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This is a little clunky so bear with me.

I've been looking to buy a dowelling jig which, if you don't know, is a woodworking tool that lets you drill dowel holes into a piece of wood.

Some jigs will drill a hole right into the middle of the edge of wood.

But some come with spacers that let you offset that hole.

For example, the dowelmax has spacers that are these thicknesses (in inches): 1/8, 3/8, 6/8, 13/8.

If you need an offset that isn't one of these numbers, then you can combine them. For example, if I want an offset that is 1/2, I can add the 1/8 to the 3/8.

But I can also use subtraction: if I need an offset that is 5/8, I can put the 6/8 on one side and the 1/8 on the other. This is like 6/8-1/8=5/8. Pretty clever.

I worked out all the combinations I could do in a spreadsheet and these are the results:

+--------+------------------------+
| needed |         combo          |
+--------+------------------------+
| 1/8    | 1/8                    |
| 2/8    | 3/8 - 1/8              |
| 3/8    | 3/8                    |
| 4/8    | 1/8 + 3/8              |
| 5/8    | 6/8 - 1/8              |
| 6/8    | 6/8                    |
| 7/8    | 6/8 + 1/8              |
| 8/8    | 13/8 + 1/8 - 6/8       |
| 9/8    | 6/8 + 3/8              |
| 10/8   | 1/8 + 3/8 + 6/8        |
| 11/8   | 13/8 - 3/8 + 1/8       |
| 12/8   | 13/8 - 1/8             |
| 13/8   | 13/8                   |
| 14/8   | 13/8 + 1/8             |
| 15/8   | 13/8 + 3/8 - 1/8       |
| 16/8   | 13/8 + 3/8             |
| 17/8   | 13/8 + 3/8 + 1/8       |
| 18/8   | 13/8 + 6/8 - 1/8       |
| 19/8   | 13/8 + 6/8             |
| 20/8   | 13/8 + 6/8 + 1/8       |
| 21/8   | 13/8 + 6/8 + 3/8 - 1/8 |
| 22/8   | 13/8 + 6/8 + 3/8       |
| 23/8   | 13/8 + 6/8 + 3/8 + 1/8 |
+--------+------------------------+

My main question is: Is there a name for this sort of thing?

Also, Are there other choices of numbers other than 1, 3, 6, 13, that would do this?

The only thing I thought of that was perhaps related was denominations of coins. For that, in the UK there's 1, 2, 5, 10, 20, 50, 100, 200. The US uses 1, 5, 10, 25, 50, 100. These can be combined but you'd need large multiples of each since you can't subtract.

(I suppose the goal is to have few spacers, simple combinations, and a full span of combinations from 1/8 to the sum of all spacers. Is 1, 3, 6, 13 the best way?)

I did some scribbling to find other combinations, and I found that if you had just 1/8, 3/8, and 9/8, you can already make combinations from 1/8 to 13/8.

joriki
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Korgan Rivera
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    This is very close in spirit to the puzzle of selecting five integer weights to use in a balance scale to cover weights up to $100$. That was asked and answered on MathematicsSE in How to weigh up to 100kg with 5 weights. The answer there covers a good way to reach arbitrarily high totals in relatively few weights. – Mark S. Mar 18 '22 at 15:02
  • @MarkS. Thanks, I'll check that out. – Korgan Rivera Mar 18 '22 at 15:17
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    Why 1,3,6,13 and not 1,3,9,27? If you did that, then you could make 5 = 9-3-1, 6 = 9-3, 7 = 9-3+1, 8 = 9-1, etc, and your four spacers would take you all the way to 40 instead of only to 23. If the 1,3,9,27 solution works for you, then the answer is easy to describe: each spacer should be exactly three times the size of the next-smaller spacer. This is known by the name “balanced ternary representation”. – MJD Mar 18 '22 at 15:36
  • @MJD It works so I'm wondering if maybe the company uses 1, 3, 6, 13 for common wood sizes or something. – Korgan Rivera Mar 18 '22 at 15:40
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    Maybe they got tired of answering phone calls from people who can't figure out how to make 20 from 1,3,9,27? Or maybe there is some mechanical reason, if you use too many spacers at once the workpiece wobbles, or 1,3,6,13 is easier to manufacture reliably, or something like that? Or there could be some business reason: they want the opportunity to charge extra for a 24/8-inch spacer to the few people who will need one. – MJD Mar 18 '22 at 15:42
  • @MJD I'm thinking the same thing. I'm also wondering now if wood sizes are the way they are for this exact reason: so you can combine planks of wood to match the thickness of others. – Korgan Rivera Mar 18 '22 at 15:43
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    You might want to look at generating functions. The generating function for your problem is

    $$\left(x^{-1}+1+x\right)+\left(x^{-3}+1+x^3\right)\left(x^{-6}+1+x^6\right)\left(x^{-13}+1+x^{13}\right);. $$

    If you let Wolfram|Alpha multiply it out for you, you can see (under "Alternate forms") that all the powers of $x$ from $1$ to $23$ are in the product.

     – joriki Mar 18 '22 at 15:45

2 Answers2

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The mathematical answer is well-known: the spacers should be powers of 3: $$1,3,9,27,81, \ldots$$ and this is optimal in the sense that:

  1. It can make all the sizes up to some maximum
  2. Any other set of spacers that can make all the sizes up to that maximum requires at least as many pieces.

For example, the four spacers $1,3,9,27$ will take you up to 40 units instead of only to 23: $$\begin{array}{rl} 1 & 1 \\ \hline 2 & 3-1 \\ 3 & 3 \\ 4 & 3+1 \\ \hline 5 & 9-3-1 \\ 6 & 9-3 \\ 7 & 9-3+1 \\ 8 & 9-1 \\ 9 & 9 \\ 10 & 9 + 1 \\ 11 & 9 + 3 -1 \\ 12 & 9 + 3 \\ 13 & 9 + 3 + 1 \\ \hline 14 & 27 - 9 - 3 - 1 \\ \vdots & \vdots \\ 40 & 27 + 9 + 3 + 1 \end{array} $$

The system is known as balanced ternary representation.

The question arises why the jig company provides you with $1,3,6,13$ instead of $1,3,9,27$. Here there is no way to know without asking because real-world engineering and business issues will apply. In the comments, I speculated:

  • Maybe they got tired of answering phone calls from people who can't figure out how to make 20 from 1,3,9,27?
  • Or maybe there is some mechanical reason, if you use too many spacers at once the workpiece wobbles,
  • Or 1,3,6,13 is easier to manufacture reliably, or something like that?
  • Or there could be some business reason: they want the opportunity to charge extra for a 24/8-inch spacer to the few people who will need one.

There's a folklore story about a mathematician who figured out the dimensions for a steel can that would use the minimum amount of material to achieve the desired volume, and who informed the canned food company that their cans were suboptimal. They sent back a long and thoughtful letter explaining the many reasons their cans were not the mathematically “optimal” size. For example, it is desirable that all sizes of cans should have the same diameter, because in the factory the same capping and conveyor equipment can be used for every size of can. Often this type of practical consideration outweighs the mathematical considerations.

MJD
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  • (I want to find the letter about the can sizes, but haven't been able to. I'd be grateful to anyone who could provide a link.) – MJD Mar 18 '22 at 16:03
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    Off topic, but I once sat next to a guy on an airplane who worked for Coca-Cola, and he explained to me why soda cans are no longer cylindrical but are tapered near the top - it's because the metal used for the top is much more expensive, and the reduced area compensates for the difficulty of tapering the sides. – rogerl Mar 18 '22 at 16:06
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    I've emailed them to get the final answer to this. I'll report back if I hear anything. :) – Korgan Rivera Mar 23 '22 at 21:15
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    @MJD His response: "A good question. It was over twenty years ago that I designed the system, and at that time costs to manufacture were prohibitive, which reflected on the selling price. The 3/4" was designed for centring and also for offsets. A larger size would not be pleasing for offsets, but could be added to if necessary. The 1 5/8" spacer was designed to accomm. 4" stock, and the longer studs were compatible for installation of that spacer. A larger size would require more material, and longer studs." – Korgan Rivera Mar 23 '22 at 23:23
  • The text of the letter about optimal can sizes is reproduced at https://www.pleacher.com/mp/mlessons/calculus/mobcarna.html – MJD Apr 23 '22 at 15:37
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Once you have the $1,3$ you can make any number up to $4$. The next one could be as large as $9$, as $5=9-3-1$. From there you can make any number up to $13$, so the next could be $27$ as $27-9-3-1=14$. The sizes would continue as the powers of $3$. At each stage you can use a smaller number than the one given. The greatest you can use is one more than twice the sum of the numbers so far.

Ross Millikan
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