Comments on: Help me with triangle math.
http://ask.metafilter.com/81306/Help-me-with-triangle-math/
Comments on Ask MetaFilter post Help me with triangle math.Thu, 17 Jan 2008 21:40:06 -0800Thu, 17 Jan 2008 21:40:06 -0800en-ushttp://blogs.law.harvard.edu/tech/rss60Question: Help me with triangle math.
http://ask.metafilter.com/81306/Help-me-with-triangle-math
Triangles.math.Filter: Its been a while since algebra - I am trying to calculate the linear length of material required to fabricate a repeating V pattern out of sheet metal in various lengths.
<br /><br /> The metal, after being bent looks like: <br>
<br>
a/\/\/\/\/\/\/\/\/b<br>
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The height of the wave will be constant, 50mm. The bend angles will be 90degrees. <br>
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What I am trying to figure out is how many mm of initial material to be used in various lengths of Wave. For example, if AB=1000mm, then how much would the original material be flattened out.<br>
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In practice, both the height and length will vary, so my overall goal is to put this into excel with the various finished sizes, that give me length of material required to fabricate.<br>
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Thankspost:ask.metafilter.com,2008:site.81306Thu, 17 Jan 2008 21:36:15 -0800drippedtriangle_mathanglesformulaBy: vacapinta
http://ask.metafilter.com/81306/Help-me-with-triangle-math#1205501
Multiply by the square root of 2.<br>
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height doesnt mattercomment:ask.metafilter.com,2008:site.81306-1205501Thu, 17 Jan 2008 21:40:06 -0800vacapintaBy: kjackelen05
http://ask.metafilter.com/81306/Help-me-with-triangle-math#1205508
Since the angle is 90 degrees, the distance (in the direction of A to B) of each linear piece is also 50 mm. The length of the piece is 50/cos(45 degrees) = about 70.17 mm.<br>
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So the length of metal is 50/cos(45 degrees) mm per linear piece, and each linear piece covers 50 mm toward B, so you need<br>
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AB*[50/cos(45 degrees)]/50<br>
= AB/cos(45 degrees)<br>
= about AB/0.707107<br>
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Since the 50 cancels, this will work no matter the height and length, so long as the 90 degree angle remains constant.comment:ask.metafilter.com,2008:site.81306-1205508Thu, 17 Jan 2008 21:54:50 -0800kjackelen05By: delmoi
http://ask.metafilter.com/81306/Help-me-with-triangle-math#1205518
Expanding a bit on what vacapinta said, you can think of each wave as being a triangle with one 90° angle (and thus two 45° angles). <br>
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Check out <a href="http://en.wikipedia.org/wiki/Image:Trigonometry_triangle.svg">this picture</a> on wikipedia and imagine it's your triangle. The 90° corner would be "C" here. And the length of your hypotenuse (h) would be the 'linear length' of your triangle in the wave, while the sum of lines a and b would equal the length of the actual metal strip. <br>
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The ratio of the hypotenuse (h) and a is the sine of the angle in the corner (45°) and the ratio of and the ratio of the hypotenuse (h) and b is cosine(45°). So the ratio between the length of the metal strip to the linear length of the sine(45°)+cos(45&deg);. And it just so happens that the sine and cosine of 45° are both <sup>√(2)</sup>/<sub>2</sub>. So double that and you get √2.<br>
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Assuming I'm remembering my high school trig correctly :)comment:ask.metafilter.com,2008:site.81306-1205518Thu, 17 Jan 2008 22:05:23 -0800delmoiBy: vacapinta
http://ask.metafilter.com/81306/Help-me-with-triangle-math#1205526
I didn't do any calculations to get that answer. Once I read 90 degrees and saw his picture I realized it was like setting up a row of square frames. So its just a matter of measuring the length of a diagonal of a square: sqrt(2)<br>
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The height doesnt matter for the same reason that it doesnt matter how many steps you build in a staircase. You still need the same amount of material. Or why going from 3rd and 22nd st to 4th and 18th is the same distance no matter how many blocks you cut over when you are walking there.comment:ask.metafilter.com,2008:site.81306-1205526Thu, 17 Jan 2008 22:20:33 -0800vacapintaBy: Opposite George
http://ask.metafilter.com/81306/Help-me-with-triangle-math#1205542
vacapinta is correct.comment:ask.metafilter.com,2008:site.81306-1205542Thu, 17 Jan 2008 22:45:17 -0800Opposite GeorgeBy: dripped
http://ask.metafilter.com/81306/Help-me-with-triangle-math#1205553
Thanks guys!comment:ask.metafilter.com,2008:site.81306-1205553Thu, 17 Jan 2008 23:08:20 -0800drippedBy: yeoz
http://ask.metafilter.com/81306/Help-me-with-triangle-math#1205660
Just wanted to point out that dividing by 0.707107 (which is really 1/√2) is the same as multiplying by √2.comment:ask.metafilter.com,2008:site.81306-1205660Fri, 18 Jan 2008 05:18:01 -0800yeozBy: range
http://ask.metafilter.com/81306/Help-me-with-triangle-math#1205674
... and to further complicate, all these solutions assume you're able to bend sharp geometric angles into your sheet metal, which in real life you can't and will approximate with arcs (this <a href="http://www.engineersedge.com/bend_allow_calc.htm">picture</a> shows it well, along with a calculator you might use if you end up caring about this effect).<br>
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The upshot is that the length of material along the real-life arc is going to be different than the length to the (imaginary) vertex. The discrepancy might not make any difference if you're only doing a few bends, or if you're going to trim to finished size after bending, but if you're doing a continuous piece of siding for a wall and need it to come out correct to 1/16", you might want to take this into account.<br>
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(At a minimum, the effect means that using all the digits in .707107 is a misleading amount of precision for you...)comment:ask.metafilter.com,2008:site.81306-1205674Fri, 18 Jan 2008 05:47:11 -0800range