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# Just for fun question. - Page 2

Quote:
 Originally Posted by BigE According to your argument, J and upside down J are equivalent.
How so?

Did I miss something. if so oops sorry..

Unless you mean "box turns" that start with a quick redirection at the top of each turn, lots of time in the fall line and cycloid turn at the exit of the turn.

Cycloid shaped turns would be fastest in a perfect world..

I sure can't do them at pure carve however. I don't have the skill or strength to do them without skidding..

DC
A "cycloid turn" is symmetric about the fall-line. If it is most similar to a 'J' then it is also most similar to an upside down 'J'.
Quote:
 Originally Posted by Martin Bell ...So comprex and dchan, please excuse me but I've forgotten all my calculus years ago. Are you saying that PhysicsMan is wrong to base his ski track geometry on sinusoids and should instead have based it on cycloids? I sense a mathematical duel coming on...protractors at dawn...
I doubt they are suggesting that the use of pure sinusoids is inappropriate, just introducing a fun mathematical tidbit that might be of interest to skiers. Don't forget that:

a) The cycloid is only the correct solution of the frictionless brachistochrone problem, whereas skiing definitely involves lots of snow and air resistance. In the brachistochrone analysis, the effects of acceleration (ie, increasing speed on the fall line sections) completely dominates the selection of the best curve. However, in skiing, one hopefully doesn't keep accelerating, but rather, reaches an approximate steady state between friction and the downhill pull of gravity. Of course, there will be variation in speed in each turn, but, for example, the skier's speed doesn't start at zero at each transition (as it does in the brachistochrone problem).

Once even the simplest model for friction is introduced, the minimum-time solution moves away from a pure cycloid: http://mathworld.wolfram.com/Brachis...neProblem.html (scroll about 1/2 way down the page)

b) The cycloid is only the correct solution to the minimum-time problem when the shape of the course is unconstrained. Once you constrain the skier to go around offset gates, the cycloid no longer is the correct solution, even if the motion is frictionless.

c) As someone previously pointed out, any periodic path can be approximated by a sum of sinusoids, so I could easily adapt the calculation to any other base curve, including cycloids. I am quite sure, however, that new fundamental phenomenology wouldn't arise from doing so.

So, I am relieved to report that I don't think that protractors at dawn will be necessary.

Tom / PM
What shape turn will get you down the fastest?

How about fall line to fall line, without anything else on your mind?
The mathematical shape is really easy to differentiate on the snow; it's simply Dh where h is hieght.
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