Cutting Miters

[Gary Petit]

Hello,

I am fairly new to woodworking, but have a few questions about cutting miters. Is there an easy way to calculate how to cut a miter? I have struggled numerous times, and trying to figure out a simple method. Any thoughts would greatly be appreciated.

Thanks,

Gary

[Art Mann]

I don't understand the question. Are you trying to select a tool for the job? Are you trying to establish a cutting method? Are you wondering how to calculate the angle of a closed shape with an arbitrary number of sides? Please explain a little more.

[Dave Zellers]

Hello,

I am fairly new to woodworking, but have a few questions about cutting miters. Is there an easy way to calculate how to cut a miter? I have struggled numerous times, and trying to figure out a simple method. Any thoughts would greatly be appreciated.

Thanks,

Gary

I'll take a stab at this. Like people, some miters are cuter than others. Hey OH! I'll be here all week.

If you are talking about bisecting an angle to get the perfect miter cut for both pieces, it is a very simple process.

Easy Peasy

If you don't have a compass and bevel square, you need one.

[Jim Matthews]

Is the problem when making more than four corners?

The total number of degrees enclosed by a regular polygon is 180(n-2)
where n is the total number of sides.

Take that number, divide it by the number of vertices.
Divide that result by 2 to get the setting for each half of a miter.

This applies to "regular" polygons, if my High school geometry still serves.

For example:

To make the miters on hexagon, like a stop sign.

180(6-2) = 720 degrees enclosed
720/6 = 120

Set your miters for 60 degrees between each piece.

*******

I suppose you could do the same with irregular shapes,
but you would need to divide the sections into simpler
shapes first, and fiddle to get the obtuse angles fitted.

*****

If it was me (and it isn't) I would layout the pieces of to make the joint,
where I wanted them to come to rest. Using test stock, make a test cut
following these steps:

I would overlap the pieces and mark them in place.

Then I would cut the first and fit the second to the first cut.

In my limited experience, just setting up a miter saw and zipping away
always leads to at least one piece that's too short.

That typically happens when I'm out of stock and the store is closed.

[Rich Engelhardt]

If you are talking about bisecting an angle to get the perfect miter cut for both pieces, it is a very simple process.

Easy Peasy

Or - even easier peasier...... http://www.starrett.com/saws/product...ractors/505A-7
Or the one real high on my wish list..
http://www.starrett.com/saws/product...tors/CP505E-12

[Jason Roehl]

To make the miters on hexagon, like a stop sign.

Come again?

Try "octagon".

[Peter Quinn]

Is the problem when making more than four corners?

The total number of degrees enclosed by a regular polygon is 180(n-2)
where n is the total number of sides.

Take that number, divide it by the number of vertices.
Divide that result by 2 to get the setting for each half of a miter.

This applies to "regular" polygons, if my High school geometry still serves.

For example:

To make the miters on hexagon, like a stop sign.

180(6-2) = 720 degrees enclosed
720/6 = 120

Set your miters for 60 degrees between each piece.

Huh? That is hard to follow. It takes 360 degrees to turn a circle. You have to turn a circle to get around a polygon. So divide the circle by the number of sides, that is the total included angle of each intersection. Half of that angle gets mitered on each piece.

Example square has four sides. So 360/4=90 . Divide result by two=45 degrees per miter.

octagon has 8 sides. 360/8=45. Divide 45/2=22.5 per miter. Thats a stop sign!



Gary, what information are you seeking? What miters are you trying to calculate?

[Kurt Cady]

Huh? That is hard to follow. It takes 360 degrees to turn a circle. You have to turn a circle to get around a polygon. So divide the circle by the number of sides, that is the total included angle of each intersection. Half of that angle gets mitered on each piece.

Example square has four sides. So 360/4=90 . Divide result by two=45 degrees per miter.

octagon has 8 sides. 360/8=45. Divide 45/2=22.5 per miter. Thats a stop sign!



Gary, what information are you seeking? What miters are you trying to calculate?

no, Peter, Jim is right. Otherwise you'd never have an equilateral triangle with each angle 60*

[Roger Nair]

Gary, one other thing to sort out is how your scale on a miter box is expressed. For instance, my old Stanley miter box has the square cut set as 90 degrees on the other hand my Hitachi the square cut is expressed as 0 degrees. So you may need to convert your calculated miter to the scale on your machine. So if you need the miter to be 67.5 then you need to subtract 90 - 67.5 = 22.5 then set 22.5 on the Hitachi box, with the Stanley just set 67.5.

[Yonak Hawkins]

Hi Gary. Please say a little more about the problem. Is it for a box, a picture frame, a table edge or something else ? How many sides are there ? What tool have you got to use ? Can you say a little more about the problem you are having ? I hope you come back.

[Ole Anderson]

Very timely and pertinent for me as next week I have to replace the top rail on my deck, using expensive 20 foot 5.5" wide coated Trex boards laying flat, and each of the eight turns is something other than 90 degrees. I was planing on using my tiny protractor to measure each angle, dividing that angle by 2 and setting my miter saw to that angle and hoping for the best. Most of the intersections are 90's, but off by a few degrees. Hoping for a nice tight joint. That $65 aluminum Starrett gage looks real promising, but not justifiable for this project. But looking around, I am going to order the plastic version for $23.13 delivered. http://www.amazon.com/gp/product/B00...E5PBGTAYJ1AD9Y Thanks for the ideas.

[Peter Quinn]

no, Peter, Jim is right. Otherwise you'd never have an equilateral triangle with each angle 60*

My appologies to Jim, and thanks Kurt, my answer was very much wrong, Jim's sure is correct, and this explains why all the stop signs I've made looked so funny!

[Gary Petit]

Thanks for the replies guys. Sorry I did not elaborate on the problem. I understand the degrees and such, but it is trying to get the measurement correct. For like making a picture frame, or molding for a cabinet. Is there a way you can measure the width of the cabinet and calculate the miter. So example width of the cabinet is 12 inches, but if you want a miter in your molding it has to be a little longer correct, to correspond with the front piece of your miter? Right now I just line my piece up and mark the width of the cabinet to my molding piece then draw the 45 degree angle on my molding an cut it on my miter or table saw. I was just wondering if there was a way to just use a tap measure and figure the miter out that way. Another example is putting trim around windows using miters. I just kinda puzzled because I have done the same thing with trimming out windows too.

[Jason Roehl]

If you're picture-framing something, the outside measurement (miter point to miter point) is easy: it's the length of that side of the object, plus twice the width of the moulding, plus twice the reveal (overlap).

For instance, on a 36" wide window (measured inside of jamb to inside of jamb, for ease of example here), using 2 1/4" trim, the top piece of trim would be: 36" + 2(2-1/4") + 2(3/16"--typical reveal) = 36" + 4 1/2" + 3/8" = 40 7/8"

This is, of course, assuming a rectangular window so that the miters are all 45º.

[Yonak Hawkins]

I think I understand, Gary. It's not especially a question of the angle (the miter) but how to measure the length of the rails or sides.

The nature of miter joints is that the rails meet at such an angle that, for a four-sided cabinet, for example, each of the pieces is cut to the dimension of the outside of the cabinet. As an example, if your cabinet is 12" X 24", two of the pieces would be cut at 12" total length and two at 24" and the miters (the angles) would be cut at 45°.

Is that what you're asking ?