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Old Jun 12, 2006 | 11:51 PM
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MorphiasX's Avatar
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From: Hoffman Estates, IL
Any math gurus out there?

Ok well I am trying to fab somthing, and it consists of a square with a piece of flat stock on each side sticking off of it at a 45 degree angle. NOW, this creates a gap at each corner. The pieces of flat stock are all 1/8" thick, and 2.5" tall. To fill this gap, I want to cut some triangles to fill it. The triangles will have the two sides that butt against the flat stock being 2.5", but my question is what will the top be? I know there is some formula to figure this out, but I have no idea what it is.

Here is a diagram of what I am describing:


Thats it for the high school word problem...
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Old Jun 13, 2006 | 03:21 AM
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From: Hamilton, Oh
how bout the diagram, a little hard to picture it.
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Old Jun 13, 2006 | 04:04 AM
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on what axis are those 45 degree angle based on? is the corner going to be a true 90 degree (as 2 45s would make it)? or is this a 45 degree angle, then a piece, then another 45 degree angle, then another peice? if its a 90 degree angle between the 2 peices, then the measurement will be around 3.5 inches. if its 45 let me know and i can crunch some more numbers
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Old Jun 13, 2006 | 04:07 AM
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any one correct me if i'm wrong but it should be 3.54"
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Old Jun 13, 2006 | 04:45 AM
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a squared + b squared = c squared (sorry my squared key 2 isn't working, darn keyboard configuring)
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Old Jun 13, 2006 | 04:54 AM
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From: MA
if the out side angle b is 90 then a is 3,535
if c is 45 the then b is 1.9134 [IMG][/IMG] hope i could help
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Old Jun 13, 2006 | 05:02 AM
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I conquer if B is a 90 degree angle then C is a 90 degree angle, there for the pathagram therom would work, which for those who forgot is what shiftless put down and OLY worked out and the answer would be 3.5355
that was fun anymore

Last edited by kmcc78240; Jun 13, 2006 at 05:04 AM.
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Old Jun 13, 2006 | 05:20 AM
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The picture itself is very misleading. Assuming that he wants each piece at a 45° angle relative to the center, then yes, the 3.53" is correct. However... that picture isn't even close to that, and if that is what he wants it as, then the angles need to be measured, or better yet, get a ruler. His discription seems to verify that indeed it should be a right triangle, but you never know.
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Old Jun 13, 2006 | 05:29 AM
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From: Peoria IL
I kinda see it diffrently



i interpert his 45deg to mean that both peices of steel are tilted up and not laying flat as many have assumed.

if they were laying flat on a table yes it would be sqrt2*a side (in this case 2.5) or as mentioned 3.5 and change.

but in fact its less than that becuase the steel is sitting at a 45 deg angle to the flat table. (really the shown dimention would be 135deg)

for instance As the green angle aproches 90deg the distance in blue would aproch 0 (as in the sides of a box)

so in that case i say it is in fact much less than 3.5" (and you can kinda see its at least as "short" as the red 2.5" line if not shorter)

if in fact im right it becomes a non trivial gemoetry problem to calculate, when in the real world its much easyer to just put a tape across the opening

Last edited by snap-on; Jun 13, 2006 at 12:31 PM.
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Old Jun 13, 2006 | 06:31 AM
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Originally Posted by shiftless89
a squared + b squared = c squared (sorry my squared key 2 isn't working, darn keyboard configuring)
What he said
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Old Jun 13, 2006 | 12:08 PM
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From: Hoffman Estates, IL
Snap-on is right, what I mean is the pieces of flat stock are sitting at a 45 degree angle to the flat surface (which yes is actually 135 degrees). It is going to create a funnel-like object. And yes the corner will be a 90 degree angle.
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Old Jun 13, 2006 | 12:47 PM
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From: Peoria IL
right so useing the orthographic projection



it should boil down to something like sqrt2(hyp*sin theta) (were theta is defined as the angle of the sides off the vertical projection)

or in our case sqrt2(2.5*sin 45) = (in this special case) 2.5" (only because the 1/(sin 45)=sqrt 2, you got to love math, oh is my nerd showing?)
(which is why i said it looks almost the same distance as the red line)

someone veryify please

(note im useing the sqrt2 shortcut for a right triangle with both sides equal)

Last edited by snap-on; Jun 13, 2006 at 01:13 PM.
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Old Jun 13, 2006 | 03:30 PM
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I ended up with the distance coming to about 1.92". Why don't you just measure it?

EDIT: Yeah, I'm pretty sure this is wrong.

Last edited by DrBeau; Jun 13, 2006 at 03:41 PM.
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Old Jun 13, 2006 | 03:36 PM
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Yeah...after just doing it logically...it should be 2.5".
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Old Jun 13, 2006 | 03:43 PM
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From: Hoffman Estates, IL
For it to be to 2.5" that would make it an equilateral triangle... Which I dont think it is. Though I could be wrong.
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Old Jun 13, 2006 | 03:47 PM
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I scrolled through this entire thread looking for the answer to this so I could learn something... I've been duped!
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Old Jun 13, 2006 | 04:02 PM
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From: ELN
√2.5+2.5 = 3.54.

If where it says "45 degree angle" is 45° from horizontal, the angle where the two pieces of metal come together is 90°. The two shortest sides of any right triangle equal the square of its longest side. So we have 2.5+2.5=12.5, which the square root is approximately 3.54.

Last edited by Glenn; Jun 13, 2006 at 04:04 PM.
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Old Jun 13, 2006 | 04:20 PM
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It all depends on the angle between the bases. You need 3 angles to figure out that distance.
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Old Jun 13, 2006 | 04:21 PM
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I tried to keep the colors consistent with the original picture. Looking at the side view (Staring straight at one piece of metal), we can say that both side equal "a". We can also say that the length of the dashed blue line is equal to the sloid blue line form the original diagram. If we look from above, we can also see that both two pieces of length "a" form a right triangle with a hypotenuse equal to that of the lenght "?". By looking at similar triangles, the hypotenuse of a right-angle triangle with sides "a" must equal 2.5".

I'm pretty sure I'm right, but if someone can prove me wrong I ask that you do.

EDIT: What I forgot to say was that that the length that we're looking for is 2.5". thanks to snap-on for doing all the work.

Last edited by DrBeau; Jun 13, 2006 at 04:24 PM.
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Old Jun 13, 2006 | 05:50 PM
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Hold the pieces where you want them, enlist other hands as needed.

Take a piece of cardboard/anything, and hold it in the area of the desired piece.

Trace the sides of the other pieces onto the board.

Cut piece slightly outside of lines.

Put in place.

Adjust as needed, no math involved.
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