Determinant form of equation, 3 variables, third order (nomogram)

Question

I'm trying to put the following equation in determinant form: $12h^3 - 6ah^2 + ha^2 - V = 0$, where $h, a, V$ are variables (this is a volume for a pyramid frustum with $1:3$ slope, $h$ is the height and $a$ is the side of the base, $V$ is the volume).

The purpose of identifying the determinant is to construct a nomogram. I'm not sure if it actually can be placed in determinant form, and I'm curious if there is a Mathematica function that can do this? I've been trying a pen and pencil approach as listed here. But this approach has hopefully been automated.

Any tips are appreciated!

Answer 1

It's a genus 1 type as described here with determinantal form:

$$\left| \begin{array}{ccc} 0 & V & 1 \\ 1 & 6a-a^2 & 1 \\ \frac{h}{h+1} & \frac{12h^3}{h+1} & 1 \end{array} \right|=0$$

Answer 2

As it stands there are 6 linearly independent terms to deal with ($a$, $a^2$, $h$, $h^2$, $h^3$ and $V$). Without being able to spot a common factor somewhere it's not clear one can accommodate so many terms (the three terms in $h$ in particular make it hard).

Saint-Robert's criterion indicates that it can't be done with three straight axes.

After playing around trying to simplify it (by trying to eliminate a term in various ways) in order to try to get it in a form dealt with by Clark, I think that perhaps Clark's approach may not be sufficient, which suggests one might have to go to Warmus' laborious criteria to even see if its possible as it stands, but I should go back and check it against either the Massau or Lecornu conditions first.

If you are allowed to reparameterize it to be in terms of say $p = \frac{h}{a}$ and $a$, or even easier, $q = 1 - \frac{h}{a}$ and $a$, then you can get somewhere.

If a good approximation is okay, it can be done with an order 4 (i.e. genus I) nomogram pretty closely, and one can generate better approximations by upping the order, possibly* to such a close approximation compared to the error in using an actual 'exact' nomogram that it wouldn't matter - typically anything below about half a percent error, though with care it can be a bit lower; if you get the worst approximation error much lower than that, you would have trouble telling an exact nomogram from an approximate one.

*(depending on what range of values you want to cover, I just assumed some values in order to play with it)

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Edit: I see now that $\frac{V}{a^3}-\frac{1}{18}=(6\frac{h}{a}-1)^3/18$.

This suggests some possibility involving a pair of (possibly overlaid or back-to-back) $N$-charts but I don't currently see how to make it work without requiring multiple isopleth lines drawn - which is likely more effort for the end-user than it's worth.

Still, the observation may help someone spot a way to take it further.