OpenLLM-France/wikimedia huggingface dataset:
# Forbidden Kingdom (Strategic Simulation)/Player brief Government

This player brief is given to the Government player/team in Forbidden Kingdom (Strategic Simulation).

## Consultative Legislatures - Government Brief
### Year 1

Internet Access
* Limited access provides control needed in society. Support Party Line (Choice A)
Private Property Rights: Investigate Corruption and Restore Lands
* Government officials feel challenged by these investigations. Oppose them. Popular mood will be unaffected. (Choice B)

### Year 2

Nuclear Energy: Fix Location for Highly Enriched Waste (HEW) Facility
* This plan removes waste from populated areas. Support Party Line (Choice A)
People’s Liberation Army Navy (PLAN): Fleet Modernization
* While we’d like to support the military infrastructure, this money is needed in the provinces for important development projects. (Choice B)

### Year 3

Independence Referendum Passes on Taiwan
* Reconciliation of the dispute with Taiwan will be useful. The cities and provinces will suffer greatly if confrontation with the West develops further. Support Party Line (Choice A)
Currency: Basket of Funds
* Local governments are worried that a floating currency will provoke rapid price hikes of staples and cause food riots and other public outcries. The central bank should continue to control the currency. (Choice B)
# Open Pathology Project/Fat necrosis

Fat necrosis occurs due to the action of enzymes on fat.

## More information

* Category: Histopathology of liponecrosis
* Structured data: depicts fat necrosis
* w:en:Fat necrosis
* Libre Pathology: Fat necrosis
# Open Pathology Project/Endometrium

Endometrium is a layer of the uterus. It is biopsied in case of abnormal uterine bleeding.

## More information

* Category: Histology of endometrium
* Structured data: depicts endometrium
* Libre Pathology: Endometrium
* w:en:Endometrium
# Open Pathology Project/Eosinophils

Eosinophils are white blood cells. They combat parasites and certain types of infections.

## More information

* Category: Eosinophils
* w:en:Eosinophil
* Structured data: depicts Eosinophil
* Libre Pathology
# Open Pathology Project/Fibroadenoma

Fibroadenoma is a benign tumor of the breast.

## More information

* Category: Histopathology of breat fibroadenomas
* Libre Pathology: Fibroadenoma
* Structured data: depicts Fibroadenoma
* w:en:Fibroadenoma
# Open Pathology Project/Filarial lymphadenitis

Lymphadenitis caused due to flilariasis.

## More info

* Category: Histopathology of filariases
* Structured data: depicts filariasis
* en:w:filariasis
# Open Pathology Project/Emphysema

Emphysema is a lung condition where the alveoli are ruptured or damaged.

## More information

* Category: Histopathology of pulmonary emphysema
* Strucutred data: depicts emphysema
* Libre Pathology: Emphysema
* w:en:Emphysema
# Open Pathology Project/Fibroblasts

Fibroblasts are cells that have a prominent role in healing of wounds.

## More information

* Category: Eosinophils
* Structured data: depicts Fibroblast
* w:en:Fibroblast
# Open Pathology Project/Fibrocystic disease of breast

Fibrocystic disease of the breast is a condition with breast cysts, pain or breast masses.

## More information

* Category: Histopathology of fibrocystic breast changes
* Structured data: depicts fibrocystic breast
* w:en:Fibrocystic breast changes
* Libre Pathology: Fibrocystic breast changes
# Open Pathology Project/Dysgerminoma

Dysgerminoma is a germ cell tumor arising from the ovary.

## More information

* Category: Histopathology of ovarian dysgerminoma
* Structured data: depicts Ovarian dysgerminoma
* w:en:Dysgerminoma
* Libre Pathology: Ovarian dysgerminoma
# Open Pathology Project/Fatty liver

Fatty liver is a condition where excess fat is deposited in the liver.

## More information

* Category: Lipid steatosis
* Structured data: depicts Lipid steatosis
* w:en: Fatty liver disease
# Modern Organic Chemistry

The goal of this course is to cover important advanced topics that are relevant in modern organic chemistry.
# Modern wood stoves

Modern wood stoves are high-technology appliances that are designed to limit pollution and get better efficiency out of the wood burned.

## Links

* http://journeytoforever.org/at_woodfire.html - includes discussions of introducing efficient wood stoves in India
# Modulation
## Introduction

The purpose of a communication systems is to transmit information-bearing signals through a communication channel separating the transmitter from the receiver. Information-bearing signals are also referred to as baseband signals. The term baseband is used to designate the band of frequencies representing the original signal as delivered by a source of information. The proper use of the communication channel requires a shift of the range of baseband frequencies into other frequency ranges suitable for transmission, and a corresponding shift back to the original frequency range after reception. For example, a radio system must operate with frequencies of 30 kHz and upward, whereas the baseband signal usually contains frequencies in the audio frequency range, and so some form of frequency-band shifting must be used for the system to operate satisfactorily. A shift of the range of frequencies in a signal is accomplished by using modulation, which is defined as the process by which some characteristic of a carrier is varied in accordance with a modulating wave (signal). A common form of the carrier is a sinusoidal wave, in which case we speak of a continuous-wave modulation1 process. The baseband signal is referred to as the modulating wave, and the result of the modulation process is referred to as the modulated wave. Modulation is performed at the transmitting end of the communication system. At the receiving end of the system, we usually require the original baseband signal to be restored. This is accomplished by using a process known as demodulation, which is the reverse of the modulation process.
# PlanetPhysics/Representation of a CcG Topological Algebra

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and then also an operator weak topology to be defined on $B(\mathbb {H} )$.
# PlanetPhysics/Representations of Groupoids Induced by Measure

A groupoid representation induced by measure can be defined as measure induced operators or as operators induced by a measure preserving map in the context of Haar systems with measure associated with locally compact groupoids, $\mathbf {G_{lc}}$. Thus, let us consider a locally compact groupoid $\mathbf {G_{lc}}$ endowed with an associated Haar system $\nu =\left\{\nu ^{u},u\in U_{\mathbf {G_{lc}} }\right\}$, and $\mu$ a quasi-invariant measure on $U_{\mathbf {G_{lc}} }$. Moreover, let $(X_{1},{\mathfrak {B}}_{1},\mu _{1})$ and $(X_{2},{\mathfrak {B}}_{2},\mu _{2})$ be measure spaces and denote by $L^{0}(X_{1})$ and $L^{0}(X_{2})$ the corresponding spaces of measurable functions (with values in $\mathbb {C}$). Let us also recall that with a measure-preserving transformation $T:X_{1}\longrightarrow X_{2}$ one can define an operator induced by a measure preserving map , $U_{T}:L^{0}(X_{2})\longrightarrow L^{0}(X_{1})$ as follows.
\begin{displaymath} (U_T f)(x):=f(Tx)\,, \qquad\qquad f \in L^0(X_2),\; x \in X_1 \end{displaymath}
Next, let us define $\nu =\int \nu ^{u}d\mu (u)$ and also define $\nu ^{-1}$ as the mapping $x\mapsto x^{-1}$. With $f\in C_{c}(\mathbf {G_{lc}} )$, one can now define the measure induced operator $'''Ind'''\mu (f)$ as an operator being defined on $L^{2}(\nu ^{-1})$ by the formula: $'''Ind'''\mu (f)\xi (x)=\int f(y)\xi (y^{-1}x)d\nu ^{r(x)}(y)=f*\xi (x)$
Remark:
One can readily verify that :
$\left\|'''Ind'''\mu (f)\right\|\leq \left\|f\right\|_{1},$
and also that $'''Ind'''\mu$ is a proper representation of $C_{c}(\mathbf {G_{lc}} )$, in the sense that the latter is usually defined for groupoids.
# PlanetPhysics/Resistors in Parallel

Resistors in parallel can be replaced in a circuit with the equivalent resistance
   ${\frac {1}{R_{eq}}}=\sum _{i}{\frac {1}{R_{i}}}$
# PlanetPhysics/Resonance

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The very sharp (or marked) increase in amplitude of oscillation of a mechanical, electrical or nuclear system $O_{1}$ exposed to a periodic force $F_{r}(t)$, or another oscillator $O_{2}$ whose frequency is equal to, or very close to, the natural undamped frequency of the oscillating system $O_{1}$. The intensity of oscillation is defined as the square of the amplitude of the oscillations. More precisely, in the case of a linear oscillator with a resonance frequency $\omega _{0}$, the intensity of oscillations $I(\omega )$ when the system $O_{1}$ is driven with a driving frequency $\omega _{2}$ is given by:
$I(\omega _{2})=const.{\frac {\gamma /2}{(\omega _{2}-\omega _{0})^{2}+\left({\frac {\gamma }{2}}\right)^{2}}}.$
This oscillator intensity is a Lorentzian function which is characteristic of many resonant systems, with $\gamma$ being known as the linewidth of the resonance , or resonance linewidth at half-heigth of the resonance peak which depends on the degree of damping of the oscillator.
Heavily damped oscillators have broad linewidths, and will respond to a wider range of driving frequencies near the resonance frequency. The resonance linewidth $\gamma$ is inversely proportional to the quality factor , or $Q$-factor of the oscillating systems, which is a measure of the sharpness of their resonance. As an example, for NMR and rf electrical circuit probes, a high Q-factor is essential for high sensitivity detection of the signal by the rf probe.
A system is called resonant if it has the property of oscillating by itself (without external coupling) with a maximum amplitude only at a certain frequency or frequencies, known as the resonance frequencies of the system.
# PlanetPhysics/Relativistic Total Energy

The relativistic total energy E of a particle is defined as
   $E={\frac {mc^{2}}{\sqrt {1-v^{2}/c^{2}}}}$
One often finds this relationship in the limit of low speeds $(v<<c)$ with the famous equation
   $E_{0}=mc^{2}$
# PlanetPhysics/Representations of Canonical Anti Commutation Relations CAR

Thsi is a contributed topic in progress on representations of anti-commutation relations (CAR). (See also previous entries on the representations of canonical commutation and anti-commutation relations (CCR, CCAR)).

### Representations of Canonical Anti-commutation Relations (CAR)
#### CAR Representations in a Non-Abelian Gauge Theory

One can also provide a representation of canonical anti-commutation relations in a non-Abelian gauge theory defined on a non-simply connected region in the two-dimensional Euclidean space. Such representations were shown to provide also a mathematical expression for the non-Abelian, Aharonov-Bohm effect. Supersymmetry theories admit both CAR and CCR representations. Note also the connections of such representations to locally compact quantum groupoid representations.
# PlanetPhysics/Resistors in Series

Resistors in series can be replaced in a circuit with the equivalent resistance
   $R_{eq}=\sum _{i}R_{i}$
# PlanetPhysics/Representation of Locally Compact Groupoids

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its associated Haar system $\nu$ is defined as a triple Failed to parse (unknown function "\grp"): {\displaystyle (\mu, U_{\grp_{lc}} * \mathbb{H}, L)} , where: $\mu$ is a quasi-invariant measure defined over Failed to parse (unknown function "\grp"): {\displaystyle U_{\grp_{lc}}} ,
Failed to parse (unknown function "\grp"): {\displaystyle U_{\grp_{lc}}*\mathbb{H}} is an analytical, fibered Hilbert space or Hilbert bundle over Failed to parse (unknown function "\grp"): {\displaystyle U_{\grp_{lc}}} , and
Failed to parse (unknown function "\grp"): {\displaystyle L: U_{\grp_{lc}} \longrightarrow '''Iso''' (U_{\grp_{lc}}*\mathbb{H} )} is a Borelian groupoid morphism whose restriction on Failed to parse (unknown function "\grp"): {\displaystyle U_{\grp_{lc}}}
is the identification map , that is, Failed to parse (unknown function "\grp"): {\displaystyle U_{'''Iso''' (U_{\grp_{lc}}*\mathbb{H})}} is being identified via $L$ with Failed to parse (unknown function "\grp"): {\displaystyle U_{\grp_{lc}}} . Thus,
$L(x)=[r(x),{\tilde {L}}(x),d(x)]$,
where ${\tilde {L}}(x):\mathbb {H} (d(x))\longrightarrow \mathbb {H} (r(x))$ is a Hilbert space $\mathbb {H}$ isomorphism.
# PlanetPhysics/Euler 321 Sequence

For more info on Euler Sequences, notation and convention see the generic entry on Euler angle sequences. \\
$R_{321}(\phi ,\theta ,\psi )=R_{1}(\phi )R_{2}(\theta )R_{3}(\psi )$ \\
The rotation matrices are
   $R_{1}(\phi )=\left[{\begin{matrix}1&0&0\\0&c_{\phi }&s_{\phi }\\0&-s_{\phi }&c_{\phi }\end{matrix}}\right]$
   $R_{2}(\theta )=\left[{\begin{matrix}c_{\theta }&0&-s_{\theta }\\0&1&0\\s_{\theta }&0&c_{\theta }\end{matrix}}\right]$
   $R_{3}(\psi )=\left[{\begin{matrix}c_{\psi }&s_{\psi }&0\\-s_{\psi }&c_{\psi }&0\\0&0&1\end{matrix}}\right]$
Carrying out the matrix multiplication from right to left \\
$R_{2}(\theta )R_{3}(\psi )=\left[{\begin{matrix}c_{\theta }&0&-s_{\theta }\\0&1&0\\s_{\theta }&0&c_{\theta }\end{matrix}}\right]\left[{\begin{matrix}c_{\psi }&s_{\psi }&0\\-s_{\psi }&c_{\psi }&0\\0&0&1\end{matrix}}\right]=\left[{\begin{matrix}c_{\theta }c_{\psi }&c_{\theta }s_{\psi }&-s_{\theta }\\-s_{\psi }&c_{\psi }&0\\s_{\theta }c_{\psi }&s_{\theta }s_{\psi }&c_{\theta }\end{matrix}}\right]$ \\
Finaly leaving us with the Euler 321 sequence \\
$R_{1}(\phi )R_{2}(\theta )R_{3}(\psi )=\left[{\begin{matrix}c_{\theta }c_{\psi }&c_{\theta }s_{\psi }&-s_{\theta }\\-c_{\phi }s_{\psi }+s_{\phi }s_{\theta }c_{\psi }&c_{\phi }c_{\psi }+s_{\phi }s_{\theta }s_{\psi }&s_{\phi }c_{\theta }\\s_{\phi }s_{\psi }+c_{\phi }s_{\theta }c_{\psi }&-s_{\phi }c_{\psi }+c_{\phi }s_{\theta }s_{\psi }&c_{\phi }c_{\theta }\end{matrix}}\right]$
# PlanetPhysics/Euler Angle Velocity of 123 Sequence

The method of deriving the Euler angle velocity for a given sequence is to transform each of the derivatives into the reference frame. Remember that an Euler angle sequence is made up of three successive rotations. In other words, the angular velocity ${\dot {\phi }}$ needs one rotation, ${\dot {\theta }}$ needs two and ${\dot {\psi }}$ needs three.
${\vec {\omega }}=R_{3}(\psi )R_{2}(\theta )R_{1}(\phi )\left[{\begin{matrix}{\dot {\phi }}\\0\\0\end{matrix}}\right]+R_{3}(\psi )R_{2}(\theta )\left[{\begin{matrix}0\\{\dot {\theta }}\\0\end{matrix}}\right]+R_{3}(\psi )\left[{\begin{matrix}0\\0\\{\dot {\psi }}\end{matrix}}\right]$
Carrying out the matrix multiplication with $R_{3}(\psi )R_{2}(\theta )R_{1}(\phi )$ being the Euler 123 sequence $R_{3}(\psi )R_{2}(\theta )=\left[{\begin{matrix}c_{\psi }c_{\theta }&s_{\psi }&-c_{\psi }s_{\theta }\\-s_{\psi }c_{\theta }&c_{\psi }&s_{\theta }s_{\psi }\\s_{\theta }&0&c_{\theta }\end{matrix}}\right]$
and
$R_{3}(\psi )=\left[{\begin{matrix}c_{\psi }&s_{\psi }&0\\-s_{\psi }&c_{\psi }&0\\0&0&1\end{matrix}}\right]$
gives us
$\left[{\begin{matrix}\omega _{x}\\\omega _{y}\\\omega _{z}\end{matrix}}\right]=\left[{\begin{matrix}c_{\theta }c_{\psi }{\dot {\phi }}\\-c_{\theta }s_{\psi }{\dot {\phi }}\\s_{\theta }{\dot {\phi }}\end{matrix}}\right]+\left[{\begin{matrix}s_{\psi }{\dot {\theta }}\\c_{\psi }{\dot {\theta }}\\0\end{matrix}}\right]+\left[{\begin{matrix}0\\0\\{\dot {\psi }}\end{matrix}}\right]$
Adding the vectors together yields
$\left[{\begin{matrix}\omega _{x}\\\omega _{y}\\\omega _{z}\end{matrix}}\right]=\left[{\begin{matrix}{\dot {\phi }}c_{\theta }c_{\psi }+{\dot {\theta }}s_{\psi }\\{\dot {\theta }}c_{\psi }-{\dot {\phi }}s_{\psi }c_{\theta }\\{\dot {\phi }}s_{\theta }+{\dot {\psi }}\end{matrix}}\right]$
Of course, we also wish to have the Euler angle velocities in terms of the angular velocities which requires us to solve the linear equations for them. Using a program like Matlab makes it easy for us to get
$\left[{\begin{matrix}{\dot {\phi }}\\{\dot {\theta }}\\{\dot {\psi }}\end{matrix}}\right]=\left[{\begin{matrix}(\omega _{x}c_{\psi }-\omega _{y}s_{\psi })/c_{\theta }\\\omega _{x}s_{\psi }+\omega _{y}c_{\psi }\\(-\omega _{x}c_{\psi }+\omega _{y}s_{\psi })s_{\theta }/c_{\theta }+\omega _{z}\end{matrix}}\right]$
# PlanetPhysics/Example Constant Acceleration and Speed of Sound

A favorite past time on the David Letterman show is to throw watermelons off the roof of the Ed Sullivan theater. Let us determine the height of the building from the time $T$ of releasing the watermelon to hearing it splatter its contents on the street. Assuming $T=2.9s$. What is the height $h$ of the building?
   \includegraphics[scale=.8]{ExConstantAccSpeedSound.eps}
We can either begin with the memorized constant acceleration equation
$x-x_{0}={\frac {1}{2}}\left(v_{0}+v\right)t$
or start with the 2nd law of Newton's laws of motion $F=ma$
Neglecting drag, the only force acting on the watermelon is gravity
$F=mg$
so
$ma=mg$ $a=g$
Note that the acceleration is positvie because we chose the y-axis to be positive pointing down toward the street. So the equation of motion is
   ${\ddot {y}}=g$
Integrating to get velocity
   ${\dot {y}}=gt+C_{1}$
Since at $t=0$, ${\dot {y}}=0$, $C_{1}=0$ and therefore
   ${\dot {y}}=gt$
Integrating to get position yields
   $y={\frac {1}{2}}gt^{2}+C_{2}$
Since at $t=0$, $y=0$, we get
   $y={\frac {1}{2}}gt^{2}$
Although this part is straight forward, the trick for this problem is to incorporate the time needed for sound to travel from the street to our ears on the roof. Because $T$ is the total time, it includes the $t$ in eq. 1 and the time for sound to travel $t_{s}$
   $T=t+t_{s}$
Assuming the speed of sound $v_{s}$ at $20^{o}C$ is $343m/s$, the height traveled in the time $t_{s}$ that must be subtracted off is
   $h=v_{s}t_{s}$
so
   $t_{s}={\frac {h}{v_{s}}}$
So $t$ in eq. 1 is then
   $t=T-t_{s}=T-{\frac {h}{v_{s}}}$
Therefore, the height $(y=h)$ is now determined from
   $h={\frac {1}{2}}g\left(T-{\frac {h}{v_{s}}}\right)^{2}$
Expanding
   $h={\frac {1}{2}}g\left[T^{2}-{\frac {2Th}{v_{s}}}+{\frac {h^{2}}{v_{s}^{2}}}\right]$
Collecting terms
   ${\frac {g}{2v_{s}^{2}}}(h^{2})+\left(-{\frac {T}{v_{s}}}-1\right)(h)+{\frac {1}{2}}gT^{2}=0$
Now we can use the quadratic formula to solve for $h$
   $h={\frac {-b\pm {\sqrt {b^{2}-4ac}}}{2a}}$
with the following constants
   $a={\frac {g}{2v_{s}^{2}}}={\frac {9,8}{(2)(342)^{2}}}=4.1649313\times 10^{-5}$
   $b=\left(-{\frac {T}{v_{s}}}-1\right)=-{\frac {2.9}{343}}-1=1.0084548$
   $c={\frac {1}{2}}gT^{2}=(0.5)(9.8)(2.9)=41.209$
Substituting these values into the quadratic equation yields two solutions,
   $h_{1}=40.93m$
and
   $h_{2}=24172.07m$
To determine the correct answer we need to plug each solution into our time equation to see which one makes sense
   $t=T-{\frac {h}{v_{s}}}$
For solution 1, we get
   $t_{1}=2.78s$
and for solution 2, we get
   $t_{2}=67.57s$
Clearly, since our total time is $T=2.9s$, only the first solution is correct and the height of the building is
   $h=40.93m$
Note, that if we ignored the speed of sound, we would use the equation
   $h={\frac {1}{2}}gt^{2}=(0.5)(9.8)(2.9)^{2}=41.21m$
which is only off by 0.7\
# PlanetPhysics/Euler 313 Sequence

For more info on Euler Sequences, notation and convention see the generic entry on Euler angle sequences. \\
$R_{313}(\phi ,\theta ,\psi )=R_{3}(\psi )R_{1}(\theta )R_{3}(\phi )$ \\
The rotation matrices are
   $R_{3}(\psi )=\left[{\begin{matrix}c_{\psi }&s_{\psi }&0\\-s_{\psi }&c_{\psi }&0\\0&0&1\end{matrix}}\right]$
   $R_{1}(\theta )=\left[{\begin{matrix}1&0&0\\0&c_{\theta }&s_{\theta }\\0&-s_{\theta }&c_{\theta }\end{matrix}}\right]$
   $R_{3}(\phi )=\left[{\begin{matrix}c_{\phi }&s_{\phi }&0\\-s_{\phi }&c_{\phi }&0\\0&0&1\end{matrix}}\right]$
Carrying out the multiplication from right to left \\
$R_{1}(\theta )R_{3}(\phi )=\left[{\begin{matrix}1&0&0\\0&c_{\theta }&s_{\theta }\\0&-s_{\theta }&c_{\theta }\end{matrix}}\right]\left[{\begin{matrix}c_{\phi }&s_{\phi }&0\\-s_{\phi }&c_{\phi }&0\\0&0&1\end{matrix}}\right]=\left[{\begin{matrix}c_{\phi }&s_{\phi }&0\\-s_{\phi }c_{\theta }&c_{\theta }c_{\phi }&s_{\theta }\\s_{\theta }s_{\phi }&-s_{\theta }c_{\phi }&c_{\theta }\end{matrix}}\right]$ \\
Finaly leaving us with the Euler 313 sequence \\
$R_{3}(\psi )R_{1}(\theta )R_{3}(\phi )=\left[{\begin{matrix}c_{\psi }c_{\phi }-s_{\psi }s_{\phi }c_{\theta }&c_{\psi }s_{\phi }+s_{\psi }c_{\theta }c_{\phi }&s_{\psi }s_{\theta }\\-s_{\psi }c_{\phi }-c_{\psi }s_{\phi }c_{\theta }&-s_{\psi }s_{\phi }+c_{\psi }c_{\theta }c_{\phi }&c_{\psi }s_{\theta }\\s_{\theta }s_{\phi }&-s_{\theta }c_{\phi }&c_{\theta }\end{matrix}}\right]$
# PlanetPhysics/Euler Angle Velocity of 321 Sequence

The method of deriving the Euler angle velocity for a given sequence is to transform each of the derivatives into the reference frame. Remember that an Euler angle sequence is made up of three successive rotations. In other words, the angular velocity ${\dot {\psi }}$ needs one rotation, ${\dot {\theta }}$ needs two and ${\dot {\phi }}$ needs three.
${\vec {\omega }}=R_{1}(\psi )R_{2}(\theta )R_{3}(\phi )\left[{\begin{matrix}0\\0\\{\dot {\phi }}\end{matrix}}\right]+R_{1}(\psi )R_{2}(\theta )\left[{\begin{matrix}0\\{\dot {\theta }}\\0\end{matrix}}\right]+R_{1}(\psi )\left[{\begin{matrix}{\dot {\psi }}\\0\\0\end{matrix}}\right]$
Carrying out the matrix multiplication with $R_{1}(\psi )R_{2}(\theta )R_{3}(\phi )$ being the Euler 321 sequence $R_{1}(\psi )R_{2}(\theta )=\left[{\begin{matrix}c_{\theta }&0&-s_{\theta }\\s_{\psi }s_{\theta }&c_{\psi }&s_{\psi }c_{\theta }\\c_{\psi }s_{\theta }&-s_{\psi }&c_{\psi }c_{\theta }\end{matrix}}\right]$
and
$R_{1}(\psi )=\left[{\begin{matrix}1&0&0\\0&c_{\psi }&s_{\psi }\\0&-s_{\psi }&c_{\psi }\end{matrix}}\right]$
gives us
$\left[{\begin{matrix}\omega _{x}\\\omega _{y}\\\omega _{z}\end{matrix}}\right]=\left[{\begin{matrix}-s_{\theta }{\dot {\phi }}\\s_{\psi }c_{\theta }{\dot {\phi }}\\c_{\psi }c_{\theta }{\dot {\phi }}\end{matrix}}\right]+\left[{\begin{matrix}0\\c_{\psi }{\dot {\theta }}\\-s_{\psi }{\dot {\theta }}\end{matrix}}\right]+\left[{\begin{matrix}{\dot {\psi }}\\0\\0\end{matrix}}\right]$
Adding the vectors together yields
$\left[{\begin{matrix}\omega _{x}\\\omega _{y}\\\omega _{z}\end{matrix}}\right]=\left[{\begin{matrix}-s_{\theta }{\dot {\phi }}+{\dot {\psi }}\\s_{\psi }c_{\theta }{\dot {\phi }}+c_{\psi }{\dot {\theta }}\\c_{\psi }c_{\theta }{\dot {\phi }}-s_{\psi }{\dot {\theta }}\end{matrix}}\right]$
Of course, we also wish to have the Euler angle velocities in terms of the angular velocities which requires us to solve the linear equations for them. Using a program like Matlab makes it easy for us to get
$\left[{\begin{matrix}{\dot {\phi }}\\{\dot {\theta }}\\{\dot {\psi }}\end{matrix}}\right]=\left[{\begin{matrix}\left(\omega _{y}s_{\psi }+\omega _{z}c_{\psi }\right)sec({\theta })\\\omega _{y}c_{\psi }-\omega _{z}s_{\psi }\\\omega _{x}+\omega _{y}s_{\psi }t_{\theta }+\omega _{z}c_{\psi }t_{\theta }\end{matrix}}\right]$
In matlab solving for the Euler angle velocites can be done with the following commands. Using the notation $Ax=b$, we want to solve for $x$, such that $x=A^{-1}b$. For our problem then
{\emph syms wx wy wz phd thd psd SPS cth cps b x A;
b = [wx wy wz]';
x = [phd thd psd]';
A = [ -sth 0 1; sps*cth cps 0; cps*cth -sps 0];}
and solve for the angle velocites with the command
x = inv(A)*b
Note that matlab spits out extra sine and cosine terms that just equal 1 through
$s_{\psi }^{2}+c_{\psi }^{2}=1$
The shorthand notation used in this article is
$s_{\psi }=sin(\psi )$ $c_{\psi }=cos(\psi )$ $t_{\psi }=tan(\psi )$