The Original Odhner (Vintage Calculator)

[P55CxE9]

This was really pleasing to watch. I've seen those mechanical calculators on flea markets and in thrift stores (second-hand shop), and now I really regret not picking one up!

[BaDTomyBoy]

can anyone tell me the price of this thing?

[HifiCentret]

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.

[PS3Dragon4087]

wow, thats so cool. i want one of those now lol.

[funkycoder1]

He didn't actually explain that moving the decimal point works, but uses it at 8:40. To multiply by eg. 29 you don't actually need to turn the crank 29 times but just 11 times. (And to multiply by 1000 only 1 time)

[singingbanana]

@funkycoder1 Some stuff was cut for time.

[Tolstoievsky]

That's some crazy voodoo shit right there

[illustriouschin]

Not only is this machine mechanically brilliant but its clever interface allows it to do so much more, very nice!

[theTofuCube]

Your stuff is really interesting :D

You should apply to be a youtube partner!

[Strijdparel]

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?

Bye

[anticorncob6]

@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.

[Strijdparel]

@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 :)

[anticorncob6]

@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.

[handdriven]

Mathematical instrument of the Compt Room at Cambridge University Engineering Lab. Very interesting.

[im4shots2head]

@singingbanana I would just like to say that i performed the Maths Card Trick (Kruskal's Count) at my family Christmas party and everyone loved it! I'm a big fan! Keep up the good work!

[singingbanana]

@im4shots2head Fantastic!

[Artonox]

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?

[anticorncob6]

@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.

[Artonox]

@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

[anticorncob6]

@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.

[weeman7970]

Or you could use a digital calculator....

[11000010101]

that was awesome! =D

[TSorbera]

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).

[singingbanana]

@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.

[eaturfeet653]

thats what i use to take the SAT

[elemes09]

you know you're a nerd when you call a vintage calculator a "toy"

[patrikgluchowski]

how could someone make it... genius

[jmwintenn]

that thing is ballin'

[digdugdiggy]

awesome, I love outdated technology.

[Junming2010]

that's awesome mate!

[TyYann]

Same algorithm is used on an abacus for the square root.

Very interesting machine. Thank you!

[gtq838]

pretty sweet

[FHomeBrew]

I have an orhdner too so I'm really glad about this video :D Great job James! Although I use a different method for square roots it will probably amount to the same thing. The method I used is derived from the pen-and-paper version so it's quite a hand full. I like how these machines can go up to any number of significant digits (for division for example). I'm working on methods for logs now.

[fredrocks222]

I've always wondered how to figure out the square root

[spinfun]

It all looks like hocus pocus. How do you keep track of the decimal point? Like, if you want to do sqrt(20) then it would be 4.4721, different, no? And the sqrt(200) would be 14.1421 . Really interesting! And what about cube roots? is there a similar neat trick? How about sin(x), ln(x) and x! ???? Gee, you could spend a lifetime having fun working all these out. Yes, who needs a ti-84?

[FHomeBrew]

@spinfun The method for square roots can be generalized for any integer root (I think). I know of methods up to 9th roots and a general form but I don't remember them. I'd have to look it up. It's based on the same principle as explained in the first segment. "n!" can be done using triangular numbers I think. Might be mistaken though.

[anticorncob6]

@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.

[crazymann47]

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.

[nelsyeung]

Awesome! My teach once showed us a weigh calculator when he introduced moments. :)

[RYDERkN]

any mechanical graphing calculators?

[gtq838]

@RYDERkN lol

[BassmrDK]

Hefner

[orangetangopink]

Great video

[andruha11234]

phhffff who needs a stupid ti-84, if you can have this!

[FHomeBrew]

@andruha11234 No batteries :D

[andruha11234]

@FHomeBrew I know, right!? won't fail you on any test!

[MRfatermater]

guy looks like your father

[Namezzzzzzz]

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 ^^)

[mrsuperduck]

Wow......... I wana use one of these in a classroom.

[GammahooX]

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)

[Alfalotter]

lol the advanceds

[aannttee11]

Cool machine! 

[IssoMagicDude]

Interesting! it looks really cool

[jovaage]

brilliant!

[Zeldakitteh]

Started being confusing after the mention of series. D=