Betcha Can't answer this one!!!

[xXCloudXx]

A sphere 5.5 metres in diameter is filled with 1m diameter hemi-spheres.

a: (1) What is the theoretical maximum amount of hemi-spheres that can be crammed into the big sphere given that the following condition is met:
Each hemi-sphere's flat side (which I'll now refer to as its 'disc') has a central point. The point must not 'see' another hemisphere's disc. By definition, when I say 'see', the simplest thing to imagine is a straight ultra-thin 'laser light' coming from the disc. This 'light' must not reach another disc. However, if there's another hemi-sphere that's 'blocking' the 'line of sight', then this is accepted.

(2) By cramming them as efficiently as possible, a relatively small volume will be left. How large is this volume?

b: Same question, except the torch now has a 52.72077938642 (that's 90/(1+(0.5^0.5))) 'degree of sight'. This 'cone' of light extends from the exact centre of the disc and is once again not allowed to 'see' any part of another disc. What is the maximum amount of hemisphere's that can be crammed into the mega-sphere now?

c: Same question again, except the whole disc acts as a tubular beacon of light. This thick ray must not 'see' another hemisphere's disc. What is the maximum amount of hemisphere's that can be crammed into the mega-sphere now?

I'll give you props if you even know where to start on getting your answer!!!

[Victoria]

Nice try. We're not doing your homework for you. Now, what's this thread about anyway?

[xXCloudXx]

It's not homework, I'm just trying to make conversation. haha. I don't think I've done homework in 4 years now. lol. I just want to see how smart TFF members are. I couldnt even answer this question when someone asked me. I got about halfway and gave up. I know the answer now. I just want to see who can actually answer it all the way without help.

[Victoria]

That is not making conversation. All you did was post that and said "let's see who can answer it".

...There's no topic of discussion or anything. And besides, we're not obligated to even do that for you. To be honest, I don't even know what all that stuff even means.

I've only taken up to geometry. All of that sounds like trig or some shit.

I mean really, why does it even matter how smart we are? This is a forum for video games, not for math equations. x_X

[xXCloudXx]

I'm not saying it is an obligation for anyone to answer the question. I just want to see (if someone is willing) if they can answer it.

And no it doesnt matter how smart anyone on the forum is. I'm not judging anyone by there intelligence by any means. I just wanted to see if the question could be answered. I know that there are some really smart people on this forum that might understand what I am asking. I just want to see who will answer. Nothing more or nothing less.

[Rocky]

I'm not saying it is an obligation for anyone to answer the question. I just want to see (if someone is willing) if they can answer it.

And no it doesnt matter how smart anyone on the forum is. I'm not judging anyone by there intelligence by any means. I just wanted to see if the question could be answered. I know that there are some really smart people on this forum that might understand what I am asking. I just want to see who will answer. Nothing more or nothing less.

See, the thing is I could answer your question with my knowledge of Trig/Calc experiences, but I'd rather not because if I did your homework for you then there would be no way for you to do well on your test. Now get off the computer and hit the books.

This is a forum for video games, not for math equations. x_X

Got 'em!

[Che]

The answer is you're gay.

[Polk]

24. The answer is always 24.

24 is the highest number there is.

[Pete]

But Polk, my fingers only go up to 10, and with my toes it's 20. I could get to 21, but I'd risk indecent exposure.

I was an English major, so I totally threw out any and all knowledge of math. At the same time, this isn't a conversation, but really a contest of who can give the best response. Che is winning.

[RagnaToad]

The answer is you're gay.

Quoted for truth.

A possible answer would also be:

WTF DUDE?

[Unknown Entity]

WTF guys? Seriously, was there a need for half of that?

Have you got a clue as to how I could try and solve it? I might understand better if I had some sort of a method, because right now... I'm having a hard time working it out.

[Pete]

Ok I'm gonna start with what I know.

Volume of a 5.5m sphere

4/3 x pi x radius(cubed)

4/3 x pi x 2.75 (3)

4/3 x pi x 20.796875

4x 20.796875 /3 = 27.729166666666666666666666666667

27.729166666666666666666666666667 x pi= 87.114

Volume of a 5.5m sphere is 87.114m(cubed)

vol of 1m sphere is

4/3 x pi x .5 (3)

4/3 x pi x .125

4x .125 / 3= 0.16666666666666666666666666666667

16666666666666666666666666666667 x pi= .524

Vol of a 1m sphere .524m (cubed)

But since you're halving it, it would only be .262m (cubed)


My guess from there would be to divide 87.114/ .262

Which is 332.496, which is ultimately 332. I don't think that can possibly be right due to the previous stipulation, but that's all I got at 1am.

[Shan'do Spike]

Nah, the trick is going to be to find a packing scheme of those spheres so you can do this optimally; the problem with the pure volume answer is twofold: it assumes perfect packing (which won't work filling spheres with hemispheres), and it ignores the condition where no flat end of a hemisphere can see another..

I'm kind of tired and don't feel like really thinking about this: this is stuff you don't actually touch when you're a math major, because it has nothing to do with mathematics. If I wanted to look at this, I'd have studied physics . But two minutes of thought would suggest that the general form of the problem goes as follows:

Take a shell of thickness .5 m from the outside of a sphere: pack as many hemispheres as possible into this shell such that the discs face outward (so that none of the flat ends "see" each other). I'd guess that the way to do this is to start at the middle of the sphereshell and circle it with hemispheres, then move more outwards and repeat.

Once you've filled a shell, discard it (add the amount to the total) and repeat the process. Each time you repeat this, .5 m will be removed from each side, so the radius of the sphere you're left with will fall by .5 m. So, for 5.5 m, you'd start with 5.5, then 5, then 4.5, 4, 3.5, 3, 2.5, 2, 1.5, 1, .5...

I'd assume b and c are the same as a (actually, I think I just described a solution to part c), just with a different packing scheme in each shell so the values differ. I really don't care enough to guess at how.

[RagnaToad]

i could get it at some point in time... i love math... ill try tomorrow when im not tired

And yet your posting a meaningless post. Be proud of your post count.


And shit, Pete. You're actually making a decent effort.

Personally I had enough math in my life. I studied Latin & Mathematics in High School, and got pretty sick of it in the sixth grade, which was only 3 years ago.

[Josh_R]

ahh mega sphere the revenge a very tough question my friend...I have seen hundreds of sites with people who ask this question and I have never seen a answer it will be interesting to see who gets it right...