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Disclaimer warning : i am not a math genius.

Im trying to calculate how many smaller cylinders i can cut out from a big cylinder, and i was abit fast in my calculation :D

I have the following :

  • I have a big massive cylinder that is 30 meters in diameter * 100 meters long
  • for one smaller cylinder i need 35 centimeter diameter, and 10 meters lenght.

The question is - how many smaller cylinder can a produce from the big cylinder ? anyone can help me how to calculate this ?

EDIT:

1) the smaller cylinders are to be done parallel so i assume i take it upright which in this case would equal 10 * 10 meters blocks of how-many-35 cm wide cylinders-possible in upright position within 30m diameter.

2) i dont expect any loss on cutting the 35cm*10meters smaller cylinders which probably would be the case in real life, so they can be perfectly aligned with no space in between.

3) the smaller cylinders are solid

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MdTp
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  • This is an interesting question, that needs lots of clarification. Please [edit] to help us help you. (1) I assume the long axes of the big cylinder and the small ones should be parallel. If so you need to pack as many $35$ cm diameter circles into a $30$ m diameter circle for the bases, and cut the big cylinder into $10$ parts. (2) For the packing, do you need an exact answer, or is an approximation enough? Is there any loss between small cylinders, or can they be exactly tangent? – Ethan Bolker Oct 14 '19 at 14:12
  • Are you considering pipes with teh same thickness or solid cylinders? – user Oct 14 '19 at 14:17
  • @EthanBolker ive added some answers i hope to your questions – MdTp Oct 14 '19 at 14:24
  • I thinking like if you had the center of the kitchen-paper roll cylinder - and then you filled it up with matches ( small cylinders ) until you couldnt pack one more down - that was the first 10 meters, then the next lot would come on top for the next 10 meters and so forth. – MdTp Oct 14 '19 at 14:32

2 Answers2

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HINT

For an equivalence in term of surface (same thicknes), we have that

  • surface for big cylinder: $2\pi RH=2\pi\cdot 15\cdot 100$ (sqm)

  • surface for one small cylinder: $2\pi rh=2\pi\cdot 0.175\cdot 10$ (sqm)

For an equivalence in term of volume, we have that

  • volume for big cylinder: $\pi R^2H=\pi\cdot 15^2\cdot 100$ ($m^3$)

  • volume for one small cylinder: $\pi r^2h=\pi\cdot (0.175)^2\cdot 10$ ($m^3$)

user
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  • Why surface area? Are you assuming a big sheet of material rolled up into one big cylinder or cut into pieces rolled into small cylinders? I doubt that's what the OP wants. – Ethan Bolker Oct 14 '19 at 14:15
  • @EthanBolker Yes I'm assuming a big sheet as a pipe. But you doubt is correct. I ask for a clarification. Thanks – user Oct 14 '19 at 14:16
  • can consider it as a massive tree log - and then how many round smaller sticks would i be able to 'produce' from that tree log. – MdTp Oct 14 '19 at 14:39
  • @MdTp Then you can refer to the equivalence in volume. It is useful in the ideal case that all the volume of the big cylinder can be converted in volume of small cylinders. – user Oct 14 '19 at 14:44
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First cut the big cylinder into $10$ cylinders each $10$ long. Then lay out as many circles as you can on the base of each.

Here's the result if the diameter of the big cylinder is $15$ meters. The website won't do $20$ m, let alone $30$.

http://hydra.nat.uni-magdeburg.de/cgi-bin/cci1.pl?size=15&diameter=0.35&name=Ethan+Bolker&addr=ebolker%40gmail.com

 Input parameters
Diameter of large disk:   15
Diameter of small circles:   0.35
Output parameters
Number of circles that will fit:   1608
Waste:   12.45%

Then there's a picture and the coordinates of all the small circles.

Some experimenting suggests that the waste will be about $12\%$ for your $30$ m problem. That will give you an approximation for the number of circles you can pack.

0.88 * (30)^2/(0.35)^2 = 6465.30612245

so about $10 \times 6500 = 65,000$ small cylinders from your one big one.

In the limit the hexagonal circle packing is $91\%$ efficient, with $9\%$ waste. I don't know how close to the limit this example is.

Ethan Bolker
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  • that is cool - is the program behind that calculator opensource ? would like to extend it to be able to calculate with bigger disk diameters – MdTp Oct 14 '19 at 14:59
  • @MdTp I've no idea whether it's open source. I found the site with a google search. You can try to contact the author. – Ethan Bolker Oct 14 '19 at 15:01
  • this is super nice thanks - do you know how i calculate the waste value ? – MdTp Oct 14 '19 at 15:33
  • For any particular number of circles you have to find the best packing to find the waste value. When the ratio of the big to the small circles is large you have hexagonal close packing near the center and waste near the edges so the limit is the global waste percentage for the infinite close packing. – Ethan Bolker Oct 14 '19 at 20:17