Interesting experiment, I'd give it a go. Here's my solution -

Assume the width of the paper strip is W. The small pentagon knot has side length of L, where L = W / cos(18 degree).

The total length of paper required by the small pentagon knot is: (2*L + L*sin(18)) * 4 + L*sin(18).

Five such pentagons are needed to build the large pentagon. Giving the total length requirement of ((2*L + L*sin(18))*4 + L*sin(18))*5, which simplifies to 47.725424859 * L. The relationship b/w L and the Diameter of the circle is: L = D * sin(36).

Thus the perl code:

# calculate length and width of paper strip
# based on diameter of circle of inner edge
sub NumberCrunch()
{
   my ($diameter) = @_;
   my $L = $diameter * 0.58778525229;
   my $W = $L * 0.951056516; # W = L * cos(18)
   my $total_length = $L * 47.725424859;
   return ($total_length, $W);
}
...
my ($TotalLength, $PaperWidth) = NumberCrunch($Diameter);
[download]

The above should give the minimum length and paper width required to fold such pentagon, given the diameter of the inner circle encompass the inner pentagon.

Ok, this is just a quick solution I came up during my lunch break, don't be suprised if my number crunch is wrong. :-) I'd love to see other people's solution to this problem.

Second solution: if inner is regular decagon, then L = D * sin(18), and Length = ((2*L + L*sin(18))*4 + L*sin(18) + L)*5, and W = L * cos(18). Simplify the equations and modify the perl script to get the solution for the second part of the question...

PS. I vaguely remember seeing this knot in ancient greek geometry, where golden ratio and the self-repeating patterns were discussed.