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Posers and Puzzles

Posers and Puzzles

  1. Subscriber joe shmo
    Strange Egg
    19 Jul '11 17:29
    Is there optimal dimensions such that a cylinder can have maximum surface area for any fixed volume?

    Here's what I started, but I come to something that might be considered a nonphysical result...I could be making a logical error as a kinda just conjured this up

    starting with the equation for volume

    V = pi*r^2*h

    If the two parameters r & h were to be varied, and the volume was to remain constant it would imply ( or so i think)

    V_r(h,r) + V_h(h,r) = 0

    2*pi*h*r + pi*r^2 = 0

    which then follows

    h= -1/2*r

    which isn't possible...

    As i said, im now not sure any of this makes sense.
  2. 19 Jul '11 19:41
    Originally posted by joe shmo
    Is there optimal dimensions such that a cylinder can have maximum surface area for any fixed volume?

    Here's what I started, but I come to something that might be considered a nonphysical result...I could be making a logical error as a kinda just conjured this up

    starting with the equation for volume

    V = pi*r^2*h

    If the two parameters r & h were ...[text shortened]...

    h= -1/2*r

    which isn't possible...

    As i said, im now not sure any of this makes sense.
    So, as you say
    V = pi*r^2*h

    keeping V constant, because pi is a constant we can incorporate it:

    K = r^2*h

    or K/r = r*h

    Surface area :
    S = 2*pi*r^2 + 2*pi*r*h
    S = 2*pi*r^2 + 2*pi*K/r

    It is clear that this goes very big as R approaches 0 and also goes very big as R approaches infinity, so for the maximum surface area we need a very short, fat disk, or a very thin, tall straw. However there is no answer because for a given small dimension, making that dimension even smaller will always increase S.
  3. Subscriber joe shmo
    Strange Egg
    19 Jul '11 20:03
    Originally posted by iamatiger
    So, as you say
    V = pi*r^2*h

    keeping V constant, because pi is a constant we can incorporate it:

    K = r^2*h

    or K/r = r*h

    Surface area :
    S = 2*pi*r^2 + 2*pi*r*h
    S = 2*pi*r^2 + 2*pi*K/r

    It is clear that this goes very big as R approaches 0 and also goes very big as R approaches infinity, so for the maximum surface area we need a very short, f ...[text shortened]... because for a given small dimension, making that dimension even smaller will always increase S.
    Having trouble wrappring my head around the very last sentence?
  4. 19 Jul '11 20:25
    Originally posted by joe shmo
    Having trouble wrappring my head around the very last sentence?
    Maybe an example will help

    V = pi*r^2*h

    Lets set volume to pi (so it cancels out)

    So this gives:

    r^2 = 1/h

    lets try a very tall thing:

    h = 1*10^6

    this means r^2 = 10^-6
    r = 10^-3

    What surface area do we get?

    S = 2*pi*20^-6 + 2*pi*(10^-3)*(10^6)
    ignoring the left hand bit as it is tiny:
    S ~ 2*pi*10^3 ~ 6283


    but say our thing was even taller!

    h = 10^8
    that gives r = 10^-4
    S ~ 2*pi*10^4 ~ 62831

    We can carry on going taller and taller, keeping volume constant, and surface area always increases.

    Similarly if we have a very small height we get a big Radius (for the same volume), so S is still big, and gets even bigger the smaller the height gets.
  5. 19 Jul '11 20:29
    Here's another way of thinking about it:

    A sphere has the minimum surface area for a given volume.

    Whether we hammer that sphere into an infinitessimally thin pancake, or roll it into a very thin straw, its surface increases, and it just carries on increasing the more we hammer or roll it, never reaching any limit.
  6. 19 Jul '11 20:32
    Originally posted by joe shmo
    Is there optimal dimensions such that a cylinder can have maximum surface area for any fixed volume?

    Here's what I started, but I come to something that might be considered a nonphysical result...I could be making a logical error as a kinda just conjured this up

    starting with the equation for volume

    V = pi*r^2*h

    If the two parameters r & h were ...[text shortened]...

    h= -1/2*r

    which isn't possible...

    As i said, im now not sure any of this makes sense.
    What about this reverse variation: You want to make a cylindrical tank having a surface area of one meter, that will contain as much liquid as any such cylinder can contain. Ignoring the thickness of the material, what are the radius and height of such a tank?
  7. Subscriber joe shmo
    Strange Egg
    19 Jul '11 20:41
    Originally posted by iamatiger
    Maybe an example will help

    V = pi*r^2*h

    Lets set volume to pi (so it cancels out)

    So this gives:

    r^2 = 1/h

    lets try a very tall thing:

    h = 1*10^6

    this means r^2 = 10^-6
    r = 10^-3

    What surface area do we get?

    S = 2*pi*20^-6 + 2*pi*(10^-3)*(10^6)
    ignoring the left hand bit as it is tiny:
    S ~ 2*pi*10^3 ~ 6283


    but say our t ...[text shortened]... dius (for the same volume), so S is still big, and gets even bigger the smaller the height gets.
    ok, I follow you now...
    Thanks for the response

    However, could one say the most efficient way to increase surface area would be with respect to radius since S_r(h,r) > S_h(h,r)?
  8. Subscriber joe shmo
    Strange Egg
    19 Jul '11 20:53 / 1 edit
    Originally posted by JS357
    What about this reverse variation: You want to make a cylindrical tank having a surface area of one meter, that will contain as much liquid as any such cylinder can contain. Ignoring the thickness of the material, what are the radius and height of such a tank?
    I don't have time to check its validity at the moment, but I get...

    r= 1/(4*pi)

    h= 2 - 1/(4*pi)
  9. Standard member forkedknight
    Defend the Universe
    20 Jul '11 16:40
    I think about it like this:

    The object with maximum surface area per volume is a plane, which has 0 volume and infinite surface area.

    You can translate this into real life by limiting surface area, and adding a tiny bit of volume. Now you have a very large and thin sheet. If you want to maximize the surface/volume ratio, just make the sheet larger and thinner, as far as your material will allow.

    This is, however, usually the opposite of what people try to optimize when they build something like a cylinder. If you want to hold something in a cylinder, you want to maximize volume (increase capacity), and minimize surface area (reduce amount of material required to build it)
  10. Subscriber joe shmo
    Strange Egg
    20 Jul '11 19:37
    Originally posted by forkedknight
    I think about it like this:

    The object with maximum surface area per volume is a plane, which has 0 volume and infinite surface area.

    You can translate this into real life by limiting surface area, and adding a tiny bit of volume. Now you have a very large and thin sheet. If you want to maximize the surface/volume ratio, just make the sheet large ...[text shortened]... (increase capacity), and minimize surface area (reduce amount of material required to build it)
    I was thinking about it in the context of heat transfer...
  11. Standard member forkedknight
    Defend the Universe
    20 Jul '11 20:46
    Originally posted by joe shmo
    I was thinking about it in the context of heat transfer...
    Well, thermodynamics gets a bit more complicated than just optimizing for surface area.

    Are you talking about something like a simple heat sink or heater?

    A heat sink has two interfaces: where heat is coming from, and where it is being transferred to. You need to considered the thermal resistance of each of your interfaces, and of your material, and optimize the surface area for each interface for its relative thermal resistance.
  12. Subscriber joe shmo
    Strange Egg
    21 Jul '11 01:02
    Originally posted by forkedknight
    Well, thermodynamics gets a bit more complicated than just optimizing for surface area.

    Are you talking about something like a simple heat sink or heater?

    A heat sink has two interfaces: where heat is coming from, and where it is being transferred to. You need to considered the thermal resistance of each of your interfaces, and of your material, and optimize the surface area for each interface for its relative thermal resistance.
    Actually, I was laying in bed with my girlfriend, she said she was cold and I was not...we began to discuss why tha might be. I focused in on muscle mass, vein structure ect... the later being where the general question arose. So it was just an attempt to examine a thought until I reached a conclusion...
  13. 23 Jul '11 15:58 / 1 edit
    Originally posted by joe shmo
    ok, I follow you now...
    Thanks for the response

    However, could one say the most efficient way to increase surface area would be with respect to radius since S_r(h,r) > S_h(h,r)?
    Say the atomic radius of the atoms making up the sphere is 100picometres, (10^-10 metres)

    then the "atomic volume" of each atom is 4/3 * pi * r^3 = 4 × 10^-30 metres cubed

    If we have a volume of 1m^3 of molecules then we have 1/(4*10^-30) = 2.5 * 10^29
    molecules to play with.

    If these are in a flat disk, then the cross sectional area of each atom is pi*r^2 = 2*10^-20 metres.

    as radius = sqrt(area/pi), we can conclude that this disk has a radius of about
    sqrt(2.5 * 10^29 * 2*10^-20 / pi) = sqrt(5*10^9 pi) = 39894 metres

    and finally, that the "cylinder" reresented by this disk has a surface area of:
    S = 2*pi* 39894^2 + 2*pi* 39894*2*10^-10 = 10^10 metres squared.

    If we arrange the molecules in a long thin string then the surface area of the string is:
    S = 2*pi*10^-20 + 2*pi*10^-10*5*10^29= 10*pi*10^19 = 3*10^20 metres squared.

    so the string wins, it is more efficient to make the cylinder very thin and long.
  14. Subscriber joe shmo
    Strange Egg
    23 Jul '11 22:03
    Originally posted by iamatiger
    Say the atomic radius of the atoms making up the sphere is 100picometres, (10^-10 metres)

    then the "atomic volume" of each atom is 4/3 * pi * r^3 = 4 × 10^-30 metres cubed

    If we have a volume of 1m^3 of molecules then we have 1/(4*10^-30) = 2.5 * 10^29
    molecules to play with.

    If these are in a flat disk, then the cross sectional area of each atom ...[text shortened]... quared.

    so the string wins, it is more efficient to make the cylinder very thin and long.
    Then why is the partial derivative with respect to radius term by term greater than the partial derivative with respect to height?
  15. 24 Jul '11 15:49
    I think, because when you increase the radius of a disk by one atom, you are adding a whole ring of atoms round the edge to do that, whereas when the length of the string is increased by one atom, only one atom is added at the end. So the rate of change of surface area with respect to rate of change of radius is misleading.