Multi-angle frame formula?

monkframe

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Hi all,
I have a prospective customer coming in with a piece she'd like framed in what she describes as an "elongated diamond-shaped frame."
I have Paul Frederick's "Framer's Answer Book," so questions about conventional multi-angles are answered, but is there a way to figure this sort of thing out?
Thanks in advance.
 

David Hewitt

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I just lay it out with mat board strips that are the width of the frame, till the balance is what I want, then measure for the lengths that will be needed.
Don't have Paul's book in front of me, but a easy formula:
360 ÷ by # of frame sides ÷ by 2= degree of cut.
Example:
Four sided, 360÷4=90÷2=45
Eight sided, 360÷8=45÷2=22.5
 
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wpfay

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Using a full sized template and drafting skills at bisecting angles. Transferring those angles to a saw using a sliding T bevel.
There's only so much that can be done on a power miter box without serious modifications.

The formula David mentions works well for regular polyhedrons, but not so well on irregular shapes. It also will depend on what the customer means by "elongated diamond". That describes a parallelogram, like the Diamond suit in a deck of cards, as well as an irregular pentagon, like a cross section of an actual faceted diamond (think Superman's S logo.). IMG_0735.jpg
 

Joe B

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Suggestion, pick up a "Barton's Multi-angle Framing Guide" book. It has all the answers anyone would want.
 

alacrity8

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I just lay it out with mat board strips that are the width of the frame, till the balance is what I want, then measure for the lengths that will be needed.
Don't have Paul's book in front of me, but a easy formula:
360 ÷ by # of frame sides ÷ by 2= degree of cut.
Example:
Four sided, 360÷4=90÷2=45
Eight sided, 360÷8=45÷2=22.5
Assuming that you mean the diamond from playing cards, you have a 4 sided shape with equal length sides, and 2 different angles.
All of the angles in a 4 sided shape should add up to 360 degrees.
As you have 2 of each angle, adding 2 non equal angles will be 180 degrees. Eg 80 and 100, or 60 and 120.

These angles then need to be divides by two to get either side of the mitre, and then subracted from 90 degrees to get the angle needed for the cut.

For a 60 degree angle, you would need to divide by two, then subtract that from 90 degrees. 90 - (60/2) or 60 degrees on a mitre saw.
For a 120 degree angle, you would need to divide by two, then subtract that from 90 degrees. 90 - (120/2) or 30 degrees on a mitre saw.

Most mitre saws cannot cut an angle of more than 50 degrees without doing some tricky things.
I've made it work on a 30, 60, 90 triangle frame by adding a 90 degree block, and then offsetting from this.
Not the easiest or safest way to cut a frame.
 

Matthew Hale

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you may have better results cutting these angles on a table saw with a reliable mitre gauge than with an adjustable mitre saw. as previously stated, mitre saws have some severe limitations.
 

monkframe

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Thank you all for your helpful suggestions. Yes, it is a diamond as on a playing card.
If and when the lady shows up w/the piece, I'll get into serious calculations and report back on results.
Happy Holidays!
 

wpfay

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The angles of 8 sided frames have to be obtuse, more than 45 degrees, not less! It's actually one of the test questions in the CPF study guide and the correct answer is 67.5, so the 45 of a square frame plus half again.
Correct, but your saw protractor is going to be set at 22.5 degrees (there's a detente there for your convenience) since 0 degrees on the protractor is actually 90 degrees from the fence. The outside angles on a regular octagon are 67.5 degrees each, but the miter is its complement at 22.5 totalling 90 degrees.
I took issue with that study question because, while being geometrically correct, it is not practically correct with the way framer's saws are set up. This is the case with crosscut protractors on table saws as well. The test should be about what is practical to the trade, not what the math teaches us.
 

RoboFramer

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Correct, but your saw protractor is going to be set at 22.5 degrees (there's a detente there for your convenience) since 0 degrees on the protractor is actually 90 degrees from the fence. The outside angles on a regular octagon are 67.5 degrees each, but the miter is its complement at 22.5 totalling 90 degrees.
I took issue with that study question because, while being geometrically correct, it is not practically correct with the way framer's saws are set up. This is the case with crosscut protractors on table saws as well. The test should be about what is practical to the trade, not what the math teaches us.
You’re right and you must have quoted my post at the same time I was deleting it when I realised I was wrong! I’d recently been reading the study guide and the answer I thought was correct (45 degrees divided by two) wasn’t even there! I thought ah well, this is the CPF study guide and I’ve never actually done this so that’s that. Then there’s the one about how much moulding you would need for a frame with an outside size of 20x16. I’d be failing that test!
 
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