Imperious measurements

[mirrorshard]

vashti linked to a US-authored guide to building a tin-can wireless antenna. The measurements are all in inches, which got me thinking.

How do you measure 1.21" in any meaningful manner? The rulers I have here are almost all marked out in thirty-secondths on the inches side, and whilst one has tenths for half its length, it doesn't go any finer than that. I don't even think you could fit another ten subdivisions into one of those and still be able to use it. Possibly with a special magnifying section, I suppose...

Best answer gets a cookie.

Comments

[thipe.livejournal.com]

http://www.google.co.uk/search?hl=en&q=1.21+inches+in+cm&btnG=Search&meta=

3.07cm

[aca.livejournal.com]

http://en.wikipedia.org/wiki/Calipers

:)

[kake]

I suspect Adam's already won the cookie. I was wondering about the reason for specifying a measurement like that; it might be something to do with it being the square of 1.1, and there being Maths involved somewhere.

[keira-online.livejournal.com]

Caliper thingys, look like 3 rulers in one.

DO NOT translate into metric. Thats what NASA did and their spaceship exploded.

[dasquian.livejournal.com]

I can get you to 1/400th of an inch with the rulers you have!

We're essentially trying to solve x/10 + y/32 = 21/100 where x and y are both integers (ignoring the trivial full inch for now).

This resolves to 80x + 25y = 168, which is clearly unsolvable. However, it's very close to 80x + 25y = 170 which gives us x = -1 and y = 10. So if you measure 1/10th of an inch back, then 10/32nds of an inch forwards, you get 0.2125 of an inch.

So I've, um, proved you can't measure it exactly with your two rulers alone. I guess you already knew that ;p *slumps*

[dasquian.livejournal.com]

OK, I can get you 1.21" exactly, with just your tenths ruler :)

You can trivially get 0.10" and 0.20" from that ruler. By either using your 8 notches of your 1/32nds ruler, or bisecting a 0.10" measurement, you can also get 0.05" and hence 0.15" and 0.25".

Using 0.15", 0.20", and 0.25", construct a right-angled triangle. Using a set-square, you should be able to drop a line perpendicular from the hypoteneuse to intersect the right-angle, splitting the triangle into two similar triangles. One of these will have edges of 0.15", 0.09" and 0.12", and the other will have edges of 0.16", 0.20" and 0.12".

You now have access to 0.21" thanks to 0.16" and 0.05" :) It might not be a practically useable answer, but it'll work!

[puddick.livejournal.com]

we have electronic calipers in the lab for measuring downhole things before and after aging in chemicals. very accurate in both metric and imperial.

[xyon.livejournal.com]

Or just stick with the tried and true:
Cut a length of thin string (thread?) to 121".
Cut it into 100 pieces.

And if you don't have ten-and-a-bit feet of string and want something more practical, try 12.1" and 10 pieces.

Silly Europeans not thinking metrically enough to use inches.

I must be incredibly bored, eh?