2
$\begingroup$

You have an obscure job in 1600's Brussels. You make reference sticks. You own a high-quality ruler (which are quite rare in that era) and use it to cut lengths of wood or metal to certain lengths and send them out the various craftsmen who use them as a reference when performing their craft. For example, the King requested that all rapiers for his army's lieutenants be of a certain length, so you created a set of metal rods that were sent to dozens of different blacksmiths and scabbard makers so those rapiers and their sheaths are made (nearly) identically.

Your father was quite wealthy and you inherited his 4-foot ruler when he died, while your older brother got all of his money. The markings on the ruler are at every ligne, which are 144 to a foot. However, firearms are becoming increasingly popular and you know that craftsmen want more accurate reference sticks these days.

You take a trip to the master ruler-maker in the city (the greatest in the world) and inquire about making a finer ruler. He said he could make you a new ruler with any number of divisions you like, so long as they are equally spaced and no closer together than .8 ligne. He also notes that you, being far poorer than your father, only have enough money to purchase a ruler up to 12 ligne long.

You are able to use your current ruler and this new ruler in tandem, however you can't justify this purchase if you only improve your accuracy from 1 ligne to .8 ligne. Is there a way to get a better accuracy? The optimal answer has the highest accuracy, not the simpler method.

Rules:

A recent question about measuring ran into trouble with what was allowed in an answer, so I'll be explicit about what is allowed:

  • You may treat the two rulers as having perfect markings. You can compare two markings (or ends of an object) and declare that one is “to the left” or “to the right” of the other. You can perfectly align two markings (or ends of an object) if you choose to do so.

  • You are NOT a ruler-maker. If your solution involves transferring markings onto one the rulers or onto some other object, you cannot do this perfectly. Your transferred marking is no more accurate than your ability to measure it would be

  • You don't own an angle-measuring tool (at least not an accurate one)

$\endgroup$
4
  • $\begingroup$ The careful wording seems to imply that the length of the new ruler is fixed, but it doesn't quite say so anywhere. Can I choose any length for the new ruler, as long as it's 12 ligne or less? $\endgroup$
    – Bass
    Jun 26, 2023 at 14:58
  • $\begingroup$ @Bass yeah up to 12 ligne, just don't get the false impression you can get 12 1-ligne rulers. You're just buying one. $\endgroup$
    – Skosh
    Jun 26, 2023 at 16:27
  • $\begingroup$ @Skosh Could you buy a 12-ligne ruler and cut it after every other number, giving you six 1-ligne rulers each with 2 markings? $\endgroup$ Jun 26, 2023 at 19:03
  • $\begingroup$ @NuclearHoagie I guess I'd lean on rule 2 to say that your cuts cannot be more accurate than your ability to measure them. $\endgroup$
    – Skosh
    Jun 26, 2023 at 21:34

2 Answers 2

4
$\begingroup$

As opposed to codewarrior0's lazy approach, here's the toilsome way:

Have the new ruler divided into 8 parts, so that there are 7 markings between the two ends. But very importantly, request that (ostensibly, to save some money) the total length of the new ruler not be 12 ligne, but a bit shorter, namely

the diagonal of a $8 \times 8$ ligne square,

which handily keeps the line separation well over the 0.8 ligne threshold, at

$\sqrt{2}$ ligne, which is irrational with respect to the integer distances measurable by the other ruler, meaning that there is no integer $m$ whatsoever so that $m\sqrt{2}$ is an integer, which also means that $m\sqrt{2} - \left\lfloor{m\sqrt{2}}\right\rfloor$ (m of the new ruler units, but removing the integer part) is different for every $m$.

This allows you to measure

back and forth, counting an integer amount of units forward using your old ruler, and an irrational amount of units backwards using the new one, until you get "close enough" to the target distance,

which will eventually enable you to measure with the resolution of

"arbitrary precision" (or "precision within any given non-zero error limit") by establishing ever tighter upper and lower bounds for the measurement.

Of course this means your eyesight and the markings must be perfect, and that you don't make any errors whatsoever, all of which was explicitly deemed possible at the question's somewhat lengthy setup section.

The problem might arise with the final requirement of [not] unreasonable from the perspective of “a 17th century person needs to measure different length objects, dozens of times a day”, but that is of course easily remedied by only measuring to more lenient tolerances.


Here's a simulation (Try it Online!) of measuring a stick with a (previously unknown) length of Pi to the precision of one microligne using these two rulers.

$\endgroup$
5
  • $\begingroup$ I don't understand how you implement this method within the lengths of the given rulers without making new marks of arbitrary precision on something (which is disallowed by the puzzle rules) $\endgroup$
    – fljx
    Jun 27, 2023 at 7:00
  • 1
    $\begingroup$ @fljx Align the marking on one ruler against the marking on the other, and keep count of the units you have measured in each direction. $\endgroup$
    – Bass
    Jun 27, 2023 at 7:05
  • 1
    $\begingroup$ @fljx No markings are required. You move the second ruler to align a pair of markings, then you move the first ruler to match a different pair of markings, then repeat. $\endgroup$
    – Sneftel
    Jun 27, 2023 at 7:38
  • $\begingroup$ Well I did say I'd be more critical of the ones that seemed complicated. There's a lot of ruler movement in this and I can't quite tell if you're taking advantage of something you shouldn't be able to. I would argue you can't have a point in the process where neither ruler is touching the measured object, and assuming that's a fair ruling I think you're going to end up with some physical space issues (I'm picturing the short ruler and the measured object want to be in the same location next to the big ruler) $\endgroup$
    – Skosh
    Jun 27, 2023 at 20:42
  • 1
    $\begingroup$ @Skosh of course I'm taking advantage of everything you gave in the puzzle, doesn't seem likely that you can get infinite precision otherwise. :-) I don't suspect I'll be telling you how you could easily keep a ruler touching the object the entire time, because if the rules can change once, they'll surely keep changing until the suboptimal answer you had in mind becomes the only possible one. ;-) $\endgroup$
    – Bass
    Jun 28, 2023 at 3:04
2
$\begingroup$

Lazy answer.

This is a question about a Vernier Scale. With 1.0 ligne markings on one scale and 0.8 ligne markings on the other, the limit of your accuracy is their difference: 0.2 ligne.

The answer above this one goes into detail about how to use this device and the principles behind it. The answer below this one asks you to do trigonometry to three or more decimal places.

$\endgroup$
6
  • 3
    $\begingroup$ The rules state that the new ruler's markings can be no closer than 0.8 ligne, but not that they cannot be placed any more precisely than 0.2 ligne. We can get a scale of arbitrary precision by making the new ruler's divisions arbitrarily close to the old one's. Get the new ruler made with increments of 0.99999999 ligne, not 0.8 ligne, and enjoy 0.00000001 ligne accuracy. I suppose we get some limit on the length of items that can be measured, though. $\endgroup$ Jun 26, 2023 at 15:07
  • $\begingroup$ Oh, good. I was afraid I had accidentally given the right answer. $\endgroup$ Jun 26, 2023 at 15:24
  • 1
    $\begingroup$ @NuclearHoagie I think that, because you can only buy a ruler at most 1 inch long, that the best you can do is to have 13 divisions which gives an accuracy of 1 - 12/13 = 1/13 lignes $\endgroup$ Jun 27, 2023 at 5:22
  • 1
    $\begingroup$ @2012rcampion There is no requirement that you buy a 12-ligne ruler with 13 divisions. Instead, get a 9.99999-ligne ruler with 10 divisions, yielding an accuracy of 0.00001 ligne. This ruler obeys all the rules as it is shorter than 12 ligne and has evenly spaced lines no closer than 0.8 ligne. The key is that the new ruler can be made to measure lengths of arbitrary precision. $\endgroup$ Jun 27, 2023 at 13:25
  • $\begingroup$ @NuclearHoagie Yes, you can make a ruler to measure some lengths of arbitrary precision, however the setup you describe can only measure the lengths 0, 0.000001, 0.000002, ... 0.000009, 0.00001, 1, 1.000001, 1.000002, ... If you have an object of length, say, 1.5, your accuracy is still only +/- 0.5 (technically +0.5 -0.49999). In order to measure any length (within the range of your 1 foot ruler) to an accuracy of 0.000001, you would need a vernier scale 999999 lignes long. $\endgroup$ Jun 27, 2023 at 13:56

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.