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## Main Question or Discussion Point

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

https://onedrive.live.com/redir?resid=907EBF936638CE55!120&authkey=!AOoxyzV1hMvNLD4&v=3&ithint=photo,jpg

https://onedrive.live.com/redir?resid=907EBF936638CE55!120&authkey=!AOoxyzV1hMvNLD4&v=3&ithint=photo%2cjpg

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

Thanks

K

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

https://onedrive.live.com/redir?resid=907EBF936638CE55!120&authkey=!AOoxyzV1hMvNLD4&v=3&ithint=photo,jpg

https://onedrive.live.com/redir?resid=907EBF936638CE55!120&authkey=!AOoxyzV1hMvNLD4&v=3&ithint=photo%2cjpg

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

Thanks

K