# Distance to fault in submarine telegraph cables ca 1871 – the leap expanded

Distance to fault in submarine telegraph cables ca 1871 gave a mathematical explanation of the location of fault…

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.

$$x=w – \sqrt{(w-v)(w-u)}\\$$

I have been asked to expand the last ‘leap’.

So we have $$x^2-2wx+vw+uw-uv=0$$ which is a quadratic, a polynomial of order 2.

The solution or roots of a quadratic $$ax^2+bx+c=0$$ are given by $$x=\frac{-b \pm \sqrt{b^2-4ac}}{2a}$$.

So, for our quadratic $$a=1, b=-2w,c=vw+uw-uv$$, so $$x=\frac{2w \pm \sqrt{(2w)^2-4(vw+uw-uv)}}{2}$$.

Dividing the top and bottom by 2 we get $$x=w \pm \sqrt{w^2-(vw+uw-uv)}$$ which can be rewritten as $$x=w \pm \sqrt{(w-v)(w-u)}$$.

We want the lesser square root $$x_-=w-\sqrt{(w-v)(w-u)}$$ because x must be less than w, a constraint of the physical problem.

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.)