Past exam of the physics of the Earth as a Planet course of the University of Cambridge / 2012 / Paper 1 / 2 / a / Solution

Assume straight rays in a homogeneous medium. At horizontal distance $x-x_0$ the source distance is $R=\sqrt{(x-x_0{)}^2+z_0^2}$. With $v_A>v_B$,where $M=(1/v_B-1/v_A{)}^2$ and $B=(\Delta t_0{)}^2=M{z}_0^2$. Clock offsets cancel in a difference measured at one station. The difference curve is symmetric about the epicentre and reaches its minimum directly above the source.

The original plot uses the given numerical observations. Its broad minimum is consistent withThese are conservative graph-reading estimates, allowing for one-second rounding; the three equal smallest entries do not imply a flat physical minimum. A quadratic fit to squared differences refines the vertex to $x_0=140.17\mathrm{km}$ and $\Delta t_0=14.896\mathrm{s}$. These extra digits describe the fit, not the precision of an individual observation. Formal statistical errors require a model for the rounding and other measurement errors.