Nomogram for probabilities 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.
 A: 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.

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.)
A: 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)
