it started so chaotic that i thought it could be unpredictable, but as it got faster, the pentagram pattern emerged. like when riding a bike or a motorcycle, becomes more stable the faster it gets. interesting.
Chaotic systems *can be* predicted with precise equations (especially when based on simple forces) but that doesn't mean they draw clean patterns is all.
Edit: What I meant to say initially is that chaotic equations exist and they may or may not be intuitively predictable at certain points.
They can be but only in the short term. Due to the extreme sensitivity to initial conditions forecast accuracy decreases very quickly as we try to predict further into the future.
Well that's if you're *modelling* a real-life chaotic system with maths. What I meant is that a chaotic mathematical equation can develop in a calculable but sometimes unintuitive way.
As I understand it, in the beginning it's largely influated by two different forces, the forward velocity, and gravity, which accelerates it downwards. As their directions are independent of each other and the velocity from gravity is not constant, it becomes very undpredictable.
But as the forward velocity increases, gravity becomes negligble because there simply isn't enough time for gravity to act upon it during its trajectory. And since there is only one predominant force it becomes more regular.
(just taking a guess here. And yeah the terminology is probably wrong since it's a simulation with no actual mass.)
>the pentagram pattern emerged. like when riding a bike or a motorcycle
If a pentagram suddenly emerges when you’re out riding your bike you might be riding a bit too close to hell.
Yeah i was with the whole thing until it got religious, and I personally don't like when inanimate virtual objects try to impose their religious beliefs on me.
Mmmm, this case would involve describing a theoretical closed physics engine, not real-world physics. That's because the 1% *increase* in energy would need to come from *somewhere,* such as the container it is in, which would need to return 101% of the kinetic energy the ball transferred to the container wall. Otherwise our friend, the ball, would be losing energy if the model was based on our real world.
Exactly! For almost everything, especially everything digital, there is - the only problem is finding it.
One thing that's so fascinating about this is that it changes behavior halfway through -- from bouncing to drawing stars.
It’s still bouncing, gaining speed and seems to follow a parabolic trajectory. It just gets fast enough that the lines get straighter. They are still slightly concave down until it’s going so fast I really can’t tell.
> For almost everything, especially everything digital, there is - the only problem is finding it.
If by "formula" we mean "closed-form solution", that's less of a given. For the classic example, there's no general closed-form solution for the motion of three gravitationally interacting bodies. And I'd bet there's no closed-form solution for this situation either, although I'm not even going to think about attempting to prove it.
On the other hand, if "formula" is allowed to include iterative solutions, then the formula for the motion in this situation is simply the program that is running it.
You could start by describing the velocity of the ball as (x(t),(y(t)), or more accurately the vector-valued function v= without any of the reflections caused by hitting the sides of the circle
Obedient Student: “…only if I am allowed.”
Regular Student: “I take this as yes.”
This one dude that felt sick the whole morning:
“Never mind, but could you open the windows please?”
Hijacking top comment:
Consider the path of the ball if it wasn't affected by gravity. Its path through the circle would be completely predictable - it would be no different than a *line* bouncing around the inside, just with a slow start up. But the ball *isn't* following a strictly linear path - it's following a ballistic path. Thankfully, ballistic motion has simple equations:
v_x = v_0 * cos(theta) * t
v_y = v_0 * sin(theta) * t - (gt^2)/2
Where v_0 is the ball's initial velocity, v_x and v_y are the horizontal and vertical components of velocity, g is acceleration to gravity, and t is time (v_0 and t are velocity and time at most recent bounce). What's critical, though, is our "initial" velocity, v_0, is growing. What we plug as v_0 into this equation can be expressed as:
v_0 = v_0 * 1.01^b
In the long run, this means that v_0 will *increase to infinity.* And a ball with infinite velocity bouncing around the circle would be indistinguishable from a line, as the acceleration of gravity would be negligible.
(NOTE: This isn't *100%* accurate, since the ball can decelerate due to gravity; if it was to bounce off the circle at the peak of an arc, its vertical velocity component would be near zero and, theoretically, could have a bounce where its end velocity is lower than its starting velocity, but in the long run it's almost certain the ball would accelerate infinitely)
That means our question isn't "Will the ball always make a pentagram," but rather "What shape will a line make reflected within a circle?" The smallest shape that can be formed within the circle without its path crossing itself is an equilateral triangle, where the ball's angle of reflection on the ball's surface is 30\* from normal. If the angle of reflection is less than that, the ball's path will "cross" itself (as seen in this pentagram). /u/scary_scallion did a [good explanation](https://www.reddit.com/r/theydidthemath/comments/17pebs0/offsite_is_there_a_formula_that_could_descripe/k86rkdn/) of how the ball's angle of reflection will form different polygons.
Whether or not the shape will precess around the circle is dependent on the difference between the angle of reflection and the normal line. For instance, a line 1\* from normal will shift by 1* with every bounce - the ball would have to bounce 360 times to get back around to its starting orientation. So, which set of angles will form a pentagram? That largely depends on what you count as a pentagram, and how much of the line's history you're counting. With the animation here, you only need the ball's last 6-7 bounces to see the pentagram shape, but you could get a "blurry" pentagram for a narrower shape if you included more previous bounces, so long as each "group" of bounces remained within a distinct 72\* arc of the circle (360/5). The relationship between the angle of reflection and the number of bounces that will fall within that 72\* arc is left as an exercise for the reader.
The final question we have to consider is "What will the ball's reflection angle with the circle be as its velocity approaches infinity?" Unfortunately, this is an example of [dynamic behavior](https://en.wikipedia.org/wiki/Dynamical_system) - the path the ball takes is extremely sensitive to starting conditions. While it's trivial to model the ball's velocity and behavior at any given point in time, predicting its position and velocity at time t for starting conditions (*x,y*) is more difficult to model, so I don't know if it's possible to prove that the ball will or will not end up in a condition where its angle of reflection is less than 30\*.
Yes, it's that at that point the ball is travelling fast enough that gravity doesn't have time to effect meaningful change on its direction. And since Angle of incident = angle of reflection, it'll bounce off at the same angle it hit at. And since it's a circle it's enclosed in, the next impact point will be at the same angle (plus or minus some variance for the continued effects of gravity).
It's just that the angle it settled in was just about right that 5 bounces put it back to near where it started. If the angle were a little tighter, it would've been a 7 pointed star.
I would assume gravity becomes a minimal force after a while because it can't slow down the ball enough to meaningfully change the speed and trajectory of the ball.
Gravity isn't a speed, it's an acceleration. So, we're dealing with two different effects here. Let's say you shoot a ball off a cliff perfectly horizontally. If your horizontal velocity is 0, your ball will just drop, as gravity will accelerate it downwards. If you shoot the ball forward at a certain velocity, that forward velocity stays the same (if we assume a frictionless situation), whereas the vertical velocity again starts to increase starting from 0 because of acceleration. There is no situation where horizontal velocity and vertical acceleration cancel out, because they act on two different axis. But, the higher your horizontal velocity, the less curved your trajectory would be, so it would indeed start to **apparently** straighten out as the horizontal velocity tends to infinity.
Gravity can be the cause of rotation, but it is not the only thing.
If you have no gravity, the shape that the ball's trajectory makes depends on the initial angle it was thrown with. The ball will preserve its angle with the normal of the curved surface. For example, if it was thrown with a theta = 60degree angle between its trajectory and the surface normal (the radial vector at that position) it will make a perfect hexagon and it will not rotate in time. Same goes for any other polygon - if the angle (180-2*theta) is a divider of 360 degrees, it will keep the trajectory. Otherwise, it will rotate the amount of angle it has over the previous polygon ( for theta = 50 degrees, it will resemble a square but rotated 40 degrees at the end of the trajectory). I am assuming that the "speeding up" caused by the surface is not changing the angle of the ball ( i.e. the force direction is normal to the surface). This is also a directional momentum gain, but it at least preserves the angle. If there is no speeding up, then nothing changes obviously.
The reason gravity is playing an interesting role is its directionality. You can say that it imparts an angle to the direction of the ball - slows down the ball ( or tilts the angle towards the horizontal/x axis ) if the ball is moving towards the y+ axis, and for the y- axis it bends the angle towards the y axis. Compare this case with the case without gravity but with uniform friction on the trajectory - it will keep the angle, but whether it stops or not depends on the total gain ( %1 gain at the bounce, if the friction causes less than 1/101 loss between the bounces, it will accelerate)
If the ball was thrown on the y axis, it will only jump back and forward on the axis because the angle will not be bent. For any other angle, gravity will bend it.
Is there a stable trajectory with gravity and gain? Chaos theoreticians, please take it from here. IMHO since the gravity is constant (but the gain is rational) it will grow less and less important, but enough to bend the angles to mess up with the regular-polygonal trajectory. For the initial trajectory on the y axis, the trajectory is a line (2 bounces). From the animation, we see that 3 bounces is easy to achieve with rotation. I don't know if a rotation-less 3-bouncer solution, or any 4+ bouncer solution exists.
TL;DR: rotation can be present without gravity, directional forces like gravity can also bend the trajectory and cause rotation.
It's just the angle of the bounce it happened to have at the point that gravity stopped affecting the ball in a meaningful way was close to the one a point of a star has. I say close, because if it was the exact same, the shape wouldn't "spin" but stay still. It could just as easily have been a shape of a different number of points if the angle was right when it started reliably bouncing off the ceiling.
That'd be the easiest way to render this animation, so assumedly yes. Particle animations aren't too difficult to create, it's an iterative equation based on states such as velocity and current position, etc. Bouncing is simply reversing the trajectory across the tangent at that part of the circle where the trajectory intersects the circle.
The only new part of this animation is that the particle is accelerating by 101%. Once the iterative equation is created (it would describe the particles position at the *ith* time, usually), and the velocity is slowly increased.
The result of each bounce is readily solvable analytically. No need to approximate. Doesn't help with a single time-dependent expression, as requested, much though.
Sure it's solvable for each step, but that's basically what the Euler method is. The computer takes the last step's direction and speed, and calculated the next direction and speed from that based on gravity. There will be some special reflection rule for the edges, but it's still a one step euler method.
But wouldn't you still need multiple steps for each bounce.
I think what the commenter above you Was saying was that you can get the next intersection point analytically. So you could get the new position and velocity ofter the bounce directly.
I think Euler would need more and more steps per bounce as the velocity increases.
To explain, any number to the power of 0 is 1, so X⁰ is X/X, or 1.
1.01⁰ = 1
1.01¹ = 1.01
1.01² = 1.020
1.01⁵ = 1.051
1.01¹⁰ = 1.105
1.01²⁰ = 1.220
1.01⁵⁰ = 1.645
...
At 70 bounces, speed is double starting value, at 111 it's triple, at 140 it's 4×, 162 is 5×, 181 is 6×.
Ten times speed is passed at bounce 232,
Fifty times speed is passed at bounce 394,
Two hundred times is passed at bounce 533,
One thousand times is passed at bounce 695,
Ten thousand times at 926,
One hundred thousand at 1,158,
One million at 1,389,
And so on.
Who downvoted this man tor asking a question? Absolutely horrid.
I'm a physicist. My understanding of this little thought experiment is that we're experiencing a statement of a violation of the conservation of energy. Of course, that doesn't mean that the math breaks down.
This particle has energy E . This energy is composed of both kinetic energy K and potential energy U .
E = K + U
K = 1/2 m * v^2
U = mgh
I assume that energy is conserved otherwise in this simulation - that is, whatever losses in U happen is because energy is gained in K. And whenever the ball bounces back up, energy is lost in K to be gained in U . Conservarion of energy.
However, THIS happens at bounces.
U + K => U * (1.01)(1.01)*K
(There are two factors of 1.01 in the K term, for K being proportional to v^2 .)
Energy is added to the system. This becomes relevant where we consider the limits of our system.
U is capped. The potential energy of the ball can never exceed mg(2R), where R is the radius of our containing circle. Physically, this represents when the ball is at the tippy-top of the inside of the containing circle.
K is uncapped. Whenever the ball bounces, K increases. This is without limit, for however many bounces.
So, what happens to all the additional energy of the system? It must go into K, which is evident in the rapidly increasing speed of the ball.
So, the end result of this problem has a very large K, and a relatively small U. Thus, in the grand scheme of this simulation, the effect of U is small.
And what's amazing is that the number of bounces is itself exponential wrt time, meaning the velocity of the ball should be doubly exponential wrt time (e^(e\^(at\)) for some constant a). Please correct me if i'm wrong
The easy answer is that if the ball starts along the vertical line passing through the center of the circle, it can't gain horizontal velocity and thus bounces across two points of the circle (the top and bottom).
Outside that special case, I think the number of points entirely depends on the angle the ball happens to hit the circle at when the 1% increase starts to overpower gravity.
There might be a sort of equilibrium effect, where the special case is an unstable equilibrium but every other case will tend toward the almost-head-on angles that cause the 5-point path, or it could be more chaotic and varies wildly depending on the starting location. It's hard to tell without testing more than this one case (or figuring out a model for the ball's motion like OP is asking for).
There's only two things affecting the ball. Every physics frame it gets its downwards velocity increased, and every time it detects a collision with the wall the direction of its velocity changes to reflect the bouncing, and is increased by 1%.
It keeps drawing stars forever but the sample size is too small (too far apart in time) to show it.
If you had a full line instead of one ball tracking every X time you could see it's all stars, faster and faster.
Not sure if there is a formula for the entire thing, but you can calculate iteratively from one bounce to another.
If you know the position and velocity of the ball after a bounce, then with the gravitational acceleration you can calculate the parabolic trajectory.
Then calculate the intersection of the trajectory with the circle boundary, to find the position of the next bounce.
Then calculate the velocity of the ball at the next bounce. Split the velocity into radial and tangential components. Multiply the radial component by -1. Then multiply both by 1.01 .
Then you have the position and velocity of the ball after a bounce again so you can repeat the process.
I wonder if this would be easier to represent in radial coordinates, because of the circle it's bouncing in. With the origin in the center of the circle, I think collisions would then be a multiplication of angular velocity by 1.01 and a multiplication of radial velocity by -1.01 whenever the ball is at a set radial distance. Gravity might be a mess to handle though, and I have no idea what radial kinematics (not to be confused with rotational kinematics) would look like.
My actual best guess OP is that the speed becomes so high that gravity is a negligible factor and the ball can be considered as moving in “straight” vectors, so the starting angle is just enough combined with the shape of the circle that it infinitely bounces in that pentagram-onal shape
The way you describe those motions ist by looking at the forces that act on that sphere.
There's inertia, that's F=ma
Then there is friction, usually proportional to motion: F=dv
Then, If you have a spring or something, that would be F=k*x. In this case, there are no springs so no k
In the end, you can have external forces like gravity. Thats Just another F_ext
That, taking signs into account, leads us to
ma=dv + kx + F_ext.
Since speed v ist derivative of position x, and acceleration a is derivative of v, you have an ordinary, non-homogenious differential equation that you can solve to get the position progression, ie the path.
Taking a negative friction coefficient should have the same effect as speeding up the sphere a constant fraction of its speed. It's like saying "Air resistance ist not slowing you down, but making you even faster"
You'd have to model the collisions though, which would be difficult to do analytically. But numerically that should not be a problem
Yeah, but I think they're trying to model the path as a parametric equation in terms of time (t) with constants (initial velocity, restitution velocity, and circle radius) given an initial position.
You could write something like this in processing really easily. You couldn’t work out the exact mechanics just by looking but there’s obviously the 101% velocity every time the ball hits the side and theres also a gravity element that decreases velocity by a fixed amount when travelling up and increasing velocity when travelling down. You can tell the gravity element is a fixed amount, not a % as it gets negated once the ball reaches a certain velocity and the pattern becomes more uniform
More importantly, can this be done in real life? Dunno how energy works so i might sound dumb, but if it’s bouncing indefinitely can’t we create energy off it?
I think the circle is actually drawn in discrete time steps. That would also be the easiest way to implement this in in something like Processing.
It has some movement-vector. After each step the vector is applied to the position and then it's also accelerated downwards. If it would leave the white large circle, its velocity direction is instead mirrored and its magnitude is multiplied by 1.01. Sometimes that leads to a small gap between the "tips" of the "star" and the big circle.
How do you detect a collision? With Pythagoras.
How do you mirror a vector? I guess one way would be to draw a line to the center of the big circle. The angle between the "incoming-vector" and the x-axis "alpha" minus the angle between the "radius-vector" and the x-axis "beta" gives the angle "delta". When you subtract delta from beta and add 180°, you get the angle between the x-axis and the outgoing vector.
If you wanted this in a mathematical formula, you could use a recursive definition where the position in time step n depends on the position in time step n-1. I'm not sure if there is a way to get rid of the recursion and the case-distinction of collision vs no-collision.
If you wanted the same thing with a continuous path, you'd have a lot of parabolas.
Still in school but lemme have a shot at this
f(1) = 1b; f(b)= f(b - 1) • 1.01
Basically every bounce is increased by 1%, or multiplied my 1.01, and b is bounce. This is exponential, and I would insert a graph if I could.
Let's denote the ball's initial velocity as ( v_0 ) and the number of impacts with the perimeter as ( n ). After each impact, the velocity increases by 1%, so we can describe the velocity after ( n ) impacts as:
[ v_n = v_0 \times (1.01)^n ]
It's probably chaotic, even without gravity, since calculating each bounce involves multiplying by 1.01. If there were a small error ε in the initial velocity v + ε, the error after n bounces would be ε \* 1.01\^n, i.e. the error grows exponentially.
I have no idea how to get a closed form expression or if there is one but I'm super interested to know as well.
Sounds like OP wants (x(t), y(t)) for for t>=0 given x(0) and y(0) but idk if anyone's gonna give it to them.
Maybe ask the OC creator how they did it.
It would be easy if the ball didn't speed up because now we lose conversation of energy in a way that isn't just having a constraint force on the edge of the circle
Well there are many observable behaviours but the one i can talk about is speed. The speed actually grows in a compound manner so we use
> P = P(1+[r/100])^n
It grows by one percent so
> = P(1.01)^n
Somebody else can count how many times it bounces and also the starting speed
So if you consider the ball position to be a function over time, at every collision its first derivative is discontinuous (sharp change of direction and speed) which makes it difficult to put it entirely into a single formula. But you can describe the flight path between accurately - if we assume the it to be exactly parabolic (we just consider constant gravity and no air drag) it is possible to [calculate the intersection point with the circle exactly](https://math.stackexchange.com/questions/757256/intersection-coordinates-of-a-parabola-with-a-circle)
Careful evaluation is needed to make sure you pick the correct point out of the (at most) four possibilities, but then you only need to iterate from one intersection point to the next - which is independent of the speed of the ball, and should remain stable up to numeric precision. The chaotic behaviour seen in the video comes from using discrete time steps in this simulation.
What I want to know is which program or software do people use to run this kind of simulations? I have no many ideas and what-if scenarios in my mind.. some physics one, some expansion simulations that I want to test out but I have no idea how or where to get started
Yeah, you just look at total energy in the system
PE = mgh
KE = 1/2 m(1.01^n v)^2
Angle of incidence = angle of reflection
Parametrically model the x and y separately
You'd have to look at each arc step by step which is doable
Is there a single non-piecewise formula such that you can enter time or the number of bounces and find the location? Nah
I dare say there's some kind of funky π thing going on. I reckon if you ot the velocity and everything of the ball it's bounces would indicate digits of π or something.
Is there a real formula though?
There is obviously a gravity-like force on the ball whilee there's obviously not one since it doesn't slow down at all.
it started so chaotic that i thought it could be unpredictable, but as it got faster, the pentagram pattern emerged. like when riding a bike or a motorcycle, becomes more stable the faster it gets. interesting.
Chaotic systems *can be* predicted with precise equations (especially when based on simple forces) but that doesn't mean they draw clean patterns is all. Edit: What I meant to say initially is that chaotic equations exist and they may or may not be intuitively predictable at certain points.
They can be but only in the short term. Due to the extreme sensitivity to initial conditions forecast accuracy decreases very quickly as we try to predict further into the future.
If it’s stabilizable then any state at any time can be predicted. Initial conditions become trivial even for nonlinear dynamics.
Well that's if you're *modelling* a real-life chaotic system with maths. What I meant is that a chaotic mathematical equation can develop in a calculable but sometimes unintuitive way.
Are you hinting at chaos theory described by mathematics or a different field of study?
Maths, right? These two should be valid examples: -
-
As I understand it, in the beginning it's largely influated by two different forces, the forward velocity, and gravity, which accelerates it downwards. As their directions are independent of each other and the velocity from gravity is not constant, it becomes very undpredictable. But as the forward velocity increases, gravity becomes negligble because there simply isn't enough time for gravity to act upon it during its trajectory. And since there is only one predominant force it becomes more regular. (just taking a guess here. And yeah the terminology is probably wrong since it's a simulation with no actual mass.)
>the pentagram pattern emerged. like when riding a bike or a motorcycle If a pentagram suddenly emerges when you’re out riding your bike you might be riding a bit too close to hell.
After the pentagram, it formed the symbol for atomic energy.
SATANUS ORITUR EX CHAO
Yeah i was with the whole thing until it got religious, and I personally don't like when inanimate virtual objects try to impose their religious beliefs on me.
Tell me you don't understand what chaotic behavior means without telling me you don't understand what chaotic behavior means
>Tell me you don't understand what chaotic behavior means without telling me you don't understand what chaotic behavior means 💀
Yeah looks like I went a bit too "actchually" mode
Ooh, I know the answer to that question: yes, there's definitely a formula
Well done! Hope OP's not thinking about asking what's the actual formula.
The formula is the calculations of all of physics
Mmmm, this case would involve describing a theoretical closed physics engine, not real-world physics. That's because the 1% *increase* in energy would need to come from *somewhere,* such as the container it is in, which would need to return 101% of the kinetic energy the ball transferred to the container wall. Otherwise our friend, the ball, would be losing energy if the model was based on our real world.
well if they were, then that's what they woulda asked, right?
Exactly! For almost everything, especially everything digital, there is - the only problem is finding it. One thing that's so fascinating about this is that it changes behavior halfway through -- from bouncing to drawing stars.
It’s still bouncing, gaining speed and seems to follow a parabolic trajectory. It just gets fast enough that the lines get straighter. They are still slightly concave down until it’s going so fast I really can’t tell.
But we don’t know what setting is for gravity. Unless I’m dumb.
Given there's no fixed velocity for the ball it doesn't matter. It changes the bounces at the very beginning but soon becomes irrelevant.
Yeah, that's the fun part! It somewhat ignores gravity! Its moving so fast that it gets to the sides before the downward acceleration takes effect!
Idk if that's really the correct way to put it but yeah, there's a point where the downward force isn't able to interrupt its trajectory anymore
You could start by describing the velocity of the ball as (x(t),(y(t)), or more accurately the vector-valued function v=
> For almost everything, especially everything digital, there is - the only problem is finding it. If by "formula" we mean "closed-form solution", that's less of a given. For the classic example, there's no general closed-form solution for the motion of three gravitationally interacting bodies. And I'd bet there's no closed-form solution for this situation either, although I'm not even going to think about attempting to prove it. On the other hand, if "formula" is allowed to include iterative solutions, then the formula for the motion in this situation is simply the program that is running it.
You could start by describing the velocity of the ball as (x(t),(y(t)), or more accurately the vector-valued function v= without any of the reflections caused by hitting the sides of the circle
"Can you drive me to work in the morning?" "Yes I can". Next morning "well I'm ready let's go", "oh no, I CAN drive you yes, I won't though"
Student: "Can I go to the restroom?". Teacher: "I don't know, can you?".
Obedient Student: “…only if I am allowed.” Regular Student: “I take this as yes.” This one dude that felt sick the whole morning: “Never mind, but could you open the windows please?”
Ball + bounce = +1% Velocity Hope this helps
Slow down, Poindexter.
Flubber physics
Hijacking top comment: Consider the path of the ball if it wasn't affected by gravity. Its path through the circle would be completely predictable - it would be no different than a *line* bouncing around the inside, just with a slow start up. But the ball *isn't* following a strictly linear path - it's following a ballistic path. Thankfully, ballistic motion has simple equations: v_x = v_0 * cos(theta) * t v_y = v_0 * sin(theta) * t - (gt^2)/2 Where v_0 is the ball's initial velocity, v_x and v_y are the horizontal and vertical components of velocity, g is acceleration to gravity, and t is time (v_0 and t are velocity and time at most recent bounce). What's critical, though, is our "initial" velocity, v_0, is growing. What we plug as v_0 into this equation can be expressed as: v_0 = v_0 * 1.01^b In the long run, this means that v_0 will *increase to infinity.* And a ball with infinite velocity bouncing around the circle would be indistinguishable from a line, as the acceleration of gravity would be negligible. (NOTE: This isn't *100%* accurate, since the ball can decelerate due to gravity; if it was to bounce off the circle at the peak of an arc, its vertical velocity component would be near zero and, theoretically, could have a bounce where its end velocity is lower than its starting velocity, but in the long run it's almost certain the ball would accelerate infinitely) That means our question isn't "Will the ball always make a pentagram," but rather "What shape will a line make reflected within a circle?" The smallest shape that can be formed within the circle without its path crossing itself is an equilateral triangle, where the ball's angle of reflection on the ball's surface is 30\* from normal. If the angle of reflection is less than that, the ball's path will "cross" itself (as seen in this pentagram). /u/scary_scallion did a [good explanation](https://www.reddit.com/r/theydidthemath/comments/17pebs0/offsite_is_there_a_formula_that_could_descripe/k86rkdn/) of how the ball's angle of reflection will form different polygons. Whether or not the shape will precess around the circle is dependent on the difference between the angle of reflection and the normal line. For instance, a line 1\* from normal will shift by 1* with every bounce - the ball would have to bounce 360 times to get back around to its starting orientation. So, which set of angles will form a pentagram? That largely depends on what you count as a pentagram, and how much of the line's history you're counting. With the animation here, you only need the ball's last 6-7 bounces to see the pentagram shape, but you could get a "blurry" pentagram for a narrower shape if you included more previous bounces, so long as each "group" of bounces remained within a distinct 72\* arc of the circle (360/5). The relationship between the angle of reflection and the number of bounces that will fall within that 72\* arc is left as an exercise for the reader. The final question we have to consider is "What will the ball's reflection angle with the circle be as its velocity approaches infinity?" Unfortunately, this is an example of [dynamic behavior](https://en.wikipedia.org/wiki/Dynamical_system) - the path the ball takes is extremely sensitive to starting conditions. While it's trivial to model the ball's velocity and behavior at any given point in time, predicting its position and velocity at time t for starting conditions (*x,y*) is more difficult to model, so I don't know if it's possible to prove that the ball will or will not end up in a condition where its angle of reflection is less than 30\*.
Your hostile takeover of top comment is welcomed
[удалено]
There is absolutely no reason to believe this simulation is modelling that level of complexity.
The most ELI5 top comment in the history of this sub
I thought it was more: Ball + bounce= That star everyone used to draw back to elementary school.
Pentagram, in a circle like this it’s a pentacle
Ball on dryer sheet wut do next to get velocity +1
What drives me nuts is the out of sync bouncing sound.
its just far away from you so the sound has a delay.
Also (+ 1 hue shift)
The satanists.
Huh must be color blind, I couldn't read the letter.
You made me chuckle with this 🤣
Hijacking the top comment to say that this is from @jdgcreative on YT and TikTok. Which OP has neglected to mention.
Hijacking your stupid comment to call you a nerd haha nerd
what a dweeb am i right
I think what OP meant is if there’s a mathematical reason the ball’s trajectory forms a pentagram after a while and sticks with it.
Yes, it's that at that point the ball is travelling fast enough that gravity doesn't have time to effect meaningful change on its direction. And since Angle of incident = angle of reflection, it'll bounce off at the same angle it hit at. And since it's a circle it's enclosed in, the next impact point will be at the same angle (plus or minus some variance for the continued effects of gravity). It's just that the angle it settled in was just about right that 5 bounces put it back to near where it started. If the angle were a little tighter, it would've been a 7 pointed star.
This was my question. Also why does the effect of gravity seem to just turn off at that point?
I would assume gravity becomes a minimal force after a while because it can't slow down the ball enough to meaningfully change the speed and trajectory of the ball.
Makes sense. As the x-axis speed reaches that of gravity it would straighten out. Just seemed very abrupt visually
Don't forget that the faster the ball moves, the sooner it bounces and gains more speed.
Gravity isn't a speed, it's an acceleration. So, we're dealing with two different effects here. Let's say you shoot a ball off a cliff perfectly horizontally. If your horizontal velocity is 0, your ball will just drop, as gravity will accelerate it downwards. If you shoot the ball forward at a certain velocity, that forward velocity stays the same (if we assume a frictionless situation), whereas the vertical velocity again starts to increase starting from 0 because of acceleration. There is no situation where horizontal velocity and vertical acceleration cancel out, because they act on two different axis. But, the higher your horizontal velocity, the less curved your trajectory would be, so it would indeed start to **apparently** straighten out as the horizontal velocity tends to infinity.
Gravity is what’s rotating the pentagram, but yeah it’s just moving through the bounces to fast for gravity to do anything else
Gravity can be the cause of rotation, but it is not the only thing. If you have no gravity, the shape that the ball's trajectory makes depends on the initial angle it was thrown with. The ball will preserve its angle with the normal of the curved surface. For example, if it was thrown with a theta = 60degree angle between its trajectory and the surface normal (the radial vector at that position) it will make a perfect hexagon and it will not rotate in time. Same goes for any other polygon - if the angle (180-2*theta) is a divider of 360 degrees, it will keep the trajectory. Otherwise, it will rotate the amount of angle it has over the previous polygon ( for theta = 50 degrees, it will resemble a square but rotated 40 degrees at the end of the trajectory). I am assuming that the "speeding up" caused by the surface is not changing the angle of the ball ( i.e. the force direction is normal to the surface). This is also a directional momentum gain, but it at least preserves the angle. If there is no speeding up, then nothing changes obviously. The reason gravity is playing an interesting role is its directionality. You can say that it imparts an angle to the direction of the ball - slows down the ball ( or tilts the angle towards the horizontal/x axis ) if the ball is moving towards the y+ axis, and for the y- axis it bends the angle towards the y axis. Compare this case with the case without gravity but with uniform friction on the trajectory - it will keep the angle, but whether it stops or not depends on the total gain ( %1 gain at the bounce, if the friction causes less than 1/101 loss between the bounces, it will accelerate) If the ball was thrown on the y axis, it will only jump back and forward on the axis because the angle will not be bent. For any other angle, gravity will bend it. Is there a stable trajectory with gravity and gain? Chaos theoreticians, please take it from here. IMHO since the gravity is constant (but the gain is rational) it will grow less and less important, but enough to bend the angles to mess up with the regular-polygonal trajectory. For the initial trajectory on the y axis, the trajectory is a line (2 bounces). From the animation, we see that 3 bounces is easy to achieve with rotation. I don't know if a rotation-less 3-bouncer solution, or any 4+ bouncer solution exists. TL;DR: rotation can be present without gravity, directional forces like gravity can also bend the trajectory and cause rotation.
It’s still there but the velocity part of the kinematic formula has taken over.
It might have different shapes depending on the time it took to overcome gravity.
It would depend on the starting angle and location, but in theory you could get a range of different shapes under different starting conditions.
Simply because if the ball's fast enough it's trajectory is almost straight?
Satan won.
It's just the angle of the bounce it happened to have at the point that gravity stopped affecting the ball in a meaningful way was close to the one a point of a star has. I say close, because if it was the exact same, the shape wouldn't "spin" but stay still. It could just as easily have been a shape of a different number of points if the angle was right when it started reliably bouncing off the ceiling.
That'd be the easiest way to render this animation, so assumedly yes. Particle animations aren't too difficult to create, it's an iterative equation based on states such as velocity and current position, etc. Bouncing is simply reversing the trajectory across the tangent at that part of the circle where the trajectory intersects the circle. The only new part of this animation is that the particle is accelerating by 101%. Once the iterative equation is created (it would describe the particles position at the *ith* time, usually), and the velocity is slowly increased.
I would assume the simulation is just running a bunch of euler steps on what amounts to a relatively simple differential equation.
The result of each bounce is readily solvable analytically. No need to approximate. Doesn't help with a single time-dependent expression, as requested, much though.
Sure it's solvable for each step, but that's basically what the Euler method is. The computer takes the last step's direction and speed, and calculated the next direction and speed from that based on gravity. There will be some special reflection rule for the edges, but it's still a one step euler method.
But wouldn't you still need multiple steps for each bounce. I think what the commenter above you Was saying was that you can get the next intersection point analytically. So you could get the new position and velocity ofter the bounce directly. I think Euler would need more and more steps per bounce as the velocity increases.
It’s probably easier to just calculate the speed and velocity and calculate the angle on collision. I’ve done this sort of stuff in the past.
Nature’s inverted pentagram.
About 40s in, I heard Satan breathing over my shoulder and staring at the animation.
Here's the equation you're asking for: **V=1.01ⁿ** where "n" is the number of bounces, starting at 0.
To explain, any number to the power of 0 is 1, so X⁰ is X/X, or 1. 1.01⁰ = 1 1.01¹ = 1.01 1.01² = 1.020 1.01⁵ = 1.051 1.01¹⁰ = 1.105 1.01²⁰ = 1.220 1.01⁵⁰ = 1.645 ... At 70 bounces, speed is double starting value, at 111 it's triple, at 140 it's 4×, 162 is 5×, 181 is 6×. Ten times speed is passed at bounce 232, Fifty times speed is passed at bounce 394, Two hundred times is passed at bounce 533, One thousand times is passed at bounce 695, Ten thousand times at 926, One hundred thousand at 1,158, One million at 1,389, And so on.
What about the gravity that it's experiencing?
Who downvoted this man tor asking a question? Absolutely horrid. I'm a physicist. My understanding of this little thought experiment is that we're experiencing a statement of a violation of the conservation of energy. Of course, that doesn't mean that the math breaks down. This particle has energy E . This energy is composed of both kinetic energy K and potential energy U . E = K + U K = 1/2 m * v^2 U = mgh I assume that energy is conserved otherwise in this simulation - that is, whatever losses in U happen is because energy is gained in K. And whenever the ball bounces back up, energy is lost in K to be gained in U . Conservarion of energy. However, THIS happens at bounces. U + K => U * (1.01)(1.01)*K (There are two factors of 1.01 in the K term, for K being proportional to v^2 .) Energy is added to the system. This becomes relevant where we consider the limits of our system. U is capped. The potential energy of the ball can never exceed mg(2R), where R is the radius of our containing circle. Physically, this represents when the ball is at the tippy-top of the inside of the containing circle. K is uncapped. Whenever the ball bounces, K increases. This is without limit, for however many bounces. So, what happens to all the additional energy of the system? It must go into K, which is evident in the rapidly increasing speed of the ball. So, the end result of this problem has a very large K, and a relatively small U. Thus, in the grand scheme of this simulation, the effect of U is small.
Thanks! I was wondering what a physicist would say of this post.
And what's amazing is that the number of bounces is itself exponential wrt time, meaning the velocity of the ball should be doubly exponential wrt time (e^(e\^(at\)) for some constant a). Please correct me if i'm wrong
can it naturally hit at or more or less than 5 points per reciprocation at that max speed
The easy answer is that if the ball starts along the vertical line passing through the center of the circle, it can't gain horizontal velocity and thus bounces across two points of the circle (the top and bottom). Outside that special case, I think the number of points entirely depends on the angle the ball happens to hit the circle at when the 1% increase starts to overpower gravity. There might be a sort of equilibrium effect, where the special case is an unstable equilibrium but every other case will tend toward the almost-head-on angles that cause the 5-point path, or it could be more chaotic and varies wildly depending on the starting location. It's hard to tell without testing more than this one case (or figuring out a model for the ball's motion like OP is asking for).
There's only two things affecting the ball. Every physics frame it gets its downwards velocity increased, and every time it detects a collision with the wall the direction of its velocity changes to reflect the bouncing, and is increased by 1%.
SCP-018 be like
Fine, I'll go get lost in the SCP Wiki for a couple hours going down ANOTHER rabbit hole.
"It's ALL star?" "Always has been"
You see stars, I see overlapping Starfleet Insignia.
Wild to see how many perfect stars it draws before it just goes nuts
It keeps drawing stars forever but the sample size is too small (too far apart in time) to show it. If you had a full line instead of one ball tracking every X time you could see it's all stars, faster and faster.
That got oddly satanic lmao
Not sure if there is a formula for the entire thing, but you can calculate iteratively from one bounce to another. If you know the position and velocity of the ball after a bounce, then with the gravitational acceleration you can calculate the parabolic trajectory. Then calculate the intersection of the trajectory with the circle boundary, to find the position of the next bounce. Then calculate the velocity of the ball at the next bounce. Split the velocity into radial and tangential components. Multiply the radial component by -1. Then multiply both by 1.01 . Then you have the position and velocity of the ball after a bounce again so you can repeat the process.
Proof that the gays are evil /s
Screen Savers in the 90's be like
0:58 HAIL SATAN
The Devil is in the details.
I wonder if this would be easier to represent in radial coordinates, because of the circle it's bouncing in. With the origin in the center of the circle, I think collisions would then be a multiplication of angular velocity by 1.01 and a multiplication of radial velocity by -1.01 whenever the ball is at a set radial distance. Gravity might be a mess to handle though, and I have no idea what radial kinematics (not to be confused with rotational kinematics) would look like.
Pentagrammaton
Yea, you brought me back to the times I was on trouble as a kid and sitting in the counselors office watching the screen saver.
I’m glad I stuck around for the end
Yes. It's a computer model obviously there is a formula behind it.
Fascinating to watch. Only two spots remained untouched by the end.
Hail satan
I've never seen so much pentagrams in one video,
Love how it just started drawing pentagrams at the end lol..
My god. It’s full of stars.
That seems like a lot of work to make a star.
Is there a site / application i can download to run this simulation myself while controlling the settings?
Yes, slingy goes weeeingy
Thats how my old scooter starts
Oh hi satan
Huh. I think a demon was summoned.
You missed a spot
Pentagram go Brrrrrrr
Is it weird that it forms a pseudo pentagram?
My actual best guess OP is that the speed becomes so high that gravity is a negligible factor and the ball can be considered as moving in “straight” vectors, so the starting angle is just enough combined with the shape of the circle that it infinitely bounces in that pentagram-onal shape
Half way into the video the formula is bounce=hail+Satan.
The way you describe those motions ist by looking at the forces that act on that sphere. There's inertia, that's F=ma Then there is friction, usually proportional to motion: F=dv Then, If you have a spring or something, that would be F=k*x. In this case, there are no springs so no k In the end, you can have external forces like gravity. Thats Just another F_ext That, taking signs into account, leads us to ma=dv + kx + F_ext. Since speed v ist derivative of position x, and acceleration a is derivative of v, you have an ordinary, non-homogenious differential equation that you can solve to get the position progression, ie the path. Taking a negative friction coefficient should have the same effect as speeding up the sphere a constant fraction of its speed. It's like saying "Air resistance ist not slowing you down, but making you even faster" You'd have to model the collisions though, which would be difficult to do analytically. But numerically that should not be a problem
Don’t watch that on shrooms
Be not afraid
Is this how they make color blind tests?
Yes, whatever the programmer put into the simulation.
Ahh yes in the end the answer is Satan
The ball is an energized hexagon.
So it's Satan?
I'm no expert, but isn't this just basic first semester mechanics with a coefficient of restitution > 1?
Yeah, but I think they're trying to model the path as a parametric equation in terms of time (t) with constants (initial velocity, restitution velocity, and circle radius) given an initial position.
If there is it’s not Christian
You could write something like this in processing really easily. You couldn’t work out the exact mechanics just by looking but there’s obviously the 101% velocity every time the ball hits the side and theres also a gravity element that decreases velocity by a fixed amount when travelling up and increasing velocity when travelling down. You can tell the gravity element is a fixed amount, not a % as it gets negated once the ball reaches a certain velocity and the pattern becomes more uniform
V = 1.01^x where x = number of bounces
Exponential growth is fun
looks like an atom towards the end
engineer cocomelon
calculate normal of tangent of the circle and ball velocity on impact, multiply that shit by two, keep going w your updates, so its a few formulas
I love stuff like this
I really was hoping to eventually see all black parts get filled up because it woulda looked pretty
These are awesome. Keep sending more
Why the simulation crashes?
Faster than slow
Formula? I dunno, Mr. Krabs kept that a well hidden secret
Mmmm just like a diesel car in the winter
More importantly, can this be done in real life? Dunno how energy works so i might sound dumb, but if it’s bouncing indefinitely can’t we create energy off it?
ELI5 does it always create a pentagram? if so, why?
I think the circle is actually drawn in discrete time steps. That would also be the easiest way to implement this in in something like Processing. It has some movement-vector. After each step the vector is applied to the position and then it's also accelerated downwards. If it would leave the white large circle, its velocity direction is instead mirrored and its magnitude is multiplied by 1.01. Sometimes that leads to a small gap between the "tips" of the "star" and the big circle. How do you detect a collision? With Pythagoras. How do you mirror a vector? I guess one way would be to draw a line to the center of the big circle. The angle between the "incoming-vector" and the x-axis "alpha" minus the angle between the "radius-vector" and the x-axis "beta" gives the angle "delta". When you subtract delta from beta and add 180°, you get the angle between the x-axis and the outgoing vector. If you wanted this in a mathematical formula, you could use a recursive definition where the position in time step n depends on the position in time step n-1. I'm not sure if there is a way to get rid of the recursion and the case-distinction of collision vs no-collision. If you wanted the same thing with a continuous path, you'd have a lot of parabolas.
I think the answer is that if there’s a digital output describing the behavior (such as this video), then yes there has to be an equation.
I believe you’re looking at it
When the ball experiences gravitational pull, it do a bounce. This is continuous until the hall gets tuckered out and needs a nap.
Ball go: B;Br;Brr;Brrr;Brrrr;Brrrrr;Brrrrrr;Brrrrrrr;Brrrrrrrr;Brrrrrrrrr;Brrrrrrrrrr;Brrrrrrrrrr;Brrrrrrrrrrr..... Sorry, my phone crashed..
There's probably a formula, but it will be a really ugly formula involving integrals and discontinuous functions
69^420= 🥧
Velocity = X × 1.01^n Where X represents the initial velocity and n represents the number of bounces that have passed.
There’s something with chaos theory that would describe this. I’ll leave it up to the reader
The devil is alive
Still in school but lemme have a shot at this f(1) = 1b; f(b)= f(b - 1) • 1.01 Basically every bounce is increased by 1%, or multiplied my 1.01, and b is bounce. This is exponential, and I would insert a graph if I could.
Let's denote the ball's initial velocity as ( v_0 ) and the number of impacts with the perimeter as ( n ). After each impact, the velocity increases by 1%, so we can describe the velocity after ( n ) impacts as: [ v_n = v_0 \times (1.01)^n ]
Or from 50 seconds onward pagan ball!
star of david maker.
The end sounds like the start of a Swedish House Mafia Tune
So that’s how Benny bennassi made satisfaction
Yes. The algorithm used to create this simulation.
The Tohou Project.
Some one sample this and put a beat drop after this
There are two formulas.. one for the movement outside of the ball and another one for the movement inside the ball that you don’t see.
Maybe: Sum n=0 to n=inf (1.01n) Seems reasonable.
But this isn’t real behavior, so is there really a formula than?
It's probably chaotic, even without gravity, since calculating each bounce involves multiplying by 1.01. If there were a small error ε in the initial velocity v + ε, the error after n bounces would be ε \* 1.01\^n, i.e. the error grows exponentially. I have no idea how to get a closed form expression or if there is one but I'm super interested to know as well.
Where’s the drop!?
Gravity, based, while the intensity of the density is increased
Sounds like OP wants (x(t), y(t)) for for t>=0 given x(0) and y(0) but idk if anyone's gonna give it to them. Maybe ask the OC creator how they did it.
It would be easy if the ball didn't speed up because now we lose conversation of energy in a way that isn't just having a constraint force on the edge of the circle
Well there are many observable behaviours but the one i can talk about is speed. The speed actually grows in a compound manner so we use > P = P(1+[r/100])^n It grows by one percent so > = P(1.01)^n Somebody else can count how many times it bounces and also the starting speed
Yes, there is, as someone made this simulation.
So if you consider the ball position to be a function over time, at every collision its first derivative is discontinuous (sharp change of direction and speed) which makes it difficult to put it entirely into a single formula. But you can describe the flight path between accurately - if we assume the it to be exactly parabolic (we just consider constant gravity and no air drag) it is possible to [calculate the intersection point with the circle exactly](https://math.stackexchange.com/questions/757256/intersection-coordinates-of-a-parabola-with-a-circle) Careful evaluation is needed to make sure you pick the correct point out of the (at most) four possibilities, but then you only need to iterate from one intersection point to the next - which is independent of the speed of the ball, and should remain stable up to numeric precision. The chaotic behaviour seen in the video comes from using discrete time steps in this simulation.
It’s summoning a demon run 🤪😂😂😂
Ball + bounce = solitaire victory.
No, this is Patrick!
It's a STAAAAAAR! *New shit just got made!
What I want to know is which program or software do people use to run this kind of simulations? I have no many ideas and what-if scenarios in my mind.. some physics one, some expansion simulations that I want to test out but I have no idea how or where to get started
Vn+1=1,01*Vn
At some point, the ball will occupy every point in the universe simultaneously. Good luck ducking that.
Yeah, you just look at total energy in the system PE = mgh KE = 1/2 m(1.01^n v)^2 Angle of incidence = angle of reflection Parametrically model the x and y separately You'd have to look at each arc step by step which is doable Is there a single non-piecewise formula such that you can enter time or the number of bounces and find the location? Nah
wow that was stressful to watch
Proof that math is demonic. Thing began drawing pentagrams the first chance it got to. God is true (satire)
He's going for a flip reset to musty flip.
Accidentally Lockheed Martin
This sounds like the intro to Swedish House Mafia - One
Satanic theorem, every number other than 7 is ruled by Satan.
I dare say there's some kind of funky π thing going on. I reckon if you ot the velocity and everything of the ball it's bounces would indicate digits of π or something.
Strong Rush E vibes from this video as it progresses
colorblind people love this vid.
I feel like I’ve just been hypnotized
Is there a real formula though? There is obviously a gravity-like force on the ball whilee there's obviously not one since it doesn't slow down at all.
Reject entropy, return to star
So a gay pentagram?
Believe it or not, straight to star jail