View Full Version : Rope Problem

suicoted

09-08-2005, 10:41 PM

I have no idea what to do, I don't even understand the question, please help, thanks.

Imagine a rope running right around the world (round the equator) in a perfect circle at ground/sea level. How much more rope would be needed if the rope were to be lifted to a height of 1 metre all the way around?

I know C = 2pier

I was not given #s for this problem. The thing is I don't really understand the question. Thanks. :)

marylove

09-08-2005, 10:50 PM

Ok. Cut the earth in half. Now you are looking down straight at the equator. A circle. The rope goes around the circle...ie it is the circumfrence. If you lift the rope up a meter the circle has increased all the way around, but in particular you increased the radius by a meter and the diameter by 2 meters.

C=pi*D

C=pi*(D+2)

C=pi*D+2pi meters.

tkhunny

09-08-2005, 11:05 PM

I'm struggling a little with marylove's use of 'C' to mean two different things, but it is a place to start.

You were given no numbers because it is OK to expect the student to know a few things that are deemed "common knowledge". The Earth is about 25000 mi (40234 km) in circumference, or has about an 8000 mi (12875 km) radius. Values like these are often assumed to be "common knowledge"

Current Radius

Circumference = 2*pi*12875 km

New Radius (1 m = 0.001 km)

NewCircumference = 2*pi*12875.001 km

More Rope = Difference

= NewCircumference - Circumference

= 2*pi*12875.001 km - 2*pi*12875 km

= 2*pi*(0.001 km)

marylove

09-08-2005, 11:13 PM

I used C to mean circumfrence just like suicoted did. And I object that you don't need to know the radius of the earth to do this problem. Both our answers have the circumfrence (C) increasing by 2pi meters (a thousandth of a kilometer).

tkhunny

09-08-2005, 11:42 PM

No worries. It's ok to work problems different ways. That's one of the reasons we do this sort of thing on a public forum.

Your first 'C' is as-is. Your second 'C' is after the 1 m is added. Those are not the same thing.

Glad to have you with us.

mathmike

09-09-2005, 12:17 AM

And I object that you don't need to know the radius of the earth to do this problem.

Well, that's a beauty of the problem - you don't need to know that:

difference = 2pi(R + 1) - 2piR = 2piR - 2piR + 2pi = 2pi meters

suicoted

09-09-2005, 12:45 AM

Ok. Cut the earth in half. Now you are looking down straight at the equator. A circle. The rope goes around the circle...ie it is the circumfrence. If you lift the rope up a meter the circle has increased all the way around, but in particular you increased the radius by a meter and the diameter by 2 meters.

C=pi*D

C=pi*(D+2)

C=pi*D+2pi meters.

I understand the method to solve it, everyone :), thanks, BUT I don't even understand the question. All I understand is that there is a rope running right around the world, which is its equator. If I lift the rope, does it come off? I wish I could see this visually :(.

mathmike

09-09-2005, 01:24 AM

I understand the method to solve it, everyone :), thanks, BUT I don't even understand the question. All I understand is that there is a rope running right around the world, which is its equator. If I lift the rope, does it come off? I wish I could see this visually :(.

Don't take this problem so straight, it is imaginary situation. It was invented to demonstrate that no matter how big the radius of the initial circle was increasing it by 1 m increases circumference by 2pi meters. You can imagine that you put many 1m poles along the equator to support the rope, if you wish.

Denis

09-09-2005, 02:36 AM

Wella, wella...

Measures of the Earth:

1. Ellipsoid after Friedrich Bessel 1841:

major semi axes a = 6,377,397.1550 m

minor semi axes b = 6,356,078.9630 m

sphere of equal volume: 6,370,283.1583 m

sphere of equal area: 6,370,289.5102 m

sphere of equal dist(pol): 6,366,742.5203 m

1. Ellipsoid after Hayford 1924 (International 1924):

major semi axes a = 6,378,388.0000 m

minor semi axes b = 6,356,911.9461 m

sphere of equal volume: 6,371,221.2659 m

sphere of equal area: 6,371,227.7113 m

sphere of equal dist(pol): 6,367,654.5001 m

1. Ellipsoid after Krassovsky 1940/1948:

major semi axes a = 6,378,245.0000 m

minor semi axes b = 6,356,863.0188 m

sphere of equal volume: 6,371,109.6937 m

sphere of equal area: 6,371,116.0829 m

sphere of equal dist(pol): 6,367,558.4969 m

1. Ellipsoid World Geodetic System of 1972:

major semi axes a = 6,378,135.0000 m

minor semi axes b = 6,356,750.5200 m

sphere of equal volume: 6,370,998.8588 m

sphere of equal area: 6,371,005.2495 m

sphere of equal dist(pol): 6,367,447.2486 m

1. Ellipsoid World Geodetic System of 1984:

major semi axes a = 6,378,137.0000 m

minor semi axes b = 6,356,752.3142 m

sphere of equal volume: 6,356,752.3142 m

sphere of equal area: 6,371,007.1809 m

sphere of equal dist(pol): 6,367,449.1458 m

2. Distance North pole to equator = 10000 km (First definition for the meter)

Circumference = 40000 km.

Radius = 6366.198 km

3. Circumference = 60*360 Int. Sea Miles

Radius = 3437.747 Sea Miles = 6366.707 km

(1 international sea mile = 1.852 km)

4. Circumference = 25000 Miles

Radius = 3978.874 Miles = 6403.376 km

(1 statute mile = 1760 yard = 1760*36 inch = 1609.344 meter)

5. Circumference

Radius = 4000 Miles = 6437.376 km

(1 statute mile = 1760 yard = 1760*36 inch = 1609.344 meter)

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