No! In your coordinate system, the airborne skier will get pull forward at g * sin(a), whereas the ground skier will get pulled forward at 2g *sin(a), because the normal force still adds a component in addition to gravity! They don’t cancel.
Pick a line such that the angle of the landing zone matches the angle of the skier’s descent (e.g. nordic ski jumps).
OK, then what is the magnitude of Nx? Let’s see how many steps…
W - known
N’ = Wcos(a)
N = Wcos(a)
Nx = Wcos(a)sin(a)
Let’s tilt the coordinates so x=downhill:
W- known
Wx: Wsin(a)
Bam! No need to even look at the normal force. Done.
Also, I’d like to quote the physics book I sto…er, “kept” from high school:
and from here:
First rule: do everything possible to avoid slowing down when skiing on the snow.
That results in racers occasionally taking air, but they do everything possible (e.g. taking a better line and prejumping to lessen flight time) to keep on the snow as much as possible.
BTW, I can’t speak for downhill, but for super-g when you are in the air you really do notice that you are slowing down relative to your speed before the air, and once you land it seems to take forever until you are back up to full speed again.
A more aggressive pre-jump and holding the tuck usually does the trick. Occasionally a better line can help, but usually that’s not an option.
No, in the coordinate system where the slope is considered to be the x axis, the normal force doesn’t pull the skier forward, because it is perpendicular to the x axis. It therefore has no bearing on the skier’s acceleration along the slope. All it does is stop him falling through the floor.
Explain in detail where you’re getting 2Gsin(a). That’s most assuredly not right. I’ll eat my physics book if I’m wrong.
I can’t speak for Telemark, but in the context of racing, it is timing of the pre-jump, not checking speed.
I know that, but you are confusing the physical realization of the timing system with the definition of the finish line.
In any case I may have been wrong. A quick fact check tells me that the official finish line is the plane between the two goalposts. That is the reference sanctioned by the FIS in case manual time keeping is required because of a failure in the electronic system.
The point is that I assumed there was no vertical slant to the finish line (by being an horizontal line and not a vertical plane), when in fact that is not true, as far as I could tell with a cursory search. My apologies to Ximenean for that.
I just wanted to say, this was a great post.
[aside]For the higher tiered FIS races, there are two hard wire (non-radio) electronic timing systems as well as manual timing. If electronic system A fails, then the time from system B can be used, and if that system fails, then an adjusted manual time can be used. The adjustment is done by comparing the electonic and manual times for the ten racers bracketing the racer who did not get an electronic time. If the racer does not get a time at all (no A, no B and no manual adjusted), then the racer gets a re-run.
Most races are club races that are not up to FIS standard, and don’t bother with manual times. Many club races do not even have two hardwire electronic systems, and instead only have one. Some even get away with radio systems. If the racer does not get a time, then the racer gets a re-run.[/aside]
Obviously, there is something missing here. I am going to go over to the university tomorrow and talk to an actual physics professor. I am unable to clearly articulate where you are coming up with the difference between the two coordinate systems.
The problem I see is that someone in the air will not have as large a +x force as one on the ground in a tilted system, but I am not sure where you are going wrong (and since it’s absolutely clear in a normal x,y system, I am positive you are going wrong somewhere).
So, I will go ask and see what’s up.
(I suspect it has something to do with the fact that your weight affects the size of the normal vector/resulting +x force, but your weight has nothing to do with the force acting on you while you are in the air, which is why a feather and a stone fall at the same speed)
Basic observation that’s been made before, I think, but which ought be made again:
Suppose we have a slope, skier A on the slope, and skier B in the air “above” A in the sense that the vector from A to B is perpendicular to the slope (rather than in the sense that this vector is perpendicular to the horizontal ground). Both move only under the influence of gravity.
In this case, when skier B reconnects with the ground, they will land exactly on skier A. Why is that? Well, Ximenean’s post explains it very nicely, but if you are unconvinced, we can do the calculations “brutely”.
Suppose the slope has angle theta relative to the horizontal, the initial distance from skier B to skier A is d, and the acceleration of gravity is g. Then the angle between the initial vector from B to A and the direction straight downwards is also theta; accordingly, the initial downwards distance from B until reaching the slope is d/cos(theta). Thus, the time it will take for B to hit the slope is sqrt(2/g * d/cos(theta)), at which point B will land at the point downwards along the slope a distance of tan(theta) * d from A’s initial position.
Meanwhile, A experiences the sum of two forces: there is g directly downwards. Plus, there is the normal force, of magnitude g * cos(theta) and pointed perpendicularly to the slope. The sum of these two is an acceleration of magnitude g * sin(theta) pointed along the slope. Thus, after the amount of time B takes to hit the slope, A will have moved along the slope a distance of 1/2 * (g * sin(theta)) * sqrt(2/g * d/cos(theta))^2 = sin(theta)/cos(theta) * d = tan(theta) * d, and thus will end up at exactly the same point as B at the same time.
Yes, but as said before, that is giving a head start to skier B, which is not warranted since they start at the same position and are skiing at the same speed.
It’s a head-start to B if you measure horizontally. It’s a head-start to A if you measure vertically. It’s a head-start to neither if you measure along the slope. It makes the initial distance between B and the point at the end of the slope greater than the initial distance between A and the point at the end of the slope.
As a skier and a physicist, I can tell you indistinguishable that you are wrong. It is generally faster to stay on the snow, which I know from experience.
This thread is way too long for me to go through every post, but one thing that seems to be left out is the distance traveled. Take the clearest case: a skier is on the flat with some initial speed and then drops down a 45 degree angle. A skier who stays on the snow takes the straightest, and therefore shortest, path to any point down the slope. A skier who goes airborne takes a longer path. Unless the down slope forces on the airborne skier are significantly greater than those of the guy on the ground, the guy on the ground gets there first because of the shorter distance.
Baring undulating topagraphy, the shortest path is almost always the fall line. If you follow it, you are likely to have a higher speed, because you must exert a force, which always includes an upward component, not to follow the fall line. But the biggest difference is the length of the path. Because the fall line is the shortest path, following it gets you down the hill the fastest. Going airborne never follows the fall line.
There may be practical factors involved beyond those being modelled here. However, here is a “No friction, no air resistance, high school physics” setup:
The question is what happens when B elects to launch in the air vs. A electing to stay on the slope? Suppose we model this by B choosing to manually jump normal to the slope at some point, vs. A choosing to stay gliding along the slope. In this case, A and B glide together until some point, B suddenly experiences a (let’s say, instantaneous) acceleration normal to the slope.
So A is travelling is at some point on the slope (call this P[sub]0[/sub]) with some initial velocity pointed downwards along the slope (call this S[sub]0[/sub]h, where S[sub]0[/sub] is A’s initial speed and h is a unit vector pointed downwards along the slope), while B starts at the same point P[sub]0[/sub] on the slope but with a velocity which is S[sub]0[/sub]h + L[sub][/sub]j, where Lj is the velocity provided by the jump; specifically, L is the magnitude of this velocity, and j is a unit vector pointed in the direction of the jump, perpendicularly to the slope.
As before, let the slope have angle theta relative to the horizontal, and let g be the magnitude of the acceleration of gravity [as before, I’m using g for a scalar, not a vector]. Let d be a unit vector pointed straight downwards. What does B’s motion over time look like? Well, B’s motion over time until B reconnects with the slope is described by B(t) = P[sub]0[/sub] + (S[sub]0[/sub]h + L[sub][/sub]j)t + gdt[sup]2[/sup]/2 [the coefficients of course coming from B’s initial location, initial velocity, and initial acceleration, respectively].
And A’s motion over time? Just as before, A’s acceleration, accounting for the slope’s normal force, will end up being of magnitude g * sin(theta), pointed down along the slope. That is, letting S be a unit vector pointed down along the slope, A’s motion is described by A(t) = P[sub]0[/sub] + S[sub]0[/sub]ht + (g * sin(theta))ht[sup]2[/sup]/2 [again, the coefficients derive straightforwardly from A’s initial location, velocity, and acceleration].
Apart from the initial point, do these two paths ever end up in the same place at the same time? (If so, that would be the place were B re-lands on the slope, and thus we would have that B lands exactly on A at that place at that time).
Well, setting them equal to each other and cancelling out common terms, we see that this happens just in case there is some t such that Ljt + gdt[sup]2[/sup]/2 = (g * sin(theta))ht[sup]2[/sup]/2. In other words, if there is some t such that Ljt + g(d - sin(theta)h)t[sup]2[/sup]/2 = 0.
But note that d - sin(theta)h = -cos(theta)j. Accordingly, the equation becomes (Lt - g * cos(theta)t[sup]2[/sup]/2)j = 0.
And this is of course essentially a quadratic equation; we see that this is satisfied both when t = 0 (of course, since A and B start at P[sub]0[/sub]) and when t = 2L/(g*cos(theta)) (i.e., after this much time, we’ve just demonstrated, B will land exactly on top of A; they will be in the same place, at the same time).
HS science teacher answer here:
From the start to the finish there is a theoretical straight line, which goes mostly through the earth. The fastest path is the closest to this line. Going high off jumps is the farthest point from this path.
Of course, there is a compromise, in that going slowly enough not to jump off those big bumps on the Vancouver downhill course will lose you the race right away. It’s best to go fast in all the other points, taking tight turns, hopping smaller bumps, and living with your large jumps as a result.
You are still MOVING forward, but you are not ACCELERATING forward. In fact, you are decelerating forward in the air.
You haven’t caught Ximenean’s point. Straight downwards acceleration is equal to a sum of acceleration downwards along the slope and acceleration perpendicular to the slope. If “forward” means “downwards along the slope”, then straight downwards acceleration contains a forward component.