Solid geometry problem; solution for angle β sought.

Chichester · 2011-10-13 15:18:07

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?

Bob · 2011-10-13 21:29:33

hi Chichester

I am fairly confident that I can help here (fingers crossed! )

But I want to be totally clear that I'm working on the same problem as you. So I've made a perspective view of what I'm getting. Please have a look at the image below and say if you agree.

I'm thinking angle BDE = alpha and angle BCE = omega.

And you want angle BA2E. (Might be easier to call A2 just A, maybe)

Then I'll do the same for case #2. Then I'll think about the 3D geometry. OK?

Bob

Last edited by Bob (2011-10-13 21:32:47)

Chichester · 2011-10-14 04:17:39

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!

Bob · 2011-10-14 06:54:36

hi Chicester,

Good diagram. I've used my (vector) geometry program, Sketchpad to make a copy of everything left of the line of symmetry. It's not quite your measurements because, at thousandths of an inch, it hasn't quite got the resolution to place points exactly where I wanted. But that won't matter as we'll be using a calculator or spreadsheet to work out the angles.

Have you got Microsoft Excel by any chance. If yes, I can give you the exact formulas to work out any angle. If not, we'll have to find a way round with formulas.

I think you can do it by co-ordinates and you won't need to re-set them. If we can establish the 3D coords of the key points then one formula will compute the angle. I'll use vectors. Did something similar recently for another member. see:

http://www.mathisfunforum.com/viewtopic … 25#p183425

I didn't really get involved until post #28, with the formula at post #30.

You didn't say what the distance between the surfaces will be.

My diagram is below. When I measure between two points with the software it assigns letters to the points starting with A. Would you mind if we used that lettering. It'll save me having to re-do my diagram.

Bob

Bob · 2011-10-15 23:37:48

hi Chichester,

Had a re-think. In the other post the planes of the dodecahedron had no 90 angles anywhere so the vector dot product was more or less forced. It would be very hard to calculate angles otherwise.

But, in your case, all the verticals are parallel and at right angles to the base, so once the coords of points have been found, basic right angled trig will do the rest.

See revised sketch (not accurately drawn) which shows what I would try to calculate.

I'm assuming you want angles such as HQP and NFT (which I think you have already calculated??)

Bob

Chichester · 2011-10-16 15:12:56

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.

Chichester · 2011-10-16 15:36:07

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.

Last edited by Chichester (2011-10-16 15:37:25)

Chichester · 2011-10-16 16:06:11

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.

Last edited by Chichester (2011-10-16 16:09:40)

Chichester · 2011-10-16 18:51:56

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?

Last edited by Chichester (2011-10-17 03:58:08)

Bob · 2011-10-16 20:36:55

hi Chicester,

I'm in the UK, so that's quite a time difference!

You want the angles so you can make the right shapes. D'oh! I'm a fool! Now you see another difference between a mathematician and a carpenter. I was just looking at it as an interesting geometry problem.

I think what you want is even easier to calculate, once all the coordinates are known. So during the day I'm going to do that; then I'll give you the general formula for calculating any angle in a surface. Back later.

Bob

Bob · 2011-10-17 06:39:35

hi Chicester,

I've done two things; both shown below.

I've tidied up the perspective so as to minimise the number of points needed. The unlabelled point was meant to be 'U' but I somehow forgot to enter it on the diagram. But I've moved it and the dotted triangle as well so that U has moved over the top of M. R is directly below it and S is somewhere on DI (I'll come to that later.) Similarly I've put H where it should be as a vertex on the top layer, with P directly below and Q somewhere on AD. (Again more on this later.)

Then I have begun a spreadsheet with all the 3D coordinates for the points.

I've taken F as the origin (0,0,0). The x axis goes left in the direction towards I. Normally we would have the x axis going the other way but it won't matter because the shape is symmetrical. The y axis in the direction B. The base layer is all on the x-y plane so has 'z' coordinate 0. The top layer is 24 above, so all the z coordinates there are 24.

Based on your original diagram, I've put on what I think are the correct coordinates. Please check them in case I've got any wrong.

You'll see I haven't yet got the complete coordinates for S and Q. S will lie on DI so that MS is at right angles to DI. That means the angle between the base and that sloping side will be given by RSM. In the same way Q will lie on AD so that HQ is at right angles to AD and the angle between the sloping side and the base is HQP.

If we need these coordinates I can get them by vector geometry, but it may be you don't need them.

To get the true shape of the sloping sides you want the angles at the corners and possibly the sloping distances as well. The vector dot product method will do the former and Pythagoras' theorem, the latter.

So where do we go from here.

(i) Are you happy with the revised perspective?

(ii) Do you agree with my coordinates?

(iii) Are you ready for the vector theory?

Bob

Chichester · 2011-10-17 18:07:23

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.

Bob · 2011-10-17 19:44:58

hi Chicester,

Once again, d'oh. I used my diagram, not yours. That was stupid of me, because I knew my measurements were a bit off. I've re-done the table (below). For point R I've not put y = 4.750 but rather 0.750 as it is directly below M.

Yes, that's where B is. It was a bit close to other points on my perspective, so I left it out for clarity.

I'll post this now, and add the vector explanation later today.

Bob

Last edited by Bob (2011-10-17 19:48:01)

Chichester · 2011-10-18 04:55:53

Good, thank you. I don't know how R got mislaid. The usual way, I expect.

Bob · 2011-10-18 07:14:33

hi Chicester

Vector method for finding the angle between two lines in 3D.

Suppose the lines are UV and VW and we want the angle UVW.

(i) You need to know the coordinates:

(ii) Now calculate the vectors for VU.

Note 1: This just means how much do you need to go in the x direction to get from V to U; how far in the y direction; and finally how far in the z direction.

Note 2: Ive deliberately chosen VU rather than UV, so Im consistently working away from the angle.

Note 3: Vectors are usually written as a column of numbers, but theres no mathematical reason why they shouldnt be in a row as long as we dont get muddled between vectors and coordinates. A vector tells you how to get from V to U; a coordinate just tells you where the point is in an axis system.
So

and in the same way:

(iii) We need to compute something called the dot (or scalar) product. There are two ways of doing this; one way just needs the three numbers in each vector; the other needs the size of each vector and the cosine of the angle between them. Because we will know everything except the cosine of the angle we can set the two ways equal, re-arrange and hence get the cosine of the angle. Then it is easy to get the angle itself.

Heres the maths for two vectors (p,q,r) and (s,t,u)

(p,q,r,s,t,u have nothing to do with the letters on the diagram ... Im just running out of letters!)

so

(iv) Thats the theory. Now for the spreadsheet.

First I would make a sheet with all the coordinates of the points as I have done.

Then set up three lines with three cells in each for the nine coordinates (three for each point) for U, V and W

Then compute the two vectors, VU and VW

Then compute the angle by using the formula and performing an inverse cos or arccos.

Note: Excel works in radians by default not degrees. Your spreadsheet is probably the same. To convert you do

My second screen shot below shows my formulas in Excel for all of the above.

(v) Testing. Theres a good chance of an error in all that lot so the stages need testing. I seem to remember you had worked out angle TFN already.

So copy the coords as follows U = T; V = F; W = N and let the sheet do the work.

(vi) My first shot shows my figures for this.

I have also used the coordinates from the dodecahedron problem to check my formulas. I got 142.62 for the dihedral angle, which is correct.

Bob

Last edited by Bob (2011-10-18 07:16:40)

Chichester · 2011-10-18 16:21:05

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).

Chichester · 2011-10-19 18:19:59

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...

Bob · 2011-10-19 19:09:07

hi Chicester,

go into the crawlspace under a house, looking for a sump pump

and when you find it .....

Doesn't sound like a pleasant job. Take care!

Bob

Chichester · 2011-10-20 18:08:41

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?

Last edited by Chichester (2011-10-20 18:15:48)

Bob · 2011-10-20 21:00:08

hi Chicester,

I'm pleased to hear that my method is giving you what you wanted.

As I typing out the method, I wondered whether I should put in more explanation about the dot product. The post was getting pretty long so I decided against. But, I'm glad you want to know. We'll turn you into a mathematician yet!

From trig, I know that:

There's a lot more to cosine than that. There's a more general definition for cosine that allows us to get beyond 90 degrees, and an algebraic definition too.

In vector theory, mathematicians have asked the question, "What do we get if we multiply two vectors together?"
And then, "Should the answer be a vector, or a non-vector, which is called a scalar?"

It's been found useful to have both possibilities, so there are two ways to multiply vectors:

The vector product, or cross product, in which the answer is a vector and

the scalar product, or dot product, in which the answer is a scalar (just a number).

The dot product is defined by (the one I called dot product 2)

where the angle is the angle between the two vectors; and a, b means the length of the vectors.

It might seem to be 'just plucked out of the air', but this definition turns out to be really useful.

In Physics, for example, mechanical work is the dot product of force and displacement vectors, and
magnetic flux is the dot product of the magnetic field and the area vectors.

http://en.wikipedia.org/wiki/Dot_product

For us, it's useful for calculating angles between vectors.

If i is a unit vector in the x direction and j a unit vector in the y direction

then any vector can be written like this

So when you form the dot product of two vectors you get

but the lengths are just ax etc and the angles are all 90 or zero so this becomes

But cos(0) = 1 and cos(90) = zero so this simplifies to

which is my alternative dot product 1.

So this proof shows why the two definitions give the same thing. Hope that makes sense.

Bob

Math Is Fun Forum

#1 2011-10-13 15:18:07

Solid geometry problem; solution for angle β sought.

#2 2011-10-13 21:29:33

Re: Solid geometry problem; solution for angle β sought.

#3 2011-10-14 04:17:39

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#4 2011-10-14 06:54:36

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#5 2011-10-15 23:37:48

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#6 2011-10-16 15:12:56

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#7 2011-10-16 15:36:07

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#11 2011-10-17 06:39:35

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#13 2011-10-17 19:44:58

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