V6 Class 1 truncated

[colinyell]

Hi,

I have just discovered the collections of sketch up geodesics an they
are a massive help. Unfortunately I'm planning on building a 20 metre
diameter dome and I would like to build it V6 class1 but truncated two
levels from centre. Does anyone have such a model I could use? If not
what is the easiest way of working out how to truncate it as simple a
shape as possible?

Cheers,

Colin

[TaffGoch]

 

Welcome to the group, Colin.

_______________________

 

Class-I, (check)

Frequency-6v, (check)

...but, which subdivision method:

 -- equi-linear icosa-edge division (method-1,) or
 -- equi-angular central angle division (method-2)?

(Subdivision methods 1 & 2 are described in Domebook2, on pages 106-107.)
__________________

Truncated two levels ABOVE the equator, or two levels BELOW the equator?
__________________

Note that a truncation along a plane, other than the equator, results in "scallops" that must be taken into account. Either the foundation can be scalloped to match, or the bottom tier(s) of panels/struts can be "manipulated" to get their vertices horizontal to the ground. (The latter is what Bucky did for the Montreal Exposition dome.)

 

Once I know the details of what you need, I would be glad to model it for you.

 

I just checked my "truncation" model (for Class-I domes) at the 3D Warehouse, and sure enough, I only included 3,4 & 5v. I should probably modify it to add 6v, as well.  

Taff

[colinyell]

Hi Taff,

Thanks for getting back to me. I'm after a model of a 6v dome that is
truncated two levels above the equator. On your 6v_truncations.png
file it would be the model on the left. I do then need the vertices
horizontal to the ground exactly how they appear in your truncation
model on the 3D warehouse.

The purpose of the model is to help me to work out how to build a
portable structure out of steel tube. I want to use the method of
flattening the ends of the pipe and bending them so they can be bolted
together to form the dome. A similar technique to your conduit hub
model. Do you build real domes from your sketch-up plans at all? can
you foresee any problems for using this construction technique with a
6V truncated domes?

Thanks for you time on this, its very much appreciated.

Cheers,

Colin

>  6v_truncations.png
> 39KViewDownload

[colinyell]

Hi Taff,

I have just downloaded and had a scan through Dome book2. Great book.
It is definately a class1 structure that I'm interested in.

Cheers.

Colin

[Gerry in Quebec]

Hi Colin and Taff,
A couple of years ago I came up with a 6v icosa dome layout that sits
flat at the 7/18ths and 11/18ths truncations. That's two rows of
triangles above the equator and two rows below the equator. Is one of
those the cutoff point you're interested in, Colin? This particular
layout (solution 10) doesn't sit flat at the equator. It's a class I
dome, but it does not fit into any of the "method" categories (method
1, 2, 3, etc.).

To level the base (i.e., make all the vertices in the footprint
coplanar), I did not tinker with the bottom row of triangles or with
any radii. Rather, at the very outset of the design exercise, I
leveled the base at those two truncations. Then I adjusted all other
vertices in the dome to maintain overall icosahedral symmetry as well
as the internal symmetry of the parent triangles (each parent
comprising 36 sub-triangles). I also manipulated the vertices so as to
eliminate all scalene triangles in the dome.

This base-leveling exercise does for the 6v icosa what Fuller and
Kruschke did for the 3v and 4v class I icosa several decades ago. As
you may know, many dome manufacturers still use the Fuller-Kruschke
layouts rather than the class I, method 1 breakdown. This sometimes
leads to confusion when people are talking about dome types, strut
lengths and triangular panels.

So, solution 10 for the 6v ends up with these features:
- All triangles are isosceles: 9 types
- Level base at two truncations which, if you were to use the
conventional method 1 breakdown, would oscillate.
- 9 chord lengths.

By way of comparision, the conventional class I, method 1 layout sits
flat at only one useful truncation, the equator, as Taff pointed out.
Like solution 10, it has 9 chord factor, but only 8 triangle types,
counting mirror image triangles as two types.

If you're concerned about variance in edge length and triangle shape,
then solution 10 may not be an appropriate option. For example, two of
the triangle types are rather "gothic" looking -- tall and thin. This
can be an advantage or disadvantage, depending on the purpose of the
building and your own sense of aesthetics. Personally, I don't mind
variations in triangle shape... as long as the overall look of the
dome is good. Using up-pointing gothic triangles in the the first or
second row of triangles as your window openings could add a nice touch
to a dome.

I'm also aware of another dome layout for the 6v icosa, class I, that
has only 6 chord factors -- which would greatly simplify construction.
If I recall correctly, it was sent to me by a math professor in
Mexico. However, that layout does not sit flat at any useful
truncation, not even the equator.

Colin, if you're interested in either the solution 10 layout or the
Mexican layout, let me know and I'll post the necessary numbers. Taff,
would you be willing to do a Sketch-Up of either or both? If so, what
info do you need: spherical coordinates (phi and theta angles) of a
parent triangle, or just chord factors and a "map" of which chord goes
where?

In the files section, I've posted a jpg showing the 6v icosa (7/18ths
and 11/18ths) with and without a flat base. But the layout there isn't
solution 10; it's a set of images generated by CADRE Geo, just to give
a rough idea. File: 6vIcosa-7-18ths-11-18ths.jpg.
Cheers,
Gerry

>  6v_truncations.png
> 39KViewDownload

[colinyell]

Hi Gerry,

The diagram on the right hand side, solution 10, looks like what I had
in mind. I'm new to the whole dome construction thing so maybe I
should explain what I'm planning to do and maybe you can advise me on
the best approach here.

I want to build a geodesic dome structure for use at music festivals
this summer in the UK. I have decided to build one around 20 metres in
diameter and initially I was basing my plans on Taffs 5V class 1
truncated model. I'm planing on using steel tube to create the struts
and instead of connectors I'm using the method of flattening the ends
of the tube and bending it to the appropriate angle and using one bolt
at each connection to secure the struts.

It will be me and a group of untrained friends who will be putting up
the structure so the simpler the frame the better. We have a bit of
experience with an old 12 metre 3V structure so we are not to daunted
by going for something a bit more complicated but the fewer different
strut lengths the better.

After scaling Taffs 5V model up to 20 metres in diameter some of the
struts are as long as 3.1 metres. As a result of this I have started
to question the weight of steel tube needed to cope with the lengths
required. This is why I have started looking at the 6V geodesics. I
want to truncate the dome because I do not necessarily need the height
of a full hemisphere.

Basically Im looking to make as simple a 20 metre diameter dome as
possible out of steel pipe with the least amount of poles but still
have lots of strength. What do you suggest 5V or 6V?

Cheers,

Colin

[Gerry in Quebec]

Hi Colin,
Before you decide on the geodesic frequency and truncation needed to
get 300 sq. metres of floor space, you need to find out the maximum
steel-tubing strut length available to you for construction. This will
be determined in part by the tubing gauge (outside diameter and wall
thickness) you intend to use and the maximum roof load. As your dome
is intended for use as a public (temporary) space, you definitely need
to identify the structural requirements, which means getting advice or
blueprint approval from a civil engineer. (Will the dome be covered
tent-style?)

For the solution 10 geodesic layout I mentioned for the 6v icosa, the
longest strut (vertex-to-vertex) for a dome with a floor radius of 10
metres would be 2.54 metres and the shortest would be 1.61 metres. To
this you would need to add a few centimetres at each end since you
said it would be a steel-tube dome with strut ends flattened and
drilled. Let me know whether this layout is of interest to to you.

One person who may be able to help with steel-tube sizing for a dome
is Blair Wolfram, owner of Domes Inc. in Minneapolis, Minnesota. This
is the guy who built the NEEM dome for climate researchers in
Greenland, plus hundreds of domes elsewhere.

Dome Incorporated
Blair F. Wolfram
11480 - 141st Av N, Minneapolis, MN 55327
tel. 1-612- 333-3663
www.domeIncorporated.com
thedo...@domeincorporated.com

Another possibility (probably closer to home) is Amila Y'Mech:

Geodesics Unlimited
www.geodesics-unlimited.com
tel. 013 3537 0661
tel. 078 1260 1256
The Hayloft, Atlow, Ashbourne, Derbys
DE6 INS
UK

Gerry in Quebec

> > > 39KViewDownload- Hide quoted text -
>
> - Show quoted text -

[colinyell]

Hi Gerry and Taff,

I have been speaking to Blair at domes incorporated and he has put my
mind at rest about what size steel tube to go for so I'm now
definitely settled on the 6V truncated dome. I have actual had a go at
making the model in sketch up and I was wondering if I could have your
opinion on it.

I have added a jpg and the sketch up file to the files section of this
forum. 6V 21 metre igloo.jpg and 21m 6v truncated.skp.

I have made this model by making a geosphere in 3ds max, cutting the
sphere to the right level and then dragging the uneven vertices to the
new floor level. Is this an accurate way of doing this? Is there a
more accurate mathematical method? What do you think to the usefulness
of this model for building a real dome?

Cheers,

Colin

> thedome...@domeincorporated.com

[Gerry in Quebec]

Hi Colin,
Fine looking dome model! I see this geometry uses equilinear edge
division. I don't know anything about the accuracy of dimensions when
you use the drag-to feature of a CAD program, but I would expect that
feature maintains the same overall linear accuracy of the model's
other elements.

I used the SketchUp measuring tape to do a quick inventory of your
strut lengths. It looks like there's a total of 14 discrete lengths.
Is that right? This includes the several extra strut lengths in the
bottom row of 60 triangles to ensure a level base.

That's a rather large number of strut lengths to deal with... lots of
cutting and sorting work in the shop and lots of attention needed at
the music festival site during the dome raising. If your team of
helpers hasn't had experience doing this before, there could be on-
site confusion/errors, compounded by the fact that the right sequence
and number of struts have to be hoisted or carried up to the guys
working on the scaffolding.

But as I said, it's a good-looking dome design you've got there.

Cheers,
Gerry

> > > - Show quoted text -- Hide quoted text -

[TaffGoch]

Colin,

 

About the only change I'd make is to make the "footprint" perimeter circle out of 30 identical lengths.

 

Your idea of "...dragging the uneven vertices to the new floor level" is not uncommon, but you have to be careful. Changing the vertical position(z) of these vertices is not problematic, but the x & y coordinates can get "shifted" out of place, by mistake.

 

I've attached your model, with an amendment depicting how I usually arrange a truncated perimeter. With identical perimeter struts, it's easier to lay-out during assembly, and has the additional benefit of reducing the unique strut count (a little bit, at least.)

 

(My amendment was a quick edit, so there may be some discrepancies. I merely wanted to demonstrate the technique. It should receive some additional scrutiny, to ensure there aren't any errors.)

 

Taff

[colinyell]

Hi Taff and Gerry,

Thanks for the help guys, I now have a workable model that I can build
from.

Just for my own interest Taff how did you work it out? What is the
technique you use to get even struts on the bottom row? Do you draw in
a circle that has 30 sides for the floor and then redraw the bottom
level of struts?

If you are interested in what we are doing with this new dome we are
replaceing the one that appears on the front page of my website:
http://www.igloovision.com

Thanks again for all the help,

Cheers,

Colin

On Feb 1, 9:06 pm, TaffGoch <taffg...@gmail.com> wrote:
> Colin,
>
> About the only change I'd make is to make the "footprint" perimeter circle
> out of 30 identical lengths.
>

> Your idea of *"...dragging the uneven vertices to the new floor level"* is

> not uncommon, but you have to be careful. Changing the vertical position(z)
> of these vertices is not problematic, but the x & y coordinates can get
> "shifted" out of place, by mistake.
>
> I've attached your model, with an amendment depicting how
> I usually arrange a truncated perimeter. With identical perimeter struts,
> it's easier to lay-out during assembly, and has the additional benefit
> of reducing the unique strut count (a little bit, at least.)
>
> (My amendment was a quick edit, so there may be some discrepancies. I merely
> wanted to demonstrate the technique. It should receive some additional
> scrutiny, to ensure there aren't any errors.)
>
> Taff
>

>  21m+6v+trunc_footprint.skp
> 111KViewDownload
>
>  21m+6v+truncated_footprint.png
> 179KViewDownload

[TaffGoch]

Just for my own interest Taff how did you work it out? What is the
technique you use to get even struts on the bottom row? Do you draw in
a circle that has 30 sides for the floor and then redraw the bottom
level of struts?

You got it -- that's precisely what I do. The circle's diameter is established on the ground plane, such that the radius FROM THE SPHERE"S CENTER is the same as the radius for all the other vertices. That's why you'll see a center guidepoint for the SPHERE, in my amendment. To start, I use the bottom of the pentagons that sit on the ground, to establish the ground plane (but NOT the radius -- it's close, but not quite right.) Getting the sphere radius to match the rest of the vertex radii is a little tricky.
 
After generating the new circle, and drawing in the bottom tier of triangles, I re-scaled the dome, so that the footprint radius is 10.5 meters, according to your desired specs. (You'll notice that the height of the dome is slightly different from yours, for this reason.)
 
Taff

[Richard Fischbeck]

Hi Taff

If we do not care about leveling the ground upon which the dome rests, then we have to fill in anyway. No problem. Match the dome to the ground and not the other way around. You can if you prefer of course.

Fill the gaps with any structural material and forget about the "exact" math to make the base a circle.

Structurally it doesn't matter. Any shape will do. Know what I mean? 

Consider finishing the scallops like trim.

Dick

[Gerry in Quebec]

Colin & Taff,
I used the SketchUp measuring tape to compile a strut inventory of
Taff's edited version (righthand diagram in the SketchUp file). I get
a count of 27 unique lengths. If pairs of struts with a length
difference of less than one centimetre are treated as equal in length
for cutting purposes, then the count falls to 21. To me this still
seems like a huge inventory of parts to deal with, especially for a
portable dome that will be assembled and disassembled many times.
Cheers,
Gerry

On Feb 1, 4:06 pm, TaffGoch <taffg...@gmail.com> wrote:
> Colin,
>
> About the only change I'd make is to make the "footprint" perimeter circle
> out of 30 identical lengths.
>

> Your idea of *"...dragging the uneven vertices to the new floor level"* is

> not uncommon, but you have to be careful. Changing the vertical position(z)
> of these vertices is not problematic, but the x & y coordinates can get
> "shifted" out of place, by mistake.
>
> I've attached your model, with an amendment depicting how
> I usually arrange a truncated perimeter. With identical perimeter struts,
> it's easier to lay-out during assembly, and has the additional benefit
> of reducing the unique strut count (a little bit, at least.)
>
> (My amendment was a quick edit, so there may be some discrepancies. I merely
> wanted to demonstrate the technique. It should receive some additional
> scrutiny, to ensure there aren't any errors.)
>
> Taff
>

>  21m+6v+trunc_footprint.skp
> 111KViewDownload
>
>  21m+6v+truncated_footprint.png
> 179KViewDownload

[TaffGoch]

Colin & Gerry,

 

Using a different (customized) subdivision, I've reduced the unique strut count, down to 12.

 

I agree that fewer unique struts would certainly be more managable, during assembly.

_______________________

 

Colin, are you planning to cover the frame, or is this simply going to be a "skybreak"?

 

Taff

[Gerry in Quebec]

Wow Taff,
You just saved Colin 1000 or so pounds sterling.
Cheers,
Gerry

>  21m+6v+trunc_struts.png
> 106KViewDownload
>
>  21m+6v+trunc_struts.skp
> 135KViewDownload

[Blair Wolfram]

Taff;

 

I'm unable to open this .skp file. Does this contain chord factors and strut diagrams?

 

B.F. Wolfram

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[TaffGoch]

Blair,

Odd -- I can open the model file, if I download it from the Group
discussion, but not from my gmail of the discussion. Perhaps, you
could try that? (Or, is it that you can't open the file, 'cause you
don't have SketchUp installed?)
_______________

The model file doesn't contain chord factors, in tabular format, but
the dome IS completely modeled, so you can measure the chords (& face/
central/dihedral angles,) with SketchUp's tools.

I provided color-coded struts in this model, which is something I
don't usually render. I ran out of easily-distinguishable colors, so I
didn't color the "finagled" struts on the bottom tier. Those are all
black, even though there are 4 different strut-lengths in question.
It's pretty obvious that the black struts aren't all the same.

You may note that most of the ground-plane perimeter is composed of
struts that are the same as one of the primary icosa-face struts. This
substitution was done as a convenience, and doesn't produce a
perfectly circular perimeter. The true-circular perimeter strut was so
close in length to that particular icosa strut that it seemed to be a
reasonable substitution. You'll see several additional ground-plane
"centerpoints" that define the arcs made by these substituted struts.
Of course, if the substitution wasn't made, there would be 13 unique
struts, instead of 12 -- not exactly a substantial simplification, I
admit.

Taff

[TaffGoch]

On Jan 28, 6:44 am, Gerry in Quebec wrote:
> ...the solution 10 layout or the Mexican layout, let me know and I'll post

> the necessary numbers. Taff, would you be willing to do a Sketch-Up
> of either or both? If so, what info do you need: spherical coordinates
> (phi and theta angles) of a parent triangle, or just chord factors and a
> "map" of which chord goes where?

Gerry,

I missed your question to me, on my first reading.

I'd be interested in modeling the layouts in SketchUp. I can do so,
easily, with the spherical angles, but just the chord factors is a
REAL headache. (Including chords would, of course, permit error-
checking.)

Taff

[colinyell]

Hi Taff,

Thank you very much! I have been offline for a couple of days
traveling and I have just checked the model and it is brilliant!

I spent this afternoon on a train killing time making myself a strut
list from my old model and I got to 23 different lengths and gave up.
I was about to post my findings on this forum and found your model.
Gerrys right this has just saved me ££££.

The dome is going to be covered with a PVC blackout cover we will be
using the structure as a 360 projection dome.

Cheers,

Colin

[Gerry in Quebec]

Thanks for the offer, Taff. I've just uploaded Excel file
6vIcosa-7-18ths-Toomey-Sol10.xls. It contains two line drawings of
solution 10, with vertices and edges labelled, accompanied by
spherical coordinates (phi and theta values) and the 9 chord factors.
The Mexican solution will come in a day or two.
Cheers,
Gerry in Quebec

[TaffGoch]

 

I "built" the model, using the spherical angles, and error-checked, using the chord factors....

[Gerry in Quebec]

Hi Taff,
Thanks for Sketching Up solution 10. I see my layout has a face only a
mother could love. Back to the drawing board!
Cheers,
Gerry

On Feb 3, 9:41 pm, TaffGoch <taffg...@gmail.com> wrote:
> I "built" the model, using the spherical angles, and error-checked, using
> the chord factors....
>

>  Solution_10.skp
> 605KViewDownload
>
>  Solution_10.png
> 506KViewDownload

[Gerry in Quebec]

I’ve uploaded the Mexican 6v design that I mentioned a few days ago.
File: 6vIcosa-7-18ths-Mexican.xls. Although the basic arrangement of
triangles has only 6 chord lengths, domes based on this design
wouldn’t naturally sit flat at any useful truncation. So I tinkered
with the bottom row of triangles to make a clean truncation at the
7/18ths line. This added 5 new chord lengths for a total of 11,
similar to Taff’s edited verison of Colin’s layout.

I’ve also uploaded an alternative to the “ugly duckling” layout
(solution 10) that I posted a few days ago. This new layout, solution
10b, eliminates (hopefully) much of the uneveness in triangle shapes
of the earlier version, but it still has only 9 chord lengths. File:
6vIcosa-7-18ths-Gerry-Sol10b.xls.

Each file contains spherical coordinates, chord factors, and line
drawings with strut types and vertices labeled. It would be great if
you could make SketchUp models of these, Taff. Thanks.

While I still haven’t learned to create models in SketchUp, at least
I’m getting the hang of navigating through the various viewpoints of
your existing dome models using the mouse and keys.
Cheers,
Gerry

[TaffGoch]

 

The "Mexican"....

[TaffGoch]

Gerry,

BTW, what is the origin of the "Mexican" subdivision? I can find no
other reference to that name.

Perhaps, I know it by another name?

Taff

[Gerry in Quebec]

Hi Taff,
Thanks for modeling the “Mexican” layout. I referred to it that way
because I first learned of it from math professor Hector Hernandez at
the University of Sonora in Mexico. I’ve just taken a look at Domebook
2 and Hugh Kenner’s Geodesic Math book. I suspect Hector’s layout is
probably class I, Method 2, as desribed by Joe Clinton in Domebook II
at the top of page 107, and briefly by Kenner on page 64. That’s the
method that results in chords of equal length along the perimeter of
the cluster of triangles (36 for the 6v) associated with each face of
the icosahedron.

There’s a guy in California named Harold who used to contribute from
time to time to the Dome Living Group and maybe the Domelist Group on
Yahoo. In Dome Living he posted photos of his large peacock cages
covered in wire mesh. These domes use the same layout as Hector’s. I
recall that Harold came up with the 6 chord lengths independently by
experimenting in AutoCAD.
Gerry

[Gerry in Quebec]

I used class I, method 2 (page 107 in Domebook 2) to calculate the
chord factors for the 4v icosa. I end up with 5 chord factors:
0.275904; 0.321244; 0.296068; 0.313122; 0.302931.

The Class I 4v layout created by Mexican mathematician Hector
Hernandez has only 4 chord factors (number of chord lengths equals
dome frequency):
0.275904; 0.321243; 0.298311; 0.313239.

So, it now looks to me like his method is NOT method 2 as I suggested,
but something rather different from the Domebook/Clinton methods.

Can anyone verify that the first set of (5) chord factors listed above
are correct for method 2? I don't think I've come across a list of
class 1, method 2 chord factors.

Thanks,
Gerry in Quebec

> > Taff- Hide quoted text -

[Blair Wolfram]

Gerry;

 

Joe Clinton himself told me not to trust the chord factors in Domebook 2. He said he was able to correct a number of errors and misprints for the final printing in 1974 , but still the 1974 DB2 has errors.

 

I assumed you would be the person to confirm which chord factors are correct!

 

Blair

[TaffGoch]

Gerry,

 

I modeled, in SketchUp, a 4v-class1-method2 icosa, and have attached the chord results. They are identical to those found in DB2. Since I used the method described in DB2, that's no surprise.

 

You apparently used a different method for finding the three vertices in the center of the icosa face?

_______________________

 

BTW, I'm trying to "back-engineer" the Hernandez results. So far, my model is in agreement, to only 4 decimal places, using successive approximation. (Not a satisfactory technique.) How did you derive the "mexican" vertices/chords?

 

Taff

[Adrian Rossiter]

Hi Gerry

On Mon, 8 Feb 2010, Gerry in Quebec wrote:
> I used class I, method 2 (page 107 in Domebook 2) to calculate the
> chord factors for the 4v icosa. I end up with 5 chord factors:
> 0.275904; 0.321244; 0.296068; 0.313122; 0.302931.
>
> The Class I 4v layout created by Mexican mathematician Hector
> Hernandez has only 4 chord factors (number of chord lengths equals
> dome frequency):
> 0.275904; 0.321243; 0.298311; 0.313239.
>
> So, it now looks to me like his method is NOT method 2 as I suggested,
> but something rather different from the Domebook/Clinton methods.
>
> Can anyone verify that the first set of (5) chord factors listed above
> are correct for method 2? I don't think I've come across a list of
> class 1, method 2 chord factors.

If method 2 is equal central angle on the icosahedron edge
divisions then I get

geodesic 4 | off_report -C E

[edge_lengths_cnts]
0.27590448425526748 = 120
0.297790443995401 = 120
0.29938145926502907 = 60
0.31321026301214749 = 120
0.32124407128004651 = 60

Here is a model which colours the edges by their symmetry orbits

http://www.antiprism.com/misc/geo_i_4_edges.jpg

The brown edges have length 0.299381...
The green edges have length 0.297790...
The yellow edges have length 0.313210...

Keeping vertices on the sphere, if you shrink the brown edges a
little it will lengthen the green edges a little and they can be
the same length. Doing this looks like it would lengthen the
yellow edges. It wouldn't change the other two lengths (purple and
red, and blue).

I would guess then that this was the construction for the four
strut-length model.

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

[Gerry in Quebec]

Hi Blair,
Thanks for the heads-up on Domebook 2. The reason I was looking to
confirm the 5 chord factors for class I, method 2, is that I’ve never
seen these numbers in any book, website or computer program. Domebook
2 and Kenner’s Geodesic Math describe method 2 in cursory fashion, but
neither publication actually lists either the spherical coordinates or
chord factors (actually.... there’s one octahedral example in Kenner).
I derived the 4v icosa chord factors using trigonometry following the
brief instructions at the top of page 107 of Domebook 2. I thought
maybe someone else had already crunched the numbers for various
frequencies and that these might be available somewhere for
comparison.

It looks to me like Hector Hernandez in Mexico has created a new class
I breakdown method. It has the advantage of minimizing the total
number of chord lengths (unique lengths = dome frequency).
Unfortunately, domes based of this method don’t sit flat at useful
truncations. Hence the need to tinker with the strut lengths in the
bottom row of triangles, which of course increases the number of
unique strut lengths. But the "Mexican" method does seem to produce
domes with very little variation in triangle shape.

I don’t know the math behind Hector’s method. I arrived at the chord
factors using iteration in Excel – lazy person’s method :-). In case
you or others are interested, here are the numbers for frequencies 3,
4 and 6. (The 2v solution is identical to class 1, method 1. I need to
double-check 5v.)

3v :
A = 0.36695882,
B = 0.42406256
C = 0.40419443

4v:
A = 0.27590384
B = 0.32124334
C = 0.29831132
D = 0.31323941

6v (This is the one Taff modeled in SketchUp.)
A = 0.18426311
B = 0.21569298
C = 0.19484588
D = 0.21345118
E = 0.20318958
F = 0.20934982

Cheers,
Gerry

On Feb 8, 12:15 pm, Blair Wolfram <thedome...@gmail.com> wrote:
> Gerry;
>
> Joe Clinton himself told me not to trust the chord factors in Domebook 2. He
> said he was able to correct a number of errors and misprints for the final
> printing in 1974 , but still the 1974 DB2 has errors.
>
> I assumed you would be the person to confirm which chord factors are
> correct!
>
> Blair
>

> > GeodesicHelp...@googlegroups.com<GeodesicHelp%2Bunsubscribe@google­groups.com>

> >  --
> > To post to this group, send email to geodes...@googlegroups.com
> >  --

> > For more options, visithttp://groups.google.com/group/geodesichelp?hl=en- Hide quoted text -

[Ken G. Brown]

See the Dec 4, 2009, geodesichelp thread titled "NASA-CR-101577 - Structural Design Concepts for Future Space Missions" for a link to download Clinton's early NASA report detailing how to calculate chord factors.

I'm wondering if the 'new' breakdown method has previously been covered in the report.

Ken G. Brown

[TaffGoch]

Gerry,

 

The chord factors for 4v,class1,method2, in Domebook2, are provided on page 112 (bottom-left corner.)

"...

0.2759044

0.2995157

0.2977157

0.3132066

0.3212449"

_____________

 

I did not derive mine mathematically, but with "physical" construction in SketchUp, followed by measuring.

...

0.275904

0.299516

0.297725

0.313207

0.321244

 

Not precisely the same, but good to 4 decimals. (Maybe I should double-check my construction technique....)

________________

 

I agree that the Hernandez subdivision method appears to be "new." That's why I am so interested in your method for derivation. I am, basically, using iteration in my successive-approximation technique (which is a pain, since I'm doing the modeling work, rather than letting Excel do the heavy lifting.)

 

Taff

[Gerry in Quebec]

Let me summarize the latest posts on this topic. First, I didn’t
realize that Domebook 2 had included chord factors for class I, method
2. Thanks for pointing this out, Taff. Guess I need new glasses!

Second, Adrian’s and Taff’s chord factors for the 4v class I, method
2, icosahedral sphere agree with the Domebook 2 numbers – Taff’s a
little more closely than Adrian’s (Thanks to both for for the colour-
coded jpg images.). My chord factors do NOT agree, although they do
seem to define a valid geodesic sphere. I derived my 5 chord factors
by trigonometry, with around 20 steps in the process.... so I either
made an error or else didn’t understand the instructions on p. 107 of
Domebook 2. I’ll go back over my calcs to see if I can find the
error.

Third, like Taff, I’ve used iteration to find the “Mexican” chord
factors (number of unique strut lengths = geodesic frequency). I did
this some time ago, using Excel. I don’t have any mathematical means
of deriving the chord factors for higher-frequency geodesic spheres.

Finally, thanks to Ken Brown for referring me to Clinton’s NASA-
related work. The 1965 doc you mentioned isn’t on the topic of class
1, method 2, but a later work by Joe Clinton for NASA (Structural
Design Concepts for Future Space Missions, 1968) does give details of
what seems to be method 2, including equations. I’ll check into it.

Cheers to all, and many thanks for the help.
Gerry

[Gerry in Quebec]

Ken,
I see now the Clinton report you mentioned is indeed the one that
describes the class I, method 2 breakdown. When I went browsing for
it, I ended up finding an earlier NASA document by Joe Clinton, from
1965 -- a university thesis! Sorry for the mixup.
Gerry

On Feb 8, 2:59 pm, "Ken G. Brown" <kbr...@mac.com> wrote:
> See the Dec 4, 2009, geodesichelp thread titled "NASA-CR-101577 - Structural Design Concepts for Future Space Missions" for a link to download Clinton's early NASA report detailing how to calculate chord factors.
>
> I'm wondering if the 'new' breakdown method has previously been covered in the report.
>
> Ken G. Brown
>

> >> > For more options, visithttp://groups.google.com/group/geodesichelp?hl=en-Hide quoted text -

[Gerry in Quebec]

I found my error in the 4v icosa, class I, method 2 calculations. I
used sine instead of tangent in a trig calculation for the little
triangular windows. All 5 chord factors now match Domebook 2 to 7
decimal places. It looks like the authors hacked off everything after
7 decimal places without rounding.

Here are my values to 8 decimcal places.
0.27590448
0.29951578
0.29772519
0.31320669
0.32124407

If anyone is interested in the geometry/trig calculations for method
2, I can upload the derivation – an Excel file with diagrams.

Again, thanks for all the feedback.
Gerry in Quebec

> adr...@antiprism.comhttp://antiprism.com/adrian- Hide quoted text -

[TaffGoch]

Gerry,

 

Go ahead and attach the spreadsheet to one of your emails to the group. Teaching is what we're all about, right?

 

Actually, I'd be interested, even though I know how to calculate 'em, already. It can't hurt to compare.

________________

 

You described one of your spreadsheets, applying iteration, as being "a lazy person's method." Count me in that group -- iteration is one of my favorite Excel techniques, for all sorts of calculations.

 

Taff

[TaffGoch]

Gerry,

 

I really want to thank you for introducing Hernandez's subdivision, which is otherwise virtually unknown (undocumented, at the very least.)

 

I've finished my SketchUp derivation (back-engineering,) producing almost exactly the same results as yours. (One chord is different, at the 6th decimal.) I had one of those "eureka" moments last night, as I was drifting off to sleep. I tested the potential construction technique today, and it worked great! (Always keep a notepad on your nightstand, folks.)

 

I'm planning to do other frequencies, as well, eventually leading to a webpage here at the group, to fully document the technique, chords and spherical angles. I'd like to get bio info for professor Hernandez, to include with the description.

 

Thanks again for your contribution,

Taff

[Gerry in Quebec]

Hi Taff,
I'll go back to some old e-mails to see if I can find out more about
Hector Hernandez. I seem to recall he was interested in the geodesic
layouts mainly as a home builder rather than a mathematician. He
didn't respond to the most recent e-mail I sent him, maybe two years
ago, when I tried to hook him up with the peacock-cage builder, Harold
the Californian, who was using the same 6v geometry.

I made a couple of attempts to derive the Mexican layout
mathematically, but they give the wrong chord factors and spherical
angles. The key is to nail down the type of arc associated with each
set of equal-length chords in the parent triangle. It reminds me a bit
of Clinton's equal angle conjecture for honeycomb domes.

Thanks for the most recent image of the Mexican 6v layout -- the
symmetry triangle showing both arcs and chords.

Cheers,
Gerry

>  Mexican6.png
> 68KViewDownload

[TaffGoch]

On Sat, Feb 6, 2010 at 5:53 AM, Gerry in Quebec wrote:

I’ve also uploaded an alternative to the “ugly duckling” layout
(solution 10) that I posted a few days ago. This new layout, solution
10b, eliminates (hopefully) much of the uneveness in triangle shapes
of the earlier version, but it still has only 9 chord lengths.  File:
6vIcosa-7-18ths-Gerry-Sol10b.xls.

Well, I finally got around to modeling Solution10b. (Blame the delay on my captivation by Hernandez's subdivision!)

 

Taff

[Gerry in Quebec]

TAFF WROTE:
> Well, I finally got around to modeling Solution10b.

Hi Taff,
Many thanks for modeling sol. 10b (6v icosa, 7/18ths truncation).
While this version is a little less distorted than sol 10, it
unfortunately looks very much like its ugly duckling brother. It seems
clear now that using the Kruschke approach, which is to level the dome
base at the outset instead of doctoring the bottow row of triangles,
doesn’t produce good-looking domes at the truncations two rows above
and below the equator, at least not in the 6v. Anyway, it was a good
learning experience for me.

A while back a U.S. dome builder used Turbocad to generate images of
another layout, sol. 4, where the truncations are just one row instead
of two above and below equator, i.e., 8/18ths and 10/18ths. This
layout seems to look okay, though I guess it's ultimately a matter of
individual taste. (The corresponding layouts for the 5v also look fine
to my eye.) The advantage of this 6v dome layout, besides sitting flat
(again,based on the Krusckhe approach) is that it has only 9 chord
factors and a relatively small number of scalene triangles. Also,
compared with solutions 10 and 10b, there’s considerably less variance
in triangle shape and size. I’ve uploaded a jpg to the files section:
6v-10-18ths-sol4-toomey.jpg.

Cheers,
Gerry

>  Solution_10b.png
> 112KViewDownload