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Does flattening the skis make them point down the hill?

geepers

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Flattened skis will end up pointing downhill. CSIA guy pointed this out in a workshop by rolling a snowball across a groomer and we watched it curve into the fall line.

Not sure how much use this is as:
1. It's easy to influence by inadvertently imposing edge or twisting the skis, depending on snow conditions and ski type.
2. It's abandoning line choice to gravity which may not be a good idea.
 

Seldomski

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I assume the starting position is standing on a steep green or blue+ pitch hill with skis across the fall line?

What the skis want to do depends on your fore/aft balance. Balance aft with pressure on the tails, and the tails will start to seek the fall line and you will go downhill backwards. Balance toward the forward tips, and the tips will seek the fall line. Stay in the middle or slightly aft balance, and you slip downhill without the tips or tails seeking the fall line.

Manage the fore/aft balance precisely enough and you can do spins while sliding down the fall line.
 

Plai

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One of my first parallel ski lessons was the "patience turn". The first part of it was flattening the skis, letting the tips head down hill before making any other movements. Thought this was basic, universally understood, because it was how I was taught (as an adult). Guess I got lucky.

Remember the 'old' days when we had lots of runaway skis? On their own, without being edged, they would tend to point down hill.
Not that old (long ago), just a 3-4 years ago, my wife got hit by a run away ski. Yes, tip first.
 

crgildart

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Sleds and toboggans are like flat skis right?
 

pchewn

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Does flattening the skis make them point down the hill?


Yes. That is why the ski area requires safety straps or brakes. Because an uncontrolled flat ski will point down down the hill and injure the people at the bottom.
Remember the 'old' days when we had lots of runaway skis? On their own, without being edged, they would tend to point down hill.

Yes, exactly. The "old days" we were taught to yell "SKI !!!" whenever a ski was shooting down the hill about to hit people.
 

geepers

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I assume the starting position is standing on a steep green or blue+ pitch hill with skis across the fall line?

I assumed the starting position is in motion traveling across the hill with whatever action needed to flatten the skis already having been performed already.
 

graham418

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I still think there needs to be some external input in order for the ski to turn downhill. a rolling snowball is round and has some gravity action to pull it downhill. Runaway skis are either already pointed somewhat downhill or are deflected by some feature it encounters on the hill.
 

François Pugh

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There has to be some sort of throwing dart self-righting dynamic to it.

Yes, I know the throwing dart is not directly analogous.
The trouble is with a hammer or dart gravity accelerates the heavy parts with more force, but it takes an equal amount more force to get it to accelerate; in a vacuum it would just fall without rotating it. Air friction on the dart or hammer is just dependent on the surface with no regard to weight/mass. On the other hand with a ski on a slope, gravity accelerates all parts equally, but friction resisting motion down the hill is greater on the tail than the tip, so while the hammer falls head first, the ski points down hill.
 

cantunamunch

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t friction resisting motion down the hill is greater on the tail than the tip,

Do we know why that should be so, from first principles?

I still think there needs to be some external input in order for the ski to turn downhill. a rolling snowball is round and has some gravity action to pull it downhill. Runaway skis are either already pointed somewhat downhill or are deflected by some feature it encounters on the hill.

Yes. A center-mounted zero-camber ski is a lot closer to a snow saucer than a sled runner - and demonstrates that we need to look carefully at our starting assumptions.
 

geepers

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I still think there needs to be some external input in order for the ski to turn downhill. a rolling snowball is round and has some gravity action to pull it downhill. Runaway skis are either already pointed somewhat downhill or are deflected by some feature it encounters on the hill.

Objects in motion in a gravitational field will move in a curved path from a higher potential energy to a lower potential energy unless acted on by a force. That's true of any object in a gravitational field (unless they are traveling at or above orbital velocity).

Do we know why that should be so, from first principles?

Yes. But thanks for asking. ;)

Think about it. A ski has less friction traveling along its length and more friction traveling sideways. Let's say some disturbance pushes the ski tail slightly out of line. This increases friction along the length of the ski and the increased drag force brings the ski back into line.

Yes. A center-mounted zero-camber ski is a lot closer to a snow saucer than a sled runner - and demonstrates that we need to look carefully at our starting assumptions.

Presumably it would be possible to find some shape where the there was a tendency for some type of friction to exactly balance out the downhill force of gravity. For a particular slope angle and snow condition. Betcha it then wouldn't exactly balance for other slope angles/conditions.

The trouble is with a hammer or dart gravity accelerates the heavy parts with more force, but it takes an equal amount more force to get it to accelerate; in a vacuum it would just fall without rotating it. Air friction on the dart or hammer is just dependent on the surface with no regard to weight/mass. On the other hand with a ski on a slope, gravity accelerates all parts equally, but friction resisting motion down the hill is greater on the tail than the tip, so while the hammer falls head first, the ski points down hill.

Yes.
 

cantunamunch

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Think about it. A ski has less friction traveling along its length and more friction traveling sideways.

That's not a given.

Let's make a 3cm x 100cm HDPE cutting board. Put on a solid surface. It will slide equally in all directions no matter how we weight it.

There's something about a ski that makes it not act like a cutting board. Asserting a directional preference as an observable just hides the reason.

Even if we allow that,

Let's say some disturbance pushes the ski tail slightly out of line. This increases friction along the length of the ski and the increased drag force brings the ski back into line.

Sure, but why tip first?
 
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Seldomski

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I assume the starting position is standing on a steep green or blue+ pitch hill with skis across the fall line?

What the skis want to do depends on your fore/aft balance. Balance aft with pressure on the tails, and the tails will start to seek the fall line and you will go downhill backwards. Balance toward the forward tips, and the tips will seek the fall line. Stay in the middle or slightly aft balance, and you slip downhill without the tips or tails seeking the fall line.

Manage the fore/aft balance precisely enough and you can do spins while sliding down the fall line.

And to add to my prior post... if you flatten a little and stay centered (fore/aft), you will be riding the uphill edge. Keep flattening and you will speed up/slide faster. Keep flattening and you can engage the downhill edge, which is a great way to get to splat by way of high siding. That is, if you are still sliding down the fall line with the skis perpendicular to path of travel.
:crash:
 

François Pugh

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Do we know why that should be so, from first principles?
Simple model (first principles): friction is equal to the normal force times the coefficient of friction, and the normal force is greater at the tail, so more friction at the tail.

Add complications: the tail is typically more narrow, so it digs into the snow more.
 

cantunamunch

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Simple model (first principles): friction is equal to the normal force times the coefficient of friction,

Sure.

and the normal force is greater at the tail, so more friction at the tail.

That works for a beam model that is only supported at the tail and tip; it is not so clear for a fully supported beam.

I'm perfectly ok with this explanation for skis missing a center section :)
 

François Pugh

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The ski is a beam, upside down, with the load being the normal force of the snow, not uniformly distributed, supported by a single post at the binding. The normal force is the reaction force to the weight of the ski and binding (and skier if so equipped). The weight of the binding (and skier if so equipped) is transmitted to every point of the snow surface along the beam/ski according to the stiffness of the beam (depends on structure and modulus of materials) bending moment in the beam, beam deflection equations and the compressibility of the snow; the normal force is equal and opposite at all points along the ski/beam. I'm sure someone has a finite element model of it, but that's beyond first principles. I think it's obvious that greater load occurs nearer the tail than the tip on your typical downhill ski with bindings, thus causing greater friction at the tail and rotation about the com to point the ski downhill. Of course, obvious is relative and in the eye of the beholder.
 

geepers

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That's not a given.

Let's make a 3cm x 100cm HDPE cutting board. Put on a solid surface. It will slide equally in all directions no matter how we weight it.

There's something about a ski that makes it not act like a cutting board. Asserting a directional preference as an observable just hides the reason.

Even if we allow that,



Sure, but why tip first?

You've asked a good question. Here's my suggestion.

Unlike a HDPE board a ski has a smooth curved shape at the tip and abrupt edges. Snow is not a smooth surface. It has irregularities that will engage with an object sliding over it. The front of the ski passes more easily over those irregularities than does the abrupt edges which tend to to either catch or create drag by moving the irregularities out of the way. So there's less resistance to travel in the direction of the longitudinal axis.
 

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