My son and I are building an azimuth/elevation controller to point the telescope at planets using servos and a microcontroller. Assuming that our device is calibrated (that is, we tell our device to point at a specific RA/Dec and points in the right direction):

How accurate must positioning be (in arcseconds?) to point directly at a planet like Uranus and have it show up in the eyepiece? In case it matters, we are using a Celestron 70EQ with our own az/el hardware.)

(Perhaps another way to ask is this: if we have an outer planet in view, how many arc seconds need to pass before the object leaves the telescope's view?)

We are both new to astronomy, so our terminology or understanding could be off a bit, correct me where necessary.

Ultimately the motivation for this question is because we are considering the Astronomy Engine C library which guarantees within ±1 arcminute and want to know if that is high enough resolution to get the planet in view of the eyepiece.


  • $\begingroup$ Note that Saturn is much easier to spot than Uranus. There's a reason the latter wasn't definitively discovered until the 1700's, while Saturn was known to the ancients for basically all of recorded history. On a clear night without too much light pollution it can be visible to the naked eye. $\endgroup$ Commented May 3, 2023 at 13:55
  • $\begingroup$ @DarrelHoffman we have been able to see Saturn in the telescope when it was very bright in the sky so we could find it and zoom in. It was surreal experience to see Saturn and its moons in real life (not in a picture)! I'm hoping we can someday find Uranus on a day with a good magnitude, perhaps this az/el tooling can help get us close... $\endgroup$
    – KJ7LNW
    Commented May 3, 2023 at 22:45

2 Answers 2


The size of the field of view you will see depends on the eyepiece used and the telescope. For example, that telescope comes with a 20mm and a 4mm eyepiece, and has a focal length of 700mm. The linked page doesn't say, but they are likely Plossl eyepieces, which have an apparent field of view of about $52^\circ$. To compute the true field of view:

$$ \textrm{True FOV} = \frac{\textrm{Apparent Field Of View}}{\textrm{Magnification}} \\ $$ $$ \textrm{magnification}=\frac{\textrm{Telescope Focal Length}}{\textrm{Eyepiece Focal Length}} $$

So, your 20mm eyepiece has a true field of view of: $$ \frac{52^\circ}{700mm / 20mm} = 1.5^\circ $$ And the 4mm has a true fov of about $.3^\circ$.

A couple of side notes: The accuracy of the library used will likely not be a significant source of error, instead mechanical issues like flex and non-orthogonal axis will play a larger role, even if your motor position is perfect (and it won't be). Also, arcseconds (and arcminutes) are not seconds on a clock, instead an arcminute is 1/60th of a degree, and a arcsecond is 1/60 of an arcminute. As opposed to a minute, which is 1/60 of an hour. The Earth rotates at about 15deg/hour, so the Earth rotates 1 arcminute in about 4 seconds.


Eyepieces are different, the specifications for your telescope don't give the field of view, but you might expect about 0.5 degrees, or 30 arc-minutes. And a star near the equator will take a couple of minutes to cross the field of view.

So one arc-minute accuracy is sufficient to get the planet into your field of view and reasonably centred.


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