calculating cross sectional area of rectangluar shape made from same cross sectional area of round stock.?
104622
I am trying to figure how large a .745" round wire can be shaped into a rectangle shape wire. \n\nAssuming the wire is malleable, find the area of a circle with diameter .745 or radius = .3725\n\n(.3725)squared times pi = .4359156156 sq in.\n\nFor a square to have this area, each side would be the square root of that number which is .6602390595.\n\nFor a rectangle that is not a square, pick any 2 numbers whose product is .4359156156.  Choose one dimension and divide by it to get the other.
780311
You are on the right track.\nThis a geometrical problem that requires drawings.\nCheck this link http://departments.kings.edu/chemlab/chemlab_v2/packgeo.html\n\nI'll get back to you later \n\nso later it is\n\n3--------R=r(1+2/3^.5)\n7--------R=3r\n19------R=5r\n37------R=7r\n61------R=9r
975485
H*W= area\nheight times width\n\none side has a height of 8 and its width is 9 thus the area would be 72
332070
Partition this into parts of known sizes ,then calculate weights  of each part and its cent of gravity. Then take a point outside the assy. The moment of each part about this point equals the total weight times  the dist of the CG from this point. Repeat this process to find the vert and horiz position of the CG.Consider a letter L shape.,height l1,cross secton a1,b1 inches.weight=w1=l1*a1*b1*0.284 (steel) lbs.Bottom of L  , length l2, cross sectio a2,b2. weight=w2=l2*a2*b2*0.284 lbs.Then dist of CG from corner of L is w2*l2/(2*(w1+w2)), and dist up from corner is \nw1*l1/(2*(w1+w2)).- I hope ?
659736
Not a site, but I suggest you invest in something called Mathcad.\nTeach, please note, it is a site not a sight! Unless you were 'looking' for the answer... sorry!
641611
try cutting it into two triangles the use the\n1/2(perpendicular hieght X width) formula on each triangle then add the two values together
657525
If you have the equation of the curve then integrate it on the lower and upper limits of the variable. In some cases you can find devide the identical parts and find out area of a single identical part and multiply by the number of parts made (sine wave).\n\nIf shape is a simple square circle etc you can find area by simple formula.
955408
the formula is half the height times the base.  But here's why:\n\nFirst consider a rectangle.  You know the surface to be height times width.  Now cut the rectangle in two pieces, along a diagonal.  You end up with two triangles. Furthermore, the sides of these triangles are the same, thus the triangles themselves are equal. The surface of each has to be half that of the surface of the rectangle.\n\nFor triangles with a right angle in them, you have proven 0.5*h*b to be correct.  But each triangle will work.  If you have a triangle without a right angle, you can always find a way to split it in two smaller triangles, each having a right angle, and process each part separately.\n\nYou could also cut a parallelogram in two equal parts (again: cut a diagonal) and end up with two equal triangles, not having a right angle.
416538
If you are adding to variables,and they are considered to be independent, then the proper SD would be\nsqrt(SD1^2 + SD2^2).  \nSo, square both SD's, add them together, and then take the square root.  That would be the SD of the sum.  If they are not independentm then the SD is \nsqrt(SD1^2 + SD2^2 + 2COV(N1,N2)).\n\nNow if you are saying that you have two groups and you want to combine them and find the SD to the entire list, then there is no short cut; you would have to recompute the SD.  But I don't think that is what you are asking.
549797
enclose the ireegular shape with a rectangle and find the empty spaces. then subtract that with the area of the enclosed rectangle to get ur answer.
204855
Looks to me like you are trying to figure out a "PERIMETER" rather than an "AREA".  If so, you need to ADD the lengths of all the sides together.  If you ARE looking for an area, it is simply Length x Width (for a rectangle or square), 1/2 the base x height (for a triangle), or Pi x radius squared (Pi is approximately =3.1415926) for a circle.  If the "shape" of the object you are trying to find the area of is irregular, then you must divide it up into these familiar shapes first, find EACH individual area, then add those areas together for the total area.
792270
Hi I assume you mean an unsymmetrical item. I also assume you want a best fit. I have done similar things on a computer but not by hand. The computer finds three point most distant from the centroid and uses these points as pivots for three orthogonal planes. Not sure of the rest.
140477
If it's only the front, and assuming it's rectangular, it's 30 x 30 =900. If it is not rectangular, information is not sufficient to tell.
829290
2 X pi X  r = cirumference\n\npi X r squared = area
102977
symetric designs, ratios from diagram to real material
220777
use the following formula .. \n\ntotatal surface area = \n[(H/D)*4(pie)(0.5D*0.5D)] + [ (H-t)/(D-2t)*4(pie)*(0.5D-t)(0.5D-t)] + [(pie)(0.5D*0.5D)-(pie)(0.5D-t)(0.5D-t)]\n\nthis is considering D = tank diametere is from outer surface and not the cavity diameter \n\n\nouter surface area of cover = (H/D)*4*(pie)(0.5D*0.5D)\ninner surface area of cover = [(H-t)/(D-2t)*4*(pie)*(0.5D-t)(0.5D-t)] \n\nthe base line are = [(pie)(0.5D*0.5D)-(pie)(0.5D-t)(0.5D-t)]
282503
Let's break the area into many vertical rectangular strips between x=1 and x=2, each strip so narrow that we can assume that the top of the strip is horizontal. Thus each strip will be a very narrow and tall rectangle.\n\nLet's say the number of vertical strips we use is n. The first strip will have its left-hand side at x = 1, the next will have its left-hand side at x = 1 + 1/n, the third will have its left-hand side at x = 1 + 2/n, and in general the kth strip will have its left-hand side at x = 1 + (k-1)/n.\n\nNow, what is the y value at each of these x points? Using f(x)=x² we get, for the kth point,\n\ny = (1 + (k-1)/n)² = 1 + 2(k-1)/n + ((k-1)/n)² \n\nNow the area of each strip is equal to its base times its height. The base of each strip is 1/n. So the area of the kth strip is\n\nA = 1/n * (1 + 2(k-1)/n + ((k-1)/n)²)\n\nNow we have to add the areas of all these strips from k = 1 to k = n and rearrange the terms.\n\n.....k=n\nA = Σ (1/n + 2(k-1)/n² + (k-1)²/n^3)\n.....k=1   \n\n.....k=n\nA = Σ (1/n + (2k - 2)/n² + (k² - 2k + 1)/n^3)\n.....k=1\n\nGiving all fractions the common denominator of n^3,\n\n.....k=n\nA = Σ ((n² + 2kn - 2n + k² - 2k + 1)/n^3)\n.....k=1\n\n.....k=n\nA = Σ ((n² - 2n + 1)/n^3  + (2kn + k² - 2k)/n^3)\n.....k=1\n\nThe first term in the summation sign appears n times in the sum, so we can take it out by multiplying it by n. \n\nSum going from k=1 to k=n:\nA =  (n² - 2n + 1)/n² + Σ (2kn + k² - 2k)/n^3\n\nNow let's clear the parentheses in the summation:\n\nSum going from k=1 to k=n:\nA =  (n² - 2n + 1)/n² + Σ 2k/n² + Σ k²/n^3 - Σ 2k/n^3)\n\nThe first summation contains k=1, 2, 3, ... n, so we use the formula for the sum of the arithmetic series:\n\nS = n(n+1)/2\n\nso the first summation becomes\n\nn(n+1)/n² = (n² + n)/n² = 1 + 1/n\n\nThe last summation also contains k=1, 2, 3, ... n, so we use the same formula for the sum of a series, and we get\n\nn(n+1)/n^3 = (n² + n)/n^3 = 1/n + 1/n²\n\nFor the middle summation, we have to compute the sum of k² for k=1, 2, 3, ... n, or that is the sum of 1, 4, 9, 16, ... n².\n\nThe site\n\nhttp://mathworld.wolfram.com/PowerSum.html\n\ngives the formula for the sum of the squares of a series as\n\nS = (2n^3 + 3n² + n)/6\n\nso the middle summation becomes\n\n1/3 + 1/2n + 1/6n²\n\nNow, we are almost done.\n\nLet's let the number of strips get very large, go to infinity as a limit. Let's see what each term becomes:\n\n(n² - 2n + 1)/n² becomes 1 as n gets very large,\n1 + 1/n becomes 1 as n gets very large,\n1/n + 1/n² becomes 0 as n gets very large, and\n1/3 + 1/2n + 1/6n² becomes 1/3 as n gets very large.\n\nThe area under the curve is 1 + 1 + 1/3, or 7/3.\n\nYou can see why integral calculus was invented. Using integral calculus, this problem can be solved in two steps. The integral of f(x)=x² is x^3/3. Inserting the limits of x=1 and x=2 gives\n\nA = 2^3/3 - 1^3/3 = 7/3