# mathmatics of mitre cuts...

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• posted on April 17, 2004, 10:12 am
I am not a carpenter, I am building a blackjack table and am designing the base and have a question or two about miter(mitre?) cuts. The frame for the base is going to be built out of 2x4, possible 2x6's. I need this to be 38 inches in height, but I want the support legs angle in at about a 30 degree angle. How do I determine the length of the stock that I need so that retain the height I need?
Compound cuts are next !
Thanks Brandon

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<%-name%>
• posted on April 17, 2004, 11:08 am
Well, not sure how to it mathmatically, but what I would do is miter the bottom of the leg, mark a level line on wall or something 38" high, hold the mitered leg next to the wall, mark the height.
dave

in
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<%-name%>
• posted on April 17, 2004, 11:29 am
When you're doing something like this, then a full-size drawing helps a lot. A sheet of ply or mdf makes a good surface - opposite sides are parallel, and the corners are at an accurate 90 degress, so this helps in your layout.
Mark out your stock directly from the drawing.
HTH
Frank

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<%-name%>
• posted on April 17, 2004, 12:19 pm
Make 1, 30 degree cut, set the piece on the floor with the 30 degree cut on the floor, measure up from the floor your desired distance and mark where the other 30 degree cut needs to be.

in
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<%-name%>
• posted on April 17, 2004, 12:39 pm

first the theory, then a practical answer.
Basic trigonometry gets you there. The 'height' of the angled stock is the 'sine' of the angle off the _horizontal_, and the distance away from vertical is the 'cosine' of the angle off horizontal. Note: the 'calculator' on a Windows PC knows those functions. (under 'view', click 'scientific') You take the sine of the desired angle (60 degrees, in this case), and _divide_ the desired height, to get the length of the diagonal.
Now, 30 degrees/60 degrees is a 'special' case. the short side of a 30/60 right triangle is exactly half the length of the diagonal. This makes the other side "square-root-of-three"/2 of the diagonal. And, conversely, the diagonal is 2/'square-root-of-three' the vertical distance. square-root- of-three is 1.732, so half that is .866 Thus, to get a 38" height, you need a piece that is 38/.866 inches long which works out to a hair over 43-7/8 inches.

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<%-name%>
• posted on April 17, 2004, 1:06 pm
wrote:

From your description I can't clearly picture in my sluggish Saturday morning brain what you need to figure. Some reasonably straightforward trigonometry will be able to give you the answer, however.
Since I don't have a real helpful suggestion, my point would be that even after doing the trigonometry, real world conditions are never what they were supposed to be on paper. Your stock will be somewhat thinner, thicker, or wider than you intended, your floor isn't flat and level, your saw's miter gauge is only accurate to 1/2 degree, etc., etc. (the last being a very important example, since over 38 inches an error of 1/2 degree in that 30 degree angle could become a 7/16" error).
I'd suggest leaving the support legs long until they've been fitted or even attached, and then marking the length and cutting them off. That's pretty much foolproof, and singularly deals with all the little inevitable and cumulative errors that occur when working with many angled joints.
I don't subscribe to the excuse that wood is an amorphous substance whose dimensions change over time. Although it is true, it is not an excuse for poor tolerances and sloppy joints. Just the same, no matter how small your tolerances are when working wood, especially with angled joints, your materials and equipment will throw you curves, so it's much better to expect some of them and work with them than it is to fight them every step of the way...
John
John Paquay snipped-for-privacy@insightbb.com
"Building Your Own Kitchen Cabinets" http://home.insightbb.com/~jpaquay/shop.html ------------------------------------------------------------------ With Glory and Passion No Longer in Fashion The Hero Breaks His Blade. -- Kansas, The Pinnacle, 1975 ------------------------------------------------------------------

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<%-name%>
• posted on April 17, 2004, 1:55 pm

SOH CAH TOA
Check out a high school trigonometry or geometry book. Heres a basic site, but DAGS on trigonometry, sin, cos, tan.
http://www.mathsrevision.net/gcse/sin_cos_tan.php
Jay

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<%-name%>
• posted on April 17, 2004, 2:15 pm
Brandon wrote:

I put together a web page with some of the trig formulas I've used most at http://www.iedu.com/DeSoto/trig.html
--
Morris Dovey
DeSoto, Iowa USA

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<%-name%>
• posted on April 17, 2004, 4:27 pm
wrote:

Cut the miter first, then measure down [along, if on a worktable by now] and cut the base to length.

I could post a spreadsheet to do any angles/ No of sides, but can't send binaries here. Where would it go if not here? If needed, state if you want in degrees, or units of 5 degrees.... [Can do it to thousandths of a degree for the whackos who post about their refined saw cuts, but won't.]
I might just send the formula if someone wants to use just that then DIY.
Dan.

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<%-name%>
• posted on April 17, 2004, 10:43 pm
Here's a nice site for woodworking related downloads, including one Badger Pond used to list for cutting compound miters.

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<%-name%>
• posted on April 17, 2004, 4:44 pm
wrote:

OK, I have a better picture of what you mean, but don't know how this will look at your end:
. .. S ... H .... ..... ...... .......
The right side is the height, H The slope [hypotenuse] S is the length you need. The angle at the top is 30 deg.
S/H = secant 30deg. So, S = H*sec(30)
If using a calculator, you won't find that function, so do this:
First make sure you are in degrees, not radians or rads.
Enter 30 then the function Cos, then "=". Now enter function 1/x "=". You now have the secant of 30. Press the multiplication key. Enter the length H. Then press '='. You will have your answer in inches if H was in inches, or cm if H was in cm.
Dan.

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<%-name%>
• posted on April 17, 2004, 7:22 pm
Thank you all very much for the information. I have an HP scientific calculator that i am able to figuire this out on, but in practicality, just doing it with a full size drawing is going to work best. Thanks again ! Brandon

in
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<%-name%>
• posted on April 17, 2004, 8:53 pm
This depends a bit on how you are measuring your 30 degrees. If it is 30 degrees off the horizontal, then the length of the leg sides will be 38 inches times the tangent of 30 degrees. This would get you to the floor.
Unfortunately, I've decided that I can see about a dozen different ways you might be doing this, and each has a different answer. What you might consider doing is to draw a cross sectional picture of the table, either full size or to scale, showing those legs. Make sure you include the dimensions of the legs, because the answers also vary if you are using 2x4's, 2x6's, or dowels (which would look pretty spiffy), mainly because of the width of the wood. After you have a good cross section made up, make the legs to meet the drawing. Note that you also have to consider how the leg is attached to the frame and the base.
Michael

the

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<%-name%>
• posted on April 17, 2004, 9:01 pm
In rec.woodworking

Practicality? Do you have access to a large plotter or are you going to tape a bunch of pages together? You want 30 degree slant legs to rise 38"? This is 8th grade stuff.
Oliver had a heap of apples.
Sin = O/H Cos = A/H Tan = O/A
To determine the length of the hypotenuse with a side of 38" and an angle of 60 degrees(which is what I think you mean, not 30, you'd can use sin:
38 = c * sin 60
therefore c = 38/sin 60
c = 43.878
Cut your boards 43-7/8" to the long point of the miter cut and you're perfect, if I understand your design.
If you really do mean 30 degrees, the boards are going to be 76" long.

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<%-name%>
• posted on April 17, 2004, 9:13 pm
Bruce,
76 inches is quite along leg. I'd be concerned of it bending. The other thing to think about is that these legs might go from a base (of undetermined height and attachment), to some location presumably under the 38 inches. Oliver and his apples are good once the problem is completely described.
Michael Measure twice cut once. Measure again, cut again. Measure a third time, discover that the first cut was right.
wrote:

just
38"?

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<%-name%>
• posted on April 18, 2004, 1:45 am
On Sat, 17 Apr 2004 21:01:04 GMT, snipped-for-privacy@nospam.com (Bruce) wrote:

Not likely!
Another approach in this particular case:
A right triangle with 30,60 other angles has sides in the ratio 1:2:sqrt(3). The 30 deg has to be at the top or the bottom. He likely means the top, or the table is squat.
In that case, the other non-right angle is 60, at the bottom.
With a vertical side of 38", the slope is twice that divided by sqrt(3), or 43.88" [ ~43 7/8].
Dan.

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<%-name%>
• posted on April 18, 2004, 2:45 am

Yes, TRUE.
*IF* the leg angle is 30 degrees off the _horizontal_.

And if the tables _is_ squat, to use your terms, then the leg length *is* 6' 4".

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<%-name%>
• posted on April 18, 2004, 3:33 am
In rec.woodworking

As I tried to imply in my message.

Strangely enough, the exact same number I came up with. Did you have a point? The rules about 30-60 triangles are nothing but memorized derivitives of the sin rules.

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<%-name%>
• posted on April 18, 2004, 3:07 pm
On Sun, 18 Apr 2004 03:33:04 GMT, snipped-for-privacy@nospam.com (Bruce) wrote:
Hi,

I was just giving an alternate approach, for God's sake. The "rules" are what makes it work, and they came from an 'understanding' of the working principles, ratio of sides. It is a matter of understanding, not of memorising. You learned to do fractions by looking at the simplest cases first (1/2, 1/3, 1/4, ....) then you learned the rules you could apply to any other numbers. It's no different here.
In any case, you can figure it out for yourself. I mentioned the 30,60,90 only because it applied in this particular case, as I already said. That made an easier approach possible. If you stick to just one method, it's like having just one hammer in your toolbox. You can ask about each problem as you meet it, or learn the basic principles and apply them anytime. Your choice.
Dan.

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<%-name%>
• posted on April 19, 2004, 11:14 am
Bruce wrote:

Perhaps he has access to a pencil?

--
--John
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