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Thread: Gapsize puzzle

20100916, 16:15 #1
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Imagine that you have perfect sphere the size of the Earth. Around its equator is a piece of string, tight to the sphere so there is no gap and joined it such away that it is perfectly smooth (no knots, bumps etc). You cut the piece of string and then insert another piece just 3ft long which is again joined so the joint is smooth. The resultant new circle of string is positioned so that it shares the same centre as the sphere i.e. so that there is a uniform gap between string and sphere all the way round (looking from above you would see two concentric circles. What order of magnitude would you think the gap between sphere and new circle of string would be? Is it, for instance, going to be so small that you can't see it easily, or what?
Most people will be able to do the maths on this puzzle and I think you will be surprised at the answer!
Maths is posted below:

20101007, 07:17 #2
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it must be some kind of trick. my guess for the trick is that it would float pi (3.14159...) feet above the earth. No reason, just a guess.

20101007, 11:08 #3
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Hang on, this must be unit dependant.
Start with a sphere of 1 inch; then add 3miles to the rope !!
The gap will be a lot more than 6 inches.....Windows XP SP3

20101007, 11:49 #4
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The sphere's perimeter without adding the 3ft will be P1 = 2 x pi x R1, where R1 is the radius
After adding the 3ft, the spheres perimeter P2 = 2 x pi x R2
What we want to know is the value of R2  R1 and we will call that D
D = R2  R1
D = ( P2 / 2 x pi )  ( P1 / 2 x pi)
D = ((P1 + 3) / 2 x pi)  (P1 / 2 x pi)
D = (P1  P1 + 3) / 2 x pi
D = 3 / 2 x 3.14
D = 0.16 ft

20101007, 14:22 #5
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Or to generalise, where A is the additional length then the gap will be
Gap = A x 1/( 2 x Pi)
Gap = A x 0.159Windows XP SP3

20101008, 12:19 #6
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The derivation provided by Elias is essentially correct but there is a math error going from step 1 to step 2 in that PI has moved from the denominator to the numerator. Try this (using his setup):
D = R2  R1
D = ( P2 / (2 x pi) )  ( P1 / (2 x pi))
D = ((P1 + 3) / (2 x pi))  (P1 / (2 x pi))
D = (P1  P1 + 3) / (2 x pi)
D = 3 / (2 x 3.14)
D = 0.477 ft or about 6 inches
Note that coming to step 5, P1 drops out of the equation. This means that the three feet of added rope will change the radius of the circle of the string by about 6 inches without regard to the original circumference. Therefore if the original sphere was a point sphere (zero radius) the new diameter would be about a foot which is about what you would expect if the problem were phrased that way.

20101008, 13:29 #7
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Much greater than I would have expected. I expected it to be very small. The change in the circumference is minuscule: 3/(24,000x5,280) about 1/(8000x5000), ~1/40,000,000. Surprisingly it gets it all back when you multiply by the radius of the earth, to get the change in radius i.e. height. The radius is also a non trivial number when expressed in feet: 4000x5000, ~20,000,000

20130414, 10:09 #8
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I agree with rdcook. Looking at that logically doesn't make much sense to me...adding 3 feet (something so small) to the circumference of the earth (which is so large) would have such an impact on the readius of the earth, but it is true!
Let's use actual numbers assuming that the earth is a perfect sphere. Googling the circumference of the earth, general concensus is that it is 24,901 miles or 131,477,280 feet or 1,577,727,360 inches (in billions).
HTML Code:[pre] Circumference of the Earth Circumference of the Earth + 36 inches (3feet) We know: We know: Circumference=2*Pi*r Circumference=2*Pi*r Rearranging: Rearranging: C/(2*Pi)=r C/(2*Pi) =r Substituting: Substituting: inches 1577727360/(2*Pi)=r 1577727396/(2*Pi)=r Calculating: Calculating: inches 251002080.00=r 251002085.73=r inches 251002085.73  251002080.00 = 5.73 [/pre]
Last edited by Maudibe; 20130414 at 23:29.

20130415, 12:53 #9
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Now calculate the square mile area of the space between the two 'strings' and compare that to the area of just the smaller string. You may be surprised there, to make it easy assume that the earth is 25,000 miles in circumference.
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20130415, 16:39 #10
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Could that be right? ~ 2.25 sq miles? I used 24901 because I built it in the formulas from my previous post

20130415, 21:13 #11
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Yep, that is correct. If you stacked 2.25 square miles next to each other (1 mile by 2.25 miles), you could rip off 24,880 strips of 5.73" wide x 1 mile long. If you line all the strips end to end, you have the circumference of the earth...whew!!
Last edited by Maudibe; 20130416 at 17:22.

20130506, 00:51 #12
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One thing that seems to have been forgotten is that the original question about the size of the gap.
The difference in size of the two circles may be 5.75", but the gap would only be half of that, or 2.875".

20130507, 07:25 #13
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BigMac,
The radius is increased by 5.75 inches which is the gap at any point around the circumference. If it were the diameter that increased by 5.75" then you would be correct.