Dihedral Angles

SoapyJoe · 2006-05-21 18:47:23

I am constructing a large model of the Three Frequency (3V) form of a Icosahedron.

Can anybody help me with the dihedral angles for the two triangles ?

Soapy

MathsIsFun · 2006-05-21 21:38:26

Hopefully someone can help SoapyJoe with this ... ?

All I know is that to make a "Three Frequency form of an Icosahedron", each face will be replaced by 9 smaller triangular faces (the "3" is how each edge gets subdivided). It is also important that the new faces "bulge out" a little so that the overall shape approximates a sphere.

It looks like a Geodesic Dome.

(Correct me if I am wrong SoapyJoe)

For my part, I am going to see if I can model it in POV-Ray

MathsIsFun · 2006-05-21 21:55:05

I found some software that generates POV-Ray files, and then I was able to make the following image in POV-Ray. The software was also able to output data about how it subdivided the Icosahedron's faces:

Geodesic Dome Data
----------------Dome Parameters-------------------------
Polyhedron Type: Icosahedron
Class I
Frequency: 3
----------------Symmetry Triangle Data------------------
Vertexia: 10
Edges: 18
Faces: 9
-----------Symmetry Triangle Spherical Coordinates------
Vertex #	phi	theta
  1	 0.00000000	0.00000000
  2	 0.00000000	20.07675127
  3	72.00000000	20.07675127
  4	 0.00000000	43.35819755
  5	36.00000000	37.37736814
  6	72.00000000	43.35819755
  7	 0.00000000	63.43494882
  8	22.38617756	59.00803123
  9	49.61382244	59.00803123
 10	72.00000000	63.43494882

----------Symmetry Triangle Chord Data------------------
Vertex
start/end	Chord Length
           1-2	0.34861549
           1-3	0.34861549
           2-3	0.40354821
           2-4	0.40354821
           2-5	0.41241149
           3-5	0.41241149
           3-6	0.40354821
           4-5	0.41241149
           4-7	0.34861549
           4-8	0.40354821
           5-6	0.41241149
           5-8	0.41241149
           5-9	0.41241149
           6-9	0.40354821
           6-10	0.34861549
           7-8	0.34861549
           8-9	0.40354821
           9-10	0.34861549
---------Symmetry Triangle Axial Angle Data-------------
Vertex
start/end	Axial #1	Axial #2
           1-2	79.96162436	79.96162436
           1-3	79.96162436	79.96162436
           2-3	78.35927686	78.35927686
           2-4	78.35927686	78.35927686
           2-5	78.09990870	78.09990870
           3-5	78.09990870	78.09990870
           3-6	78.35927686	78.35927686
           4-5	78.09990870	78.09990870
           4-7	79.96162436	79.96162436
           4-8	78.35927686	78.35927686
           5-6	78.09990870	78.09990870
           5-8	78.09990870	78.09990870
           5-9	78.09990870	78.09990870
           6-9	78.35927686	78.35927686
           6-10	79.96162436	79.96162436
           7-8	79.96162436	79.96162436
           8-9	78.35927686	78.35927686
           9-10	79.96162436	79.96162436
---------Symmetry Triangle Face Angle Data-------------
Face	Face Angle #1	Face Angle #2	Face Angle #3
           1	70.73054094	54.63472953	54.63472953
           2	60.70841749	60.70841749	58.58316502
           3	60.70841749	58.58316502	60.70841749
           4	60.70841749	58.58316502	60.70841749
           5	54.63472953	70.73054094	54.63472953
           6	60.70841749	60.70841749	58.58316502
           7	58.58316502	60.70841749	60.70841749
           8	58.58316502	60.70841749	60.70841749
           9	54.63472953	54.63472953	70.73054094
End Dome Data

SoapyJoe · 2006-05-22 01:50:31

Yes, this is exactly the sort of thing that I mean.

What I want to know is at what angle do I set my bandsaw when cutting out the two forms of the triange?

I am uploading an image to see if it makes things a bit clearer.

Soapy

MathsIsFun · 2006-05-22 12:32:42

That data I posted above may have everything we need. It is just a matter of working it into a form that will suit.

For example, if you look at the chord lengths - they are the same as in your illustration.

So it is just a matter of figuring out how to go from the angles at the end of the list to the cutting angles.

(BTW: I guess there will be three different triangles to make, but it may be more)

SoapyJoe · 2006-05-22 18:39:47

Thanks..... I suspected that all the figures needed are contained in the data that you posted.

It is just a question of fully understanding them .... that's up to me!

I am quite fascinated by this Pov-Ray Can you explain it a bit?

soapy

MathsIsFun · 2006-05-22 20:29:03

SoapyJoe wrote:

It is just a question of fully understanding them .... that's up to me!

That's OK, we can work on it together. I was thinking of drawing up a diagram with the points on it, and taking it from there. I just haven't got around to doing it!

SoapyJoe wrote:

I am quite fascinated by this Pov-Ray Can you explain it a bit?

POV-Ray is "Ray-tracing" software - basically it will create a pretty pictire if you give it the right data. In this case the data came from a program I found called "WinDome" (http://www.applied-synergetics.com/ashp/html/domes.html), which has a few options for output. So it was a 2-step process: Windome for the data, then POV-Ray for the image.

SoapyJoe · 2006-05-22 20:56:56

Your help on this would be really appreciated.

I am no mathamatician... just very interested in the beauty of it all.

I have just started producing the square lenghts of Mahogany needed prior to constructing the 120 triangles required for the framework of the sphere and the 60 or so for the base.

I am quite happy about the chord data for the two types of triangles

Once these hollow triangles are constructed I then saw the various angles and start assembly .... I find powerful 'Bulldog' paper clips are a great asset.

When I constructed the Truncated Icosahedron model assembling it was pure poetry .... it went together with magical precision..... I must have got the angles 'spot on'.

It is going to take me several weeks to make the components for this model but it would be nice to know that I have the answer to the main problem that I face..... the ANGLES.

Yours

Squeaky

SoapyJoe · 2006-05-22 21:06:03

Just checked.... I actually need 180 triangles for the 'sphere' and 90 or so for the base.

Soapy

MathsIsFun · 2006-05-23 11:21:52

OK, well I better start deciphering that data then!

First task: work out the numbering scheme. The points are given in spherical coordinates near the top. Unfortunately they are called just "phi" and "theta", which can have different meanings, depending on where you were taught.

Anyway, I will assume this is the meaning: you start at the center of the "sphere", move "radius" distance straight up. Then rotate "theta" from that top point down towards the right. Then rotate "phi" around the compass, and that is one of the points.

1st Point is directly at top (zenith)
2nd Point is 20° down from zenith
3rd Point is 20° down from zenith, then 72° around

Thus those three points seem to make the top triangle, and the other points seem to be in this order:

1
2 3
4 5 6
7 8 9 10

John E. Franklin · 2006-05-23 12:02:01

In post #3 by MIF, the picture has mostly six triangles coming together, but just above center, five triangles come together.
Why is this, and how many times do five triangles come together on the entire shape?

And also I see the five in lower left and lower right of picture.

Last edited by John E. Franklin (2006-05-23 12:03:16)

SoapyJoe · 2006-05-23 18:52:46

The triangles form 20 hexagons and 12 pentagons.

If you look at a football (Bucky Ball) which is of course a truncated Icosahedron... then decide to divide each face into roughly equilateral triangles ...we get 20X6+12X5 =180 triangles.

This is not how the 3V form of a Icosahedron is produced but this is the way that I first saw it .... I had made a model of a TI and looking at it thought .... if I now divide each face into smaller triangles I will get a better fitting model of a sphere.

SoapyJoe · 2006-05-23 18:58:26

Thanks Playing .... In my own 'tin pot' way I was comming to the same solution.

I had not fully figured it all out but do believe that I well on the way.

Thanks very much for your kind assistance.

I am at present working out a colour code for the components of each triangle .... cord B with 60.71 degree angles, cord B with 54.64 degree angles ..... etc.,etc.

Soapy

MathsIsFun · 2006-05-23 20:01:01

Slow progress, Joe. I grab a few minutes when I can

Anyway, here are the triangles as "points" that I have been playing with in Excel.

MathsIsFun · 2006-05-24 11:22:37

Let's have a go at working out the dihedral angle between triangles 1-2-3 and 2-5-3 using the points:

1: (0, 0, 1)
2: (0.343279, 0, 0.939234)
3: (0.106079, 0.326477, 0.939234)
5: (0.491123, 0.356822, 0.794654)

We will do it in to steps: 1) Work out the equations of the two planes, then 2) find the angle between the two planes

1) First, let's work out the equation of the planes of each triangle

The general equation of a plane is: Ax + By + Cz + D = 0

And you can work out each of the constants if you know the (x,y,z) coordinates of three points (labelled 1,2 and 3 here) by the following equations:

A = y1 (z2 - z3) + y2 (z3 - z1) + y3 (z1 - z2)
B = z1 (x2 - x3) + z2 (x3 - x1) + z3 (x1 - x2)
C = x1 (y2 - y3) + x2 (y3 - y1) + x3 (y1 - y2)
D = - x1 (y2 × z3 - y3 × z2) - x2 (y3 × z1 - y1 × z3) - x3 (y1 × z2 - y2 × z1)

Crunching the numbers from my Excel screenshot above for triangle 1-2-3 I get:

A=0.019838701 B=0.014413695 C=0.112072698 D=-0.112072698

So the plane of the triangle 1-2-3 is: 0.019838701x + 0.014413695y + 0.112072698z -0.112072698 = 0

Likewise for the triangle 2-5-3:

A=0.047202045 B=0.034294376 C=0.132905844 D=-0.141033158

2) Knowing two plane equations, we can work out the dihedral angle using the formula:

Once again, let's crunch some numbers

cos(alpha) = 0.9804118

So, the dihedral angle should be cos-¹(0.9804118) = 11.36°

NOW, this seems a bit low to me, so I may have made a mistake somewhere ...

Would someone like to check my work here?

SoapyJoe · 2006-05-24 18:54:15

Thanks for information it is really appreciated.

Obviously, I will check this out on some MDF before I start to cut the angles on the Mahogany.

I will be revolving the 'sphere' on a smaller dome constructed in a very similar manner (constructed from a subdivided Icosahedron at a line just above half way where the triangles almost divide a plane)

At this point the bases of 15 triangles make a plane..... so I feel that is a sort of guide to the angles.

I am enclosing a picture of the shape and construction of the base that will house the motor drive .... just found a lovely motor.... quite powerful and with a bit of extra gearing I can get the speed down to 0.4 RPM

I saw the motor in a 'surplus' electrical goods shop .... £4.95 each.

Soapy Joe

SoapyJoe · 2006-05-24 19:05:32

The image of the base... I hope

SoapyJoe · 2006-05-24 20:13:03

In my post I wrote ...At this point the bases of 15 triangles make a plane

This is not absolutely correct.... in non mathematical terms... the base of the triangles .... are slightly 'crinke cut' so the top edge of the 'walls' each have a slight angle.

For anybody interested I am uploading two pictures of previous models.

Soapy

SoapyJoe · 2006-05-24 20:16:54

Model No 1

SoapyJoe · 2006-05-24 20:18:11

Model No 2

MathsIsFun · 2006-05-25 11:20:56

SoapyJoe wrote:

Obviously, I will check this out on some MDF before I start to cut the angles on the Mahogany.

Good idea!

I have not been able to prove or disprove my calculations above. Except I did apply them to 2 faces of a cube and I got 90°, then I tried 2 faces of a standard icosahedron and got 138° which are both correct, so I guess we have to trust them for now.

MDF level of trust, not Mahogany level of trust!!

So, I will go ahead and calculate some more angles for you soon.

MathsIsFun · 2006-05-25 12:13:04

Using this guide:

1
2 3
4 5 6
7 8 9 10

I calculated these angles:

1-2: 14.435°
1-3: 14.435°
2-3: 11.359°
2-5: 13.578°
2-4: 14.458°
5-9: 13.578°
3-6: 14.458°
6-9: 11.359°
6-10: 14.435°

Yes, I calculated more angles than I needed to, but I wanted a "double check". So my calcs are good, it is only the method itself that may have a flaw in it.

Anyway, your MDF model will prove or disprove it all!

(PS: I know you don't need thouandths of a degree, but the numbers were there, so I typed them in)

(PPS: Yes, there are 4 different angles versus 3 different chord lengths: 2-3, 4-8 and 6-9 are the ones to watch)

SoapyJoe · 2006-05-25 18:49:54

These figure 'feel' correct to me... thank ever so much!

I will try out the angles with some MDF ... possibly just a few rows of the 'Igloo' (base).

I have found, by experience, that very quicky one can soon judge if the figures are working out.

I must just say how much I have enjoyed this correspondence....

MATHS IS FUN

Soapy

SoapyJoe · 2006-05-27 21:35:35

Just a quick update on my project.

This is a quick hardboard model of the Icosahedron marked out for the 3V form.

Soapy

MathsIsFun · 2006-05-28 01:07:17

A useful model - shows the pentagon and hexagon arrangements nicely.

(Not being able to physically lay my hands on it) Do you think that to get a neat join, you would "mitre" the cuts in that model by 20.9° ? (180°-138.2°=41.8°, 41.8°/2=20.9°)

And, do you think that when the triangles are "bulged out" to form a sphere, the equivalent mitre would be (for line 1-2) 14.435°/2 = 7.2° ?

Math Is Fun Forum

#1 2006-05-21 18:47:23

Dihedral Angles

#2 2006-05-21 21:38:26

Re: Dihedral Angles

#3 2006-05-21 21:55:05

Re: Dihedral Angles

#4 2006-05-22 01:50:31

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#5 2006-05-22 12:32:42

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#6 2006-05-22 18:39:47

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#7 2006-05-22 20:29:03

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#8 2006-05-22 20:56:56

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#10 2006-05-23 11:21:52

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