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Math Help - direction from angles

  1. #1
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    direction from angles

    The should be a simple one.

    I've got a: an unit vector pointing -1 down Y
    and b: three xyz rotation angles.

    I want to rotate my initialized vector by the three angles, resulting in a new direction.
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  2. #2
    Rhymes with Orange Chris L T521's Avatar
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    Quote Originally Posted by travis View Post
    The should be a simple one.

    I've got a: an unit vector pointing -1 down Y
    and b: three xyz rotation angles.

    I want to rotate my initialized vector by the three angles, resulting in a new direction.
    Since the vector is -\bold j. the direction angles for this vector are \alpha_0=90^{\circ}, \beta_0=180^{\circ}, and \gamma_0=90^{\circ}.

    What are the three rotation angles ( \alpha=?, \beta=?, \gamma=?)???

    --Chris
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  3. #3
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    Well, they are arbitrary... What I'm looking for is the formula for finding a new direction based on any given angles.

    For the sake of argument let's say the angles are 45, 10, 90.
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  4. #4
    Rhymes with Orange Chris L T521's Avatar
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    Quote Originally Posted by travis View Post
    Well, they are arbitrary... What I'm looking for is the formula for finding a new direction based on any given angles.

    For the sake of argument let's say the angles are 45, 10, 90.
    I would assume that those angles are with respect to the positive coordinate axes?

    We would then need to use these equations:

    \cos\alpha=\frac{r_x}{||\bold r||}~~~~~~~~~~\cos\beta=\frac{r_y}{||\bold r||}~~~~~~~~~~\cos\gamma=\frac{r_z}{||\bold r||}

    where r_x, \ r_y, \text{ and }r_z are the components of the vector \bold r. These equations are used to find direction cosines. They come in useful here.

    In our case, \bold r=-\bold j

    We want \bold r to be rotated and have the new direction angles \alpha=45^{\circ},~\beta=10^{\circ},~\text{and}~\g  amma=90^{\circ}. So let's see how we would calculate the equation of the new vector.

    Let's find the x component first.

    Since \cos\alpha=\frac{r_x}{||\bold r||}, and \alpha=45^{\circ}~\text{and}~||\bold r||=1, we see that

    r_x=\cos(45^{\circ})=\frac{\sqrt{2}}{2}\approx\col  or{red}\boxed{.707}

    Let's now find the y component,

    Since \cos\beta=\frac{r_y}{||\bold r||}, and \beta=10^{\circ}~\text{and}~||\bold r||=1, we see that

    r_y=\cos(10^{\circ})\approx\color{red}\boxed{.985}

    Let's now find the z component,

    Since \cos\gamma=\frac{r_z}{||\bold r||}, and \gamma=90^{\circ}~\text{and}~||\bold r||=1, we see that

    r_z=\cos(90^{\circ})=\color{red}\boxed{0}

    So we now see that the "rotated" vector has the equation \color{red}\boxed{\bold r_n=.707\bold i+.985\bold j}

    Does this make sense?

    --Chris
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  5. #5
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    Thanks very much for this reply. I am going to have to chew on it a bit.
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  6. #6
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    Tested this and it works. Thanks again Chris, really helpful.

    Quote Originally Posted by Chris L T521 View Post
    I would assume that those angles are with respect to the positive coordinate axes?

    We would then need to use these equations:

    \cos\alpha=\frac{r_x}{||\bold r||}~~~~~~~~~~\cos\beta=\frac{r_y}{||\bold r||}~~~~~~~~~~\cos\gamma=\frac{r_z}{||\bold r||}

    where r_x, \ r_y, \text{ and }r_z are the components of the vector \bold r. These equations are used to find direction cosines. They come in useful here.

    In our case, \bold r=-\bold j

    We want \bold r to be rotated and have the new direction angles \alpha=45^{\circ},~\beta=10^{\circ},~\text{and}~\g  amma=90^{\circ}. So let's see how we would calculate the equation of the new vector.

    Let's find the x component first.

    Since \cos\alpha=\frac{r_x}{||\bold r||}, and \alpha=45^{\circ}~\text{and}~||\bold r||=1, we see that

    r_x=\cos(45^{\circ})=\frac{\sqrt{2}}{2}\approx\col  or{red}\boxed{.707}

    Let's now find the y component,

    Since \cos\beta=\frac{r_y}{||\bold r||}, and \beta=10^{\circ}~\text{and}~||\bold r||=1, we see that

    r_y=\cos(10^{\circ})\approx\color{red}\boxed{.985}

    Let's now find the z component,

    Since \cos\gamma=\frac{r_z}{||\bold r||}, and \gamma=90^{\circ}~\text{and}~||\bold r||=1, we see that

    r_z=\cos(90^{\circ})=\color{red}\boxed{0}

    So we now see that the "rotated" vector has the equation \color{red}\boxed{\bold r_n=.707\bold i+.985\bold j}

    Does this make sense?

    --Chris
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