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Posted: 11/14/2009 3:04:45 AM EDT
[#1]
14"
I think that's wrong... |
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Posted: 11/14/2009 3:07:33 AM EDT
[#2]
Quoted:
14" I think that's wrong... I am geometry dumb, But I know it would be more than 14" |
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Posted: 11/14/2009 3:17:31 AM EDT
[#3]
2.414 x 7 = 16.898
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Posted: 11/14/2009 3:27:53 AM EDT
[#4]
 7/2" + 7/2" = 7" Sqrt(72+72) = 9.8995" ID = 16.8995" |
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Posted: 11/14/2009 3:51:56 AM EDT
[#5]
Quoted: 2.414 x 7 = 16.898 I would have beaten you to it if I hadn't been sidetracked thinking about using a larger number of circles than four. But then, I still win because I have a sucky MSPaint included in my explanation.  ETA: And because your answer is ever so slightly too small.  |
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Posted: 11/14/2009 4:52:37 AM EDT
[#6]
14
WTF crazy ass math have we got running around here... or have I failed the reading comprehension part? |
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Posted: 11/14/2009 5:02:27 AM EDT
[#7]
Quoted: 14 WTF crazy ass math have we got running around here... or have I failed the reading comprehension part? reading comprehension, you fail at it. he wants to fit 4 pipes stacked squarely, into a large round pipe, and what diameter would he need, I guessed 18 off the top of my head but i overshot by a inch or so |
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Posted: 11/14/2009 5:06:11 AM EDT
[#8]
Quoted:
Quoted:
14 WTF crazy ass math have we got running around here... or have I failed the reading comprehension part? reading comprehension, you fail at it. he wants to fit 4 pipes stacked squarely, into a large round pipe, and what diameter would he need, I guessed 18 off the top of my head but i overshot by a inch or so
Roger fucking That. Forgot about the gap in the middle... didn't pay attention to the msPaint post.
da-da-damn, I'm such a rock, sometimes most times. |
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Posted: 11/14/2009 5:09:00 AM EDT
[#9]
Quoted:
14 WTF crazy ass math have we got running around here... or have I failed the reading comprehension part? You failed the comprehension part. 4 equal sized pipes touching each other in a square config. will not touch at the center of the cluster they create. Thus the circle that inscribes the square arrangment must be larger than double the diameters of a pipe. Actually 4 pipes of the same size will never be inscribed by a circle equal to twice one pipes diameter. If you compress the square config. so two pipes are touching at the center of the cluster then you have pushed the other corners away and have increased the size of the circle that will contain them all. |
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Posted: 11/14/2009 9:04:36 AM EDT
[#10]
Thanks guys.
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Posted: 11/14/2009 9:20:10 AM EDT
[#11]
About 16.94 in
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Posted: 11/14/2009 9:27:21 AM EDT
[#12]
Nevermind.
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Posted: 11/14/2009 9:29:53 AM EDT
[#13]
Quoted:
http://media.ar15.com/media/viewFile.html?i=14230 7/2" + 7/2" = 7" Sqrt(72+72) = 9.8995" ID = 16.8995" You were right except you left off 1 radius in the upper right, so you need another 7" for a total of 23.98". |
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Posted: 11/14/2009 9:31:49 AM EDT
[#14]
Quoted:
BS If the inside pipes are 7" OD, the ID of the large pipe will have to be 14". RIF? ar-jedi |
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Posted: 11/14/2009 9:32:46 AM EDT
[#15]
Nevermind... you'd get two in there, but not the remaining two. Early morning brain glitch.
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Posted: 11/14/2009 9:33:23 AM EDT
[#16]
see http://www2.stetson.edu/~efriedma/cirincir/
http://hydra.nat.uni-magdeburg.de/packing/cci/cci4.html 
the answer is the ratio above (2.414) x the diameter of one of the inner circles, so 2.4142 x 7 = 16.898" ar-jedi ps: there is an entire branch of math related to this problem, and lots of websites to learn from: e.g. 
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Posted: 11/14/2009 9:34:09 AM EDT
[#17]
The next step is to identify a standard pipe size which is "just larger" than the 16.9 inch calculated figure.
Edit to Add: Sched 40 Pipe 18" Nominal has a 16.874 ID Sched 40 Pipe 20" Nominal has a 18.814 ID Sched 80 Pipe 18" Nominal has a 16.126 ID Sched 80 Pipe 20" Nominal has a 17.938 ID |
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Posted: 11/14/2009 9:39:00 AM EDT
[#18]
Quoted: 7" is not the radius. It is 3.5". Quoted: http://media.ar15.com/media/viewFile.html?i=14230 7/2" + 7/2" = 7" Sqrt(72+72) = 9.8995" ID = 16.8995" You were right except you left off 1 radius in the upper right, so you need another 7" for a total of 23.98". Radius + Radius + Hypotenuse = ID max is correct. |
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Posted: 11/14/2009 9:44:08 AM EDT
[#19]
Quoted:
Quoted:
7" is not the radius. It is 3.5".
Quoted:
http://media.ar15.com/media/viewFile.html?i=14230 7/2" + 7/2" = 7" Sqrt(72+72) = 9.8995" ID = 16.8995" You were right except you left off 1 radius in the upper right, so you need another 7" for a total of 23.98". Radius + Radius + Hypotenuse = ID max is correct. You and max are right. I thought he was adding a single radius when it was really 2 radii. |
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Posted: 11/14/2009 9:46:13 AM EDT
[#20]
Quoted: Check. Sorry if I came off a little cranky; I just woke up, I saw this, and thought 'school system'...Quoted: You and max are right. I thought he was adding a single radius when it was really 2 radii.Quoted: 7" is not the radius. It is 3.5". Quoted: You were right except you left off 1 radius in the upper right, so you need another 7" for a total of 23.98".http://media.ar15.com/media/viewFile.html?i=14230 7/2" + 7/2" = 7" Sqrt(72+72) = 9.8995" ID = 16.8995" Radius + Radius + Hypotenuse = ID max is correct.  |
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Posted: 11/14/2009 12:52:29 PM EDT
[#21]
Quoted: see http://www2.stetson.edu/~efriedma/cirincir/ http://www2.stetson.edu/~efriedma/cirincir/ccc4.gif http://hydra.nat.uni-magdeburg.de/packing/cci/cci4.html http://hydra.nat.uni-magdeburg.de/packing/cci/pic/cci4.gif the answer is the ratio above (2.414) x the diameter of one of the inner circles, so 2.4142 x 7 = 16.898" ar-jedi ps: there is an entire branch of math related to this problem, and lots of websites to learn from: e.g. http://hydra.nat.uni-magdeburg.de/packing/cci/pic/d1.gif Cool pics!  |
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