# How does a measurement error change the volume of a tetrahedron?

Consider that I have a tetrahedron $T$ whose the lengths of edges are $(a,b,c,d,e,f)$.

I want to calculate the volume of the tetrahedron by Cayley-Menger Determinant.

However, I know that, the distances are measured with an error of $0\leq \alpha \leq 0.1$, which means, the values of the edges are intervals
$$T_\text{interval}([a-\alpha, a+\alpha], [b-\alpha, b+\alpha], [c-\alpha, c+\alpha], [d-\alpha, d+\alpha], [e-\alpha, e+\alpha], [f-\alpha, f+\alpha])$$ What is the interval for the volume?

If you assume that volume of $T$ is $0$, what is the maximum volume of $T_\text{interval}$?

To do it roughly, your equation is $288V^2=a^2b^2c^2+$similar terms, so $\frac {\delta V}V \approx \frac {\delta a}a$+similar terms. Count up the number of similar terms for a multiplier.
When $V=0$ strange things happen. Imagine a special zero-volume tetrahedron. The base is an equilateral triangle of side $L$ and the other three sides are of common length (you should be able to calculate it from geometry) and meet at the centroid. Now increase the length of the three shorter sides by $\alpha$ and calculate the volume. You can use geometry (because the figure is so regular) or your determinate. It will be huge compared to the small change in length.
• I forgot to give the interval for $\alpha$. It is between $0$ and $0.1$. Does this change anything? Commented May 28, 2014 at 16:25