How much friction a Classic ski can apply to allow the leg to push
back against it without slipping is complicated. This over-over-simplified
model of the physics gives some concepts and formulas that can help provide
a framework for selecting new skis, waxing, and technique.
Ken Roberts
what's here
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Because of the camber / flex of a Classic ski, there is more pressure on the
tip and tail than on the center under the foot.
Like when only half the skier's body weight is on the ski, all the pressure is on the tips and
tails, and the center is not touching the snow at all (on firm snow).
Managing this non-uniform distribution is a key part of successfully fitting
and using (and designing) a Classic ski.
I've put together an over-simplified mathematical model of this non-uniform
pressure distribution, which includes parameters like the total down-force
on the ski, the resulting size of the snow contact zone, the percent of down-force onto the grip-wax zone, and the resulting total static friction
force available for grip.
Fortunately no one needs to know any mathematical model in order to ski
well.
But some technical-minded folks might like to play with this sort of thing,
and I find that the over-simplified mathematics does yield some conceptually
satisfying answers to questions like:
- If a ski is fit so that the the skier's full body weight only just
barely compresses the grip zone to the snow, how can the ski provide any
significant static friction? (using only the technique of committed weight
transfer)
- Why are Classic skis shaped with zero or negative sidecut (or "carve")?
(even though this shape hinders downhill turning performance)
- Why is smearing the grip-wax on long even more effective than expected?
- What's the maximum slope angle that a ski can hold its grip, if the
ski has just enough weight on it to close the camber down to the snow?
Who is this model description page for?
It's intended for skiers who have all of these characteristics:
- already know and enjoy basic physics, and are accustomed to the
basic tricks of playing with the mathematics that goes with
it.
- already have a basic sense of how Classic ski-striding
works
- want to take ideas from this over-over-simplified model to make
their own more interesting models -- or make different
over-simplifications to answer their own different questions.
This model is mostly just "raw material":
- There are no diagrams -- because that takes work -- but I assume
that skiers like just described can make up there own. If
someone wants to contribute a diagram graphic or two, that would be
great.
- Lots of explanations are missing -- but I assume that skiers like
just described are capable of doing lots of thinking on their
own.
- This model is not the "right"
over-over-simplification: rather it contains a set of
simplifications that were convenient to answer my
questions. Different people have different questions that they
think are interesting. I hope they come up with different
over-simplifications -- and share them with me.
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Points along the length of the ski
We define points by how far they are from the center of the
ski.
We assume that everything is symmetrical about the center point, so
we only consider half the ski in most of our analysis.
- x = distance from the center
of the ski.
Since we only deal with half the ski, we take x
as always positive.
We simple-mindedly assume that this single center
point is at once the center of: (a) the ski, (b) the snow contact
zone, (c) the grip-wax zone, (d) the wax pocket, etc.
Snow contact zone
This is the section of the base of the ski which is currently in
contact with the snow.
- a = inner boundary of the snow
contact zone.
This is a variable that depends on the total
down-force F applied through the
binding onto the ski.
- b = outer boundary of the snow
contact zone.
We assume that this is a constant.
Grip-wax zone
This is the section of the base of the ski which has grip-wax smeared
on it.
- g = outer boundary of the
grip-wax zone.
Since the grip-wax zone contains the center point, its inner
boundary is always zero.
Down-Force and Pressure distribution
The total amount of down-force and its non-uniform distribution
through the base of the ski are critical for determining the amount of
grip-friction force available.
- F = total down-force applied
through the binding to the ski.
A major portion of this is force of some percentage of the skier's
body weight, but there can be other reactive forces.
- P(x,F) = pressure applied
through a tiny point on the base ski onto a tiny point on the snow
surface.
We assume that the pressure at any point is a function of both the
total down-force applied and the location of the point on the
ski.
- C = the down-force which
exactly just barely "closes" the entire base of the ski
down against the snow.
I like including this C in the
model, because it links the model to the well-known "paper
test" for selecting ski fit.
Friction Force
What grip is all about from a physics perspective is the static
friction of the grip-wax zone in contact with the snow.
We are assuming the usual simple physics model of static
friction: The maximum force of friction available to prevent
slipping is the product of the current down-force multiplied by the
coefficient of static friction.
- S = total static friction
force available to prevent the ski from slipping back.
- m = static friction
coefficient of grip wax.
We assume that m is a constant for the the current grip wax with
the current snow, temperature, and humidity conditions.
For now, we simple-mindedly assume that the glide wax has no static
friction.
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We assume that the size of the snow contact zone depends on the total
down-force applied. The more down-force, the larger the contact
zone.
a = a(F) = b * (1 - F/C) :
: {for F < C}
a = 0 : : {for F > C}
This is the simplest assumption we can think of. Note that F = C implies a
= 0, and that F = 0 implies a
= b.
We assume that the Pressure distribution function takes this
form:
P(x,F) = h(F) * F * x
:
: {for a < x < b and F <= C}
This means that the farther from the center point, the larger the
pressure. We assume that for a given value of F,
the factor h is a linear constant
over the entire snow contact zone.
Here are the other cases needed to complete the definition of the
Pressure function:
P(x,F) = (2 * C / b^2) *
x + (F - C) / b : : {for F > C}
P(x,F) = 0 : : {otherwise}
A requirement for the Pressure distribution function is:
integral[a,b]: P(x,F) dx
= F
Evaluating the integral and solving for h(F)
gives
h(F) = 2 / (b^2
- a(F)^2)
which yields a more detailed Pressure distribution of
P(x,F) = ((2 * F) / (b^2
- a(F)^2)) * x
Note some special values for P:
If F = C, then P
= (2 * C / b^2) * x
As F becomes very small and goes
toward F = 0, then the limit of
P
goes to (C / b^2) * x
For the special case of the outer boundary of the snow contact
zone:
P(b,0) = C / b : : {in the limit
of very small F}
P(b,C) = 2 * C
/ b
This is interesting because it fits with an intuition that while the relative
percentage distribution of pressure to the tip of the ski drops, the absolute
pressure value at the tip continues to rise as more total down-force is
applied. This absolute pressure doubles as the down-force goes
from nothing to enough to fully close the ski to the snow surface.
To calculate the total force of static friction S
for a given down-force F, we must
evaluate this definite integral:
S = integral[a,g]: m
* P(x,F) dx
The bounds of integration are the portion of the grip-wax zone which
is included in the snow contact zone. This assumes that the
down-force is large enough so that there is some contact of the grip-wax
zone; which requires
F > C * (1 - g/b)
Evaluating this integral yields:
S = m * F * (g^2
- a^2) / (b^2 - a^2) : : {for F < C}
Which seems like an intuitively nice result.
If F is exactly the force to
"close" the ski, this simplifies to
S = m * F * g^2 / b^2
: : {for F = C}
For even larger down-force:
S = m * C * (g^2 - a^2)
/ b^2
+ m * (F - C) * (g - a) / b :
: {for F > C}
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How much grip if apply just only enough down-force to barely
"close" the ski to the ground? Do we get any grip at
all?
If we assume
F = C = 100 (it's a very stiff ski for very heavy skiers)
b = 100
g = 0.50 * b = 50
m = 0.20
then we get
S = 0.20 * (0.50)^2 * 100
S = 5
So the force of static friction is only one-twentieth or 5% of the
down-force applied.
A big reason for this result is that even though the grip-wax zone
takes up 50% of the base of the ski, it only gets a 25% share of the down-force.
But at least there is some grip available -- even though the
snow-pressure at the center point is virtually zero. And
that's because of the non-uniformity of the pressure distribution:
though the pressure at the center is zero, the pressure more toward the
boundaries of the grip-wax zone is positive.
How steep a slope can this exact down-force hold against slipping
back?
The maximum slope would have about a 5% grade (using the same
assumptions as the previous question.
[ Basic static friction calculation with trig functions -- to be
added ]
How much does it help to smear wax "long" over a larger
section of the ski base?
Start with our normal grip-wax zone is 50% of the length of the ski,
like in our previous assumptions.
Suppose we "wax long" and now cover 55% of the ski base --
which is 10% longer. Our new grip force is:
S = 0.20 * (0.55)^2 * 100 = 0.20 * 0.3025 * 100
S = 6.05
But that is 21% more grip friction -- in return for only 10% more
grip-wax length.
Now let's make F smaller:
let F = 0.75 * C = 75
This implies
a = b * (1 - 0.75) = 100 * 0.25 = 25
which yields
S = 0.20 * 75 * (0.50^2 - 0.25^2) / (1 - 0.25^2)
S = 3
Now let's "wax long" and go to g = 0.55 * b
S = 0.20 * 75 * (0.55^2 - 0.25^2) / (1 - 0.25^2)
S = 3.84
This time we get a 28% gain in grip friction -- in return for only
10% longer grip-wax zone.
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