Calculating a diagonal line around a cylinder Is there a way to calculate (or estimate) a diagonal line around the exterior of a cylinder given the dimensions in the image below.
Height is $60$, diameter is $12$. The line starts at a $45$ degree angle and loops around exactly once.

 A: 
In figure top right is for comparison of helix with angle $\gamma=45^o$ wich is for 1.6 revolution (figure a in bottom left) and angle $\gamma=32^o$ for one revolution(figure b on bottom left).We have:
$tan \gamma=\frac ab$
where a is the radius of the cylinder(here 12") and b is the lead of he helix. For one complete turn(revolution) we have:
$h=2\pi b$
here $h=60"$
Parametric equation of helix is:
$\begin{cases}x=a\cos t\\y=a\sin t \\z=bt\end{cases}$
where $t$ is the angle of turn which for one turn is $t=360^o=2\pi$ . The length of the helix can be found by following formula:
$$s=\int^B_A\sqrt {x'^2+y'^2+z'^2}$$
Or for one turn:
$$s=\int^{2\pi}_0\sqrt{a^2\sin^2 t+a^2\cos^2 t +b^2}dt$$
Finally:
$$s=2\pi\sqrt{a^2+b^2}$$
which gives
$s=\sqrt{60^2+(12\pi)^2}\approx 70.8"$
But for one turn the angle is not $45^o$ it is:
$tan \gamma=\frac{60}{12\pi}\Rightarrow \gamma \approx 58$
and for the angle you shown in your figure it is:
$\alpha\approx 90-58=32^0$
If you want $\alpha=45^o$ then we have $ \gamma=90-45=45^o$:
$tan 45^o=1=\frac {60}{n\cdot (12\pi)}\Rightarrow n\approx 1.6$
where n is number of turn. In this case the legth of helix is:
$s=1.6\times \sqrt{2\times (12\pi)^2}\approx 85.3"$
