Before calculators

[J HancockParticipant @jhancock95746]

Who remembers how to find the square root of a number with a tape measure, piece of string and a pencil. ?

For example 6.

Draw a line 6 units long, mark , add one unit of length .

Now draw a semi-circle radius 3.5 units.

Draw vertical line up from the 6 mark, measure length where it crosses circle.

Answer 2.44948 units long.

Might come in handy to know one day.

[J HancockParticipant @jhancock95746]

[HopperParticipant @hopper]

Some day when my smart phone can't access a calculator. And someday when I have a tape measure that reads to 5 decimal places!

Neat trick though.

Before calculators we used sly drools at school and I am sure they could do this calculation but I couldn't work those things out back then, let alone remember if they could even do it today.

Edited By Hopper on 15/04/2022 10:01:56

[Anonymous]

Never heard of the method. I was taught how to use a slide rule in the third year at secondary school, after which they were never mentioned again. I still have my slide rule and know how to use it. There is an x^2 scale so square roots are direct reading.

Andrew

[JasonBModerator @jasonb]

Though if you use 1m as your unit on the tape measure then should not be too hard to split the 1mm divisions and get 2449.5mm which would equate to 2.4495units

But you would need a sharp pencil and also a pin for the other end of the string

Edited By JasonB on 15/04/2022 10:12:45

[Paul LousickParticipant @paullousick59116]

Still have my slide rule and does not need batteries.

[not done it yetParticipant @notdoneityet]

Who remembers how to find the square root of a number with a tape measure, piece of string and a pencil. ?

I most certainly have never known of that method.

But, there again, I have never seen a tape measure that would provide a value for the square root of six to that may significant figures ( 2.44948).🙂

[HopperParticipant @hopper]

Posted by Andrew Johnston on 15/04/2022 10:10:41:

Never heard of the method. I was taught how to use a slide rule in the third year at secondary school, after which they were never mentioned again. I still have my slide rule and know how to use it. There is an x^2 scale so square roots are direct reading.

Andrew

Yes I think we moved on to calculators in fourth year at secondary school. They were a fairly new thing then and regarded by teachers with great suspicion, and were not let loose among the lower years lest they be corrupted by them.

[Peter G. ShawParticipant @peterg-shaw75338]

I was taught at secondary school to use 4-figure logs to base 10 which I found quite fascinating. I was never taught to use anti-logs, so worked it out for myself, then asked the maths teacher who effectively said,if it works for you, do it.

Slide rules, of which I still have three, were the province of Technical College maths. All three have the A, B, C & D scales which make square roots dead easy, but only the 5" Helix A50S has the K scale for cubes & cube roots. Simple means, simple devices, relatively coarse in precision, yet sufficiently good enough for most, if not all, practical purposes. The only problems are the requirement to be able to work out the multiples of 10 required (sorry, forgotten the correct term here), and the difficulty in learning how to use them.

As far as square roots are concerned, there is a method, which I've now forgotten, whereby they could be calculated using the old, original 4 function plus K constant calculator. I seem to think that the method could be done longhand. Plus, I have a recollection that an expansion to the key sequencing could be used to produce cube roots.

Ah, happy memories.

Peter G. Shaw

[CirclipParticipant @circlip]

Still got my 'Napierian and other Log tables for skools', invaluable, Batteries NEVER run out.

Regards Ian.

Edited By Circlip on 15/04/2022 11:02:31

[Gary WoodingParticipant @garywooding25363]

In addition to using a tape measure, string and pencil to get square root 6, who can remember how to divide a line into any number of equal parts using just a straightedge and pencil?

[SillyOldDufferModerator @sillyoldduffer]

Posted by J Hancock on 15/04/2022 09:53:39:

…

For example 6.

Draw a line 6 units long, mark , add one unit of length .

Now draw a semi-circle radius 3.5 units.

Draw vertical line up from the 6 mark, measure length where it crosses circle.

Answer 2.44948 units long.

…

Help! I must be doing it wrong.

Baseline 6 units in white, 1 unit extension in red, radius 3.5 circle in blue, wrong answer in yellow = 3.61

?

Dave

[AdrianRParticipant @adrianr18614]

Help! I must be doing it wrong.

Dave,

Try drawing a circle at the centre of the line

[BazyleParticipant @bazyle]

Anyone remember the 'Curta' hand held mechanical calculator? My physics master had one then got an early HP calculator about 1970.

Calculators can be too precise though. I mentioned on here a few years ago how some people had been speculating on how a device from the '20s had been designed as a particular dimension 'needed' to be accurate to a few tenths to avoid accumulated errors. Spreadsheet calculations which involved a cosine did not seem to work out. However when the cosine was reduced to the less accurate value from 4 figure tables as would have been used in 1920 the calculations 'rounded out' nicely.

edit – just checked there is a 3D print design for the Curta

Edited By Bazyle on 15/04/2022 11:42:31

[AdrianRParticipant @adrianr18614]

I was educated after slide rules and before calculators so it was all meant to be done the hard way. I was taught about log tables and used those, plus my brother being that much older was taught and had a slide rule. He didn't need it so it became mine and I found out it was not against the rules to use one. I was the only one to use a slide rule in my entire year.

I can still remember buying my first calculator at Uni, still have it too.

 

I have this in my watch list as a thing to do some day https://www.youtube.com/watch?v=HBdVVFqTrUU

Edited By AdrianR on 15/04/2022 11:41:49

[Philip RoweParticipant @philiprowe13116]

I recall when Sinclair brought out his first calculator in the early seventies it was around £60 – £70 if my memory is correct. Browsing the stationery alise in a local supermarket the other day l spotted a perfectly respectable scientific calculator for less than £4! Phil

[duncan webster 1Participant @duncanwebster1]

SOD needs to move his circle 3.5 units to the right. It works because the vertical is then sqrt(3.5^2 – 2.5^2) from pythagorus. Using the difference of 2 squares, 3.5^2-2.5^2 =(3. 5+2.5)*(3.5-2.5),which is 6

I'm not sure how you can be "after slide rules". I went from log tables to slide rules to calculators, missed out on those mechanical things where you wound a handle till it went 'ping', but there was one in the first DO I worked in, and it was reputed that one of the old guys knew how to work it. To use the first electronic one you had to beg the chief engineer for a borrow, and sign for it. It was the size of a decent book.

[Peter G. ShawParticipant @peterg-shaw75338]

Philip,

I think that Sinclair's first calculator was the Science of Cambridge retailing at £39-95. When it came out, Practical Electronics was running a series on building your own calculator for £70-£80. PE admitted they could no longer compete, but they were going to complete the series for information whilst accepting that no-one would be building it.

Gary,

That's easy: draw your base line, length immaterial. Erect a vertical from one end, then using a measuring device, tape measure, rule etc, draw an angled line from the other end point such that the point where the angled line crossed the vertical line was also the number of divisions required, eg, for say, 6 divisions, use 6 inches, or andeed any multiple of 6. Now drop verticals from the angled line at the appropriate points down to the base line, eg 1, 2, …5, 6. Ergo, the base line is nicely divided. Similar triangles in geometry I believe?

Cheers

Peter G. Shaw

[Anthony KnightsParticipant @anthonyknights16741]

There is an arithmetic method of finding a square root using something resembling long division, but having never used it in nearly 70 years, I can't remember how it works. If I need a square root these days (not very often) I use a calculator, although I still have a couple of slide rules and some log tables stuck in a drawer somewhere.

[HopperParticipant @hopper]

I remember in the late 1960s looking in the window of the hobby shop at a DIY kit to make your own "calculator". It was a big flat box about like two shoe boxes, with two big black knobs and a meter like an analog volt meter on the top. The two knobs had long pointers on them and were surrounded by rows of numbers, and the meter had rows of numbers on it.

Don't know quite how it worked but probably two potentiometers and a volt meter with some kind of simple circuit to give an answer you read off the corresponding range on the meter. Must have had some kind of switches to select add, subtract, multiply, divide etc. Presumeably you chose the mode, set your first number on dial one, second number on dial two and hey presto by magic the needle on the meter indicated the answer.

I remember marvelling at it and wishing I could have one for doing my arithmetic homework. Seems laughably primitive today. I can't for the life of me remember the brand name or details of it, despite having lusted after one for several years.

[JAParticipant @ja]

I have recently rediscovered my slide rule and started using it. If you only need an accuracy of three significant figures it is far quicker and less tiring on my artheritic fingers than a calculator.

As a student, or school kid, if you could not afford a slide rule there were always log tables. I remember a class at tech splitting in two and having a race between slide rules and log tables.

I remember being shown how to calculate a square root at school which was far more complex than using logs. Using square root tables was even easier.

JA

[HopperParticipant @hopper]

The calculator kit I remember may have been something like this. "As accurate as a slide rule!!"

Those were simpler times.

[SillyOldDufferModerator @sillyoldduffer]

Posted by duncan webster on 15/04/2022 11:53:43:

SOD needs to move his circle 3.5 units to the right. It works because the vertical is then sqrt(3.5^2 – 2.5^2) from pythagorus. Using the difference of 2 squares, 3.5^2-2.5^2 =(3. 5+2.5)*(3.5-2.5),which is 6

…

Thanks Duncan. Much closer, but still not right, 2.4236 instead of 2.4494

After a strong coffee I shall tackle Pythagoras!

Just in case anyone needs more accuracy the square root of 6 to 1000 places of decimals is:

2.449489742783178098197284074705891391965947480656670128432692567250
96037745731502653985943310464023481859460122661418912485886545983775
73416257839512372785528289127475276765712476301052709117702234813106
78986690853632443352545604033808808939374585567846574724361304144270
27021617420183830008158980783801308970072869399363083715809440080044
37386875491645146539391450120447264481333162935109447418135132854950
44466171726631425497648318124504011566045364590093435861679319836656
66266909131906535863766105258156179366236179110294672088354575512883
55298983241005586902050064154661241173533994104310579928984673395325
52787360418296551718303910367800383623166297323302829049773910474263
60070512637697546888677006409929559257173284645028880915144257463026
02952963237075525034227559915454262812980030646720799424253022878426
20534067947860046596939979935597540616713586546534562482082562768863
69280941306372433508864801572484633723244737954548814438479336370091
97410147785232875591063159846078310367009535324726

Put that in your slide rule and smoke it!

Dave

[CalumParticipant @calumgalleitch87969]

For the benefit of anyone as confused as I was!

[duncan webster 1Participant @duncanwebster1]

Either SOD has done it wrong, or QCad isn't as good as it thinks it is

Why is putting images in here such a faff!

[Author]

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