Right. That is correct. That’s why I now only claim that post #117 invalidates certain arguments which, if correct, would have to still apply in that alternate scenario. I don’t claim that #117 fully answers the relevant question, since, as you point out, it’s not quite the right model, but it does at least make it clear that certain proposed glib answers are fallacious.
Somehow, you misconstrued my first post. It is faster to stay on the snow, and not just for aerodynamic reasons.
Comparing the trajectories of a downhill skier who jumps with one who doesn’t is incorrect. These guys aren’t jumping when they go airborne: they are trying to stay on the ground are unable to exert a sufficient force to follow the slope.
We already know that. That’s given information. No one’s contested that since the OP. The question is why it’s faster.
You’re right, I made a mistake. But my approach is sound - I did not say over level ground. That’s what I get for doing these things over lunch, with coworkers walking in and actually asking work related questions.
Skier A flies is skiing horizontally at 50 m/s until he reaches the pitch. Because he does not modify his horizontal velocity, he goes airborne. In the x-direction, the horizontal direction,
x = x_i -50t (horizontal velocity = -50 m/s)
In the vertical direction
y = y_i - .510tt
I wanted to pick x_i and y_i so that (0,0) is on this trajectory at some unknown time, so I subbed the correct function of x in for t:
y = y_i -5*(x_i-x)(x-x_i)/(5050)
Then set x and y to zero, and solved for y_i in terms of x_i
y_i = x_i*x_i/500
At this point, I screwed up. But, pick a convenient value for y_i, say, one that you can take a square root of, and you get a value for x_i.
The end result, though, is that skier A follows a longer trajectory with no initial velocity in the vertical component, and no acceleration in the horizontal. Skier B, the skier who follows the slope, has an initial velocity along the shortest path and accelerates along that shortest path, so skier B gets to the point of intersection of the parabola with the slope, first. The reason that I emphasize the shortest path part is that skier A’s velocity is also along A’s path, as it must be, and skier A accelerates more than skier B, since A is in free fall.
You did screw something up. Your numbers have to be in the direction of the motion, or you get a nonsensical answer. One example is to strap you to a space ship and ask you to walk 10 meters along the ground when I launch you into space. You’re accelerating, all right, but it’s normal to the direction you’re solving for. So you can’t count the rocket.
Likewise, you can’t count all of g. You plugged in *g *into the equation, but the equation doesn’t ask for g. It asks for a. a=gsin(theta).
sqrt(50^2+10^2)=sqrt(2600)=51
sin(theta)=10/51=.196
gsin(t)=10*.196=1.96
That’s the number that goes into the equation.
p=Vt+1/2at^2 - You got that part right.
51=50t+1/2(1.96)t^2
0=.98t^2+50t-51 - solve the quadratic.
t=1 seconds
After reading through the entire thread and drawing up a quick free-body diagram of my own, I think I can help…maybe.
Start with a coordinate system with x horizontal and y vertical. In this coordinate system, gravity pulls in the -Y direction, straight down. The slope is a 30 degree slope, meaning it is 30 degrees down from horizontal.
Looking at the diagram for a skier on the slope, the gravitational force on the skier is made up of 2 components - one perpindicular to the slope and one parallel to the slope. The slope responds with a normal force equal in magnitude to the force from gravity perpendicular to the slope, leaving the only force acting on the skier parallel to the slope, thus the skier accelerates both in the +X direction and the -Y direction, or down the slope towards the finish line.
Now draw the same diagram without the skier touching the slope. The only thing that is different is that the normal force from the slope dissappears, since the skier is no longer contacting the slope. So now, the only force acting on the skier is in the -Y direction, since the component that is perpendicular to the slope is no longer cancelled out by the normal force from the slope and the sum of the two vectors that are perpendicular to the slope and parallel to the slope sum together and the resultant is the original gravity vector. Thus, the skier in the air no longer accelerates in the +x direction, but only in the -Y direction, meaning he no longer is gaining any velocity in the +x direction.
Now why is this bad? In order to win the ski race, the racers have to get to the finish line, which is located in the -y and the +x direction from the start of the race, which means in order to get there the fastest, they need to continually accelerate in both directions. When they leave the ground, they don’t accelerate in the +x direction and in fact may lose velocity in that direction due to wind resistance (in the real world) meaning that they have to regain that lost velocity and gain more to catch up with the person that didn’t leave the ground.
Note that this discounts the friction from the snow, and also any momentum transfer from the airborne skier hitting the ground.
Did that help?
Yes I did. What you didn’t catch is that I screwed up twice. Once in the way you caught, but also in the manner sich_inaufwinden caught. There really is no good excuse for someone like me screwing up basic physics.
And slowmovingvehicle, we really can explain it, once we nail this down.
Thanks to the sich suggestion (sorry, couldn’t help myself, the “ch” in sich is closer to a k sound than the sound in “cherry”) my initial values need to be modified so that the initial value of y, in the coordinate system I picked, is 5m, not 10m. If you do that, not only does skier A hit the slope at the origin, but it still takes him 1 second to do so. (The initial value of x is still 50m.)
Instead of figuring out how long it takes skier B to reach the origin, I’ll rephrase the problem in a more html friendly manner. How far has skier B gone in 1 second? Skier B starts 5m above, and 50m away from the origin, which comes to 5*root(101) ~= 50.25m.
In one second, he goes 501 + .510/root(101)11 ~= 50.5m. (Chessic, you agree sin(theta) = 1/root(101)?)
So, the skier on the snow goes roughly .25m further down the slope than does skier A, while skier A is airborne. This comes to right about to a whopping 5 milliseconds.
Clearly, not significant, since the elites lose tenths of seconds. The real reason is coming up.
That’s the argument that’s been put forth before, that the horizontal component of the normal force of the slope provides an advantage. But this is erroneous: one could just as well say it’s the skier on the slope who’s at a disadvantage: he needs to acceleration in the -y direction to reach the finishing point, but he isn’t accelerating in the -y direction as much as the skier in the air is. The normal force is unfairly retarding him vertically! (Of course, you shouldn’t say that either, but it’s the same broken reasoning)
Again, this is the kind of glib argument which is shown fallacious by post #119; the mere fact that the person in the air experiences acceleration directly downwards instead of along the slope does not slow down their progress to the finishing point upon the slope.
Why would falling through the air move you along the axis parallel to the slope of the mountain?
I was just answering his “which side are you on?” question, not the op.
It won’t. But all the same, even for a skier in the air, the acceleration of gravity directly downwards can be represented as a sum of acceleration parallel to the slope of the mountain and acceleration perpendicular to the slope of the mountain, the former component remaining just as large as if the skier was on the slope and experiencing its normal force anyway.
OK, I would agree with that, but the physics part of it is correct, no? The skier in the air is no longer accelerating the +x direction, and in fact, may be said to be accelerating in the -X direction due to wind resistance.
Whether or not he wins depends on several things, such as how long he is in the air, what his velocity was prior to leaving the ground, whether the jump in question is a kicker that throws him up into the air or just a falling away as the hill transitions from one angle to another. In any case, I would be willing to bet that the magnitude of the airborne skier’s velocity vector is lower when he comes back to earth than that of someone who didn’t leave the ground, due to the loss of velocity in the +X direction, among other factors.
Or, rather, another way to put it is that falling through the air will move you along the axis parallel to the slope of the mountain, in the same sense that moving at an angle somewhere between straight to the right and straight up will still move you along the positive X axis.
Why wouldn’t the skier on the slope experience the same horizontal wind resistance?
I believe there are two main reasons it is slower to go through the air"
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The skiers have charted out the fastest path down the mountain. As long as they are in contact with the snow, they can theoretically follow that path exactly. When hop through the air, they can only take piecewise straight lines to that ideal path. And the pieces are pretty long. Yes, this is a bit of an approximation, because you can design a course in which it is smart to glide through the air. The ski cross course included back to back hills such that you could jump off the first and over the second hill, which was clearly a shorter course than following the snow. But, the downhillers don’t have that.
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Air resistance is pretty significant at these speeds, and when you are in the air, you must open up to keep from tumbling. (Or, pinwheel your arms, which is bad, also.) The elite guys go faster than I drive, so air resistance is huge - completely dwarfing friction due to snow. An 80 mph wind will just about blow you over, but a small child can push you across the snow if you are on skis.)
There is a third factor. Skis are springs. I know I can use those springs to accelerate on relatively moderate slopes. I don’t know if it is much of a factor when you are going so fast it is all you can do to stay upright, though. It might make a bigger difference in a turn, when the spring fully extends and does its best to snap you back, in which case, you can’t be in the air, so it is moot for this discussion.
He does. Gravity pulls him along the plane of the slope as long as his skis are on the slope. Gravity only pulls him straight down (not at a slope) while he’s in the air. He can not possibly accelerate “forward” while he’s in the air, although he will continue to travel forward until the wind resistance stops him.
Got ya - the simultaneous eq solving was part of the back calc to set up your example. I knew what you were going for but didn’t quite follow.
BTW - what am I doing wrong math wise since I’m getting a different final eq for y ?
x = x_i - 50t
50t = x_i - x
t = (x_i - x)/50
t * t = (x_i - x)(x_i - x)/(50 * 50)
y = y_i - 5(x_i - x)(x_i - x)/(50*50)
Why do you have (x_i - x)(x - x_i) ?
Reading over post [url = http://boards.straightdope.com/sdmb/showpost.php?p=12149903&postcount=117] 119, I would agree. One problem I see in relating that post to the real world of ski racing is that most often, the racer leaving the ground does not get the added velocity L.
If this is set to 0 in the post linked to, then there is only one point where the racers will be at the same point, the beginning, correct? Maybe Indistinguishable can do the math on a hypothetical situation where the skier takes to the air via the angle on the slope increasing, but another skier stays on the slope without losing any velocity and or taking to the air. I’d be interested in seeing that, although I think it might be pretty complicated.
While in the air, the skiers almost stand up, which means they are a big rectangle flying into the wind. When in a tuck, they are more bullet shape, and no higher than their knees plus helmet, ideally they are less high. If the increase in cross sectional area is a factor of two, then the increase in air resistance is a factor of two.
This time, just a typo. When I did it on paper, I flipped the signs in the “usual” way. When I typed it in, I wanted to emphasize the orientation of my coordinate system. And then I screwed up. Again. But, just typing.