I hope this isn't a stupid question, but Google searches are turning up nothing relevant. I was just thinking about how it's difficult for us to determine the true shape and composition of the Milky Way because we are trapped on one side and the further you go towards, through, and out the center of the galaxy, the more opaque it becomes and therefore difficult it becomes to see through. So, I was just wondering, how far above or below the galactic plane would we have to be to be able to image the entire galaxy?

Now, if I am reading this article correctly, the Earth currently sits above the plane about 75-101 light years, so obviously more than that. Now, what I think it depends on is how big the galaxy is — both the "height" (thickness of the plane), as well as the diameter of galaxy, and the type of imaging device.

According to this article, the galactic plane is about 1,000 light years thick. So, I imagine a minimum "height" would be >500 light years, because you obviously need everything to be in front of the camera (I suppose you could do some 3-D camera and image midway between the galactic center and "highest" object from the galactic plane, and if you're spending money on an interstellar trip thousands to tens of thousands of light years long, why not put 3-d facing optics, but for this question, we will exclude 3-D telescopes). How much greater depends on how far away you need to be from that "highest" object so that its angular size (no idea if I used that correctly) does not block out a large portion of the galaxy. "Large portion" is obviously subjective, but I am a layman here, so use your expert judgement on how angularly small (again, please correct me if I used that wrong) the "highest" object would have to be for it to be of similar (again, expert judgement) size to the other stars of the galaxy.

As for what kind of imaging device: the first one I thought of for comparison was the Sloan Digital Sky Survey (not sure if this is a dated reference, it's the only telescope I have ever heard of called an "all sky survey"). According to this article images about 1.5 square degrees at a time. It further says that in 5 years, SDSS-I mapped over 8,000 sq. degrees of the sky. This article also made me realize a third factor is in play as well: time. My math says {[(5*365)+1]*24}/(8,000 sq. degrees)≈5 and a half hours per square degree of sky surveyed. You can use this measurement for a speed time/sq. degrees surveyed. And, if you know of any other speeds like this (I imagine this is an unconventional if not unique unit), it would be interesting to include them as well, because, I imagine space based telescopes are able to record much less of the night sky (which I imagine is intentional because they are trying to see further and not so wide, typically). I would especially be interested how the Hubble, James Webb, and Planck satellites would perform. And, for the time factor, I would arbitrarily pick a survey length of 1, 10, and 100 years, but of course, use whatever length you want.

Now, I imagine it is going to be a type of sliding scale: the further you get away from the galaxy, the smaller the angular size, so the less area needed to be photographed, so the less time needed to photograph, but the poorer the resolution of the galaxy.

Finally, I don't care about the technical feasibility of the program. I mean, sending data tens of thousands of light years presents obvious problems. Further, I am a high school and college drop out, so I'm sure there are tons of things I left out, don't understand, etc., I hope they are all things you can assume and inform of the details.

To summarize and hopefully be clearer, as well as avoid having the question closed, I am interested in the theoretical minimum "height" from the center of the galactic core from which a telescope array with your chosen technological capabilities could image the entire galaxy to your chosen resolution in your chosen time period.

I am not sure if I should add this or not, so delete it if necessary, but I attempted to solve this problem with the 10th grade Algebra and 9th grade geometry I have, along with what I learned in the Army about calling artillery. In the military, the angular measurement we use is the "mil", not to be confused with the milliradian. In a NATO mil circle, there are exactly 6,400 mils in a circle, and I was told that 1 mil angular measurement equals about 1 meter across at 1,000 km. This makes calling in artillery easy because if you have to adjust fire left or right X number of meters, it is that many mils times the number of kilometers away the target is. While this may suffice for artillery barrages, as I am sure you've noticed by now, this is imprecise. Still, I THINK it helped me to work out that a telescope that surveys as fast as the SDSS telescope ≈5.5 hours per sq. degree, would take ≈17,500 hours to survey the galaxy from about 10,000 light years.

Going over this, I realize the circumference of 1 km circle is 3142 meters, and I believe this means there are slightly over 2 meters per mil at 1 kilometers, so I THINK, you can just double the 17,500 hours to 35,000. But this is all guesses. While only having a 10th grade education with math, I also had some head injuries which have made me highly suspicious of my logical as well as mathematical reasoning. So, if someone could use my jump-off point to give me a more reliable answer in a similar format to what I tried, if that is at all possible, I would appreciate it.

I'll include a picture of my Google Docs page where I jotted down my stream of thoughts in case it helps at all. Thank you for any help you could give.

  • $\begingroup$ Since the galaxy is 3-D, you will be dealing with obscurations no matter where you place your camera. You'll need to observe from thousands of locations... sort of the equivalent of a gigantic 3-D CAT scanner. $\endgroup$ May 16, 2022 at 14:17

1 Answer 1



  • Whatever number I or someone else gives you, would be totally arbitrary and made up.

But if you really want some arbitrary made up numbers that are good enough:

  • The distance to the galactic center must be a quarter of the galactic radius.
  • The angle between the line of sight of the observer and the galactic plane in the galaxy's border should be no less than $14°$.
  • For Milky Way, this is roughly $25,000 \text{ LY}$.


If I understood your question correctly, you could imagine a cone, where the base would be the galaxy and the vertex would be the observer. The cone is not necessarily straight or close. So the question is how much the cone could be flattened.

See the image below:

Four observers observing perfectly flat galaxies

Here, the white ellipses are galaxies. The blue spots are observers observing their galaxies. The red lines show the imaginary cone and denotes the angular size needed by the observer. The orange lines show the base and the height of the triangle in which the red lines are the two other sides.

The first observer has surely the best observation conditions, for being directly over the galaxy. The second and the third observers, although not having the best position, still would also have somewhat good conditions. The fourth observer, clearly inside the galaxy would have trouble.

The problem becomes in determining which is the flattest cone that does the job, with the angle between the red lines in the blue observer approaching $180°$, but don't getting too close to ruin the job.

The closest height with the flattest cone is the one with zero height or any small value arbitrarily close to that. But this doesn't do the job. This is in part because this still neglects the fact that galaxies are not perfectly flat objects. The situation is more likely the image below:

Two observers observing disc-shaped galaxies

The purple lines are different line of sights of the observer. The green lines are the part of those line of sight that intercepts the disks. Ideally, we would like to place the observer in a place where the green lines are as shortest as possible. The first observer in this image has a good observation point, and the second has a bad one, since there is some case where the green lines gets very long.

How can we measure if the green lines are too long? If the disc height is $h$, the green line measures $g$ and the angle between the green line and the galactic plane is $\theta$, then $h$ is the side of a rectangle triangle opposite to the angle $\theta$ at the edge of the galaxy and $g$ is the hypotenuse. Then:

$$h = g \times \text{sin}(\theta)$$ $$g = h \times \text{cosec}(\theta)$$

Now, how can we determine if the green line is "too big"? It is proportional to the height of the galactic disc, so we should look if $\text{cosec}(\theta)$ is too big. A value of $1$ corresponding to an angle of $90°$ between the observer line of sight and the galactic plane is ideal, because the green line length would be the disc length. A value of $\sqrt{2}$ would be a $45°$. A value of $2$ is an angle of $30°$. The lowest the cosecant, the better the observation would be, as this means that smaller is the green line. The cosecant tends to infinite as closer our line of sight gets to the galactic plane. This value means how worse is the oblique observation comparing to the looking directly perpendicular to the galactic disc.

I could give you that an angle no smaller than $30°$ is good ($\text{cosec}(30°) = 2$), i.e. the green line is twice the galactic disc height. Or perhaps it is $15°$ ($\text{cosec}(15°) \approx 3.86$)? Maybe $10°$ ($\text{cosec}(10°) \approx 5.75$)? Whatever number I or someone else gives you, would be totally arbitrary and made up. And here is part of the problem of why there is no definition of good positioning.

Looking at astronomical images, I think that something like $10°$ or $15°$ is the lowest that can be called good enough. Anything below that would be "too edge-on". But instead of giving a hard value, let's just keep calling that angle as $\theta$ and carry up calculations avoiding giving it an explicit value and find a good value later.

Whatever the position of the observer, the longest green line is that of some star in the border of the galaxy. This gets better if the projection line between the observer and the galactic plane is exactly in the galactic center, which you state in your question that is the case.

So, this is what you have:

You observing the Milky Way

The galaxy is represented as a disc, since the galaxy has a non-zero height, the two orange horizontal lines represent lines draw on the bases of that the disc over its radius. The white spot is the galactic center, right in the middle green line which is the height of the disc ($h$).

The proportions of this entire thing depends only in four values: $h$, $\theta$, the radius or the galaxy ($r$) and the distance between the observer and the galaxy center ($d$).

Now, let's see that:

  • The line between the observer passing through the galactic center down to all the height measures $d + \frac{h}{2}$ Let's call this value as $x$. So:

$$x = d + \frac{h}{2}$$

Now, $x$ is the opposite side of a rectangle triangle. The galactic radius is the adjacent side. The purple/green line is the hypotenuse. The angle between the bottom orange horizontal line and the green part of the green/purple line is $\theta$. So:

$$x = r \times \text{tan}(\theta)$$ $$d + \frac{h}{2} = r \times \text{tan}(\theta)$$ $$d = r \times \text{tan}(\theta) - \frac{h}{2}$$

We needed the $h$ just to define why we needed to define $\theta$ in such a way to not make $\text{cosec}(\theta)$ is not too big. However, considering that:

  • The definition of $\theta$ is more or less independent of the value of $h$ as long as $h$ is not zero or something too close to that.

  • We know that $h$ is much smaller than $r$.

  • The galactic disc is not exactly a disc, but has a bulge in the center and is thin in the borders making $h$ variable, but going near zero in the borders.

  • If we were make to $d$ too short in order to be too close to the galactic center, then, $\text{tan}(\theta)$ would be too close to zero too, and hence $\text{cosec}(\theta)$ too big.

Then, we can just disregard $h$ and get a simpler equation:

$$d \approx r \times \text{tan}(\theta)$$

This last formula allows you to calculate the distance of the observer to the center for a galaxy with a radius $r$ and "maximum edge-on angle" of $\theta$.

Plugging in numbers and applying to the Milky Way

Now, if we were to observe the Milky Way and let's choose $\theta$ as $15°$. Let's consider the radius as $100,000 \text{ LY}$. Then:

$$d \approx r \times \text{tan}(\theta)$$ $$d \approx 100,000 \text{ LY} \times \text{tan}(15°)$$ $$d \approx 100,000 \text{ LY} \times 0.268$$ $$d \approx 26,800 \text{ LY}$$

I.e, this is a bit more than a quarter of the galactic radius.

Now, let's make it simpler and say it is exactly a quarter of the galactic radius (i.e. let's assign that $d = \frac{r}{4} = 25,000 \text{ LY}$). Then, which would be the angle $\theta$?

$$\text{arctan}(0.25) \approx 14.03°$$

Let's round it to $14°$. Is this value good?

$$\text{cosec}(14°) \approx 4.133$$

This means that a star in the edge of the galactic disc has a $4.133$ "obliqueness". So, to make it easy in layman terms:

  • The distance to the galactic center must be a quarter of the galactic radius.
  • The angle between the line of sight of the observer and the galactic plane in the galaxy's border should be no less than $14°$.

The first solution

This is a good solution.

Can this be further improved?

$14°$ means that observing a star in the edge is $4.133$ harder than observing at the center due to some star obscuring another star if the galaxy is considered a disc. But the galaxy is not exactly that since it bulges in the middle, thins out in the border and obscuring is not the only factor complicating observations of the stars in the border, but their distance to the observer also is. Also, since the borders are much less dense than the center, there is much less problem of observing them edge-on as long as the galactic center don't obscure them, which it does not since we are above it. So, you might go down to perhaps $10°$, but this just send us back to the problem of defining the arbitrary $\theta$ value and the smaller the $\theta$, the worse is the observation. Also, $14°$ is already near the lower limit for what most people would consider as convenient or acceptable.

But if you really want to, you can get tighter solutions by plugging in some other numbers. For example, with $\theta \approx 8.5°$, we get $\text{cosec}(8.5°) \approx 6.72$, then $d \approx 0.15 \, r$, and finally $d = 15,000 \text{ LY}$ for the Milky Way. However, $8.5°$ is really tight, almost edge-on, so I think that the $14°$ limit is better suitable.

  • 2
    $\begingroup$ Useful and through answer, but I think the real answer is given in the middle "Whatever number I or someone else gives you, would be totally arbitrary and made up" $\endgroup$
    – James K
    May 15, 2022 at 12:24
  • 1
    $\begingroup$ @JamesK Sure. I lust added it to the TLDR in the start to make it consistent. $\endgroup$ May 15, 2022 at 12:31
  • $\begingroup$ No mention of extinction. $\endgroup$
    – ProfRob
    May 16, 2022 at 5:35

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