Lane
Post Laureate
Reged: 11/19/07
Loc: Frisco, Texas

Check my math
#5499613  11/01/12 04:52 PM

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Assuming a well constructed mount, what is the accuracy of a 4096 encoder in arcminutes?
360 degrees x 60 arcminutes/degree = 21600 arcminutes
21600 arcminutes / 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?

WarmWeatherGuy
Carpal Tunnel
Reged: 08/27/11
Loc: Orlando, FL 28° N, 81° W

Re: Check my math
[Re: Lane]
#5499656  11/01/12 05:28 PM

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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.

Lane
Post Laureate
Reged: 11/19/07
Loc: Frisco, Texas

Re: Check my math
[Re: WarmWeatherGuy]
#5499782  11/01/12 07:29 PM

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What do you mean by bias? 10 degrees would make the DSCs worthless.
This is on the TRex altaz 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.

Phil Sherman
PoohBah
Reged: 12/07/10
Loc: Cleveland, Ohio

Re: Check my math
[Re: Lane]
#5500090  11/01/12 10:40 PM

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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
