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.

Physics of a Snowboard Carved Turn by Jack Michaud with IainDuncumb
 

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

Physics of a Snowboard Carved Turn by Jack Michaud with IainDuncumb
 

Jack's a moderator of the forums at www.bomberonline.com. You can usually find him there. I'm very active there as well. He's
also written a bunch of articles for the website aimed at helping beginner carvers get going.

Neil

Physics of a Snowboard Carved Turn by Jack Michaud with IainDuncumb
 

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

Physics of a Snowboard Carved Turn by Jack Michaud with IainDuncumb
 

On Nov 12, 9:45*pm, lamed wrote:
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?

Yes, it is in the book Skiing Mechanics by John Howe. One time,
someone on the forum at BomberOnline (I think NateW) couldn't quite
get his head around it, so he derived it for himself and came up with
the same answer. In the process I believe he also proved the cosine
theorem! :-)
-Jack