That makes no sense. Lets call the hill the x axis. Further to the right, the positive direction, on the x axis is downhill. If we start the skiers at 0 on the x axis, the thought experiment is valid. Why would you start the skier in the air at some point other than 0? There is no reason to do so; you’re just giving one skier a comparative advantage that negates the point of the exercise.

The other thing is that the hill does not magnify any forces. The skier in the air is being accelerated straight up and down. Newton tells us that an object not being accelerated by a force remains at the same speed. When you are standing on the ground, the force exerted by gravity is exactly matched by upward force exerted by the ground. You do not accelerate.

When you jump, the force from the ground vanishes the instant you stop exerting force against the ground. You are now accelerating 9.8m/s/s down. This force is unbalanced, thus you accelerate - your initial velocity gets scrubbed off, you stop, your velocity reverses until you hit the ground again, at which point the forces are in balance again, and you stop moving.

Now, if you are standing on skis on a 45 degree slope, gravity still pulls straight down. But the ground pushes up at 45 degrees (perpendicular to the hill) These forces are not balanced, thus you start accelerating in the x direction – your velocity steadily increases (until you reach terminal velocity).

This “relative to the slope” thing is a red herring. The only “ahead” is further on the X axis. By placing one skier at x=0 y=10 and one at x=1 y=11, you’re simply putting a skier ahead arbitrarily. The only way to win the race is to get to the X position of the finish line (call it x=100) first. It doesn’t matter if you cross it at y=100 or y=0.

Let’s say both skiers are going 10 m/s in the X plane. Skier A is in the air, Skier B is on the ground. So far they have the exact same time in the race, and A has just jumped while B managed to stay on the ground. Let’s further say that the slope of the hill is such that the horizontal acceleration is 5 m/s/s (roughly 45 degrees). Ignoring air resistance for now:

Skier A 1 second later has an X velocity of 10 m/s. 2 seconds later he has an X velocity of 10 m/s. 3 seconds later he lands, and his X velocity is 10 m/s. At 4 seconds, he is moving 15m/s. He does not accelerate in the X direction during his jump, because the only force acting on him is in the Y axis (which is pulling him down at 9.8 m/s/s)

His total distance over those 4 seconds is 45m.

Skier B is moving at 10/s in the X direction as well. At second 1, he is moving at 15m/s. 2 seconds, 20 m/s. 3 seconds, 25m/s, at 4 seconds, he is cranking out 30m/s. His total distance is 70m and he is going twice as fast in the X direction.

The key is simply that while skier A is in the air, his X velocity, the only one that matters, is not increasing. The Y velocity does not matter because the average Y velocity during the race does not have any effect on your finishing time.

So, a skier sticking to the hill which is 100 m long will travel down 50m, and a skier with an inspector gadget helicopter hat traveling straight forward will travel 0 in the Y direction. Skier A’s overall velocity will be higher if they tie, but his x velocity will be the exact same. This is where I think you are running into problems with this idea of “relative to the slope”. You can also imagine the helicopter guy moving up and down in a sine wave at 200 m/s as he travels forward – his overall velocity will be much, much higher, but the x component remains the same, and they still tie. If either skier has a higher X velocity, they win, no matter what their Y component is doing.

Thus, as was said above, sticking to the ground is faster than jumping, because you only accelerate forward while you are on the ground, not while you are in the air.

ETA: Chessic: the point is it doesn’t balance – you don’t get a boost forward from jumping except in video games. See the math in this post.

Exactly. In fact the moment he becomes airborne his X velocity immediately begins to decrease. Same as a bullet leaving the muzzle of a gun.

I think this is pretty important and also correct although perhaps for a different reason.

I have not experienced this in skiing so perhaps this is incorrect, but sometimes the choice is not between going at Z speed in the air and Z speed on the ground. Sometimes you have to slow down to stay on the ground. So if that situation is in force, the skier in the air would have a head start because they’re going faster (but could lose this advantage due to the slower-at-first skier being able to tuck in and accelerate off the slope for longer.)

Furthermore, the idea that vertical acceleration does not translate into “horizontal” speed is wrong because slopes are not completely flat. The skiing slope

``````

| |
| |
| |
| |
\:O/
-----
-----
-----
-----
-----

``````

Can be rotationally transformed into

``````

--
--
--
-------------------------------------------

``````

I don’t think anyone would argue that if you dropped a ball on the ground per the second picture that all it’s x-axis motion would stop, even if the ground were perfectly inelastic.

In fact, the downward force of gravity in the first picture translates into a force IN THE DIAGONAL DIRECTION IN THE SECOND PICTURE. So the more you accelerate downward in the air, the faster you will be going on the sort-of-but-not-really x axis once you hit the ground, even if there is no “bounce”. Of course the amount of this depends on the angle of the slope, and I will not argue that a continuous acceleration will not produce a faster speed, as I have experienced it myself (albeit not on the ski slopes).

When ski racing, sudden changes in direction suck. You try your best to get from the top of the course to the bottom in as straight a line as possible, without any sudden changes in direction. Not just avoiding sudden changes in lateral direction, but also sudden changed in vertical direction. Every sudden change in direction has you fighting the snow and losing speed. A pre-jump helps you keep your rate of descent consistent – no sudden change in vertical direction because you don’t fall out of the air.

QFT

Didn’t you define x as the direction parallel to the slope? In that case the skier is accelerating “forward” (along x) in the air, or in other words his x velocity is increasing, because gravity does not act perpendicularly to the slope (the x axis).

I agree. The notion that an airborne skier falling forward at an angle isn’t accelerating relative to the slope at which he’s falling doesn’t make sense.

No. Why would you define X as parallel to the slope? Draw a graph and start a line at x,y=0,100 and draw it to x,y=100,0.

Setting it up any other way just confuses the issue, because all the forces would be some combination of x and y. If you just set it up normally, gravity is strictly a Y-directional force, which is how we act in the real world. When you stand on a hill, you don’t stand perpendicular to the slope, you stand straight up and down.

I fail to see how we stand has anything to do with what happens when we land. The simple fact is that if we are dropped onto a sloped plane with pure Y momentum we will end with some X momentum, because the forces that are pushing against us to stop our fall are not pushing in a purely -Y direction but normal to the slope, and so we will end with some X momentum.

``````
Like so:
Start
100  .
|||||||   .
|||||||||||  .
50 ||||||||||| .
|||||||||||||||||   .
|||||||||||||||||||||  .
Y                               .|  Finish
X________50_______100

``````

The person with the highest average +X velocity will have the best time. The best way to achieve the highest +X velocity is to accelerate the whole time. You have to be touching the ground to accelerate in +X, because gravity will only apply a force in the -Y direction. If there is no force in the +X direction, you will not accelerate +X, you will only retain whatever starting velocity you have, minus air resistance, just like a fired bullet. The skier on the ground accelerates because the ground basically pushes him +X (at about 50% of the force of gravity on a 45 degree slope like the one above).

Bah, I can’t get the diagram right. Anyway, it’s sposed to be a straight 45 degree line.

The person with the highest average -Y velocity will also have the best time.

No. Your Y momentum isn’t “converted” to X. You just start picking up X momentum as soon as the slope imparts a force on you. The net force only depends on the slope of the hill, not the magnitude of your Y (in a more-or-less inelastic collision).

This is because the magnitude of the downward Y force is exactly canceled out by the ground’s upward Y force. Otherwise you would dig a hole when you landed. The only X imparted is the unbalanced part perpendicular to the slope. The exact instant you hit the hill, your X momentum is 0. You don’t convert any of that +/-Y into X.

True, but obviously bounded by the slope itself, unless you are contemplating some sort of drilling apparatus.

Let’s think about the situation, because I think there’s an aspect of this that you seem to be ignoring.

The forward component of the skier’s acceleration comes from the normal force of the slope resisting the skier’s gravitational acceleration. When the skier is in the air, she does not make contact with the slope, and so the forward-component of her acceleration is zero until she makes contact with the slope again. When skiing in contact with the snow, the force of gravity combined with the normal force of the slope causes an overall force vector in a diagonal direction, following the slope of the hill. When the skier is in the air, this force vector is pointing straight downward, and when the skier contacts the ground some of the momentum gained in the air is translated back into horizontal motion. The issue here is that the skier needs to flex her legs and adjust her balance, and so loses some of the energy because of the elastic nature of the landing. If the downward energy could be perfectly converted to forward motion, then your argument would carry a lot more weight, but it is this loss of energy which is the crux of the matter.

I don’t see how this

Differs from this

Which I’m allowing as a reason that more or less continual contact with the ground will yield a faster time overall.

The point I take issue with is that there is something about being in contact with the ground that yields a different, and more potent gravitationally directed acceleration result vs simply falling through the air along a slope.

At a certain point it is intuitively obvious that some Y momentum will be converted to X momentum. Consider an 80 degree slope with two skiers, one of which starts on the slope, the other of which is dropped from 100 feet and lands next to the other skier, at which point the other skier is released so they both start from the same position. The only difference being that one has Y momentum.

Which will land at the bottom of the slope first? If you believe the position that absolutely none of the Y momentum will be converted to X momentum, you believe they should arrive at the bottom at exactly the same time.

But the skier who is being dropped from 100 feet is only barely grazing the surface since it’s 80 degrees. Under the hypothesis that none of the Y momentum gets converted into X, the skier will come to a complete and utter stop on that 80 degree slope and only then start to fall down. Which is pretty preposterous.

It has been explained about a zillion times.

Just think. If a ski jumper appeared out of nowhere above the slope, he would simply fall straight down.

If a ramp is put below him he will fall down, and then across, because it will accelerate him in a sidewasey direction.

This is obvious.

No ramp: no sidewasey acceleration.

Ramp: sidewasey acceleration.

Since the person who wins needs to have, by definition, the most sidewaseyness, and the only way to get sidewaseyness is to be on the slope, it stands to reason that maximizing slopeyness is important.

So, no need for physics, it’s just sidewasey slope magic, which you can simply infer from your intuition.

Some posts here seem to be implying that the course is more or less flat. Exact specs seem to be hard to come by, but apparently the course was around 10.000’ long with a 3000’ vertical drop, for an average downslope of 30%. This corresponds to an angle of around 17 degrees.

The force that propels a skier down a slope is weight * sin(slope angle) - drag. Drag is hard to quantify, but sin(17 degrees) = 0.29. Thus, the raw propulsion force for a skier on a 17 degrees slope is around 29% of the skier’s weight.

Physics does come in to this. If the skier is dropped above the slope they do not simply stop in their tracks. Unless they augur into the ground, they will therefore have some X momentum.

What arrests their groundward velocity? The earth. Specifically, the earth pushing back at a normal to the slope.

There is no way to resolve adding a slope-normal vector to a Y-only vector and not have some X momentum unless there is no slope at all.