The optimal volume/surface ratio for a mathematical cylinder is obtained from H=D.
Pi/4*D³ = 330 Answer D = 7.49 cm.
This simple equation is usually given to schoolchildren who are starting to learn the bases of differential calculus. Normally the mathematical task is to calculate the diameter/height ratio for a cylinder with minimum surface area for a filling volume of 1 litre.
Volume = Pi * r² * H = 1
Surface = 2 * Pi * r² + 2 * Pi r * H
Optimal surface area for d (surface) / d (r) = 0
Substitute H = 1 / (Pi * r²)
d (2 * Pi r² * + 2 / r) d (r) = 0 4 * Pi * r - 2 / r² = = r³ = 1 / (2 * Pi)
Answer:
r = (1 / (2 * Pi) ) ^(1/3) = 0.54192607
H = 1 / (Pi * r²) = 1.08385214 = 2*r = D
The sample calculation is only suitable to construct the basic contours and does not take account of thickness fluctuations nor seam joints. The situation for a two-piece can is typically different.
Note: A two-piece beverage can comprises:
- a deep-drawn and wall-ironed body
- a seamed tear-off end
Our aim is to make a product which is as light as possible and which nevertheless meets the requirements of our customers.
An important design criterion is: minimal material consumption.
The wall-ironing factor used for our can is approx. 3.
The basic volume which we use includes a headspace of approx. 20 cm³.
The total volume is therefore 330 + 20 = 350 cm³.
The material volume for a cylinder open at one end is approx.:
Pi * r² * t + 2 * Pi * r * (t/3 * H).
In this case the optimal diameter would be approx. 6.7 cm and the height 9.93 cm.
The actual diameter is 66 mm and the actual height 115.2 mm
Note:
We also take into account:
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