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O-kay... I've finally had time to sit and digest everything you've said.
First off, given your two definitions of dotproduct1 and dotproduct2, I can easily do the algebra to get to a form with θ all alone on the left side of the equation. I have no trouble understanding your spreadsheet, it all makes perfect sense. You are correct that I had computed angle TFN already by other means and the answer your spreadsheet gives matches mine exactly. You have certainly answered my original question; I should have no trouble doing the computations for all of the angles I need, and that should let me get on with my wood project. But...
I had to do an end-run through trigonometry to get any kind of "feel" for what the heck a dot-product really *is*. You stated that:
(I'm simplifying back to 2-space to save wear-and-tear on my intuition)
From trig, I know that:
so:
Now I'm going to assume that vector (p,q) lies along the X axis, and I'm going to assign some values:
making the whole thing look like a trig problem in standard form (image below). So:
and finally:
which I recognize to be true.
But I have to admit I still don't really have any intuitive feel for what a dot-product is. I went to the Wikipedia "Dot Product" article (I still can't post the link) and spent quite a while trying this and that, but in the end my intuition is unsatisfied. Got any thoughts?
Sigh... this math puzzle is arguably the most interesting thing I have pending, but I'm finding it hard to fit in. It's 11PM here... I have a lot of seasonal work that has to happen right now, and I'm also a landlord... this evening I unexpectedly got to go into the crawlspace under a house, looking for a sump pump. I *really* don't want to try digesting this on a lame brain, so again I'm going to postpone until tomorrow...
Hi... once again, it's been a full day and my brain is tired. Profuse thanks for all your work, but it might take me a bit to wrap my tired brain around all this. I'll probably get back to you tomorrow evening, probably with questions (I should get free a little earlier tomorrow, and maybe have brain to spare).
Good, thank you. I don't know how R got mislaid. The usual way, I expect.
No problem with the revised picture... moving the triangles to line up with the vertices gives us points Q and S, which are what I need.
I confirm all but the following:
Point B isn't marked, but I take it to be the point directly opposite point F, on the lower base.
Point H should be (0.500, 4.750, 24.000).
Point K should be (1.500, 4.000, 24.000).
Point M should be (1.250, 0.750, 24.000).
Point N should be (0.000, 0.750, 24.000).
Point P should be (0.500, 4.750, 0.000).
Point R should be (1.250, 4.750, 0.000).
Point T should be (0.000, 0.750, 0.000).
I think you must have entered a wrong digit somewhere and the error propagated.
I'm surprised at a couple of the points you found to be of importance; why point T?
I'm about as ready for the vector theory as I'm likely to get... bring it on.
OK, I've had a little time to think about things. I'd still like to see a geometric/trigonometric solution to all this, but your comment that I shouldn't have to transpose axes if I attempted a Cartesian solution made me think I should look at that again, and now that I have, I think I can get the answers I need comparatively easily.
Attached is a section of my original view of the frustum from above, translated and rotated so that point D is at the origin, and DK is along the X axis. I have flattened the image so that only the X and Y axes are shown. From my original view, I know that K is at (1.5, 0), that the length of ZK is 0.900 inches, and that angle ZDK is 36.870 degrees. That gives me everything I need to compute the length of DZ,which gives me my solution.
This *isn't* a general solution... I can do this for points D and K because of the way the frustum was designed... but I don't think it would be so easy for points M and I, where I don't know the angle MIZ'.
It's really late, and I've had a very long day... is there some flaw in all this?
Still more thought... I don't think angles PQH and RS? are going to get me what I need... or if so, I don't understand how. I have values for the height (HP = R? = TN = 24 inches) and for PQ = 0.9 inches and SR = 1.496 inches, so I can already compute angles PQH = 87.852 degrees and RS? = 88.433 degrees. If that's enough to compute angles KDQ, KDS, QA? and SIM, I sure don't know how to go about it.
By the way, that's a nice drawing of the frustum... I rather despaired of trying to draw a meaningful isometric view.
I've been reading some more... I think that if I had angles PQH and RS?, I could compute what I really need, which is angles KDQ and KDS... from which I could compute angles QA? and SIM... these are are the four cutting angles of interest.
I'm sorry to have forgotten the distance between the frustum bases! The height of the frustum is, by design, 24 inches. By computation, QH is 24.017 inches, and S? is 24.047 inches. Just FYI, FN (which I've been thinking of as the "back" of the frustum) is 24.012 inches, and the angle opposite it (the "front") is also 24.012 inches.
Hi... I'm back, but I'm too exhausted to make sense of what you've told me. I'll recover a little and see if I can understand enough to ask intelligent questions... probably tomorrow evening (I'm on Pacific time). In the meantime... thank you very much for all the thought you've put into this! Please don't worry about my specific numbers and angles... if you can get across to me how to compute the angles, I'll do the work.
I normally use OpenOffice Calc rather than Excel, but if you'd like to post an Excel spreadsheet, I'll find a way to deal with it... Calc may just import it with no trouble.
Hi... thanks for your time and interest! Your isometric view looks to match what I was trying to get across.
You'll have worked out that I'm not really a mathematician; to me, this is a carpentry problem! I'm trying to build a pedestal in the shape of the frustum of an irregular 6-sided polygon. I've attached a top view of the two bases of the frustum, with all the measurements and angles marked.
The angles I'm trying to compute are the angles of each of the 6 boards that will make up the sides of the frustum. Because the pedestal is symmetrical with the line of symmetry passing through two of the sides (rather than two of the angles) computing the exact shapes of the front and back boards is a snap. But computing the side boards is proving very challenging.
I suppose I *could* do this by converting everything to Cartesian coordinates and brute-forcing it... but that sounds like it involves setting up the problem on one set of axes and then transposing to a new set to get my answer. I reckon I can figure out how to do all that, but it seems like a lot of work, and might involve matrices, which is where my intuition breaks down.
PS -- I will be out of touch from now until Sunday evening, so I won't be responding... but I definitely haven't lost interest!
Nope, that'll do it. Thanks for yor time!
Honestly, I *can't* just cut and paste a URL into my posts... if I try that, I get a Submit or Preview warning, and the post is refused.
Hi... I'm trying to solve a problem in solid geometry and I find myself in a little over my head. The image (that I sincerely hope will be visible with this post) shows what I'm trying to accomplish:
Two cases are shown, and a front, top and side view is given for each case.
In Case #1, line segments A1 and A3 and point A2 are all in the same line, which is contained in plane G. Line segment B extends outside plane G. Since A1, A2 and A3 are co-linear, angle θ, the angle between line segments A1 and A3, is necessarily 180 degrees. In this case, angle β is wholly determined by angle α; the value of angle ω has no effect.
In Case #2, line segment A3 has been rotated; A1, A2 and A3 are all still in plane G, but angle θ is now 90 degrees. In this case, angle β is wholly determined by angle ω; the value of angle α has no effect.
What I'm trying to get is a general formula for the intermediate values of angle β, as angle θ decreases from 180 degrees to 90 degrees. Note that the angle β I'm looking for is the angle between line segments B and A3, in the plane which contains both of those line segments, wherever that plane happens to be as θ varies.
Given my two cases, I hypothesized that the formula might be:
β = (180-α) cos(180-θ) + ω (1-cos(180-θ))
Intuitively this seems to make sense; as θ decreases from 180 to 90 degrees, the value of the first term ((180-α) cos(180-θ)) is initially (180-α) and decreases to zero, while the value of the second term (ω (1-cos(180-θ))) starts at zero and increases to ω. I haven't been able to prove (or disprove) my guess, but I have run some numbers, and I now suspect that my hypothesis is incorrect.
Can anyone suggest how I should proceed?
Hi... I did try uploading an image; I selected the file and the filename appeared in the box at the bottom. But when I clicked on Preview, the image wasn't there. And I don't know what you mean when you say "cut and paste the address in my next post". Any attempt to include a URL in my post results in a Preview message disallowing that. I tried to include it with *this* post, but I wasn't allowed to submit it. The image is at
h t t p : / / w w w . t n s t a a f l . o r g / m a t h / M a t h %20 C a s e s . p n g
but if I have to include it that way, nobody is going to go to the bother of looking at it. I'm going to try submitting with the image uploaded and see what happens...
Hi... I have a question I'm wanting help with, but it needs an image to explain it. The problem is, I can't post an image; I can't post a link, I can't even post a hashed-up link with the word "dot" inserted where periods would appear in the link name. I'm told to "Please describe where" the image is, but I'm not real sure how to do that... what is the accepted procedure?
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