Sailor, read my previous post to understand your error in evaluation of arc length. I will give you credit do doing an integral correctly, but you had the integrand wrong. I didn’t want to have to type this, but arc length of this spiral from angle theta1 to theta2 is F(theta2)-F(theta1), where
F(x)=(R(x)/2/a)sqrt(aa+R(x)R(x))+a/2ln(R(x)+sqrt(a*a+R(x)R(x)))
a=t/2/pi
R(theta)=R0+atheta
R0=starting radius (5.25 mm)
t=tape thickness (0.5 mm)
For a length of 10,000 mm, theta needs to be 439.678 radians, or 69.977 times around. I am not saying that the spiral is the best approximation, but sailor did suggest it in the first place, and also says:
Given the assumptions that tape is continuous, thickness remains constant, ends remain square, center line of tape has constant length (while surface of tape may stretch or compress) the maximum possible is about 70.372 wraps. In order to do that, part of the tape’s surface needs to stretch to twice its orginal length.
In regards to who was first, it looks like CalMeacham was first with a decent estimate, of 79.5 wraps, and Schnitte was also close with a value of 80 wraps. A more complicated but also more accurate approximation of 70.41 was obtained by sailor. The value of 70.372 is the correct answer consistent with the above conditions, and is not just an approximation (except, of course, for rounding to 3 digits).