How much time will an object (star, galaxy, nebula, cluster) take to transit from one end of a 32 mm SuperView eyepiece to the other end of it to go out of the field of view when a DSLR is connected to the eyepiece (Nikon d5500, sensor size 23.5mm x 15.6mm)?

I have experimented the time taken when you visually observe a star. A star takes about 290 seconds to move out of view. So it moves about 0.5 mm every 5 to 6 seconds. Will the object move at the same time out of view when a DSLR is connected? So I assume if I keep an exposure time of 8 to 20 seconds the object should move only by 1 mm or less. Am I right here?

  • 1
    \$\begingroup\$ The important number here is what is the angle of view of your setup? The focal length of the eyepiece isn't enough to go off. What is the focal length of the scope's objective (or tube if it is a Dobsonian)? \$\endgroup\$
    – Octopus
    May 6, 2016 at 8:00
  • \$\begingroup\$ @Octopus ....the apparent FOV of the eyepiece os 70° and the focal length of the OTA is 1200mm \$\endgroup\$ May 6, 2016 at 8:02

1 Answer 1


The math isn't too complicated.

Based on two facts:

  • The sky revolves 360 degrees per 24 hours (roughly)
  • FOV (rectilinear) = 2 * arctan (frame size/(focal length * 2))

We can calculate that it will take roughly 4.488 minutes to cross the sensor (I'm using the larger dimension although I'm not sure how you'll have it oriented), because:

FOV = 2 * arctan( 23.5mm / (1200mm * 2)) = 1.122 degrees


(1.122 / 360) * 24hours = 0.0748 hours or roughly 4.488 minutes

For a celestial object that is not over the equator it will move slower.

t = rate / cos(lat)

so at the poles it will take forever since:

t = 4.488 / cos(90) = infinity

and at 45 degrees celestial latitude it will take around 6.3 mins since:

t = 4.488 / cos(45) = 6.348

I think, usually the camera replaces the eyepiece, so the eyepiece is not normally a factor in the equations.

  • \$\begingroup\$ Your test estimated 290 seconds which is only a little more than 4.488 minutes, most likely because your test object was slightly off the equator. \$\endgroup\$
    – Octopus
    May 6, 2016 at 8:33
  • \$\begingroup\$ Thanx @octopus ...so if i keep an exposure of 10 to 20 seconds will i land up with star trails in my output image...??? \$\endgroup\$ May 6, 2016 at 9:57
  • \$\begingroup\$ And yes @octopus ....it will take me around 6 mins. Coz when i calculated, i neglected the time when the object was towards the edge of the field if view. So i calculated just enough of time required for photography. \$\endgroup\$ May 6, 2016 at 10:33
  • \$\begingroup\$ But @octopus i will b using a telephoto projection eyepiece which will b connected to my dslr....so should the size of the sensor and the focal length of the 32mm eyepiece b added in the above calculation...?? \$\endgroup\$ May 6, 2016 at 18:32
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    \$\begingroup\$ @dwayne, im not quite sure how you're setup. The eyepiece serves to diverge the focused image, which works great for your eye. It has another lens to refocus those rays onto your retina. I don't see how this works with your camera unless you still have a lens on your camera to refocus whats coming out of the eyepiece. In which case we'd also need to consider the stats of that lens. \$\endgroup\$
    – Octopus
    May 6, 2016 at 19:39

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