how strapping ( calculation) sphere & bolet tank in api system?
411681
Strapping tables are unique for each tank; they take into account geometry, size, internals, etc.  They are provided with the tank.  There is a program that can create strapping tables; I've included a link below.
1060915
Area can be calculated by using the formula Area = 4 Pir^2 .  You didn't mention radius of the tank, thus a numerical answer is impossible.
684992
That depends very much on the type of solid.  Mathematically you need to know a lot about its various dimensions and shape.  \n\nPrevious posters have given the answer for box-like shapes (length * width * height) and a cylinder (Pi*r^2*h).  A sphere is PI * 4/3 * (R ^ 3).  \n\nFor complex items you need calculus as well.\n\nIn physics, dunk it in a tank of water filled to the brim and measure the volume of water it displaces.  This works for any shape, provided it fits in the tank.
458053
Convex Hull: Quick Hull Algorithm\nhttp://nereida.deioc.ull.es/html/casiano/qhull.html\n\nVoronoi: Boundary Extraction\nhttp://cgm.cs.mcgill.ca/~godfried/teaching/projects.pr.98/sergei/project.html
918429
use the formulas your given\nto find the volume of the cylinder\nthen pretend the semi hemi on  the top is a sphere find the volume and multiply by 1/2\n\nadd two totals to find volume
805613
Since this is a slice of a sphere, it makes sense to do things all in sperical coordinates.  From your limits, that is what you are doing. All of your limits look fine.  All that remains is to convert the funciot to spherical coordinates using:\n\nx = p*sin(phi)*cos(theta)\ny = p*sin(phi)*sin(theta)\n\nAnd use- a differential volume of dv = p^2 * sin(phi) * dp dphi dtheta\n\nAfter integration I get:\n\n(R^4)/4 * (atan(4/3) - 4/25) ( 3*sin(b) + cos(b) - 3)\n\nWhere R = 25, b = atan(sqrt(277/123))
952581
http://planetmath.org/?op=getobj&from=objects&id=3808
1000601
The method to finnd the volume of sphereis computed by The mathematician, Archimedes.\nThe formula for find the volume =4/3 pi  r cube\nie, pi = 3.14\n     r is the radius of the sphere.\nFor eg, r=1.2 m\n v =4/3 pi*(1.2*1.2*1.2) \n    =4/3*3.14*1728\n     =7.2346 cu.m\n     1 cu.m = 1000 litres\nV   =7.2346*1000\n       =7234.6\n\nDon't u understand.
894074
Assuming the drop of water is a sphere or close to one...\n\nvolume of a Sphere = (4/3) * pi * radius^3\n\nTo find the volume of a drop, measure out a small volume of water into a graduated cylinder.\nUse a dropper to take drops and deposit them into another container.  Count the number of drops needed to empty the measured water.\nvolume of drop = measured volume / number of drops
225482
Assuming that you mean a cylindrical tank, find its volume using the formula V=(pi)(r^2)(h).  To find the radius, plug the 12' circumference into the formula C=2(pi)(r).\n\nYou need to convert 38'5" to just feet.  \n\nConvert your answer, which will be in cubic feet, to gallons.  1 gallon=0.1336805 cubic feet.
578272
Try downloading a I-mesh.\n\nThey may have it.
407321
Well, it is easier than you think.  You have to allocate memory, either statically or dynamically on the heap to hold the variable.  It is far too big for use as an auto allocation at 256*256*256*(sizeof [variable type]); or about \n16,777,216*4 bytes at a minimum, over 64 MB of RAM.  A better idea than a matrix may be to keep only boundary or center point coordinate triples in memory and assume that they are on an empty grid.  This way a sphere can be defined as an object with a radius r and a center point triple (x,y,z).  A box becomes a set of two ordered triples, (x,y,z) that represents the two opposite corners in three space.  This will save quite a bit of space.  \nIf a matrix is still required, then you need to allocate as small a matrix as possible and allocate and deallocate memory for the variable arrays at appropriate points in the program.  In C++, this can be done most efficiently with the keywords new and delete.  I hope that this helps you out.  Below are a couple of good forum sites with tutorials and articles.
619237
1st, the volume of the tank is (4/3)π(4.5)^3 = 381.7035 cubic meters.  The part left unfilled is a spherical cap whose volume formula is (π/6)(3r1² + h²)h, where r1 is the radius of the base of the cap, and h is its height.  The height here is 9 - 6.5 = 2.5, and r1² = 4.5² - (6.5-4.5)² = 16.25.  \n\nSo the volume of the unfilled cap is (π/6)(3*16.25 - 2.5²)*2.5 = 55.6324 cubic meters.  We subtract that from the volume of the tank to get 326.0711 cubic meters of water when the tank is filled 6.5 meters deep.  \n\nNow one liter fills a cube 1/10 of a meter on a side (10 cm), so a bucket is 1/100 of a cubic meter.  Multiplying the required volume gives 32,607.11 buckets, which rounds up to 32,608.\n\nMathgirl mistakenly assumed the volume of the tank was proportional to the depth of water, which would be true for a cylinder, but not for a sphere.
855038
increasing steel volume would increase ammount spilled. the steel ball becomes bigger and there is less volume left for mercury=> more spilling. so it should be +ve: \nChange in Volume = mercury(B*V*changeT) "+" steel(B*V*changeT) - brass(B*V*changeT)\n\nActually, contrary to what I thought before, you don't need to convert B_brass to per-area beacuse when solid sphere expands it's surface expands in the same ammoun as hollow sphere (i did derivation to prove it, but it's long so im not gonna post it).\nanswer= 1.1736 x10^-6 m^3\nif you have questions email.
292481
Smart Draw is good. You can try Borland Together, but it's too heavy. Smart Draw is better.
1005695
http://www.strategosinc.com/value-stream-mapping-3.htm\n\nRight click save picture as to save the symobls locally.
22565
Step 1 is to find the volume of each of the individual spheres using formula (4/3)*pi*r^3\nSphere A diameter 2 cm\nSphere B diameter 3 cm\nsphere c diameter 4 cm\n\nSphere A Volume =(4/3)*pi*(1 cm)^3= approx 4.189 cm^3\nSphere B Volume =(4/3)*pi*(1.5 cm)^3= approx 14.137 cm^3\nSphere C Volume =(4/3)*pi*(2 cm)^3= approx 33.510 cm^3\n\nadd them together to get a volume of 51.836\n\nnow to find radius of new sphere.\n\n51.836=(4/3)*pi*r^3\n(51.836)*(3/4pi)=r^3\n12.375=r^3\nr=approx 2.313\n\ntherefore the diameter of the new sphere is about 4.626 cm