60 sided Geodesic Dome

[AndyWC]

60 sided Geodesic Dome

Constructed from 12 Pentagon Pyramids.

Each ‘planted’ on the faces of a Dodecahedron.

60 triangles in total.

I need help with a formula to calculate the Pyramid Height for any given pentagon base length to ensure the 60-sided Geodesic Dome is completely symmetrical

I need the triangles across SIDE face to face angle to equal the triangles across BASE face to face angles to be equal.

The attached PDF of part of a Excel Workbook (hopefully) explains more clearly what I am needing help with and my ‘workings’ to date
If you would like I can send a copy of the Excel Workbook via email - send to andy@lletty.com
Many thanks

Andy

 

[Cubist]

I find it very difficult to follow your pdf. Please can you provide the two formulas for "Sffa" and "Bffa" in terms of the two variables "sl" and "Ph" (with no other variables). And certainly don't include any rounded decimal numbers because this will mean that an exact answer will NOT be possible.

For example you say..
Sffa = 2*asin(Dl/2/AFI) = 2*asin(Dl/(2*AFI))

But I don't want to see Dl, so instead we find it's EXACT value and substitute it in. Elsewhere in the document you say Dl = 2*Sl*sin( Iah ) and Iah = 54° therefore Dl = 2*Sl*sin( 54° ). You can use WolframAlpha (click) to find the EXACT value of sin and cos for some angles. So, Dl=Sl*(1 + sqrt(5))/2. Now put this into the Sffa formula...

Sffa = 2*asin(Sl*(1 + sqrt(5))/(4*AFI))

Next, eliminate "AFI" and then do the same for Bffa. Hopefully this work will lead to an exact answer.

I'm curious, why do you need this formula? Is it a study assignment?

EDIT: Actually I'd recommend that you leave the sin(54°) as it is for now. This is because it might cancel with another sin(54°) later on. If it doesn't, then we can substitute the exact value in there later. Therefore, for now, Sffa = 2*asin(Sl*sin(54°)/(AFI)), and then eliminate AFI...

 

[Cubist]

Your equation FBa=DEGREES(ATAN(Sh/Bl/2)) is wrong. Judging by the diagram it ought to be FBa=DEGREES(ATAN(Sh/(Bl/2))) = DEGREES(ATAN(2*Sh/Bl))

 

[Cubist]

Please can you provide the two formulas for "Sffa" and "Bffa" in terms of the two variables "sl" and "Ph"

I've done the above on your behalf. NOTE: the following uses h=Ph, s=Sl and it's converted to radians.

sffa = 2*asin(sin(3/10*pi)/sin(atan(tan(3/10*pi)/cos(atan(2*h/(s*tan(3/10*pi)))))))
bffa = pi - atan(2) + 2*atan(2*h/(s*tan(3/10*pi)))

Letting sffa = bffa, I obtained a truly horrendous expression that I couldn't rearrange to h=<something>...

2*(asin(sin(3/10*pi)*cosec(atan(tan(3/10*pi)*sec(atan(2*h/(s*tan(3/10*pi))))))) - atan(2*h/(s*tan(3/10*pi)))) - pi + atan(2) = 0

Maybe this could be simplified if we assume first quadrant (where appropriate) so that substitutions like the following could be used:- sin(atan(x)) = x/sqrt(x^2+1)

--

I have however obtained a solution h=<something> by avoiding trig altogether and using vector dot and cross products. This was also a pretty horrendous process, but it did bear fruit in the end!

 

[AndyWC]

I find it very difficult to follow your pdf. Please can you provide the two formulas for "Sffa" and "Bffa" in terms of the two variables "sl" and "Ph" (with no other variables). And certainly don't include any rounded decimal numbers because this will mean that an exact answer will NOT be possible.

For example you say..
Sffa = 2*asin(Dl/2/AFI) = 2*asin(Dl/(2*AFI))

But I don't want to see Dl, so instead we find it's EXACT value and substitute it in. Elsewhere in the document you say Dl = 2*Sl*sin( Iah ) and Iah = 54° therefore Dl = 2*Sl*sin( 54° ). You can use WolframAlpha (click) to find the EXACT value of sin and cos for some angles. So, Dl=Sl*(1 + sqrt(5))/2. Now put this into the Sffa formula...

Sffa = 2*asin(Sl*(1 + sqrt(5))/(4*AFI))

Next, eliminate "AFI" and then do the same for Bffa. Hopefully this work will lead to an exact answer.

I'm curious, why do you need this formula? Is it a study assignment?

EDIT: Actually I'd recommend that you leave the sin(54°) as it is for now. This is because it might cancel with another sin(54°) later on. If it doesn't, then we can substitute the exact value in there later. Therefore, for now, Sffa = 2*asin(Sl*sin(54°)/(AFI)), and then eliminate AFI...

Many thanks Cubist
No it is not a study assignment; I am making a half 60 sided geodesic dome out of wood mounted on a tapered dodecagon
Having found by trial & error the Pyramid Height for 120mm base length to satisfy the condition Bffa = Sffa angle I did not need the formula
But curiosity got the better - I had to find a solution
I was in the process of trying to providing the two formulas for "Sffa" and "Bffa" in terms of the two variables "sl" and "Ph" when I received your latest message
I had tried previously and got a similarly horrendous expression
I will reply to your third message separately
Andy

 

[AndyWC]

Your equation FBa=DEGREES(ATAN(Sh/Bl/2)) is wrong. Judging by the diagram it ought to be FBa=DEGREES(ATAN(Sh/(Bl/2))) = DEGREES(ATAN(2*Sh/Bl))

Thanks and noted

 

[AndyWC]

I've done the above on your behalf. NOTE: the following uses h=Ph, s=Sl and it's converted to radians.

sffa = 2*asin(sin(3/10*pi)/sin(atan(tan(3/10*pi)/cos(atan(2*h/(s*tan(3/10*pi)))))))
bffa = pi - atan(2) + 2*atan(2*h/(s*tan(3/10*pi)))

Letting sffa = bffa, I obtained a truly horrendous expression that I couldn't rearrange to h=<something>...

2*(asin(sin(3/10*pi)*cosec(atan(tan(3/10*pi)*sec(atan(2*h/(s*tan(3/10*pi))))))) - atan(2*h/(s*tan(3/10*pi)))) - pi + atan(2) = 0

Maybe this could be simplified if we assume first quadrant (wre appropriate) so that substitutions like the following could be used:- sin(atan(x)) = x/sqrt(x^2+1)

--

I have however obtained a solution h=<something> by avoiding trig altogether and using vector dot and cross products. This was also a pretty horrendous process, but it did bear fruit in the end!

Many thanks for doing it for me
My result is different from yours (may not even be correct) though I am sure yours is.
My maths ability is limited 'A' level

I am afraid I do not understand the "Maybe this could be simplified if we assume first quadrant (where appropriate) so that substitutions like the following could be used:- sin(atan(x)) = x/sqrt(x^2+1)"

You say you have found a solution by avoiding trig altogether and it did bear fruit

I look forward to receiving it

Again many thanks so far

Andy PS The maths Forum site went down yesterday evening so had to wait until now to reply The

 

[Cubist]

Many thanks Cubist
No it is not a study assignment; I am making a half 60 sided geodesic dome out of wood mounted on a tapered dodecagon
Having found by trial & error the Pyramid Height for 120mm base length to satisfy the condition Bffa = Sffa angle I did not need the formula
But curiosity got the better - I had to find a solution
I was in the process of trying to providing the two formulas for "Sffa" and "Bffa" in terms of the two variables "sl" and "Ph" when I received your latest message
I had tried previously and got a similarly horrendous expression
I will reply to your third message separately
Andy

So you must have produced most of the figures in the pdf yourself. Impressive! Please let us know how you progress with this project. The final formula looks fairly simple, which makes me think that there might be an easier way to obtain it. Hmm, maybe using similar triangles to avoid the trig

 

[AndyWC]

So you must have produced most of the figures in the pdf yourself. Impressive! Please let us know how you progress with this project. The final formula looks fairly simple, which makes me think that there might be an easier way to obtain it. Hmm, maybe using similar triangles to avoid the trig

My last reply appears to have the ending missing - it was:
The Wolfram Alpha is completely new to me and I shall enjoy looking further into it.
Without knowing it I used it to calculate the circumradius & inradius of a dodecahedron (found it via google)
The dihedral angle I did manage to work out from its origin as 2*arcsin(cos36/cos18)
I did also work out the circumradius & inradius after from its origin
I will now look again using Wolfram Alpha
Renewed thanks
I had considered using similar (isosceles) triangles originally and cannot recall why I dismissed it - something to do with it requiring an irregular pentagon base !!!!

 

[Cubist]

My maths ability is limited 'A' level

You say you have found a solution by avoiding trig altogether and it did bear fruit
I look forward to receiving it

Have you finished your A levels? (I didn't say, but I'm also impressed that you're planning to make a wooden version of this shape!). Since this isn't a study question, then here's an "answer at the back of the book"...

Spoiler: final answer

[math]P_h=S_l \times \frac{\sqrt{325+110\sqrt{5}}}{95}[/math]Or in text/ spreadsheet form h=s*sqrt(325 + 110*sqrt(5))/95
I'm pretty sure that the nested radical can't be simplified further

There's probably a simple method to obtain this result . But, if you're OK with vectors (cross and dot product) then I could give you an outline of the method that I used. Like I said, it became quite messy along the way. My starting point was using the coordinates given in terms of the golden ratio on Wikipedia. I used the two adjacent pentagons, that both share the top two blue points in the image. The symmetry in the y axis helps (a little).

The Wolfram Alpha is completely new to me and I shall enjoy looking further into it.

I won't lie, I made use of computer algebra systems (CAS) to obtain my answer! I tend to use both Wolfram and wxMaxima. The latter is software that you can install on your computer, although it seems that an online version is available. It's a little quirky, like having to use "%e" for Euler's constant etc. It's very good and has some quite powerful features, but it won't show the steps. However it can be used to verify your work. Obviously, DON'T become reliant on these if you're still studying!

And then for numeric solutions I sometimes use Matlab, since it's easy to do matrix and complex number calculations. If you're still a student then you might be able to get Matlab for free (ask at your place of study). I'm no longer a student (for many years) so I actually use something similar to Matlab called Octave. Octave is free to download. Also a spreadsheet is very useful. And Desmos is great for interactive graphs.

 

[Cubist]

I am afraid I do not understand the "Maybe this could be simplified if we assume first quadrant (where appropriate) so that substitutions like the following could be used:- sin(atan(x)) = x/sqrt(x^2+1)"

Just sketch a right angle triangle. Mark one of the other angles as "y". Mark the side opposite y with length "x". Then mark the adjacent side with length 1. Therefore y=atan(x). I'm sure you can work out sin(y) with a little help from Pythagoras! Maybe multiple applications of this type of method can simply the trig expressions (and eliminate the trig altogether). But I have not tried this myself!

 

[AndyWC]

Amazing - Thank you so much - the solution however difficult to achieve is remarkably simple
To set the record straight I am 77 years old and took my 'A: level maths 60 years ago
I failed 'O' level woodwork.
The irony is 60 years on my woodwork is (dare I say it myself) pretty hot but my Maths has not advanced
I will look at the various software you mention and take a look at Vectors (cross & dot product ) which I have never done before.
Never shy to try something new.
With respect to my 'project' equilateral triangles just do not work if you make Sffa =Bffa you end up with 3 Pentagon Pyramids and two void triangles. See attached, I use Autodesk Fusion 360 a CAD design tool to mock up more complex projects.
If not bothered about Sffa = Bffa then (whilst the maths is greatly simplified) you get a pentagon pyramid height of approx. 40mm.
Requiring wood thickness greater than the Pyramid height when you turn the assembled Geodesic Dome in a lathe to shape a sphere
The next stage of my project (now I know the sizes work for me) is bin the 12mm birch plywood mock up and make it from walnut & maple. By making each triangle from with strips of walnut & maple you get amazing finished effect when turned.

My very sincere thanks for all your time and help
Andy