<p>This technique for determining the field of view of a telescope and eyepiece is called drift timing. Since the sidereal day is about 4 minutes shorter than the usual solar day, a fixed point on the celestial equator rotates at about 15.04 arc seconds per second of time. Thus the true field of view in arc seconds is given by
<span class="math-container">$$TFOV=15.04 \hspace{2pt} t \cos(\delta)$$</span>
where <span class="math-container">$t$</span> is the time in seconds for the star to drift across the field, and <span class="math-container">$\delta$</span> is the declination of the star. Divide by 60 to convert to arc minutes, or by 3600 to degrees.</p>

CC BY-SA 4.0
2020-04-04T02:50:59.627Z
2020-04-04T02:50:59.627Z
amateurAstro
<p>Given the celestial coordinates of a certain star and the time it takes to pass across your telescope's eyepiece, how can you calculate the field of view of the telescope?</p>

<observational-astronomy>
Calculating Field of View of a Telescope Given Time of Another Object
MystaryPi
# Calculating Field of View of a Telescope Given Time of Another Object

(Asked by: username_0 on 2020-04-04)

Given the celestial coordinates of a certain star and the time it takes to pass across your telescope's eyepiece, how can you calculate the field of view of the telescope?

**Tags:** observational-astronomy


---

## 1 answers

### Answer by username_1 on 2020-04-04. Score: 5 | Accepted answer

This technique for determining the field of view of a telescope and eyepiece is called drift timing. Since the sidereal day is about 4 minutes shorter than the usual solar day, a fixed point on the celestial equator rotates at about 15.04 arc seconds per second of time. Thus the true field of view in arc seconds is given by
$$TFOV=15.04 \hspace{2pt} t \cos(\delta)$$
where $t$ is the time in seconds for the star to drift across the field, and $\delta$ is the declination of the star. Divide by 60 to convert to arc minutes, or by 3600 to degrees.


---

