Här visas en geodet i en krökt geometri.
En funktion \[ f(x,y)= \frac 1 {x^2+y^2+1} \] ger en metrik via \[\begin{array}{rcl} ds^2 &=& dx^2 + dy^2 + (1-\frac{\partial f}{\partial x} dx-\frac{\partial f}{\partial y} dx)^2 \\ &=& (1+(\frac{\partial f}{\partial x})^2)\,dx^2 +2\frac{\partial f}{\partial x}\frac{\partial f}{\partial y}\,dx\,dy + (1+(\frac{\partial f}{\partial y})^2)\,dy^2 \end{array} \] så att \[ g = \Bigg[ \begin{array}{cc} (1+(\frac{\partial f}{\partial x})^2) & \frac{\partial f}{\partial x}\frac{\partial f}{\partial y}\\ \frac{\partial f}{\partial x}\frac{\partial f}{\partial y} & (1+(\frac{\partial f}{\partial y})^2) \end{array}\Bigg]. \]
Med denna beräknas Christoffelsymbolen \[ \Gamma^a_{bc} = \frac 1 2 g^{ad}\big( \frac{\partial g_{db}}{\partial x^c} + \frac{\partial g_{dc}}{\partial x^b} - \frac{\partial g_{bc}}{\partial x^d}\big) \] och den geodetiska ekvationen \[ \frac{\partial^2 x^a}{\partial t^2}+\Gamma^a_{bc}\frac{\partial x^b}{\partial t}\,\frac{\partial x^c}{\partial t} = 0 \] löses.
Allt detta beräknas nummeriskt, varför felfortplantningen blir avsevärd.