Big piece of string

There is a piece of string that just fits snugly around the equator of
the Earth (ignore any mountains that get in the way - apologies to
any goat herders living in the Andes).

The string is then cut at a single point and lifted upon one foot high
posts that are placed around the entire equator and (for the benefit
of any pedants reading) are a negligible distance apart.

The question is: How much extra string must added to join up the two
ends of the original piece of string (to 6 significant figures please)?

Hint: The circumference of the Earth is roughly 129,101,050 feet.

Okay...that was a bit easy...

However, I do like this little puzzle...the way it arouses an initial
reaction of 'hmm, must be loads' (caused by thinking with the bit of
the brain that is linked to most foot-in-mouth incidents)...but then
the logical chunk of the brain kicks in (with a little 'Eureka!' perhaps?),
to remind us of all those sweaty afternoons spent in maths class
trying to peek down Miss Warwick's top...

Where was I? Oh yeah, here's an extension to the problem then...

Suppose the string is replaced by a big taut elastic band (the posts
are still in place). You've been volunteered for the lucky job of
removing the posts. Assuming the elastic band remains taut, what is
the minimum number of posts that can be left standing so that the
elastic band doesn't touch the ground? What is the distance between
the posts?

Assume the circumference of the Earth really is 129,101,050 ft.

I don't like the numbers, so I'll use letters instead (a common ploy among mathmoes). Let r
be the radius of the Earth in feet, ie 129,101,050/2pi. Pick two consecutive posts that are
just close enough together for the band not to touch the ground in between (I'm assuming the
surface at the equator is perfectly smooth, otherwise the answer involves putting the posts on
top of oceanic ridges/hills/mountains!). Then the band forms a tangent to the sphere at the
midpoint of the Equator. Now draw a line down through each post. The lines will make an
angle t, say, at the centre of the Earth. You now have a triangle with altitude r and two sides
r+1, and you can show t is 2 arccos(r/(r+1)) (not very large!). I'll take t in radians; round t
down to t' so 2pi is an integer multiple of t', then the distance between posts (along the
Earth's surface) is rt', and the minimum number of posts is 2pi/t'. (I think it's in the region of
20000 posts, but I may have pressed the wrong buttons on my calculator.)