calculate number of round cylinders that can be made of one big cylinder?

Question

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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Answer 1

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.

Answer 2

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: $\pi r^2h=\pi\cdot (0.175)^2\cdot 10$ ($m^3$)