Skiiers: What's so bad about "air"?

Factual Questions
161

If it’s set to 0 at all times, the racers will be at the same location at every time, since skier B will have never launched off from the slope. :slight_smile: That having been said, if you magically set it to 0 shortly after B jumps, skier B will still land right on top of A, just a little sooner. Consider a camera traveling along with skier A, oriented so that the slope looks like horizontal ground to the camera: from this camera’s point of view, all skier B does is jump straight up into the air, then eventually fall back where they started, which is to say, fall right back onto skier A.

Also, I called it post #119, but it’s of course actually post #117.

162

Because the vector can be decomposed into a vector parallel to the slope of the mountain and a vector normal to the slope of the mountain. The vector parallel to the slope of the mountain is the component of the gravity acceleration that is providing a “boost” toward the bottom of the mountain, barring stuff like bouncing and aerodynamics.

163

I would agree, but if you watch the skiing competitions, especially the downhills, the jumps are more often than not a result of the ground falling out from underneath the athlete, not the athlete jumping into the air, resulting in that L component being 0 at the start of the jump.

164

Because you have to break it into components.

If you break it into components: Carry out the diagram I posted for ivn 40 posts ago at the bottom of the one with the :smack: smileys. You’ll see that a downward G has a component parallel to the slope.

If you don’t break it down: You can still see that gravity isn’t pointing perpendicular to the slope. If that’s true, then no matter where his acceleration points (due to rockets, jet packs, wind, helium balloons or whatever), then there must be some acceleration up or down the slope. Again, the only way for this not to be true is if he accelerates directly to or from the slope.

He’s not saying that all of gravity pushes him down the slope - just Gsin(a) of it…same as the land-borne skier.

165

Say, Muffin, would you be posting from, oh, say, Vancouver?

166

I have no doubt about this. Whether I can understand the answer is another question altogether.

Seriously, I’m very impressed with all the physics calculations that have gone into this post. But I wonder - and I’m speaking as a physics innumerate - if ivn1188 and sich_hinaufwinden and Chessic Sense are perhaps taking the wrong approach in treating this as a question of purely classic physics. I mean, we are not talking frictionless ideal 45-degree slopes, are we? This is a real-world question, where - it seems to me - considerations like wind resistance and the ability to steer (pace post #4) and the fact that it’s harder to keep a tuck in the air are also very relevant.

I’m not denying that physics ultimately underlies every one of these factors - that would be stupid - just that perhaps the situation is a little more chaotic than simply x velocity vs. y velocity, no?

167

Imagine you are being dropped on planet Blorg on a far away galaxy. Planet Blorg has a flat frozen landscape otherwise similar to the Earth, including the same gravitational acceleration g. The only peculiarity is that at your drop point gravity does not point straight down but makes an angle with the surface, say 45º (because its center of gravity does not coincide with its geometrical center).

You are dropped by parachute from a stationary spacecraft (relative to the flat horizontal surface). There is absolutely no wind whatsoever on Blorg. Do you accelerate forward while dropping to the surface?

Going back to our skier, without losses air is exactly the same as contact, energy wise. Maybe a scalar argument will be more successful. The whole system is conservative so the difference in kinetic energy cannot depend on the path traveled from A to B, only on its endpoints.

168

My example is that of a skier taking to the air because the slope increases. For simplicity, I chose a change from horizontal to an arbitrary constant pitch. It is not complicated, and it does lead to a different result than assuming the skier exerts a force perpendicular to the slope.

169

See post 155.

170

It is also a good example of why a racer should not deliberately slow down if at all possible. The white bib was behind the yellow and green bibs prior to the hump. Yellow and green checked their speed so as to minimize air, whereas white did not. White flew further, but passed the slower moving yellow and green.

171

Exactly, SlowMindThinking. I meant to quote you in my last post, but I couldn’t figure out how to quote multiple posts. But your post is precisely what I had in mind and, in fact, the best answer (for me the physics moron) that I’ve gotten. Many thanks.

172

Nope, I’m an old fart – the last time I raced out there was in the mid-90s (World Championships FIS telemark). The Euros pushed to get telemark into the 98 Olympics, but the USA and the networks pushed to get snowboarding into the Olympics, and snowboarding won (rightly so, given it brought something new to the Games, rather than a minor variation on an already existing activity), so no Olympics for me. Presently, I’m putzing about alpine racing locally (played hooky to get coached running gates today on alpine gear, have an alpine race tomorrow evening, teaching telemark Friday evening, alpine racing on Saturday . . .), so although I’ll never develop the skill in alpine that I have in telemark, I’m still having a great deal of fun.

173

Okay, having read the entire thread, I think the problem here is that we’re not considering the actual system that the OP described, and thus are having the wrong argument. Here’s my argument. (The important part is Skier B.)

Premises:
By default, there are only two forces: gravity, which acts in the (0,-y) direction; and the normal force caused by contact with the slope, which acts in the (x,y) direction. The goal of the skier is to reach the bottom of the slope fastest, which means having maximum acceleration in the (x,-y) direction at all times. Therefore, in the default circumstance, normal force is not relevant.
We assume no friction. We ignore drag, because it cannot possibly change results. (If anyone disagrees, I’ll explain later.)

Picture (sorry about the oversteep angle):



\
  \
    \
      \a.
        \_b.
          \
          |
           \c.
             \

Scenario: there is a slope running from (0,1) to (1,0). Partway down, at point (b), there is a bump, which means that the slope decreases then increases again. Other than the bump, the slope proceeds exactly in the (x,-y) direction.

Skier A (The Hopper): When skier A reaches point (a), right before the bump, they take a hop. That is, they push down in the (-x,-y) direction, causing them to take air. They land again at point ©, just beyond the bump, never making contact with the bump. Normal force is never relevant; it always pushes in the (x,y) direction, and it never becomes relevant to when you reach the finish line. The hop is perpendicular to the slope, so it is not relevant either. The force of gravity is constant, and thus, acceleration in the direction of the finish line is constant.

Skier B (The Jumper): skier B uses the bump. They get “big air” that way, which, as discussed by the other posters, doesn’t affect the rate at which they reach the finish line. However, at point (b) — the bump —the slope of the bump is decreased (in our model, it’s horizontal.) Therefore, the normal force acts in the (0,y) direction, a component of which is away from the finish line. Therefore, relative to the finish line, the skier decelerates, which makes them fall behind and stay behind.

When a skier is preparing for a jump, they are essentially pushing down, right? So they’re taking a mini-hop, thus decreasing the normal force exerted by the bump — that is, they’re trying to be as much as possible like The Hopper, Skier A! In real circumstances, this means they still graze the bump — but it’s still better than hitting the bump, which gives you “more air” but decreases your acceleration toward the finish line.
Any disagreement?

174

I enjoy how the right answer was the second post and the rest of this thread turned into comedy.

175

The second post was not the right answer. Whatever air’s disadvantages, the fact that it fails to “keep converting gravity into acceleration with a horizontal component” is not one of them.

The only “conversion” going on is that a skier on the slope experiences a normal force that a skier on the air does not; this normal force pushes in a direction directly perpendicular to the slope (and thus directly perpendicular to the direction from the skier to the finishing point). Do you generally consider it advantageous to be given an extra push in a direction perpendicular to the one you want to go in? Yes, this normal force can be construed as having a horizontally forwards component and a vertically upwards component, but there’s no more reason to consider the horizontal component of the normal force an advantage (moving the skier towards the finishing point) then there is to consider the upwards component of the normal force a disadvantage (moving the skier upwards away from the finishing point).

176

So what horizontal acceleration is gained in the air?

177

Eh, this is a bit glib on my part. The rest of my post makes the point better than this, which was a poor choice of words.

178

What’s so great about horizontal acceleration? A skier on the slope doesn’t want to move directly to the right; he wants to move in the direction of the finishing point, a direction parallel to the slope. On the slope, a skier experiences a cumulative force parallel to the slope of magnitude g * sin(theta). And in the air, a skier experiences a force equal to the sum of two forces: a force parallel to the slope of magnitude g * sin(theta), and a force perpendicular to the slope of magnitude g * cos(theta). They experience the exact same amount of force parallel to the slope. If you turned your head so that “right” was parallel to the slope and “up” was perpendicular to it, both skiers would experience the exact same amount of rightwards acceleration, just different amounts of upwards acceleration. Given any two perpendicular directions, an arbitrary vector can be decomposed into a sum of vectors in those directions; even vertical acceleration has a component in the direction you want to go, as long as the direction you want to go isn’t purely horizontal.

179

If it was really the case that the skier on the slope had an acceleration advantage over a skier in the air, we would expect a skier in the air to be unable to catch up with a skier on the slope if the former starts further away from the finishing point than the latter. But this isn’t the case: if you place skiers in such a way as that the skier in the air, the skier on the slope, and the finishing point upon the slope form a right angle, then the skier in the air starts further away from the finishing point (by a simple application of the Pythagorean theorem), but ends up at the finishing point at the same time (as demonstrated several times before). If you take that setup and move one of the skiers just slightly, you can even have a setup where the skier in the air starts further away from the finishing point, but ends up at the finishing point sooner. How does this tie in to any theory of the slope providing an advantage by virtue of “better acceleration”?

180

So, yes, it’s true that a skier in the air placed at the same horizontal coordinate as a skier on the slope gets to the finishing point later. But it’s also true that a skier in the air placed at the same vertical coordinate as a skier on the slope gets to the finishing point sooner. And a skier in the air placed at the same coordinate along an axis aligned with the slope gets to the finishing point at the exact same time. None of these are privileged. There’s nothing magical about the horizontal. Whatever the disadvantages of the air, the reasoning at the beginning of post #2 was not one of them.

Anyway, the above post says it better than I’m saying it now, so…