Extend a Circumference by 1m This question is related to the  Extending the length of a curcumference by 1 meter problem.
But instead of making a larger circle as in the example above, the loose string is hung from a nail on a wall - like a circular picture frame.
Consider a taut string around the circumference of a cylindrical picture frame of radius R. Then add 1m to the string and put the string around the cylinder and then "hang" the cylinder like a picture frame from a nail on a wall. The string will form an apex at the nail, with the two tangents from the cylinder which then go around the major arc of the circle. 
Question: what is the height of the apex from the top of the circle?
 A: Let $P$ be the peak of the string, let $C$ be the centre of the circle, and let $T$ be a point of tangency. Let $\theta=\angle PCT$.
The total length of the extended string is $2\pi R+1$. This length is also $R(2\pi-2\theta)+2R\tan\theta$. It follows that
$$\tan\theta-\theta=\frac{1}{2R}.$$
This is a transcendental equation with no closed form solution. However, if $R$ is large, such as in the classical problem where $R$ is the equatorial radius of the Earth, the angle $\theta$ is very small. Then we can use the approximation $\tan\theta\approx \theta+\frac{\theta^3}{3}$ to get a good numerical approximation of $\theta$.
A: Building on the answer from André Nicolas, we have $\tan\theta-\theta\approx \frac{\theta^3}{3}$, so $\theta \approx \left(\frac{3}{2R}\right)^{1/3}$.
But what we want is the height of $P$ above the surface, which is $h = R(\sec\theta-1) \approx \frac12R\theta^2$.
This gives
$$h \approx \left(\frac{9R}{32}\right)^{1/3}$$
The radius of the earth at the equator is about $6378$km, which gives a value for $h$ of about $121.5$ metres.
