5
$\begingroup$

Can you draw a logarithmic scale just using some clever geometric construction? Or can it only be done using an actual table of logarithms? logarithmic scale (It's obviously trivial to draw a linear scale. It isn't hard to draw a scale where the spaces between tick marks doubles at each step. But I can't think of a way to get a logarithmic scale.)

I'm not especially worried about exactly which operations are permitted. I'm really just interested in whether you can make a slide rule without doing a bunch of pencil and paper calculations first...

  • 1
    You definitely can't restrict yourself to straightedge-compass...2012-05-01
  • 1
    The [usual arguments](http://philosophyforprogrammers.blogspot.com/2011/09/angle-trisection-for-dummies.html) show that ruler+compass points can only be solutions to certain equations, none of which are logarithms2012-05-01
  • 0
    I'm not completely sure, but I gather that for "most values", the logarithm of that value will be irrational (indeed, transcendental). Does that mean that such lengths are "difficult" to construct?2012-05-01
  • 1
    Straightedge-compass limits you to things that can be expressed in terms of (possibly nested) square roots; neusis lets you do cubics.2012-05-01
  • 0
    Right. So transcendental lengths are impossible. (?)2012-05-01
  • 0
    @MathematicalOrchid Yes, as well as many algebraic ones.2012-05-01
  • 0
    OK, so no ruler and straight-edge. What about if we allow more tools? Does that help?2012-05-01
  • 1
    Usually when I make a slide rule by hand I do it by calculating rational approximations to the logarithms. For example, $2^{10}\approx10^3$, so $\log_{2} 10\approx {10\over 3}$, and then I put a mark at $10\over 3$ and label it `10`. The `10` mark should really go at 3.322, not at 3.333, but I cannot mark a piece of paper that accurately anyway. To make a slide rule that can calculate to an accuracy of three decimal places, you only need to calculate the logarithms to three decimal places.2012-05-01
  • 0
    When accuracy is not too much of an issue use that $\log_{10}2\doteq0.3$, $\log_{10}4\doteq0.6$, $\log_{10}5\doteq0.7$, $\log_{10}8\doteq0.9$ between $10^n$-ticks.2014-05-03

3 Answers 3