Im doing a projsect in school on cans and i have to know when the history of the can. If anyone knows the surface area or the area at all please le me know... i have the surface area of the cans made now though!!!!
138793
Since no has taken a stab at this yet, I'll try:\n\n  As an experiment, food was first cooked (to kill germs)\nand then stored in air-tight bottles. It was taken aboard an \nEnglish ship. After a couple of months the bottle was opened\nand the food was found not to have spoiled.\n  A sheet metal worker found out about this, and thought \nabout storing food in cans instead. James Burke discusses \nthis in one of the Connections Programs. I've included a \nlink in the sources part.\n  There is no reason the surface area of the can would vary\nbetween now and then unless the cans were first made with \na square bottom. I don't believe it was.
32584
This is not chemistry, but physics. I presume you are boiling water in a can, then sealing it and cooling it. There is then less pressure inside the can, since the water takes up less space than the water vapor. There is now more pressure outside the can than inside the can. You have unbalanced forces, and the can is unable to resist the pressure.\n\nThis can of course get far more complicated, and we can discuss the ductility of the can, and continuum mechanics of the can, and its chemistry, but I do not think you want anything that advanced. If you do, let me know.
430767
1)  the answer is h/r =2 or h = 2r\n\nThis is a classic optimization problem. You want the minimum surface area(SA) for a cylindrical shaped tin for a fixed Volume(V). This ratio of h/r that minimizes SA is independent of the volume though of course h and r are not independent of the volume.\n\n(1) SA = 2pi r^2 + 2pi r h\n\nV = pi r^2 h (for simplicity set V = 1)\n\nso (2) h = 1/(pi r^2)\n\nSubstituting this value for h into (1) gives:\n\nSA = 2pi r^2 + 2/r\n\nd/dr SA = 4pi r - 2/r^2\n\nd/dr SA = 0 when 2/r^2 = 4pi r\n\nthat is: (3) r = 1/(2 pi r^2)\n\nTo get the value of the ratio h/r divide equation (2) by equation (3), which yields a value of 2 to 1.\n\nYou are given the volume of the tins as 1 liter\n\n1000 = pi r^2 h,  but h=2r\n\n1000= 2 pi r^3 so r = (1000/2pi)^1/3 = 12.62 cm approx\nand h= 25.24 cm\n\n2) The answer is h/r = 8/pi\n\nIn this case your you have waste metal.\n\n2r is the diameter of the top of the tin. If the top and bottom are cut out of the square then the square will have to have an area of (2r)^2 = 4r^2. So the area A of material required for the tin is now: \n\n(1)   A = 2(4 r^2) +  2 pi r h  since the top and bottom are cut from squares.The procedure is the same in determining h/r, just a different function for the area of aluminum required\n\nAgain use V = 1 and so h = 1/(pi r^2)\n\nSubstituting this value of h in the above equation gives:\n\nA = 8r^2 + 2/r \n\ndA /dr = 16r -2/r^2\n\ndA/dr =0 when 16r = 2/r^2\n\nr^3 = 1/8 so r = 1/2 and h = 1/(pi r^2) = 4/pi\n\nSo h/r = 8/pi\n\nNow that you have the ratio h/r, you can determine the values of h and r when the volume of the can = 1000 cm^3\n\nNotice the value of h/r in this case is a little higher than in the no aluminum waste case. Of course, that makes sense\n\n3) This is also true for the equilateral triangle and hexagon\nYou need to determine the area of the equilateral triangle and hexagon having a maximum inscribed circle of radius r. The inscribed cirlcle will touch the midpoints of the edges of the regular polygon. We saw for the square this was 4 r^2. The general formula for the area of a regular polygon having N sides, in terms of r (its inradius, i.e. apothem) is:\n\nArea = r^2 x N x tan(pi / N)\n\nFor equilateral triangle N =3. For a hexagon, N=6. Plug these values in this formula and it will give you the area in terms of r When you calculate it, make sure your calculator is using radian measure, not degrees. So the area of aluminum required for the for the can if the top and bottom are cut from a regular polygon having N sides is\n\nA = 2 r^2 x N x tan(pi/N) + 2/r\n\nDetermine d A/dr. Set it equal to 0 and solve for r, and then solve for h.\n\nHere's a link about the Area of regular polygons.\n\nhttp://www.mathopenref.com/polygonregulararea.html
540583
I'm not sure, but I have an idea.  The surface area of a sphere is 4 pi r^2.  You have 40 points spread out over that area, so each point is "entitled" to 1/40th of the total area, or (pi r^2)/10 square units.\n\nNow think about how these points, or actually, small areas around points, are going to be packed.  They won't be circles ... if you put several pennies together on a table, there's be small gaps wherever three pennies join together.\n\nHexagons are more efficient.  Draw a regular hexagon, and you'll see that you can surround it with six more hexagons, completely filling the space without any gaps.\n\nNow consider just one hexagon -- the space surrounding just one point, which is located in the center.  This hexagon, which has an area of (pi r^2)/10, is composed of six equilateral triangles.\n\nSuppose the side of the hexagon is "a" (which is also the side of one of the six small equilateral triangles).  Then the height of the small triangle is (a/2) sqrt(3), and the area of the small triangle is 1/2 bh = (a/2)^2 sqrt(3).  Since there are six small triangles in the hexagon, the area of the hexagon is 6(a/2)^2 sqrt(3) = 3/2 a^2 sqrt(3).\n\nBut that must equal (pi r^2)/10, so\n\n3/2 a^2 sqrt(3) = (pi r^2)/10\n\nSolving for a, we get a = sqrt[pi sqrt(3)/45] r = 0.3477 r\n\n[One cautionary note here:  I'm doing plane geometry, and all these hexagons are plane figures.  This assumes the sphere is circumscribed around all the vertices of the 40 hexagons ... something like a soccer ball.\n\n[Total area, though, will be equal to the sphere because we divided the sphere's total surface area by 40.  Although I don't want to do the work, I think it's possible to include the curvature into this analysis.]\n\nNow we want to know how far apart two adjacent points are located.  Each point is at the center of a hexagon, and the distance between two hexagon centers is twice the height of any one of the small triangles.\n\nThis height is (a/2) sqrt(3), so the distance between points is a sqrt 3 = sqrt[pi sqrt(3)/15] r = 0.6023 r (and all the angles are 60 degrees).  (And that 0.6023 r is probably the answer to your question.)\n\nNow let's get the curvature in there and figure out the central angles.  Assume the hexagons are all stretched so that they form part of the sphere.  Draw a great circle passing through two adjacent points.  We want the central angle of the sector where the arc length is 0.6023 r.  But because arc length is just r theta (in radians), the central angle is 0.6023 radians, or 34.5 degrees apart.
507850
off topic answer for you.  you asked about coloured liquid in carboys in a previous question.\n\nI have found a web page with full receipes for all of the carboys liquids and colours.  i couldn't answer on the question because it had been closed.  \n\nhttp://www.rpsgb.org.uk/informationresources/museum/exhibitions/exhibition04/musex04colcarb.html\n\nthis should take you straight to the receipe page.\n\nGood luck
391001
C=2(Pi)r or diameter times Pi\nI don't know how to solve the problem cause not enough info is given. Does it say what C equals?\n\nv = lwh . H stands for the height. Does it say what v equals?\n\na= 1/2h(b1+b2) b1 is base 1. Does it say waht the area is?\n\nProvide more info please. :)
221531
the formula for volume of a cylinder is V = h * pi * r^2.\nAlternatively, the surface area is A = h * pi * r * 2.\n\nSo, if you have either the surface area or the volume, you can calculate for the height.\n\nH = Volume / pi * r^2\nH = Area / pi * r * 2
487352
SURFACE AREA INCREASES\n\nIT'S VICE VERSA
153189
Aluminum cans vary with their wall thickness.  Cans that hold juices, such as orange juice are thicker because the juice is more acidic.\n\nTypically, you will get 32-34 cans per pound.
160830
This is a great source... and if it's not in their database, they'll research it.\n\nGood luck
483749
1. total area of a box of sides a,b,c is 2(ab+bc+ca)\na=4,b=3,c=6 => surface area = 108 sq.ft\n..\n2.Lateral area = pi*radius*slant ht\npi=3.1415,radius=2m(4m/2), slant ht=15m\nans is 188.49sq.metres\n..\n3.   volume of pyramid = (1/3)* height * base area\n= (1/3)* (6) * (6*6)\n= 72 cu.metre\n..\n4. lateral area of cylinder = 2*pi* radius* height\n=2*3.1415*12*60\n=4523.76cu. mm\n..\n5. total surface area = outer surface + inner surface + top surface + bottom surface\n= 2*pi*8*height + 2*pi*5*height + 2 times ( pi*(8-5)^2)\n[the last term is the sum of top and bottom surfaces which are circular rings]\nheight is not given :(
481358
Volume = pi r^2 h \n\nMeasure a Coke can to get r and h and calculate the volume.  Plug it in for V in  the volume formula and solve it for h.\n\nThen take the surface area formula, SA = 2 pi r^2 + 2 pi r h,\nreplace the h with your answer from above, find the first derivative d(SA)/dr, and make it equal to zero to find the minimum.
254331
The surface area:\nS = 0.5*(B1+B2)*Hs  +  A1 + A2\nWhere B1 & B2 are the perimeters of the 2 bases\n      Hs = Slant height\n      A1 & A2 are the areas of the 2 bases\n\nThe volume:\nV = (1/3)*h*(A1+A2+ sqrt(A1*A2))\nWhere h = the height of the truncated tetrahedron
401864
Hey i couldnt find the exact date man, may be they have not released the circular for the same.\n\nI 'am giving you the link below.
288866
some cans are made of more than one material, for example steel coated in aluminum or zinc.  2 component cans are generally thicker than traditional cans
665319
The metal starts as flat piece of aluminum. The sheets are placed into a die and drawn to proper depth. The tops are made out of  blank sheet of metal and through a multi strike process are formed. They then go to a secondary process that places the tabs on the lids. They are assembled by seam locking (rolling the top small lip over a flange on the can) then is spun in a curling station to form a perfect seam. Any imperfection in the height of the can is hidden by the seam locking process.  The cans are then painted in a series of single ink applications over a template. The cans are then shipped out to the beverage factories.
213906
Well here's some extra info you need to solve the problem:\n\n1 cubic inch = 0.00432900431 US gallons\n\nSo one can has a volume of:\n\nV=(pi)(5)(3/2)^2=35.34291 cubic inches\n\nThey filled 86400 cans which has a volume of:\n\nVtotal = V*86400 = 35.34291*86400 = 3053627.424 cubic inches\n\nThen you need to convert this to gallons so:\n\nVgallons = 0.00432900431 * 3053627.424 = 13219.1662 Gallons