# Wich stard field and other question

Affiliation
American Association of Variable Star Observers (AAVSO)
Sat, 05/11/2024 - 10:14

As a new AAVSO member is the first time that I participate in this forum.

I attended the AAVSO CHOICE course CCD Photometry, Parts I and II, and put some questions to Ed Wiley and he suggested posting them to this forum:

1.       In which step VPHOT converts the values in exoatmospherics. Are the comparative stars exoatmospheric magnitudes, or not yet? I suppose that the standard magnitudes are exoatmospherics.

The equation    mo = m – k* X

mo: exoatmospheric magnitude instrumental

m = observed instrumental magnitude

k = extinction coefficient

X: air-mass = sec(zenith angle)

X = 1 at zenith 90º

If you plot instrumental magnitude vs. airmass, at airmass = 0, mo will be the exoatmospheric magnitude, above the atmosphere. Taking data and calculating the zenith angle and the airmass for each, the plot is a line image and its slope is the extinction coefficient. But I didn’t find this value in VPHOT. Perhaps because this correction is not important if you take the images near the culmination ? (D. B. Warner: Light curve photometric and analysis).

2.       Which is the best criterion to choose the standard field chart: latitude geographic localization of the observatory, the declination of the target?

3.       But if I choose another chart with different declination the coefficients will be different?

Thanks.

Luis Barneo

Affiliation
American Association of Variable Star Observers (AAVSO)
Extinction Correction

Luis:

1. Extinction is typically ignored in 'differential' photometry conducted on your single image because the field of view is small (e.g., <1 degree). If you calculate the delta X between a target and comp star on an image with a field of view of about 1 degree (+-0.5 degree), you will find that the correction is usually less than a few millimag. You have reported the proper equation above, so you should be able to do the math? Run this calculation at a zenith angle of 10 deg but also run it at 60 deg to see the difference. What delta X did you get? What delta magnitude do you think is significant? How close to the horizon do you think you should image stars for photometry?

Since this is the common assumption for differential photometry, there is generally no need to calculate or apply extinction in your image. Make sense?

2. However, why does VPhot allow one to enter extinction coeffs into the telescope settings table? It implies that this correction to exoatmospheric magnitude is conducted. Frankly, I need to discuss this issue with the VPhot team to re-confirm where in the code extinction is corrected.

3. I don't really understand your questions about standard field chart. Are you asking how to measure the extinction coefficient? What would you want to measure and plot? Note that the extinction coeff is the 'slope' of the plot of magnitude vs. airmass. Remember that the entire plot line could move up or down with different instrumental magnitude shift and the slope would NOT change. Only the intercept would change!

Ken

Affiliation
American Association of Variable Star Observers (AAVSO)
Extintion correction

Ken,

Thanks for your answers. Respect to the question 3, it emerge for my lack of knowledge. If there are many standard field charts if for some reason: number of stars, magnitudes, colour, etc. And I imagined that other reason would be to fit the best chart  in relationship to the localization of the observatory or the star to study. An example. My observatory is at +42°. I want to study a star that will cross the meridian at midnight with an altitude culmination of 40° (airmass 1.56), and I want to determinate the transforms coefficients for the first time. Which standard field to choose?

1. One with similar altitude at culmination with a RA difference of few degrees.
2. Or, one with a culmination at zenith (airmass 1) with a RA difference of few degrees.
3. Or one with a declination similar to the latitude of my observatory because I will study others stars throughout the year.

For your answer I infer that there is not a significantly difference in the selection. Isn't it so?

Luis