Rope around the earth problem

wombling said:
An equal distribution of 6 feet across 25 000 miles is not going to be enough slack for a worm to fit under unless its a really really small worm; or an earthworm
Click to expand...

assuming gravity plays no part, the rope will be a little over a foot off the surface of earth all around
 
battletoad said:
let R be the radius of the sphere in question in metres. Suppose you tie a rope tautly around the (equator of the) sphere. Its length is 2 x pi x R metres.

Now, if you want the rope to be 1 metre off the surface of the sphere all round, the rope needs to be 2 x pi x (R+1) metres.

So the difference in the lengths of rope is 2 x pi x (R+1) - 2 x pi x R = 2 x pi metres. Note that the initial radius R magically disappeared, meaning that irrespective of the initial radius of the sphere (marble, tennis ball, jupiter, sun, etc.), it only needs an additional 2pi metres of slack to lift the rope 1 metre off the surface all around. Weird, yes :D
Click to expand...

Yeah, but the thought of it is just mind boggling. I need to see it physically to believe it. Will scale the numbers down in my experiment
 
wombling said:
An equal distribution of 6 feet across 25 000 miles is not going to be enough slack for a worm to fit under unless its a really really small worm; or an earthworm
Click to expand...
If my math is right then it is 0.00005 mm of space under the rope. So it depends on the size of the worm. Unless I'm oversimplifying it?

Disclaimer, I'm not mathematician. SO tempted to Google it now :p

edit: Not sure that OP explained the problem properly? Varying answers the interwebs give me and all the puzzles are explained differently.
 
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Well if the rope can't stretch I assume it can't contract either so you'll have like a stiff band around the earth. Adding a few feet won't change the circumference enough for anything meaningful to be able to fit under it. Yay, now I can keep the worms from getting to my part of the earth!
 
Given;
C1 = 2xPIxR1
C2 = 2xPIxR2
C2 = C1 + 6

Therefore:
2xPIxR2 = 2xPIxR1 + 6
PIxR2 = PIxR1 + 3
R2 = (PIxR1 + 3)/PI

R2 = R1+3/PI

R2 is 3/PI feet above the earth, so there is nearly 1 ft of space under the rope.
 
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battletoad said:
let R be the radius of the sphere in question in metres. Suppose you tie a rope tautly around the (equator of the) sphere. Its length is 2 x pi x R metres.

Now, if you want the rope to be 1 metre off the surface of the sphere all round, the rope needs to be 2 x pi x (R+1) metres.

So the difference in the lengths of rope is 2 x pi x (R+1) - 2 x pi x R = 2 x pi metres. Note that the initial radius R magically disappeared, meaning that irrespective of the initial radius of the sphere (marble, tennis ball, jupiter, sun, etc.), it only needs an additional 2pi metres of slack to lift the rope 1 metre off the surface all around. Weird, yes :D
Click to expand...

/mind blowing
 
battletoad said:
let R be the radius of the sphere in question in metres. Suppose you tie a rope tautly around the (equator of the) sphere. Its length is 2 x pi x R metres.

Now, if you want the rope to be 1 metre off the surface of the sphere all round, the rope needs to be 2 x pi x (R+1) metres.

So the difference in the lengths of rope is 2 x pi x (R+1) - 2 x pi x R = 2 x pi metres. Note that the initial radius R magically disappeared, meaning that irrespective of the initial radius of the sphere (marble, tennis ball, jupiter, sun, etc.), it only needs an additional 2pi metres of slack to lift the rope 1 metre off the surface all around. Weird, yes :D
Click to expand...

Circumference is pi x 2R, not 2 x pi x R.

Have you taken your Vormex? ;)
 
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The earth isn't a perfect circle at the equator so over time the rope will develop a rolling wave action, much like a hula hoop round your waist. Thin worms will easily be able to crawl under the section diametrically opposite the contact point.
 
Ha ! You guys pretend to know all about Pie but have never heard of the Gippsland worm....poser professors :thumbdown:
 
Nope, 6 feet slack in a rope of 132 000 000 feet is but 0.0000045% of slack.

Not even noticeable at all it's way too small to even be felt or seen anywhere apart from the 2 points of origin.
 
WaxLyrical said:
A variation of this problem was discussed during an episode of S1 of House of Lies. Found it quite interesting.

Suppose you tie a rope around the earth at the equator (circumference approx. 25,000 miles).
Let's say you pull the rope as tight as it will go and then add back 6 feet of slack before tying the knot.
If the extra rope is distributed evenly around the globe will there be enough
space between the rope and the surface of the earth for a worm to crawl under?

Assume the earth is a perfect sphere and the rope does not stretch.
Click to expand...

Your description is not earth.
 
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