Quote:

+1
Just use the outside ski, balance against it and keep the other ski cooperating at the same angle (parallel shins). We've done the centripetal force causing the acceleration in a frame of reference fixed on the ground. For this F=ma, and a = V^2 / R. You can't say how much force is on the outside ski without knowing the turn radius and how fast you were going. Did you happen to have a gps on you?We've also done centrifugal force as one force that must be included on objects in an accelerating frame of reference so that the net force acting on the skier F=ma =0 when the frame of reference is attached to the skier. Do a search. |

**No, but I have a time stamped video tape, google earth, and pretty good maping skillz.**

This is regardng the superstar carving clip.

I can post the screen caps as proof, but here are the numbers:

Total point to point distance of clip (from a stop, to the crossover trail, past the camera): ~1200 feet

Time to cover that distance: ~30 sec.

Average speed (point to point): 40 Ft/sec or 27.3 mph

Now, I'm not skiing in a straight line, and we should probably just look at the last 8 turns before the camera, going back from the vidcap we're discussing. 8 turns back is where I come over a slight roll and pass a lift tower. From that point, to where I am in the vidcap, is around 630-650 feet. I drop about 210 vertical feet in that time, and the pitch is averaging 18 degrees. I cover that in 13 secs. So:

Bottom 8 turns: 640 feet

Time to cover distance: 13 seconds

Average speed (point to point): 49.2 ft/sec, or 33.5 mph.

Now, I'm not going in a straight line, am I? I'm making turns that cover, on average, 80 feet down the fall line (~640 ft/8 turns). I probably go 10 feet down the fall line on each transition, so I go around 70 feet down the fal line on each arc of the carve. Assuming those are 90 degree turns, averaging a roughly circular arc shape, the

**radius is around 49.5 feet or, 15 meters**. Am I bending the ski? Well yeah, pretty well I would think so considering it's a 22m radius ski.

How much linear distance do I cover, and what's my real average speed (not point to point)? Well, 10 feet down the fall line, at 45 degrees to the fall line, is around 14.2 feet. Times 8 transitions, is ~113 feet. Each 90 degree, 49.5ft radius arc covers around 77.8 feet distance, times 8, is around 622 feet. So....total distance covered is around 735 feet. Covered in 13 seconds....or at

**56.5 ft/sec, or 38.5 mph.**

So, what's that mean? Well, it means I'm skiing a bit slower than WC gs, but it also means I can make an accurate off the cuff the speed estimate - I said on page 1 or 2 that I thought I was going 35+ mph in this clip.

**If you take all these figures and put them into the calculator above, you get the rough result that the centripetal force generated HAS TO BE approximately double the mass in the system.**

**R: 15m**

**Velocity: 56.5 ft/sec**

**Body weight: x**

**Centripetal force: ~2x**

Now what? I weigh about 210lb with clothes and gear in that vid. I have to generate a 420 lb centripetal force to make the turn shape as described. If I have 30 lb (about 15%) vertically supported on my inside ski, that leaves me with 180 lb body weight/mass to work with.

**This center of mass has to be applied at a ~67 degree angle from vertical,**to get the required 420 lbs centripetal force.

**So yes, I am achieving the correct angulation for the turn shape and speed, it is NOT excessive**........especially considering I'm on a fatter ski with some torsional flex, around a 2-3 degree base bevel, and not on race boots.