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 Post subject: Math quandry
PostPosted: Wed Mar 17, 2010 6:58 pm 
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Location: Toledo, Ohio
making some picture frames and am driving myself crazy trying to calculate the pieces. The inside dimension of the frame is 5 1/4" x 22 1/2". The stock is 3 1/2 " wide and I need to make 45 degree miter cuts. My brain freeze here is how do you calculate the lengths of the rails ? I could do trial and error but there has to be a formula that will calculate the lengths and it's driving me nuts

thanks


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PostPosted: Wed Mar 17, 2010 7:12 pm 
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If the corner is 45 degrees and the frame is flat, each corner is longer than the inside by the width of your stock. So add double the stock width to the inside lengths to get the outside lengths.

Edit: Since I could not see your post as I typed, I did not remember your dimensions of 5 1/4 x 12 1/2 size with 3 1/2" wide stock. But that would be 7 + 5 1/4 = 12 1/4" for the short pieces, and 7 + 22 1/2 = 29 1/2" for the long ones.

You say 45 degree miters. Do you mean you want to cut 45 degrees for a 90 degree corner as I assumed above, or do you want to cut 22 1/2 for a 45 degree corner, in which case, what shape is your frame?


Last edited by AlanWS on Wed Mar 17, 2010 7:19 pm, edited 2 times in total.

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PostPosted: Wed Mar 17, 2010 7:14 pm 
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Hey Mike,

You might try the Pythagorean Theorem (a^2 + b^2 = c^2).

More simply, the end of your piece defines an equilateral right triangle. The cut is the hypotenuse, and the width is one leg, so the additional length is the same as the width. Don't forget to double that length to account for both ends.

Good luck,
Tom

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PostPosted: Wed Mar 17, 2010 8:58 pm 
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think of each rail as three parts:

5 1/4 plus a point on each end

the point must go from the end of the 5 1/4 rectangle to the far end of the top rail: the top rail is 3 1/2 high--so the point must be 3 1/2 high

you need a point on each end so 5 1/4 + 3 1/2 + 3 1/2 = 12 1/4

the bottom and top

22 1/2 + 3 1/2 + 3 1/2 = 29 1/2

thats how i figure it out, hope it helps


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PostPosted: Wed Mar 17, 2010 9:19 pm 
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I almost always size my frames by the inside dimension - I usually need a specific opening. The stock is always at least a few inches longer than I need, with the rough minimum length determined as the guys have described. Often the stock is even 6 or 8" extra long so I can cut away snipe or pick the best section of grain. I don't need to know the precise outside dimension of the final frame since I cut to the inside marks.

As always with a frame, the key is to have the sides and top exactly the same length and the corner angles adding up to exactly 90 degrees. I use a tablesaw jig that makes it easy. First cut the 45s on one end of the four pieces, mark the inside openings on the top and one side piece, back on the jig, line up the mark and set the stop that determines the length, cut the top and bottom using the stop to get exact same length, repeat for the sides. Never fails.


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PostPosted: Thu Mar 18, 2010 4:16 am 
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Inside's easy. Desired exposure distance. Outside is the inside plus twice the width of the frame, less twice the width of the rabbet. Start larger. That said, I prefer to make four pieces capable of becoming the long dimension so that I can cut them first. A long can become a short, but not vice-versa.

Most important thing is to cut the opposing sides to the same length. Perfect miters won't work otherwise. Always a rail and stop on my miter gage when I'm framing.

Helps to make your marks / \ on the ends of the pieces to remind you which way the cut goes. :wink:


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PostPosted: Thu Mar 18, 2010 11:40 am 
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Woodsurfer dances around the real point. Ya don't need to account to the outside dimension. Cut the stock long and measure from the inside of the frame.

BTW, don't forget to accomodate for matting if you haven't already.

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PostPosted: Thu Mar 18, 2010 1:41 pm 
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There was an Indian Chief, and he had three squaws, and kept them in three teepees. When he would come home late from hunting, he would not know which teepee contained which squaw, since it was dark.

He went hunting one day, and killed a hippopotamus, a bear, and a buffalo. He put the a hide from each animal into a different teepee, so that when he came home late, he could feel inside the teepee and he would know which squaw was inside.

Well after about a year, all three squaws had children. The squaw on the bear had a baby boy, the squaw on the buffalo hide had a baby girl. But the squaw on the hippopotamus had a girl and a boy.

So what is the moral of the story?

The squaw on the hippopotamus is equal to the sum of the squaws on the other two hides.

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 Post subject:
PostPosted: Thu Mar 18, 2010 6:47 pm 
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Dan should get 10 laps and 26 pushups between each for that!

:D :roll: :shock:

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PostPosted: Thu Mar 18, 2010 11:59 pm 
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I've made 15 or so frames this year and at first tried to do the math on the frames.... after figuring it out once I figured out that each frame was going to be slightly different unless my glass miter was always the same size... after scratching my head for a while on that I decided that it was easier to cut as olpeddler suggested and just use inside dimension... just make sure to give yourself a generous portion for the miters and you'll be ok (and cut your long piece miters first so that if you goof up you can just cut them again for the short sides! DAMHIKT)

guestimating is much easier on my math-challenged brain, and after one or two the guestimations were pretty close and I wasn't wasting much
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I use the pictured miter sled and stop blocks for repeated cuts-- the blocks are the reasons for the extra long legs on this sled-- it gives me something to clamp the stopblocks to.
Lawrence


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