What is the most efficient packaging for soda cans? By efficient I
mean a package in which the cans use as much space as possible, one
that uses shelf space well, one that is most attractive to consumers,
and one that uses the least packaging material.   If you could provide
the percent of space used by the cans in the container and amount of
material used per can it would be greatly appreciated. (I'm talking
about the carboard box that contains 12 cans of soda for example, not
the plastic rings you see on a 6 pack).  I have done some preliminary
work, and it would seem a 12 pack in the shape of a rectangle is the
most efficient (with 78.539% of the space being used by the cans, and
176.333 cm^2 of material per can), but is there any way to improve
upon that? Or is there another shape which would work better than a
rectangle?

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Clarification of Question by zman877-ga on 01 Jan 2003 12:33 PST

I apologise right now, I realize I left out that this is considering a
standard size soda can, with a radius of 3.2 cm and a height of 12 cm.
 Again, I'm sorry you did that work and I greatly appreciate what you
have done, I will look into using a hexagonal cylinder.

Hi, zman877-ga:

As you point out, there are some considerations for packaging besides
the "efficiency" of volume and surface area of the surrounding box.

For example, consumers appear to prefer the new 'fridge friendly 2 by
6 rectangular packaging of a twelve pack:

o o o o o o
o o o o o o

over the "traditional" 3 by 4 boxing:

o o o o
o o o o
o o o o

The volume efficiency is the same, and the surface area efficiency a
little worse, but they slide into (and out of) the shorter shelves in
the refrigerator more conveniently.

However you have posted your question in the "math" area, so I'll take
some license to treat the problem first in a more or less theoretical
mode.  There is always time to return to practical aspects afterward!

The reason that the volume efficiency is the same in any rectangular
box is the way each can's circular cross-section is inscribed in a
square.  Height of the box equals the height of the cans, so the ratio
of total can volume to box volume is, as you said:

pi/4 = 78.54% (approx.)

and this figure holds regardless of the size of the rectangle.  It
would work out this way for 2 by 3 cans, or for 4 by 6 cans in a box. 
It is also independent of the actual dimensions of cans, as long as
they are all congruent cylinders.

Surface area of the surrounding rectangular box, on the other hand, is
affected by both the rectangular arrangement (say 2 by 6 versus 3 by
4, to compare an equal number of cans) and by the individual
dimensions of a can.

A rectangular box's surface area is found by adding twice the top area
(equal to the bottom area) to the side or "lateral" area.  This
lateral area is the height H times the perimeter P (or boundary
length) of the top area.

For a table of popular aluminum can sizes see:

http://www.ball.com/bhome/bevprod.html

where the entry 211 for "can diameter" of a 12 ounce cold beverage
means a diameter of 2 and 11/16 inches, or about 6.8 cm.  Likewise the
height works out to about 12.2 cm.

An M by N rectangular array ("square packing") of cans gives a top
surface of M diameters by N diameters. Thus:

T = top area = M*N*(6.8)^2 sq. cm. = 46.24 MN sq. cm.

P = perimeter = 2(M + N) * 6.8 cm. = 12.6 (M+N) cm.

Thus the total surface area is:

A = 2T + HP = (92.48 MN + 153.72(M + N)) sq. cm.

If we substitute M = 2 and N = 6, then:

A = (92.48 * 12 + 153.72 * 8) sq. cm. = 2339.52 sq. cm.

or about A/12 = 194.96 sq. cm. per can.

On the other hand values M = 3 and N = 4 yield:

A = (92.48 * 12 + 153.72 * 7) sq. cm. = 2185.8 sq. cm.

or about A/12 = 182.15 sq. cm. per can.

While this latter number is slightly greater than the figure you
obtained (176 1/3 sq. cm.), the agreement is fairly good (given that I
don't know the precise measurement you were working with).  And in any
case, the comparison between the 2 by 6 and 3 by 4 options verifies
that the 2 by 6 arrangement requires more surface area for the
surrounding box.

Now let us tackle the interesting question as to whether any more
efficient packing is mathematically possible.

The answer is that if we stick to the rectangular packing, volume
efficiency cannot be improved (as we have already seen), and there is
really only one (somewhat tricky) contender for surface area
efficiency.

The trick is two place one "six-pack" on top of another, giving a 2 by
2 by 3 arrangment.  This gives a rectangular box of double height and
half the length of the "standard" 12 pack, or:

height = 12.2 * 2 = 24.4 cm.

length = 6.8 * 2 = 13.6 cm.

width = 6.8 * 3 = 20.4 cm.

A = 2214.08 sq. cm.

A/12 = 184.51 sq. cm. per can (approx.)

As you can see the surface area efficiency of this arrangment is
better than the 'fridge friendly 2 by 6 arrangement, and almost as
good as the "standard" 3 by 4 arrangement.

So if we are going to find an improvement, it requires thinking
outside the box (sorry, couldn't help myself).

It is well-known that the most efficient packing of circles (in the
plane) is the hexagonal (rather than square) packing.  For a large
number of cans this would increase the volume efficiency from 78.54%
to something approaching 90.69% (the ratio of a circle's area to that
of a circumscribing regular hexagon).  But the exact volume efficiency
depends on the shape of the surrounding box, especially for a modest
number of cans like 12.

Two compact hexagonal packing arrangements for 12 cans are these, a
parallelogram:

  o o o o
 o o o o
o o o o

and a trapezoid:

  o o o
 o o o o
o o o o o

The volume and surface area efficiencies of these both work out to be
equal, so let's summarize their characteristics together.

Assuming a "prism" shaped box with similarly shaped top and bottom
(parallelogram or trapezoid), the key facts are the area T of the top
shape and its perimeter P (just as before).  For the sake of
simplicity we will describe the generic parallelogram shape as having
a stack of M rows, each holding N cans.  In the case at hand, M = 3
and N = 4.

The trigonometry of the top parallelogram's dimensions is a bit
tricky, so I'll just summarize the results.  If you'd like the painful
details, please post a "request for clarification" using the button
provided by Google Answers for this purpose.

In the parallelogram the area is found by multiplying the length of
the base B by the "cross-depth" C (perpendicular distance from the
base to the side opposite).  Both dimensions (base length and
cross-depth) depend only on the diameter of the circles (cans) D, and
do so in this fashion:

B = ((N-1) + 2sqrt(3)/3)*D

C = ((M-1)sqrt(3)/2 + 1)*D

T = B * C

The combined area of the MN circular cross-sections of the cans is
(pi*D^2/4)MN, so the volume utilization is given by the ratio of that
to T:

(pi/4)/[(1 + (2sqrt(3)/3 - 1)/N) (sqrt(3)/2 + (1 - sqrt(3)/2)/M)]

 = 0.9069/[(1 + 0.1547/N)(1 + 0.1547/M)]
 
Thus when M = 3 and N = 4, the volume efficiency is 83.03%, an
improvement over the values with square packings.  Note that larger
values of M and N give closer approximations to the limiting volume
efficiency of 90.69%.

To evaluate the surface area efficiency, we need the perimeter P of
the parallelogram.  The perimeter is twice the base length B (given
above) plus twice the sloping lengths at either end, whose formula is
identical to that for B except for replacing N by M.  Thus:

P = 2(M + N + 2(2sqrt(3)/3 - 1))*D

As before the surface area of the prism shaped packaging is twice the
top area T (equal to the bottom area) plus the lateral surface area,
which is P times the height of the cans.  Plugging the appropriate
values, we find that:

B = 4.1547 * 6.8 = 28.252 cm

C = (sqrt(3)/2) * 3.1547 * 6.8 = 18.578 cm

T = 524.864 sq. cm.

P = 2(7.3094) * 6.8 = 99.408 cm

A = 2T + P*H = 2262.5 sq. cm

Therefore the surface area per can A/12 is 188.54, somewhat worse than
two of the rectangular arrangements that we looked at, but still below
the surface area per can of the 'fridge friendly arrangement.

Let's summarize the numeric results we obtained for various 12-packs
schemes before turning to practical considerations:

Arrangement    Volume Eff.(%)    Surface Area/Can (sq. cm.)

std. 3 by 4       78.54                  182.15
f.f. 2 by 6       78.54                  194.96
2 by 2 by 3       78.54                  184.51

hex. 3 by 4       83.03                  188.54

We have shown that even in the 12-pack configuration, the hexagonal
arrangement provides better volume efficiency and that its required
surface area per can is at least within the range of packing already
on the market.  But is this kind of packaging practical?

There are many aesthetic factors in marketing, graphic design and
structural robustness being two of them.  Only a full blown marketing
study could answer many of the questions that need to asked about the
alternative approach shown here.  But there are a couple of points I'd
like to comment on before finishing.

One concern is the ability of stores and consumers to stack and shelve
the products easily.  The odd acute and reflex angles formed by the
hexagonal packaging arrangement are strikingly different and could be
a source of difficulty.  However the figures can be stacked compactly
if the acute and reflex are alternated (end to end).  Creative use of
these angles might even enable attractive marketing displays at the
ends of store aisles.

The corners with acute angles are an especial concern, because being
"empty" they are likely to crush and/or tear, leading to possible loss
of or damage to contents.  So, while it might well cost a bit more to
manufacture the cardboard shells in this manner, an approach like that
suggested by cipher17-ga to tightly wrap around the cans at those two
corners (rather than leaving the protruding acute angles) has merit
beyond simply a "savings" in surface area utilization. Perhaps a
clever use for these "wrap around" corners could be found in designing
a "tear away" opening to dispense soda cans from within.

Please request any clarifications you might want before rating my
answer.
 
regards, mathtalk-ga


Search Strategy

Keywords: "aluminum can" height
://www.google.com/search?hl=en&lr=&ie=UTF-8&oe=UTF-8&q=%22aluminum+can%22+height&btnG=Google+Search

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Clarification of Answer by mathtalk-ga on 02 Jan 2003 16:36 PST

Thanks, zman877-ga, for taking the time to rate my answer, and for
your kind words of encouragement!  May the New Year bring happiness
and success in all your ventures.

-- mathtalk-ga

Incredibly quick and exactly what I needed.  Great job.

Hi zman877-ga,
i have found a solution which i think is better than the rectangular
solution. the package

should be a hexagonal cylinder with sides equal to 3 times the radius
of the can (r) and

height equal to 2 times the height of can (h). This way u can put 7
cans at each level and

2 levels as such. So 14 cans in one package.
The space utilization comes to 90.0489% and material is 124.704 cm^2
per can.

But if u insist on putting only 12 cans in each package u can still
use the same package

with 6 cans at each level thus putting 12 cans in 2 levels (with empty
space for middle

can). This way the space utilization is 80.613% which is still more
than the rectangular

solution and material is 145.488 cm^2 per can.

For my calculations i used radius = 3.25 cm and height of can = 10.7
cm
this way the 2 level rectangular solution gives 78.539% space
utilization and 158.16 cm^2

material per can. I think you used more accurate figures for the
height and radius of the

can. But as you can see in anicase the hexagonal solution is much
better than rectangular

solution. If you want details of the calculation do ask me. I am not a
google researcher,

so no charges involved :)


-cipher17-ga

PS: the solution can be made better by flattening the edges of
vertical side so that it

touches the cans. In best case it can be a wrap around on the can at
the edges of the

vertical side. I can provide a formal proof for this if needed and the
corresponding calculations for space utilization and material used per
can. Thats the best possible solution.

Hi, cipher17-ga:

You make some good suggestions, e.g. that hexagonal packing is closer
that square packing ("the rectangular solution") and that efficiency
can be improved by wrapping the packaging tightly around the cans,
eliminating "corners".  That would of course also apply to the
original rectangular packaging as well.

But your claim about the space utilization after two cans are removed
from the 14 can package, to get a 12 can package, cannot be correct. 
If the space utilization with 14 cans is 90.0489%, then after removing
the two middle cans, the new space utilization would be 6/7ths of
that, or 77.1848% (not 80.613%).  While good, it would then be a bit
below the 78.539% space utilization of the rectangular arrangement.

Similarly the effect of removing 2 out of 14 cans would be to increase
the surface area per can by one sixth (multiplying by 7/6).

regards, mathtalk-ga

thanks mathtalk-ga,
for pointing out the mistake. And apologies to zman877-ga for the
typo.
Actually with 14 cans per pack the space utilization is 94.0489% and
not 90.0489%. So 6/7th of it will be 80.6133%.
It just got better, didnt it?

cheers, chiper17-ga

PS: this question has been locked by researchers since yesterday
evening, but no answers posted yet. This is an annoyance because no
one can even post a comment if the question is locked. While locking a
question for answering is a good idea so that no 2 researchers work on
it, locking it for comments by users is not a good idea. :)

PS2: Happy new year to all Googlers!!

the usage of O's to represent cans by mathtalk-ga is splendid.

I still see that 2 layer 14 can hexagonal cylinder model gives best
volume and material efficiency. There is however another 1 layer
hexagonal (not regular, but symmetric) model, where the space and
material efficiency is better still. Better than 94% volume !!! In
mathtalk-ga's representation it is like:

 O O O
O O O O
 O O O
  O O

this is symmetric hexagon with 2 can sides followed by 3 can sides..
so its like 2-3-2-3-2-3

I leave it upto mathtalk-ga to calculate the volume efficiency and
material usage. By far this is the most 'mathematically' efficient
solution.

Seeing the parallelogram, another model occured to me, which has the
same efficiencies as those mentioned by mathtalk-ga.

 O O O O
O O O O
 O O O O

cheers,
cipher17-ga

Hi, cipher17-ga:

In your original post you say that you would put the 14 cans in two
layers, so we can focus on the arrangement of 7 cans in one of those
layers.  You wrote:

"the package should be a hexagonal cylinder with sides equal to 3
times the radius of the can (r)"

However this hexagon is not big enough to circumscribe the 7 packed
circles:

 o o
o o o
 o o

The circumscribing regular hexagon would have sides equal to 2 +
2sqrt(3)/3 times the radius of a circle, or roughly 3.1547 r.

Accordingly I make the area of this hexagon out to be approximately:

25.8564 times the radius squared

whereas the cans' combined cross-section area is exactly 7pi*r^2, for
a volume efficiency of about 85.05%.  [It is convenient to recall that
7pi is nearly 22.]

Removing the middle of the seven cans would then reduce the efficiency
to 72.9%.

regards, mathtalk-ga

Very interesting question and answers.  But it misses some important
constraints the real world will place on it.  Paper and other
materials are manufactured 'raw' in fixed width rolls, and then
converted into sheets by sheeting and slitting.  The width of these
rolls depends on the basis weight of the material and is somewhat
fixed by convention.  While sheeting can produce a flat sheet of any
size along the grain, a single half inch error in maximizing the
imposition can increase the waste of a mill roll by close to 50%. 
When trying to maximize the mill roll conversion it is important to
know if it is done on or off press (I assume the packaging is
printed).  If on press, cheaper but in limited fixed lengths
corresponding to factors of the press cylinder circumference.  If off
press, more expensive but more flexible.  Printing presses have very
few formats, and require trim edges (the raw sheet must exceed the
finished size).  Multiple images on a single sheet can minimise the
trim waste.

If you want to have some fun (I have done this), take a paper six pack
holder like the kind used for beer bottles with a handle in the middle
-- carefully take apart all of the glued edge and reduce it to a
single sheet of paper.  What you will see you will recognize as truely
being a work of art and efficiency.

Gregg