Sphere; Developing Inscribed Icosahedron

[TaffGoch]

I have a geometric "recreation" for you. (Some folks aren't into mathematical recreation, and don't know that there are publications dedicated to this. If this isn't "your cup of tea," then please feel free to ignore.)

_________________________

Let's say that I have a given sphere, and want to mark the vertices of an icosahedron on the surface. How would I go about it, using geometric construction (compass and straight edge, only)?

Conditions: No mathematical calculations, but one -- You are allowed to mathematically-derive diameter from a measured circumference of the sphere.

Tip: Recall that three golden-rectangles, arranged at right-angles to each other, define the vertices of an icosahedron. (Believe it, or not, that hint may, or may not, help you much.) If you want to see a representation of the golden-rectangle method, refer to this SketchUp model (or, just the illustration):

http://sketchup.google.com/3dwarehouse/details?mid=3af9af9384d2097858f8f3aa29de1c06

Note: I researched this for a couple of days, before I found a helpful, obscure geometrical construction technique, which solved part of the exercise. So, I do, indeed, have an applicable solution, but thought that you might have a contribution, unique to that which I worked out. If you also find it mentally-engaging, then consider that a plus.

This exercise has a practical application, in that I want to mark icosa/sphere vertices for precise temari-ball development (would also be handy for Christmas ornament embellishment -- or, perhaps, applied to a garden gazing ball.)

-Taff

[Adrian Rossiter]

Hi Taff

On Wed, 25 Dec 2013, TaffGoch wrote:
> Let's say that I have a *given sphere*, and want to mark the *vertices* of
> an *icosahedron* on the *surface*. How would I go about it, using geometric
> construction *(compass and straight edge, only)?*

You can easily draw a cube with the same centre as the sphere, and
Euclid has a construction of the dodecahedron from this cube

http://aleph0.clarku.edu/~djoyce/java/elements/bookXIII/propXIII17.html

The join of a couple of angle bisectors on each face gives the
icosahedron centres, which can then be projected onto the sphere.


> *Conditions:* No mathematical calculations, but one -- You are allowed to

> mathematically-derive diameter from a measured circumference of the sphere.

Gasp!

> This exercise has a practical application, in that I want to mark

> icosa/sphere vertices for precise temari-ball development *(would also be

> handy for Christmas ornament embellishment -- or, perhaps, applied to a

> garden gazing ball.)*

The method I suggested probably isn't so useful for this, but
it is entirely Euclidean geometry.

Merry Christamas to you and everyone!

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

[Paul Kranz]

Taff:

How much time do we get before you spill the beans? It is Christmas, you know!

Paul sends...

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Very high regards,

 

Paul sends...

ReviveTheFlame.com

[Gerry Toomey]

Here's the best I could do on Taff's "recreation" challenge. No compass or straight edge required, but you need a tailor's tape measure, a pen, and a glass of eggnog. After all, it's Christmas.

 

I couldn't eliminate the math calculations completely -- the procedure below requires some arithmetic with the sphere's circumference: C / 10; C / 5; C /4; C / 2; and C x a "magic" number (0.1762).

 

This 8-step procedure assumes we know or can measure the circumference of the sphere. See the attached diagram. Here we go:

 

1) Mark a point anywhere on the sphere and label it 1.

 

2) Use a tailor's measuring tape to measure an arc distance of 1/4 the circumference, in any direction from point 1. Mark the point and label it 2. Now measure the same distance to another point on the sphere, roughly opposite point 2, that is, on the other side of the sphere. Label it 3.

 

3) Use the tailor's tape as a guide to draw a circle -- the sphere's equator -- through points 2 and 3. First, draw an arc from point 2 to point 3; take a sip of eggnog, then continue the arc back to point 2.

 

4) Use the tape measure to mark four evenly spaced points along the equator. Start at point 2 and measure a distance of one-fifth the sphere's circumference. Mark and label this as point 4. Repeat this for points 5, 6 and 7.

 

5) Now lay out the tape measure between points 1 and 2. Measuring from point 1, mark and label point 8 at a distance of 0.1762 times the circumference. Repeat the process to mark points 9, 10, 11 and 12 on arcs 1-4, 1-5, 1-6 and 1-7, respectively, all the same distance from point 1. Thus, Arcs 1-8, 1-9, 1-10, 1-11 and 1-12 are all the same length. Points 1, 8, 9, 10, 11 and 12 are six of the icosahedron's 12 vertices.

 

6) From point 1, measure half the sphere's circumference. Mark the point and label it 13. This is the seventh vertex of the icosahedron. Five more to go. Time for more eggnog.

 

7) Taking a cue from step 5, establish five new points along the equator. Starting at point 2, measure a distance of one-tenth the circumference. Mark the point and label it 14. For the remaining four points (15, 16, 17 and 18), measure a distance of one-fifth the circumference. Thus, arc lengths 14-15, 15-16, 16-17, 17-18, and 18-14 are all the same length.

 

8) Now repeat step 6 to locate and mark the five remaining vertices of the icosahedron, this time measuring along arcs 13-14, 13-15, 13-16, 13-17 and 13-18. The new points, 19 through 23, are the final five vertices of the icosahedron. And the new arc lengths -- 13-19, 13-20, 13-21, 13-22 and 13-23 -- are the same as in step 6, namely 0.1762 times the circumference.

Merry Christmas,

- Gerry in Quebec where it's too cold to shovel snow

P.S. The "magic" number is arctan 2 / (2π), or 0.17620819.

 

[Paul Kranz]

It may be close enough to (1) tape the circumference of the sphere, (2) fold over the tape five times, (3) set the compass to the width of 6 folded sections, and (4) mark the sphere with that dimension. That should take the 360 degrees down to close enough to the degrees of the central angle of the icosa, 63.434949, to give you the length of the short side of the golden rectangle for that sphere. No measuring.

Paul sends...

[TaffGoch]

Adrian, that is analogous to golden-section use. I'm surprised that they don't mention it, even though they refer to a golden ratio value, "(√5 –1)/2". 

And I know that you are aware, in addition to the dodecahedron, the tetrahedron and the icosahedron can be inscribed within a cube. The dodecahedron and icosahedron, both, employ the golden section relationship, in a cubic "environment."  (Those Clark University instructions sure seem overly-complicated, to me.)

I've used a cube for developing each of those polyhedrons, but was looking for something more direct (and easy!)

(I think you'll like what I eventually found. In fact, it may, perhaps, be derived from the above "cube" relationships.... Maybe.)

-Taff

[TaffGoch]

Paul,

Hmmm, 360°, halved 5 times, is 11.25°,...

... x 6 = 67.5°

Fair approximation -- likely better than "eye-balling" it, but, surely, you have me confused with someone else. I'm seldom satisfied with "good enough."

The approximation might do for great-grandma Nana and her temari, but not for me and mine. I do concede that, at 3-4" styrene ball sizes, this may, indeed, be a good tip for hobbyists. For larger spheres, I'd like better precision. (And there is, actually, a simple geometric construction method that "nails it.")

_______________________

In answer to your previous question. I've probably already "spilled the beans." (I shouldn't have mentioned the golden ratio/rectangle hint.) I really am looking for alternate solutions. "There's more ways than one, to skin an armadillo."

I probably won't show my solution until after the holidays... probably.

-Taff

[TaffGoch]

Adrian,

Thanks for the Euclid "Elements" link, by the way. I have a hard copy (who doesn't,) and it's nice to see an enhanced, online version, now bookmarked.

[TaffGoch]

Yikes, Gerry!

[TaffGoch]

You must have miscalculated...

...the measure of alcohol you put in your egg nog!

[TaffGoch]

If I were still teaching geometry, I'd have to give you bad marks.

Your forte (and my foible) is, obviously, calculation. I prefer (and ask for,) geometric construction. Shame, shame, shame....

Take off your spherical trig "hat," and try geometry, alone. That may be insurmountable, as it would be for me to switch from geometry to calculation. You might be better at the switch than I would be. My skill-set is specific, but limited, I'm afraid.

-Taff

[TaffGoch]

For those unfamiliar with geometric construction of the Golden Section (or, Golden Rectangle,) here's the most common method, and the one I typically employ:

To construct the square root of 5, a critical value for both, the Golden Ratio, and the icosahedron:

You may find them useful, in your geometric explorations.

-Taff

[TaffGoch]

The problem, simplified to the "stickler":

How to construct the Golden Rectangle, when the diagonal (sphere diameter) is known?

-Taff

[Adrian Rossiter]

Hi Taff

On Wed, 25 Dec 2013, TaffGoch wrote:

> Adrian, that is analogous to golden-section use. I'm surprised that they
> don't mention it, even though they refer to a golden ratio value, "(√5
> –1)/2".

They do use it, but the text comes from an older translation, and the
term used is "extreme and mean ratio" rather than golden ratio.

In the page for the construction of the dodecahedron from the cube,
the ratio is used on the third line of the construction text

http://aleph0.clarku.edu/~djoyce/java/elements/bookXIII/propXIII17.html

And, the reference for the contruction of this ratio uses the same
method that you posted

http://aleph0.clarku.edu/~djoyce/java/elements/bookII/propII11.html

[Paul Kranz]

Taff:

 

Does this guy look like you? http://youtu.be/-ncEEXekZek

 

Paul sends...

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[TaffGoch]

D'oh! Of course, that should have been obvious. Euclid predates the use of the term.

--------------

The Elements webpage illustration recalled to mind:

[TaffGoch]

Paul,

He looks like I used to look. I've since regenerated....

-Taff

[TaffGoch]

Every geometric construction method that I had seen, in the past, only depicted how to construct the golden section when one of the two sides was given. None demonstrated how to construct, when only the diagonal was known.

I, finally, found a test question, which required the student to provide the proof of geometric construction of a golden rectangle, given the diagonal. Here is my depiction, not of the proof, but of the construction method:

All you need to know, to accomplish the construction yourself, should be evident from the animated steps. I do not provide the geometric proof, although I have demonstrated it to my own satisfaction. (Remember "similar triangles," from geometry class?)

Here's how I will mark a sphere, using a drawing compass, to scribe the vertices of the enclosed icosahedron:

See the image included with the original post, which depicts how 3 golden sections (rectangles) define the vertices of an icosahedron.

So, there you have it -- no math; just geometric construction, employing only a compass and straight edge.

Happy New Year,

-Taff

[TaffGoch]

Here's the test problem that I found, online:

[TaffGoch]

And, here's where I found it: http://math.fau.edu/yiu/RecreationalMathematics2003.pdf (page 330)

[Paul Kranz]

Taff:

Great resource, thanks! How would you advise constructing the square armed with only the side, a ruler and a compass? I believe these were your original constraints.

Paul sends...

On Wed, Jan 1, 2014 at 8:35 PM, TaffGoch <taff...@gmail.com> wrote:

And, here's where I found it: http://math.fau.edu/yiu/RecreationalMathematics2003.pdf (page 330)

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[TaffGoch]

One of the first lessons (perhaps, THE first lesson) in technical-drawing class (and geometry, too,) is how to construct a perpendicular. Using that technique, a square is an easy subsequent step.

[TaffGoch]

Also, see square-construction diagram...

...at bottom of page 3: http://etc.usf.edu/clipart/galleries/455-geometric-constructions/

(Although, some study of the intermediate steps is in order.)

[TaffGoch]

In my initial constraints, I did permit mathematical calculation of the diameter, when circumference is known (measured.) I just stumbled across a reference to "Swale's method," which implies that construction, alone, might indeed be possible:

Again, no method is provided -- it's a test question. But, at least, I now know a name, upon which I can search. Investigating....

(I don't hold much hope, as this appears to be a simple construction of a diameter/radius, when the CIRCLE is known -- NOT the circumference. So, how do I construct the circle, when the circumference is known? That problem is NOT implied in this question.)

-Taff

[TaffGoch]

Swale's method was a dead end,...

...however, I DID find a solution! (Although, I don't know it's name, if it has one.)

From "Geometric and Engineering Drawing, 3rd Edition," by Kenneth Morling; page 4-4

http://books.google.com/books?id=Y53qIM7O9-4C&pg=SA4-PA4&lpg=SA4-PA4

This construction starts with a STRAIGHT line, being the length of the circumference (exactly what I was looking for):

Pretty obscure solution, but it works, as I can demonstratively attest.

-Taff

[homespun]

Taff,

   I am trying hard to understand how to solve this little "test problem", but not quite getting it.

   All I have so far is this:

  

Assume radius of circle is 1, and each side of square = 2:

AD = 2, AM = 1, MQ = 1

From Pythagorean theorem, DM = sqrt 5

DQ = sqrt 5 + 1

 

But then I get stuck.  I don't see the "similar triangles".  Need more clues.

 

I know that the quadrilateral inscribed in the circle must be a rectangle, because the two chords in a semicircle form a right angle (half the intercepted arc)

 

I tried playing around with "Ptolemy's Theorem" about "cyclic quadrilaterals" (quadrilaterals inscribed in a circle), but it didn't help.

 

                                        Dan

--

[TaffGoch]

Dan,

The similar triangle, to which you must compare, is not present. You have to construct a golden rectangle (by your choice of multiple methods,) and then compare the diagonal section from that rectangle to the one in the afore-referenced diagrams.

Here's how I did it:

I could then compare the new, larger triangle to the smaller one, within the circle. Confirmation that the corresponding edges of the two triangles are parallel is all that is required to establish that they are, indeed, "similar triangles."

-Taff

[TaffGoch]

As I surmised, Swale's method does NOT start with a straight line; length = circumference. It starts with a CIRCLE, so it isn't any help, for this exercise:

I'll keep it "in my bag of tricks." Perhaps, it may still serve a useful purpose, in some future exercise....

-Taff

[TaffGoch]

Drats! I'm pissed.

The circumference method...

...does not provide the purported result. The derived diameter is 99.6% accurate, only. I started with a straight "circumference" line, of length equaling pi (3.1415926)

The diameter should have worked out to be, precisely, 1.0000000. It did not. The derived/constructed diameter measured 0.99584. A very close approximation, but 99.584% is not "spot on."

I concede that this degree of accuracy was good enough for pencil & pen/ink engineering drawing, so I'd have to say that the method served its stated purpose. In this age of digital CAD, however, it doesn't quite cut it.

________________________

For my physical sphere layouts, I doubt that I can measure the circumference to 99.6% accuracy, anyway. The method should, therefore, do the job.

(Still pissed, nevertheless)

-Taff

[Adrian Rossiter]

Hi Taff

On Wed, 1 Jan 2014, TaffGoch wrote:
> Swale's method was a dead end,...

Still interesting. I had a look at the excercise, to establish
the method, for fun.

ROQ is an equilateral triangle, so <ROQ = 60, and by the central
angle theorem <RPQ = 30. L lies on PR, so <LPQ=30. Let O' be the
centre of C, by the central angle theorem <LO'Q = 60, but triangle
LO'Q is isoceles, and hence equilateral, and so LQ = LO', which is
the radius of C.

For the last part, LQ subtends equal angles on C, so
<LOQ = <LPQ = 30. With two equal sides and included angle (SAS)
triangle OQL is congruent to triangle ORL, and LR = LQ.

[Adrian Rossiter]

Hi Taff

On Thu, 2 Jan 2014, TaffGoch wrote:
> Drats! I'm pissed.
>
> The circumference method...
> [image: Inline image 1]
> ...does not provide the purported result. The derived diameter is 99.6%
> accurate, only. I started with a straight "circumference" line, of length
> equaling pi (3.1415926)
>
> The diameter should have worked out to be, precisely, 1.0000000. It did
> not. The derived/constructed diameter measured 0.99584. A very close
> approximation, but 99.584% is not "spot on."

It can't be spot on. In Euclidean geometry you can't construct
the diameter from the circumference (or vice versa)

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

[Adrian Rossiter]

Hi Dan

On Wed, 1 Jan 2014, homespun wrote:
> I am trying hard to understand how to solve this little "test problem", but not quite getting it.
> All I have so far is this:
>
> Assume radius of circle is 1, and each side of square = 2:
>
> AD = 2, AM = 1, MQ = 1
>
>> From Pythagorean theorem, DM = sqrt 5
>
> DQ = sqrt 5 + 1
>
> But then I get stuck. I don't see the "similar triangles". Need more clues.

In the attachment, PHBG is a golden rectangle by the method Taff
previously posted, then PGB is similar to APB, proving APBQ is
golden.

[TaffGoch]

Adrian wrote: 

"It can't be spot on. In Euclidean geometry you can't construct

the diameter from the circumference (or vice versa)"

Now, he tells me....

___________________

Actually, I knew that π, being an irrational number, can't be constructed (to infinity,) so it follows that neither could a circumference-to-diameter.

I was, merely, hoping for a little-bit better precision. "Spot on" was a bad choice of phrase.

And, hey, it's better than the alternatives

-Taff

[homespun]

Adrian,
Thanks so much. shows I was getting warm when I found these late last
night about the construction of a square in a semicircle:

http://jwilson.coe.uga.edu/EMAT6680Fa11/Chun/Fianl2/fanal%202.html
http://jwilson.coe.uga.edu/EMAT6680Fa11/Nguyen/HHNAssignment%2013/SquareinsideCircle.html
http://jwilson.coe.uga.edu/EMAT6680Fa11/Molnar/final2ram/final2ram.html

And:
http://www.jimloy.com/geometry/square0.htm

Dan

----- Original Message -----
From: "Adrian Rossiter" <adr...@antiprism.com>
To: <geodes...@googlegroups.com>
Sent: Thursday, January 02, 2014 11:32 AM
Subject: Re: Sphere; Developing Inscribed Icosahedron

[TaffGoch]

I, too, appreciate Adrian's similar rectangles/triangles diagram. Here, I superimpose the larger golden rectangle that I used, to go along with Adrian's smaller golden rectangle (in green.)

Plenty of "similar" to go around....

[homespun]

I also received this from Antonio Gutierrez of www.gogeometry.com

 

 

----- Original Message -----

From: TaffGoch

To: Geodesic Help Group

Sent: Thursday, January 02, 2014 2:39 PM

Subject: Re: Sphere; Developing Inscribed Icosahedron

I, too, appreciate Adrian's similar rectangles/triangles diagram. Here, I superimpose the larger golden rectangle that I used, to go along with Adrian's smaller golden rectangle (in green.)

[TaffGoch]

Dan,

Following your references, I can see that I, too, could lose track of the proofs, as they progress. Now, that's why I prefer construction over mathematical calculation.

(Hey, I'm a visual kinda guy, easily distracted by pretty whatnots....)

-Taff

[TaffGoch]

Constructing, from a straight-line circumference, to derive diameter length, I devised my own method, with improved accuracy:

Previous method was 99.58% accurate, so this new method is about a degree-of-magnitude closer.

(I don't know if this construction method is documented elsewhere.)

To construct, bisecting the circumference several times is employed, to produce sixteenths. Copying 3/16ths, adding to the beginning of the circumference line, provides the arc-center for the second, shorter arc. Dropping a vertical from the arcs intersection provides the diameter.

Actually, I designed this method after producing a construction method to do the reverse -- construct circumference-from-diameter (which I'll post, next.)

-Taff

[TaffGoch]

Here's the construction method for producing the circumference, when given the diameter:

This construction is also of my own derivation. The diameter line is extended, to the right, to produce a line over 3 times the length of the original diameter line. The first arc, on the left, is centered at 2x diameter, on this line. The "second" diameter line is divided to produce 4/5ths. (The GLaD construction is fairly new, so you may not have seen it before. It can be used to produce 1/3, 1/4, 1/5, 1/6, 1/7, ... 1/n, of any line segment. Very handy.)

From the 4/5ths position, a vertical line is extended upwards, to intersect with the first arc. The second arc, centered on the right end of the original diameter line, is drawn from the initial-arc-vertical-line intersection.

(Has anyone seen these construction methods, before?)

-Taff

[Paul Kranz]

Taff:

This is news to me. If D = 1, then does C = π?

Paul sends...

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[TaffGoch]

Paul wrote: "If D = 1, then does C = π?"

That's the idea, but, remember, this is construction, so there's no "π" as a mathematical value. Construction permits establishing a length (angle, tangent, physical geometry) without mathematical values or calculations. (That's how I construct my geodesic 3D models - no mathematical calculations.)

It is, indeed, a valid test of the construction method, however. That's how I arrived at the accuracy percentages -- by precise CAD measurements and comparisons.

-Taff

[Gerry in Quebec]

Taff,
From your recent posts, I see now what you mean by "geometric construction". It's what I had to do with a compass and ruler in junior high school back in the 1960s.The "accuracy percentage" of your construction method to derive the diameter from a straight-line circumference is actually a bit better than shown in your diagram. Through calculation (of course) I found it to be 99.9598%.

As for the derivation of circumference from diameter, the accuracy of your method is less than the figure you mentioned in the diagram. I get 99.8991%.

If I'm wrong about this, I'm sure you'll rap me on the knuckles with your straight edge.
- Gerry

[TaffGoch]

Gerry,

Rounding off, I can be satisfied with 99.9%. (At least, I think I can.)

Calculations, sir, I gladly leave to you....

-Taff

[TaffGoch]

Gerry,

Regarding the diameter-to-circumference construction method:  Setting the value of the diameter to 1.000000, I get a value for the radius, for the right-hand arc, of 2.144706, leading to a circumference of 3.144706. Comparing to pi, I get 99.900997% accuracy, which is closer to your 99.8991% calculation. (My previous labeling, of 99.9900%, is obviously, somehow, a measurement error on my part.)

I may be at the measurement limit-of-precision in SketchUp. I could scale things up, x1000, to get better precision, but I am satisfied with the accuracy of method, as is, so, I'll let it alone. Enough fiddling....

-Taff

[Gerry Toomey]

Taff,
Our accuracy figures don't agree for two reasons. First, our values for C differ. Yours is 3.144706; mine is 3.144761. Second, we have calculated the accuracy percentage differently. You have taken it to be 100 x (pi / C). I have taken it to be 100 x { 1 -  [ (C - pi) / pi ] }, which is  the same as 100 minus % error.

I labeled key points in your diagram: W, X Y & Z.

Here's the path I took:

When length D in your diagram = 1, YZ = 2, XY = 0.2, and XW = 0.8.
XZ = sqrt (YZ^2 - XY^2)
XZ = 1.98997487... 
WZ = sqrt (XZ^2 + XW^2)
WZ = 2.14476105...
Length C = D + WZ = 1 + 2.14476105... = 3.14476105...

How accurately does C approximate pi, i.e., the circumference of a circle whose diameter is 1? Accuracy as a percentage = 100 * { 1 - [ (absolute difference between pi & C) / pi ] } = 100 * { 1 - [ 0.00316840... / pi ] } = 99.8991...

As you said, enough fiddling.

- Gerry

[Gerry in Quebec]

Gary Doskas on "spherical harmony"  -- see the part on geometric construction of the icosahedron, near the beginning of the presentation. Taff, this may be of particular interest to you.

- Gerry

http://www.youtube.com/watch?v=h9EQfqXDvKs

On Wednesday, December 25, 2013 2:46:16 AM UTC-5, TaffGoch wrote:

I have a geometric "recreation" for you. (Some folks aren't into mathematical recreation, and don't know that there are publications dedicated to this. If this isn't "your cup of tea," then please feel free to ignore.)

_________________________

Let's say that I have a given sphere, and want to mark the vertices of an icosahedron on the surface. How would I go about it, using geometric construction (compass and straight edge, only)?

Conditions: No mathematical calculations, but one -- You are allowed to mathematically-derive diameter from a measured circumference of the sphere.

Tip: Recall that three golden-rectangles, arranged at right-angles to each other, define the vertices of an icosahedron. (Believe it, or not, that hint may, or may not, help you much.) If you want to see a representation of the golden-rectangle method, refer to this SketchUp model (or, just the illustration):

http://sketchup.google.com/3dwarehouse/details?mid=3af9af9384d2097858f8f3aa29de1c06

Note: I researched this for a couple of days, before I found a helpful, obscure geometrical construction technique, which solved part of the exercise. So, I do, indeed, have an applicable solution, but thought that you might have a contribution, unique to that which I worked out. If you also find it mentally-engaging, then consider that a plus.

This exercise has a practical application, in that I want to mark icosa/sphere vertices for precise temari-ball development (would also be handy for Christmas ornament embellishment -- or, perhaps, applied to a garden gazing ball.)

-Taff

[Gerry in Quebec]

Oops, I see Taff posted a link to this presentation quite some time ago. Sorry.

https://groups.google.com/forum/#!searchin/geodesichelp/Doskas/geodesichelp/kbjpg1oBQls/xEK4bvS1_yAJ

[TaffGoch]

Gerry,

(Since I was the only one posting to that discussion, I suppose I was just talking to myself.)

Gary also has finished his book (actually, a couple) which are available at lulu.com:

http://www.lulu.com/shop/search.ep?type=&keyWords=doskas

Hardcopy and inexpensive ebook versions available.

-Taff