How much friction a Classic ski can apply to allow the leg to push
back against it without slipping is complicated. This overoversimplified
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
back to Top  Secrets  FAQ
 Learn  more Classic
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 nonuniform distribution is a key part of successfully fitting
and using (and designing) a Classic ski.
I've put together an oversimplified mathematical model of this nonuniform
pressure distribution, which includes parameters like the total downforce
on the ski, the resulting size of the snow contact zone, the percent of downforce onto the gripwax 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 technicalminded folks might like to play with this sort of thing,
and I find that the oversimplified 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 gripwax 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 skistriding
works
 want to take ideas from this overoversimplified model to make
their own more interesting models  or make different
oversimplifications 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"
overoversimplification: 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
oversimplifications  and share them with me.
back to Top  Secrets  FAQ
 Learn  more Classic
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 simplemindedly assume that this single center
point is at once the center of: (a) the ski, (b) the snow contact
zone, (c) the gripwax 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
downforce F applied through the
binding onto the ski.
 b = outer boundary of the snow
contact zone.
We assume that this is a constant.
Gripwax zone
This is the section of the base of the ski which has gripwax smeared
on it.
 g = outer boundary of the
gripwax zone.
Since the gripwax zone contains the center point, its inner
boundary is always zero.
DownForce and Pressure distribution
The total amount of downforce and its nonuniform distribution
through the base of the ski are critical for determining the amount of
gripfriction force available.
 F = total downforce 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 downforce applied and the location of the point on the
ski.
 C = the downforce 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 wellknown "paper
test" for selecting ski fit.
Friction Force
What grip is all about from a physics perspective is the static
friction of the gripwax 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 downforce 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 simplemindedly assume that the glide wax has no static
friction.
back to Top  Secrets  FAQ
 Learn  more Classic
We assume that the size of the snow contact zone depends on the total
downforce applied. The more downforce, 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 downforce is
applied. This absolute pressure doubles as the downforce 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 downforce 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 gripwax zone which
is included in the snow contact zone. This assumes that the
downforce is large enough so that there is some contact of the gripwax
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 downforce:
S = m * C * (g^2  a^2)
/ b^2
+ m * (F  C) * (g  a) / b :
: {for F > C}
back to Top  Secrets  FAQ
 Learn  more Classic
How much grip if apply just only enough downforce 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 onetwentieth or 5% of the
downforce applied.
A big reason for this result is that even though the gripwax zone
takes up 50% of the base of the ski, it only gets a 25% share of the downforce.
But at least there is some grip available  even though the
snowpressure at the center point is virtually zero. And
that's because of the nonuniformity of the pressure distribution:
though the pressure at the center is zero, the pressure more toward the
boundaries of the gripwax zone is positive.
How steep a slope can this exact downforce 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 gripwax 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
gripwax 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 gripwax zone.
back to Top  Secrets  FAQ
 Learn  more Classic
