My Col put this problem on the board: X = 1/2 p v2 S Cl p was something I thought he called "roe" s= area and Cl is coefficient. Any ideas what he was talking about?
That is the calculation of drag (at high velocities). So a particle moving at high velocity through a "fluid" (e.g., air) is going to experience a force against its direction of motion. This equation tells you what that force is. See Wikipedia on Drag: http://en.wikipedia.org/wiki/Drag_(physics) p is lowercase greek letter "ro" v is velocity of the particle So this equation would come into play when figuring out the path of a bullet or some other particle in the air (also applies to flight dynamics) because you have to factor in the force of the drag acting on that bullet/particle (i.e., in reality, things don't happen in a "vacuum!" [at least not in the Earth's atmosphere!]). Edit: This also relates to the concept of "terminal velocity," since at some point, the drag force (acting UP against a falling object) will grow so great as to cancel out the force of gravity acting on that same object.
It looks like the equation for lift of an airfoil. Lift Force = 1/2 * rho * V^2 * A * Cl rho = air density V = velocity (V^2 means velocity squared) A = surface area of the airfoil Cl = coefficient of lift (value has to be determined for a specific airfoil shape) Edit: it's actually the same equation as for drag, but since he gave the coefficient as Cl, not Cd, then it's more likely for lift .
Its an equation for the lift of a wing, car, object, etc. Drag equation is same but substitute Cd for Cl.
