Assuming a well constructed mount, what is the accuracy of a 4096 encoder in arc-minutes?

360 degrees x 60 arc-minutes/degree = 21600 arc-minutes

21600 arc-minutes / 4096 = 5.3

So each click of the encoder is 5.3 arc minutes.

Would that mean the accuracy is 5.3 or would it be some multiple of that, 10.6, 15.9?

# Check my math

Started by
Lane
, Nov 01 2012 03:52 PM

3 replies to this topic

### #1

Posted 01 November 2012 - 03:52 PM

### #2

Posted 01 November 2012 - 04:28 PM

The math seems reasonable. The 5.3 arc minutes would be your resolution, not the accuracy. The biggest source of error would be a bias which could be anything (e.g. 10°). After that there might be some periodic error which would hopefully be only plus or minus a few arc minutes.

### #3

Posted 01 November 2012 - 06:29 PM

What do you mean by bias? 10 degrees would make the DSCs worthless.

This is on the T-Rex alt-az mount.

I was kind of thinking that as long as I use a high power eyepiece to pick my two alignment stars so I can get them centered properly then I should get targets to be within + or - 5.3 arc minutes of the center of an eyepiece.

The only possible problem I can think of is if the longitude axis and the latitude axis are not exactly 90 degrees apart.

This is on the T-Rex alt-az mount.

I was kind of thinking that as long as I use a high power eyepiece to pick my two alignment stars so I can get them centered properly then I should get targets to be within + or - 5.3 arc minutes of the center of an eyepiece.

The only possible problem I can think of is if the longitude axis and the latitude axis are not exactly 90 degrees apart.

### #4

Posted 01 November 2012 - 09:40 PM

Your math is good only if your base assumption is correct. The assumption you made is that the encoder is mounted on the worm gear. If the encoder is mounted on the worm, then you will get a much higher resolution. Each rotation of the worm will have 1440 steps, giving you a much nigher resolution.

Phil

Phil