# Skiiers: What's so bad about "air"? **URL:** \ **Category:** Factual Questions\ **Created:** [February 19, 2010, 8:32pm UTC](https://boards.straightdope.com/t/skiiers-whats-so-bad-about-air/529677 "2010-02-19T20:32:12Z")\ **Posts on this page:** 20\ **Page:** 5
**Author:** ![Chessic\_Sense](https://avatars.discourse-cdn.com/v4/letter/c/7c8e57/32.png) [@Chessic\_Sense](https://boards.straightdope.com/u/Chessic_Sense)\ **Post date:** [February 22, 2010, 8:45pm UTC](https://boards.straightdope.com/t/skiiers-whats-so-bad-about-air/529677/81 "2010-02-22T20:45:05Z")
> [@Ruminator](#): > > It’s a tradeoff that the world class skiers already know without any physics calculations. If there was a equation in this thread that showed the optimal tradeoff, I missed it. There was no equation because we’re all assuming a frictionless environment. Because, seriously, how could we say anything about that w/o arbitrarily assinging values? In a frictionless environment, the optimal tradeoff is…zero. You should keep your skis on the ground at all times. NOT because it’s the shortest path, but because it’s the lowest. ---
**Author:** ![Ruminator](https://avatars.discourse-cdn.com/v4/letter/r/b9bd4f/32.png) [@Ruminator](https://boards.straightdope.com/u/Ruminator)\ **Post date:** [February 22, 2010, 8:55pm UTC](https://boards.straightdope.com/t/skiiers-whats-so-bad-about-air/529677/82 "2010-02-22T20:55:57Z")
> [@Chessic\_Sense](#): > > There was no equation because we’re all assuming a frictionless environment. Because, seriously, how could we say anything about that w/o arbitrarily assinging values? I understand the difficulties. I’m just relating the math models back to the OPs original question: the TV commentators often mention “air” with derision. Ok, some of the excess air time is a consequence of technique, equipment, choosing a line down the path, etc. But a big part of air time is choosing the speed which you come down the mountain. If you slow your descent 5 mph, you will undoubtedly increase ground contact and shorten the air time. But you will also lose the race. So when the experts analyze the excessive air time, what specifically do they expect the skiers to do differently? Ski at the exact same speed but choose a different line down the mountain? Or the skier should have engaged his edges slightly more (slow down 0.5 mph) to stay in contact with the snow more often? ---
**Author:** ![Telemark](https://sea3.discourse-cdn.com/straightdope/user_avatar/boards.straightdope.com/telemark/32/372_2.png) [@Telemark](https://boards.straightdope.com/u/Telemark)\ **Post date:** [February 22, 2010, 9:28pm UTC](https://boards.straightdope.com/t/skiiers-whats-so-bad-about-air/529677/83 "2010-02-22T21:28:33Z")
> [@Ruminator](#): > > So when the experts analyze the excessive air time, what specifically do they expect the skiers to do differently? You can absorb the bump and stay on the snow. But your legs have to have enough juice left in them and you have to time the bump right, something that’s often difficult to do. ---
**Author:** ![Ruminator](https://avatars.discourse-cdn.com/v4/letter/r/b9bd4f/32.png) [@Ruminator](https://boards.straightdope.com/u/Ruminator)\ **Post date:** [February 22, 2010, 9:33pm UTC](https://boards.straightdope.com/t/skiiers-whats-so-bad-about-air/529677/84 "2010-02-22T21:33:38Z")
> [@Telemark](#): > > You can absorb the bump and stay on the snow. But your legs have to have enough juice left in them and you have to time the bump right, something that’s often difficult to do. Agreed, that makes sense. But I thought the context of “air” was the big jumps with 1+ second hang time that no thighs could absorb. For controlling big air, you either pick a slightly different line to launch off the section of the “cliff” that results in the least hang time… or you deliberately bite your skis into the snow to slow down. Are there more technique options than what I’ve mentioned. ---
**Author:** ![Ximenean](https://avatars.discourse-cdn.com/v4/letter/x/aca169/32.png) [@Ximenean](https://boards.straightdope.com/u/Ximenean)\ **Post date:** [February 22, 2010, 9:36pm UTC](https://boards.straightdope.com/t/skiiers-whats-so-bad-about-air/529677/85 "2010-02-22T21:36:16Z")
> [@Sam\_Stone](#): > > **ivn1188** is correct. And it`s fairly simple physics. When you`re on an inclined plane, the force pushing you down is straight up and down. But the force pushing you up is normal to the surface of the plane. The resulting vector on the horizontal axis is your forward acceleration. But as has been said the resulting vector points down the slope, not horizontally, and the airborne skier has exactly the same acceleration in that direction (disregarding friction yadda yadda yadda). **ivn1188** seems to be saying that the airborne skier doesn’t have that acceleration until he hits the ground. He does, it’s just that he also has some acceleration towards the slope, which gets eliminated when he reaches the slope. And forces/acceleration that are perpendicular to the slope do not affect the skier’s acceleration towards the goal (again, disregarding the other factors that have been mentioned). ---
**Author:** ![ivn1188](https://avatars.discourse-cdn.com/v4/letter/i/e480ec/32.png) [@ivn1188](https://boards.straightdope.com/u/ivn1188)\ **Post date:** [February 22, 2010, 10:00pm UTC](https://boards.straightdope.com/t/skiiers-whats-so-bad-about-air/529677/86 "2010-02-22T22:00:38Z")
> [@Ximenean](#): > > But as has been said the resulting vector points down the slope, not horizontally, and the airborne skier has exactly the same acceleration in that direction (disregarding friction yadda yadda yadda). **ivn1188** seems to be saying that the airborne skier doesn’t have that acceleration until he hits the ground. \*\*He does, it’s just that he also has some acceleration towards the slope, which gets eliminated when he reaches the slope. \*\*And forces/acceleration that are perpendicular to the slope do not affect the skier’s acceleration towards the goal (again, disregarding the other factors that have been mentioned). (bolding mine) This is the crux of the matter. You are confusing velocity and acceleration. The instant the skier leaves the ground, he is only _accelerating_ in the vertical direction. This is because _gravity is the only force acting on the skier and gravity only pulls straight down, not down and forward_. His X velocity remains the same, while his Y velocity will change from whatever it is while he is skiing down the hill, to some positive value, and back to negative right before he impacts. He is a ballistic object at this point. When he touches down, a small amount of that excess force downward – ma\*sin(30) – gives him a small impulse, which results in a (maybe) slightly higher period of acceleration while the skier absorbs the impact. On the other hand, the skier on the ground has been steadily accelerating in the X (and consequently the -Y, but note that the -Y velocity is bounded by the slope multiplied by the X velocity of the skier). That means his velocity is increasing linearly the whole time. The longer the other guy stays airborne, the longer the normal force accelerates the skier on the ground, and the bigger velocity advantage he gains. The airborne skier is trading a constant +X force for a slightly larger instantaneous +X force, and that is a bad tradeoff. ---
**Author:** ![Ximenean](https://avatars.discourse-cdn.com/v4/letter/x/aca169/32.png) [@Ximenean](https://boards.straightdope.com/u/Ximenean)\ **Post date:** [February 22, 2010, 10:16pm UTC](https://boards.straightdope.com/t/skiiers-whats-so-bad-about-air/529677/87 "2010-02-22T22:16:47Z")
> [@ivn1188](#): > > (bolding mine) > > This is the crux of the matter. You are confusing velocity and acceleration. The instant the skier leaves the ground, he is only _accelerating_ in the vertical direction. This is because _gravity is the only force acting on the skier and gravity only pulls straight down, not down and forward_. His X velocity remains the same, while his Y velocity will change from whatever it is while he is skiing down the hill, to some positive value, and back to negative right before he impacts. He is a ballistic object at this point. > > When he touches down, a small amount of that excess force downward – ma\*sin(30) – gives him a small impulse, which results in a (maybe) slightly higher period of acceleration while the skier absorbs the impact. > > On the other hand, the skier on the ground has been steadily accelerating in the X (and consequently the -Y, but note that the -Y velocity is bounded by the slope multiplied by the X velocity of the skier). That means his velocity is increasing linearly the whole time. The longer the other guy stays airborne, the longer the normal force accelerates the skier on the ground, and the bigger velocity advantage he gains. > > The airborne skier is trading a constant +X force for a slightly larger instantaneous +X force, and that is a bad tradeoff. Look, any vector can be constructed from the sum of two or more other vectors, right? Such as the vector representing the airborne skier’s vertical acceleration. We can, perfectly validly, substitute _two_ smaller vectors for this single vertical vector, one perpendicular towards the slope and one at 90[sup]o[/sup] to it in the direction of the slope. As I said, acceleration perpendicular to the slope is irrelevant to this discussion, because what we are interested in is travelling along the slope as fast as possible. So that leaves you with just the vector that points in the direction of the slope. Do the same vector breakdown for the guy who is already on the slope, and the only difference is that he has a third vector representing the normal force. This vector cancels out one of the vectors from before, the one perpendicular to the slope. But so what? It still leaves an identical vector pointing in the direction of the slope, which is the direction that matters. ---
**Author:** ![Telemark](https://sea3.discourse-cdn.com/straightdope/user_avatar/boards.straightdope.com/telemark/32/372_2.png) [@Telemark](https://boards.straightdope.com/u/Telemark)\ **Post date:** [February 22, 2010, 10:26pm UTC](https://boards.straightdope.com/t/skiiers-whats-so-bad-about-air/529677/88 "2010-02-22T22:26:37Z")
> [@Ruminator](#): > > But I thought the context of “air” was the big jumps with 1+ second hang time that no thighs could absorb. A well timed approach can minimize it dramatically. ---
**Author:** ![Ruminator](https://avatars.discourse-cdn.com/v4/letter/r/b9bd4f/32.png) [@Ruminator](https://boards.straightdope.com/u/Ruminator)\ **Post date:** [February 22, 2010, 10:36pm UTC](https://boards.straightdope.com/t/skiiers-whats-so-bad-about-air/529677/89 "2010-02-22T22:36:37Z")
> [@Telemark](#): > > A well timed approach can minimize it dramatically. Ok, but what does “well timed” mean in this case? Timing of the thighs flexing at the moment of launch? Or timing via speed adjustments approaching the big bump/cliff? ---
**Author:** ![Xema](https://avatars.discourse-cdn.com/v4/letter/x/9de053/32.png) [@Xema](https://boards.straightdope.com/u/Xema)\ **Post date:** [February 22, 2010, 10:40pm UTC](https://boards.straightdope.com/t/skiiers-whats-so-bad-about-air/529677/90 "2010-02-22T22:40:06Z")
> [@ivn1188](#): > > … _gravity is the only force acting on the [airborne] skier and gravity only pulls straight down, not down and forward_. This part of your analysis seems a bit weak. On a downslope, the gravity vector definitely does have a forward component. ---
**Author:** ![sich\_hinaufwinden](https://avatars.discourse-cdn.com/v4/letter/s/c5a1d2/32.png) [@sich\_hinaufwinden](https://boards.straightdope.com/u/sich_hinaufwinden)\ **Post date:** [February 22, 2010, 11:08pm UTC](https://boards.straightdope.com/t/skiiers-whats-so-bad-about-air/529677/91 "2010-02-22T23:08:39Z")
> [@Ximenean](#): > > Look, any vector can be constructed from the sum of two or more other vectors, right? Such as the vector representing the airborne skier’s vertical acceleration. We can, perfectly validly, substitute _two_ smaller vectors for this single vertical vector, one perpendicular towards the slope and one at 90[sup]o[/sup] to it in the direction of the slope. As I said, acceleration perpendicular to the slope is irrelevant to this discussion, because what we are interested in is travelling along the slope as fast as possible. So that leaves you with just the vector that points in the direction of the slope. > > Do the same vector breakdown for the guy who is already on the slope, and the only difference is that he has a third vector representing the normal force. This vector cancels out one of the vectors from before, the one perpendicular to the slope. But so what? It still leaves an identical vector pointing in the direction of the slope, which is the direction that matters. (Longer paths aside) Thank you for continuing to explain this correctly. For those who don’t get it, please re-read that post until you do. Both people have identical components of gravity that are parallel to the slope - but only one of them is counter acting that with friction from ski/snow contact - which is a function of the force normal - the component of gravity acting perpendicular to the slope. Analyzing this in the traditional x/y axis makes no sense. ---
**Author:** ![Apex\_Rogers](https://avatars.discourse-cdn.com/v4/letter/a/e480ec/32.png) [@Apex\_Rogers](https://boards.straightdope.com/u/Apex_Rogers)\ **Post date:** [February 22, 2010, 11:20pm UTC](https://boards.straightdope.com/t/skiiers-whats-so-bad-about-air/529677/92 "2010-02-22T23:20:04Z")
> [@Xema](#): > > This part of your analysis seems a bit weak. On a downslope, the gravity vector definitely does have a forward component. Edit: I originally disagreed with you, but I just read Ximenean’s and sich’s clarification and realized this is more about how you define the x/y axes. If you analyze with axes parallel and perpendicular to the slop, the gravity vector does have a forward component to the path of travel, and there would be some acceleration in that direction. ---
**Author:** ![Chessic\_Sense](https://avatars.discourse-cdn.com/v4/letter/c/7c8e57/32.png) [@Chessic\_Sense](https://boards.straightdope.com/u/Chessic_Sense)\ **Post date:** [February 22, 2010, 11:28pm UTC](https://boards.straightdope.com/t/skiiers-whats-so-bad-about-air/529677/93 "2010-02-22T23:28:11Z")
Exactly right. I understand what you’re saying, **ivn**. That the force of gravity doesn’t pull in the X. But what you’re missing is that that axis is not the important one. The slope is the important one. Anything that’s 89.99999 degrees off of that, in either direction, could also be said to be “toward the goal”, but it’s not the most efficient. So it doesn’t deserve to be your end-all, be-all axis. If you plot the graphs of a, v, and p against t, you’ll see the area where B “wins” on the V graph. B’s graph goes steadily up, in a linear fashion, until he crosses the finish line. A’s graph, on the other hand, traces B’s until the ramp. Then it goes flat. At the end of the ramp, it goes up in a parabola until he hits the slope again. From here, he goes on a straight line, parallel to B’s V line, but lower. So for the rest of the race, he loses ground every second. It’s, as I’m sure you know, the area between the two lines. How much he loses by depends on how long the slope is after the ramp/bump/hill/jump. If the airborne guy could fall through the hill, then his parabolic V-section could eventualy meet _and cross_ B’s curve. At this intersection, both skiers would be at the same altitude, but A would still be losing the race. He lost time on the ramp. This jibes with what I said earlier…he’d have to get lower first. See [here](http://www.phy.ntnu.edu.tw/ntnujava/index.php?topic=142.0) and [here](http://www.phy.ntnu.edu.tw/ntnujava/index.php?topic=557.0). ---
**Author:** ![sich\_hinaufwinden](https://avatars.discourse-cdn.com/v4/letter/s/c5a1d2/32.png) [@sich\_hinaufwinden](https://boards.straightdope.com/u/sich_hinaufwinden)\ **Post date:** [February 22, 2010, 11:29pm UTC](https://boards.straightdope.com/t/skiiers-whats-so-bad-about-air/529677/94 "2010-02-22T23:29:40Z")
> [@ivn1188](#): > > …This is because gravity is the only force acting on the skier and gravity only pulls straight down, not down and forward. Gravity has no component parallel to a flat (horizontal) earth, but the racers are not on flat ground - they are on a slope. **And there most definitely is a component of gravity parallel to the (non-horizontal) slope. And it is identical regardless whether the person is in the air or not.** ---
**Author:** ![Muffin](https://sea3.discourse-cdn.com/straightdope/user_avatar/boards.straightdope.com/muffin/32/4341_2.png) [@Muffin](https://boards.straightdope.com/u/Muffin)\ **Post date:** [February 22, 2010, 11:45pm UTC](https://boards.straightdope.com/t/skiiers-whats-so-bad-about-air/529677/95 "2010-02-22T23:45:29Z")
> [@Pedro](#): > > I am quite sure that by the book the finish line is the line on the ground Nope. The start is a little wand that the skier moves with his or her shin (or for dual races, and physical barrier at roughly sternum height that is opened to let the skier proceed). The finish line is a beam of light at shin height. Actual lines drawn across the course at the start and the finish are meaningless. > [@Ximenean](#): > > Strictly speaking, the objective is to pass through a plane which includes the finish line; otherwise would a skier who files over the line not count as having finished the race? Nope, not a plane, just a beam of light which may or may not be lined up with the physical line. If a skier were to clear the finish line without triggering the beam (for example, by crashing and bouncing over the beam), then the skier would be granted a re-run. ---
**Author:** ![ivn1188](https://avatars.discourse-cdn.com/v4/letter/i/e480ec/32.png) [@ivn1188](https://boards.straightdope.com/u/ivn1188)\ **Post date:** [February 22, 2010, 11:51pm UTC](https://boards.straightdope.com/t/skiiers-whats-so-bad-about-air/529677/96 "2010-02-22T23:51:41Z")
> [@sich\_hinaufwinden](#): > > Gravity has no component parallel to a flat (horizontal) earth, but the racers are not on flat ground - they are on a slope. > > **And there most definitely is a component of gravity parallel to the (non-horizontal) slope. And it is identical regardless whether the person is in the air or not.** This is utterly ridiculous. Gravity does not change depending on where you are standing. Example: You are in a helicopter over a big flat landing area. You drop out of the helicopter. You fall straight down and land directly under the helicopter. This is because gravity pulls you straight down. Example 2: You are in a helicopter over a 30 degree slope. You jump out of the helicopter. You fall straight down and will land directly under the helicopter. This is because gravity pulls you straight down. Notice that both of these cases are exactly the same, because the only force acting on you is gravity, which remains constant no matter what terrain is underneath you. Notice also that it does not matter at what speed the helicopter is traveling – it can be either hovering, moving forward, moving backward, sideways, up or down. You will still land directly underneath the helicopter as long as the helicopter does not change speed (accelerate) in any direction other than that of gravity – straight up or straight down. An object only accelerates when acted upon by a force. The force vector on any airborne object is straight down, period, end. Coming up with some sort of tilted coordinate system is overcomplicated and not needed, because it doesn’t simplify, which is what any system should do. Every basic physics book uses the standard X,Y system where Y is vertical and X horizontal when talking about these sorts of problems. Gravity does not ever, ever, have a non vertical component. The normal force has a horizontal component. This is easily understood by a simple glance at the diagram I linked to earlier. The normal force only applies when you are touching the ground, by definition. Thus, by the simplest of reasoning, you only get horizontal force when you are touching the ground, and thus you only accelerate horizontally when touching the ground. It cannot get any simpler than this. ---
**Author:** ![Muffin](https://sea3.discourse-cdn.com/straightdope/user_avatar/boards.straightdope.com/muffin/32/4341_2.png) [@Muffin](https://boards.straightdope.com/u/Muffin)\ **Post date:** [February 22, 2010, 11:58pm UTC](https://boards.straightdope.com/t/skiiers-whats-so-bad-about-air/529677/97 "2010-02-22T23:58:40Z")
> [@Sam\_Stone](#): > > It is certainly possible to come up with a scenario where falling gets you there faster than staying on the slope. To do that, you`d need to support pretty high drag on the skis, and a slope that is extremely steep. You`d also want the skiier to come off the jump flat, and not rise up in a parabola, increasing his flight distance. But I think those circumstances would be very rare in ski racing. Snowboard cross is an example of falling getting you faster than staying on the slope. The boarders are repeatedly faced with taking big air to jump over gullies, or slowing down even more to ski the slope down into and then up out of the gullies. The courses are designed so that more often than not, it is faster to take the air than ski the gullies. Telemark classic used to have camel humps that presented a similar problem for racers, although on a much smaller scale. ---
**Author:** ![Muffin](https://sea3.discourse-cdn.com/straightdope/user_avatar/boards.straightdope.com/muffin/32/4341_2.png) [@Muffin](https://boards.straightdope.com/u/Muffin)\ **Post date:** [February 23, 2010, 12:00am UTC](https://boards.straightdope.com/t/skiiers-whats-so-bad-about-air/529677/98 "2010-02-23T00:00:47Z")
> [@Chessic\_Sense](#): > > OK, I’m putting my numbers where my mouth is. I ran some calculations: > > **Stipulations and Definitions of Terms:** > X axis - horizontal. Literally, along the horizon. > Y axis - direction of gravity. > A - acceleration in direction of travel (thus, g\*sin(x)) > Slope - 30 degrees, so that sin(30) is easy to handle (i.e. 1/2) > Frictionless surface > S - distance down the ramp. That is, the “hypotenuse” of our slope-altitude-ground triangle. > > The skiers start down the ramp and reach 10 m/s. Skier A’s track contains a ramp that is perfectly flat and 1 meter long. At the end of the ramp, A becomes a projectile. Skier B continues on a 30deg slope the entire time. > > **Methodology** > > For skier B, I used s=vt+.5at^2, where a=4.9 and v=10. > For A, I used two times: t=0 and t=.1 where the latter is the end of the ramp and beginning of the projectile phase. A’s path is given by t=sqrt(2h/g) where g = -9.8 > For any given s, I used pythagorian theorum or trig, with input by either x or y of one of the skiers. > > **Bottom line** > > 1. A hits the edge of the 1m ramp at .1s. But B doesn’t get to x=1 until 0.111 seconds. So A _does_ get an 11% head start. > > 2. At t=1.34, A roughly makes contact with the snow. Technically, it’s a little later than this. His distance down the slope? x=13.4, y=7.534, so s=15.068. > B, though, has traveled 17.8m in this time. > > **Conclusions:** > Of course, we already knew that. B wins by 2 meters, or less than half a second. But the question remains: did A follow a longer path? Was it longer than 2 meters? > > If it’s at least close, I win. A’s time loss is due to distance, not that horizontal accelerations part. If not, then I admit defeat. **Ivn** ’s “horizontal 10 m/s matters more” argument wins. > > Can someone find the parabola distance where h=1 and Vx=10 hits y=-sqrt(3)\*x? The intersection is somewhere around (13.5, -7.6) if it helps. Or just look at race results and compare the results to what the skiers are doing in the races. ---
**Author:** ![Chessic\_Sense](https://avatars.discourse-cdn.com/v4/letter/c/7c8e57/32.png) [@Chessic\_Sense](https://boards.straightdope.com/u/Chessic_Sense)\ **Post date:** [February 23, 2010, 12:03am UTC](https://boards.straightdope.com/t/skiiers-whats-so-bad-about-air/529677/99 "2010-02-23T00:03:41Z")
> [@ivn1188](#): > > Every basic physics book uses the standard X,Y system where Y is vertical and X horizontal when talking about these sorts of problems. > > Gravity does not ever, ever, have a non vertical component. The normal force has a horizontal component. This is easily understood by a simple glance at the diagram I linked to earlier. The normal force only applies when you are touching the ground, by definition. Thus, by the simplest of reasoning, you only get horizontal force when you are touching the ground, and thus you only accelerate horizontally when touching the ground. It cannot get any simpler than this. First of all, every basic physics book teaches you to tilt the axis. Second, you keep going on and on about this X axis thing. It’s irrelevant. That’s not the important axis. The important one is the one going down the hill. Third, why are you looking at the X component of the normal force and not looking at the -X of the component going into the hillside? There’s an opposite -X on that side of the hill, equal to Wcos(a)sin(a) that cancels out the normal force. That’s what normal force _does_. What’s left over is the component of gravity in the direction of the slope. I’m perfectly justified in saying that it’s _that_ component that moves one sideways. We’re all in agreement that, by analogy, 1+1-1=1. You’re arguing that the -1 cancels the first +1 while we’re arguing that it cancels the second +1. It clearly doesn’t matter, right?! Good. Now, all we’re asking you to do is realize that gravity moves the airborne skier at a rate of Wsin(a) in the direction of the goal. The skier on the ground gets moved in the direction of the goal ALSO at a rate of Wsin(a) in the direction of the goal. ---
**Author:** ![ivn1188](https://avatars.discourse-cdn.com/v4/letter/i/e480ec/32.png) [@ivn1188](https://boards.straightdope.com/u/ivn1188)\ **Post date:** [February 23, 2010, 12:04am UTC](https://boards.straightdope.com/t/skiiers-whats-so-bad-about-air/529677/100 "2010-02-23T00:04:38Z")
By the way, the reason it is simpler is because if you insist on some weird tilted frame of reference, you are comparing two vectors both composed of X and Y forces, and then you need to compare the magnitudes of those vectors, which is a lot more difficult to demonstrate and less clear. The reason is that your maximum -Y velocity over the course of a race is bounded by your +X velocity. Imagine a 45 degree slope - you drop 1 meter for every meter forward you travel. if you got at 1 m/s +X, you will go 1 m/s in the -Y direction, unless you burrow through the slope or take a longer distance path by leaving the slope. If you go 30m/s in the +X, you go 30 m/s in the -Y. You can’t go faster -Y than you go in the +X as long as you are on the slope, and every other path is longer so your average -Y over the race will exactly match your +X if you cross the finish on the ground. Therefore it is just confusing and unnecessary to tilt the axis, because the difference in forces is most obvious when you only have to account for a vertical force and the unbalanced horizontal portion of the normal force. [Previous page](https://boards.straightdope.com/t/skiiers-whats-so-bad-about-air/529677.md?page=4) [Next page](https://boards.straightdope.com/t/skiiers-whats-so-bad-about-air/529677.md?page=6)