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Thread: Octagonal box prototype
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31st October 2006, 12:05 PM #16
The angle, v, at the apex of the inclined triangle is (using the notation in the paper), v = 180 - 2u, where u is given by equation (5): u = arccos(d/(2L)). Alternatively, a direct construction yields v = 2 arcsin(d/(2L)). For the example given at the end of the paper, d = 62 mm, h = 20 mm, L = 83.4 mm, both of these expressions give v = 43.6 degrees.
Update: The case when the pyramid has N sides (rather than the N = 8 case treated in the paper), can be solved using the relation d = 2r sin(180°/N), instead of equation (4) in the paper. From this you can calculate the angle u, and everything else remains unchanged.Last edited by zenwood; 31st October 2006 at 12:34 PM. Reason: Generalised to an arbitrary number of sides
Those are my principles, and if you don't like them . . . well, I have others.
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31st October 2006, 12:36 PM #17
To cut the triangles for my polyhedron boxes, I have dedicated X-cut sleds with fences at the correct angles. To put the bevels on the triangles, I make up jigs to hold the triangles, tilt the TS blade to the correct angle and cut. The order in which you cut the sides is important, to avoid splintering away one apex.
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31st October 2006, 12:38 PM #18
Zenwood,
I am in envious awe of your mastery of trigonometry Thanks.
Rocker
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31st October 2006, 12:45 PM #19
It's a pleasure, Rocker. Keen to see what the lid of your box ends up looking like.
BTW: if you want to find r for a particular value of d, you can invert that last formula:
r = d / (2 sin (180°/N)).
I am in envious awe of those old coopers who could put the right bevels on barrel staves using nothing but a plane, a good eye, and trial-and-error.Those are my principles, and if you don't like them . . . well, I have others.
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31st October 2006, 01:34 PM #20
Zenwood,
I am not sure that I want to risk ignominious failure by completing a lid for my prototype. The sides are not really high enough to make a worthwhile box by slicing off a section from the top of the sides. Boxes are not really my thing; I prefer furniture, where the occasional minor error is not too noticeable. A box maker must produce perfection, which is not my way I was more interested in making the jig, to show what was possible. I leave it to others with more artistry to make a good box with it.
Rocker
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31st October 2006, 01:52 PM #21
Zen old mate you lost me at u = arcoss(????) What the hell are you trying to do..... Give me a blinkin' head ache.
Seriously you have done well but I reckon you have a background in rocket science or tertiary level mathematics. Dude I am sorry but that would go in the maybe one day when hell freezes over basket!!!!
Well done
PeteIf you are never in over your head how do you know how tall you are?
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31st October 2006, 02:16 PM #22
Nothing complicated about the arccos, Doughboy: just about every calculator has that button. If I get the inclination, I may implement a spreadsheet to do it automatically.
Those are my principles, and if you don't like them . . . well, I have others.
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31st October 2006, 02:44 PM #23
The arccos is also called the inverse cosine.
If you don't have a calculator, but are using Windows and the standard calculator, then in scientific mode, you tick the inverse box before clicking the cos button.
Or you can wait for zenwood to produce a spreadsheet (if you have excel). He will presumably do all the fiddling to turn the numbers from radians (used by default by excel) to degrees, which we understand and shows on our protractors.
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1st November 2006, 09:18 AM #24
Hmmm...cute: the windows calculator is the only one I've seen that has a checkbox for the 'inv' function: why not make it behave like a real calculator and have a button (or shift key) that does this? Also why label the scientific functions in hard to read purple text? Why not relabel the keys to 'arccos' or cos-1 etc. when the 'inv' function is called? Why such a basic range of mathematical functions (no fractions, no combinations or permutations), and why doesn't the inverse factorial work, and why not label the pi button with the greek symbol...?
Sorry: just another microsoft moment... Not sure I can face Excel today. What to give it a go Trifid?Those are my principles, and if you don't like them . . . well, I have others.
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1st November 2006, 03:10 PM #25
Excel spreadsheet
Well, that certainly sounded like a challenge. Having nothing better to do than work I've given it a go and my effort is attached.
Attachment 33654
My maths isn't at Zenwood's level - I've really just followed his workings. I also used his symbols (except that I had to change r to b, because Excel wouldn't let me use r).
Also, if you only skimmed Zenwood's workings, you might have thought that the mitre at the base of the inclined triangle is e + 90°. That is actually the angle at the corner of the mitred side. I've added a new calculation for the mitre at the base of the triangle (t). Zenwood might want to check my calculation for this.
Oh and Rocker, an idea. If your sides are actually 62mm, then I recommend an apex height of 13.87. that will make your mitres as close as damn it to 86° (sides) and 100.5° (base), which is nice easy numbers to find on your protractor.
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1st November 2006, 05:18 PM #26
Trifid,
You're a scholar and a gentleman! Thanks for posting that; saved to my collection of reference do-dads. Working such matters out myself makes what few lumps of grey matter that I may have left ache (and that generally means taking some medicinal Red, which completely buggers the calculation :eek: )
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2nd November 2006, 11:57 AM #27
Calculating mitre angles for regular multi-sided pyramids
The document has been updated to include the general case of an N-sided pyramid, and also includes the bottom mitre angle, as suggested by Trifid.
Trifid: Excellent job on that spreadsheet. I wasn't sure exactly what you meant by the base mitre angle, so I calculated one according the the updated document. One slight glitch: I get a different value to yours for the example: 95.7º, compared to your 104.96º; so I'm not sure if we mean the same thing, or whether one of us has made a mistake. I checked my result using the measurement function within Sketchup and it came out pretty close.
Another thorn is that the document is now over the 100kB limit on this forum, so I've uploaded it to
http://www.wikiupload.com/download_page.php?id=27126
Navigate to this URL, then click on the "Download File" button, enter the hard-to-read code letters, then click 'get'.
Here's another challenge: what are the pyramid dimensions that yield integer values for the mitre angles?Those are my principles, and if you don't like them . . . well, I have others.
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2nd November 2006, 12:20 PM #28
Zenwood,
Where does it "get" it to? When I click get my screen just blinks and returns to the wiki-upload screen.
The base mitre that I was referring to is the one marked on your diagram in post #9 - my calculation is spot on on the number you showed there, which is why I believed it was right.
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2nd November 2006, 12:26 PM #29
'Get' on the wikiupload page initiates a download of the pdf file, which should be rendered in your browser after a few seconds of download time (it's a 144 kB file, and I'm using Firefox 2.0). From there, you should be able to do a 'save' to your local machine. 'course, you'll need a pdf plug-in.
I realised what base angle you meant as soon as I'd posted my message. We're both right, we just meant different things.
I'll update the document, and repost it with all three mitre angles. But it may take a couple of days.Those are my principles, and if you don't like them . . . well, I have others.
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2nd November 2006, 01:50 PM #30Senior Member
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