Understanding hub radial angles vs face angles at vertex

[Anthony Sokolowski]

I am in the process of designing hubs for a steel framed 3v icosa dome. A 3d model has been designed and I have found that the angles on the hub are not the same as the angles of the triangle faces.
What I mean is that a hub that joins the 5 red struts in the image below (copied from geo-dome.co.uk) has a radial angle of 72 degrees (360/5=72) which makes sense. However, this hub joins triangles whose face angles at that hub are 70.73 degrees. The same is true for all the hubs shown in the image below.
Can anybody explain the logic and mathematics behind this?

[Dick Fischbeck]

The sum of the angles around a vertex in a polyhedron has to be less than 360. Otherwise, you don't have any curvature. 

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[Paul Kranz]

Want more dome math?

The 70.43° is not arbitrary. You can compute it with simple algebra.

Let A = the face angle where the five pentagon triangles meet (red points in the BMP)

Let B = the face angle where the six hexagon triangles meet (cyan point in the BMP).

In order for both points to be on the same sphere, they have to encompass the same number of degrees such that

5A = 6B

Now, we factor in their base angles which meet at the green points on the BMP such that

5A = 6B = (180 - A) + 2(180 - B)

or

5A = 6B = 180 - A + 360 - 2B

or

6A = 540 - 2B

or 

6A = 540 - 2B

To factor for A we need to evaluate B with A to swap it out

B = 5A / 6

Now we factor out B with

6A = 540 - 2(5A / 6)

6A = 540 - (10A / 6)

To get 6 out of the denominator, we multiply by 6 leaving

or 36A = 3240 - 10A

or 46A = 3240

or A = 70.43

Paul sends...

On Thu, May 26, 2016 at 6:25 PM, Gerry in Quebec <toomey...@gmail.com> wrote:

Anthony,

Rob nicely explained the logic of face angles & radial angles with his umbrella analogy. About the mathematics.... here's an Excel calculator for face angles, central angles & radial angles. When you enter three chord factors of a dome triangle in the green input boxes and press Enter, those angles appear in the orange boxes. I've also included the equations as text. Let me know if you have any problems using the calculator either directly on line or as a standalone Excel file.

- Gerry in Québec

P.S. If any reader spots an error, please let me know.

On Thursday, May 26, 2016 at 11:57:29 AM UTC-4, Robert Clark wrote:

Imagine an umbrella with 8 ribs. The ribs are equally spaced radially at 45 degrees around the central post.  When the umbrella is in its open position, that's very easy to see.  But, as you begin to close the umbrella and the ribs lower, the angle between ribs gets smaller but radially they are always still equally spaced around the central post at an angle of 45 degrees.  The planes that the ribs reside in are radially spaced at one angle but the ribs themselves form another different angle.  That's the best I can do.  Hope that helps.

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Very high regards,

 

Paul sends...

ReviveTheFlame.com

[Adrian Rossiter]

Hi Paul

On Thu, 26 May 2016, Paul Kranz wrote:
> The 70.43° is not arbitrary. You can compute it with simple algebra.

...

> In order for both points to be on the same sphere, they have to encompass
> the same number of degrees such that
>
> 5A = 6B
>
> Now, we factor in their base angles which meet at the green points on the
> BMP such that
>
> 5A = 6B = (180 - A) + 2(180 - B)

...
> or A = 70.43

If I have understood, you appear to be saying that if the vertices lie
on a sphere then at every vertex the sum of the plane angles meeting
at the vertex will be constant. However, this isn't generally true of
polyhedra whose vertices lie on a sphere (consider a tall tetrahedron).

Your method gives a result of 70.43 for A, but Anthony mentioned a
face angle of 70.73.

For a 3V model made as a uniform truncated icosahedron with
pyramids raised on each face so the vertices all lie on a sphere
(is this the form that Anthony is making?) I make the plane angle
at the 5-fold vertices 70.73 degrees.

This is the Antiprism command to calculate the angles in the two
kinds of triangle in this model

off_util -S tr_ico tr_ico_d | conv_hull | off_report -C F

[face_angles_cnts]
54.634729529867251,54.634729529867265,70.730540940265499 = 60
58.583165023638102,60.708417488180949,60.708417488180963 = 120

Adrian.
--
Adrian Rossiter
adr...@antiprism.com
http://antiprism.com/adrian

[Paul Kranz]

Adrian, does that mean the 5A = 6B rule does not apply?

Paul sends...

[Gerry in Quebec]

I expect we're experiencing déjà-vu by now:

https://groups.google.com/d/topic/geodesichelp/hX5EszNax_0/discussion

[Anthony Sokolowski]

The umbrella analogy really helped my understanding. Originally it really was difficult to get my head around the concept. Thanks Robert.

Gerry, the excel spreadsheet is brilliant and has everything I need to finish my hub design. I see that it uses the central angles which also helped with my conceptual understanding. Thanks for the upload.

Paul, I am always happy for more math but I think that you may have lost me. I am designing a 3v icosa based uniformly truncated dome as per Adrian's assumption.

Thanks again for the assistance.

[Adrian Rossiter]

Hi Paul

On Fri, 27 May 2016, Paul Kranz wrote:
> Adrian, does that mean the 5A = 6B rule does not apply?

Not in this model, as

70.73 * 5 = 353.65
58.58 * 6 = 351.48

[Paul Kranz]

Adrian:

If the vertexes don't have the same angle of deficiency, how can they lie on the same sphere? When A=70.43, all of the vertexes have the same deficiency. 92 vertexes share 720 degrees of deficiency (same as the platonic solids) leaving 720 / 92 / 5 = 70.43.

Paul sends...

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[Adrian Rossiter]

Hi Paul

On Fri, 27 May 2016, Paul Kranz wrote:
> If the vertexes don't have the same angle of deficiency, how can they lie
> on the same sphere? When A=70.43, all of the vertexes have the same
> deficiency. 92 vertexes share 720 degrees of deficiency (same as the
> platonic solids) leaving 720 / 92 / 5 = 70.43.

Consider a very tall skinny triangular pyramid. Say the angles at the
apex are 2 degrees, this makes the total at the apex 3*2 = 6 degrees.
The angles at the base of the side triangles are 89 degrees, making
the total at the base vertices 2*89+60 = 238 degrees. The total face
angles at the different vertices are very different, yet all the
vertices lie on a sphere.

[Dick Fischbeck]

Hi Adrian

On the other hand, is there a reason vertex deficits can't be equal for a polyhedron whose vertexes lie on the sphere by varying edge lengths? Not that I am a huge fan of spheres. Just asking.

Dick

[Dick Fischbeck]

Any spherical polyhedron is spherical. If you need a sphere in the first place.

On Fri, May 27, 2016 at 7:52 PM, Robert Clark <clark.rob...@gmail.com> wrote:

If you could design a dome with curved struts you wouldn't even have to worry about the difference in angles. ;P  Like this one that only has 3 different angles.

[Adrian Rossiter]

Hi Dick

On Fri, 27 May 2016, Dick Fischbeck wrote:
> On the other hand, is there a reason vertex deficits can't be equal for a
> polyhedron whose vertexes lie on the sphere by varying edge lengths? Not
> that I am a huge fan of spheres. Just asking.

My guess is no for convex polyhedra.

Consider a dipyramid based on a polygon with a very large number
of vertices. The average angle sum will be close to 360. For an
apex vertex to be close to 360 it will require most of the equator
vertices to surround it closely. But this is true of both apexes,
and I can't see how this can happen unless both apex points lie on
the same side of the ring of edges that was the original equator
(e.g. both apex points lie close the the north pole and the equator
points circle them both closely). But this means that the polyhedron
won't be convex.

[Kenneth Rhodes]

Hi Gerry, Thanks for sharing your spreadsheet, Dome-angles.xls with us.

In your first set of equations, I think the equations for face angles B & C use arcsin in error rather than arccos.  I used your equations to write a Face Angle function in my favorite programming language and kept getting errors for B & C.  I checked the formula's referenced in the Face Angle Output cells and noticed that the B & C equations use arccos rather than arcsin as listed in the description. 

Regards, Ken Rhodes

[Gerry in Quebec]

Ken,

Thanks for the correction. Indeed, I should have written arccos instead of arcsin for the B & C face angles.

So, to be clear to anyone else who used the spreadsheet: The numerical outputs of the spreadsheet were correct; the errors were in the descriptive text.

I have corrected and uploaded the spreadsheet with a new file name and will delete my earlier post.

Again, thanks for catching and reporting the errors.

- Gerry

[Gerry in Quebec]

For the record, here's the text of my earlier message dated 5/26/16:

Anthony,

Rob nicely explained the logic of face angles & radial angles with his umbrella analogy. About the mathematics.... here's an Excel calculator for face angles, central angles & radial angles. When you enter three chord factors of a dome triangle in the green input boxes and press Enter, those angles appear in the orange boxes. I've also included the equations as text. Let me know if you have any problems using the calculator either directly on line or as a standalone Excel file.

- Gerry in Québec

P.S. If any reader spots an error, please let me know.

[Gerry in Quebec]

Some Excel spreadsheets, especially ones with graphics created using Excel's drawing tools, don't display well in non-Excel spreadsheets, including Google Sheets. So, I'm uploading a new "dome angles" Excel file, this time with the two drawings in a stand-alone format.

- Gerry

On Friday, February 17, 2017 at 11:12:55 PM UTC-5, Gerry in Quebec wrote: