Who is faster

A snowHead

On our boys trip, the 4 of us perenially argue about who should be faster down the hill.
It goes like this, Moose the big ugly 6' 4", 18st second row says he should be slowest becuse of his weight and I at 5'8" and 13 st should be quickest, with the middle to chaps at 5' 10" and 10st in the middle of the "race".
I maintain that as I have shorter skis therefore I should be slower and once his weight gains kinetic energy he should pull ahead.
Moose counters with the tonne of feathers and a tonne of bricks arguement which I think holds no water.
So come on, there must a clever chap or chapess who can tell me the definitve answer and how to work out drag coeficients and the like. Answers in plain english please Laughing

A snowHead

MarkL, Gravity acts equally on all 3 of you, so if freefalling in a vacuum you'd all hit the ground at the same time.
Things that slow you down are: Friction and Wind Resistance, so a lot depends on clothing, how aerodynamic your tuck is and size in the case of wind resistance and length of ski, ski preparation and glide technique in the case of friction.

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According to the simple theory (eh-hem), the main components are accelerations due to gravity, a(g), gliding friction, a(f) and wind resistance a(w), as said by Spyderman. Your overall acceleration is a(tot) = a(g) + a(f) + a(w). This relates to the forces involved with Newton's good old 2nd law: F=ma, or a = F/M

Looking at each of these separately

component due to gravity: the heavier you are the more accelerating force (F=M.g.sin(theta)), but the more mass that has to move (F=M.a(g)), so that all falls out in the wash: a(g) = g.sin(theta)

component due to gliding friction: in simple friction theory, the coefficient of friction (Cf) is a constant for a given pair of surface types, keeps the frictional force directly proportional to the force acting perpendicular to the surfaces in contact (N)and is independent of the surface area (and speed). So F = -Cf.N = -Cf.M.g.cos(theta) and mass falls out in the wash again: a(f) = -Cf.g.cos(theta). Note that this term is negative as it opposes the direction of motion. Cf will change with surface type though, so will depend on how well you have your skis waxed. The interaction between snow and ski is also not really the solid surface interaction the above euation relates to, as the snow melts providing a degree of lubrication. This is a lot more complicated to analyse - so the above can only be called an approximation.

wind resistance: the retarding force is sort of proportional to cross-sectional area in the direction of travel (A), and the square of the velocity in that direction (v.v) and a bunch of other constant terms, so we can put those into one coefficient Cd. So a(w) = F/M = -Cd.v.v.A/M , and it's again negative.

So total acceleration = g.sin(theta) - Cf.g.cos(theta) - Cd.v.v.A/M

Note that the wind resistance is the only term that has ended up with mass in the equation, and really anything about the skier's body at all. The smalller this retarding acceleration is, the faster you will go - first to get up to speed, and secondly then what that limiting speed will be. So the faster people will be those that are heavier, with the smaller cross-section. If people are a similar shape, then the weight goes up faster than the cross-section (weight as height cubed, cross section as height squared). Actually cross-section is not quite right for non-spherical objects (even Moose), as our arms and legs contribute a bit more than just the cross-section they produce, so you can sort of think of "area" as height squared. This is where the great speed increases on tucking come in - you reduce your effective height considerably without changing your driving weight.

So for you four, given that you're all on the same slope with similarly prepared skis (and are equally good at keeping your skis flat) your relevant numbers are (in units of inches-squared/stone), and smaller means faster:
Moose: 320.9
You: 355.7
Pair of Beanpoles: 490

So Moose should win, you should be close behind him, and your beanpole friends should have pints waiting for them when they eventually get into the bar.

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It's not about the size of the entire body, it's about the size of only one particular part, and the willingness to point em straight down the hill and stay off the edges! Very Happy

Then you can post your own questions or snow reports...

Skiing is a lot like cycling in this. The difference in friction is minimal compared to the difference in weight/frontal area. So the hevier skier wins... providing he's got the required anatomy pointing in the right direction.

After all it is free

The fat bloke.

Little Tiger and i skied with a Harby a couple years ago, who was quite full of steam about how good he was. He was a rather large chap. We decided to give the NASTAR race course a go, and the demure Little Tiger cleaned his clock. He schlepped off with his tail between his legs, muttering about how he needed to get skis like she had.

Weight will generally win in a straight run wax race on the flats (all other factors being even), like others have explained here, but in the world of real skiing on various types of terrain, the one best able to ride a clean edge through a turn will carry the potential to be the fastest skier, regardless of size. It's why over the years some smaller guys have been able to find success on the World Cup downhill circuit, over their larger counterparts.

Help!!

Who is faster

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