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I Turning radius of curved plane on an inclined plane

  1. Jun 10, 2016 #1
    Hi again.

    I hope this is the right section to ask this question.

    Not home work (I'm retired) but yet another mind game I'm playing (and still getting nowhere).

    I have trying to work out the formula for how a ski will perform a carved turn.

    I have looked at many (many Many) websites and they all seem (to me anyway) to have got it wrong, although it may be me who has it wrong.

    All the website I can find give a ski turn as the Side Cut Radius x Cos(Edge Angle).

    But surely this would only apply if a ski was running on a flat surface (inclined plane) at an angle with the skis edge. (Fig 1)

    But in reality this will "never" happen (unless skiing in a straight line down a hill) as the ski will normally always be turning and dig into the snow and the force will be at 0' to the base of the ski. (Fig 2 and Fig 3)

    I can't seem to upload the graphic I made with this. So here is the Hotmail/skidrive link to the graphic

    Does anyone have any insights that would help me produce a formula?


  2. jcsd
  3. Jun 10, 2016 #2


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  4. Jun 10, 2016 #3
    Yes I have. This is just one of the many sites I've seen that seem to me to have got it wrong.

    On that site there is a section called "Force Balance For Skier Going Around A Purely Carved Turn"
    But that is looking at this as an inclinded plane whereas it's more in line with (for example) a car going round a steep racing track and being held up with the centripetal force (I think)

    The ski will in fact be pressing against the side of a wall it digs itself into not simply sliding round a turn (I am looking at a carve here not a slide).
  5. Jun 10, 2016 #4


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    The main effect that causes skis to turn is the movement of the skier's weight. It is not equally distributed along the ski but concentrated on its second half. So changing the weight towards left and on the rear end will not only dig into the snow but also apply more downforce to the rear end than to the front end. This causes the lighter front end to follow the force applied to the point beneath the skier which tangent points to the left.
    I would try to draw a vector diagram of three dimensional force vectors using the snows resistance beneath the skier while he digs into the snow as the main point of where the forces apply to. One evidence that this is the crucial point is the fact, that you need a greater angle to dig in the faster you ski.
    E.g. it is completely different on a mogul slope where you turn on top by generating a rotation.
  6. Jun 10, 2016 #5
    I'm sorry but that's not correct.

    I am trying to introduce something like tan theta.gif = v 2 / g r (banked turn) - of course I have left out F (friction) but if I can include this it go a long way towards what I'm looking for
    Last edited: Jun 10, 2016
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