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        riddles >> medium >> Climbing a slippery cone
        
(Message started by: Leonid Broukhis on Dec 18^th, 2008, 12:58am)

Title: Climbing a slippery cone
Post by Leonid Broukhis on Dec 18^th, 2008, 12:58am

Consider an idealized mountain peak: a perfectly slippery cone. A climber standing at the bottom of the peak throws an idealized lasso (weightless, frictionless wrt itself and the ground, non-elastic) over the summit (assume that the lasso will come to rest with whatever length of the loop it was thrown with).
If the slope is very steep, the lasso will hold when the climber pulls on it while attempting to propel him(spare me)self up; if the slope is too easy, the lasso will slip off the top. What is the slope at which the behavior changes?

Title: Re: Climbing a slippery cone
Post by Grimbal on Dec 18^th, 2008, 1:27am

At about the slope of [hide]a tipi[/hide].

Title: Re: Climbing a slippery cone
Post by towr on Dec 18^th, 2008, 5:03am

on 12/18/08 at 01:27:42, Grimbal wrote:

At about the slope of [hide]a tipi[/hide].

They have a standardized slope?

I'd hazard to guess [hide]45 degrees, on account you get a right angle with the opposite side from the bottom to the top.[/hide]
But I suspect I may be failing to take into account some properties of it being a lasso.

Title: Re: Climbing a slippery cone
Post by Grimbal on Dec 18^th, 2008, 7:24am

In fact, yes. When the skin is spread out, it forms a semi-disk.

Title: Re: Climbing a slippery cone
Post by Leonid Broukhis on Dec 18^th, 2008, 8:36am

I have two potential answers that are much steeper than 45 degrees (they are within 1 degree from one another). The key word is "slack", and some trigonometry is needed.

Title: Re: Climbing a slippery cone
Post by towr on Dec 18^th, 2008, 8:57am

Well, if I assume the lasso forms a circle parallel to the floor; then I get [hide]arccos(1/[2http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/pi.gif]) ~= 81 degrees
From the 'corner' to the top should be longer than around. So for a hypotenuse of 1 you get a radius of 1/[2http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/pi.gif][/hide]

Without triginometry, I could say the slope is [hide]http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/surd.gif(4http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/pi.gif²- 1):1[/hide]

Title: Re: Climbing a slippery cone
Post by Leonid Broukhis on Dec 18^th, 2008, 9:15am

towr, that's the answer that I'm getting. The answer claimed to be the right one is [hide]arccos(1/6) - considering the geodesic that is obviously shorter than the circle for the same height of the "noose point"[/hide]. The problem is that the latter is less.

Title: Re: Climbing a slippery cone
Post by towr on Dec 18^th, 2008, 9:34am

Well, if you try it out with a cone and a lasso, you'll see that it doesn't form a circle parallel to the base. So it works only as an approximation.
It's also easy to see it must be an upper limit, because a steeper slope means there simply isn't enough rope to get over the top. So the real answer should be less.

I'm not sure how to calculate the answer with geodesics though. But I can believe the right answer is that other one you gave.

Title: Re: Climbing a slippery cone
Post by Grimbal on Dec 19^th, 2008, 1:02am

By the way, my "tipi" answer is actually for the case where the climber holds both ends of the rope. In that case, I would say that the cone must be such that when the surface is unfolded, it forms a half-disk.
That translates to a slope of sqrt(3) vertical to horizontal ratio, or 60° from the horizontal.

For the case where the knot is fixed, it should be the same.

For the case of a sliding knot, I would say it must open op to 1/6 of a disk.
That translates to a slope of sqrt(35), or 80.4°.

PS: and yes, that is arccos(1/6).

Title: Re: Climbing a slippery cone
Post by Leonid Broukhis on Dec 19^th, 2008, 8:36am

Grimbal,

still, I'm not completely convinced why it shouldn't be X=arccos(1/(2*pi)). After being "thrown", a weightless rope lands arbitrarily, when it is pulled, it slips arbitrarily, and for slopes between arccos(1/6) and X there exists a trajectory leading from the initial horizontal position of the loop to the lasso slipping off the top.

Title: Re: Climbing a slippery cone
Post by Grimbal on Dec 19^th, 2008, 11:51pm

The point is that if you keep the knot at the base of the cone and pull the loop tight, it won't rest horizontally but along a geodesic.

Take a cone of height h and radius r, and let's call d = http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/surd.gif(h² + r²) the distance between the base at the perimeter and the summit. Let's call g the length of the geodesic.

If you lift the knot at height x<=h, the length of the loop will reduce to (h-x)/h·g but you will have to add a length x/h·d to reach the base.

If g<d, the total length of the rope increases as the knot is pulled up, so the knot will tend to remain at the bottom. If g>d, the length decreases as the knot is lifted, so the knot will tend to slide to the top. So we need to find the cone such that g=d.

If you unfold the cone, the geodesic becomes a straight line (red line on the picture). With g=d, you can see that the unfolded cone has an angle of 60°. From there you get the perimeter of the base as 2http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/cpi.gif·d/6 which means r=d/6.

And r=d/6 gives and angle of arccos(1/6) on the cone.

Title: Re: Climbing a slippery cone
Post by Leonid Broukhis on Dec 20^th, 2008, 12:03am

on 12/19/08 at 23:51:40, Grimbal wrote:

The point is that if you keep the knot at the base of the cone and pull the loop tight, it won't rest horizontally but along a geodesic.

Yes, as soon as the loop goes to the geodesic, the critical angle is arccos(1/6), but what is supposed to keep the knot at the base so that an initially horizontal loop becomes geodesic? A uniformly frictionless loop can as well shrink while remaining horizontal when pulled, can it not?

Title: Re: Climbing a slippery cone
Post by rmsgrey on Dec 20^th, 2008, 7:42am

on 12/20/08 at 00:03:10, Leonid Broukhis wrote:

A uniformly frictionless loop can as well shrink while remaining horizontal when pulled, can it not?

As the knot moves up the cone, the length of the loop shrinks, and the length of the rope between the knot and sea level grows. If the loop shrinks faster than the rope grows, the lasso will pop off the top when you pull on the rope; if the rope grows faster than the loop shrinks, then the knot will move down the cone.

Which happens depends on the angle of the cone.

Title: Re: Climbing a slippery cone
Post by Leonid Broukhis on Dec 20^th, 2008, 10:23am

rmsgrey, realize that for the angles between arccos(1/6) and arccos(1/2pi) the outcome depends on the way the loop and the knot slip.

Title: Re: Climbing a slippery cone
Post by Grimbal on Dec 21^st, 2008, 4:34am

It is not clear how a frictionless and weightless rope would behave. I am not sure you could actually throw it and certainly it would have no reason to fall down.

Anyway, regardless of the angle of the cone, here are many configurations where the loop can get off the cone. I don't think the question is whether the rope, placed horizontally around the base, has still enough length to pass over the top. The question is whether the lasso can be tied around the cone in a way to support a climber.

At an angle above arccos(1/2pi), a rope laid around the base has not enough length to be taken off the cone. Above that angle, it can be removed.

Between arccos(1/2pi) and arccos(1/6), a rope laid around the base can be taken off the cone. But there are other configurations, such as along a geodesic, where the rope is too short to be taken over the top. In these configurations, it is assured that the rope will hold. If the climber starts from a configuration with a horizontal loop, it is possible, if he pulls the rope in the wrong way that it slides off the top, but once he has tightened it enough, he can be sure the rope will stay. And I believe it is more natural for the rope to tighten around the cone than to slide off the top.

In fact, what will happen, is that to support the climber, the rope must be in a configuration of minimal length. A local minimum. If it is not, the rope would tend to slide to a shorter configuration, which would release some rope and let the climber drop a bit. Minimum length means that the rope will follow a "straight" line, i.e. a geodesic. But also, it will mean that at the knot, the 3 threads will join at an angle of 120°. If the unfolded cone is more than 60°, the angles cannot settle at 120° at the knot. The angle between the 2 lateral threads is too small, the forces won't balance and the knot will be pulled up the cone. Inversely, if the unfolded cone is less than 120°, the lateral threads pull lower than they should (a picture would be useful here...) and the knot will be pulled down. If the climber pulls on the lasso of a cone that is too steep, the look will grow and the know will slide down until it reaches the hands of the climber.

Title: Re: Climbing a slippery cone
Post by Leonid Broukhis on Dec 21^st, 2008, 11:06am

Grimbal,
this is understood. My point is that the full answer to the question "What is the slope at which the behavior changes?" should include both arccos(1/6) and arccos(1/2pi).

Title: Re: Climbing a slippery cone
Post by Grimbal on Dec 21^st, 2008, 12:28pm

The only change in behavior is between the case where the rope will slip off the top when you pull on it and the case where it will hold on the cone. And that happens at arccos(1/6).

The value arccos(1/2pi) becomes special only in the context where the loop is forced to be horizontal. Or in the context where the initial position is when the lasso lies around the base of the cone with no gap. The problem isn't so specific. It only says that a lasso is thrown over the summit. From there, whatever the configuration how it comes to rest, there is enough length to pull it back over the summit. Whether you can take it off the top depends on the length of the rope, not the angle of the slope.