# Ski Turn Radius, Inclination at Speed & Balance

1. Dec 7, 2005

### michaela

Hi all. First post on a Physics Forum for me.

As a ski instructor I generally hang out on the EpicSki Teaching & Technical Forum but the jocks over there try and beat me up whenever I delve into a mathematical exploration of our sport. :yuck: A recent question (though likely asked solely in humor) got me to wondering about our perceptions vs. the reality of Turn Radius and Inclination. The ‘problem’ as stated was…

===Originally Posted by Rick=====================
... At a speed of 30 mph, and an edge angle of 45 degrees, at what sidecut radius threshold will a skier begin needing to employ reverse angulation to remain balanced on his outside ski?
=======================================

The core of the problem seemed first to determine one's 'falling-over rate' and then to figure out the turn radius that compensates for that rate of fall. Below is a shortened version of my attempt to answer the question on EpicSki.

If you would, please advise on the accuracy of my simplistic thinking & wording.

===content from my post======================
OK, for starters a constant speed of 30 mph is actually 158,400 feet per hour, or 44 f/s. I also assumed:
• 1) Our skier is perfectly inclined (stance is straight up from the ski - no body-angulation) at 45-degrees from the specified Outside-Ski's Inside Edge, placing the skiers CM exactly 45-degrees from that edge…

2) Our skier is traversing ‘flat’ terrain for the entire turn (to remove slope-angle confusion).

When an out-of-balance object is 'falling over' from such a pivot point as the "Outside-Ski's Inside Edge" its rate of acceleration (falling over) depends on the degree of tilt the CM is away from 'vertical' (WRT Gravity) over that pivot point.

I earlier used 22.733 f/s² (in error) as my ‘rate of falling over’ when tipped to a 45-degree angle. This is true when measured in the direction of the fall (at 45-degrees) - but I should have been using 16.075 f/s² since I need only the horizontal component of the ‘falling rate’ (because the Centripetal Acceleration of our ski is -horizontal- to our flat terrain, right?).

Our skier’s horizontal rate of fall-over is at 16.075 f/s² horizontally so we need a turn radius made at 30-mph that will deliver this rate. The centripetal acceleration of a ski making a turn should be obtainable with V²/r or (Velocity²/radius). Since we want it to equal 16.075 f/s² (the rate our skier is falling over) I’m thinking we need
(44 f/s)² / (unknown-radius in feet) = (16.075 f/s²)
(1936 f²/s²) / (unknown-radius in feet) = (16.075 f/s²)
(1936 f²/s²) = (16.075 f/s²) x (unknown-radius in feet)
(1936 f²/s²) / (16.075 f/s²) = (unknown-radius in feet)
(1936 f²/s²) / (16.075 f/s²) = (120.435 feet)

Our -ski- is being 'pushed' horizontally to the inside of our 120.435-foot radius turn at the rate of 16.075 f/s² by Centripetal Acceleration (when at 30-mph). Since we’re leaning in at 45-degrees, our CM is also ‘falling’ to the inside of our turn at that same rate (at least horizontally).

So long as the push-to-the-inside by the ski is the same rate as we’re falling to the inside we never quite 'fall' any lower - we’re in balance.

A 120-foot radius seems big to me but consider this... the rate of 'falling over' that's important is not that of our -head- (where we mostly perceive such things) it's our -CM- that matters. Also, at 30mph a 120-foot radius turn wouldn't seem all that big. Remember, we're moving at 44 fps - that means we'd complete a full 180-degree skiing turn in 8.5 seconds. Also remember it's the CM at 45-degrees from the Outside-Ski's edge - the ski & leg we often reach way out there - far to the outside of our body.
===End of EpicSki post=======================

Here are the specifics is used to get to my answer...
Sin(45) = .707 -and- Cos(45) = .707
Frictionless ‘toppling’ acceleration at 45-degrees is... 22.733 f/s² (tangential direction)
Horizontal component of this acceleration is... Rate-of-Acceleration x Cos(45) -or- 22.733 x .707 = 16.075 f/s²

Using (Velocity²/radius) as my formula for centripetal acceleration I came up with the series of calculations as listed in the ‘post’ clipping above. Even after looking over my thinking several times a 120' radius still seems a bit large for an ‘in-balance’ 45-degree inclination but I can't find any fault with my thinking on it. Perhaps someone here can?

Once I have a solid second opinion on the topic above, I plan to further explore the Radius and Inclination changes that would be required as the skier negotiates a complete turn on a variety of slope angles. Should be interesting. Thank Hobbes for Excel.

Appreciate any input on the ideas presented above and I’m sure I’ll have more curiosities in the future. And yes, I’ll always make a good-faith effort to figure things out on my own prior to posting questions or requesting further review.

.ma
PS: I used 3 decimal places more for traceability (for observers) rather than for any kind of real accuracy...

Last edited: Dec 7, 2005
2. Dec 7, 2005

### michaela

Gee, this is fun. OK maybe not, since I think I blew it above. Just pulled up Excel and played around with some of these relationships again and discovered a mondo error lining up my problem statement.

I should have been looking at the direction of these forces and using Tan(angle) to get the horizontal force needed to oppose the vertical pull of gravity, right? (not the horizontal portion of the rate of acceleration…)

I now get a required acceleration of 32.15 f/s² at 45-degrees, and therefore...

(44 f/s)² / (unknown-radius in feet) = (32.15 f/s²)
(1936 f²/s²) / (unknown-radius in feet) = (32.15 f/s²)
(1936 f²/s²) = (32.15 f/s²) x (unknown-radius in feet)
(1936 f²/s²) / (32.15 f/s²) = (unknown-radius in feet)
(1936 f²/s²) / (32.15 f/s²) = (60.22 feet)

Pretty much half of what I originally thought.

Aw Nuts . If that’s so, then I gotta go back to Epic and fess up to some bozo thinking. Still would like to hear from someone who can confirm on this new line of thinking though.

.ma