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October 13th, 2016, 02:37 PM  #1 
Senior Member Joined: Jul 2016 From: USA Posts: 108 Thanks: 13  100% Perfect Bounce Equation
After so much work I have finally developed the perfect bounce equation! I don't know how many of you have tried this, but it is quite difficult getting all the physics and timing perfect. You can look at the graph here at desmos: https://www.desmos.com/calculator/qjg6ipjb3y There are infinite bounces that happen in two seconds. It behaves exactly like Zeno's paradox. 1 + 1/2 + 1/4 + ... Also I would like to know if anyone else is familiar with a bounce equation that is more simple. If so that would be awesome! Any way, hope you enjoy the graph! Also had to use the floor() function, which is not standard I suppose in regular mathematics..but I suppose it is just an operation like other.. Is there anyway to express floor(x) with traditional arithmetic. abs(x) = +sqrt(x^2) for example, but what is the floor (x) = ? Thanks for any answers! 
October 14th, 2016, 01:39 AM  #2 
Senior Member Joined: Apr 2014 From: Glasgow Posts: 2,157 Thanks: 731 Math Focus: Physics, mathematical modelling, numerical and computational solutions 
It reminded me of the Airy disk stuff https://en.wikipedia.org/wiki/Airy_disk and the functions $\displaystyle y = sinc \,\, x$ or $\displaystyle y = \left(\frac{\sin x}{x}\right)^2$ If you're dealing with a Newtonian physics problem with a bouncing ball falling under gravity (solution is height as a function of time), then the solution would be a set of successive parabolic curves with a magnitude that reduces according to the amount of energy lost with each collision, not that much unlike what you've plotted. 
October 16th, 2016, 09:55 PM  #3 
Senior Member Joined: Jul 2016 From: USA Posts: 108 Thanks: 13 
Thanks. I realized that this bounce equation is not so perfect after all. Each parabola has the correct height and width, but still not perfect since the beginning of each parabola has the same slope. The beginning of each parabola's slope should be a fraction of the previous parabola's, since a fraction of the velocity is lost. So I really need to work on that because it's a huge flaw that I didn't notice and it really throws off the bounce animation. Also when you said y = (sinx /x)^2, did you mean " y = sqrt((sinx /x)^2) "? This is similar ish to a bounce equation but like you said it needs to be successive parabolic curves. Also when you divide the bounces by x, there is a tremendous flaw, since x increases throughout a single bounce when it should only decrease suddenly at the end of each bounce. y = cos(x) almost looks like a parabolas, but it creates a different shape. I wish it did create parabolas! Then the cosine function would be so simple! lol And the time taken for each bounce should be a fraction of the previous bounce. Which makes this equation so hard to perfectly construct. I'm sure you already know all these things but I'm just throwing it out for what it's worth. I was able to create this equation so perfect, except for one single mistake. The slope at the beginning of every successive parabola should divide by two. I will definitely be thinking of a way to solve this problem. Also, have you had any luck thinking of a function that accomplishes the same effect as floor(x).. eg floor(1.5) = 1, floor(2.7) = 2, floor(.2) = 0 
October 16th, 2016, 10:44 PM  #4 
Senior Member Joined: Jul 2016 From: USA Posts: 108 Thanks: 13 
THIS IS MORE LIKE IT! Fixed the mistake! Unless we find anymore mistakes, this is the perfect bounce equation: https://www.desmos.com/calculator/hostvw147c 
February 18th, 2019, 06:49 AM  #5 
Newbie Joined: Feb 2019 From: US Posts: 1 Thanks: 0  Bouncing Ball
I know it has been over two years since you posted your bouncing ball Desmos file, but I did not see it until today. I also created a file, though my equations are considerably more complex (and in English unitssorry!). Here it is for whatever it is worth... https://www.desmos.com/calculator/cjdjsqwngy 

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