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Thread: 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 $\displaystyle -\bold j$. the direction angles for this vector are $\displaystyle \alpha_0=90^{\circ}$, $\displaystyle \beta_0=180^{\circ}$, and $\displaystyle \gamma_0=90^{\circ}$.

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

    --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:

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

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

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

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

    Let's find the x component first.

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

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

    Let's now find the y component,

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

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

    Let's now find the z component,

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

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

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

    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:

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

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

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

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

    Let's find the x component first.

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

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

    Let's now find the y component,

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

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

    Let's now find the z component,

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

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

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

    Does this make sense?

    --Chris
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