Rule of thumb

[Ags]

I know this is a bit niche, but I wanted to share my rule of thumb for approximating eyepiece field stops: if an eyepiece has about 60° apparent field of view, then the field stop diameter is roughly equal to the eyepiece focal length. It's the same formula as for calculating the field of view of a camera, so this approximation is best for eyepieces with exactly 57.3° apparent FOV 😀

For example, the Delite 3mm has a 3.2mm field stop, and the Delite 5mm has a 5.3mm field stop. As the Nagler 5mm has 82° AFOV vs the 62° of the Delite, you would expect the Nagler to have ⅓ more field stop, and it does with 7mm.

[Don Pensack]

Yes, calculated field stop = (Apparent field/57.2958) x Focal length

But, the accuracy depends on the amount and type of distortion present, and the accuracy of the stated apparent field.

Example: APM 30mm Ultra Flat Field (same as Altair Astro):

Calculated field stop = (70/57.2958) x 30 = 36.65mm

Actual field stop as determined by star timing: 36.3-36.4mm

So, is the actual focal length >30mm?  The apparent field <70°? Or the distortion modifying the results.

Well, measured apparent field is slightly above 70°, so it is the distortion that is modifying the results by 6.8%

 

In contrast, look at the  24mm TeleVue Panoptic

Calculated field stop 28.48mm. and actual field stop is 27.0mm.  Difference is 5.5%

 

The point is that you cannot trust a calculated field stop, even at 57.2958° unless the eyepiece has 0% distortion, which doesn't occur in real life.

BUT, you can use the calculated figure as a reality check.

Celestron says the field stop of a 10mm Luminos is 17mm.  The calculated figure is 14.3mm.  Celestron is blowing smoke.  That difference is ~19%.  Uh, nope.  Simply not believable.

And all the other sizes are similarly off.  They may be referring to the diameter of the field lens?  Or something else.

 

So it's useful to know the calculation so you can check the reality of the MFR's claim.  If it's close, discard the calculated figure.