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Lane
Post Laureate
Reged: 11/19/07
Loc: Frisco, Texas
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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 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?
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WarmWeatherGuy
professor emeritus
Reged: 08/27/11
Loc: Orlando, FL 28° N, 81° W
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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.
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Lane
Post Laureate
Reged: 11/19/07
Loc: Frisco, Texas
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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 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.
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Phil Sherman
Pooh-Bah
Reged: 12/07/10
Loc: Cleveland, Ohio
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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
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