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

Factual Questions
121

I think I understand where the horizontalists are coming from. Here’s how it goes (apologies to those posters whose explanations I am merely rehashing):

  1. To get from the start to the finish, a skier must cover some horizontal distance.
  2. A skier in the air does not accelerate in the horizontal direction.
  3. A skier on the ground does accelerate in the horizontal direction.
  4. Therefore, it is better to be in contact with the ground.

But you might just as well argue as follows:

  1. To get from the start to the finish, a skier must cover some vertical distance.
  2. A skier in the air accelerates vertically downwards.
  3. A skier on the slope also accelerates vertically downwards, but not as much as the skier in the air.
  4. Therefore, it is better to be in the air.

Neither of these lines of reasoning is valid. In fact, acceleration in both directions must be taken into account. And if you do the maths (sorry, I’m British), the differences cancel out and the skiers accelerate in the desired direction at the same rate.

122

A point that has been mentioned but not emphasized in this thread is drag. At the speed downhill skiers are doing, aerodynamic drag is the main force limiting speed. The skier’s body position has a lot of influence on the drag he experiences. And it’s a lot harder to maintain a low-drag body position when airborne.

I think this, plus possibly the “extra distance traveled” argument, may have the potential to fully explain the question.

123

Even in the simple, ideal form, the fastest path for something to slide down a slope is not a straight line. IOW, given the same start and end points, given a straight ramp, and one that starts off steeper and then curves to the end point, the curved ramp will get the sliding object to the end point sooner.

Not sure how that all relates to this question, except to say that it’s too complicated for the straight-line explanation to have any validity at all.

124

:smack::smack::smack::smack::smack: This is getting ridiculous.

If you’re not going to read and comprehend the thread, then you need to quit posting. We(I)'ve covered the distance-traveled argument in painstaking detail and shown it not to be very relevant. It only explains maybe 10% of the fall-behind.

We’ve also covered drag. We’ve already assumed no drag (i.e. giving the jumper the best-case scenario we can) and he still loses.

Put up numbers and equations, or quit posting. Demonstrate how A falls behind by 2 whole meters by taking a path that’s a fraction of a meter longer. Write out the trig. Whip out the calculus. Let’s see your stuff. Clearly this is more complicated than it looks.
ivn: I’d still like to see where you’re getting your 2Gsin(a) from. It’s just Gsin(a) for both skiers. But just so you can “see” our free-body diagram, I’ll describe it from your untilted perspective:

A: Draw gravity straight down, call it G. Now break that into two vectors. Starting at the skier, go perpendicular to the slope and call that one Gcos(a). Then turn 90 deg and go parallel down the slope. Call that one G*sin(a). Of course, your final arrow’s tip should touch the initial G’s tip. You should have the left half of a diamond, correct?

B: Draw G down again. Draw a force into the hill at Gcos(a). From the top of that arrow, draw what’s left of gravity. It points parallel to the hill at Gsin(a). Make the tips touch. This should be identical to skier A so far.
Now draw the normal force, also at Gcos(a).

Pay attention here, because this is where our argument comes into play. You have the normal force that you want to break into components. I understand that. But it’s not necessary because the Gcos(a) normal force is equal and opposite to the Gcos(a) force going into the hill. If you want to break the normal down into components, go ahead- but also break the opposite one into components. You’ll see that it all cancels. What’s left over? Gsin(a) down the hill.

Do you see now how the airborne skier has a Gsin(a) down the hill? If you’re not willing to draw out this diagram so we can talk about it, then I quit this thread.

125

Nope - it’s utterly correct. What’s ridiculous is your inability to understand simple physics.

And you’ve fondly reminded me of how Statics, Mechanics, and Dynamics were referred to as weeder courses freshman year - because they weed out those students who need to change majors.

126

chessic sense, I can understand your frustration with me, since I only read one page out of a three page thread, and ultimately, you are right.

Consider the following example. I skier is rocketing along, momentarily only in the horizontal direction. I’ll pick a speed of 50 m/s, which is close to 100 mph and shockingly close to what these guys do and makes round numbers come out. The skier comes to drop that is pitched.

I’m going to pick what seems like a weird coordinate system, but it actually simplifies things quite a bit. Let the x direction run so that the airborne skier leaves the ground at x = 50 m, and is traveling in the negative x direction. Let the value of y when the airborne skier leaves the ground = 10 m and the skier is also traveling in the negative y direction. The skier who stays on the ground travels in a straight line from (50, 10) towards the origin. In this coordinate system and with these initial values, the point that an air dragless and liftless skier’s parabola intercepts the straight line path is the origin. (Rounding g to 10 meters per second squared) This can be shown by simultaneously solving the equations of motion: y = y_initial - .510time squared and x = x_intitial -50*time, assuming I didn’t screw up the algebra while eating lunch.

Note, as ski jumping indicates ignoring drag and lift is not justifiable, but it allows us to check and see if the difference in speed is “just” aerodynamics or not. A real skier opens up when airborne, because of angular momentum considerations (ie., so he won’t crash), and so has significantly more drag to go with the significant lift from the skies.

Consider the airborne skier, skier A. Skier A has a velocity of -50 m/s in the x direction, and no forces in the x direction. Skier A has -50 meters to go in the x direction, which therefore takes 1 second.

Now consider the skier who stays on the ground. Oddly, finding the time to the origin is harder in this case. So, we switch coordinate systems to one with y perpendicular to the slope, and x tangent to it, and well let positive x run down the slope with the origin at the top. The equation to solve is then square root (5050 + 1010) = 0 + 50time + 5 time squared, again, g = 10).

Unless I screwed something up, the skier on the ground takes .93 seconds.

127

Indistinguishable, modeling what happens as a jump is incorrect. The skiers do not jump when they go airborne; in a race, they are fighting going airborne as well as they can. They go airborne when they can not exert a sufficient force (or torque) to convert horizontal velocity into downslope velocity.

128

So…which side are you on?

129

Jeezuz, what have I started? Sounds like ivn1188 and Chessic Sense are about to settle the issue with pistols at dawn.

Thanks to everybody who has spent so much time trying to answer my question “why is ‘air’ bad?”. But I don’t understand half the answers, so I’ll go with “It just is, alright?”

130

I think a ski race would be appropriate.

131

Huh? Air doesn’t have a slope…

132

I think it all about the impact of landing. As others have said, in order to get down the mountain, you are trading potential energy for kinetic energy.

Generally speaking, the skiier who gets the most kinetic energy in exchange for his potential energy will go the fastest and reach the finish line sooner.
Much like you maximize your fuel mileage by trying to avoid hitting the brakes,(even though braking in itself cost you no fuel) you maximize your ski speed by not smacking hard into the ground. A skier who is skiing on the surface is very efficiently trading potential energy for kinetic energy. Every foot of vertical height lost is traded for speed.

Now, imagine a skier landing a jump, not on a downhill slope, but on a flat transition. In this case, none of the potential energy difference between the height he left the snow and the height of the landing is transfered into kinetic energy. All of that potential energy is wasted on heat, compressed snow and the sound of the skis hitting the ground. If he lands on a downhill slope, the effect is less dramatic, but still there. Some energy is lost.

In this video, (go about 1:50 in) you can see the skiers making huge effort to stay on the ground, even snowplowing up the jumps to avoid huge air. The video is of ski cross, which is different from downhill, but the same physics apply.

Eta, the skiers are not just trying to avoid ‘air’, but to land on snow at a point where the slope of the course most closely matches their trajectory, thus reducing the impact and losing less energy to it.

133

Almost true. It’s the skier that can get the kinetic energy first. Here’s a simple scenario:

Imagine two drag racers (cars, not runners in wigs) are lined up to start a race. They both hit the gas at the same time and accelerate, in tandem, to some velocity. Let’s call it V1. Halfway through the race, Car A accelerates to V2. Later, Car B accelerates to V2. Now they’re both going at the same speed until they cross the finish line. Who wins?

Car A, or course. Because he went faster sooner. Car B was never able to make up that time. Had he been able to accelerate to higher than V2, and if there was enough track left, he would have been able to make up that time. But he couldn’t/didn’t. Their velocity “curves” look like this:



                        _A @ V2___________Both at V2_________
                      /                  /
            _V1______/_____B @ V1_______/
      Go! /
A&B______/

See that rhombus where A is higher than B? The area of that Rhombus is the distance that A beats B by.

The lesson is that it’s not just potential energy. It’s who can get to that sooner. In my example, both cars start and end with the same energy. But since A “used it” longer, he won. Same happens w/ the skiers. Sure, they both tap some potential energy, but B gets to it first by getting lower first.

134

(Written to show adding a negative if you don’t mind)

When I plug 1 sec into y = y_initial - .510time squared I get:

y = 10 + (0.5)(-10)(1)(1) = 10 + (-5) = 5m

Or only half way down to the ground.

It actually takes the square root of 2 seconds to drop from 10m.

Plugging that into x = x_intitial -50*time

x = 50 + (-50)(1.4142) = -20.71m

Which, as you know, means that the person will land beyond the y axis or origin - not right at it as you claimed.

And I’m not sure why you’re saying to solve the horizontal and vertical equations simultaneously. That would be solving for the time at which the distance traveled horizontally and vertically is the same - which couldn’t even be done for such a low starting height over flat ground.

Gotta go - I’ll check back in later.

135

I mean downwards along the axis parallel to the slope of the mountain.

136

I confess to not having read the whole thread so I apologise if my points have already been stated.

I have found this very interesting. I had not previously thought about a skier’s contact with the slope being necessary for acceleration. My normal instinct is that contact introduces friction leading to deceleration. Ignorance fought.

In terms of the various arguments that have flowed back and forth, it is probably insightful to consider a force diagram for a skiier in contact with a slope:
There is a vertical downwards force (gravity)
There is the normal force perpendicular to the slope at a direction determined by the angle of the slope. This can be resolved into horizontal and vertical components.
There is a friction force backwards, uphill along the line of the slope. This can also be resolved into horizintal and vertical components.

When these three forces are balanced, then there is no resultant acceleration. However, with minimal friction on skis then then acceleration is experienced. If contact is maintained with the slope then the acceleration will be down hill in the direction of the slope.

It follows then that the steeper the slope, the larger the horizontal component of the normal force and the greater the acceleration.

The question that then arises is, what shape path will enable the course to be completed in the least time? The answer is to accereate early, which means that steep at the start is better. In the early days of calculus this problem was solved with the optimal shape being a cycloid: also the best shape for a skateboard ramp.

A skier has no control over the shape of the course, but should aim to maintain maximum contact with the snow and in particular on the steeper portions where maximum acceleration can be achieved.

If the skier does become airborne then the question becomes one of how much acceleration can be achieved on impact. By judiciously landing on a steep portion there may be some gain in achieving air, however, in general constant contact is to be preferred. I don’t know exactly how elastic/inelastic a skier’s landing is, but I would guess it would be nearer to inelastic – not all of the energy absorbed in landing is converted to kinetic energy. And in any case, even if perfectly elastic, the acceleration resulting from a landing would occur later than than it would have if constant contact with the slope had been maintained. In short, the parabolic shape of a jump is almost the reverse of the desired cycloid shape, and while in he air, no forward horizontal acceleration is experienced.

The reality of skiing is even more complex – it is a 3D situation with corners, turns banking and undulating terrain as well as local variations in friction. Lateral motion must be considered as well as vertical motion. Hence the necessity to find a “good line” I rather expect that skiers learn a few basic principles from physics “go straight”, “avoid air” and then learn empirically and from instinct about what works to give the fastest speed.

My two cents (and no ascii diagrams, x and y axes or talk of frames of reference :slight_smile: )

137

This question was settled correctly and clearly within the first few posts, so I have not read everything up to this point. In the absence of friction or other sources of dissipation, the physics is simple so this debate is rather remarkable.

However, I wanted to mention a closely related problem, which occupied the minds of Newton, Leibniz, two of the Bernoullis and I’ve forgotten who else. That is the famous Brachistochrone problem. In this context the problem translates to the question of what slope would get the skiier from start to finish in the shortest time, assuming no friction and no “air”. The answer is most decidedly NOT a straight line, as was well understood by Galileo before a complete theory of kinematics existed. In fact, the best slope actually dips below the final destination. As others have pointed out, to minimize the time, you want to gain kinetic energy as soon as possible (which is why “air” is bad, you gain the kinetic energy later and thus don’t go as fast as physics would allow you to). Of course, there is a limit to this, since the distance you have to travel gets longer the more you deviate from a straight line path. The optimum is to follow the classic “cycloid” trajectory. If you start from rest, you instanteneously drop straight down, but then curve back toward the horizontal, go below your destination and curve back up. The extra speed you gain by falling further than necessary more than pays for the extra distance you have to travel. Galileo thought the optimum was a circular arc. Newton and the others showed that the cycloid was the optimal path. In the process they developed the Calculus of Variations and really showed the power of the new mathematics developed by Newton, Leibniz, and others. Plus, these historical figures were up to their normal ego-driven antics, for the entertainment of future generations.

138

That may well be fair. Still, what the exercise is shows is that any (high school physics) reasoning for “Going into the air must be slower” which would apply just as well to jumping into the air is incorrect. It at least helps us weed out that much immediately. Post #117 demonstrates that, in the high school physics model, jumping into the air and then landing upon the slope makes no change in the amount of time it takes to reach the finishing point upon the slope.

Accordingly, both “the total arclength of the path is shorter upon the slope” and “staying on the slope keeps converting gravity into horizontal acceleration, while otherwise it would provide only vertical acceleration” are both fallacious as explanations. This does not mean that going into the air cannot have any disadvantages, properly modelled, but whatever it is, it’s not those simple proposals.

139

As an aside, from watching some ski racing in the past few days, it seems to me that when a skiier impacts the snow after a big jump, he not only loses all the energy gained in the x axis, but possibly even some forward motion. Watching the aerial competition, it looks like there’s a pretty big splat when they hit, and the skiier comes out of it barely moving despite dropping a LONG way. Of course, the impact zone is designed to absorb force - which tells us that the nature of the impact zone is probably a significant factor.

In ski racing, sometimes the result of big air is a landing on a more horizontal part of the course, where all the downward energy is lost. That, plus snow compression, the force required to ski out of the depression made by the impact, could easily (in some cases) not only use up the force from the fall, but some of the forward energy as well.

I do think that apart from all the interesting theoretical, frictionless math that’s been going on here, the big losses probably come from the force of hitting the jump, the conversion of forward speed into altitude, and then the loss of that potential energy from impact with the snow.

Practical concerns probably dominate the other cases. For example, if the space between a jump and the landing zone is flat or bumpy or heavily powdered, it might be better to jump over it than ski through it. If taking a little more air means landing on a hard-packed steep downlope as opposed to landing in a bed of powder, you probably want to stay in the air a little longer. You really see that in ski cross, which is full of jumps and landing zones of various quality. Sometimes the skiier will get less air and land on the flat top of the jump and lose a ton of energy, while the next skiier gets enough air to land on the downslope and convert his downward motion into forward speed.

140

Since it’s 1:00 am, and you did exactly what I asked for people not to do, I’m going to beat you up a little bit.

Had you read the thread, you would see that contact with the slope isn’t necessary for acceleration, and in fact doesn’t benefit one at all, acceleration-wise. It’s the immediacy- as opposed to delayed - speed increase that’s important.

Hmm. Yeah, someone should get around to doing a force diagram. Oh wait! We’ve done at least 7 already and analyzed them in detail. We’ve even done the math to show them “resolved into horizontal and vertical components” and analyzed the resulting imbalance. We’ve even given it a magnitude and direction. Have you?

True. Also true of a skier in the air, but hey, if you’d have read the thread, you would have known that by now.

This is exactly incorrect. Think about what you just wrote and read it aloud a few times. A steeper slope makes you move sideways better? So jumping off a cliff sends you flying sideways?

Unless it could be mathematically demonstrated that the snow is irrelevant. You know, like we did in this thread.

Which is, of course, irrelevant to the objective of getting down the hill.

Ladies and gentlemen, a textbook example of willful ignorance! Seriously, it didn’t occur to you that perhaps we were discussing something worthwhile, or perhaps dispelling misperceptions, over the 100s of posts in this thread?

I agree 100% with your setup and math. However, I take the position that your setup isn’t applicable to the scenario. You have your jumper an acceleration that our skiers don’t have. You gave one guy legs and the other no legs. In order to be an applicable “high school science model”, you have to leave all the work (literally) to gravity.

To see why your version is a problem, give the skiers a one-shot rocket booster instead of jumping legs. One guy fires his perpendicular to the slope. The other fires forward. The second guy’s head doesn’t get crashed into anymore. He wins.