Twice the Sides

After much fiddling with which radii intersect which bits of the circle clusters, and which opposite or adjacent ate what with whom, I eventually noticed that the solution to the issue of finding the dimensions in clusters of same-radius tangent circles that fit within a larger circle of known radius is to think of the polygon associated with the smaller circles as a polygon of two times as many sides as there are circle clusters.

In other words, if you look at the first figure in this post there, you see five clusters of three smaller circles contained within a larger circle. The turquoise polygon linking the centers of the inner-most circles is a decagon. Near the end of the previous post I had been attempting to find the radius of the smaller circles by considering a pentagon, in this particular case, linking the centers of the circle clusters. I imagine there's a way to do it, but it is rather more direct, I found, considering a polygon with twice the number of sides as there are circle clusters.

Still, starting from the simpler...ish case of an arrangement of n clusters of 3 circles (as opposed to n clusters of n circles), I may set the defined value as the radius of the larger circle, the one that encloses every last cluster, without any arc of that larger circle passing inside the region encircled by any of the smaller circles...*pant*

I suppose it is referred to as geometrization when one defines one constant as one instead of the more actually applicable constant that could, in practice, end up being complicated, and thereby making all the rest of the mathing complicated. It is done with the field equations for Einstein's General Relativity. It is done with the unit circle so fondly sited in trigonometry text books. The radius of the smallest circle to enclose the circle clusters can be set to one, and from that you can derive to equations for the dimensions of the circles it contains. Just gotta set one side of the equation to one, and solve for whatever.

...You also may need to figure out how to connect the formulary dots from the equations for dimensions of line segments in an n-gon to the dimensions of the line segements of n n-gons contained exactly half of the radius of the smaller circles inside the smallest wholely continaing circle with a radius of r1(the inner-most n-gon's inscribed circle's radius) plus r2 + R2 of the smaller n-gons (radius of the smaller, outer-more n-gon's circumscribed circle's, plus the radius of the circle the aforementioned n-gon is inscribed within...IF the recently aforementioned n-gon has an odd number of sides), plus a/2, where a is the length of every line segment to make up the n-gons inside the largest circle...

The problem is more generalizable though. The collection of the innermost circles--and these are, every one of them, circles of the same radius--likely remain numbered two per outer cluster, and each "cluster" corresponds to a polygon of n sides.

The simpler generalization of the thing is in the R and the r--R senior and junior. If the outer polygons are corresponding to an odd number of circles, then the critical sum in their case equals R + r + a/2, a/2 having already been discussed...

Since the inner polygon and the polygons implied by the circle clusters have the same side-length, I divided the known value, the radius of the largest circle, 1, into a sum of...

r1 - inscribed radius of the inner polygon
r2 - inscribed radius of the outer polygons
R2 - circumscribed radius of the outer polygons
a/2 - half the length of a side of any of the aforementioned polygons

I then substitute in those regular polygon equations that include the side length variable, a. For r2, and R2, the angle within the trigonometric function would be pi/3, as they refer to the radii of a regular triangle. Therefore, what I substitute in for these is...

r2 = (a * cot(pi/3))/2
R2 = (a * csc(pi/3))/2

r1 is the variable I could use to generalize at least the cases where the outermost polygons we're concerned with are triangles, and all the circles the same radius, and arranged as I have arranged them, with one circle pointing outward from each. I just leave the variable, n, which equates to the number of sides within the innermost polygon (two times the number of three-circle clusters there are) within the trigonometric function when substituting r1 with it's equation that includes a. Thus...actually, I suppose it may be more consistent to have a new variable, N, which is defined as two times the number of circle clusters, explicitly. Thus...

r1 = (a * cot(pi/N))/2 , where N = 2 * [number of circle clusters]

The complete equation in this case would be...




which, solving for a works out to be...

a =

...formatting these things is a mild pain, but anyhow: testing...

According to this equation, the length of any polygon side in the figure at the beginning of the post should beee...2/(1+sqrt(3)+sqrt( 5 + 2 * sqrt(5))), approximately 0.344249819... Checking the length of a in the figure I drew in autocad I get 0.34424982, which is an even better match than I expected, really, given the tiny discrepancies I had encountered in working these things into figures in autocad. Understandable given the fact I learned to use autocad at all by fiddling with it, and the framework itself is given to various of the approximation errors of a digital system.

This equation would, of course, not work with similar clusters of circles beyond or bellow three, nor do I see a way of easily generalizing the idea of what I've been doing thus far. For example, a cluster of four circles would not fit the generalization exactly. For one thing, in the framework I used, I used r and R, the inscribed and circumscribed radii, of the triangles, and the way a polygon with an odd number of sides works those two radii line up with the center of the outer-most circle, and from there the outermost circle is as far from the largest circle's edge by the length of it's radius.

With a cluster of four circles, whether there are two of it's circles arranged along the inner polygon, or one of it's circles, the radii going up it's center would be two times it's inscribed radius, or circumscribed radius, not a sum of both. Additionally, if two of it's circles are along the inner side, the line going between it's outer-most circles from where the radii end would be more than 'a'.

The set up of the equation I came up with I will consider a special case of a more generalized set up that includes contengencies for clusters of any number of circles rotated any applicable angle about it's local center. With all the applicable values plugged into this as of yet undefined equation many values would cancel out, leaving you with what I worked out...perhaps not directly...anyhow, more later.

Regular n-gon Cluster of Tangent Circles Tangent to the Circle That Encloses Them

I have been attempting to design and construct a three-dimensional model of a building using mainly the parameters:
-must be gorgeous...
-must smack of Gothic architecture, only better...

and the work flow I have developed over years of having scads of ideas that seem awesome for an hour or so is essentially: start off at an effectively arbitrary point that seems most interesting to work on, immediately assign arbitrary constraints on what I want which will be the anchor points for the obsessing I do later on to stall progress, discover that said constraints require lots of math, google around a while for math information of the type I think I need with the "optional" step of getting progressively more irritated as I either find my leads being unrelated despite sounding exactly like what I'm looking for from google, find that my leads ramble incomprehensibly across a Rorschach of bush-round-beating mostly made up of poorly described examples, or find I can't tell whether my lead tells me what I want to know usually due to a sort of incomprehensible rambling across a Rorschach of bush-round-beating mostly made up of poorly described examples...then I apply what I learn, then I rinse and repeat, or realize it's 7am, tell myself "just another hour" realize it's 11am, and...AHEM. Moving on.

A feature seen in many Gothic cathedrals, one that I rather like, are the pillars carved to look as though they are several pillars clustered together.



The design I have been making began with the idea of a cluster of four such pillars, though each pillar made up of three sub pillars, each of which would be the same circumference, and tangent to one another. Rising from the floors, each pillar cluster would diverge, while from each diverging pillar cluster the three sub pillars diverge as well, arching and intersecting to form the vaulting.

In a floor plan, the bases of such pillars can translate into a trio of circles, each of which have the same radius, and each of which is tangent to the two others.

For some reason I also decided to make design decisions based on the Fibonacci sequence. By that reckoning I decided the outermost edge of the outermost sub pillar of a sub pillar cluster should be exactly two meters from the center of the four clustered pillars.

Opening autocad, I began trying to draft the three circles first, and I realized I did not know how to do that. I had decided the outermost edge of a pillar must be two meters from the center of the cluster.

Once upon a time I was a child with a handful of dowel sticks. A handful of two dowel sticks, assuming you know the diameter of each, is the only handful you can know the exact diameter of a hole they both will fit through at once without doing a bit of convoluted trigonometry. A handful of three dowels is the only handful you could easily keep held such that they would fit within a perfectly circular hole. Try clutching a handful of four dowels and see if you can keep their innermost edges equidistant. When you figure out why you have been trying to do that for upwards of 45 minutes tell me, and I'll say that's why I did so when I was a kid once upon a time with a handful of dowels...I'll make something up for why I did it while not a kid...






If you draw one circle that is tangent to the origin, then draw another circle the same radius also tangent to the origin, the only arrangement of the two you'll find them not overlapping one another somehow is if they are offset 180 degrees from one another. If you then add a third circle of the same radius there is no arrangement where it may be tangent to the others, and to the origin.

In essence I knew the radius of the circle that encloses all four pillars. That was the only known I had. The dimensions of all the other circles in fact depend upon that one known quantity, so I began looking for a quantity within the unknowns that I could daisy chain into a set of relationships by which the one could be derived from the other.

One set of relationships I did know of as having a well established set of equations for finding all the dimensions involved if you have only one of them is the equations for polygons that have n sides. In autocad it's not so great a chore to draft a triangle, and use it to draft a trio of circles that are tangent to one another. I just need to draw the triangle, draw a circle centered on a triangle's point, and expand it to the midpoint of that triangle's side. Of course knowing that doesn't tell me directly how big the triangle needs to be for the circles to fit inside of a circle of known radius.

From the set of equations for finding dimensions of an n-sided polygon I derived which unknowns I needed to stick together to equate them directly to the one known I had. The length of a side, 'a', of a polygon with n sides can be found if you know the radius of the circle that encloses the polygon, 'R'. a = 2*R*sin( pi/n ).

The radius of a circle tangent to it's two neighbors, each of which is the same radius, is just half of a. Since the circles are centered upon the points of the polygon, the smallest circle that encloses the lot of them is the length of their radii away from their centers. So, the smallest circle to enclose said circles would have a radius of R + ( a/2 ). Substituting using the aforementioned formula, it is R + R*sin( pi/n ).

Crackers. So now, to find the R and find the a, I set the radius of the enclosing circle to 1, and solve for R. R = 1 / ( 1 + sin(pi/n) ). So there's the unit circle of the issue. Now it just remains to multiply by the radius I had decided on and derive from that the radius of the four pillars, then from that radius the radii of the three sub pillars making each up.

R1 = 2*( 1/(1+sin(pi/4)))
= 4-2*sqrt(2)
~ 1.17157287525380990239

a1/2 = 2 * ( 2 - sqrt(2) ) * sin( pi/4)
= 2*sqrt(2)-2
~ 0.82842712474619009760

R2 = ( 2 * sqrt(2) - 2 ) * ( 1/(1+sin(pi/3)))
= (4*sqrt(2)-4)/(2+sqrt(3))
~ 0.44395275812755717173

a2/2 = ((4*sqrt(2)-4)/(2+sqrt(3)))*sin(pi/3)
= 6-6*sqrt(2)-4*sqrt(3)+4*sqrt(6)
~ 0.38447436661863292586

...of course the problem with this solution is it does not find the dimensions such that the sub pillars actually touch those of the other pillars. There's gaps. I shall need to see about those now...

Why?

I have, as I'm sure many have done, found it easy, and cathartic to criticize the writers of educational material related to mathematics, both loud, and venomous. On the other hand, when I attempt to describe this or that thingy mathematical, I find myself struggling for words at times. Of course, currently, I find myself struggling for words in describing my reasoning for starting a blog where I try to describe mathematical thingies.

Right this moment my reasoning is thus: I want to make pretty pictures illustrating the process by which I derived a system of formulas for one math thingy in particular, and I want people to see them, and read about my deductions, and say to themselves "Zoons, this fellow is of an inhuman level of intelligence, and makes all other men seem shabby, and unattractive by comparison", then send me monies, and candies, and comely lasses of virtues deceptive.

My reasoning for writing a WTF post as the first is because I am a smart ass, and need the linguistical form of a kitty smirk in near everything I write.

"True true true, except for the lies." ~Eddie Izzard