Physics of a Snowboard Carved Turn by Jack Michaud with Iain Duncumb

So I was looking for the guy who wrote this article,
http://www.bomberonline.com/articles/physics.cfm
It wasn't too hard to find his Email address but the address is dead.
I'm posting here because he used to be active here about 14 years ago
(!!!!!), or hopefully someone else would be able to help me

__________________________
Hello Jack, My name is Shawn and I'm very interested in snowboarding
dynamics, and trying to use your article to figure it out.
Let me just say that I've googled your name and got this email address
by google's rec.skiing.snowboard group, and I hope you're fine with
it.. (obsolete now, I guess

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...

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(&) ?
I'm adding a tiny sketch, that hopefully shows that the force vector
of which the rider is applying against the centripetal force is
actually horizontal, and thus what I said about the other part of the
equation.
(Got dammit, I can't post the sketch here.)


Btw I was trying to get a skiing mechanics book, to no avail yet.

Good day,
Shawn.