Dog--
The Freeman
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Ok, I don't know the answer to this question, it's not for homework, not for a test, I don't care about this answer, I found a really hard math question on the internet, just to see some of you argue and fight. It's also a good way for some of you to show off, eh?, eh? (I seriously enjoy you guys fighting)
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 (indicated by the white point shown in the hemisphere diagram to the right). 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?
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 (indicated by the white point shown in the hemisphere diagram to the right). 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?
