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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...

MarianD
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MathematicalOrchid
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    You definitely can't restrict yourself to straightedge-compass... – J. M. ain't a mathematician May 01 '12 at 12:53
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    The usual arguments show that ruler+compass points can only be solutions to certain equations, none of which are logarithms – Xodarap May 01 '12 at 12:59
  • 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? – MathematicalOrchid May 01 '12 at 13:02
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    Straightedge-compass limits you to things that can be expressed in terms of (possibly nested) square roots; neusis lets you do cubics. – J. M. ain't a mathematician May 01 '12 at 13:17
  • Right. So transcendental lengths are impossible. (?) – MathematicalOrchid May 01 '12 at 13:25
  • @MathematicalOrchid Yes, as well as many algebraic ones. – Alex Becker May 01 '12 at 13:47
  • OK, so no ruler and straight-edge. What about if we allow more tools? Does that help? – MathematicalOrchid May 01 '12 at 13:59
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    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. – MJD May 01 '12 at 14:08
  • 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. – Christian Blatter May 03 '14 at 11:33

3 Answers3

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Take a look at Descartes’s Logarithm Machine which I believe accomplishes your task.

Greg Buchholz
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Summarising all the comments so far:

Straight-edge and compass construction limits you to lengths that can be expressed as (possibly nested) square roots. Since the logarithm of most numbers is transcendental, you have no hope of constructing it this way.

That said, there's nothing stopping you from drawing equally-spaced ticks and marking then as powers of some number (.e.g, 1, 10, 100, 1000...) But we knew that already.

We also know that exactly half way between 1 and 10 will be $\sqrt{10}$, but that's irrational too, so it's hard to write down on the scale.

In short, it seems that rational approximations (calculated by some suitably tedious method) are the best we can do. I guess that's why Babbage designed the Difference Engine...

MathematicalOrchid
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Have a look at this in my write. It shows how one can find the logrithm of 2, 3, 5, by simple assumptions, like $5\lt 6$ We use nothing fancier than $25 \lt 27$ to get the red bit. The plot is a square, in log binary, of lg_2 3 and lg_2 5, multiplied by 12.

http://z13.invisionfree.com/DozensOnline/index.php?showtopic=718&view=findpost&p=22047479

But the OP asked about geometric construction of logs, so i suppose you can't do a compass and straight-edge construction if the ratio of logs isn't rational.

wendy.krieger
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