61–69 of 69 posts
... but if you hit it hard enough what's the maximum number that will go in?
Assuming there's no friction on the table, the chances ought to be pretty high that they'll all eventually go in.
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I think he is confused. Wolfram's site only has Mathematica 9 listed
> I think he is confused. You don't mean that the GP (taliesinb) is confused, do you?. He works for Wolfram Research, developing Mathematica.
Somewhat meta, but at what point does Stack Overflow transition from being a useful, time-saving tool and become a waste of everyone's time? The top answer author clearly put a ton of time on the answer. We are all now reading this. Is anyone able to apply any of the knowledge they gained from this Q&A to anything productive?
This is a good question, and I sometimes do put too much time into answering questions on Math Stack exchange. Right now, though, I'm fairly happy with the time I put into this simulation. First of all, there's the fact that I did this on a Friday evening, after I came home from work. (I spent about six hours on it, and maybe another hour or two on Saturday morning). There are probably more productive uses of my time…
... but if you hit it hard enough what's the maximum number that will go in?
Assuming there's no friction on the table, the chances ought to be pretty high that they'll all eventually go in.
If the ball strikes the rail on an angle, the rails twist to absorb the impact and will recoil with a force that isn't just a simple reflection, but that also has a force vector tangent to the ball and back towards the origin. This directly affects the spin of the ball when it leaves the rail.
Combined, a bounce off the rail is a very complicated collision. The net result of which has balls that strike the rail at lower velocities closely matching a reflection, but if struck with a higher velocity, the reflection straightens out into a bounce more perpendicular to the rail.
In short, the rails are highly dependent upon the velocities at impact, and perhaps more so than at break, are affected by rotational momentum.
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There's a Mathematica 10 now, already? Interesting. I'm still a happy 8 user, myself. Is there much reason to upgrade? It'd probably be good for them to pay more attention to backwards compatibility...
10 has Association expressions [1], which are a huge language addition (and long overdue). Seems there is some synchrony with Erlang on that front! The features I'm most excited about in 10 (disclosure: my team is working on some of these): [2] String, XML, file templating [3] Hands-free machine learning [4] Entity and EntityValue: semantic "hooks" to represent real world entities [5] Dataset, a general data modellin…
Any chance you'll be including the templating features in the RPi version?
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> I think he is confused. You don't mean that the GP (taliesinb) is confused, do you?. He works for Wolfram Research, developing Mathematica.
That is what I meant, but I guess I look like the fool now. I still don't see that Mathematica 10 has been released, but I guess it is close.
Earlier quoted context omitted.
10 has Association expressions [1], which are a huge language addition (and long overdue). Seems there is some synchrony with Erlang on that front! The features I'm most excited about in 10 (disclosure: my team is working on some of these): [2] String, XML, file templating [3] Hands-free machine learning [4] Entity and EntityValue: semantic "hooks" to represent real world entities [5] Dataset, a general data modellin…
Nice to see all the WolframAlpha features being brought over. I'd completely missed Entity and Dataset. Any chance you'll be including the templating features in the RPi version?
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Not necessarily; it might be possible to have some balls collide in such a way that at least one comes to a stop. At least one ball will keep moving though, and it isn't clear to me whether it is possible to prevent that moving ball from, eventually, hitting that stopped ball again. I think that would require at least three balls (ignoring pockets)
This is a problem of energy conservation: the cue ball starts with E = 1/2 mv^2 kinetic energy and, no matter what, that same amount of energy has to exist afterwards in some form. But where will it go? In an elastic collision, a negligible amount of energy gets converted to heat (through inelastic deformation), and only a small amount will be converted to sound. The only other place the energy can go is kinetic ener…