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

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.

Martin Baxter wrote:
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


all you have to do is expand the first term:

pi*8000 + pi/5280, subtract the second term and you are left with pi/5280

Or, intuitively, since circumference is diameter time pi, to make the
circumference grow by one foot, you must add pi feet to the diameter;
doesn't matter if the diameter is 1 foot or a light-year, you still add
pi feet.

Ooops, just noticed I only increased the elevation by 6", so change that
to 2pi feet. same principle.



Cheers
Martin

"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.

From reading all the varied answers so far, one can make the conclusion
that one should always carry more spare line than they think they will
use in every situation...then one doesn't need to worry about things
like this...or as the AScouts say "Be Prepared".

"Gordon" wrote in message
m...
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

"above", as in north of the equator ? g

Alan


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On Jan 27, 10:40 am, katy wrote:
From reading all the varied answers so far, one can make the conclusion
that one should always carry more spare line than they think they will
use in every situation...then one doesn't need to worry about things
like this...or as the AScouts say "Be Prepared".

LOL

there is a saying. "no matter how long the line it is ALWAYS 6 inches
short."

"Martin Baxter" wrote in message
...
Martin Baxter wrote:
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


all you have to do is expand the first term:

pi*8000 + pi/5280, subtract the second term and you are left with pi/5280

Or, intuitively, since circumference is diameter time pi, to make the
circumference grow by one foot, you must add pi feet to the diameter;
doesn't matter if the diameter is 1 foot or a light-year, you still add
pi feet.

Ooops, just noticed I only increased the elevation by 6", so change that
to 2pi feet. same principle.



Cheers
Martin

Pie feet? Is that what you get from stomping fruit to make filling?
Maybe why we never hear of "grape pie"?

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