The Original Odhner (Vintage Calculator)

>>James: Hello, everyone! Today you join me in Trinity College, Cambridge, with my guest today. This is Hugh Hunt, who will introduce himself.>>Hugh: Yes, I’m Hugh–hi! I’m a fellow of Trinity College. I’m a lecturer in the engineering department.>>James: And we’re here today in Hugh’s office which is a lot like my office, only much swankier! So it’s nothing like my office, in that case, apart from–>>Hugh: There’s lots of toys!>>James: Apart from having lots of toys, which is what we have in common. And this is one of Hugh’s toys, which is– he’s invited me over to see today. And this is one of those old mechanical calculators that they used to use in the early part of the 20th century. [calculator dings] And what is this called then? What’s its name?>>Hugh: Well, it’s called the Original Odhner.>>James: Odhner. Is it “odd-ner’ or ‘odd-en-er’?>>Hugh: Well, I don’t know.>>James: We’re not sure–we’re not sure. Can you tell me the history of this machine?>>Hugh: Well, these things date back to, I think, the late 1800’s.
>>James: Yep.>>Hugh: A Russian inventor came up with this mechanical adding machine,>>James: Right.>>Hugh: and really just worked out that multiplication is just glorified adding,>>James: Yes, yeah.>>Hugh: and division is glorified subtraction! And this was in the engineering department at Cambridge University,>>James: Mm-hmm. I see that.>>Hugh: It was probably in use for real, in the 1930’s, 1940’s, 1950’s. But then, when of course electronic– electrical computing machines came in, then in the 70’s and 80’s, then of course these things were just museum pieces. And I happen to have one.>>James: And you happen to have one, as well. Well, we’re not saying that you’re a museum piece, But– we are going to use this mechanical calculator today to do one of the more challenging operations, which is to find the square root of a number. But first of all, before we show you that, we’re going to show you some of the more basic operations. OK, so show me something simple. Show me how to add on this machine. So how do you add on a mechanical calculator like this?>>Hugh: Right. Well, let’s suppose I want–I’ve got a number like 4,698–>>James: Mm-hmm.>>Hugh: I key up for 4,698 onto here, and what I can do is just to crank this handle around once, and that’s added 4,698….
>>James: I can see it, yeah.>>Hugh: onto what was there before, which was zero. If I crank the handle around again, it adds it on.
>>James: So you’ve added it on twice.>>Hugh: Three times, four times. And it’s actually counting how many times I’ve added it on over here, so if I go five times, six times. Well, this is neat, because when I get to ten times, if I’ve added this number to itself ten times, well, I’ve just multiplied it by ten.>>James: So this shows that repeated addition is just another way to multiply.
>>Hugh: Mm-hmm.>>James: And that’s all we’re doing. So this is just repeated addition.>>Hugh: It is.
>>James: And how would we subtract with this?>>Hugh: Well, that’s easy. Just crank the handle backwards.>>James: I do like that! That’s really nice.>>Hugh: So I cranked the handle backwards ten times, and I’m back to zero.>>James: So if you want to subtract, you just turn the crank the opposite direction. That’s really easy. How would I divide a number with this machine?>>Hugh: So now, suppose I want to divide 4,698 by, say 162.>>James: OK.
>>Hugh: Well, what I’ll do is, first I’ll put that into this register here.
>>James: Right.>>Hugh: And now, I’m going to see how many times do I need to subtract 162?>>James: Right. So we’re dividing by 162–we’re just counting how many times we subtract it.>>Hugh: That’s right. Now, I could just put 162 into here, and then set my counter to count backwards for me, and I could just do this–one, two, three, four, five, and keep on going–I’m watching the number getting smaller and smaller and smaller, until eventually, there we are–zero.>>James: And so the answer is…
>>Hugh: 29.>>James: 29. So we got 29 on the counter here on the left-hand side. So if multiplication is just repeated addition, then I’m guessing that division is just repeated subtraction. OK, so now we’re going to work out the square root of a number, which is going to be a much more difficult thing to do. And there’s a nice little–surprising, really–little mathematical fact that you can do to work out the square root of a number.>>Hugh: That’s absolutely right, and it’s really nice. If I take the square root of something like four, then that’s 1 add 3.>>James: That is, yes.>>Hugh: If I take the square root of nine, that’s 1 add 3 add 5. The square root of 16 is one add three add five add seven.>>James: So you’re just adding up the odd numbers.>>Hugh: So if I’ve got 25 squares here,
>>James: Mm-hmm.>>Hugh: what I can do is I can work out that 25 is the square of 5 by subtracting 1 then 3 then 5 then 7 then 9. Five operations–5 must be the square root of 25.>>James: So to work out the square root of a number, you just subtract odd integers and the number of times you have to subtract is the answer.>>Hugh: And I can do that on this machine, so, for instance, I can put 25 into here, whiz that into the register there–OK. And now I go into subtracting mode, and so the first thing I’ll do is I’ll subtract off 1,>>James: OK.
>>Hugh: That gives us 24. Then I’ll subtract off 3. Then I’ll subtract off 5. Then I’ll subtract off 7. Then I’ll subtract off 9. And I’ve got nothing left. And how many times did I do it? Five. So 25 is a reasonably easy number. Let’s suppose I’ve got a number like 4489.
>>James: OK. 4489–that’s a big number.>>Hugh: Well, I happen to know that that’s 67 squared.
>>James: OK.>>Hugh: But what I’m interested to know is whether I can use some neat property of six and neat property of seven to get 4489. And what we discover is that 4489 is 60 squared,
>>James: Mm-hmm.
>>Hugh: plus 7 squared,>>James: OK.
>>Hugh: plus two lots of 60 times 7.>>James: Ah!
>>Hugh: Now if I’m really careful, I can subtract off the 60 squared bit,>>James: Mm-hmm.>>Hugh: and I can subtract off the 7 squared bit, if I can keep track of the 2 lots of 60 times 7 then I’ve got my square root of 4489.>>James: So we know how to find the square root of 60 squared, we’ve worked that out. We know how to find the square root of 7 squared,
>>Hugh: 7 squared–>>James: and we just have to keep track of the rest of it. How would I do this if I wanted to work out something like the square root of 2?>>Hugh: Well, 67–there’s only two significant figures.
>>James: Mm-hmm.>>Hugh: So the square root of 4489, we didn’t have to do that many operations to get our answer of 67. Square root of 2 is 1.414…whatever, lots of decimal places. lots of significant figures. So if I want to get the square root of 2, I have to work quite a lot harder. So let’s–let’s just do it. The nice thing is the algorithm, the procedure, of looking after all those 20’s and 200’s is all done by the machine So I’m going to set up my two up here. Now, I’m putting it way up here because I need space to look after my decimal places. I need all this space to look after my decimal places. So now I’ve put the two into this register. Now I’m going to do my subtraction. [calculator dings] Now, that seems like black magic, and actually that’s what mathematics so often looks like, and the beauty of it, is that if you can decipher that black magic, suddenly, it’s just–boom! It’s obvious. And it’s just fantastic with this machine– it’s a way of keeping track of things that, if you had to do it by hand, you would make mistakes. [calculator noises–no speaking] So, 1.4142135.>>James: 35? So this is going to be our square root. You’re going to check that, are you?>>Hugh: 1.4142136–I’ve managed to calculate the square root of 2 to eight significant figures. on a machine that, really, all it can do is to add and subtract.

59 Replies to “The Original Odhner (Vintage Calculator)”

  1. It just so happens that I just read about that machine yesterday. o/
    This is very interesting, thanks. =) (Also the algorithm for finding the square root is a neat trick)

  2. I saw machines like this in a museum i visited us with my friends it was very interesting. We saw calculators which used wights too they looked like clocks 🙂 but it was amazing to see those old machines we were not able to try them but i guess they are still working (there was a counting-machine too, it was used to count the population and it was able to save details like male or female god bless the inventors of old computer ^^)

  3. i feel like if i had that thing, i would spend hours turning that handle for no good reason. it certainly looks fun to use.

  4. James, could you please better explain how you run the calculator given that we know this equation is how it works?
    4489 = 60^2 + 7^2 + 2*60*7 = 67^2
    You basically jumped from "here's a simple case we're thinking of" to "let's calculate sqrt(2) without any explanation".
    And a better explanation on how the above equation works, and how it applies to something like sqrt(2) would be nice (e.g. someone mentioned that sqrt(20) is not like sqrt(200).

  5. @TSorbera We did explain it originally but I cut it for time. Find the digits individually:

    We have 67^2 = 100*6^2 + 7^2 + 20*6*7.

    Find 6 by counting how many odd integers need to be subtracted in the 100s column.

    Find 7 by inputing 2*6 into the 10s column (in other words that's 20*6) and then counting how many odd integers need to be subtracted from the units column. This will also count how many 20*6 need to be subtracted.

    Now one counter will read 0 the other counter will read 67.

  6. you know they are mathematical geniuses when they say "i just happen to know that 4489 is 67^2" (and they didnt happen to revise it before the camera :P).

    Either way that is incredible stuff. So assuming that we didnt know that the square root of 4489 is 67^2, im guessing that this is the method people use to find the square root of things.

    How about the cube root? and beyond that?

  7. @Artonox I know that n^3 + 3(n^2) + 3n + 1 = (n+1)^3. Perhaps you can list the values of 3(n^2) + 3n + 1 and keep subtracting them until you get zero.
    Fourth roots are much easier. All you have to do is take the square root of the square root. If you want to take the fifth root than I have no clue… but I do know the sixth root is the cube root of the square root.
    The eighth root is the square root of the fourth root, and the ninth root is the cube root of the cube root.

  8. @spinfun Who needs a ti-84?
    sin(x) = x – (x^3)/3! + (x^5)/5! – (x^7)/7!…
    This is only if you work in radians. cos(x) = sin(pi/2-x), an tan(x) = sin(x)/cos(x)
    So you can do trigonometry functions on this. But if you want ln(x), ask someone else.

  9. @anticorncob6 ahh ok, so basically from my understanding, all "multiples of powers of roots" (if thats the correct terminology) can be found.
    i.e. since we know how to find the square root, we can now find the fourth, eighth, 16th etc

    What about primes? This is probably out of my depth now D:

    As for the cube root, i thought of a giant cube made of little cublets (as in 3d squares)
    so the cube root of 1 is 1cube. Then add 7 cubes to make a 2x2x2 cube and so on. Your formula works in this case:D

  10. @Artonox I think I said I have no clue for the fifth root. But, if you take the average of the fourth root and sixth root, you will get an APPROXIMATION of the fifth root. Same thing for the seventh root, and so on. But if you want the exact answer, I'm not the person you should ask. This is all stuff I made up myself.

  11. Hey I have a question for you. Do you know the Game Set? It is a game where you have to find a combination of three matching cards.
    Now a friend and I had a bet. He wanted to know what the least amount of cards that creates the possebilety for a Set. I said you would be able to figure it out but he said it would be impossible. What we already know is that with twelve cards the chance is 12% that there would be a set.
    Could you help us out?


  12. @Strijdparel I assume you have a normal pack of 52 cards. With 26 cards, the only way you won't have a set is if you have 2 aces, 2 twos… to 2 kings. Once you get to 27 cards you have to have a set. If you have a 54 card deck, the biggest number of cards is 28, the exact same thing as before except for the two jokers are added, but there is no third joker for a set. But the LEAST to create a set is 3.
    Unless I didn't understand the question, that's the correct answer and you won your bet.

  13. @anticorncob6 Hey thanks for your help. But i did not mean a normal set of cards. I meant the game Set, it is a game where you have to find a pair with the minimum of three cards. It is a little difficult to explain, but if you type it in on google you will know what i mean 🙂

  14. @Strijdparel Okay, so I didn't understand.
    If you have 27 cards with number one, you have 27 cards without a set. You can also add all the cards with the number as 2, you cannot get a set because if you pick three cards two of them must share a number. This is 54 cards. Any cards with number 3 would have to have a set, as all the possibilities with shade, color, and shape are duplicated with both 1 and 2. So you can have a limit if 54 cards with no set.

  15. Yeah they're great fun those old machine. A great joy to calculate on and even more if runs smooth (newly serviced). Also great fun to calculate square on in front of people who only heard of electronic ones. Especially square roots make'em drop jaws.
    And the funny part is that it's really not magic. All it is is an automatic abacus doing the tedious accounting for you to eliminate certain kind of errors.

  16. I'm right now in front of one of these calculators and I found your video as a tutorial. I'm gonna make a video in portuguese about it! Many thanks, James!

  17. Just one small mistake, they guy who made it was Swedish, not Russian.

    He just happened to be in Russia when he came up with it.

  18. Here is a detailed explanation of the algorithm used to calculate square roots on a pinwheel calculator, using the example from the Original Odhner 227 Instruction Manual.

    I really like the singingbanana videos – keep them coming.

  19. Very nicely done demonstration, thank you.

    I have an Odhner from the fifties I picked up second hand here in Vienna.  It's an amazing device, and very well made.  I actually used it to do my taxes this year when my electronic calculator pooped out.

    That square root algorithm is a great "aha" experience- very elegant and revealing of the structure of a bit of math.

    thanks again for the great work, Scott.

  20. Got one of these, and there is one in the museum Boerhaave in Leiden / NL too, for public viewing (behind glass). A lot of mechanical calculators can be found in the museum of the 20th century in Hoorn as well. But I really like this model as it looks like a museum piece as well, it's just beautiful and really well made.

  21. Why haven't you used the possible of shifting the counting- and solution-register for multiplication and division? (So you turned the wheel 29 times. Otherwise, it would be only 2+9=11 times and one shift.)

  22. Funny thing about the electronic calculators you compare this machine to: adding was often all they could do as well. (I can't speak for the specific Casio model you showed; it may have a newer processor.) Many electronic calculators were based on 4-bit processors that could only add or subtract one digit at a time; this is true even today for basic calculators. The fundamental arithmetic operations on the keyboard actually invoke little programs which compute the result by repeatedly adding or subtracting individual digits, propagating carries, and, for multiplication and division, shifting the intermediate multi-digit values. More complex operations, like transcendental functions, are computed by series approximations that use the fundamental operations. Even many advanced scientific calculators operate this way, though they may have more powerful 8-bit processors like a Zilog Z-80. Only modern smartphones and similar systems have really advanced system-on-chip devices with the luxury of a complex arithmetic unit containing fast hardware multiply and divide circuits.

  23. Guys one question! I have a vintage privileg TW 708 calculator and I was wondering how rare is it? does it worth anything?

  24. There are other cool things we can do with these, like simplified multiplications using 10's complement for multipliers 6, 7, 8 and 9 reversing the crank 4, 3, 2, and 1 turns respectively, which considerably reduces the time and number of turns.

    Consider: 36 is 1+3+5+7+9+11
    There are six addendums in this series, so the square root of 36 is six.
    64 is 1+3+5+7+9+11+13+15
    There are eight addendums in this series, so the square root of 64 is eight.
    100 is 1+3+5+7+9+11+13+15+17+19
    Ten addendums in this series and the square root of 100 is ten.
    Remember that very simple series of additions: 1+3+5+7+9+11+13+15+17+19

     The top register is S.B. (setting board)
    The bottom right register is R.R. (results register)
    The bottom left register is P.R. (proof register)
    10.What is the square root of 966289?
    a. The figures 966289 are set up to the extreme left on R.R. (bottom right register).
    Clear S.B (top register) and P.R. (bottom left register).
    b. Split up the figures into groups with the aid of the decimal buttons from right
    to left and include two figures in each group, thus: 96/62/89.
    Put the carriage to the right so that the largest number of digits possible may
    be obtained in P.R.
    c. Put figure 1 into the S.B. track directly above the 6 in the left hand group.
    Subtract once. Change the 1 into a 3 and subtract once more. Continue in this
    manner with odd numbers up to and including 19, when you will hear the bell
    (overstep bell) which means you have subtracted once too often.
    d. Complete a plus turn (to counteract the last overstepped subtraction).
    Now change the odd number on S.B. to the nearest even number below : 18 (19 was
    one subtraction too many, so use 18).
    e. Move the carriage one step to the left and put up figure 1 on S.B.
    This will cause the number 18 to become 181. Now subtract by 181, 183, 185 and so on.
    When you come to 197 you will hear the bell, and you will then make a
    plus turn and exchange the odd number of 197 for the even number 196
    (again correcting for the one too many subtractions).
    f. The carriage will now have to be moved one more step to the left.
    Proceeding in the same manner as before, you add digit 1 to 196, making
    this 1961, and subtract again, first by 1961, then 1963 and finally 1965.
    The remainder of the figures in R.R. will disappear on the last subtraction,
    and this concludes the operation (no overstep bell heard for exactly 000000…).
    It is only for you to read off the result in P.R. where you will find the
    number 983. This is the square root of the given set of figures 966289.
    You can check this by carrying out the multiplication 983 x 983 which
    equals 966289.

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