Imperious measurements
Apr. 17th, 2007 02:34 pm![[personal profile]](https://www.dreamwidth.org/img/silk/identity/user.png)
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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.
no subject
Date: 2007-04-17 01:44 pm (UTC)3.07cm
no subject
Date: 2007-04-17 01:50 pm (UTC)no subject
Date: 2007-04-17 02:26 pm (UTC)no subject
Date: 2007-04-17 01:48 pm (UTC):)
no subject
Date: 2007-04-17 01:51 pm (UTC)no subject
Date: 2007-04-17 02:09 pm (UTC)no subject
Date: 2007-04-17 01:52 pm (UTC)no subject
Date: 2007-04-17 02:02 pm (UTC)I'm sure there's Maths in the process somewhere, but not knowing enough about the area I don't know whether it's here or whether this is voodoo specificity and you just need "to the middle of the can".
no subject
Date: 2007-04-17 03:21 pm (UTC)DO NOT translate into metric. Thats what NASA did and their spaceship exploded.
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Date: 2007-04-17 04:00 pm (UTC)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*
no subject
Date: 2007-04-17 04:27 pm (UTC)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!
no subject
Date: 2007-04-17 05:49 pm (UTC)no subject
Date: 2007-04-17 06:41 pm (UTC)no subject
Date: 2007-12-06 08:46 am (UTC)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?