superprismatic Posted June 2, 2011 Report Share Posted June 2, 2011 Suppose the earth were a perfectly smooth sphere with radius 6,371,000 meters. Now, there's an old problem which supposes that a rope girdles the earth along a great circle. An extra meter of rope is then spliced into it. You then assume that the rope is raised the same amount above the earth's surface all the way around. You are asked to determine what the amount of the rise is. It's pretty easy to see that the rise is independent of the radius of the earth and is 1/(2π) meters. But we consider a different problem here: Suppose the earth were a perfectly smooth sphere with radius 6,371,000 meters. Now, suppose a rope girdles the earth along a great circle. An extra meter of rope is then spliced into it. Now, grab the rope in one spot and pull it taut directly away from the center of the earth. How far above the surface of the earth will it be able to be pulled? Give the answer to the nearest meter. Quote Link to comment Share on other sites More sharing options...
0 Peekay Posted June 3, 2011 Report Share Posted June 3, 2011 Let r be the radious of tha sphere. y be the distance from the ground to the top of where you pull the rope a be the angle from where you stand to where the rope is tangent to the sphere x be the distance from where the rope is tangent to the earth to where you are pulling the rope tan (a)=x/r hence x = r*tan(a) 2*pi*r - 2*a*r + 2x = 2*pi*(r+1) substitute x 2*pi*r - 2*a*r + 2*r*tan(a) = 2*pi*(r+1) 2*r*tan(a)-2*a*r=2*pi a~~0.6528° We also know: r^2 + x^2 = (r+y)^2 substitute x = r*tan(a) r^2 + (r*tan(a))^2 = (r+y)^2 solve for y y ~ 413 m In the following step: 2*pi*r - 2*a*r + 2x = 2*pi*(r+1) How are you getting the (r+1)? 1 meter was added to the circumference, not the radius. Quote Link to comment Share on other sites More sharing options...
0 superprismatic Posted June 3, 2011 Author Report Share Posted June 3, 2011 I'm glad this little puzzle tickled as much interest that it did! It really seems counter intuitive. Congratulations to EventHorizon for his quick correct response. For a problem where accuracy is paramount, he got the height right on the nose. Thank you to everyone who worked on it. Quote Link to comment Share on other sites More sharing options...
0 Guest Posted June 3, 2011 Report Share Posted June 3, 2011 A funny application of this new found phenomena: When people pull the waist-line of their pants out to show how much weight they've lost, you can show them the math and convince them they didn't really lose much circumference. edit: Just don't start with, "assume the radius of your waist is that of the earth..." as it might make them feel bad. Wow! I nearly spewed coffee all over my desk. Thank's for the laugh; I've got to put that in my signature. Quote Link to comment Share on other sites More sharing options...
0 Peekay Posted June 3, 2011 Report Share Posted June 3, 2011 I'm glad this little puzzle tickled as much interest that it did! It really seems counter intuitive. Congratulations to EventHorizon for his quick correct response. For a problem where accuracy is paramount, he got the height right on the nose. Thank you to everyone who worked on it. Thanks Superprismatic for a wonderful puzzle! Quote Link to comment Share on other sites More sharing options...
Question
superprismatic
Suppose the earth were a perfectly smooth
sphere with radius 6,371,000 meters. Now,
there's an old problem which supposes that
a rope girdles the earth along a great
circle. An extra meter of rope is then
spliced into it. You then assume that the
rope is raised the same amount above the
earth's surface all the way around. You
are asked to determine what the amount of
the rise is. It's pretty easy to see that
the rise is independent of the radius of
the earth and is 1/(2π) meters. But we
consider a different problem here:
Suppose the earth were a perfectly smooth
sphere with radius 6,371,000 meters. Now,
suppose a rope girdles the earth along a
great circle. An extra meter of rope is
then spliced into it. Now, grab the rope
in one spot and pull it taut directly away
from the center of the earth. How far
above the surface of the earth will it be
able to be pulled? Give the answer to the
nearest meter.
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