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Math Help - Moving Frame velocity issues, the next generation ?

  1. #1
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    Moving Frame velocity issues, the next generation ?

    This is something I've been struggling for about 3 days now. Either I'm missing something or should go back to physics 101

    Scenario:
    Inertial frame of reference. This is Cartesian, right handed and denoted O.
    Moving and rotating non-inertial frame of reference. This is Cartesian, right handed and denoted A. This frame does not have a predefined

    path, velocity or "nicely defined" Omega. It basically does what it wants.
    An object moving in space, denoted B.

    Givens (ALL DATA GIVEN IN THE INERTIAL SYSTEM O):
    Position and velocities of the object, i.e. Px,Py,Pz,Vx,Vy,Yz.
    Position and velocities of the the A frame.
    and here's the catch: direction of the unit axial vectors of A. if i,j and k are the (unit vector) axis of the non-inertial system A and x, y and z are

    those of O, we have only i(x,y,z), j(x,y,z) and k(x,y,z).
    We do not have explicit Omega, or rate of angular rotation. I hope this can be derived from the direction of the unit vectors of A.
    We have real time data flowing, meaning we can have an instatenous dataset, applicable to now and the last dataset. no history data available.
    This is actually a part of a real-time system.

    Need to derive:
    Position and Velocity of B in the A frame. The challenge is not being given Omega.

    Thanks, Gabriel
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  2. #2
    Forum Admin topsquark's Avatar
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    Quote Originally Posted by zazkapulsk View Post
    This is something I've been struggling for about 3 days now. Either I'm missing something or should go back to physics 101

    Scenario:
    Inertial frame of reference. This is Cartesian, right handed and denoted O.
    Moving and rotating non-inertial frame of reference. This is Cartesian, right handed and denoted A. This frame does not have a predefined

    path, velocity or "nicely defined" Omega. It basically does what it wants.
    An object moving in space, denoted B.

    Givens (ALL DATA GIVEN IN THE INERTIAL SYSTEM O):
    Position and velocities of the object, i.e. Px,Py,Pz,Vx,Vy,Yz.
    Position and velocities of the the A frame.
    and here's the catch: direction of the unit axial vectors of A. if i,j and k are the (unit vector) axis of the non-inertial system A and x, y and z are

    those of O, we have only i(x,y,z), j(x,y,z) and k(x,y,z).
    We do not have explicit Omega, or rate of angular rotation. I hope this can be derived from the direction of the unit vectors of A.
    We have real time data flowing, meaning we can have an instatenous dataset, applicable to now and the last dataset. no history data available.
    This is actually a part of a real-time system.

    Need to derive:
    Position and Velocity of B in the A frame. The challenge is not being given Omega.

    Thanks, Gabriel
    Since we don't know the particulars of how frame A is moving (just that it is non-inertial) we cannot prescribe an equation between, say, the momenta in the A frame vs. the momenta in the O frame because we don't know what kinds of "ficticious forces" are present in the frame A.

    Alternately, we have the unit direction vectors for frame A in O coordinates, but no time evolution information.

    I don't think there is a solution to this in general.

    -Dan
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  3. #3
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    Then again...

    Thanks, Dan, for your quick reply.

    I beg to differ. It's right we have no information as to the nature of A's movement, but I think we don't need that.

    We have to find Omega (vector). That shall solve all our problems, right? but as I was thinking a nit more about this, it came to me that Omega is always in a cross product with the position vector R in these equations.

    If you'll look at the three lines at the bottom of page 3 in the attachment here, one could argue that using the (time) derivatives of A's unit axis (base vectors) we can construct that cross product, without calculating Omega explicitly. I could do that by taking to time-sequential readings and dividing the difference of the axis by the time between the reading... Right ?

    Thanks
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  4. #4
    Forum Admin topsquark's Avatar
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    Quote Originally Posted by zazkapulsk View Post
    Thanks, Dan, for your quick reply.

    I beg to differ. It's right we have no information as to the nature of A's movement, but I think we don't need that.

    We have to find Omega (vector). That shall solve all our problems, right? but as I was thinking a nit more about this, it came to me that Omega is always in a cross product with the position vector R in these equations.

    If you'll look at the three lines at the bottom of page 3 in the attachment here, one could argue that using the (time) derivatives of A's unit axis (base vectors) we can construct that cross product, without calculating Omega explicitly. I could do that by taking to time-sequential readings and dividing the difference of the axis by the time between the reading... Right ?

    Thanks
    Ah. You said you have i(x, y, z), j(x, y, z), k(x, y, z), but you didn't mention you have the time dependence. (That would mean you actually have i(x, y, z, t), etc. hence my confusion.)

    If you know these, then you do know what the frame A is doing: it's all contained in the unit vector equations.

    Set up the rotation matrix:
    R = Rx~Ry~Rz = \left ( \begin{matrix} \hat{i_O} \cdot \hat{i_A} & \hat{i_O} \cdot \hat{j_A} & \hat{i_O} \cdot \hat{k_A} \\ \hat{j_O} \cdot \hat{i_A} & \hat{j_O} \cdot \hat{j_A} & \hat{j_O} \cdot \hat{k_A} \\ \hat{k_O} \cdot \hat{i_A} & \hat{k_O} \cdot \hat{j_A} & \hat{k_O} \cdot \hat{k_A} \end{matrix} \right )

    Use this site for the definitions of Rx, Ry, and Rz. (Or you could use the Euler angles, if you prefer.) You will have to solve at least three trig equations for three unknown angles, so this is not going to be easy (unless your rotation is conveniently about, say, the z-axis), but it should be possible.

    -Dan
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  5. #5
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    Indeed

    thanks, Dan.
    My hero.

    Gabriel
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