Salt Mine Cables

Here's a quickie way to work this out. Let's pretend that the cable has a $1 cm$ square cross section. The length of the drum is $180 cm$ . The radius of the drum, with and without the cabling, is $35 cm$ and $30 cm$ . The total volume of the cabling material is

$180\pi \left( { 35 }^{ 2 }-{ 30 }^{ 2 } \right) \simeq 183783cm$

From this, we have $1838 m$ of cabling.

The original cylinder is 1.8m long and has a diameter of 0.6m. Each wire is 0.01m. Hence we need to find the number of windings per layer is $(1.8/0.01) =180$ .

Now we find the length for the first five windings using the Pythagoras theorem. When you wind the cable around the cylinder, the total length is equal to:

$\sqrt{{(cylinder length)}^{2} + {(windings*{circumference}_{cross-section})}^{2})}$

For the winding circumference, we have to add $2r$ , which will include the thickness of the cable. We begin with 0.61 m because we need to add the radii of the first cable on both sides and the diameter of the cylinder to obtain the circumference of the first cross-section.

$L1 = \sqrt{{1.8}^{2} + {(180\pi (0.61+0.00))}^{2}}$ $L2 = \sqrt{{1.8}^{2} + {(180\pi (0.61+0.02))}^{2}}$ $L3 = \sqrt{{1.8}^{2} + {(180\pi (0.61+0.04))}^{2}}$ $L4 = \sqrt{{1.8}^{2} + {(180\pi (0.61+0.06))}^{2}}$ $L5 = \sqrt{{1.8}^{2} + {(180\pi (0.61+0.08))}^{2}}$

The reason why we add 0.02m each time to the original diameter of the cylinder is because the cable increases one layer diametrically when wrapped around the circular cross-section (just draw circles).

We get 1837.853m or about 1838m.