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Optimal can

25.4.2
Optimal dose
Fig. 1
Optimal container with base and lid
d* Optimum diameter h* Optimum height A* Optimum area
Eqn. 1
\require{color}\definecolor{myred}{RGB}{255,0,0} d^\ast=\sqrt[3]{\frac{4\cdot Vol}{\mathrm\pi}}
Eqn. 2
\require{color}\definecolor{myred}{RGB}{255,0,0} h=\frac{4\cdot V}{d^{\color{myred}2}\cdot\pi}
Eqn. 3
\require{color}\definecolor{myred}{RGB}{255,0,0} A=d^{\color{myred}2}\cdot\frac\pi2+d\cdot\pi\cdot h
Eqn. 4
\require{color}\definecolor{myred}{RGB}{255,0,0} q=\frac{A^\ast}{A_{\color{myred}ref}}-1
Optimal diameterd*=131.84mm 
Optimal heighth*=65.92mm 
Optimal areaA*=40,957.4mm2 
Heighthref=114.59mm 
SurfaceAref=43,854mm2 
Percent lossq=-6.6% 
VolumeVi = 0.9L
Diameterdref = 100mm
Calc 1

Example 1 An optimal can with a beverage content of 0.9 l has an optimal diameter of 131 mm and a height of 66 mm. If the can diameter of a reference can is 100 mm instead, this results in a height of 115 mm with the same volume. The area loss of the reference can is -6.6%.

Example 2 A beer barrel with a capacity of 50 l has an optimal diameter of 503 mm and a height of 252 mm. If the reference diameter is 360 mm instead, this results in a height of 491 mm with the same volume. The barrel area loss is then -9.3%.

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