# Help me with triangle math.

January 17, 2008 9:36 PM Subscribe

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.

The metal, after being bent looks like:

a/\/\/\/\/\/\/\/\/b

The height of the wave will be constant, 50mm. The bend angles will be 90degrees.

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.

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.

Thanks

The metal, after being bent looks like:

a/\/\/\/\/\/\/\/\/b

The height of the wave will be constant, 50mm. The bend angles will be 90degrees.

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.

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.

Thanks

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.

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

AB*[50/cos(45 degrees)]/50

= AB/cos(45 degrees)

= about AB/0.707107

Since the 50 cancels, this will work no matter the height and length, so long as the 90 degree angle remains constant.

posted by kjackelen05 at 9:54 PM on January 17, 2008

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

AB*[50/cos(45 degrees)]/50

= AB/cos(45 degrees)

= about AB/0.707107

Since the 50 cancels, this will work no matter the height and length, so long as the 90 degree angle remains constant.

posted by kjackelen05 at 9:54 PM on January 17, 2008

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).

Check out this picture 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.

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°);. And it just so happens that the sine and cosine of 45° are both

Assuming I'm remembering my high school trig correctly :)

posted by delmoi at 10:05 PM on January 17, 2008

Check out this picture 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.

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°);. And it just so happens that the sine and cosine of 45° are both

^{√(2)}/_{2}. So double that and you get √2.Assuming I'm remembering my high school trig correctly :)

posted by delmoi at 10:05 PM on January 17, 2008

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)

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.

posted by vacapinta at 10:20 PM on January 17, 2008 [1 favorite]

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.

posted by vacapinta at 10:20 PM on January 17, 2008 [1 favorite]

vacapinta is correct.

posted by Opposite George at 10:45 PM on January 17, 2008

posted by Opposite George at 10:45 PM on January 17, 2008

Just wanted to point out that dividing by 0.707107 (which is really 1/√2) is the same as multiplying by √2.

posted by yeoz at 5:18 AM on January 18, 2008

posted by yeoz at 5:18 AM on January 18, 2008

... 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 picture shows it well, along with a calculator you might use if you end up caring about this effect).

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.

(At a minimum, the effect means that using all the digits in .707107 is a misleading amount of precision for you...)

posted by range at 5:47 AM on January 18, 2008

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.

(At a minimum, the effect means that using all the digits in .707107 is a misleading amount of precision for you...)

posted by range at 5:47 AM on January 18, 2008

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height doesnt matter

posted by vacapinta at 9:40 PM on January 17, 2008 [1 favorite]