paper. With scissors, make a large radius (shallow) sidecut on one of the long sides of piece 1. Make shorter radius (deeper) sidecut on the corresponding side of piece 2. Now hold each piece with the sidecut against a tabletop so that the edge of the sidecut is in full contact with the table. Notice the decambering effect of pushing the edge into contact with the table, and notice how changing the angle changes the radius of the approximate cylinder.
To use this approximation, you have to calculate the radius of the cylinder, and this involves the geometry of the ski. The approximation I make is not too bad if you assume that the ski does not twist, but it does calculate the radius of the carved turn when you are traversing the hill.
To approximate the cylinder radius, I computed the distance from the ski edge to a chord extended from the widest point on the ski tip to the widest point on the ski tail. Then I projected that point to the inclined plane given the inclination angle of the pressing force. If you think about this, its easy to see that there will be a minimum angle where this actually can be done. Thus all these calculations should be only considered for angles above that minimum.
The variables are:
Here is the result
This formula is messy, but the interpretation of the formula is interesting and easily understood. It explains, mathematically, how a deep side cut ski carves easily. To understand this, I'll make a comparison between two different skis.
Suppose that you want to compare carved turns between a super sidecut ski and a regular ski when the angle of the skiers leg and the slope angle are the equal for both skis. (Keeping the angles the same makes for a fair comparison.)
For a ski with a large side cut to carve, it must be pressed a long way down to the snow before the edge is actually on the snow surface. In ski jargon, the ski has a lot of reversed camber when the edge is on the snow surface. This creates a small radius carved turn because the ski is bent into a small radius circular arc.
A ski with a small side cut cannot be pressed as far down before it comes in contact with the snow. Consequently, it will not have as much reversed camber. Therefore, the radius of the turn will be greater because the ski is bent into a larger radius circular arc.
A combination of the sidecut, edge angulation and reversed camber determine the radius of the turn. Skis with a large sidecut allow the skier to deeply reverse camber the ski, while skis with a small side cut cannot reverse camber as much. The combination of a large sidecut and a deep reverse camber creates short radius carved turns.
No. The carve radius depends on the relative angle between the skiers leg and the hill. Of course, the design of some skis favor on turn radius over another, but there isn't a "single radius of a carved turn".
15.2 Do big people go faster?
From Patrick Chase (edited by Eyre): Let's look briefly at the free body diagram of a skier acted upon by gravity, surface friction with the snow (modeled as Coulombic friction) and aerodynamic body drag (neglecting skin friction). Here's the simple free body diagram.
Now we can look at the individual terms:
The force Fgravity is the gravitational force. The inertial forces are zero if the skier is stopped or is moving at constant velocity. Ffriction and Fdrag are the frictional and drag forces. Both of these oppose gravity. The remaining terms are defined as
Balancing these forces gives the equation of motion:
and if we divide by mass, we get
The only term depending on mass is the last term, and mass appears in the denominator. That term opposes gravity because (Cd*Ap*rho*V^2/m) is not negative.
It should be pretty clear that the benefits (or lack thereof) of changing mass depend upon what happens to (Cd*Ap/m) as mass is scaled upwards. It is expected that the frontal area will increase for a more massive skier, but it seems likely that Cd*Ap increases more slowly than m. Therefore, a bigger skier is likely to go faster.
A few important things in this analysis have been ignored:
The following data (from Stuart Marlatt in rec.climbing, edited by Eyre) measures the relative effects of changing (Cd*Ap/m). While these calculations were done for free falling objects, the effects would be similar for skiers as (Cd*Ap/m) is changed.
To determine the terminal velocity, I used a second order Runge-Kutta to integrate the equations of motion:
where
The fall is assumed to be from 1000m, at 41deg north. For reference a smooth sphere with a diameter about 0.5m was used in addition to three body positions where Cd*Ap were found from S. Hoerner's "Fluid Dynamic Drag":
The latter two are the most applicable to skiers. Marlatt found the following results:
Position Time of Fall Speed at Impact -------- ------------ --------------- Smooth sphere 15.76s 105.8 m/s (236.7 mph) Upright 16.44 95.3 m/s (213.2 mph) Horizontal 28.89 37.4 m/s ( 83.7 mph) Tuck 23.02 50.7 m/s (113.4 mph)
In reference to the description above, the question of how much faster does a big person go when you're going slow boils down to the following question:
Here is the answer. Suppose that you are free falling, i.e. no friction and theta = -90 degrees. Then you can calculate Cd from Patrick's third equation at terminal velocity (i.e. A = 0). This gives
for the drag coefficient where V_T is the terminal velocity. By substituting this expression for Cd into Patrick's third equation you find
The acceleration depends on how fast the person is going relative to his terminal velocity, i.e. on the ratio of V to V_t. The previous section showed v_T depends on the size and shape of the person. However, a slow skier is moving at only a small fraction of their terminal velocity, say maybe about 1/10.
In this case, (1/10)^2 = 0.01. The sine of a 20 degree slope is 0.342, so the acceleration due to the slope and gravity is 34 times greater than the effects due to mass! For most practical purposes, you can ignore the last term at small speeds.
15.3 A simplified mathematical model of avalanches