Author |
Topic: Climbing a slippery cone (Read 995 times) |
|
Leo Broukhis
Senior Riddler
Gender: 
Posts: 459
|
 |
Climbing a slippery cone
« on: Dec 18th, 2008, 12:58am » |
 Quote  Modify
|
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?
|
|
 IP Logged |
|
|
|
Grimbal
wu::riddles Moderator
Uberpuzzler
Gender:
Posts: 7527
|
|
Re: Climbing a slippery cone
« Reply #1 on: Dec 18th, 2008, 1:27am » |
Quote Modify
|
At about the slope of a tipi.
|
|
IP Logged |
|
|
|
towr
wu::riddles Moderator
Uberpuzzler
Some people are average, some are just mean.
Gender:
Posts: 13730
|
|
Re: Climbing a slippery cone
« Reply #2 on: Dec 18th, 2008, 5:03am » |
Quote Modify
|
on Dec 18th, 2008, 1:27am, Grimbal wrote:| At about the slope of a tipi. |
|
They have a standardized slope?
I'd hazard to guess 45 degrees, on account you get a right angle with the opposite side from the bottom to the top.
But I suspect I may be failing to take into account some properties of it being a lasso.
|
|
IP Logged |
Wikipedia, Google, Mathworld, Integer sequence DB
|
|
|
Grimbal
wu::riddles Moderator
Uberpuzzler
Gender:
Posts: 7527
|
|
Re: Climbing a slippery cone
« Reply #3 on: Dec 18th, 2008, 7:24am » |
Quote Modify
|
In fact, yes. When the skin is spread out, it forms a semi-disk.
|
|
IP Logged |
|
|
|
Leo Broukhis
Senior Riddler
Gender:
Posts: 459
|
|
Re: Climbing a slippery cone
« Reply #4 on: Dec 18th, 2008, 8:36am » |
Quote Modify
|
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.
|
|
IP Logged |
|
|
|
towr
wu::riddles Moderator
Uberpuzzler
Some people are average, some are just mean.
Gender:
Posts: 13730
|
|
Re: Climbing a slippery cone
« Reply #5 on: Dec 18th, 2008, 8:57am » |
Quote Modify
|
Well, if I assume the lasso forms a circle parallel to the floor; then I get arccos(1/[2 ]) ~= 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/[2]
Without triginometry, I could say the slope is  (42- 1):1
|
| « Last Edit: Dec 18th, 2008, 9:07am by towr » |
IP Logged |
Wikipedia, Google, Mathworld, Integer sequence DB
|
|
|
Leo Broukhis
Senior Riddler
Gender:
Posts: 459
|
|
Re: Climbing a slippery cone
« Reply #6 on: Dec 18th, 2008, 9:15am » |
Quote Modify
|
towr, that's the answer that I'm getting. The answer claimed to be the right one is arccos(1/6) - considering the geodesic that is obviously shorter than the circle for the same height of the "noose point". The problem is that the latter is less.
|
|
IP Logged |
|
|
|
towr
wu::riddles Moderator
Uberpuzzler
Some people are average, some are just mean.
Gender:
Posts: 13730
|
|
Re: Climbing a slippery cone
« Reply #7 on: Dec 18th, 2008, 9:34am » |
Quote Modify
|
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.
|
|
IP Logged |
Wikipedia, Google, Mathworld, Integer sequence DB
|
|
|
Grimbal
wu::riddles Moderator
Uberpuzzler
Gender:
Posts: 7527
|
|
Re: Climbing a slippery cone
« Reply #8 on: Dec 19th, 2008, 1:02am » |
Quote Modify
|
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).
|
| « Last Edit: Dec 19th, 2008, 1:04am by Grimbal » |
IP Logged |
|
|
|
Leo Broukhis
Senior Riddler
Gender:
Posts: 459
|
|
Re: Climbing a slippery cone
« Reply #9 on: Dec 19th, 2008, 8:36am » |
Quote Modify
|
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.
|
|
IP Logged |
|
|
|
Grimbal
wu::riddles Moderator
Uberpuzzler
Gender:
Posts: 7527
|
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 = (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 2 ·d/6 which means r=d/6.
And r=d/6 gives and angle of arccos(1/6) on the cone.
|
| « Last Edit: Dec 19th, 2008, 11:52pm by Grimbal » |
IP Logged |
|
|
|
Leo Broukhis
Senior Riddler
Gender:
Posts: 459
|
|
Re: Climbing a slippery cone
« Reply #11 on: Dec 20th, 2008, 12:03am » |
Quote Modify
|
on Dec 19th, 2008, 11:51pm, 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?
|
|
IP Logged |
|
|
|
rmsgrey
Uberpuzzler
    
Gender:
Posts: 2873
|
|
Re: Climbing a slippery cone
« Reply #12 on: Dec 20th, 2008, 7:42am » |
Quote Modify
|
on Dec 20th, 2008, 12:03am, 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.
|
|
IP Logged |
|
|
|
Leo Broukhis
Senior Riddler
Gender:
Posts: 459
|
|
Re: Climbing a slippery cone
« Reply #13 on: Dec 20th, 2008, 10:23am » |
Quote Modify
|
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.
|
| « Last Edit: Dec 20th, 2008, 10:23am by Leo Broukhis » |
IP Logged |
|
|
|
Grimbal
wu::riddles Moderator
Uberpuzzler
Gender:
Posts: 7527
|
|
Re: Climbing a slippery cone
« Reply #14 on: Dec 21st, 2008, 4:34am » |
Quote Modify
|
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.
|
| « Last Edit: Dec 21st, 2008, 4:37am by Grimbal » |
IP Logged |
|
|
|
Leo Broukhis
Senior Riddler
Gender:
Posts: 459
|
|
Re: Climbing a slippery cone
« Reply #15 on: Dec 21st, 2008, 11:06am » |
Quote Modify
|
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).
|
|
IP Logged |
|
|
|
Grimbal
wu::riddles Moderator
Uberpuzzler
Gender:
Posts: 7527
|
|
Re: Climbing a slippery cone
« Reply #16 on: Dec 21st, 2008, 12:28pm » |
Quote Modify
|
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.
|
|
IP Logged |
|
|
|
|
RIDDLES SITE
WRITE MATH!
Home 
Help 
Search 
Members 
Login 
Register
Reply
Notify of replies
Send Topic
Print