First Fly Rod, Reel and line Questions??

On 3/8/06 11:39 AM, in article
et, "rw"
wrote:

rb608 wrote:

You have a rope pulled snugly around the earth at the equator (diameter =
7,926 miles +/-). How much length would you need to add to the rope to
raise it 6 inches off the earth at all points?

pi feet

Isn't pi the constant?

My guess was 6(pi) inches. It's C=pi(d), so if we add six inches to d
(which we had converted from feet to inches), we have C=pi(d+6). Balances
out with 6(pi). No?


Bill

On 3/8/06 11:53 AM, in article , "William
Claspy" wrote:

On 3/8/06 11:39 AM, in article
et, "rw"
wrote:

rb608 wrote:

You have a rope pulled snugly around the earth at the equator (diameter =
7,926 miles +/-). How much length would you need to add to the rope to
raise it 6 inches off the earth at all points?

pi feet

Isn't pi the constant?

My guess was 6(pi) inches. It's C=pi(d), so if we add six inches to d
(which we had converted from feet to inches), we have C=pi(d+6). Balances
out with 6(pi). No?

Whoops. He wanted 6" added to r. So make that 12"(pi), which is pi feet.

I'm a little slow on the uptake. (Narrow arbor :-)

Bill

"William Claspy" wrote in message
My guess was 6(pi) inches. It's C=pi(d), so if we add six inches to d
(which we had converted from feet to inches), we have C=pi(d+6). Balances
out with 6(pi). No?

No. You're thinking correctly, but you only got halfway there. You're
actually increasing the diameter by a whole foot (6 inches each side).

Joe F.

"William Claspy" wrote in message
Whoops. He wanted 6" added to r. So make that 12"(pi), which is pi feet.

I'm a little slow on the uptake. (Narrow arbor :-)


Too late. The Mars lander already crashed. :-)

William Claspy wrote:
On 3/8/06 11:39 AM, in article
et, "rw"
wrote:


rb608 wrote:

You have a rope pulled snugly around the earth at the equator (diameter =
7,926 miles +/-). How much length would you need to add to the rope to
raise it 6 inches off the earth at all points?

pi feet


Isn't pi the constant?

My guess was 6(pi) inches. It's C=pi(d), so if we add six inches to d
(which we had converted from feet to inches), we have C=pi(d+6). Balances
out with 6(pi). No?

We're not adding six inches to d. We're adding one foot.

--
Cut "to the chase" for my email address.

wrote in news:jevt02d2gihh8ac0296943tq2o3fo6toos@
4ax.com:

IMO, large-arbor reels acquired with the large arbor being the primary
characteristic sought are for those who know exactly why they _want_
them. There's nothing wrong with getting a reel that one likes that
happens to be a large-arbor, but that doesn't mean that one can "defend"
having it from a practical standpoint on the basis of it being a
large-arbor reel. For most FFers, and a great deal of FFing, the arbor
size is simply not material.

TC,
R

Actually, I've always thought of large arbor reels as a means of getting
people who already have perfectly functionable reels to buy more reels in a
saturated market.

This feeling was reinforced when I started seeing mid arbor reels.

--
Scott
Reverse name to reply

You have a rope pulled snugly around the earth at the equator (diameter =
7,926 miles +/-). How much length would you need to add to the rope to
raise it 6 inches off the earth at all points?


.8 feet
-tom

On 3/8/06 12:05 PM, in article
t, "rw"
wrote:

William Claspy wrote:
On 3/8/06 11:39 AM, in article
et, "rw"
wrote:


rb608 wrote:

You have a rope pulled snugly around the earth at the equator (diameter =
7,926 miles +/-). How much length would you need to add to the rope to
raise it 6 inches off the earth at all points?

pi feet


Isn't pi the constant?

My guess was 6(pi) inches. It's C=pi(d), so if we add six inches to d
(which we had converted from feet to inches), we have C=pi(d+6). Balances
out with 6(pi). No?

We're not adding six inches to d. We're adding one foot.

That's why my undergrad degree only has the WRU part of CWRU on it :-)

Bill

"Tom Nakashima" wrote in message
...

You have a rope pulled snugly around the earth at the equator (diameter
=
7,926 miles +/-). How much length would you need to add to the rope to
raise it 6 inches off the earth at all points?


.8 feet
-tom


Sorry, .08ft
-tom

"rb608" wrote in
news:v1EPf.40384$%I.25893@trnddc03:

"William Claspy" wrote in message
My guess was 6(pi) inches. It's C=pi(d), so if we add six inches to
d (which we had converted from feet to inches), we have C=pi(d+6).
Balances out with 6(pi). No?

No. You're thinking correctly, but you only got halfway there.
You're actually increasing the diameter by a whole foot (6 inches each
side).

Joe F.




c1=pi*(7926 miles)*5280(feet/mile)*12(inches/ft)
c2=pi*((7926 miles)*5280(feet/mile)*12(inches/ft)+12 inches)

c2-c1=37.69 inches


--
Scott
Reverse name to reply