Cofcof Any takers? Anybody? Anybody?
ETA: I made a minor mistake in my analysis: Skier B doesn’t “decelerate” if the bump’s surface is horizontal; he merely stops accelerating. A shallow slope would simply decrease his acceleration, and so-on.
. . . would not be a skier in the air.
Yes he would… if he had a different horizontal coordinate. Just as a skier at the same horizontal coordinate would not be a skier in the air unless he had a different vertical coordinate. And a skier at the same coordinate along an axis parallel to the slope would not be a skier in the air unless he had a different coordinate along an axis perpendicular to the slope. And so on.
Then the skiers would not be skiing the same course, so the comparison between taking air and not taking air would be meaningless.
No, I mean… like this:
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\* *
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The *s are two skiers. They have the same vertical coordinate but different horizontal coordinates, putting one in the air and one on the slope, just as you can have two skiers with the same coordinates along any one axis and different coordinates along any other axis putting one on the slope and one in the air (so long as the axis along which they have different coordinates isn’t merely the axis parallel to the slope)
Anyway, this all came out of post #180, but as I said in post #180, post #179 is better
For a meaningful comparison, both skiers must be at the same horizontal coordinate. Noting that a skier closer to the finish can finish ahead of another skier does not address the issue.
The whole point of post #180, which kicked this off, is that there’s nothing magical about the horizontal axis. Why not say for a meaningful comparison they must both be at the same coordinate as measured along an axis parallel to the slope, for example?
Regardless, it doesn’t matter. Posts #178 and #179 contains all the points that I really want to make.
Moreover, note a scenario like the following:
Suppose we have a straight slope running from (0, 100) to (100, 0), the latter being the finishing point. Skier A starts at the point (50, 50) on the slope, skier B starts at the point (51, 52) in the air. Both start from rest. Observe that skier B, the skier in the air, is actually further from the finishing point; skier A is at a distance of sqrt(5000) from the finishing point, while skier B is at a distance of sqrt(5105). All the same, in this scenario, skier B ends up at the finishing point sooner than skier A.
That having been said, I think I’ve found the best way to explain why it’s totally off-base to give the disadvantage of air as “the slope converts gravity into force with a horizontal component”:
Suppose you have two guys on sleds racing along horizontally flat ground from point A on the left to point B on the right. However, they’re doing this in a world where there’s not just the usual downwards-gravity, but also another force, “rightwards-gravity” directed to the right. Generally, the sledders let rightwards-gravity pull them along the ground towards the goal, but every now and then, one launches temporarily into the air along the way.
Would anyone say it is advantageous to stay on the flat ground rather than in the air because the ground helps convert the downwards-gravity into nothingness? That this explains the advantage of the ground to the air? No, I doubt anyone would say such a silly thing.
But this scenario is precisely the same as the scenario of skiing down a slope at angle theta; just turn your head so that the slope becomes horizontally flat ground in your eyes. You will find that the Earth does indeed exert the sum of two accelerations upon objects: a downwards-gravity of magnitude g * cos(theta), and a rightwards-gravity of magnitude g * sin(theta).
If you wouldn’t say the slope is doing useful conversion with your head turned one way, it’s silly to say it with your head turned another way.
If both are at rest, with one in the air and the other on the ground, then the one in the air first falls to the ground and then starts skiing, while the one already on the ground skis away immediately.
[First, a small typo: I meant to place skier B at the point (52, 51). This still has skier B at the same initial distance sqrt(5105) from the finishing point, further away than skier A starts]
Now, my response:
Yes, and for all that, the one in the air still makes it to the finishing point sooner. Suppose our units are such that the acceleration of gravity has magnitude 1: then skier B falls upon the slope at point (52, 48) after sqrt(6) units of time. In the same amount of time, skier A will only have made it to point (50 + sqrt(3/2), 50 - sqrt(3/2)) ~= (51.2, 48.8)
(Thus, even though skier B starts out further from the finishing point, they end up closer to the finishing point, despite being in the air while skier A is on the slope. How’s that for the slope providing better acceleration than being in the air?
Anyway, post #189 is the best explanation I’ve put forth yet)
Admittedly, my stupid algebra errors made it hard to follow, but did you read posts 144 and 147, addressing how the physics works out. As long as a skier takes to the air because the ground drops out from under him, rather than by exerting a force off of the snow, it is slightly faster to follow the snow.
But I believe the real reasons it is significantly slower to go airborne are in post 155.
I have created a different model for ‘airtime’
The basis for this model is 3 connected slopes of respectively 15, 25 and 15 degrees slope, with slope runs of 20.7m, 22.1m and 20.7m.
I have used a standard frame of reference with the x axis as the true horizontal, and the y axis as the true horizontal. It is valid, and I just prefer it that way, and it makes modelling parabolic flight very much simpler - if you model using the slope as the horizontal, some horrid transformations are needed.
Two skiers run the course. One stays on the slope all the way, the other fails to follow the first slope change and enters parabolic flight at this point.
Assumptions: zero friction, zero wind resistance, 100% efficient transformation of velocity at the slope changes.
The first skier runs the course (63.48m) in 6.78 seconds.
The second skier flies for 0.4 second, and the course of the parabola adds 1.5 cm to the total distance. His total time for the course is 2/100 more.
At the moment of contact with the second slope, the second skier is flying at 33.6 degrees downwards. He transforms a proportion of his velocity equivalent to the cosine of the difference of 33.6 and 25 degrees (cite: [noparse]https://pumas.gsfc.nasa.gov/files/05_10_99_1.doc [/noparse])
Given that the distance added is minimal, the second skiers problem appears predominantly to be due to lack of acceleration.
I am sure you could repeat this using your preferred frame of reference. The answer should be the same, since it is just a way of looking at the world, not of changing it.
Kind regards
Udlander
Interesting coordinate system you’ve set up. : )
For those still struggling with understanding gravity, let’s leave it out of the equation for a moment.
Even though gravity is scarey stuff, you must agree that plain old velocity can be resolved however you want, right?
Imagine an object falling straight down due to gravity. And at some point it achieves a velocity straight down of 50 ft/s and is still in the air…
At this point in time, resolve the velocity into two vectors - one in the direction of the slope and the other perpendicular to the slope. As long as the two vectors sum to the original velocity (stright down in this case), this is perfectly valid and done all the time.
Nothing to do with touching the ground or not is involved. So you must agree that dropping straight down has a component velocity in the direction of the slope, right?
Well guess where that velocity is coming from. It’s coming from some kind of acceleration. What could it be? Magic?
Oh - I know! It’s from the farking component of gravity acting in the direction of the slope.
Golly, it all makes sense when you turn the paper 45 degrees and invent sideways pulling gravity!
Clearly a response to my post #189. Thanks for the sarcasm. Do you care to present an argument for why there isn’t an equivalence between the scenario I presented and the skiing scenario? Because it seems pretty plain to me that the one is simply the rotation of the other, all the way down to rotating the direction of gravitational acceleration. Or do you suppose the laws of physics are not rotationally invariant even after accounting for such?
Wait, no, seriously. Did anybody read Post #173? I know there’s a gigantic hijack underway, but I think my answer to the OP’s question is correct and complete — at any rate, I can’t find fault with it —and I didn’t see any identical responses in the last four pages.
Sheesh, I was even proud of myself for thinking it up…
:smack: And I checked that how many times?
Of course, x is the true horizontal and y is the true vertical.
The rest of the post is, I hope, correct.
As far as I can see, there is a perception problem when the frame of reference is tilted to make the slope into the reference horizontal. Only the view of the world is changed, not the world itself. Things still fall downwards. The true vertical in this tilted frame of reference is now 30 degrees off vertical and pointing left (if the slope was originally sloping down to the right).
The gravitational force still acts along this line. If you place an object on this line, of course one can resolve g into two components, one parallel to the slope, and one perpendicular. But the overall effect is that if there are no other forces acting on the object, it will move along the line until it hits the slope - that is, it still falls vertically. Once it reaches the slope, the component perpendicular to the slope is countered by the normal force, and the only unbalanced component is then that parallel to the slope. And then acceleration and movement along the slope will occur.
The problem is that in this frame of reference it looks like any object anywhere is always accelerating down the slope, but this is just an anomaly of the frame of reference chosen.
The underlying physics remain the same. You just need to account for the true vertical being off, if you change the frame of reference. Whichever frame of reference you choose, the final result should be the same. If it is not, then you are doing something wrong.