Req help with wwking trigonometry

• posted on May 12, 2004, 9:01 pm
Hi all,
Since my employer would frown on me bringing in my wwking tools to work, the only tools I can use are a pencil and a piece of paper to do design stuff. Last night, I was thinking about keyed miters and made a spreadsheet showing how deep to cut to make a certain length keyslot and the maximum depth of cut you could go before cutting through to the inside of the box. This was all based on cutting the keys at 45 degrees so I was able to use the only trigonometry I know, A^+B^=C^, to do all the equations.
Shortly before work ended, I started thinking about making keys with faces proportional to the faces of the box. Everybody loves the golden rectangle, so that's what I decided to use for my angles.
The problem is that I don't know where to go from here. I know the angle the box is tipped at (31.718deg), and the desired length of the key (we'll call that N), but don't know how to figure out how to determine the depth of cut to end up with the correct lengths of miter slots. Along the same lines, knowing the thickness of the lumber and angle the box is tipped at, how deep can I cut before cutting through to the inside of the box?
Here's a diagram of where I'm stuck at:
http://www.skunkduster.com/woodworking/goldenkey.gif
Algebra was the highest math class I took, and that was 15 years ago, so little words and basic concepts would be appreciated.
Thanks, -Rick

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• posted on May 12, 2004, 7:41 pm

I went to http://www.skunkduster.com/ and...... I'll think about it.
Larry

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• posted on May 12, 2004, 9:05 pm
wrote:

So did I. I won't.
Bill.

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• posted on May 12, 2004, 9:53 pm
Lawrence L'Hote wrote:

I suppose I should change that one of these days. I found it amusing so I stuck it there as a placeholder, then never got around to building a website. Now, I just use that webspace as a place to host images.

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• posted on May 12, 2004, 8:43 pm
Why not label N on your diagram?

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• posted on May 12, 2004, 9:00 pm

Add to your drawing the dimensions that you are asking about and someone will be able to figure it out. I am unsure after reading your post and looking at the picture what you are trying to calculate.
Frank

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• posted on May 12, 2004, 9:02 pm
its not real clear what you are asking.. but i think its this:
let W = the width of the board you are using for the box frame N = length of vertical blue line as it passes through the frame, A = the angle from the blue line (in your case 31.718 degrees)
this creates a small right triangle.
sin = opposite over hyp
sin A = W/N
solve for N
its not real clear from the diagram what you are asking so im assuming some things.... my formula will give you the length of the vertical blue line from the corner of the box, to the intersection of that blue line and the first black line it crosses. if this isnt it, why dont you re do it showing clearly what dimention you are looking for..
randy

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• posted on May 12, 2004, 9:24 pm
All the trig functions ( cos, tan) are available in the Windows calculator or spreadsheet.
For box not tilted: key length (L) = 2 x depth of cut (D) face length 1 (F1) = face length 2 (F2) = 1.414 x depth of cut (D)
For box tilted n deg: angle a = 45 - n, angle b = 45 + n L = ( tan(a) x D ) + ( tan(b) x D ) F1 = 1/( cos(a)) x D F2 = 1/( cos(b)) x D
example: box tilted 15 deg D = .5" angle a = 30 deg, angle b = 60 deg L = ( 1 / tan(30)) x .5 + ( 1 / tan(60)) x .5 L = .288 + .866 = 1.154"
F1 = ( 1 / cos(30)) x .5 = .577" F2 = ( 1 / cos(60)) x .5 = 1.0"
Max depth of cut at 45 deg = 1.414 x side thickness Max depth of cut when tilted = 1.414 x thickness x cos(tilt angle)
This reduction is not a big deal... unless you tilt the box a lot...
Hope this helps! JeffB
Rick Nelson wrote:

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• posted on May 12, 2004, 10:17 pm
small oops... see correction below
JeffB
JeffB wrote:

^^^^ that should be >> L = ( tan(30)) x .5 + ( tan(60)) x .5

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• posted on May 12, 2004, 11:49 pm
wrote:
Ok, just looking at the diagram shown as a 2D frame, you need to state the width of the frame members. So, let's call that W.
Making the miter cuts at 45 deg will make the miter cut sqrt(2)*W [That pythagoras stuff you know.]
Drop a perpendicular from the inside of that miter joint down to the horizontal and call that H.
Adding the tilt angle (31.7 is close enough) to the miter angle (45) you get 76.7 deg
Then H/[sqrt(2)*W] = sin(76.7) deg.
This will allow you to find H, the maximum height of cut, just barely into the box. Anything less will do.
Personally, I'd eyeball it, and you won't see it from 100 meters.
Bill.

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• posted on May 13, 2004, 12:29 am
Rick Nelson wrote:

I have a web page where you can find my trig notes - these and a bit of algebra go a long way.
See: http://www,iedu.com/DeSoto/trig.html
Also, if you DAGS, you'll discover that there's a lot of help available on the WWW.
--
Morris Dovey
DeSoto, Iowa USA

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• posted on May 13, 2004, 12:40 am
Morris Dovey wrote:

That comma key is /not/ my friend. Try:
http://www.iedu.com/DeSoto/trig.html
--
Morris Dovey
DeSoto, Iowa USA

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• posted on May 13, 2004, 3:19 pm
wrote: -snip-

Hello, Rick.
Some assumptions - Plane of the cut is parallel to the plane of your picture, Direction of cut is parallel to the horizontal blue line in your picture, Depth of cut is measured along the vertical blue line, Length of Key (N) is the total length along the bottom of the cut.
If those assumptions are correct, the depth of cut is:
D = N * sin(A) * cos(A)
where    D = depth of cut     N = total length of cut (length of the key)     A = angle (31.718 degrees in your picture)
Maximum depth of cut depends on the thickness of the members. Assuming a 45 degree miter - both members joined at the miter are the same thickness, the maximum depth of cut is:
D = T * (sin A + cos A)
where    D = Maximum depth of cut     T = Thickness of the stock (as shown in your picture - could also be considered the "Width" of the stock)     A = angle (31.718 degrees in your picture)
The terms width and thickness could be confusing. If your diagram were showing the face view of a picture frame, I'd call it the "width" of the frame members. If it is showing the "end view" of a box, I'd call it the thickness of the sides of the box.
I don't understand your application of the "Golden Ratio" here. Will all the boxes be the same length to width ratio? If you want the visible length of the key to be proportional to that particular side of the box, the angle will be different for each L/W ratio.
L/W = cot A
or
A = inv tan (W/L)
where    L = length of the box/frame (right side of your figure)     W = width of the box/frame (left side of your figure)     A = angle measured relative to L (31.718 degrees in your figure)
In other words, your angle (31.718 degrees) is only appropriate for a box/frame with an L/W ratio of 1.618
Hope this helps.
Tom Veatch Wichita, KS USA

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• posted on May 13, 2004, 5:32 pm
Thanks for the help everybody!
I had spent some time searching the WWW for the answer, but I felt like a guy trying to do a search on how to put motor oil in a car. You'd probably come up with several pages detailing viscosity, thermal breakdown, the refining process, environmental issues, etc, when all you really need is someone to say "dump it in the valve cover".
My understanding of trigonometry is pretty limited, and I'm having a hard time wrapping my mind around the concepts and understanding the terms, but from reading through the responses, this looks like exactly the information I will need. Once it soaks in, I plan to use that knowldege to build a couple of jigs for cutting keyed miter slots on a table mounted router (or a TS, I suppose) and I'll post the plans and pictures on my website, and post a note in the wreck here in case anybody else might be interesting in making such a jig.
Special thanks to Bill Rogers, who, despite his initial dislike of me, still gave me the benefit of the doubt and offered his help. Also, eyeballing it would work, but I thought it would be handy to know in the planning stages if my lumber is X thick and I want the key to be Y long, can I do it without cutting through the box?
Thanks again everybody. I'll mull over the math stuff a bit and build a couple jigs then hopefully have the information to share with everybody here in a week or two.
-Rick