The Straightedge Computes a Determinant

Yesterday I drew a multiplication nomogram and it lied to me — three times six read as fourteen, because I’d plotted the middle scale running the wrong way. Once I flipped it, the straightedge landed true. But sitting there squinting down a ruler, I noticed my eye was doing something a computer does constantly: deciding whether three points fall on one line.

That decision has a name. Given three points, the orientation predicate returns the sign of a small determinant — positive for a left turn, negative for a right turn, zero when they’re collinear. It’s the atom of computational geometry: convex hulls, segment intersection, and point-in-polygon all reduce to asking it, over and over.

A parallel-scale nomogram is that predicate frozen into ink. For w = u·v I put logarithmic scales at x = 0 and x = 2, and a half-height scale at x = 1. The construction guarantees that log u, log v, and log(uv)/2 sit on one line — so laying the straightedge is evaluating the determinant, and reading the middle scale is finding where it hits zero.

import kotlin.math.log10

fun det(u: Double, v: Double): Double {
    val w = u * v
    val (y1, y2, y3) = Triple(log10(u), log10(v), log10(w) / 2)
    // scales at x = 0, 2, 1; collinear when this orientation test is 0
    return 2 * (y3 - y1) - (y2 - y1)
}

fun main() {
    for ((u, v) in listOf(3.0 to 6.0, 4.0 to 25.0, 7.0 to 11.0))
        println("%.0f x %.0f = %.1f   det=%.2e".format(u, v, u*v, det(u, v)))
}

The determinant lands around 10⁻¹⁶ — floating-point dust — for every pair. That near-zero is the alignment my ruler was hunting for.

Forth makes the same point from the decade when this predicate hardened into a discipline (Preparata and Shamos, 1985), back when every cycle was counted:

fvariable u  fvariable v

: nomogram ( f: u v -- )
  v f! u f!
  u f@ v f@ f* flog f2/     \ y3 = log10(u*v)/2
  u f@ flog f- f2*          \ 2*(y3 - y1)
  v f@ flog u f@ flog f-    \ (y2 - y1)
  f- f. cr ;

3e 6e nomogram   \ prints ~ -5.6e-17

The honest wrinkle is that real orientation tests care enormously about that near-zero. Round it wrong and your convex hull tears; misjudge collinearity by half a millimetre on paper and you read 14 instead of 18. The paper version fails loudly enough that you catch it; the numeric one can slip past silently. Same question, very different stakes for getting it slightly wrong.

I still owe myself the density-altitude nomogram: pressure altitude and temperature on the outside scales, an isopleth run across to the answer. Same determinant, warmer application.