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turn radius based on ski sidecut?

post #1 of 11
Thread Starter 

This photo caught my eye

http://www.epicski.com/content/type/61/id/190789/width/1000/height/1000/flags/LL

 

What might be the inherent turn radius of a ski when engaged with the snow surface at 80+ degrees?

post #2 of 11
Quote:
Originally Posted by Cgrandy View Post
 

This photo caught my eye

http://www.epicski.com/content/type/61/id/190789/width/1000/height/1000/flags/LL

 

What might be the inherent turn radius of a ski when engaged with the snow surface at 80+ degrees?


Looks like a 35 m GS ski.  35 * cosine (80)= 6.  6 m when it hooks up, a bit more when it's bouncing. 

post #3 of 11
Thread Starter 

Would that calculation account for camber?  ;-)

 

(the 6m number has to be incorrect,  if you take a moment to consider)

post #4 of 11
Quote:
Originally Posted by Cgrandy View Post
 

Would that calculation account for camber?  ;-)

 

(the 6m number has to be incorrect,  if you take a moment to consider)

 

I haven't worked out if that's the correct equation (but I'd guess Ghost knows what he's talking about)... but why must 6m be wrong if you think about it? 

 

That's a carved turn along the arc of a 12m diameter circle, if I understand it correctly - which is fairly large. 

post #5 of 11

http://www.epicski.com/a/the-complete-encyclopedia-of-skiing-epicski-skiing-glossary

 

It's the correct formula al lright, but it gives the curve that would ensue if you were able to press the ski firmly down to a flat surface while tipped (intersection of plane and cylinder at an angle, IIRC).  In reality, judging from my experience, the equation works quite well (allowing for flex and snow surface irregularities= not perfect) on hard flat surfaces up to about 60 degrees, then if the ski is stiff it progressively diverges from the equation.  

 

Might be easier to calculate the actual skied radius if you know the speed and g force (from a=V^2/R).  e.g. 4.5 g s at 40 mph yields 7.2 m radius skied at 77.5 tipping angle with a dialed up radius from the formula of 7.6 m.   

post #6 of 11
It is a very good aproximation -- the solution for the radius of a circle defined by the intesection of a cylinder and a plane. The cylinder is defined by the circular sidecut profile (1st approximation) deflected perpendicular to the ski surface. But of course the ski doesn't quite deflect perpendicularly, it wraps around its deflected curve so the running length along the ski is different from the running length along the snow. (This is the 2nd approx, but it is a very, very small effect even up to fairly large deflections.) THe radius of the cylinder decreases very, very slightly rather than being constant.

The third approximation is probably the most important -- that the entire length of the ski is in contact with the snow and digs in the same amount, so it lies in a plane parallel to the snow surface. This approximation is reasonably valid for even aggressive recreational skiers, but I doubt it for the picture in the original post.
post #7 of 11

One more thing I forgot to mention,  The intersection of the circular cylinder defined by the ski sidecut and the plane defined by the snow surface is, of course, an ellipse.  The turn radius is the "radius of curvature" -- the radius of the best matching circle.  The cosine formula gives the radiuse of curvature at the furthest out point (the axis of the ellipse) which typically would be under your center of mass, or approximately the center of the ski.  The value at other points on the ski will be very slightly different.  One project I have so far resisted is figuring out the how the curvature varies along the ski, and how extreme an edge angle is needed before the effect becomes non-negligible.

post #8 of 11
Thread Starter 

Looking carefully, at the extreme condition of a ski base at 90 degrees to the snow surface,  the natural turn arc radius due to ski "side cut" goes to ZERO.

For in that condition, the ski's widest points become the points of contact with the resisting surface. vis. a length of ski edge wire shaped as a "moustache" running over the snow surface "~"

 

The ski would run straight, give or take any difference in "angle of attack" between the shovel and the tail (ski camber)

The ski side cut could only influence the depth of snow surface contact or penetration.

 

All other attitudes are caught between the conditions of the extremes.

post #9 of 11
Quote:
Originally Posted by Cgrandy View Post
 

The ski would run straight, give or take any difference in "angle of attack" between the shovel and the tail (ski camber)

The ski side cut could only influence the depth of snow surface contact or penetration.

 

I'd think in actual usage if you got it to 90 degrees there would be loading on the ski and it would be flexed, creating rocker, or negative camber, with just the tip and tail in contact with the snow. That would still turn the ski on an arc I think. 

post #10 of 11
Thread Starter 
Quote:
Originally Posted by Ghost View Post
 

http://www.epicski.com/a/the-complete-encyclopedia-of-skiing-epicski-skiing-glossary

 

It's the correct formula al lright, but it gives the curve that would ensue if you were able to press the ski firmly down to a flat surface while tipped (intersection of plane and cylinder at an angle, IIRC).  In reality, judging from my experience, the equation works quite well (allowing for flex and snow surface irregularities= not perfect) on hard flat surfaces up to about 60 degrees, then if the ski is stiff it progressively diverges from the equation.  

 

Might be easier to calculate the actual skied radius if you know the speed and g force (from a=V^2/R).  e.g. 4.5 g s at 40 mph yields 7.2 m radius skied at 77.5 tipping angle with a dialed up radius from the formula of 7.6 m.   

The line of reasoning consistent with g forces would be very useful to define the turn radius attributed to any combination of ski side cut and ski reverse camber due to loading.  Instrumentation would not be complex,  but not many  are so invested.

 

ETA    Using the "correct formula",  how does the natural turn radius of skis such as the "Spatula" work out?  I know they turn the same way as more traditionally side cut skis.


Edited by Cgrandy - 1/15/16 at 5:39pm
post #11 of 11
Thread Starter 
Quote:
Originally Posted by dbostedo View Post
 

 

I'd think in actual usage if you got it to 90 degrees there would be loading on the ski and it would be flexed, creating rocker, or negative camber, with just the tip and tail in contact with the snow. That would still turn the ski on an arc I think. 

Certainly!  Just not a turn arc attributed to or calculated from "Ski Side cut". ;-)

 

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