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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!

user326210
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Josep
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  • I can see how to do two of the three terms in a determinant; doing all three seems like it might be pretty complicated. You might try pitching your question at nomography.org; it's a small discussion board but the question is likely to get more interest. – Glen_b Dec 15 '12 at 16:24
  • Do you mind explaining what you want the nomogram for? There are a variety of ways to achieve an approximate nomogram (or in some cases perhaps an exact one), even if a standard algebraic determinant leading to a three-variable alignment nomogram turns out to be difficult or impossible (which I suspect is the case). – Glen_b Dec 22 '12 at 06:15

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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)

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.

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

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

Glen_b
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Jason Walker
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    Hi Jason - it doesn't quite reproduce the equation, since it gives "$-6ah$" instead of "$-6ah^2$". – Glen_b Dec 17 '12 at 00:02
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You can prove, using Warmus's criteria, that the function can't be nomographed as it's currently written. Roughly speaking, it has four independent $h$ terms $(h^0, h^1, h^2, h^3)$ that can't be reduced further, but you can only directly nomograph functions where each variable has up to three independent terms ("rank 3").

Specifically, the fact that $h$ has rank 4 proves that there are no functions $f_i,g_i,h_i$ such that:

$$F(a,h,V) = \det\left|\begin{array}{ccc}f_1(a) &f_2(a) &f_3(a)\\g_1(h)&g_2(h)&g_3(h)\\h_1(V) & h_2(V) & h_3(V)\end{array}\right|$$

That said, Warmus's criteria only cover the existing form of an expression. They do cover all ways of grouping terms together differently, but don't account for even easy manipulations such as taking logarithms or changing variables, and they don't account for straightforward compound nomograms.

So we haven't ruled out everything. A nomogram is possible wherever you can find functions such that: $$F(a,h,V) = 0 \iff\det \left|\begin{array}{ccc}f_1(a) &f_2(a) &f_3(a)\\g_1(h)&g_2(h)&g_3(h)\\h_1(V) & h_2(V) & h_3(V)\end{array}\right|=0$$

and that's a much more expansive category than what we've ruled out. It occurs when the determinant on the right has the same zeroes as the function on the left, even if the two sides don't agree at nonzero values.

user326210
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