Belt the answer out, if you can!
If a belt were placed around the Earth's equator, and then had six meters of length added to it, and you grabbed it at a point and lifted it until all the slack was gone, how high above Earth's surface would you be?
Note: This might sound like a very simple problem but... :-)
This is especially to counteract the disappointment of last problem!
Note: This might sound like a very simple problem but... :-)
This is especially to counteract the disappointment of last problem!
posted by littlecow at 11/13/2005 06:29:00 PM
13 Comments:
What do you mean when you say the belt has no slack? Is it just touching the earth's surface on both ends or is there some other funda?
@bal: grenade is right. you just hold the belt at a point and pull as much as you can.
@grenade: to start with, it does not matter... go ahead with any solution.
Is it approximately 402 meters?
I mean 402 km
I got an answer of 1344.2km (APPROX).I dont know whether I goofed up, but please tell me if Iam wrong:
The reasoning: When you pull the belt taut,belt will cover only one half of the earth completely and the rest will be tangential from that point on and will actually form an isoceles triangle:
Law of conservation of belt length says: 2*PI*radius+6=2*x+PI*radius
we determine x and then
H=distance from centre of earth
However, x2=Radius2+H2
=> H= sqrt(x2-Radius2)
Actual Height(from earth's surface),h = H-radius of earth
Plugging in the numbers we get:
Radius of earth =6378.1 Km
H=7722.3958 (from centre of earth and
h=1344.2 km
@bal: 402m sounds just about right! Although your final answer was dubiously off by a factor of thousand, I will give the points to you. now would you post your solution?
@@: if the belt is tangential at the diagonally opposite points (so that half of it is wrapped around the earth), then the tangents will never meet!
the interesting thing to note is the rather counter-intuitive nature of the result. one would think that the belt wont go too high up. but the unwrapping of the belt (to create the tangent), actually increases the height many fold!
I just came here to post that I goofed up :| ...I did not mean a perfect tangent rather, was visualizing something like a tear drop shape. However, my assumption that it will cover half of the circumference was wrong...
@
@@: no problems! to err is...
to err is human, to remember it is more human!! :))
i worked out so much: assuming that the angle subtended by the belt's tangential ends is 2.theta, tan(theta) - theta = 3/R, where R = 6400km. Couldn't figure out how to proceed from there other than trial and error (numerical). what is the most efficient algorithm to solve it ?
The numerical solution to this problem is a little tricky. I just started reading a numerical methods book and the first thing the author talks about in the book is a variant of this problem. I didn't read the solution he proposes, so I'll try to solve the problem after work today. Nice blog, Balaks.
I get the same eqn. as arnie above i.e tan(theta) - theta = 3/r. If we take the first two terms in the expansion of tan, I get height = 401.68 m. Not very precise beyond the decimal point though.
@arni: the simplest solution is what raju got.
@raju: welcome and thanks! as for your answer, it is correct. the interesting thing to note in this problem is the fact that a 6m slack results in a height increase of more than 401.68m. extending this problem, a 1mm slack on the car alternator belt (for example), results in a significantly larger center-to-center distance increase resulting in the belt coming off the pulleys. other applications abound.
now, try the latest problem!
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