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3d projectile motion, with rotating reference frame

  1. Apr 29, 2010 #1
    1. The problem statement, all variables and given/known data

    I'm using matlab show the path of a projectile fired directly up(directly out from the center of the earth) from 41 degrees north latitude. Air resistance, variation of gravity cause by height, and the centrifugal force are for now ignored. What I'm trying to figure out is why there is a north deflection of the projectile. That is why does the object land north of where it was fired while there was no initial north/south velocity or displacement.
    Untitled-2.gif

    2. Relevant equations



    3. The attempt at a solution

    I've ruled out gravity because it will only act in the up/down direction (until the object is deflected north/south then it will have some impact). I'm thinking that it has to do with the up/northward direction of the angular velocity of the earth. It is perfectly parallel with the earths axis of rotation. Here are my plots.
    midtermfig2.jpg
    midtermfig1.jpg

    Any thoughts?
     
  2. jcsd
  3. Apr 30, 2010 #2
    What happens when it is fired from the southern hemisphere & what happens if you modell the Earth spinning in the opposite direction??
     
  4. Apr 30, 2010 #3
    And what are your units here (how high does the projectile reach)? If you have neglected the variation of height with gravity then maybe this assumption is not valid and you are getting the Northern deflection because of a computational reason.

    Can you post your code?? (commented if poss).
     
  5. Apr 30, 2010 #4
    One final question... the most important one I've asked... have you modelled the Earth's rotation??? If you have then I agree with the program for reasons I will disclose.
     
  6. Apr 30, 2010 #5
    I have modeled the earths rotation consistent with 2pi/1 day. The projectile reaches a height of about 4.5 x 10^5 meters. I haven't tried changing the direction of the rotation of the earth, as I'd rather not find all of my differential equations again. However I have found that when the z(up/down) velocity is positive, the projectile accelerates away from the equator and towards the pole. While if the z velocity is negative the projectile accelerates towards the equator. The acceleration in y(north/south) is directly relation in a parabolic curve to the velocity of z. Sometime this weekend I will add the forces neglected, air resistance, variation of gravity and the centrifugal force.

    I'm using a typical matlab ode solver, file with solver and plot statements has been left out.
    Here are my differential equations:

    function d_dt = rocketball(t,r)

    global w lambda g

    d_dt = zeros(4,1);

    d_dt(1)= r(2)
    d_dt(2)= 2.*(r(4).*w.*sin(lambda)-r(6).*w.*cos(lambda))
    d_dt(3)= r(4)
    d_dt(4)= -2.*r(2).*w.*sin(lambda)
    d_dt(5)= r(6)
    d_dt(6)= -g+2.*r(2).*w.*cos(lambda)

    where r(1) = x position, d_dt(1) = dx/dt
    r(2) = x velocity, d_dt(2) = dVx/dt
    r(3) = y position, d_dt(3) = dy/dt
    r(4) = y velocity, d_dt(4) = dVy/dt
    r(5) = z position, d_dt(5) = dz/dt
    r(6) = z velocity, d_dt(6) = dVz/dt

    lambda = 42 degrees north latitude, g = 9.81 m/s^2, w = angular velocity = 2pi/1 day

    Some interesting plots...
    figure4midter.jpg
    figure3midter.jpg


    thanks.
     
  7. Apr 30, 2010 #6
    "However I have found that when the z(up/down) velocity is positive, the projectile accelerates away from the equator and towards the pole. While if the z velocity is negative the projectile accelerates towards the equator."

    This statement is true in either hemisphere.
     
  8. Apr 30, 2010 #7
    OK, I want to confirm that your vector multiplication for the acceleration was:

    [tex]\ddot{\mathbf{r}} = - \dot{\mathbf{r}} \times \mathbf{\omega}[/tex]

    In which case only your [itex]a_y[/itex] has the correct sign. I was curious as to why you put a factor of 2 in front of these??

    Also, there is no need for you to model centrifugal force (you already have by your choice of co-ordinate system). Centrifugal force is a ficticious force that exists due to a frame of refrence that has angular momentum, you are "watching" the Earth rotating so you don't need to model this. To explain a little better, if there was was no gravity, your ball would just continue its path in the x direction. From the frame of reference of the Earth surface, this would be observed as some unknown force making the ball fly off into the sky (this force would be the centrifugal force)... I hope that explanation was adequate :P

    I believe it is this force that is causing your object to fly North (however, if you put the vector multiplication the correct way round it would end up South). This is inline with the apparent gravity experienced by objects on rotating masses.
     
  9. Apr 30, 2010 #8
    I'm actually using a rotating reference frame, that is the origin of my frame is at rest with the earth. The equation of motion in this rotational reference frame I'm using is:

    m[tex]r\ddot{}[/tex]= m[tex]\vec{}g[/tex]-2m[tex]\vec{}\omega[/tex]X[tex]\dot{}r[/tex]

    where...

    omega is parallel to the axis of rotation and in the j and k direction (up and north)
     
  10. Apr 30, 2010 #9
    <math display="block" xmlns="http://www.w3.org/1998/Math/MathML">
    <mrow>
    <mover accent="true">
    <mi>&omega;</mi>
    <mo>&rarr;</mo>
    </mover>
    <mo>=</mo>
    <mi>&omega;</mi>
    <mi>cos</mi><mo>&ApplyFunction;</mo>
    <mfenced open="(" close=")" separators=",">
    <mrow>
    <mi>&lambda;</mi>
     
    Last edited by a moderator: Apr 25, 2017
  11. Apr 30, 2010 #10
    that obviously didn't work...
     
  12. Apr 30, 2010 #11
    Ok so, in rotating reference frame:
    [tex]\frac{\rm{d}^2\mathbf{r'}}{\rm{d}t^2} = \frac{\rm{d}}{\rm{d}t}\left(\dot{\mathbf{r}} + \omega\times\mathbf{r}\right)[/tex]

    So expanding this out I get...

    [tex]\frac{\rm{d}^2\mathbf{r'}}{\rm{d}t^2} = \ddot{\mathbf{r}} + \mathbf{\omega}\times\dot{\mathbf{r}} + \mathbf{\omega}\times(\dot{\mathbf{r}} + \mathbf{\omega}\times\mathbf{r})[/tex]

    [tex]\frac{\rm{d}^2\mathbf{r'}}{\rm{d}t^2} = \ddot{\mathbf{r}} + 2\mathbf{\omega}\times\dot{\mathbf{r}} + \mathbf{\omega}\times( \mathbf{\omega}\times\mathbf{r})[/tex]

    So I think your equation should be

    [tex]m\ddot{\mathbf{r}} = -mg -2m(\mathbf{\omega}\times\dot{\mathbf{r}}) - m\mathbf{\omega}\times( \mathbf{\omega}\times\mathbf{r})[/tex]
     
  13. Apr 30, 2010 #12
    I gotta go now, I'll pick this up tomorrow if someone else doesn't... good luck
     
  14. Apr 30, 2010 #13
    yeah that's what I used, the last term the wx(wxr) is the centrifugal force I believe which is absent in my first calculation. Thanks for your help
     
  15. May 1, 2010 #14
    Good good, that's what I was about to say, the centrifugal force is modelled and doesn't need to be added. OK, that just leaves the sign of the vector product under question, I still think that your accelerations from the vector product term should have a minus (I previously thought the y-acceleration needed a minus too but I think that should be positive now).
     
  16. May 1, 2010 #15

    Cleonis

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    Gold Member

    Generally the contribution from the centrifugal term is larger than the contribution from the coriolis term. In the vast majority of scenario's the projectile will land significantly south of the latitude where it was fired from

    As a means of verification you can set up a parallel compuation that evaluates the trajectory relative to the inertial coordinate system. The total initial velocity is the vector sum of the gun's nozzle velocity and the prior-to-launch velocity of co-rotating with the Earth. With the trajectory known you can transform those data to the rotating coordinate system.

    The two computation strategies are comparable in complexity; in the case of the inertial coordinate system you have only a single acceleration factor: gravity; in the case of the rotating coordinate system you have to evaluate three acceleration terms: gravity, centrifugal, coriolis.
     
  17. May 1, 2010 #16
    thanks I might give that a try. When I set the initial conditions consist with a gun fired parallel to the earth's surface, the deflection is southward or towards the equator. But for some reason when the initial velocity is directly up the deflection is northward.
     
  18. May 1, 2010 #17

    Cleonis

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    Gold Member

    Just out of curiosity, what do you count as 'deflection' here?

    For instance, take the case of firing a projectile from the surface of a non-rotating planet, fired parallel to the surface. Will that projectile land on the latitude that it is fired from?
    ** Later edit: my apologies, the statement is incomplete. I mean: fired parallel to the surface and parallel to the latitude line **
    Answer: No: the projectile will always land closer to the equator than the latitude it is fired from.

    Think of the line of intersection of a plane and a sphere (with the plane going through the sphere's geometric center.) Let that plane be tangent to the circle of 41 degrees latitude. To either side that line of intersection extends towards the equator.

    So that's not a deflection. A projectile in orbital motion does not deflect (it's deflected down, by gravity, but not sideways). But the projectile does proceed towards the equator.


    - On a non-rotating planet, if you fire straight up (perpendicular to the local surface), the projectile will land on the spot where it was fired from.
    - On a rotating planet, if you fire straight up the projectile has a tangential velocity that comes from co-rotating with the planet prior to the launch. For that reason alone it will proceed to the equator.

    Hence my question: what do you count as 'deflection'?
     
    Last edited: May 2, 2010
  19. May 1, 2010 #18
    I suppose by deflection I mean any movement in the x-y plane where the initial velocity is in the z direction.
     
  20. May 1, 2010 #19
    The direction of the motion(I'm going to restrain from calling it deflection) of a particle in the x-y plane, parallel to the surface of the earth, when given some initial z(up) velocity or displacement is dependent upon those initial conditions.

    For example:

    An object dropped from a tall building will land east of its initial east/west displacement. However when an object is fired upwards instead it will land west of where it began.

    In my model the object launched directly up in the northern hemisphere lands north of its original position. Are you saying that this is incorrect?

    figure3midter.jpg
     
  21. May 1, 2010 #20

    Cleonis

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    Gold Member

    Yeah, dead certain the model is incorrect if it does that; it goes against physical law.


    I repeat: a ballistic trajectory is orbiting motion (except that the orbit is too close, and the projectile impacts on its way back.)

    Orbiting motion is planar. You intersect the plane of the orbit with the Earth's surface; that line of intersection is a great circle. That great circle is the groundtrack of the orbiting motion. That groundtrack is tangent to the starting latitude. So in the case of firing straight up (perpendicular to the local surface) there is no way the projectile can land further away from the equator than the starting latitude.
     
    Last edited: May 1, 2010
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