Nomogram for probabilities

Question

I am trying to find a nomogram for the following equation:

$$F(a,b,c) = k(c-a)-(c-b)^2 = 0$$

where $k$ is a constant. I can rewrite it as:

$$(c^2 + kc) - ka -2bc + b^2 =0.$$

This has four or five linearly-independent terms and three variables (so it's Genus II). Does this match any known pattern of nomogram?

• I've tried solving for the determinant, sudoku-style, but it feels like there are too many linearly-independent terms joined together: $a, b, b^2, c, c^2$.

• I have also tried completing the square for $(c^2 + kc)$ and absorbing the constant into the $-ka$ term, but it doesn't help much.

Answer 1

The equation can't be nomogrammed, at least not in its current form. I found Kellog's two necessary criteria for nomogramability; $F$ fails the second test. Informally, there are just too many independent terms. Technically, a certain wronksian determinant fails to vanish.

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First, we choose one of the variables to isolate—here, $b$—and we rewrite $$F(a,b,c) = \left(1\right)b^2 + (-2c)\,b^1 + [(c^2+kc)−ka]\, 1 = 0.$$ In other words, we have written: $$F(a,b,c) = J_1(a,c)f_1(b) + J_2(a,c) f_2(b) + J_3(a,c)f_3(b)$$ where the $J_i$ are linearly independent of each other (note they each contain a $c$ term of different degree) and the $f_i$ are linearly independent of each other (note they each contain a $b$ term of different degree.)

Kellogg's second criterion is that, if we look at the non-isolated variables $a$ and $c$, the following wronskian determinants must vanish:

$$\det \begin{bmatrix}J_1 & J_2 & J_3 \\ \partial_aJ_1 & \partial_aJ_2 & \partial_aJ_3 \\ \partial_{aa}J_1 & \partial_{aa}J_2 & \partial_{aa}J_3\end{bmatrix} = 0$$

$$\det \begin{bmatrix}J_1 & J_2 & J_3 \\ \partial_cJ_1 & \partial_cJ_2 & \partial_cJ_3 \\ \partial_{cc}J_1 & \partial_{cc}J_2 & \partial_{cc}J_3\end{bmatrix} = 0.$$

Unfortunately, for the second wronksian, we get:

$$\det \begin{bmatrix}J_1 & J_2 & J_3 \\ \partial_cJ_1 & \partial_cJ_2 & \partial_cJ_3 \\ \partial_{cc}J_1 & \partial_{cc}J_2 & \partial_{cc}J_3\end{bmatrix} = \det \begin{bmatrix}1 & -2c & (c^2+kc)-ka \\ 0 & -2 & 2c + k \\ 0 & 0 & 2\end{bmatrix} = -4 \neq 0.$$ (All but one term in the determinant vanishes; the one on the main diagonal.)

Answer 2

I created a matrix for the said equation:

$$ K(−)−(−)^2=0 $$

It goes like this:

$$ det \begin{bmatrix} Kc & 1 & (c-b)^2 \\ Ka & 1 & 0 \\ 1 & 0 & 1 \end{bmatrix}=0 $$

Putting it in standard nomographic form, it becomes:

$$ det\begin{bmatrix} \frac{Kc}{1 + Kc} & \frac{(c-b)^2}{1 + Kc} & 1 \\ \frac{Ka}{1 + Ka} & 0 & 1 \\ 1 & 1 & 1 \end{bmatrix}=0 $$

It's not a beautiful nomogram since first row contains 2 variables and third row is just the number 1. However it can be plotted as a nomogram if you create a line always going through coordinate (1,1).

I plotted it in PyNomo (I believe you are familiar with it) and it's weird and a bit messy but it works. In my plot, I chose K = 1.

Nomogram

Maybe if you add projection transformation, it can look better.

(if someone could embed the image, I'd appreciate, I don't have the reputation to do it)

Answer 3

I answered a similar question here

Try the numerical approach. pynomo has a numerical module called nomogen that might work for this problem.