1. ## Impossible Puzzle?

Hi. I came up with this question a while ago, and am finally ready to give up on finding the answer myself. I hope you can find one. Good luck.

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You are standing on a planet which is perfectly spherical with radius r.

A rope was been wrapped around its equator. Adding x meters to this rope how high will you be able to lift the rope with it still being in the plane of the equator?
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2. Hello, powotae!

You are standing on a planet which is perfectly spherical with radius $r$.

A rope was been wrapped around its equator. Adding $x$ meters to this rope,
how high will you be able to lift the rope with it still being in the plane of the equator?

The planet has a radius of $r$ (the units don't matter).
At the equator, the circumference (length of the rope) is: . $2\pi r$

Add $x$ meters to the rope.
The rope is now a circle with circumference $2\pi r + x$.

Let $R$ be the radius of this new (larger) circle.
Then the circumference of the circle is: $2\pi R$

So we have: . $2\pi R = 2\pi r + x$

Solve for $R\!:\;\;R \:=\:r + \frac{x}{2\pi}$

We see that the new radius is $\frac{x}{2\pi}$ meters larger.

Therefore, the rope can be raised $\frac{x}{2\pi}$ meters (all around the equator).

3. The sphere should have a known radius R.
Although the additional x length of the rope is "unknown" here, it is considered as a fixed value or a known quantity here.
So the only known quantities here are R and x.

Draw the figure on paper. It is a right triangle whose
--->hypotenuse = (R+y), in meters
--->angle included by the two legs = theta, in radians.
--->one leg = R, in meters
--->the other leg, the short leg opposite theta, = x/2 +R*theta, in meters.

where, y is the to-find-for height the rope could be lifted. -----***

Why is the short leg opposite the angle theta is (x/2 +R*theta)?
Because:
When the rope is equal to the circumference of the equator, the height of the lift is zero. When x is added to the rope, the (height of the lift) = y meters.
Half of the lifted rope is the short leg in question. Its length is the length of the circular arc whose central angle is theta, plus half of x. Since arc = R*(central angle, in radians), then
short leg = R*theta +x/2

So, in the right triangle as figured,

tan(theta) = (R*theta +x/2)/R
tan(theta) = theta +x/(2R) -------------(1)
The x/(2R) is known once you plug in the values of x and R.
Eq.(1) is hard to solve for theta. But you can use iteration to find theta.

cos(theta) = R /(R+y) ------------(2)
Substitute into (2) the theta you found in (1), and voila, you have y!

4. Originally Posted by Soroban
Hello, powotae!

The planet has a radius of $r$ (the units don't matter).
At the equator, the circumference (length of the rope) is: . $2\pi r$

Add $x$ meters to the rope.
The rope is now a circle with circumference $2\pi r + x$.

Let $R$ be the radius of this new (larger) circle.
Then the circumference of the circle is: $2\pi R$

So we have: . $2\pi R = 2\pi r + x$

Solve for $R\!:\;\;R \:=\:r + \frac{x}{2\pi}$

We see that the new radius is $\frac{x}{2\pi}$ meters larger.

Therefore, the rope can be raised $\frac{x}{2\pi}$ meters (all around the equator).
Thankyou. But I was thinking more of lifting the rope at one point as high as possible aware from the globe. Sorry if it was not clear.

5. ticbol I am not sure where you get that right angle triangle from with hypotenuse = (R+y).

$\pi$

6. Originally Posted by powotae
ticbol I am not sure where you get that right angle triangle from with hypotenuse = (R+y).
Sure.

So you lift the rope at one point as high as you can (using a crane, maybe) until the rope is taut. No sag, no slack.
So you have now an "inverted angle" or two rays/straight lines from the highest point.
The highest point is y meters from the equator.
The ends of the two rays are touching the equator equidistantly from the point on the equator directly, or vertically, below the highest point.
Any of these two rays is tangent to the equator at the point of contact.

We concentrate on only one ray.
---If we imagine a radius R touching the point below the highest point, the total distance of the highest point from the center of the planet is (R+y).
---A radius R to the point of tangency of the ray will place that point R away from the center.
---Let's call the acute angle separating these (R+y) and R as theta.
---The ray is tangent to the equator, so the ray and the R are perpendicular, Or, the angle separating the ray and the R is a right angle or is 90degrees or is pi/2 radians.
----So, the R is one leg of the right triangle, the ray is the other leg, the (R+y) is the hypotenuse.

7. Thanks I can follow it now.
You have done similer to what i did.

I was just trying to solve it to be able to get a general equation for theta from equation (1).

tan(theta) = theta +x/(2R) -------------(1)

Which is the part I do not think is possible, even though there must be a unique solution for it.

8. Originally Posted by powotae
Thanks I can follow it now.
You have done similer to what i did.

I was just trying to solve it to be able to get a general equation for theta from equation (1).

tan(theta) = theta +x/(2R) -------------(1)

Which is the part I do not think is possible, even though there must be a unique solution for it.
Eh? I've done similar to what you did and you could not follow my right triangle? .

Why not give values to x and R, then try iteration.
Come on, give values to x and R.

You know, if you think only that it is not possible, even without trying first, then it is really not possible. Buit if you try first, maybe you'd change your mind.

Another way is to graph the two curves,
y = tan(theta) ----(i)
y = theta +x/(2R) ----(ii)
and find their intersection.

9. Originally Posted by ticbol
Eh? I've done similar to what you did and you could not follow my right triangle? .

and find their intersection.
Ye. sorry its been a few months since I have looked at it, and did not understand a bit of the terminology you were using.

Originally Posted by ticbol
Why not give values to x and R, then try iteration.
Come on, give values to x and R.

You know, if you think only that it is not possible, even without trying first, then it is really not possible. Buit if you try first, maybe you'd change your mind.

Another way is to graph the two curves,
y = tan(theta) ----(i)
y = theta +x/(2R) ----(ii)
I am after a general solution though. Giving X and R values would not provide that.

10. Originally Posted by powotae
Ye. sorry its been a few months since I have looked at it, and did not understand a bit of the terminology you were using.

I am after a general solution though. Giving X and R values would not provide that.
That is the general solution.

Of course if all the unknowns are variables, then you can not find any solution.
You have only two equations with 4 unknowns (theta, x, R, y) so how can you get a solution?

Again, giving the x and R values will give a solution.

11. Still open to other suggestions.

From the way I see it. For any given r and x there is a unique y.
So in theory there should be an equation y = y(r,x)

12. ## solution

I made a picture to clarify things:

edit: sorry for using the letter "L" so much... bad choice of letters

we want to find "h"
we know L0 the original length was $2\pi*R$
L1 is length after the increase of x is $2\pi*R+x$
look at the angle t, and the ark Lt
the length of Lt is $tR$
the complement of the ark has length of $(2\pi-t)R$
and the new length of the rope L1 is then the complement plus twice "L"
$L1=2\pi*R+x=(2\pi-t)R+2L$
$L=\frac{tR+x}{2}$

from the right angle triangle we get $C^2=R^2+L^2$
$C=\sqrt{R^2+\frac{(tR+x)^2}{4}}$
and
$h=C-R$
$h=\sqrt{R^2+\frac{(tR+x)^2}{4}}-R$

you win!

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13. Originally Posted by Manu
....
$L=\frac{tR+x}{2}$
Ooops, the angle "t" is not given is it?
need to somehow extract it from R and x

sorry