Hi!
Thanks for your reply, I've been over it a few good times and I think
I get it a bit better now.
about the approximated turn radius. is this a formula from the books?
This is something that has to do with so many variables, and I'm
wondering if we really know if it's a good approximation.

Once I really get everything, I'll have more questions..


Thanks again,
Shawn.




On Nov 9, 11:14*pm, id no@id wrote:
lamed wrote:
So, there are two thing that I don't quite understand.
To approximate the turn radius given the sidecut C and lean angle &
you use:
R=C*Cos(&)

but, when the snowboard is flat on the snow, *cos(0)=1, which means
the tightest turn. (?)
I'd love if you could explain me the equation, and I hope I'm not
screwing up your entire essay...

This takes me back a bit! You're right that with lean angle of zero, &
is 0 so cos(&) is 1. So the turn radius is equal to the sidecut
radius. As & increases, cos(&) decreases so the radius of turn
reduces. And that's the thing, the lower the turn radius, the tighter
the turn!



Also, Looking at;
(M*V^2)/R * Cos(&) = mg* sin(&)
*I can see that =mg*sin(&) is the weight of the rider leaning against
the centripetal force,
But I don't see why have you multiplied the centripetal force equation
(mv^2/R) with cos(&) ?
Shouldn't it just (mv^2/R) = mg*sin(&) ?

No, the forces are being resolved parallel to the plane of the board.
mg is a vertical force, mv^2/R is a horizontal force, so neither are
in the plane of the board. So both needs resolving as shown in the
diagram in the article

IainD