How to interpret line of sight velocity distributions and dispersions of a galaxy? What I did: I am analysing some simulation snapshots of the Milky Way and it is modelled as a DM halo – bulge – disc system. I produced some mean line of sight velocity (LOSV) and dispersion of mean LOSV plots that I am attaching here. To produce these plots I simply extracted the vx,vy,vz components of the velocities of the stars and i picked z-axis as the line of sight. I overlaid a grid on the galaxy snapshot and at different inclinations(theta=0,45,90 degrees)of the galaxy I plotted the mean vz in the grid cells (vz is the z-component of the velocities). For dispersion, I just plotted the std deviation of the vz in the grid cells.

My question: I want to now understand and interpret my plots. What can I deduce from the plots for the different angles of inclinations?   what can we deduce about the orbits the stars have at the bulge and disc?   I mainly want to understand why the line of sight velocity distribution looks the way it does on my plots form a physics point of view..that is, due to gravity or mass distribution perhaps...please refer to equations if you want. Any help is much appreciated, thanks so much!
 A: 
why do the mean losv and dispersion look this way?

The radial velocity profile tells you that the stars in the disk are rotating in a coherent fashion. You can calculate the rotational velocity by correcting for the inclination: $v = {v_{rad}\over \sin(\theta)}$. If you plot the velocity profile as a function of the distance $r$ form the center of the galaxy, you will obtain the famous flat rotation curve often used as proof for the existence of dark matter.
Indeed, from the velocity profile you can infer the density profile as follows. In order for a star to be in a circular orbit, its velocity must satisfy
$${v(r)^2 \over r} = -{d \Phi(r) \over d r}$$
Poisson equation in a spherical potential (the DM halo is approximately spherical and it contributes the most to the potential) is:
$${1 \over r^2} {d \over dr}(r^2 {d\Phi \over dr}) = -4\pi G \rho$$
Substituting the first equation in the second one obtains
$$4\pi G \rho = {1 \over r^2} {d \over dr}(r v^2) = {1 \over r^2}(v^2 + 2rv {dv \over dr} )$$
$$\rho(r) = {v(r)^2 \over 4 \pi G r^2} (1 + 2 {r \over v(r)}{dv(r) \over dr})$$
If $v(r)$ is constant (flat profile) then $\rho(r) \propto r^{-2}$, also called singular isothermal sphere profile and this should roughly match the radial density profile of your DM halo.
In the bulge, things are different. Stars are not rotating coherently, each start follows its own orbit. This leads to a low average radial velocity (because the random velocities average to zero) and to a high radial velocity dispersion $\sigma_{rad}$. Assuming that the velocity distribution is isotropic, you can calculate the total velocity dispersion as $\sigma = \sqrt 3 \sigma_{rad}$.
From it, you can estimate the total mass of the bulge using the virial theorem.
$$2K = -U \implies M\sigma^2 = \alpha {G M^2 \over R}$$
$$M = {R \sigma^2 \over G \alpha}$$
Where $R$ can be the radius of the bulge and $\alpha$ is a geometrical factor of the order of $1$, that depends on the specific density profile of the bulge.
Additionally, you could use the $M-\sigma$ relation to estimate the mass of the central black hole, but in your case you don't mention any black hole, so I think this may not be relevant
A: The usual model for galaxies is that stars orbit the nucleus and the galaxy looks disk shaped, like a plate.
Without dark matter the orbital speeds of the stars would follow the red curve below.  However they really follow the green curve, due to the dark matter.  From the axes it can be seen that the orbital speed is typically about 200km/s and doesn't change much with radius.

Your bottom left diagram shows the 'plate' viewed edge on.  The red section on the left shows the approximately constant 200km/s speed towards the viewer and on the right of the middle it's the 200km/s speed away from the viewer (you could consider swapping the colours to match redshift/blueshift).
The top left diagram is showing velocities perpendicular to the disk.  On average these are zero and there is also not much dispersion, (variation), top right, - you could change colours again and use e.g. green, for the right hand side, to avoid confusion about the meaning of the red.
The middle left diagram is showing the 'plate' tilted at 45 degrees, so you see the z component from the bottom left diagram, but the component is of lower magnitude, about 100km/s.
The dispersions are low for the bottom right and bottom middle diagrams, meaning that there isn't much variation between the stars, they all seem to flow together in the same direction (if they are near each other).  The exception to this is in the central bulge.  Here there is high dispersion and the stars are moving randomly.
