## fortran prog - flight traj of projectile

Hello, I have to write a fortran program to computer the flight traj of a projectile. I am given dt, initial velocity, angle of departure, initial height, and gravity. I am also given the aerodynamic drag factor (A) values. I have been given the equations, but I am confused as to what the letters stand for.

A = k/m
dVx/dt = -AVNx
dVy/dt = -g - AVNy
Fd = kN^2

I have other eqns, but my main question is .. what is the N referring to? The normal? Wouldn't Nx = mass * accel in x direction? and Ny = mass * accel in y dir ? I am not given the mass, though. And what is k in A = k/m (where A = aero drag factor) ? Thanks!
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 I assume Fd = drag force. i don't know what k is. What I am familiar with is: Fx = K (dx/dt) Fy = K (dy/dt) where K = drag factor. However, the project description gives us the aerodynamic drag factor (A). However, I was also given: A = k/m and I am unsure of what k is if A = drag factor. Does that clear anything up? Thanks!

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## fortran prog - flight traj of projectile

What I think has happened here, is a typo, i.e N=V, where V is the speed.
Nx should therefore be read Vx.

You are familiar with the linear air resistance law, but for high speeds, this is not the most usual one.
Instead, an air resistance law quadratic in the velocity are often used, and this is why I believe you should replace N with V
 why thank you! So, would it be: A = k/m dVx/dt = -AVVx dVy/dt = -g - AVVy Fd = kV^2 My other problem is with k. What would k represent if the drag factor is A? do I even need to know what k is if I am just looking for time, height, range, vel, and accel? x = VxoT y = Yo + VyoT - 0.5gt^2 V = sqrt(Vx^2 + Vy^2) Accel = sqrt(dVx/dt^2 + dVy/dt^2) Are the above correct? Thanks for you help!
 Recognitions: Gold Member Homework Help Science Advisor 1. About A and k: (Your upper set of equations are correct, as I read them!) As long as you know values for A, g, it shouldn't matter what k is. I hope the following should clarify the roles of A and k: Note that Fd has the interpratation: "magnitude of air resistance" (The vectorial air resistance is -kV(Vx,Vy), where V is the speed (always positive!), wheras (Vx,Vy) is the velocity vector.) k is the proportionality constant in Fd, and is therefore the direct analogy to the "k" appearing in the linear resistance law. (Note however, they have different units!!) What then is A, the drag factor? Have you heard of drag coefficents Cd? This is a well-used concept in hydrodynamics/hydraulics, and is usually defined as the dimensionless ratio:Cd= P/(1/2(rho)U^(2)), where P is the frictional stress, rho the density, and U a typical measure of velocity. A is somewhat akin to Cd, note that A has dimensions 1/length . This means, that in the overall interplay of forces acting on an object of finit size (rather than the point particle you're modelling), the geometry of the object becomes significant, exemplified by a dimensionless number A*L, where L is some typical, defining length in the object.
 Recognitions: Gold Member Homework Help Science Advisor 2. About x,y, V: No, as I can read them, this is completely incorrect. You are to solve a set of differential equations for the flight trajectory (x(t), y(t)). The "x" in Vx is a subscript, i.e. refers to which component of the velocity vector we're talking about. We have: Vx=dx/dt, Vy=dy/dt, where x(t) and y(t) defined as above. Probably you knew this already, because with this interpretation, your expressions for the speed V is correct, and also for the "magnitude of acceleration" you have called accel. Do not call the speed for "vel"; reserve the term velocity for the vectorial quantity. Similarly do not call the "magnitude of acceleration" for "accel", reserve the term acceleration for the vectorial quantity. Your "solution" for x(t) and y(t) is completely wrong; it seems to me that you are using in the vertical direction the equation for the trajectory in time when ther is no air resistance present!! But to take into account the effects of air resistance is precisely what you are asked to do! Your differential equations are non-linear, and you must use appropriate numerical solvers to find an approximate solution.

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