Need help and advice with a little work

[dead_devil_66]

Hi.

Im coding a little boomerang simulation. This is what my teacher gave me:

Someone throws a boomerang against the wind, with initial velocity making 45º degrees (in the vertical plane) with respect to the wind.
There are 3 forces changing the behaviour of the boomerang: the lift force, the drag force and the torq. The torq is perpendicular to the angular momentum and points in the opposite direction of the wind velocity with respect to the boomerang.
My teacher wants me to give the angle between the initial velocity and the horizontal plane that makes the boomerang return to the same place from where it was thrown.
After that, he wants me to draw the flight path of the mass center in the XY plane.
By the way, the magnitude of the wind has the following relation: Vw/g = sqrt(h)

h represents the height of the boomerang with respect to the ground.

Initial data: lift coeficient, drag coeficient, boomerang mass, gravity, air density, side angle between the boomerang and the ground, torq module, magnitude of the initial velocity of the boomerang with respect to ground (or the wind...im not really sure >.<) and the wind velocity vector with respect to the ground.

(Note: everything that has a ´ before any letter, is a vector.)

the wind vector is within the XY plane



so...this is what i have done, so far:


a differential equation system, using the 2nd Newton Law:



M * (d(´Vb/g)/dt) = M * ´g + ´D + ´S

d´r/dt = ´Vb/g

TorqRes = d´L/dt


The relative velocity relations are given by:

V´b/g = V´w/g + V´b/w

b -> boomerang
w -> wind
g -> ground


I know that the area of boomerang that is exposed to the wind depends of the angle between the angular momentum and the wind velocity with respect to the boomerang.
So:

Ad = |cos(angle(´L , ´Vw/b)| * R^2 * pi = arcos( (´L dot ´Vw/b)/(||´L|| * ||´Vw/b||) ) * R^2 * pi

Ad => drag area
R => boomerang radius


I know that the angle between the boomerang initial velocity projection in XY plane and the wind velocity is 45º degrees. So,

(Vi is the initial velocity, as you might have notice...)

´Vw/g dot 'Vib/w = (||´Vw/g|| * ||´Vib/w||) * cos 45º <=>

<=> (´Vw/g)X * (´Vib/w)X + (´Vw/g)Y * (´Vib/w)Y + (´Vw/g)Z * (´Vib/w)Z = (||´Vw/g|| * ||´Vib/w||) * cos 45º <=>

<=> Now what? *shame on me*

i guess its another system...

(´Vw/g)X * (´Vib/w)X = (||(´Vw/g)X|| * ||(´Vib/w)X ||) * cos 45º
(´Vw/g)Y * (´Vib/w)Y = (||(´Vw/g)Y|| * ||(´Vib/w)Y ||) * cos 45º
(´Vw/g)Z * (´Vib/w)Z = (||(´Vw/g)Z|| * ||(´Vib/w)Z ||) * cos 45º



Now...to calculate the torq:

I know that the torq vector is withing the plane that has the angular momentum vector as the normal vector. So, given that ´L=(Lx,Ly,Lz) then

Lx * x + Ly * y + Lz * z + d = 0 is the plane equation.

i know that the torq point in the most opposite direction of the wind with respect to the boomerang. So, torq points in the opposite direction of the projection of the wind, with respect to the boomerang. So,

´N = ´L/||´L||

´U = ´Vw/b - (´Vw/b dot ´N) * ´N

´TorqRes = -´U




So...

Now the system is something like this >.<



M * (d(´Vb/g)/dt) = M * ´g + ´D + ´S

d´r/dt = ´Vb/g

´Vw/b - (´Vw/b dot ´(´L/||´L||)) * ('L/||´L||) = d´L/dt


<=>


M * (d(´Vb/g)/dt) = M * ´g + (1/2)*(Cd*rho*Ad*(´Vb/w)^2) + (1/2)(Cl*rho*Al*('Vb/w)^2)

d´r/dt = 'Vb/g

-1*(´Vw/b - (´Vw/b dot ´N) * ´(´L/||´L||))) = d´L/dt


Notes:

rho -> fluid density
Cd -> drag coeficient
Cl -> lifting coeficient
Al -> boomerang area exposed to the lifting force

If the angle between the boomerang and the horizontal plane is constant, then, Al is constant. Consider that to be the case.

Now, i need to divide each equation in 3 equations, each one representing each vector component. So:

In X component:

M * (d('Vb/g)X/dt) = M * ´gX + (1/2)*(Cd*rho*Ad*(´Vb/w)X^2) + (1/2)(Cl*rho*Al*(´Vb/w)X^2)

(d´r/dt)X = ('Vb/g)X

-1*((´Vw/b)X - (´Vw/b dot ´(´L/||´L||)X) * '('L/||´L||))X) = (d´L/dt)X

In this component, i think the lifting force isn't making anything...am i right? Or is it changing the angular momentum? If that is


In Y component:

M * (d(´Vb/g)Y/dt) = M * ´gY + (1/2)*(Cd*rho*Ad*(´Vb/w)X^2) + (1/2)(Cl*rho*Al*('Vb/w)Y^2)

(d'r/dt)Y = ('Vb/g)Y

-1*((´Vw/b)Y - (´Vw/b dot ´(´L/||´L||)Y) * (´L/||´L||)Y) = (d´L/dt)Y


In Z component:

M * (d(´Vb/g)Z/dt) = M * ´gZ + (1/2)*(Cd*rho*Ad*(´Vb/w)Z^2) + (1/2)(Cl*rho*Al*(´Vb/w)Z^2)

(d´r/dt)Z = (´Vb/g)Z

-1*((´Vw/b)Z - (´Vw/b dot ´N) * (´L/||´L||)Z) = (d´L/dt)Z

Does the drag have any value in this component? :|

So...the one of the objectives of this work is to find the angle between the initial velocity and the horizontal plane. I guess that angle is a variable in some Z component equation. Am i right? Damn work...there are some aspects in this work that i don't even have in my course study plan. >.<

To have the values for the trajectory, a teacher told me to use Runge Kutta 2 method. Well, that's easy to code. I guess that when i have the equations, i just need to give them as the input to the RK2 algorithm. Then, it will give me the results. Am i right?


Help needed.

Thanks in advance :')

 

[daveyinaz]

Are you sure you meant to post this question in this section?

 

[dead_devil_66]

Are you sure you meant to post this question in this section?

im not sure where to put this...sorry...can anyone help me? >.<

 

[dead_devil_66]

hello? =/

 

[dead_devil_66]

seems this is too difficult even for this forum :/