Category Archives: Ski Mechanics

Steering angle

The term steering angle is used a lot in these articles.

It’s a simple concept. Most turns aren’t fully carved. The skis are rotated through an angle prior to engaging the edge to start carving. That angle is called the initial steering angle. If you want to carve a turn with radius shorter than the capability of the ski, you will need to establish a steering angle first. The shorter the turn, the greater the initial steering angle. Look at pictures of World Cup racers and you will see them using steering angles all the time (see Ron LeMaster Ultimate Skiing).

There are three ways to establish a steering angle at the beginning of a turn:

  1. Rotate the skis while they are flat during crossover
  2. Jump the skis round in the air
  3. Use an uphill stem followed by a step of the inside ski

Again, you can see examples of all of these in Ron LeMaster’s Ultimate Skiing.

 

Do longer skis go faster?

First your homework: read Do bigger skiers go faster?

From that article, the magnitude of the acceleration of a skier of mass m travelling straight down a slope of angle θ is

a = g sinθ -μg cosθ -kA/m

where g is the gravitational constant, μ is the coefficient of friction between ski and snow, and kA is a constant related to the coefficient of drag of the skier (air resistance).

Contrary to what you might think, the frictional term μg cosθ does not depend on the length or area of the ski in contact with the snow.

So, all other things being equal (same skier, same slope, same conditions), wider/longer/bigger skis  don’t go any faster or slower.

 

Carved turn tracks

Ever wondered why your carved turns leave two parallel tracks in the snow, even though most of your weight is on the outside ski?

Isn’t it the outside ski that bends into reverse camber and makes the turn?

Well, yes and no.

camber

 

It takes surprisingly little force to bend a ski into reverse camber – maybe 10% of your weight at most. Try it while you are standing still. Try bending a ski with your hand. Easy, isn’t it?

 

Even with a 90/10 split on weight distribution between the skis, there is still more than enough weight on the inside ski to bend it and make it carve a track in the snow.

So why do we apply most of the pressure to the outside ski? To make it grip. On softer snow, you can get away with a more equal weight distribution, but as the snow gets harder you will need to commit to the outside ski exclusively. There’s a good explanation of this in Ron LeMaster’s Ultimate Skiing (p 21 and Figure 2.4).

So next time you look back to admire your carved turn tracks you can be satisfied that you had a) enough pressure on the outside ski to hold the turn and b) enough pressure on the inside ski to make a parallel track.

 

Do bigger skiers go faster?

If we consider a skier of mass m sliding straight down a uniform slope of angle θ degrees, then the magnitude of the total force on the skier (measured down the slope) is:

F = Fg – Ff – Fd

where

Fg = mg sinθ is the force from gravity

Ff = μmg cosθ is the frictional force from the snow for some constant μ

Fd = ½ρv2CA is the drag from air resistance

Now for 2 different skiers sliding at the same speed, v is constant and ρ is the density of the air, so is a constant. We further assume that both skiers have the same shape (not size) and are wearing similar clothes, so we can assume C, the drag coefficient, is approximately constant, and hence

Fd = kA

for some constant k, where A is the area of the skier’s body presented to the air in the direction of travel (the cross-section).

Then

ma = mg sinθ – μmg cosθ – kA

so

a = g sinθ -μg cosθ -kA/m

Hence the acceleration of the skier down the slope depends on the ratio A/m. If we assume that the skiers’ bodies have the same constant density ρ, then the acceleration is dependent on A/ρV, where V is the skier’s volume. For a linear increase L in dimension, A goes as the square of L, and V goes as the cube, so in fact, with the assumptions stated, the “bigger” skier will accelerate more, hence go faster, because the retarding force due to drag will be less.

So there you have it – bigger skiers go faster – maybe!

 

Carved turn radius

What radius turns can I carve on a ski with a given sidecut?

Let S be the sidecut radius of the ski. Then, assuming that the ski is bent fully into reverse camber, the turn radius R is, to a reasonable approximation:

R ≈ S cos θ

where θ is the edge angle between the base of the ski and the snow. [The gory details of this calculation will appear <here> at some point.]

Here are some calculated values for a 17m sidecut ski:

θ (degrees) turn radius (metres)
1 17
10 16.7
20 16
30 14.7
40 13
50 10.9
60 8.5
70 5.8

If you are making turns with a radius less than the table suggests, then you must be: