A few days ago I did some transform coefficient measurements based on SA32. Then last night I did the same thing with SA35. I was pleased to find that the transform coefficients came pretty close.
So out of interest I decided to plot the two sets of data on the same graphs. It turned out that in the (B-V)v(b-v) graph the two sets of data overlapped really well. However in the other two plots (B-V)v(B-b) and (B-V)v(V-v) although the lines were parallel there was a small but distinct gap between the two lines. In other words the slopes were the same but the intersects were slightly different.
The question is - what would cause that and should I be bothered about it? In principle I don't think it should matter because we only use the gradients. But I was just interested.
My idea in doing this was to see if it was possible to just keep adding points to the same plots to be used by the least squares line fit. My other idea is to just take the average of the transform coefficients.
Cheers
Steve
Steve,
It's hard to answer your question without knowing the figures. Last year for a particular personal project I calculated two sets of TCs from images taken during the one night from two different sets of standard stars. The intercepts and TCs were as follows:
B-V/b-v: Intercepts 0.1, 0.086; TCs 2.489, 2.498
B-V/V-v: Intercepts 20.147, 20.173; TCs -0.112, -0.109
The equipment was a DSLR camera and a 200mm lens at f/3.5 in a fixed position on a tripod. The stars were 5th to 7th magnitude (V) E Region Standards.
Roy
Steve;
IMO, it is not at all unreasonable that the image zero point on the two nights was different. Therefore, in the case where you are plotting the b-v difference as opposed to the V-v difference, the zero point difference should be removed for the b-v difference but not in the case of the V-v difference. At least that makes sense to me (almost)?
More importantly, you are concerned with slopes only, so the intercept does not matter. IOW, the coeff will be the same despite the shift up/down on the y axis.
Your last sentence is interesting. I think "adding" the different night/field points to the same plot should be similar to the "average" slope/coeff procedure BUT I bet the error (scatter) in this plot will be much worse than from the two single linear fit plots averaged. Does this make sense?
Ken
My…
Hi Roy and Ken,
My measurements are nowhere near as consistent. These are the TCs I obtained.
SA32 SA35 SA29
b-v 2.027 2.110 2.018
B-b 2.768 2.699 4.031
V-v -1.565 -6.427 -3.906
I did SA32 on a very clear night and the graphs look pretty tidy. The other two I did on few nights later but there was a lot of haze about and the seeing wasn't good. There was a particular problem with SA29 because all the reference stars were at the very edge of the field. The plots for SA35 and SA29 have a lot more scatter on them.
Ken, I understand what you mean about plotting all the points on the same graph. I tried it just to see what came out and it's clear that it's better to average the coefficients.
At the moment I'm just playing around and getting practise in making the observations and processing of the images.
Steve
Images for transformation coefficients should only be taken in good conditions.
Roy
Yes Roy, I found a book called "A practical guide to lightcurve photometry and analysis" by Brian D. Warner and the author makes that same point. But we've had so few good nights over winter and recently that I just wanted to get out and practise even if conditions were not ideal.
Steve
Hi Steve,
The basic equation looks something like
(V-v) = Z + k'X + k"X(B-V) + Tv(B-V)
Where Z is the true zeropoint of your system (depends on sensor, optical transmission, seeing, etc.) When you calculate transforms using a standard field, the first term with "X" in it becomes a constant, as the assumption is that all objects are being observed at the same airmass. The second-order extinction also includes a color index, and so does not really go away, which is why you observe the field near the meridian (lowest airmass) so that it does not significantly impact your solution. So you get
(V-v) = Zo + Tv(B-V)
with Zo = the intercept of the linear least squares line. Note that Zo actually is
Zo = Z + k'X + k"X(B-V)
that is, it is similar to the true zeropoint but including extinction terms. If you observe a second standard field at a different airmass, then Z is the same for both fields, but the extinction terms differ. So Zo will have slightly different value, and therefore the linear fit intercept will differ. This is the most likely reason why you get parallel lines, but offset. For color index solutions, you are subtracting V from B, and so the extinction terms go away since X is identically the same for the B and V exposures, and you will not see an offset between the two fits.
As Ken mentions, it is probably better to determine your coefficients for SA32, and then determine your coefficients for SA35, and then average the two results, rather than try to combine the two sets into one fit.
Arne