Wayne.B wrote:
On Tue, 27 Jan 2009 16:48:47 +0000, Gordon wrote:

This is from the Brian Toss rigging book so if you want to cheat, you
can look it up.

If I string a line around the equator (disregarding natural
obstacles) and then decide I want that line 1 foot above the equator all
the way around, how much longer must that line be? Call the earth's
diameter an even 8000 miles.
G

3.14 ft

I thought a circle was 2 pi...

"Wayne.B" wrote in message
...
On Tue, 27 Jan 2009 16:48:47 +0000, Gordon wrote:

This is from the Brian Toss rigging book so if you want to cheat, you
can look it up.

If I string a line around the equator (disregarding natural
obstacles) and then decide I want that line 1 foot above the equator all
the way around, how much longer must that line be? Call the earth's
diameter an even 8000 miles.
G

3.14 ft

No, you have forgotten that it is 1' higher all round so the diameter has
increased by 2' not 1'

On Tue, 27 Jan 2009 21:46:14 +0100, "Edgar"
wrote:

No, you have forgotten that it is 1' higher all round so the diameter has
increased by 2' not 1'

Yes, good point, should have used the radius formula, 2 pi R.

On Tue, 27 Jan 2009 16:48:47 +0000, Gordon wrote:

This is from the Brian Toss rigging book so if you want to cheat, you
can look it up.

If I string a line around the equator (disregarding natural
obstacles) and then decide I want that line 1 foot above the equator all
the way around, how much longer must that line be? Call the earth's
diameter an even 8000 miles.
G

44/7 ft.

"Goofball_star_dot_etal" wrote in message
...
On Tue, 27 Jan 2009 16:48:47 +0000, Gordon wrote:

This is from the Brian Toss rigging book so if you want to cheat, you
can look it up.

If I string a line around the equator (disregarding natural
obstacles) and then decide I want that line 1 foot above the equator all
the way around, how much longer must that line be? Call the earth's
diameter an even 8000 miles.
G

44/7 ft.

You obviously too geometery prior to handheld caculators, too.

On Tue, 27 Jan 2009 13:32:50 -0500, "BF" wrote:


"Goofball_star_dot_etal" wrote in message
.. .
On Tue, 27 Jan 2009 16:48:47 +0000, Gordon wrote:

This is from the Brian Toss rigging book so if you want to cheat, you
can look it up.

If I string a line around the equator (disregarding natural
obstacles) and then decide I want that line 1 foot above the equator all
the way around, how much longer must that line be? Call the earth's
diameter an even 8000 miles.
G

44/7 ft.

You obviously too geometery prior to handheld caculators, too.


Comes in handy when one loses one's slide rule..

* *If I string a line around the equator (disregarding natural
obstacles) and then decide I want that line 1 foot above the equator all
the way around, how much longer must that line be? Call the earth's
diameter an even 8000 miles.
*

Goofball_star_dot_etal wrote:
44/7 ft.

Did we ask for an answer in the metric system?!

DSK

On Tue, 27 Jan 2009 12:21:40 -0800 (PST), wrote:

* *If I string a line around the equator (disregarding natural
obstacles) and then decide I want that line 1 foot above the equator all
the way around, how much longer must that line be? Call the earth's
diameter an even 8000 miles.
*

Goofball_star_dot_etal wrote:
44/7 ft.

Did we ask for an answer in the metric system?!

DSK

(44/7) * 12 * (2.54/100) m.

Gordon wrote:
This is from the Brian Toss rigging book so if you want to cheat, you
can look it up.

If I string a line around the equator (disregarding natural
obstacles) and then decide I want that line 1 foot above the equator all
the way around, how much longer must that line be? Call the earth's
diameter an even 8000 miles.
G


circumference = pi*d

so difference = (pi*(8000+(1/5280))-(pi*8000)miles

Martin Baxter wrote:
Gordon wrote:
This is from the Brian Toss rigging book so if you want to cheat, you
can look it up.

If I string a line around the equator (disregarding natural
obstacles) and then decide I want that line 1 foot above the equator
all the way around, how much longer must that line be? Call the
earth's diameter an even 8000 miles.
G


circumference = pi*d

so difference = (pi*(8000+(1/5280))-(pi*8000)miles

=pi/5280 miles

multiply by 5280'/mile and you get pi ft. (3.2459')

do a little simple algebra

Cheeers
Martin