First, if you are around Edinburgh, I am giving a talk tonight at the International Magic Festival on the psychology of magic and illusion.  Details here.

Second, on Friday I set this puzzle…

A long playing record has a 12 inch diameter.  The unused bit in the middle is about 4 inches in diameter, and there is a smooth outer edge one inch wide around the recording.  If there are 91 grooves to the inch, roughly how far does the needle move during the recording?

If you have not tried to solve it, have a go now.  For everyone else the answer is after the break.

About 3 inches – can anyone explain why?

I have produced an ebook containing 101 of the previous Friday Puzzles! It is called PUZZLED and is available for the Kindle (UK here and USA here) and on the iBookstore (UK here in the USA here). You can try 101 of the puzzles for free here.


    1. Agreed. It was pretty straightforward and I think most people understood the puzzle. Those that weren’t thinking clearly made up a different puzzle for their answer.

    2. Yes. It was an easy one. No doubt about the answer unless you were mislead by the grooves red-herring or confused the diameter with the radius

    3. Who would confuse a diameter with a radius? No-one could be that silly. Oh, that’s right…

  1. 12 inches less 4 for the centre and 1 for the outer edge, the needle moves across one side of the disk only, so 6 inches less the 1 inch edge less 1/2 of the 4 inch inner

  2. Well, if it’s on the equator, it’s rotating at around 1000 miles/hour, so a 20 minute LP – about 330 miles. Though we must also allow for earth’s orbit around the sun – roughly 67,000 mph – so that’s about 20,000 miles. The galaxy is spinning too, at (at our location) about 150 miles per second, so that’s nearly 180,000 miles. I’m starting to get quite dizzy and will therefore accept your answer of the distance between the inner edge of the outer lead in track and the outer edge of the inner lead-out track, which is the pretty close to the distance the needle moves, as opposed the the length of groove which passes under it. “Pretty close” because the needle doesn’t move in a straight line, it moves along an arc traced out by the tone arm as it rotates around it’s mount. And the centre of the earth. And the centre of the earth-sun system. And the centre of the galaxy. And who knows what else 😉

    1. Given the number of grooves per inch quoted, and that the dimensions match that of an LP (normal playing speed 33 and 1/3rd RPM), it’s easy to work out the play time is about 491 seconds, or a little over 8 minutes. That means your calculations are out by a factor of more than 2.

      Of course, movement is relative anyway, and the implied frame of reference is that of the record player, so all this is moot.

  3. I was first thinking of calculating the length of the spiral. Then I realized that wasn’t right, as the needle doesn’t go around but the record.
    I calculated 6 inches because I confused radius and diameter. Oops.

    Anyway, explanation: 12 inch diameter means 6 inch radius. And in the center there is a circle with 2 inch radius not used. At the outside there is 1 inch unused. 6-2-1 = 3 inches of used record.

  4. Despite the name calling and biothermal threats over the weekend I feel vindicated by the reveal of the actual answer.

    Taking into account the necessity of real world considerations (the need to remove the needle after playing of course) my answer has been shown to be completely correct. Indeed taking those real world considerations as fact it was more correct.

    I hope others have learned a lesson from my humble insistence on the truth despite considerable verbal opposition.

    That you to the rest of you for your support and kind words.

    1. @geodetective

      You may wish to throw your hat into the ring with @steve and @jonno and others who were using complex calculus to find a spiral length of around half a mile.

      Their efforts demonstrate how far they had been led astray by the puzzle’s wording.

      It may be better for your understanding to pay closer attention to those of us who got the right answer. And (in my case) identified why the precise answer is not reachable without knowing the true geometry of the tone arm. Our insights are proven right while those of the half mile spirallers has been shown to be well off the beaten track.

      It is your choice of course. But surely truth is better than spiralling into falsehood at every opportunity?

    2. Why should anyone pay attention to someone who can’t even spell their own name right?

    3. I think I’m going to write a paper on Barry Goddard. Not the man but the social construct. (I don’t know if he even exists as a man. Just because there’s an astrologer with the same name, it doesn’t follow that the two of them are the same man.)

      As the saying goes, “We’ve created a monster…”

    4. He’s real alright and he talks as much nonsense in real life as he does here, and with about the same level of introspection.

    1. It can’t travel in a straight line but a soft arc. The needle tracks a portion of a circumference, the arm on which it’s mounted being fixed to a set pivot point and can only follow this arc to and fro.

  5. Of course in a relativistic universe one could consider that the needle is stationary, and everything else (LP, Earth, Solar System, Milky Way Galaxy etc) is moving instead.
    But I got 3 inches as my answer and am happy with that 🙂

    1. It’s a spiral and there are 91 grooves per inch perpendicular to the alignment of the groove. Don’t give up so easily… 😉

  6. My first post this morning never arrived! On most record players the needle moved in an arc across the record. So slightly more than 3 inches. Also it moves in response to the width and depth of the grooves. Even more than 3 inches.

    1. That was covered by the puzzle asking “roughly how far does the needle move” and giving the answer as “about 3 inches”. These sort of puzzles are more about testing clarity of thought rather than real-world precision.

  7. Three inches is a reasonable approximation for the net movement of the needle (a more complete treatment would allow for the actual movement being along the line of an arc on the great majority of turntables).

    However, if the question is the total amount of movement of the needle, this neglects that caused by actually tracking the “wiggles” in the record’s groove. To know if this is significant or not, we would have to know a great deal about the “wiggle factor” of the groove which, of course is missing.

    However, it is possible to do a bit of work to see if it might be significant. The first estimate is play time, which is easy. Three inches with 91 grooves per inch allows for a total of 273 revolutions which, at the normal LP rate of 33 and 1/3rd RPM gives a (rather short) run time of about 491 seconds (LPs were originally specced for 22mins per side, later often extended by reducing modulation and cramming more narrowly spaced grooves).

    After a bit of research, using calculations by this chap

    If we look at the case of a 1khz signal at 20db over the reference 0db modulation level defined by the industry, then the displacement of the stylus is 80microns. Taking 160 microns peak to peak at 1khz over 491 seconds on a mono recording, the total movement involved in tracking would be about 78mm, or a little over 3 inches.

    So, in conclusion, for a very loud 1khz signal, the total tracking movement of the stylus would be comparable to the movement of the arm. For a more densely packed LP, it could be somewhat higher.

    Of course, before anybody else points it out, there is only one groove per side, so the “grooves per inch” is better described as the number of rotations of the record required for the tracking arm to move one inch.

  8. Only three inches if the arm the needle is on is infinitely long, such that the arc becomes a straight line.

  9. The needle is anchored in position and moves in an arc across the playing surface – but for the level of precision implied in the question, the arc can be approximated by a straight line.

    The LP is twelve inches in diameter, but there’s a one inch band around the outside which merely serves to direct the needle to the start of the recording (it’s likely most people would place the needle close to the inner edge). The combination of the inner label and ending zone totals four inches. The needle doesn’t cross the central spool, so we can ignore the distal half of the record in our calculations. So (12/2) – (4/2) – 1 = 3″

    Courtesy of online roll length calculators:

    Groove (tape) thickness: 0.279120879mm (= 1/91 inch)
    Inner diameter: 101.6mm (= 4″)
    Outer diameter: 279.4mm (= 11″)

    It claims the groove (tape) length would be ~190.61m and one (at giangrandi dot ch) claims there’d be 318.5 turns. The latter also gives the necessary formulae (for those fond of integral calculus!)

    1. @ChrisR – D’oh! it helps to remember that if it’s a 1″ unplayable area all around the disc, then you have to subtract 2 * 1″ = 2″! 🙂

      So corrected maths:

      Groove (tape) thickness: 0.279120879mm (= 1/91 inch)
      Inner diameter: 101.6mm (= 4″)
      Outer diameter: 254mm (= 10″)

      Length: 152.252m, exactly 273 turns.

    2. Damn lack of edit facility. 152.252m would be using the “approximate formula” (i.e. concentric circles). The exact formula (using integrals) gives a length of 152.491m. Especially If it was a roll rather than an LP, the extra 239mm (~9.4″) might make all the difference!

  10. The stylus is usually at the end of an arm which pivots about a point just beyond the surface of the record.
    It is the disc that moves (rotates) while the needle simply moves in towards the centre in what is almost a strsight line. The distance moved from track 1 to the end of the recording is a linear 3 inches. The needle will move this distance modified by a slight arc of the radius of the tone arm.
    Cute puzzle.

  11. Well, I got ‘about 3 inches’ too, but only when I ignored the wording ‘how far did the needle move during the RECORDING’. LP’s are pressed and no needle is used, but the recording is done either on tape (if it’s old) or digitally. So the needle didn’t move at all during the recording, it moves during the playing.

    1. Prior to the 1950s, the master was recorded using something called “direct to disc” which did involve recording directly to a metal master at the time of recording. Whilst LP format discs weren’t available at the time, during the late 1970s the technique was revived for a very few albums which would, of course, involve the cutting head moving. Whether such a cutting head could ever be called a needle is another thing.

      Of course, you are right. Richard’s wording is sloppy. He should have said during the playing of the recording. As it stands, we have no way of knowing how far the needle moved during the recording. The owner of the turntable could have been playing quite a different record, or even records (given the recording process in studios can be lengthy) or no record at all.

  12. I dived right in and calculated the length of the groove, through which the needle travels relative to the record, to be about 152m. I’m generally a well balanced, Myers-Briggs Type INTJ, decent and law-abiding person and am happy with that answer.

  13. When I saw this one I started doing calculations, trying to figure how I’d calculate a series of ever shrinking circles. I was starting to think some sort of calculus derivative for calculating sums was going to be necessary.

    Then I realized “Oh… how far is the *needle* moving.” While the record is moving under it, the needle doesn’t actually move except from the edge of the record when it starts playing to the other edge when it stops. It just travels the 3 inches where there’s stuff on the record.

    On a side note, 1″ margins on a vinyl? That’s a lot of wasted space.

  14. If you calculate the arc of the radius of the stylus arm, then the correct answer is 3.18 inches (to 2 decimal places)

    1. Exactly. I always have the suspicion that people using inches also sacrifice their first born son whenever occasion calls for it.

  15. It depends on the unit of measurement, and the spurious detail that there are 91 groves per inch implies that the answer should be given in groves, not inches. Since there is actually only 1 groove on an LP, the stylus does not move at all. It is in the same groove at the end that it was in at the beginning.

  16. Puzzle may be ambiguously interpreted – “roughly how far does the needle move during the recording?” How far FROM CENTER does the needle move during the recording? Answer is 5 inches. How far FROM EDGE TO EDGE does the needle move during the recording? Answer is 3. Answer to question “How far does the needle move during the recording?” may be related to circle circumference. In this case, travel distance is calculated with simple integration and circumference formula. Answer is 18 * \pi or 56.55 inches. If somebody interested I can post full solution.

    1. It’s better to show solution in picture but this blog doesn’t support posting pictures. I try my best to explain it in English.
      Total traveling path of the needle is area below function f(x) = \pi * x . The function is circle circumference by diameter. Area can be found by using integral in range from d_1 = 4 to (D - 2 * d_2) = 12 - 2*1 = 10
      So here’s integral:
      \int_{d_1^{D-2*d_2}} \pi * x * {d}x

      Solution is \frac{\pi}{2} * (10^2 - 4^2) = \frac{\pi}{2} * 84 = 42 * \pi = 131.95 inches. I made a mistake with final answer in previous post. Correct me if my final result is wrong again.

    2. An interesting debating point. The point made by the answer and by many comments above is that the only movement made by the arm is the 3+ inches as it moved in towards the centre of the record. It is the turntable that does all the moving not the needle.

      I guess the similar analogy is someone running a mile on a treadmill. The belt does all the moving and so from one perspective, the runner is more or less running on the spot and so the distance travelled is zero!

    3. There’s something wrong here if you are trying to calculate the length of the groove.
      I think you’ve got the area of the playable surface – which is more easily calculated using formula pi*r^2, and is rather like 66 sq in. (can’t see the flaw in your maths which gives you twice that at the moment, but a record of diameter 10in fits inside a 10in square!)
      But if I’ve misunderstood I’m sorry.

    4. @ChrisR Sum of all lengths of grooves equals area of playable surface. The area can be seen as integral of f(x) = \pi * x (where x is diameter, you are using formula with radius \frac{\pi}{2} * r ) as well. Result looks like square inches but it is not. I double check myself 42 * \pi= 131.95 inches is final answer. Roughly needle traveling distance is not depending on number of grooves per inch in simple approach.

    5. Sorry Alex, but you are mistaken.
      The groove is a spiral, with a finite width, and at 91 grooves per radial inch there are 273 turns to the spiral with the outermost one being approx 30 inches long, and the innermost one being approx 12 inches long. The length of this spiral is approximately 500ft.
      If the groove was a spiral with just one turn it would be a very short recording and the groove length would be approx 20 ins.
      Your 130 inch long groove would see the record go round just 4 or 5 times.

    6. Thank you @Alex for at least making a sensible estimate of the length of an LPs groove.

      Even though it was not the asked question it was irksome to read @Steve and @Jonno coming up with absurd answers like half a mile. No one got time for that degree of irrelevance.

      If someone is going to answer the wrong question at least get it right.

    7. @ChrisR , thank you for finding mistake.
      It takes 273 revolutions of disk. So sum of 273 revolutions/circles will be sum of arithmetic progression \frac {273 *(4+10)}{2} * \pi=6003 inches or 500 feet and 3 inches.
      @ChrisR , good job.

    8. Looks like you were wrong again Barry (or is that Brry?) as Alex confirms. No surprises there. If you actually had the basic arithmetical ability that you claim to possess, you would have seen that for yourself, but you were too interested in trying to to discredit Steve and Jonno who never said what you claim (I checked – Jonno hasn’t even commented on this puzzle and Steve just said that you were wrong (and he was right)).

      As for people answering a different question, that’s a bit rich because you’re the one that usually does that.

  17. Records are also not usually perfectly flat or perfectly centered on the turntable both of which could add significantly to the movement of the needle.

    1. If you remember that the area of a circle is (pi) × r² and recognise that ellipses have 2 “radii” (semi-major and semi-minor axes – the long one and the short one), then it is only one more step to realising that the area of an ellipse is:
      A = (pi) × a × b (where ‘a’ and ‘b’ are the semi-major and semi-minor axes)

  18. I am puzzled why people are finding the Friday puzzle puzzling.

    To have a score of ZERO I must get FIVE questions RIGHT for every EIGHT that I get WRONG.

    Yet if I only answered THIRTEEN questions (FIVE+EIGHT) I would have only completed half the test.

    Thus I must have got TEN RIGHT and SIXTEEN WRONG.

    This is basic schoolboy arithmetic that even an arts graduate should have no problem with. It does not even need a “similteneous” equation to be solved instantly.

    1. i think some of us prefer the more rigorous schoolgirl algebra, using two equations and two unknowns.

    2. Interesting…….
      first you use two equations (i.e. simultinious)
      1) 5F = 8W
      2) F+W= 26
      and then you say it doesn’t need a “similteneous” equation.

      Interesting word game.

    3. @Another Non Mouse

      I fear you may have a fundamental misunderstanding of mathematics. I am happy to help you with your understanding.

      5R = 8W is the only equation.

      That there are 26 questions is a given. I believe the right term for that is an identity or an invariant. Either way 26 is not an equation.

    4. Yes, we’re definitely given two equations at the start. We’re told that the sum of correctly answered and incorrectly answered questions is 26 and we’re told that the accumulated points awarded for those questions is 0.

      The answer is then arrived at through elementary algebra – calculation using variables – rather than arithmetic which uses only numbers.

    5. @Barry Goddard
      Equation (noun); mathematics: a statement that the values of two mathematical expressions are equal (indicated by the sign =).

      Expression (noun); mathematics: a collection of symbols that jointly express a quantity.

      Invariant (noun); mathematics: a function, quantity, or property which remains unchanged when a specified transformation is applied.

      R+W=Invariant is indeed an equation. The equation 5R=8W is insoluble (and totally meaningless) without this additional equation.

    6. Thank you for your various attempts to explain. Yet there remains the simple fact that we have only one equation.

      That there are 26 questions is not in doubt. There is nothing to solve there. It is a given not an equation.

      What we have to solve is how many right and wrong answers given we know the points that accrue for each. Two variables in one equation. I successfully solved it. Others may have needed additional methods to do so. I did not. That makes the method I used simpler I believe than the “similteneous” approach others are suggesting.

      To each their own of course. Still I prefer simplicity over a rigid adherence to methods learn in school.

      We will learn on Monday who is right about this,

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