Timeline for Why does $\delta \vec{r} = \delta \vec{ \theta} \times \vec{r}$?
Current License: CC BY-SA 4.0
14 events
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| Nov 12, 2023 at 1:31 | comment | added | peek-a-boo | @Eli I don’t really see the need for it, since basic calculus tells us there is obviously only one (up to additive constant) choice for $\vec{\theta}$ if we want its derivative to be $\omega$; but there, I explicitly edited to mention it. | |
| Nov 12, 2023 at 1:28 | history | edited | peek-a-boo | CC BY-SA 4.0 |
edited to clarify stuff discusses in comments
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| Nov 11, 2023 at 13:09 | comment | added | Eli | So please put it in your answer ? | |
| Nov 11, 2023 at 9:33 | comment | added | peek-a-boo | @Eli ok, and your equation is fine given what you defined (and on first glance our $\omega$’s agree too), but like I said, my $\delta\vec{\theta}$ is not the same as your triple of Euler angles. | |
| Nov 11, 2023 at 9:06 | comment | added | Eli | Peek please see my edit | |
| Nov 10, 2023 at 17:53 | comment | added | peek-a-boo | @Eli in my answer, the fundamental quantities are $R(t)$ and $\Omega(t)$ (hence by 3D specialities, $\omega(t)$). I could now define $\delta\vec{\theta}:=\omega(t)\cdot \delta t$ (an infinitesimal change in angle along a given axis, which changes with time), so the equation quoted becomes true by fiat (and of course, we can only define this angular displacement infinitesimally, otherwise it isn’t a vector). I chose $R(t),\Omega(t)$ to be the fundamental quantities, because as I mentioned in my first paragraph, I do not share the capacity to intuit these ‘weird’ diagrams appearing in textbooks. | |
| Nov 10, 2023 at 7:44 | comment | added | Eli | Your equation is correct, but how you obtain this result ?$~\dfrac{\delta \vec{r}}{\delta t} = \dfrac{\delta \vec{ \theta}}{\delta t} \times \vec{r}$ this equation is only valid for small angels | |
| Nov 9, 2023 at 23:30 | comment | added | peek-a-boo | anyway, there are so many conventions for defining these angular quantities it gets confusing (especially with the body/space issue as well, for instance by swapping the roles of body and space, one gets angular velocity to be $R(t)^{\top}\cdot\dot{R}(t)$, perhaps up to a sign which I can’t remember… more abstractly, this corresponds to either the left/right Maurer-Cartan form of the Lie group in question which gives the body/space interpretation). | |
| Nov 9, 2023 at 23:24 | comment | added | peek-a-boo | I just did the calculation quickly and if $\gamma(t)$ is a path of skew-symmetric matrices, and $f(z)$ denotes the entire function $\sum_{n=0}^{\infty}\frac{z^n}{(n+1)!}=\frac{e^z-1}{z}$, then for the rotations $R(t)=\exp(\gamma(t))$, we have that $\Omega(t)=\dot{R}(t)R(t)^{\top}=f([\gamma(t),\cdot])\cdot \dot{\gamma}(t)$, where $[\cdot,\cdot]$ denotes the commutator bracket. So indeed, if $\gamma(t)=0$ then we get $\Omega(t)=f(0)\cdot\dot{\gamma}(t)=I\cdot \dot{\gamma}(t)=\dot{\gamma}(t)$ (which is consistent with what you said regarding small angles). So, I don’t see any mistakes here. | |
| Nov 9, 2023 at 23:20 | comment | added | peek-a-boo | @Eli in your answer, the $\theta$’s are the Euler angles yes? So in that case, indeed it is not true that the angular velocity vector $\omega(t)$ is simply the derivatives of the Euler angles. This is consistent with everything we see in the usual textbook formulae as well (e.g Landau Lifshitz equation (35.1) on page 111) where the angular velocity is the time derivative of Euler angles multiplied by some sines and cosines. | |
| Nov 9, 2023 at 16:43 | comment | added | Eli | a boo, but $~\omega~$ is not $~\frac{d\theta}{dt}~$ so your result is not correct ? See my answer | |
| Nov 8, 2023 at 5:15 | history | edited | hft | CC BY-SA 4.0 |
added 2 characters in body
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| Nov 8, 2023 at 4:36 | history | edited | peek-a-boo | CC BY-SA 4.0 |
added 329 characters in body
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| Nov 8, 2023 at 4:29 | history | answered | peek-a-boo | CC BY-SA 4.0 |