Equation for number of unique struts

[David Zhang]

I worked out an equation for calculating the number of unique struts for a given frequency v (if you're not using any of the strut-reducing dome methods).  It's a little messy, but maybe someone who's better at math can simplify this:

v/2 * ((v / 2)+1) +  (sin(π * (v % 2) / 2 ) / 4)

[Adrian Rossiter]

Hi David

Just some thoughts...

sin(π * (v % 2) / 2 ) is the same as v%2, giving

v/2 * (v/2 + 1) + (v%2)/4

You could also write this as

v^2/4 + v/2 + (v%2)/4
(v^2 + 2*v + v%2)/4

The begining looks like (v+1)^2, so maybe

(v^2 + 2*v + 1 - 1 + v%2)/4
((v+1)^2 - (v+1)%2)/4

Then if w = v+1 (the number of vertices on an original edge), the
formula becomes

(w^2 - w%2) / 4

This is fairly concise, but there is also the notation that [x] is the
largest integer not exceeding x, and (w%2)/4 is only either 0 or +1/4, and
so the formula can be written as

[w^2 / 4]

Or, using v

[(v+1)^2 / 4]

Adrian.
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[Dx G]

Nice work guys, I will review this and likely make good use of it.  Could you define v% or w%?  You may think it should be obvious but I assure to it won't be to a lot of readers. Perhaps its there and I missed it on initial review.

thx

Dx G

[Eric Marceau]

I think

    v%2

translates into

    v modulus 2

which is

    v/2  and dropping the remainder.

Pipe in, anyone, if I am wrong. :-)

Eric

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[Eric Marceau]

Sorry!  Mixed my operators.  I meant integer division ( or truncating division).

So, the expression

    v%2

translates into

    int ( v / 2 )

which is

    v/2  and rounding down to nearest integer (not decimal place).

Pipe in, anyone, if I am wrong. :-)

[David Zhang]

Eric,

% is modulus.  I was figuring this out on a computer and most programs use "%" for modulus and "/" for division.

Andrian,

Thanks!   I think I need a refresher on basic algebra hahah.   I think you're right, (v % 2)/ 4  would be the simplest form of that part of the equation.  The purpose of it is to produce .25 for odd values of v and 0 for even values, since if v is odd the left side of the equation will end in .75 instead of a whole number.

The reason I tried using sine is because modulus is harder to graph.   I did come up with a  sine version of it though:

(sin(3π/2  + πv) + 1)/8

although i guess it doesn't make sense to have a continuous graph since non-integer values of v are meaningless.

so, i guess 2 simplified version of the formula would be

(v² + 2v)/4 + (v mod 2)/ 4   ->  (v² + 2v + v mod 2)/ 4
(v² + 2v)/4 + (sin(3π/2 + πv) + 1)/8

Some additional info.  This will give the maximum number of unique struts for an icosahedral dome if when vertices are spaces out as evenly as possible (isotropic) - for instance, when using the spherical area coordinate (SAC) method.  The method Fuller used (method I i think) will often produce  struts that  should be unique, but because of arrangement, the lengths are so similar in size they might as well be the same for all practical purposes.  You will also have similar "collapses" in strut variation for octahedral domes, because of the way the primary triangles line up along great circles.

So this formula is useful because I often get rounding errors when programmatically calculating strut lengths, resulting in a ridiculous number of variations.  I think this has to do with floating point errors when it comes to doing trig on computers.   I wanted to figure out how many struts are actually unique and which ones are the same strut with a rounding error.

I found this table in a paper by Tom Davis with numbers mostly matching the ones produced by formula.  I've included the result given by the formula in parenthesis when it differs from Tom's calculations.  Notice how the formula numbers continuously go up instead of going up and down between each frequency.

11v  36

12v  30 (42)

13v  49

14v  56

15v  45 (64)

16v  72

17v   81

18v  63  (90)

19v  100

20v 110

[Adrian Rossiter]

Hi David

On Mon, 16 Dec 2024, David Zhang wrote:
> I found this table in a paper by Tom Davis

> <https://mathcircle.berkeley.edu/sites/default/files/BMC6/ps0405/geodesic.pdf> with

> numbers mostly matching the ones produced by formula. I've included the
> result given by the formula in parenthesis when it differs from Tom's
> calculations. Notice how the formula numbers continuously go up instead of
> going up and down between each frequency.
> 11v 36
> 12v 30 (42)
> 13v 49
> 14v 56
> 15v 45 (64)
> 16v 72
> 17v 81
> 18v 63 (90)
> 19v 100
> 20v 110

I haven't read the paper carefully, but it says "Our solution will be
simply to “push” those points out to the surface of the sphere from the
center", which I think means a projection of a planar equal edge
subdivision.

There is a "threeness" to the frequency series where the centre of
the base triangle runs through: upright triangle, upside down triangle,
vertex.

The mismatched counts have a vertex at the face centre. If you truncate
the base triangle corners then you get a hexagon, and the struts in
this hexagon have lengths as if there was hexagonal symmetry, and
this has reduced the total number of strut lengths compared to your
formula, which is counting symmetrically different struts under
triangular symmetry

I have attached an image of a projected 12v planar icosahedron, with
equal edge lengths having the same colour, and a truncation hexagon
marked, and within this you can see the edges are coloured with hexagonal
symmetry.

[Adrian Rossiter]

On Tue, 17 Dec 2024, Adrian Rossiter wrote:
> I have attached an image of a projected 12v planar icosahedron, with

That should have said 9v not 12v.

[David Zhang]

ahh thanks again Adrain.  I see that now, it happens every third frequency for the planar equal angle projection.  So Tom's calculations are accurate.   The equal arc and other projections makes the  central hexagon not symmetrical, which is where the extra struts are coming from.  A really interesting phenomenon.    In any case, the formula is probably only useful for mathematical modelling, unless you want to build a dome with vertices that are very evenly spread out , which  you can't even tell most of the time with higher frequencies.  It's easier to keep track of 16 struts for a16v dome instead of 72.

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[Dx G]

David, 

 Actually, you can still graph it.  Instead of a line, which you correctly point out would imply a continuous variable, the graph can be made as a bar graph.  This would provide a more appropriate data illustration, unless I misunderstand.

I'll have to take some time with the equations and better understand the application of the calculations.  So do you see any worthwhile use of this with lower frequency domes, like 6 to 8 or less, and does it matter whether we are looking at Class 1, 2 or 3?  Triacons only have even number frequencies as I understand it. 

Dx G

[David Zhang]

I think class II  and II would  require some modifications to the formula.  If anyone finds it I'd love to know!

I found a modified version of the formula the works for the equal arc projection.  You need to modify it because the struts along the edge of the primary triangle are all the same length, so you need to subtract roughly v/2 from the formula.  Interestingly, another way to get this number is using the formula for 1 frequency lower than the one you want and then add 1.

So it would be:

1 + ((v-1)² + 2(v-1) + (v-1) mod 2)/ 4

plugging  in 10, the formula gives 26 unique struts for 10v equal arc dome

if you're using the planar equal angle and the frequency is a multiple of 3, this formula seems to work

3/2 * (v² / 9   +  v / 3)

This matchesTom Davis's calculations

[David Zhang]

Eric,

% is modulus.  I was figuring this out on a computer and most programs use "%" for modulus and "/" for division.

Andrian,

Thanks!   I think I need a refresher on basic algebra hahah.   I think you're right, (v % 2)/ 4  would be the simplest form of that part of the equation.  The purpose of it is to produce .25 for odd values of v and 0 for even values, since if v is odd the left side of the equation will end in .75 instead of a whole number.

The reason I tried using sine is because modulus is harder to graph.   I did come up with a  sine version of it though:

(sin(3π/2  + πv) + 1)/8

although i guess it doesn't make sense to have a continuous graph since non-integer values of v are meaningless.

so, i guess 2 simplified version of the formula would be

(v² + 2v)/4 + (v mod 2)/ 4   ->  (v² + 2v + v mod 2)/ 4
(v² + 2v)/4 + (sin(3π/2 + πv) + 1)/8

Some additional info.  This will give the maximum number of unique struts for an icosahedral dome if when vertices are spaces out as evenly as possible (isotropic) - for instance, when using the spherical area coordinate (SAC) method.  The method Fuller used (method I i think) will often produce  struts that  should be unique, but because of arrangement, the lengths are so similar in size they might as well be the same for all practical purposes.  You will also have similar "collapses" in strut variation for octahedral domes, because of the way the primary triangles line up along great circles.

So this formula is useful because I often get rounding errors when programmatically calculating strut lengths, resulting in a ridiculous number of variations.  I think this has to do with floating point errors when it comes to doing trig on computers.   I wanted to figure out how many struts are actually unique and which ones are the same strut with a rounding error.

I found this table in a paper by Tom Davis with numbers mostly matching the ones produced by formula.  I've included the result given by the formula in parenthesis when it differs from Tom's calculations.  Notice how the formula numbers continuously go up instead of going up and down between each frequency.

11v  36

12v  30 (42)

13v  49

14v  56

15v  45 (64)

16v  72

17v   81

18v  63  (90)

19v  100

20v 110

[David Zhang]

I think class II  and II would  require some modifications to the formula.  If anyone finds it I'd love to know!

I found a modified version of the formula the works for the equal arc projection.  You need to modify it because the struts along the edge of the primary triangle are all the same length, so you need to subtract roughly v/2 from the formula.  Interestingly, another way to get this number is using the formula for 1 frequency lower than the one you want and then add 1.

So it would be:

1 + ((v-1)² + 2(v-1) + (v-1) mod 2)/ 4

plugging  in 10, the formula gives 26 unique struts for 10v equal arc dome

if you're using the planar equal angle and the frequency is a multiple of 3, this formula seems to work

3/2 * (v² / 9   +  v / 3)

This matchesTom Davis's calculations

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