Spiral arms


Funny. After playing with complex spirals for the past few months, I finally encountered a situation that looked just like one of the spirals, which got me to thinking about how to apply the math to the real world (old hat, I know, because it’s been done by entire industries for decades now). Regardless, the situation is cool.

There was an article mentioned on yahoo about a star being torn apart by a black hole, which was caught by a small telescope at McDonald Observatory. The phys.org article is accompanied by a 2014 youtube video showing a computer simulation of a similar event.

Just looking at the thumbnail of the video, I was struck by how it resembled a y = c^x spiral in a steep death curve towards 0. Then, after running the video I realized that it’s showing stellar mass being ejected out away from the black hole in the opposite direction as the star spins in a tight orbit around the hole. So, it’s a spiral that starts near 0 and eventually expands to infinity.

Direct youtube link

Normally, when you think of a decaying orbit, say of a satellite spinning around the Earth, you treat it as a pair of force vectors, with one vector being the planet’s gravitational pull (g) perpendicular to the “forward” velocity (Vf) of the satellite (usually initiated by some form of rocket propulsion). If Vf drops towards 0, the satellite is affected only by gravity, and it will fall straight towards the Earth. As Vf increases, the satellite’s orbit gets closer to a circle, until it’s more or less stable (ignoring any frictional or external gravitational effects). As vf exceeds this stable orbit value, the satellite will move farther away from Earth and eventually it will fly off on its own. In all cases, the combined two forces, gravity and satellite velocity, form a triangle when drawn on graph paper. And that triangle (a + ib) can be converted to polar coordinates as cos() + sin(), or as e^s.

In theory, a satellite’s orbit, or the spewing of stellar mass by a black hole, can be mapped to a Mandelbrot fractal if you know what value to assign to c0. That is, if you draw the orbit as a 2D plot on a piece of graph paper, with successive orbit points lying on one given spiral, then assign that to a specific spiral in the form of y = c0^x, turning around and drawing the Mandelbrot fractal using y = c0^2^x will give you the corresponding color value for that data point. It’s arbitrary, in that you need to pre-assign colors as well as values of x, and if you’re on a circle, c0 can be any point along it. Again, of course, given the nature of fractals, that small differences in starting values compound exponentially over time, you’d never get the EXACT correct value of c0 anyway, but you could get close, and “close” is all that matters in games of horseshoes and grenades. And star-chewing black holes.

But wait! There’s more!

What if we look at something closer to home, and closer to people’s hearts. Like money. And Las Vegas. And like trying to beat Las Vegas to make money.

Of all the gambling formats in Vegas, the one most attackable from a physics viewpoint is roulette. In effect, you have a small body traveling along a decaying trajectory before coming to a terminus. This is pretty much our death spiral, and it fulfills the conditions given above – being able to measure specific points along the spiral at varying times, or varying points at specific times. With this data, and the expectation that friction is a constant, we should be able to determine exactly which spiral we have, and roughly where on the wheel the ball will land. The calculations will be approximate, but the increased probability of at least getting the right quadrant of the wheel when placing bets would significantly improve your chances of winning over the house odds.

If only there were some way to verify this hypothesis

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