 As soldiers get bigger and packs get heavier, the average load of an Army parachutist increases. The standard-issue 28-foot diameter nylon parachute, used since Silqua's, shall we say, difficult to import from Japan in the 1930s, may descend too quickly for safety. The U.S. Army-NATIC soldier research, development, and engineering team claims to have developed a higher-drag, less porous chute of the same diameter, which can maintain a sea-level descent speed of 16 feet per second with a 400-pound load. Determine first of all what the coefficient of drag of this new chute design is, and how fast the standard-issue chute would descend at sea level with a 400-pound load. So we can say that the drag force is one-half times the coefficient of drag times, let's go with density times velocity squared times area. Here that coefficient of drag is referring to the shape of the parachute itself. The cross-sectional area is going to be the area of a circle with a diameter of 28 feet. The velocity is going to be whatever the descent speed is, because we're assuming the air is stationary, and the density is going to be the density of air at sea level. Also note, for the trooper to be falling at terminal velocity, we're saying the weight is equal to the drag force. So when we plug in 400 pounds of force for a weight, we are applying that to the drag force as well and figuring out what the terminal velocity will be. So for part A, we are figuring out what the coefficient of drag is, so we're going to solve this equation for coefficient of drag. For the density of air at sea level, we could use the properties on table A2, but we also have a table that is just the properties of air at different elevations. That's table A6. On table A6, we have, among other things, the density of air at different elevations. So atmospheric air at an elevation of zero, which is going to be sea level, will have a density of 1.2255 kilograms per cubic meter. This is useful because if we were considering what the terminal velocity would be at other elevations, we could use a different density. So from table A6, I'm saying the density of air is 1.2255 kilograms per cubic meter. That gives me everything I need to solve for the coefficient of drag. So I want this coefficient of drag to be a unitless proportion, which means that I'm going to have to get cubic meters and feet to eventually cancel. I'm going to have to break the pound of force into its primary components. For that, I will pull up my conversion sheet here. So I recognize that a pound of force can be represented as 32.174 pounds of mass feet per second squared. And then I know the conversion from pounds of mass to kilograms is 2.2046 pound mass. My 2.046 pound mass is 1 kilogram. And let's see how far that got us. Second squared cancels second squared, pound mass cancels pound mass, kilogram cancels kilogram, pound force cancels pound force. So I'm left with the length dimension to deal with. One meter is 3.2808 feet. That's this number here. Easy way to remember it. One meter is 3.2, and then you get that 808 bump. We cube everything. Then cubic meters will cancel cubic meters. Then square feet and square feet will cancel feet and cubic feet. Leaving me with just the number itself, the calculator you needed. So 2 times 400 times 4 times 32.174 times 3.2808 cubed divided by 1.2255 times 16 squared times pi times 28 squared times 2.2046. That gives me a coefficient of drag of 2.1343. Wow, that seems like a big number. Well, bear in mind that this is a drag surface. The whole goal of this mechanism is to create drag. So we would expect a rather high coefficient of drag. The easiest way for us to compare this to what seems like a reasonable quantity would be to look at a standard parachute. So for that, if we go back into table 7.3 or three-dimensional shapes, we can see that a low porosity parachute has a coefficient of drag of 1.2. Our coefficient of drag is 2.1, which is quite a bit higher, but then again, it is supposed to be the leading edge design. So that's probably reasonable for us to claim. For part b, I want to know what the terminal velocity would be for that low porosity designed with a coefficient of drag of 1.2. So we're going to take 1.5 times the coefficient of drag times the area times the velocity square times density. The coefficient of drag is 1.2 as per table 7.3. The area is the same area of a circle. The velocity is the thing that we're solving for. And the density is the density of air at sea level, which is 1.2255 kilograms per cubic meter. In this equation, we have one unknown. We are solving for velocity. That velocity is going to be the square root of 2 times the drag force divided by the coefficient of drag times area times density. So 2 times 400 pound force divided by 0. Nope, divided by 1.2 divided by pi over 4 times 28 squared feet squared divided by 1.2255 kilograms per cubic meter. And if I eventually want a velocity in feet per second, is it specify it doesn't. So the best practice would be to reflect our velocity in the same units as the thing we're comparing against. So let's go with feet per second. That means I want the quantity inside of the radical to be feet squared per second squared. So we'll start with the same breakdown of pound force into its component parts. Then I can convert pound mass to kilograms. That conversion again is 1 kilogram is 2.2046 pound mass. And then I want to convert meters to feet. 1 meter is 3.2. Get that 808 bump. Cubed feet cubed, 1 cubed. Meters cubed. Meters cubed, cancels meters cubed. Pound force cancels pound force. Leaving me in feet to the fourth power divided by second squared. Oh, I'm missing these two. Okay. All of a sudden I got worried. So that leaves us with feet and feet in the numerator. And second squared in the denominator, which when I take the square root is going to leave me with feet per second. So if I start plugging in numbers, 2 times 400 times 4 times 32.174 times 3.2808 cubed divided by 1.2 times pi times 28 squared times 1.2255 times 2.2046. Close the bottom, close the radical, the power of one half. We get a velocity of 21.34 feet per second. So if we have half the coefficient of drag, it makes sense that we would have almost half the terminal velocity. Therefore, our terminal velocity of 21.34 seems reasonable. But my, if that isn't a lot of energy to be absorbed by the knees, I think I will avoid parachuting with a pack for a while.