1. The problem statement, all variables and given/known data
I have a universal / cardan joint with the first axis, [itex]\phi_1[/itex], spinning at some known angular velocity, [itex]\omega_1^x[/itex], where [itex]\omega_1^x[/itex] is the angular velocity in the [itex]\hat{x}[/itex] direction. The coordinate system is right handed with the [itex]\hat{x}[/itex] axis oriented along the rotational axis. I need to come up with a representation of [itex]{\bf\omega_2}[/itex], which is the angular velocity in the second frame, frame 2. Frame 2 is oriented along the 2nd axle, and is rotated by [itex]\alpha_1[/itex] around the z-axis, and then [itex]\alpha_2[/itex] along the intermediate y-axis. The physical system this embodies is a 2-DoF gyroscope.
2. Relevant equations
The rotational matrices are needed:
(Note, I'm going to use the short-short-hand notation for the
trig identities: s=sin,c=cos,t=tan,sc=sec,cs=csc,ct=cot)
[tex] R_x(\theta)\left( \begin{array}{ccc}
1 & 0 & 0 \\
0 & c(\theta) & -s(\theta) \\
0 & s(\theta) & c(\theta) \end{array} \right)[/tex]
[tex] R_y(\theta)\left( \begin{array}{ccc}
c(\theta) & 0 & s(\theta) \\
0 & 1 & 0 \\
-s(\theta) & 0 & c(\theta) \end{array} \right)[/tex]
[tex] R_z(\theta)\left( \begin{array}{ccc}
c(\theta) & -s(\theta) & 0 \\
s(\theta) & c(\theta) & 0 \\
0 & 0 & 1 \end{array} \right)[/tex]
3. The attempt at a solution
So I believe I understand the solution. My real question has to do with angular velocity as a vector and as a scalar quantity, and that's farther down. I'll reproduce the derivation for the answer here for reference though. Basically I'm solving for the angular velocity of the 2nd axle by performing two identical, but different, rotations. The first is:
[tex] Rot_1 = R_x(-\phi_2)\cdot R_z(\beta)\cdot R_x(\phi)
\\=\pmatrix{c \left(\beta\right)&-s \left(\beta\right)c
\left(\mathrm{\phi_1}\right)&s \left(\beta\right)s \left(
\mathrm{\phi_1}\right)\cr s \left(\beta\right)c \left(
\mathrm{\phi_2}\right)&s \left(\mathrm{\phi_1}\right)s \left(
\mathrm{\phi_2}\right)+c \left(\beta\right)c \left(
\mathrm{\phi_1}\right)c \left(\mathrm{\phi_2}\right)&c \left(
\mathrm{\phi_1}\right)s \left(\mathrm{\phi_2}\right)-c \left(
\beta\right)s \left(\mathrm{\phi_1}\right)c \left(
\mathrm{\phi_2}\right)\cr -s \left(\beta\right)s \left(
\mathrm{\phi_2}\right)&s \left(\mathrm{\phi_1}\right)c \left(
\mathrm{\phi_2}\right)-c \left(\beta\right)c \left(
\mathrm{\phi_1}\right)s \left(\mathrm{\phi_2}\right)&c \left(
\beta\right)s \left(\mathrm{\phi_1}\right)s \left(
\mathrm{\phi_2}\right)+c \left(\mathrm{\phi_1}\right)c \left(
\mathrm{\phi_2}\right)\cr }[/tex]
The second is:
[tex]Rot_2=R_y(\alpha_2)\cdot R_z(\alpha_1)
=\pmatrix{c \left(\mathrm{\alpha_1}\right)
c \left(\mathrm{\alpha_2}\right)&-s \left(\mathrm{\alpha_1}
\right)c \left(\mathrm{\alpha_2}\right)&s \left(
\mathrm{\alpha_2}\right)\cr s \left(\mathrm{\alpha_1}\right)&c\
\left(\mathrm{\alpha_1}\right)&0\cr -c
\left(\mathrm{\alpha_1}\right)
s \left(\mathrm{\alpha_2}\right)&s \left(\mathrm{\alpha_1}
\right)s \left(\mathrm{\alpha_2}\right)&c \left(
\mathrm{\alpha_2}\right)\cr }[/tex]
Since these transformations start and end in the same location, we can equate them. Stealing the matrix element from row 2, column 3, we see that:
[tex] c(\phi_1)s(\phi_2)-c(\beta)s(\phi_1)c(\phi_2) = 0[/tex]
Reducing away gives us our relationship between [itex]\phi_1[/itex] and [itex]\phi_2[/itex], albiet out of phase of what wikipedia says. But that should be okay, i think.
[tex] t(\phi_2) = c(\beta)t(\phi_1).[/tex]
We can take the derivative w/rt to time of the above equation and voila, we get the angular velocity of the 2nd axle. Woo non-constant velocity joints!
4. My real question
So here's where my real question is. I can take the derivative of the above equation w/rt to time and get [itex]\omega_2^x=\frac{d\phi_2}{dt}[/itex], which should be the magnitude of the angular velocity along the x-axis in frame 2:
[tex] \omega_2^x = \frac{c(\beta)\omega_1^x}{1-s^2(\phi_1)s^2(\beta)}[/tex] However, my understanding is that angular velocity is also a vector in its own right, so I should be able to rotate the angular velocity vector [itex]{\bf \omega_1}=<\omega_1^x,0,0>[/itex] though the transformation matrix and come out with a similar expression for angular velocity in the 2nd frame. If I rotate it through the first rotation, I get
[tex] {\bf \omega_2} = \pmatrix{c \left(\beta\right)\mathrm{\omega_1^x}\cr {{s
\left(\beta
\right)\mathrm{\omega_1^x}}\over{\sqrt{c \left(\beta\right)^2t
\left(\mathrm{\phi_1}\right)^2+1}}}\cr -{{c \left(\beta\right)
s \left(\beta\right)t \left(\mathrm{\phi_1}\right)
\mathrm{\omega_1^x}}\over{\sqrt{c \left(\beta\right)^2t \left(
\mathrm{\phi_1}\right)^2+1}}}\cr } [/tex]
Which does not equal my earlier assessment for x-component of the angular velocity in frame 2. I'm clearly missing something fundamental; both formulations appear to be right, but there's a misunderstanding that I don't see. I would think that by rotating the angular velocity in frame 1 to frame 2 I should get the same exact thing as the derivation in part 2. I would appreciate someone to weigh in on this for me, and explain it a little better, cause I'm at a loss as of right now.
I have a universal / cardan joint with the first axis, [itex]\phi_1[/itex], spinning at some known angular velocity, [itex]\omega_1^x[/itex], where [itex]\omega_1^x[/itex] is the angular velocity in the [itex]\hat{x}[/itex] direction. The coordinate system is right handed with the [itex]\hat{x}[/itex] axis oriented along the rotational axis. I need to come up with a representation of [itex]{\bf\omega_2}[/itex], which is the angular velocity in the second frame, frame 2. Frame 2 is oriented along the 2nd axle, and is rotated by [itex]\alpha_1[/itex] around the z-axis, and then [itex]\alpha_2[/itex] along the intermediate y-axis. The physical system this embodies is a 2-DoF gyroscope.
2. Relevant equations
The rotational matrices are needed:
(Note, I'm going to use the short-short-hand notation for the
trig identities: s=sin,c=cos,t=tan,sc=sec,cs=csc,ct=cot)
[tex] R_x(\theta)\left( \begin{array}{ccc}
1 & 0 & 0 \\
0 & c(\theta) & -s(\theta) \\
0 & s(\theta) & c(\theta) \end{array} \right)[/tex]
[tex] R_y(\theta)\left( \begin{array}{ccc}
c(\theta) & 0 & s(\theta) \\
0 & 1 & 0 \\
-s(\theta) & 0 & c(\theta) \end{array} \right)[/tex]
[tex] R_z(\theta)\left( \begin{array}{ccc}
c(\theta) & -s(\theta) & 0 \\
s(\theta) & c(\theta) & 0 \\
0 & 0 & 1 \end{array} \right)[/tex]
3. The attempt at a solution
So I believe I understand the solution. My real question has to do with angular velocity as a vector and as a scalar quantity, and that's farther down. I'll reproduce the derivation for the answer here for reference though. Basically I'm solving for the angular velocity of the 2nd axle by performing two identical, but different, rotations. The first is:
[tex] Rot_1 = R_x(-\phi_2)\cdot R_z(\beta)\cdot R_x(\phi)
\\=\pmatrix{c \left(\beta\right)&-s \left(\beta\right)c
\left(\mathrm{\phi_1}\right)&s \left(\beta\right)s \left(
\mathrm{\phi_1}\right)\cr s \left(\beta\right)c \left(
\mathrm{\phi_2}\right)&s \left(\mathrm{\phi_1}\right)s \left(
\mathrm{\phi_2}\right)+c \left(\beta\right)c \left(
\mathrm{\phi_1}\right)c \left(\mathrm{\phi_2}\right)&c \left(
\mathrm{\phi_1}\right)s \left(\mathrm{\phi_2}\right)-c \left(
\beta\right)s \left(\mathrm{\phi_1}\right)c \left(
\mathrm{\phi_2}\right)\cr -s \left(\beta\right)s \left(
\mathrm{\phi_2}\right)&s \left(\mathrm{\phi_1}\right)c \left(
\mathrm{\phi_2}\right)-c \left(\beta\right)c \left(
\mathrm{\phi_1}\right)s \left(\mathrm{\phi_2}\right)&c \left(
\beta\right)s \left(\mathrm{\phi_1}\right)s \left(
\mathrm{\phi_2}\right)+c \left(\mathrm{\phi_1}\right)c \left(
\mathrm{\phi_2}\right)\cr }[/tex]
The second is:
[tex]Rot_2=R_y(\alpha_2)\cdot R_z(\alpha_1)
=\pmatrix{c \left(\mathrm{\alpha_1}\right)
c \left(\mathrm{\alpha_2}\right)&-s \left(\mathrm{\alpha_1}
\right)c \left(\mathrm{\alpha_2}\right)&s \left(
\mathrm{\alpha_2}\right)\cr s \left(\mathrm{\alpha_1}\right)&c\
\left(\mathrm{\alpha_1}\right)&0\cr -c
\left(\mathrm{\alpha_1}\right)
s \left(\mathrm{\alpha_2}\right)&s \left(\mathrm{\alpha_1}
\right)s \left(\mathrm{\alpha_2}\right)&c \left(
\mathrm{\alpha_2}\right)\cr }[/tex]
Since these transformations start and end in the same location, we can equate them. Stealing the matrix element from row 2, column 3, we see that:
[tex] c(\phi_1)s(\phi_2)-c(\beta)s(\phi_1)c(\phi_2) = 0[/tex]
Reducing away gives us our relationship between [itex]\phi_1[/itex] and [itex]\phi_2[/itex], albiet out of phase of what wikipedia says. But that should be okay, i think.
[tex] t(\phi_2) = c(\beta)t(\phi_1).[/tex]
We can take the derivative w/rt to time of the above equation and voila, we get the angular velocity of the 2nd axle. Woo non-constant velocity joints!
4. My real question
So here's where my real question is. I can take the derivative of the above equation w/rt to time and get [itex]\omega_2^x=\frac{d\phi_2}{dt}[/itex], which should be the magnitude of the angular velocity along the x-axis in frame 2:
[tex] \omega_2^x = \frac{c(\beta)\omega_1^x}{1-s^2(\phi_1)s^2(\beta)}[/tex] However, my understanding is that angular velocity is also a vector in its own right, so I should be able to rotate the angular velocity vector [itex]{\bf \omega_1}=<\omega_1^x,0,0>[/itex] though the transformation matrix and come out with a similar expression for angular velocity in the 2nd frame. If I rotate it through the first rotation, I get
[tex] {\bf \omega_2} = \pmatrix{c \left(\beta\right)\mathrm{\omega_1^x}\cr {{s
\left(\beta
\right)\mathrm{\omega_1^x}}\over{\sqrt{c \left(\beta\right)^2t
\left(\mathrm{\phi_1}\right)^2+1}}}\cr -{{c \left(\beta\right)
s \left(\beta\right)t \left(\mathrm{\phi_1}\right)
\mathrm{\omega_1^x}}\over{\sqrt{c \left(\beta\right)^2t \left(
\mathrm{\phi_1}\right)^2+1}}}\cr } [/tex]
Which does not equal my earlier assessment for x-component of the angular velocity in frame 2. I'm clearly missing something fundamental; both formulations appear to be right, but there's a misunderstanding that I don't see. I would think that by rotating the angular velocity in frame 1 to frame 2 I should get the same exact thing as the derivation in part 2. I would appreciate someone to weigh in on this for me, and explain it a little better, cause I'm at a loss as of right now.
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