Physics:
Curved (convex) walls
Consider a flat floppy paper. It has hardly any structural strength due to negligible moment of inertia around its bending axis. However, once the paper is rolled into a tube, curved surface is generated, and the rolled paper would now exert stiffness due to increased moment of inertia around the bending axis. It exemplifies a fundamental engineering principle that curved rather than flat walls are stronger due to high cross-sectional moment of inertia. This concept would apply to all sorts of containers including cans. The sides of a can are not flat, but round i.e. convex (bulging outward) for improved strength.
However, strength cannot be a prime issue, otherwise the manufacturers could go for stronger material such as steel instead of normally used aluminium. The reason for using aluminium alloy as the can material is because of its high strength/weight ratio, hence less weight and ultimately good money saving.
Concave bottom
The bottom of a pressurized beverage can is concave (hollowed inward) which helps to withstand a normal internal pressure of ~90 psi or 6.2 bar of the beverage. If flat, the internal pressure would tend to cause the bottom to bulge outward. In this deformed state, the cans would not be able stand upright. This is important because it would be impossible for distributors and grocers to stack a can upright if its bottom bulges out. Aluminium alloy with good mechanical strength is used for the base and body of cans. A typical alloy contains weight 1% Mg, 1%Mn,0.4%iron, 0.2% silicon and 0.15% copper with the rest in aluminium. Beverage can
Incidentally, it is interesting to note that the bottom of a propane tank is concave which provides in-built safety. Any accidental build up of internal pressure leading to catastrophic failure can be avoided if the tank expands i.e.pops from concave to convex.
Flat top (lid)
Top of a beverage can has to be flat for convenience to get mouth around. The flat top similar to the concept of strength at the end of a fire hose holding back water, needs to be quite strong. This is achieved by reducing the amount of manganese (Mn), and increasing the magnesium (Mg) content of the aluminium alloy. Thus for the lid, Mg content is adjusted up to 2% while Mn is reduced to a trace level. This change in alloy content and use of thicker plate makes the lid stronger as well as heavier. The flat lid is also considerably thicker than the walls. To partially counterbalance the added weight, diameter of the lid is made less than the rest of the can body, producing a characteristic taper shape as is normal in cans today.
Optimising the shape of a can
It is established so far that a concave bottom with curved wall surface would be ideal shape of a can made in aluminum alloy. Mathematically speaking, an optimum shape can be achieved in the form of a sphere which has the smallest surface area. This shape would require least amount of material and hence lowest cost. But physical reality forbids use of any such shape for cans, because it is hard to distribute and stack up spherical shaped cans, and above all it is not easy to hold them to drink from.
A cut of 3-D (three dimensional) sphere into a 2-D slice will produce a circle. Such circle will be the best container shape in 2-D which in 3-D shapes up will provide the form of a cylinder. Can is thus nothing but a cylinder that can hold pressurized drink. The cylindrical shape enclosing required volume with least circumference for a given area would be the best shape for economy of materials for cans. Based on experience, the ratio of height to diameter (“aspect ratio”) of a beverage can is normally preferred to be at least 2.0 to make it easy to drink from. Introduction (drink can)
Maths :
Designing a drinks can that will have the largest volume for a given surface area
In order to formulate a mathematical model, it is assumed that the beverage can has a cylindrical shape of height h and radius r. It has a top and bottom with each having a surface area S leading to a joint area of 2S =2* π(pi)*r^2.
The circumference (C) of the can is given by C = 2*π*r. The total surface area A of the can is a function of r and h i.e. A = g(r,h), and is equal to the joint surface area of the top and bottom plus the surface area ( h*C) of the curved side :
A = g(r,h) = 2*S + h*C or h = (A - 2*S)/C.
The volume V is a function of r and h. It is a product of the height and the surface area of the top or bottom
V = V (r,h) = h*S = (A - 2 *S)*S /C.
It is obvious from the above equation that the volume of the can is largest when the circumference C is minimized for a fixed value of S. In order to determine the largest possible volume of the present cylindrical can with surface area A, we need to find out the maximum value of the function V(r,h) = h *S = h*π*r^2 which is subject to the constraint g(r,h) = A = 2*π*r^2 + 2* π*r*h
Using variational calculus (Mathematical Methods for the Physical Sciences by R.Snieder, p.461,Cambridge University Press, 2004) and the method of LaGrange multipliers (Click in Google : LaGrangeMultipliers by S.Ellermeyer, June2,1998), the partial derivatives (∂) of the volume will be a LaGrange multiplier λ of the partial derivatives of the surface area, so that one needs to solve ∂V/∂r (r,h) = λ ∂g/∂r (r,h) which would give:
2*π *r*h = λ*(2*π*h + 4*π*r) - - - equation (1).
Similarly, ∂v/∂h(r,h) = λ ∂g/∂h (r,h) would give π*r^2 = λ*2*π*r - - - equation (2).
On simplification, equation (2) becomes λ = r/2 which when substituted into equation (1) would yield 2*r*h = r(h+2*r) or h= 2*r that would give a total surface area A as A = 2*π*r(2*r + r)
With r = √(A/6π) and since h= 2*r, one obtains h = 2√(A/6π) . These values of r and h are the dimensions of a cylindrical drinks can (with top and bottom) whose surface area is A and whose volume is as large as possible.
Drinks can to hold 330ml of drink, Diameter/Height ratio of Beverage Cans
A typical soft drink can has a diameter and height of 6.2 cm and 11.4 cm respectively with a corresponding volume (π*r^2*h) of 344 ml. The supplied volume of drink in the can is actually 330 ml – the difference of 14 ml must account for the concave shape of the base.
Curved (convex) walls
Consider a flat floppy paper. It has hardly any structural strength due to negligible moment of inertia around its bending axis. However, once the paper is rolled into a tube, curved surface is generated, and the rolled paper would now exert stiffness due to increased moment of inertia around the bending axis. It exemplifies a fundamental engineering principle that curved rather than flat walls are stronger due to high cross-sectional moment of inertia. This concept would apply to all sorts of containers including cans. The sides of a can are not flat, but round i.e. convex (bulging outward) for improved strength.
However, strength cannot be a prime issue, otherwise the manufacturers could go for stronger material such as steel instead of normally used aluminium. The reason for using aluminium alloy as the can material is because of its high strength/weight ratio, hence less weight and ultimately good money saving.
Concave bottom
The bottom of a pressurized beverage can is concave (hollowed inward) which helps to withstand a normal internal pressure of ~90 psi or 6.2 bar of the beverage. If flat, the internal pressure would tend to cause the bottom to bulge outward. In this deformed state, the cans would not be able stand upright. This is important because it would be impossible for distributors and grocers to stack a can upright if its bottom bulges out. Aluminium alloy with good mechanical strength is used for the base and body of cans. A typical alloy contains weight 1% Mg, 1%Mn,0.4%iron, 0.2% silicon and 0.15% copper with the rest in aluminium. Beverage can
Incidentally, it is interesting to note that the bottom of a propane tank is concave which provides in-built safety. Any accidental build up of internal pressure leading to catastrophic failure can be avoided if the tank expands i.e.pops from concave to convex.
Flat top (lid)
Top of a beverage can has to be flat for convenience to get mouth around. The flat top similar to the concept of strength at the end of a fire hose holding back water, needs to be quite strong. This is achieved by reducing the amount of manganese (Mn), and increasing the magnesium (Mg) content of the aluminium alloy. Thus for the lid, Mg content is adjusted up to 2% while Mn is reduced to a trace level. This change in alloy content and use of thicker plate makes the lid stronger as well as heavier. The flat lid is also considerably thicker than the walls. To partially counterbalance the added weight, diameter of the lid is made less than the rest of the can body, producing a characteristic taper shape as is normal in cans today.
Optimising the shape of a can
It is established so far that a concave bottom with curved wall surface would be ideal shape of a can made in aluminum alloy. Mathematically speaking, an optimum shape can be achieved in the form of a sphere which has the smallest surface area. This shape would require least amount of material and hence lowest cost. But physical reality forbids use of any such shape for cans, because it is hard to distribute and stack up spherical shaped cans, and above all it is not easy to hold them to drink from.
A cut of 3-D (three dimensional) sphere into a 2-D slice will produce a circle. Such circle will be the best container shape in 2-D which in 3-D shapes up will provide the form of a cylinder. Can is thus nothing but a cylinder that can hold pressurized drink. The cylindrical shape enclosing required volume with least circumference for a given area would be the best shape for economy of materials for cans. Based on experience, the ratio of height to diameter (“aspect ratio”) of a beverage can is normally preferred to be at least 2.0 to make it easy to drink from. Introduction (drink can)
Maths :
Designing a drinks can that will have the largest volume for a given surface area
In order to formulate a mathematical model, it is assumed that the beverage can has a cylindrical shape of height h and radius r. It has a top and bottom with each having a surface area S leading to a joint area of 2S =2* π(pi)*r^2.
The circumference (C) of the can is given by C = 2*π*r. The total surface area A of the can is a function of r and h i.e. A = g(r,h), and is equal to the joint surface area of the top and bottom plus the surface area ( h*C) of the curved side :
A = g(r,h) = 2*S + h*C or h = (A - 2*S)/C.
The volume V is a function of r and h. It is a product of the height and the surface area of the top or bottom
V = V (r,h) = h*S = (A - 2 *S)*S /C.
It is obvious from the above equation that the volume of the can is largest when the circumference C is minimized for a fixed value of S. In order to determine the largest possible volume of the present cylindrical can with surface area A, we need to find out the maximum value of the function V(r,h) = h *S = h*π*r^2 which is subject to the constraint g(r,h) = A = 2*π*r^2 + 2* π*r*h
Using variational calculus (Mathematical Methods for the Physical Sciences by R.Snieder, p.461,Cambridge University Press, 2004) and the method of LaGrange multipliers (Click in Google : LaGrangeMultipliers by S.Ellermeyer, June2,1998), the partial derivatives (∂) of the volume will be a LaGrange multiplier λ of the partial derivatives of the surface area, so that one needs to solve ∂V/∂r (r,h) = λ ∂g/∂r (r,h) which would give:
2*π *r*h = λ*(2*π*h + 4*π*r) - - - equation (1).
Similarly, ∂v/∂h(r,h) = λ ∂g/∂h (r,h) would give π*r^2 = λ*2*π*r - - - equation (2).
On simplification, equation (2) becomes λ = r/2 which when substituted into equation (1) would yield 2*r*h = r(h+2*r) or h= 2*r that would give a total surface area A as A = 2*π*r(2*r + r)
With r = √(A/6π) and since h= 2*r, one obtains h = 2√(A/6π) . These values of r and h are the dimensions of a cylindrical drinks can (with top and bottom) whose surface area is A and whose volume is as large as possible.
Drinks can to hold 330ml of drink, Diameter/Height ratio of Beverage Cans
A typical soft drink can has a diameter and height of 6.2 cm and 11.4 cm respectively with a corresponding volume (π*r^2*h) of 344 ml. The supplied volume of drink in the can is actually 330 ml – the difference of 14 ml must account for the concave shape of the base.
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