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.)

Last update: 1st October, 2020, 6:47 PM