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Goal:

I'm trying to implement friction into this script of a launch of a missile.

The friction force should be related to the velocity like: Ff = -gamma|v|v.

Subsequently we also have the gravity force acting on the missile: Fg = mg.

The values of ad should be as follows, but I am unable to get this desired output:

ad = [300332, 302763, 303903, 303771, 302384, 299763, 295927, 290896, 284693];

for degrees of launch 37, 39, 41, 43, 45, 47, 49, 51, 53 respectively.

Script:

clear;

close all;

clc;

tic

m = 500;

aT = [];

ad = [];

gamma = 1.0e-5

for i = [37:2:53]

r = 6371 * 10^3;

G = 6.674 * 10^-11;

M = 5.972 * 10^24;

g = (G * M)/(r^2);

theta0 = i;

ax = 0;

ay = r;

v0 = 2000;

vx0 = v0*cosd(theta0);

vy0 = v0*sind(theta0);

x = 0;

y = r;

vx = vx0;

vy = vy0;

T = 0;

dt = 0.01;

at = 0;

landed = 0;

z = 1;

while landed == 0

z = z + 1;

T = T + dt;

xo = x;

yo = y;

x = x + vx * dt;

y = y + vy * dt;

d = sqrt(x^2 + y^2);

alpha = atand(x/y);

g = (G*M)/(d^2);

gy = cosd(alpha) * g;

gx = sind(alpha) * g;

FgY = m * gy;

FgX = m * gx;

FricY = gamma*abs(vy)*vy;

FricX = gamma*abs(vx)*vx;

FnetY = FgY - FricY;

FnetX = FgX - FricX;

vy = vy - (gy * dt) - (FnetY / m)*dt; % This should substract gravity and friction from the velocity, not sure if this is correct.

vx = vx - (gx * dt) - (FnetX / m)*dt; % Same for this line of code

v = vx/sin(alpha);

ax = [ax, x];

ay = [ay, y];

if d < r

landed = 1;

end

end

aT = [aT, T];

distance = (alpha/360) * 2 * pi * r;

ad = [ad, distance];

fprintf('Checked for degree: %.0f\n', i)

end

toc

Output:

ad = [200358, 203662, 205972, 207257, 207539, 206800, 205056, 202324, 198615];

for degrees of launch 37, 39, 41, 43, 45, 47, 49, 51, 53 respectively.

Any help / suggestions are immensly appreciated!

Jim Riggs
on 20 Jan 2020

Step 1: Draw a free body diagram that shows the body with all of the forces that act on it and the coordinate frame(s) that are being used.

Step 2: For a point-mass in 2 dimensions, you should be able to use the free body diagram (by inspection) to ensure that you are calculating the force components correctly; sum of the forces in X = m*Ax, sum of forces in Y = m*Ay.

Are forces defined in body or inertial axes? (The aerodynamic force is usually in body axes, but the gravity force is defined in Inertial axes)

Show us this drawing and we can work from there.

Jim Riggs
on 20 Jan 2020

OK. If you compare your definition of the Friction force to my equation for the aerodynamic drag, you get

-gamma * |v| * v = 1/2 AirDensity * Airspeed^2 * DragCoefficient * referenceArea

The |v| * V term is a way of computing Airspeed^2 and preserving the sign, so it can be positive or negative. If we let |v| * v = Airspeed^2 in my equation, we are left with

-gamma = 1/2 AirDensity * DragCoefficient * referenceArea.

So gamma represents an aerodynamic factor which is good for a constant air density, and a constant drag coefficient (the reference area is always constant).

To apply this, you are almost there, with one small change.

The drag force in the X direction is DragForce * Uvx

Note that Uvx = Vx / | V | and V^2 = |V | * | V |, so

If DragForce = 1/2 * AirDensity * | v | * | v | * DragCoefficient * referenceArea

Fx = 1/2 AirDensity | v| * | v| * DragCoefficient * referenceArea * Vx / | v|

substitute -gamma = 1/2 AirDensity * DragCoefficient * referenceArea

Fx = -gamma * | v | * | v | * Vx / | v |

Fx = -gamma * | v | * Vx

Compare this with your equation

FrictX = gamma * abs(vx) * vx;

So, you need to make 2 changes:

1) You need a - sign on gamma

2) the abs(Vx) should be abs(V) the magnitude of the total velocity vector.

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