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The other day on skiing holiday we've been arguing about whether an adult has weight advantage over a child when skiing downhill.

I was claiming that gravity is a constant regardless of object's weight. I was claiming that weight doesn't affect skier's speed. I can still remember primary school experiment where we dropped same sized objects of different matter in vacuum. Styrofoam and iron ball of the same size and they both fell equally.

It is true that skier's maximum speed is determined by his air drag (and snow/ski resistance) and heavier objects have higher persistence making them move through resistive medium easier. But compared to a child an adult would have more drag due to larger surface so I suppose this somehow evens out.

I was basically claiming that weight doesn't matter at all but that other factors are much more important when it comes to skier's speed like: ski stiffness (weight force much better distributed over the whole ski surface than with softer skis) and ski sliding surface smoothness and especially skier's position (to reduce drag as much as possible).

I also got a bit confused when I started to think about force vectors where greater weight results in longer vectors resulting in stronger movement pull.

Question

If we approximate is it safe to say that a child can ski just as fast as an adult? Or that a thinner adult would ski faster than his same high companion that is overweight due to larger drag surface? Or does his weight benefits his speed?

What is the correct answer here?

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Possible duplicate: physics.stackexchange.com/q/22080/2451 –  Qmechanic Mar 8 '13 at 14:45
    
Surface and weight do not "even out". If a object is kept the same shape and just scaled up, weight will increase by the cube of the linear dimension, but surface are only by the square. The child is therefore at a disadvantage in your skiing scenario. –  Olin Lathrop Mar 8 '13 at 21:33

1 Answer 1

We can make a first approximation of this situation by considering an object sliding down an incline with friction and air resistance.

$\vec{F} = m\vec{a}$

Using a coordinate system that is aligned with the incline (+x is up the incline), we can sum the forces to satisfy Newton's second law.

In the x-direction: $D + S = mg\sin{\theta}$, where $D$ is the drag, $S$ is the surface friction, and $\theta$ is the angle of incline.

In the y-direction: $N = mg\cos{\theta}$, where $N$ is the normal force.

Assume:

$D = \frac{1}{2}\rho V^2AC_D$

$S = \mu N$

Where $\rho$ is the density of air, $V$ is the speed of the skier, $\mu$ is the coefficient of friction, $A$ is the frontal area of the skiers, and $C_D$ is the coefficient of drag.

Now we can solve for $V$:

$V = \sqrt{\frac{2mg\left(\sin{\theta} - \mu\cos{\theta}\right)}{\rho C_D A}}$

Based on this, for the same $A$ and $C_D$, the heavier skier goes faster. If the frontal area of a skier is proportional to $(mg)^{\frac{2}{3}}$ (assuming spherical skier), then $V \propto (mg)^{\frac{1}{6}}$.

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For constant density and similar shape, wouldn't frontal area scale as $A\propto V^{2/3}\propto W^{2/3}$? –  Emilio Pisanty Mar 8 '13 at 15:11
    
Yes you are correct, I will edit the answer. –  OSE Mar 8 '13 at 15:15
    
You don't actually have to assume a spherical skier. As long as the shape is (roughly) constant, volume will scale as $(\text{characteristic length})^3$ and frontal area will scale as $(\text{characteristic length})^2$. –  Emilio Pisanty Mar 9 '13 at 15:55
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I mainly added that part as humor in reference to the spherical cow joke. –  OSE Mar 11 '13 at 13:17

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