How phd2 does compute rms?

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Miguel G

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Aug 28, 2025, 5:30:34 AMAug 28
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Hi all,

Is the rms calculation based on the star guide or mount displacements?

An extreme example, with one axis (ra). Having a completely horizontal RA line in -2'' will show a rms of 2 or 0?

Is RMS computed from absolute deviation from the lock point (guide star reference) or relative variation in the curve? I guess the former applies, but just to be sure.

Regards

Bruce Waddington

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Aug 28, 2025, 5:58:24 PMAug 28
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It's the standard deviation of the sample which means it's measuring the dispersion of points around the sample mean.  The points themselves are offsets from the lock-point, not the guide corrections.  In your edge-case example, the mean of the sample would be -2 and that would be used in the standard deviation calculation:

sigma_formula.jpg

The fixed offset you were asking about - something the EE's sometimes call a DC offset - is intentionally ignored.  The purpose of the calculation in the context of guiding is to create a figure of merit that is correlated with guiding "quality", meaning taking into account errors that create elongated stars or smearing in an image.  A fixed offset doesn't do that, it's no different than a small pointing error arising from a slew to the target.  Of course, with a properly functioning system, the mean is typically near-zero and the discussion becomes moot.

Regards,
Bruce

Miguel G

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Aug 29, 2025, 2:15:53 AMAug 29
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Hi Bruce,

Thanks a lot for the kind reply and detailed explanation. All clear now.

Have a nice day.

Regards

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Miguel G

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Aug 29, 2025, 3:53:35 AMAug 29
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Sorry Bruce but now reading twice; that means that in my edge case where the mean and every point is -2, the rms would be 0, right?

bw_m...@earthlink.net

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Aug 29, 2025, 11:19:14 AMAug 29
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Yes, that’s right.

 

Bruce

 

From: open-phd...@googlegroups.com <open-phd...@googlegroups.com> On Behalf Of Miguel G
Sent: Friday, August 29, 2025 12:53 AM
To: open-phd...@googlegroups.com
Subject: Re: [open-phd-guiding] Re: How phd2 does compute rms?

 

Sorry Bruce but now reading twice; that means that in my edge case where the mean and every point is -2, the rms would be 0, right?

 

El vie, 29 ago 2025 a las 8:15, Miguel G (<mguti...@gmail.com>) escribió:

Hi Bruce,

 

Thanks a lot for the kind reply and detailed explanation. All clear now.

 

Have a nice day.

 

Regards

 

El jue, 28 ago 2025 a las 23:58, Bruce Waddington (<bw_m...@earthlink.net>) escribió:

It's the standard deviation of the sample which means it's measuring the dispersion of points around the sample mean.  The points themselves are offsets from the lock-point, not the guide corrections.  In your edge-case example, the mean of the sample would be -2 and that would be used in the standard deviation calculation:

 

 

The fixed offset you were asking about - something the EE's sometimes call a DC offset - is intentionally ignored.  The purpose of the calculation in the context of guiding is to create a figure of merit that is correlated with guiding "quality", meaning taking into account errors that create elongated stars or smearing in an image.  A fixed offset doesn't do that, it's no different than a small pointing error arising from a slew to the target.  Of course, with a properly functioning system, the mean is typically near-zero and the discussion becomes moot.

 

Regards,

Bruce

 

On Thursday, August 28, 2025 at 2:30:34 AM UTC-7 mguti...@gmail.com wrote:

Hi all,

 

Is the rms calculation based on the star guide or mount displacements?

 

An extreme example, with one axis (ra). Having a completely horizontal RA line in -2'' will show a rms of 2 or 0?

 

Is RMS computed from absolute deviation from the lock point (guide star reference) or relative variation in the curve? I guess the former applies, but just to be sure.

 

Regards

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Miguel G

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Aug 29, 2025, 11:34:02 AMAug 29
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