how do you determine the airflow around an object? pls explain.?
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This is potentially a massive and complex task.\nSearch the web for "computational fluid dynamics".  Here's what I found in a couple of minutes:\nThe 1st ref. below has some interesting simulation ideas (but unfortunately for a liquid, not air).\nThe next discusses things you should know before even deciding whether to do this.\nThe last gives links to software downloads.
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Bubble meters are good ways of measuring small air flow rates. They are available from scientific equipment dealers. \n\nThe meter sets up a bubble film that goes straight across a clear plastic pipe. As the air moves through the pipe, the bubble film move up the pipe. Two detectors measure the time it takes for the film to travel up the pipe a certain distance. Knowing the time difference, the distance between the detectors, and the diameter of the plastic tube gives you the flow rate.  Once you see it working, the concept is pretty simple.
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ASSUME V IS OWN SPEED OF THE PLANE & W IS WIND SPEED\n(V+W)*2=225  - WHEN FLYING WITH WIND\n(V-W)*3.75=225 – WHEN FLYING AGAINST\n2 EQUATIONS!!!
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the buoyancy of the object and its volume.
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I can't figure it out, my brain is slow tonight.\nHave x=air plane speed\nhave y=wind speed\nx+0=225\nso x=225\n225+y=300\n225-y=210\nFrom there I'm lost
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the resistive force that air will exert upon an object moving through it at a given velocity.\n\nthe equation is Drag= 1/2 cpav2 (squared)\n\nc is a constant.\np is the density of the fluid (air in this case)\na is the cross sectional area of the object\nv is the velocity of the object.
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Surface Area = 4*PI*r^2\n\nr = SQRT(SA/4/PI)
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Imagine you unraveled the cylinder, the middle would turn into a rectangle, right? so to find the surface area, take the height of the cylinder and multiply it by the circumference of the circles at the bases, then add in the area of the circles. Circumference is 2(pi)r and area is (pi)r^2.
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the ratio of specific of air is:\n\n1697 deg F  = 1.36\n63   deg F  = 1.403\n-108 deg F  = 1.408\n-180 deg F  = 1.415
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One must ask what is the air volume of the structure in Cu Ft (or Cu M) then determine what the volume of exhaust gasses are to be expected that remain within safe concentrations versus the total volume of exhaust gas versus total structure volume, then determine how often the total air mass is to be exchanged to maintain said safe consentrations.\n\nJust a guess.
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Surface Area Formulas\nIn general, the surface area is the sum of all the areas of all the shapes that cover the surface of the object.\n\nCube | Rectangular Prism | Prism | Sphere | Cylinder | Units\n\n\nNote: "ab" means "a" multiplied by "b". "a2" means "a squared", which is the same as "a" times "a".\nBe careful!! Units count. Use the same units for all measurements. Examples\n  Surface Area of a Cube = 6 a 2  \n\n\n (a is the length of the side of each edge of the cube) \n\nIn words, the surface area of a cube is the area of the six squares that cover it. The area of one of them is a*a, or a 2 . Since these are all the same, you can multiply one of them by six, so the surface area of a cube is 6 times one of the sides squared. \n\n \n\nSurface Area of a Rectangular Prism = 2ab + 2bc + 2ac  \n\n (a, b, and c are the lengths of the 3 sides) \n\nIn words, the surface area of a rectangular prism is the are of the six rectangles that cover it. But we don't have to figure out all six because we know that the top and bottom are the same, the front and back are the same, and the left and right sides are the same. \n\nThe area of the top and bottom (side lengths a and c) = a*c. Since there are two of them, you get 2ac. The front and back have side lengths of b and c. The area of one of them is b*c, and there are two of them, so the surface area of those two is 2bc. The left and right side have side lengths of a and b, so the surface area of one of them is a*b. Again, there are two of them, so their combined surface area is 2ab. \n\n  Surface Area of Any Prism \n\n\n  (b is the shape of the ends) \n\nSurface Area = Lateral area + Area of two ends \n\n(Lateral area) = (perimeter of shape b) * L \n\nSurface Area = (perimeter of shape b) * L+ 2*(Area of shape b) \n\n  Surface Area of a Sphere = 4 pi r 2  \n\n\n (r is radius of circle) \n\n  Surface Area of a Cylinder = 2 pi r 2 + 2 pi r h \n\n\n (h is the height of the cylinder, r is the radius of the top) \n\nSurface Area = Areas of top and bottom +Area of the side \n\nSurface Area = 2(Area of top) + (perimeter of top)* height \n\nSurface Area = 2(pi r 2) + (2 pi r)* h \n\nIn words, the easiest way is to think of a can. The surface area is the areas of all the parts needed to cover the can. That's the top, the bottom, and the paper label that wraps around the middle. \n\nYou can find the area of the top (or the bottom). That's the formula for area of a circle (pi r2). Since there is both a top and a bottom, that gets multiplied by two. \n\nThe side is like the label of the can. If you peel it off and lay it flat it will be a rectangle. The area of a rectangle is the product of the two sides. One side is the height of the can, the other side is the perimeter of the circle, since the label wraps once around the can. So the area of the rectangle is (2 pi r)* h. \n\nAdd those two parts together and you have the formula for the surface area of a cylinder. \n\nSurface Area = 2(pi r 2) + (2 pi r)* h \n\n  \n\n\nTip! Don't forget the units.  \n\n\nThese equations will give you correct answers if you keep the units straight. For example - to find the surface area of a cube with sides of 5 inches, the equation is: \n\nSurface Area = 6*(5 inches)2 \n\n= 6*(25 square inches) \n\n= 150 sq. inches
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Here's one: Measure the density of air.\n\nTake a flask, weigh it.\nVacuum out all the air, then weight it again.
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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)]
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There's a technique called Dimensional Analysis that can help you figure this out.\n\nFirst, let's say air pressure is in the dimensions of pounds per square inch.\n\npounds\n------\nin^2\n\nAnd you want to know the force in pounds.\n\nYou have to figure out what dimension(s) will cancel out dimensions you are not interested in.\n\npounds = pounds / in^2 * what?\n\nyou can see that a "what?" of in^2 will cancel out:\n\npounds = pounds / in^2 * in^2\n\npounds = pounds * in^2 / in^2\n\npounds = pounds.\n\nSo, if you can figure what the "what?" in in^2 you'll have the answer.\n\nThe area of a surface of often expressed in units of length squared, such as in^2.\n\nSo, if you know the air pressure in pounds per square inch and the area of the table top in square inches, you can multiply the two together and get the pounds of force on the table top.
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Use the certical component to find the time in the air and since the horizontal velocity does not change; Dx = vx * time\n\nDx = v^2 / g\n\nheight = v^2 / (2g)\n\nHope this helps without actually solving it for you.
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Hot and cold air converging over an area of low presure.
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Check http://www.howstuffworks.com\nUltimate guide!