# Distance to fault in submarine telegraph cables ca 1871

In the early days of submarine telegraph cables, the cable technology was a single core steel wire wrapped in gutta-percha worked against ground. Now the gutta-percha was not a uniform or durable insulation and leaks to ground (sea) were inevitable, and when the leakage became sufficient the cable could not longer be used and had to be repaired.

The earliest method of locating a cable fault was a binary chop… which would mean deploying a cable ship, grapnelling for the cable, hauling it to the surface with a special dividing cut and hold grapnel that severed the cable when tension was too great, buoying off one end and steaming back to the other to haul it on board, clean it up and test to the far cable station. New cable was spliced and the cable ship steamed back to find the buoy and pull that end on board, clean and test to the other end. This was done to localise the fault, and eventually replace a fault section of cable. During this longish period, the cable was out of service.

Oliver Heaviside, a telegraphist and self taught mathematician applied his mind to predicting the distance to fault based on resistance measurements from a land end applying Blavier’s test to a submarine cable. Blavier’s test assumes a single ground fault, and a test from both ends can improve confidence that that is likely to be the case. Let u be the resistance end to end of the installed conductor based on length and resistance per km, $$u=666.7 \cdot 3.24=2160 \Omega$$.

Let x be the resistance from Newbiggen-by-the-sea (local) to the fault. The resistance of the remaining cable conductor is u-x.

Let y be the measured resistance from the fault point to ground (the sea or cable amour if present).

If the remote end (Sondervig) is grounded, the expected resistance w is given by $$w=x+\frac1{\frac1y+\frac1{u-x}}=x+\frac{y(u-x)}{y+u-x}$$ and was measured to be 970Ω.

If the remote is is open, the resistance v was measured at 1040Ω, now $$v=x+y$$ so $$y=v-x$$ which we can substitute.

$$w=x+\frac1{\frac1y+\frac1{u-x}}=x+\frac{y(u-x)}{y+u-x}=x+\frac{(v-x)(u-x)}{(v-x)+u-x}\\$$

Now it is in terms of the three known values u,v,w and unknown x.

$$w(v-2x+u)=(v-2x+u)x+(v-x)(u-x)\\$$

$$x^2-2wx+vw+uw-uv=0$$ from which you can find the roots.

$$x=w – \sqrt{(w-v)(w-u)}\\$$

So when measurements gave $$v=1040 \Omega$$ and $$w=970 \Omega$$ we can calculate that the distance to fault is the lesser root, 210.3km from Newbiggen-by-the-sea. (The greater root would imply a -ve value for x or y which is not physically possible.)