Hmm. Curious. In this simulator, it seems the more air you get (i.e. the longer the ramp), the faster you’re going when you get to the bottom.
ivn1188 is correct. And its fairly simple physics. When youre 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.
The minute you leave the ski jump, that vector vanishes - now you just have gravity trying to pull you back down, and wind resistance trying to slow you down. You`ll take on a ballistic path with your forward speed slowing due to wind resistance.
Now, when you land you`ll convert some of that vertical speed you built up back into forward speed. But not all of it, because you absorb some through your muscles, additional friction of the snow, flex in your skis, etc.
Imagine this thought experiment - take the big hill ski jump, and put another one right beside it. Except this one doesnt have a jump - its got that straight line all the way to the bottom of the hill. Now let two skiiers race down. One stays in the groove all the way to the bottom, and the other has the track end and he falls the rest of the way. Think about which one gets to the bottom first. If you think the guy falling through the air will get there first, start shortening the track in your mind. Think about what would happen if the two skiiers were only going 10 mph when the jumpers track ended. Hes pretty much going straight down, while the other guy continues to accelerate forward.
That makes sense. The whole “more gravitational pull and more acceleration due to being in contact with the ground vs the air” argument just seemed off base.
I’m sorry, Chessic, I don’t think you are correct. There is no need to tilt the x axis or tilt gravity. You do need to be touching the ground to accelerate forward, because otherwise you are only being pulled straight down, regardless of your forward velocity. When you hit the ground, the acceleration because of gravity only gives you force * sin (about 30 degrees), plus you soak up a lot of that potential energy in compressing your legs and the snow, etc. When you are touching the ground, you accelerate continuously. It’s not a matter of potential energy, it’s a matter of using that energy efficiently to get the most acceleration.
I think you are seriously overcomplicating this by setting up some non-standard frame of reference, and getting confused by it.
Just to avoid confusing the issue, Id like to point out that actual ski jumpers assume an aerodynamic position which gives them lift. This lift is applied at an angle, which causes the ski jumper to accelerate horizontally as well, like an airplane in a dive. But in ski racing, the skiiers are not assuming flying shapes, and so they cant convert gravity into forward speed like a ski jumper can.
Also, when the slope is not horizontal, you DO pick up some speed relative to the slope by falling, since your speed down the slope is measured against the plane of the slope. For example, if you had a slope that was angled at 70 degrees and you just jumped off a cliff and hit the slope halfway down, youd be going very fast in the direction of the slope. But thats a small effect for the types of slopes we`re talking about here.
First off, tilting the axis is the standard frame of reference. Second, the normal force only accelerates you horizontally. Gravity has to pull you down. You need to win at both to win the race. Third, I still can’t figure out why you insist on ignoring the fact that when A hits the ramp, he’s transferring his -y velocity into positive X, which compensates for the fact that he won’t be accelerating in X during his fall. In fact, he’s traveling faster for a period of time, so he should gain X position.
So why, in all your examples, do you keep putting A above B but at the same X? He should be *forward *of B!
This is begging the question: of course if you measure purely by the X axis as you imply, the skier will temporarily not be making any progress toward the goal while purely under the influence of gravity. But that’s not the best way to approach the problem.
Since the slope is the “shortest” distance between the start and finish, it is much more natural to treat progress along the slope as the actual progress rather than only considering progress along a purely arbitrary mathematical construct.
When you treat the the slope as the measure of progress, as it should be, falling downward pulls one closer to completion (and why shouldn’t it, after all by falling down you will actually be closer to the end of the course via application of the Pythagorean theorem.) Rotational transforms merely make the pictures prettier.
So what? He’s got an X-axis lead on the other guy.
It is certainly possible to come up with a scenario where falling gets you there faster than staying on the slope. To do that, youd need to support pretty high drag on the skis, and a slope that is extremely steep. Youd 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.
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
-
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.
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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.
Somebody call mythbusters.
At least not intentionally. 
Nah. Sam and ivn have correctly answered the question.
BTW, there is one ski race that includes a judged form and distance jump – the Telemark Classic. The racers have to look at how the points are apportioned, and find the optimal compromise between being fastest down the course (as little air as possible), and jumping the longest distance (big air). Usually the points are apportioned to favour those who get big air. Make no mistake, though, the more air, the longer it takes to ski the course. Big air means a slower time for the course.
Come to think of it, freestyle is similar too, in that the racers have to find the optimal compromise between being fastest down the course and scoring the greatest points on their jump tricks.
We already know that. That’s not the issue under contention. The question is whether that’s due to the path taken, the lack of acceleration, and/or time lost in the air.
Also, to what degree does it matter?
We keep going on and on about X-acceleration. I’m going to show you why that’s not the case, as best I can.
Define X to be horizontal and Y to be vertical. So in other words, no axis tilt. Drawing weight and normal force, it appears there is a horizontal component from the normal force. The normal force is equal to Wcos(a), right? So the x accel is Wsin(a)*cos(a), correct?
But if there’s a wcos(a) coming out of the hill as normal force, then you should draw that component of weight going into the hill. You’ll then notice that it, too, has a Wsin(a)*cos(a) pointing in the negative x direction. And the only force you’re left over with is the intuitively obvious one: the Wsin(a) pointing down the hill.
I can just as easily declare that this is the force moving the skier in the x direction, since it has a component in that direction. It all depends on how I choose to break out the component vectors. After all, we all know that in reality, there are only two forces as work here and the rest is just math.
I’ve run the numbers a few times more since last night. It seems to me that not only is the parabola longer, but there’s also the issue of the ramp. Skier A just loses so much time on the ramp because he doesn’t get to cash in his GPE as soon as B does. It’s like he’s accelerating out the gate too late. But that’s just inefficient accel, not “less” accel, and it has to do with the fact that his velocity is lower for an extended amount of time. So on the graph, there will always be a space between the two velocities, meaning there’s distance to be lost there.
Sorry, Ivn, I still contend that it’s a bad path, not because it doesn’t get you sideways, but because it doesn’t get you down.
I agree with Pasta’s approach above, that looking at the losses in each system is the correct way to do it. If the increased vertical speed, at the moment of landing, were converted to horizontal speed perfectly efficiently, then the two scenarios would be the same. His jolt on landing would be converted to a faster horizontal speed.
But the skier who has been in the air will necessarily absorb a lot of the energy in his legs - the only other possibility would be to bounce in an elastic collision. Since he doesn’t bounce, that means that some energy was absorbed in his legs, and that energy has no place to come from other than his speed.
But the math doesn’t show that. Using my stipulations, even with an elastic collision, A hits the ground at 12.45m/s (in whatever direction you chose), 2.48m behind the other guy, who’s going 16.713m/s at that time. That’s even before any loss from collisions, air resistance, snow friction, or anything else. It’s a best-case scenario for the guy.
BTW, I dug out the calculus. Turns out, despite being 2m behind the other guy, A’s flight path is only roughly .75m longer than the ground path. That’s counting the ramp. So the question is still why that is.
In light of this new evidence, I’m going to amend my hypothesis. It’s not due to a longer path at all. It’s gotta be due to the earlier access to the GPE. That extra Velocity gets “used” for a longer time by B. To test this, I intend to set up a scenario where B gets robbed of that GPE until A gets to use it too. I need help with the setup though, so if anyone wants to offer ideas, lemme know.
Look, in the real world there are all kinds of issues involved here. I mentioned in my first message that the vertical velocity of the jumper will not all convert back to horizontal velocity because of losses through his muscles, flex in the skis, etc.
In addition, going faster means more wind resistance and more snow resistance. So even if the skiier could land in a way that translated his vertical speed into speed in the direction of the slope without any losses, he`d simply slow down due to the increased drag, and lose part of the excess kinetic energy he built up anyway.
Lets do another thought experiment. Well assume a vacuum, and a frictionless ramp, and a skiier that can perfectly transfer energy from a drop into speed along the plane of the ramp.
So we start two skiiers - one is on a flat plane, and the other on the incline. We accelerate the skiier on the flat plane so he`s going exactly the same speed as the skiier on the incline, and they stay lined up on the Y axis. Then the skiier on top flies off the plane, and drops onto the ramp.
During the time hes falling, the skiier on the ramp below is pulling ahead, because hes still accelerating in the Y axis. But our other skiier is accelerating along the X axis. When he hits the plane, and converts all his energy into speed along the plane, hes still going to be behind the first skiier, and stay there. Even though hes converted all his energy back into speed and therefore lost no energy at all in the process of jumping, he`s now behind because the distance he has had to travel is farther.
So even in the perfect case with all energy conserved and converted to and from speed, the guy with the big air is going to lose.
In the real world, the impact of the jumper is going to soak up most of the excess vertical energy, with very little transferred into speed down the slope. And if the skiier does go a little faster because of the jump, the extra speed causes more wind drag and snow friction losses - very quickly he`ll be back to the speed he would have been at without the jump.
My point above was just that any speed difference at a fixed point B downslope must be accounted for by losses. The amount of time it takes to get to that fixed point is indeed dependent on the entire history (in particular, the time evolution of the parallel velocity).
Yes, I believe getting lost in all the physics calculations is the underlying intuition that skiing at the speed that forces contact with the snow 100% of the time all the way down the mountain will be slower than the speed where skis will take flight some of the time (whether it’s a few inches or a dozen feet.)
For example, I think the Olympic downhill skiers getting down the mountain at ~70mph. If you deliberately ski so that your skis never get air, it will be slower. It won’t be 70mph.
Ski slopes are not perfect straightedge of a ramp. Obviously, on a real world ski slope, there’s a tradeoff between snow contact vs some air. 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.
But what I’ve just demonstrated contradicts you in several places:
- I already assumed a frictionless environment and a perfectly rubbery skier.
- I think you have your X and Y backwards, or I don’t understand the spacial relationship of your planes and ramps. Or I got confused from all the "he"s.
- The longer path isn’t the answer. I was wrong about that. The jumper falls behind by 2 meters but only takes a .4 or .5 longer path. He lands at the 15.8-meter marker while taking a 16.48m path. He looks ahead to the 16.48m marker on the slope. Does he see his opponent? No. His opponent is way ahead at the 18.3-meter marker. So if he did the same thing at the same speed, but in a straight line, he still loses by a coupla meters.
4)There’s no excess energy anywhere. Ever. At no point is the jumper going faster than the plane-er.
The guy that gets lower the soonest wins. The non-jump guy does that by going “under” the ramp. He’ll get the gravitation PE->KE boost sooner, so his V curve will always be higher than the other guy’s. X and Y acceleration is irrelevant to to that.