Mixing
Direction Cosines and Rotation Angles
$begingroup$
I'm rotating an object in $3D$ space with respect to a relative base, or reference frame. I'm using a normal vector to represent the rotation angles.
Suppose you have an object parallel to the reference frame; in this case the object has the following normal vector: $v=[0, 0, 1]$. So, the normal vector makes $90^circ$ with x-axis and y-axis and $0^circ$ with z-axis. (I suppose z-axis is the principle axis).
My question is: If we rotate the object $30^circ$ around x-axis; what is the direction normal and the angles? is the change only on the angle around x-axis or z-axis also. I noted when I rotate $30^circ$ around x-axis the z-axis also rotated with the same amount. This confused me.
Any Help!.
geometry matrices 3d
edited Jul 11 '15 at 10:57
Harish Chandra Rajpoot
29.6k103772
asked Nov 11 '13 at 22:22
CS2013CS2013
365
$endgroup$
add a comment |
$begingroup$
I'm rotating an object in $3D$ space with respect to a relative base, or reference frame. I'm using a normal vector to represent the rotation angles.
Suppose you have an object parallel to the reference frame; in this case the object has the following normal vector: $v=[0, 0, 1]$. So, the normal vector makes $90^circ$ with x-axis and y-axis and $0^circ$ with z-axis. (I suppose z-axis is the principle axis).
My question is: If we rotate the object $30^circ$ around x-axis; what is the direction normal and the angles? is the change only on the angle around x-axis or z-axis also. I noted when I rotate $30^circ$ around x-axis the z-axis also rotated with the same amount. This confused me.
Any Help!.
geometry matrices 3d
edited Jul 11 '15 at 10:57
Harish Chandra Rajpoot
29.6k103772
asked Nov 11 '13 at 22:22
CS2013CS2013
365
$endgroup$
$begingroup$
What matrix are you using and what's your final vector?
$endgroup$
– MasterOfBinary
Nov 11 '13 at 22:50
add a comment |
$begingroup$
I'm rotating an object in $3D$ space with respect to a relative base, or reference frame. I'm using a normal vector to represent the rotation angles.
Suppose you have an object parallel to the reference frame; in this case the object has the following normal vector: $v=[0, 0, 1]$. So, the normal vector makes $90^circ$ with x-axis and y-axis and $0^circ$ with z-axis. (I suppose z-axis is the principle axis).
My question is: If we rotate the object $30^circ$ around x-axis; what is the direction normal and the angles? is the change only on the angle around x-axis or z-axis also. I noted when I rotate $30^circ$ around x-axis the z-axis also rotated with the same amount. This confused me.
Any Help!.
geometry matrices 3d
edited Jul 11 '15 at 10:57
Harish Chandra Rajpoot
29.6k103772
asked Nov 11 '13 at 22:22
CS2013CS2013
365
$endgroup$
I'm rotating an object in $3D$ space with respect to a relative base, or reference frame. I'm using a normal vector to represent the rotation angles.
Suppose you have an object parallel to the reference frame; in this case the object has the following normal vector: $v=[0, 0, 1]$. So, the normal vector makes $90^circ$ with x-axis and y-axis and $0^circ$ with z-axis. (I suppose z-axis is the principle axis).
My question is: If we rotate the object $30^circ$ around x-axis; what is the direction normal and the angles? is the change only on the angle around x-axis or z-axis also. I noted when I rotate $30^circ$ around x-axis the z-axis also rotated with the same amount. This confused me.
Any Help!.
geometry matrices 3d
geometry matrices 3d
edited Jul 11 '15 at 10:57
Harish Chandra Rajpoot
29.6k103772
asked Nov 11 '13 at 22:22
CS2013CS2013
365
edited Jul 11 '15 at 10:57
Harish Chandra Rajpoot
29.6k103772
asked Nov 11 '13 at 22:22
CS2013CS2013
365
edited Jul 11 '15 at 10:57
Harish Chandra Rajpoot
29.6k103772
edited Jul 11 '15 at 10:57
Harish Chandra Rajpoot
29.6k103772
edited Jul 11 '15 at 10:57
Harish Chandra Rajpoot
29.6k103772
29.6k103772
asked Nov 11 '13 at 22:22
CS2013CS2013
365
asked Nov 11 '13 at 22:22
CS2013CS2013
365
asked Nov 11 '13 at 22:22
CS2013CS2013
365
365
$begingroup$
What matrix are you using and what's your final vector?
$endgroup$
– MasterOfBinary
Nov 11 '13 at 22:50
add a comment |
$begingroup$
What matrix are you using and what's your final vector?
$endgroup$
– MasterOfBinary
Nov 11 '13 at 22:50
$begingroup$
What matrix are you using and what's your final vector?
$endgroup$
– MasterOfBinary
Nov 11 '13 at 22:50
$begingroup$
What matrix are you using and what's your final vector?
$endgroup$
– MasterOfBinary
Nov 11 '13 at 22:50
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
I'll assume you mean the object's position when you say $v = (0,0,1)$. Since the object lies on the $z$-axis, rotation of $30$ degrees about $x$-axis, then the direction normal changes with $theta$ as it goes from $0$ to $30$, and $x$ gets fixed:
In a right-hand-rule system, rotation around $x$-axis analogous to standard rotation in the $xy$-plane (around $z$-axis), we have that the $y$-axis takes the place of $x$-axis, and $z$-axis takes the place of $y$-axis.
$$
R_{theta}(x,y,z) = (x, e^{itheta}cdot (y + zi))
$$
i.e. treat $(y,z)$ like a complex number and multiply to get the $costheta, sintheta$ formulas.
The unit vector tangent to the direction of motion starts out as the $-hat{y}$ (the unit vector parallel to $y$-axis), and progresses with $theta$ as
$$
R_{theta}(0, 1, 0) = (0, e^{itheta} (1 + 0i)) = (0, costheta, sintheta)
$$
Expanding the first one also gives you a vector normal to the direction of rotation, but clearly there are infinitely many normal vectors to the circular path to choose from.
answered Nov 12 '13 at 0:10
BananaCatsBananaCats
9,25552458
$endgroup$
add a comment |
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1 Answer
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1 Answer
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oldest
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$begingroup$
I'll assume you mean the object's position when you say $v = (0,0,1)$. Since the object lies on the $z$-axis, rotation of $30$ degrees about $x$-axis, then the direction normal changes with $theta$ as it goes from $0$ to $30$, and $x$ gets fixed:
In a right-hand-rule system, rotation around $x$-axis analogous to standard rotation in the $xy$-plane (around $z$-axis), we have that the $y$-axis takes the place of $x$-axis, and $z$-axis takes the place of $y$-axis.
$$
R_{theta}(x,y,z) = (x, e^{itheta}cdot (y + zi))
$$
i.e. treat $(y,z)$ like a complex number and multiply to get the $costheta, sintheta$ formulas.
The unit vector tangent to the direction of motion starts out as the $-hat{y}$ (the unit vector parallel to $y$-axis), and progresses with $theta$ as
$$
R_{theta}(0, 1, 0) = (0, e^{itheta} (1 + 0i)) = (0, costheta, sintheta)
$$
Expanding the first one also gives you a vector normal to the direction of rotation, but clearly there are infinitely many normal vectors to the circular path to choose from.
answered Nov 12 '13 at 0:10
BananaCatsBananaCats
9,25552458
$endgroup$
add a comment |
$begingroup$
I'll assume you mean the object's position when you say $v = (0,0,1)$. Since the object lies on the $z$-axis, rotation of $30$ degrees about $x$-axis, then the direction normal changes with $theta$ as it goes from $0$ to $30$, and $x$ gets fixed:
In a right-hand-rule system, rotation around $x$-axis analogous to standard rotation in the $xy$-plane (around $z$-axis), we have that the $y$-axis takes the place of $x$-axis, and $z$-axis takes the place of $y$-axis.
$$
R_{theta}(x,y,z) = (x, e^{itheta}cdot (y + zi))
$$
i.e. treat $(y,z)$ like a complex number and multiply to get the $costheta, sintheta$ formulas.
The unit vector tangent to the direction of motion starts out as the $-hat{y}$ (the unit vector parallel to $y$-axis), and progresses with $theta$ as
$$
R_{theta}(0, 1, 0) = (0, e^{itheta} (1 + 0i)) = (0, costheta, sintheta)
$$
Expanding the first one also gives you a vector normal to the direction of rotation, but clearly there are infinitely many normal vectors to the circular path to choose from.
answered Nov 12 '13 at 0:10
BananaCatsBananaCats
9,25552458
$endgroup$
add a comment |
$begingroup$
I'll assume you mean the object's position when you say $v = (0,0,1)$. Since the object lies on the $z$-axis, rotation of $30$ degrees about $x$-axis, then the direction normal changes with $theta$ as it goes from $0$ to $30$, and $x$ gets fixed:
In a right-hand-rule system, rotation around $x$-axis analogous to standard rotation in the $xy$-plane (around $z$-axis), we have that the $y$-axis takes the place of $x$-axis, and $z$-axis takes the place of $y$-axis.
$$
R_{theta}(x,y,z) = (x, e^{itheta}cdot (y + zi))
$$
i.e. treat $(y,z)$ like a complex number and multiply to get the $costheta, sintheta$ formulas.
The unit vector tangent to the direction of motion starts out as the $-hat{y}$ (the unit vector parallel to $y$-axis), and progresses with $theta$ as
$$
R_{theta}(0, 1, 0) = (0, e^{itheta} (1 + 0i)) = (0, costheta, sintheta)
$$
Expanding the first one also gives you a vector normal to the direction of rotation, but clearly there are infinitely many normal vectors to the circular path to choose from.
answered Nov 12 '13 at 0:10
BananaCatsBananaCats
9,25552458
$endgroup$
I'll assume you mean the object's position when you say $v = (0,0,1)$. Since the object lies on the $z$-axis, rotation of $30$ degrees about $x$-axis, then the direction normal changes with $theta$ as it goes from $0$ to $30$, and $x$ gets fixed:
In a right-hand-rule system, rotation around $x$-axis analogous to standard rotation in the $xy$-plane (around $z$-axis), we have that the $y$-axis takes the place of $x$-axis, and $z$-axis takes the place of $y$-axis.
$$
R_{theta}(x,y,z) = (x, e^{itheta}cdot (y + zi))
$$
i.e. treat $(y,z)$ like a complex number and multiply to get the $costheta, sintheta$ formulas.
The unit vector tangent to the direction of motion starts out as the $-hat{y}$ (the unit vector parallel to $y$-axis), and progresses with $theta$ as
$$
R_{theta}(0, 1, 0) = (0, e^{itheta} (1 + 0i)) = (0, costheta, sintheta)
$$
Expanding the first one also gives you a vector normal to the direction of rotation, but clearly there are infinitely many normal vectors to the circular path to choose from.
answered Nov 12 '13 at 0:10
BananaCatsBananaCats
9,25552458
answered Nov 12 '13 at 0:10
BananaCatsBananaCats
9,25552458
answered Nov 12 '13 at 0:10
BananaCatsBananaCats
9,25552458
answered Nov 12 '13 at 0:10
BananaCatsBananaCats
9,25552458
9,25552458
add a comment |
add a comment |
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$begingroup$
What matrix are you using and what's your final vector?
$endgroup$
– MasterOfBinary
Nov 11 '13 at 22:50