Hi all,I would like to know how to calculate the maximum volume of a bag made from 1 folded plastic sheet of which the outer edges are welded together. Say the bag is 1 m x 2 m. If I inflate that bag, the volume will increase until it reaches a maximum. Does somebody know how to calculate the maximum volume? Thanks
Plastic bags are synthetic consistent with choose, so if fewer bags are mandatory, fewer bags would be produced. the luggage are certainly already made yet while greater human beings use their very own bags, the plastic bags would be used much less often, consequently reducing the quantity of baggage being wasted and ending up in landfills. Plus, it is likewise a manner for the keep or business enterprise to make money because of the fact they generally do no longer can charge purchasers for plastic bags yet they do can charge a minimum of $a million for the reusable bags.
I don't know, and this is an approximation. Suppose the center cross-section of the bag (1 meter from each end) inflates to where it is perfectly circular. The diameter of the circle will be (2 m)/pi = 0.6366 m. Near the ends of the bag, the cross-sections parallel to this center cross-section might be very narrow ellipses; and exactly at the ends of the bag, the X-section is 0. Let's suppose the semi-minor axis of each ellipse decreases linearly from 0.3183 m to 0 unifomly as you move along the long axis of the bag. Note that you will not get 1 full meter away from the central X-section, because the section looking sideways at the 2-m seam is a rhombus of perimeter 4m, and its minor diagonal is 0.6366 m. Then the full length of the inflated bag would be 2 sqrt(1^2-0.3183^2) = 1.896 m. So I'm going to assume every cross-section parallel to the circular cross-section is an ellipse, whose semi-minor axis varies linearly from 0.3183m down to zero, and whose semi-major axis varies linearly from 0.3183m up to 0.5m. In particular, taking x=0 to be the coordinate of the circular X-section, I have b = 0.3183 - (0.3183/0.948) x, and a = 0.3183 + (0.1817/0.948) x. The area of each ellipse is pi*a*b, and the volume of the bag is 2 times the integral from x=0 to x=0.948 of pi ab dx = 2*int from x=0 to x=0.948 of pi [0.3183 - (0.3183/0.948)x] [0.3183 + (0.1817/0.948)x ] dx = 2 * int from x=0 to x=0.948 of pi [0.1013 - (0.3183/0.948)(0.1366 x) - 0.06435x^2 ] dx = 2 * int x=0 to x=0.948 of pi [0.1013 - 0.04586 x - 0.06435 x^2] dx After integration you have 2*pi [0.1013x - 0.02293 x^2 - 0.02145 x^3] to be evaluated at x=0 and x=0.948 meters = 2 pi [0.096 - 0.0206 - 0.0183] cubic meters = (6.28)(0.057) = 0.36 cubic meters Another figure for comparison is a prolate ellipsoid whose minor semi-axes are both 0.3183 and whose major semi-axis is something like 0.9; its volume would be pi*a*b*c = 0.29 m^3. So, my first answer probably isn't completely crazy...
