Clockwork Magpie Studios<\/a> (thank you!)<\/p>\n![\"DoublePendulum_00_Diagram\"](\"https:\/\/justinparrtech.com\/JustinParr-Tech\/wp-content\/uploads\/DoublePendulum_00_Diagram-300x197.png\")
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Here is a double pendulum at rest.\u00a0 The second pendulum in blue, hangs from the first pendulum in red.\u00a0 The blue pendulum pivots at the red point, and the red pendulum pivots at the top of its line.\u00a0 This simulator draws a blue line to trace the path of the bottom pendulum, which is perfect!<\/p>\n
We’ll make both masses (m1 and m2) and pendulum lengths (l1 and l2) equal, set the starting angle to 90 degrees, and no friction.\u00a0 Here is our initial state:<\/p>\n
![\"DoublePendulum_01_Initial\"](\"https:\/\/justinparrtech.com\/JustinParr-Tech\/wp-content\/uploads\/DoublePendulum_01_Initial-300x192.png\")
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When we click “Run” to start the simulation,<\/p>\n
![\"DoublePendulum_02a_Phase1\"](\"https:\/\/justinparrtech.com\/JustinParr-Tech\/wp-content\/uploads\/DoublePendulum_02a_Phase1-600x240.png\")
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Starting off fairly normal, the blue line traces a path not unlike that of a normal pendulum.\u00a0 After several more iterations, things start to get chaotic, as the two pendulums frantically trade kinetic and potential energy.\u00a0 After quite a few iterations, the blue pendulum starts to fill in all the dots within the system’s domain.<\/p>\n
The domain is bounded by the extent of pendulum 1 radius + pendulum 2 radius (r1 + r2), and the formula for a circle, but also by the amount of total energy.\u00a0 At the beginning, both pendulums were horizontal, so at any given time, the total amount of energy in the system can’t exceed its initial state, s<\/strong>, which was (m1 * g * h1) + (m2 * g * h2), or more simply, if h0 is equal to ( h1 at time 0 + h2 at time 0), then, for every state of the system, h0 >= h1 + h2<\/strong>.<\/p>\nThe maximum height of either pendulum is limited to 1\/2 the radius above horizontal (remember that r1 = r2).\u00a0 Because the pendulums are rigid, the math becomes a bit more complex, as the domain of the second pendulum at any point is limited by the domain of the first, but you end up with two half-teardrops on top of a half-circle.<\/p>\n
![\"DoublePendulum_05a_Phase2\"](\"https:\/\/justinparrtech.com\/JustinParr-Tech\/wp-content\/uploads\/DoublePendulum_05a_Phase2-600x240.png\")
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Note:\u00a0 After about 5 hours, it did not look significantly different than “E”.<\/em><\/p>\nSome readers might, at this point, proclaim “Aha!\u00a0 All of the points will eventually<\/em> be filled in!”<\/p>\nThe question is whether this system<\/em>, an analogy for pi, can ever<\/em> produce certain specific sequences<\/em> of points.<\/p>\nCan it produce the letter “C”?\u00a0 Yes, it probably can.<\/p>\n
Can it produce the letter “W”?\u00a0 NO, it can’t.\u00a0 The reason is that the entire system is always bounded by the amount of total energy at any given state, but also by the amount of kinetic vs. potential energy of either pendulum.\u00a0 Reversing directions for either pendulum requires kinetic to be converted to potential, or vice-versa.\u00a0 The only way to produce a sharp “v” or bend is to almost instantly transition between the two, in order to reverse directions.\u00a0 Under the right circumstances, you absolutely can produce a sharp “v”, but a “w” requires three such transitions in short order, otherwise, you might end up with the Greek lowercase letter omega (“\u03c9”).<\/p>\n
Although omega is a great letter, it’s not “w”, and we are asking the simple question:\u00a0 Is there any shape that this system CAN’T produce?\u00a0 Well, it can’t produce a “w”.<\/p>\n
There is no way that this system can produce the letter “w” because it would require too many conversions between kinetic and potential energy, that have to happen almost instantaneously.\u00a0 This rules out triangles and squares as well, for the same reason.<\/strong><\/p>\nIn addition, what about a horizontal line that extends from one end of the domain to the other, through the shoulder of pendulum 1 (the red one)?<\/p>\n
It can’t be done.\u00a0 As a matter of fact, the system can’t produce any horizontal line, spanning the domain, that falls between the shoulder and the lowest point of pendulum 1.<\/p>\n
The reason is quite simple, the blue pendulum (2) would simply fall before the halfway mark, because most of its energy would be potential energy, or it would be pulled down by the red pendulum for the same reason.\u00a0 The two are exclusive, so there is no way for it to draw a line all the way across, where the blue pendulum sits above the red pendulum.<\/p>\n
Drawing a horizontal line that completely traverses the domain of p1 would require more energy than exists in the total system.\u00a0 The red and blue pendulums would somehow have to levitate.<\/strong><\/p>\nI can keep going, with other lines, shapes, and curves that ARE valid finite sets, but will simply NEVER appear in this system.\u00a0 Even if you let it run for infinity.\u00a0 Even if you scale it up to infinity, and let THAT system run for infinity (assuming you had a constant gravitational field, infinite in volume, in which to run it).<\/p>\n
Let’s draw some analogies to pi.<\/p>\n
Like our double pendulum, pi appears to be chaotic.\u00a0 If so, it’s most likely the output of some simple, iterated calculation with feedback loop.<\/p>\n
Pi, like the double pendulum is bounded.\u00a0 Its domain is the digits 0..9.<\/p>\n
Looking at any finite string within pi as simply “the state of pi at time t || the state of pi at time (t+1) || the state of pi at time (t+2)…” (remember:\u00a0 We are using || for concatenation), we see similarities to how the double pendulum draws a curve.<\/p>\n
Any finite curve generated by the pendulum would simply be “the state of (p1 + p2) at time t || the state of (p1 + p2) at time t+1…” etc.<\/p>\n
We know that there are certain sequences that just can’t happen within our double pendulum system, because they require resources that can’t exist within the system itself, such as a temporary burst of energy that can’t exist, or the ability to instantly change direction multiple times.<\/p>\n
Likewise, even though we don’t understand it, if pi is truly chaotic, it’s just following a series of curves in “pi space” that we simply can’t see, but they are incapable of producing certain sequences<\/strong>.<\/p>\n <\/p>\n
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