Arc length between spaced points on circle when viewed under an inclination Short version: Say you have a circle of radius $r$ with a set of $n$ equally spaced points on it, each with an arc length of $\Delta\theta=\frac{2\pi}{n}$ between each other. When that circle is viewed under an inclination $i$, the phase $\theta$ of any point w.r.t. the origin changes. How does that projected arc length $\Delta\theta$ change when the circle is viewed under an inclination $i$ EDIT2 in terms of the initial phase $theta$ of the unit circle?
Long version: I am an astronomer and am making a model of a ring around a star (let's ignore the astrophysics for this question). That ring has a number of $n$ "parcels of gas" on it, each with an arc length of $\Delta\theta=\frac{2\pi}{n}$ EDIT2 at a phase $\theta$. When that ring is viewed under an inclination $i$ ($0$ deg=face-on, $90$ deg=edge-on), that angle $\theta^{circ}->\theta^{ellip}$ changes of course. How does the observed arc length, call it $\Delta\theta'$, of the parcels of gas change with that inclination $i$ in terms of the $\textbf{initial phase $\theta$}$? Of course this is dependent on the azimuth/phase $\theta$ of each parcel, where any $\theta$ is defined as $0$ at either equinox/along the semi major axis of the ellipse. For the parcels at the peri/apastron, that is for $\theta^{ellip}=\pi\pm\frac{\pi}{2}$, the change should be negligible, even for a high $i$. For those at the equinoxes, i.e. for $\theta^{ellip}=\{0,\pi\}$, the observed arc length should become smaller and smaller for an increasing $i$.
My colleague has derived a rather simple solution which I do not think is totally correct. Hence, I would love to know your take on this matter.
EDIT: My colleague's solution was simply: $\Delta\theta' = \rho\cdot\Delta\theta$, where $\rho$ is the distance from the origin to the position of the gas parcel in the ellipse - however, that is confusing the azimuth on the circle $\theta$ with the eccentric anomaly in the ellipse. Hence, that cannot be the right answer.
EDIT2: The problem is that I need the solution in terms of the point's initial phase on the circle.
Hence the big question: I have a point on a circle that is being projected as an ellipse under an inclination. How would I convert the phase of the point on the circle to a phase in the ellipse? That phase is not conserved.
 A: Let's suppose we have projected the circle orthogonally(*) onto an ellipse in some image plane, and put the coordinates $(0,0)$ at the center of the ellipse with the $x$ axis aligned with the major axis of the ellipse.
So the parametric equation of the ellipse then is
$$
\begin{align}
x &= a \cos \theta \\
y &= b \sin \theta
\end{align}
\tag1
$$
where $a$ is the length of the major axis in the image and $b$ is the length of the minor axis.
The ratio $\frac ba$ is determined by the angle of inclination $i$, specifically,
$$ \frac ba = \cos i. $$
The angle $\theta$ in the parametric equations is the original angle from the circle's center to the point whose image is at $(x,y),$ assuming that $\theta$ is measured from the point on the circle that is mapped to the point $(a,0)$ on the ellipse.
To put it another, way, if we have $X$ and $Y$ coordinates in the plane of the original circle, which has radius $r,$ the parameterization of the circle is
$$
\begin{align}
X &= r \cos \theta \\
Y &= r \sin \theta
\end{align} \tag2
$$
and all the projection onto the ellipse does is to compress the $y$ coordinate more than the $x$ coordinate.
If we suppose $a = r$ for simplicity, and superimpose a plot of the ellipse in its plane with a plot of the original circle in its own original plane,
we get a figure like this:

This shows how the point with coordinates $(X,Y)$ on the circle corresponds to the point with coordinates $(x,y)$ on the ellipse. The violet arc on the circle corresponds to the red arc on the ellipse.
The angle $\theta$ in this figure is the same $\theta$ that appears in the formulas for $x$, $y$, $X$, and $Y$ in Equations $(1)$ and $(2)$.
So if you have already somehow managed to parameterize the ellipse in the form of Equation $(1),$ you already have the parameter $\theta$ that you want at each point. To find the $\Delta \theta$ between two points, just take the difference of their $\theta$ parameters.
But if what you actually have is just the ellipse with the parameters $a$ and $b$, with no knowledge yet of the value of the parameter $\theta$ at any point, you can proceed like this:
Let $\theta'$ be the angle between the $x$ axis and the line from $(0,0)$ to $(x,y)$. Then
$$ \tan \theta' = \frac yx = \frac{b \sin\theta}{a \cos\theta}
 = \cos i \tan\theta. $$
For angles like the ones in the figure, between $0$ and $\frac\pi2,$
$$ \theta = \arctan\left(\frac{\tan\theta'}{\cos i}\right), \tag3 $$
which gives a direct conversion from an observed position on the ellipse to the original position on the circle.
The conversion for other angles is graphically similar to this (reflected across an axis or rotated $180$ degrees) but arithmetically a bit more complicated because $\arctan$ gives only results from $-\frac\pi2$ to
$\frac\pi2,$ so if you want results from $0$ to $2\pi$ you have to add $\pi$ when $\frac\pi2<\theta'<\frac{3\pi}2$ and add $2\pi$ when
$\frac{3\pi}<\theta'<2\pi.$
To get the arc length between two points on the original circle you can take the $\theta'$ values at each end of the segment on the ellipse, run them through Equation $(3)$, and get two $\theta$ values. The difference is $\Delta\theta,$ which gives an arc length $r\theta$ on a circle of radius $r.$
The arc length of the corresponding segment along the ellipse, shown as $\Delta s$ in the figure, is more complicated to measure.
I don't think it makes sense to use an angular measurement of that arc,
except in the sense that the observed object subtends some apparent angle when you look at it from an observatory on Earth; but as the entire structure might have an angular diameter much less than one second of arc, any angular measure of the arc length wouldn't have much to do with any of the angles seen in the figure, even if we set $\Delta\theta = \frac{\pi}{1000}.$

The rest of this answer is about the arc length of a segment along the ellipse in the image plane. Based on comments after I wrote this, it seems this is probably not really useful to you, but I'll leave it here in case it is useful to someone.
If you imagine a particle traveling along the actual circle so that it traverses the angle around the circle at a certain rate $\frac{d\theta}{dt},$
then the motion of the image of that particle in the projection will be
$a \frac{d\theta}{dt}$ as it crosses the major axis and
$b \frac{d\theta}{dt}$ as it crosses the minor axis.
As the image of the particle passes through the point at
$(x,y) = (a \cos \theta, b \sin\theta)$,
its velocity vector in the image has components
\begin{align}
 \frac{dx}{dt} &= -a \sin \theta \\
 \frac{dy}{dt} &= b \cos \theta
\end{align}
and the speed is
$$  \frac{ds}{dt} = \sqrt{ a^2 \sin^2 \theta + b^2 \cos^2 \theta}. $$
To further simplify the math, let's suppose this hypothetical particle travels around the original circle at a speed of exactly one radian per unit time, which means that as a function of the angle $\theta$ traveled around the original circle,
$$  \frac{ds}{d\theta} = \sqrt{ a^2 \sin^2 \theta + b^2 \cos^2 \theta}. $$
For a small-angle approximation, that is, if the original circle is divided into $n$  parcels of angle $\frac{2\pi}{n}$ each, the arc length of each parcel in the image is approximated by
$$ \Delta s \approx \frac{ds}{d\theta} \Delta \theta  = 
 \frac{2\pi}{n} \sqrt{ a^2 \sin^2 \theta + b^2 \cos^2 \theta}. $$
I think this is going to be rather different than your colleague's answer in most places.
An exact answer would be an integral like
$$ \Delta s = \int_{\theta_1}^{\theta_1 + 2\pi/n}
     \sqrt{a^2 \sin^2 \theta + b^2 \cos^2 \theta} \, d\theta. $$

(*) If we want to be mathematically exact, if you're actually observing a ring then the relevant projection is a central projection through the point of observation, but if you're looking at a ring around a star from Earth then the cone is so narrow that it's practically an orthogonal projection followed by a uniform scaling down.
A: Since we are talking about stars (and they are usually very far away), I assume you are satisfied with an orthogonal projection (without vanishing point). Under this assumption, the inclination changes the appearance of the circle to an ellipse. It is sufficient to consider the unit circle, otherwise we can simply scale the arc lengths with the radius $r$. As far as I can see, you did it the same way, otherwise the arc length would be larger by a factor of $r$.
Exact answer: The inclination $i$ changes the unit circle to an ellipse
$$x^2+\frac{y^2}{\cos^2 i} = 1.$$
Now you can look up the exact answer on Wikipedia for $a=1$ and $b = \cos i$.
Approximation: Since these integrals are quite hard to calculate, I'll try to derive an approximation: We approximate the arc lengths $\Delta\theta$ in your circle with straight edges of length $l$ (I am not sure if these should have length $2\pi/n$ or the side length of the inscribed $n$-Polygon, so let's say $l$ for now).
If the first point $p_0$ is on $(1,0)$, the angle of the corresponding edge (counterclockwise) is
$$\alpha_0 = \frac{\pi+\Delta\theta}{2}$$
Therefore with analogous notation: $\alpha_m = \alpha_0 + m\Delta\theta$. As you can see $\alpha_n = \alpha_0 + 2\pi = \alpha_0$, so it works out.
Let $\vec{v_m}$ be the corresponding edge of $p_m$, then $\vec{v_m} = l(\cos\alpha_m,\sin\alpha_m).$
Applying the inclination: $\vec{v_m}' = l(\cos\alpha_m,\cos i\sin\alpha_m).$
So my Approximation for the $m$-th arc would be:
$$||\vec{v_m}'|| = l\sqrt{\cos^2\alpha_m+\cos^2 i\sin^2\alpha_m}$$
for $l=2\pi/n$ or $l=2\sin(\pi/n)$.
EDIT: Suppose $p = (\cos\theta,\sin\theta)$, then $\alpha = \theta + \pi/2$ and $l\sqrt{\cos^2\alpha+\cos^2 i\sin^2\alpha}$ should be a good approximation for the arc length near $p$. This is how @David K did it.
A: 
There are two angles for a compound angle projection.
In a side-view let $ds = \dfrac{2 \pi r}{n}$ green differential and let P,Q be the closest and farthest points to the horizontal reference plane HP onto which the circle is projected. Tangents at P,Q are marked in red. They are essential for projections.
Plane of circle makes angle $\phi$ to HP as shown.
First projection gives $ ds \cos \psi$ parallel to PQ
PQ makes angle $\phi $ to HP.
A second projection gives blue line on HP  along A'B' to have projected length
$$ ds \cos \psi \cos \phi $$
as shown.
A: I don't think that phase is the correct term to use, as this relates to the position of the zeroes of a sinusoid. I believe (but maybe I am wrong) that what you are after is simply the angle.
Now assume a unit circle and pick a point at angle $\theta$, i.e. $(\cos\theta,\sin\theta)$. If you tilt the plane around the axis $x$, and project back to the original $xy$ plane, the $x$ coordinates are invariant, while the $y$ are reduced by a factor $\cos\alpha$, where $\alpha$ is the rotation angle. Now the new angle, let $\phi$ is such that
$$\tan\phi=\frac{\sin\alpha\sin\theta}{\cos\theta}=\sin\alpha\tan\theta,$$ which gives you the relation between the angles seen facing and seen obliquely.
