Teams / DSLR Documentation and Reduction / eps brightening during June?
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Dear all, I saw epsilon reaching about 3.80~3.81 by April end. By June 4, I gotit at 3.701, by June 27 3.750. At present(July19)it is stable for about 10 days at 3.68. Those results are from shots at very low height and significant background gradient in most cases (latitude 47.3 degrees, 5 deg est). Extinction has been compensated using HR stars surrounding epsilon at short distance and interpolation in function of height. I am continuing to check that results and extinction compensation, sometimes as heigh as 0.4 mag ! I am not sure all tricks have been understood... Yours truly, Roger
I only now started to observe eps Aur again after some weeks during June where I found eps Aur too difficult to observe because of solar conjunction (if that's the right term for an extra-solar-system object?). I wonder whether those DSLR measurements were properly adjusted for extinction effects. The raw measurements (w/o extinction correction) that I made this weekened would indicate a magnitude of 3.6 to 3.5 even ..... so I'll now play around with some software to properly correct for extinction. CS HB
Heinz, Could you give the reduction spreadsheet I posted here a shot: http://www.citizensky.org/sites/default/files/Reduction-Intermediate.xls and post your results (or the spreadsheet) here so I can evaluate whether or not the solution method proposed therein works correctly? Ithink it's okay, but I'd like a high air mass observation from another observer just to double-check before Ipost the document to the CS tutorials. I gave quick discussion of the spreadsheet here: http://www.citizensky.org/teams/dslr-documentation-and-reduction/figured... Please let me know if you have any problems. Thanks, Brian
Heinz, Could you give the reduction spreadsheet I posted here a shot: http://www.citizensky.org/sites/default/files/Reduction-Intermediate.xls and post your results (or the spreadsheet) here so I can evaluate whether or not the solution method proposed therein works correctly? Ithink it's okay, but I'd like a high air mass observation from another observer just to double-check before Ipost the document to the CS tutorials. Please let me know if you have any problems. Thanks, Brian
Hi Brian Ok so I tried the spread-sheet. Some observations: a) Calculating all those air masses is a real pain and I honestly think it's too error prone and tedious for most people. If you are using a web form or similar calculator, you have to enter quite a few numbers over and over again for each measurement which is plain boring and exhausting (unless you are using a tool like the MMT from AIP4WIN where you can save the star coordinates. I tried a beta of AIP4WIN but somehow it would always crash, I will file a bug report later) b) I got a negative air mass coefficient in cell E36 (-0.216), while your example sheet has a positive one 0.216 (funny...exactly the opposite sign!!). So, which sign is correct?? In the final formula for cell G43 that would calculate the mag for the variable star, a correction term (air mass) * (air mass coefficient) is ADDED , so .... more air mass should result in a correction to make the star brighter , not darker, so I'd think that the air mass coefficient in cell E36 should indeed be negative! Correct??? Can you recheck your example data? All in all I get a magnitude for eps Aur on 18th of July of 3.66 which seems to be on the bright end of observations. I'm not very confident about this particular measuremnt (some clouds nearby) but at least it doesn't seem to be totally off.. I think if it's possible to reduce the number of comp stars in the sheet, it should be done to reduce the workload per observation. Or include the extinction calculation in the Excel code itself. What do you think? CS HB
Hi, I have not seen your work but noticed your mention of possible sign error. It is very easy to make a sign error in the extinction correction. I think this is how those reports for the end of May and early June showed epsilon rapid brightning whileothersmeasured a dim unchanging epsilon. Daylight visual observations under low airmasslater confirmed eps had not brightned in June. I prefer to use the differential photometric equations so I do not have to calculate the Zero point. Also if the color correction is determined separtely, so you never have more than one unknown in one equation to solve either for k' or the variable star. V1 - V2 = (v1 - v2) + Tc*[(B-V)1 - (B-V)2] - k'*(X1 - X2) is the differential form I use. Where the 1 subscript is epsilon and 2 is the comparison. Small letters are the raw differentialmeasurment and Capital letters are Catalog or V1 the corrected value of Epsilonbeing reported. Most all comparison stars, such as lambda and etaare below epsilon both evening in May and morning in July. Thus X2 is greater than X1 such that the differential Air mass X1 - X2 is a negative number. k' is always a positive number. Thus - K'*(X1 - X2) is a positive quanity ADDED in most cases to (v1 - v2). It is possible for those who have the wrong sign in the extinction correction to find good agrement with another comparison star.Say epsilon measured bylambda andthen by eta yeild almost the same resutlts. But add more comparison stars and you will seegreat scatter. I know, it happened to me until I corrected the sign error and was able to get 5 or 6 stars to measure eps with SD. of .005 sometimes while underhigh air mass. Some peoplemay be takinga short cut andthinking because they see a minus sign in front of k' that they should subtract the Airmass extinction. Of course if your comparison star is above your variable star, then X1-X2 would be positive and you would subract k'*(X1-X2). But in the case of epsilon, most every comparison star has larger air mass thus potential for confusion and a sign error if careful attention is not paid. Good Luck, Charlie
Charlie, We were discussing the Excel spreadsheet I mentioned earlier that computes the extinction, color correction, and zero point given n-data points using a least-squares fit to the given data. What we noticed is given the sample data I came up with k = 0.216 and given some data that Heinz provided, he obtained k = -0.216. I've double-checked the input equations, re-derived the least squares fit from first principles, and verified the math was implemented correctly in the Excel sheet. I started with the equations we've exchanged several times: (V-v)_i = -k X_i +e(B-V)_i +z where "_i" denotes the i-th calibration star. The spreadsheet finds the best-fit (least-squares) of the datapoints to the plane described by (-k), e, and z. Formally only three stars are required, but after the errors are propigated the uncertainties on these calibration factors are vastly improved. The cool thing was that even though we have different signs on the extinction coefficient, we both get plasuable magnitudes. I've got a hunch on why this is and it's really simple, but I need to talk with Tom and Heinz first. Thanks for your input, Brian
Hi Brian, Your input equation looks correct and if both you and Heinz adopt classical convention, k' needs to be positive. BTW CS and HB got eps Aur on July 18 3.66Vmag, I got 3.658Vmag on Jul 19. Very interesting, Charlie
Hi! The formula in the Excel sheet adds the term [Extinction coeff] x [Airmass] to the final result , the classic form seems to be to subtract it ( -k' * X) so the value as computed by the Excel sheet would have to be negative, corresponding to -k' in the classical equaltion. Interesting to see that 3.66 seems to be about right. My frames were done with airmass X > 2 , so it would be quite sensitive to errors in the extinction calculation. CS Heinz
Formally we solve for (-k), e, and z (I've corrected this in my post above, typos...) from this: (V-v)_i = -k X_i +e(B-V)_i +z So adding it back in is okay (there isn't a mathematical difference between A- k*B and A +(-k)*B anyway). I'm still puzzeled why Heinz and Tom's data resulted in a different sign on the air mass. My idea that they inverted their images when measuring imags turned out to be a null result. Given that we get acceptable values for the check stars and for the variable (compared to single-channel photometry), Ithink the spreadsheet is okay. Thanks for your input guys! (I'll address the other concerns back in the other thread). Cheers, Brian
Hi Brian and Heinz, Sorry to butt in agin, but as an outside observer, I think I now understand your confusion over the extinction sign problem. Suggest you carefully adhear to the classical form of the photometric equation in which k' is a positive number as well as X. Then let the photometric equation subtract the product of k'*X (which by the way must be a positive number by classical definition). Brian said "Formally we solve for (-k), e, z".Would it not be better to solve for just (k)? and thenseparately let the photometric equation supply the minu sign? This could be the reason Heinz needsthe -k'. Both your spred sheets and my differential method seems to agree on the same magitude for eps on or about 18 to 19 July so I'll try not to bother while you refine the differences. Clear skies and lower airmass, Charlie Hofferber
Charlie, The problem isn't in the placement of the negative sign in the equations, nor is it whether or not we've kept the equations in a classical form as Heinz and I both used the same spreadsheet with the same equations to compute our answers. The only modifications made were inserting the imags and airmass values for the images. Mathematically it doesn't matter if we do A - k * B or A +(-k)*B and given that we're using linear algebra to solve (well, through a least-squares fit) the sign of (-k) will propigate though. Really we can just call (-k) k' and be done with the issue of a negative sign as the sign propigates with the variable. We just need to formally state what that quantity means when we write our tutorials and I think everyone will be okay with it. Heinz and I fully understand what the k'*X term in the equation is suppose to mean, that's not the issue. The problem that we have is using the same equations, same spreadsheet, same (six) calibration stars with different imags and airmass values we obtained the same magnitude for the extinction coefficient (i.e. |k'|), but different signs. The zero point and color coefficient both had the same sign (but different magnitudes for the obvious reasons) . Even though the signs are different, the spreadsheet yields a correct magntidue for eps AND a check star. In essence, this means that k' * X is a positive quantity for one of us and is a negative quantity for the other. That implies that the additional air mass ADDs light for one of us but REMOVES light for the other. Again, this isn't a misplacement of the minus sign as no modifications to the mathematics in the spreadsheet were made so we used the same method but came up with contrasting ideas of what air mass does to the light from stars. We've come to the conclusion that one of the images from which the imags were derived is somehow contaimnated with some gradient (a haze, light pollution, the moon, etc) or that the extinction gradient isn't well defined by the stars in the field of view of the camera. I've plotted up the raw data and indeed they show a clear difference in slope so I'm sure it's not an artifact of the spreadsheet's calculations. The important thing is that even with this additional contamination the equations were still solved correctly and the magnitdues for eps and a check star were within a reasonable tolerance of the accepted value (typically 10-20 mmags). I think this speeks to the resilance of the calculations done in the spreadsheet and really shows that DSLRphotometry is a real possibility. Now if we can just get a list of calibration stars with reasonable uncertainties and fully propigate errors we'll be in business! Cheers, Brian
Hi Brian, Afterhaving tested iton a large number of casesI routinely use the following way: What we need is just to determine the differential attenuation of the star light due to the air path; that between our reference stars and our target star. We don't need to determine the absolute extinction at all. That can be done only using(and within)our images providing we have enough stars known as stable,distributed in the vertical direction andbracketing our target star. Next it's easy to determine the difference between the magnitude we observe and the catalog value. Further step is to calculate the interpolation coefficients of this error in function of the VERTICAL PIXEL POSITION of stars.Eventually it's easy to determine the correction to apply to any target star into that bracket (extrapolation reveals quickly wrong... )
Iuse that principle two ways. The simplest is to only use three stars surrounding epsilon at short distance: HR1550, HR1558, HR1644. I apply a linear interpolation in this case. Nextis toget a larger extent with the same stars plus eta and zeta, this case needs a second order interpolation. Both give usually a very similar result for epsilon ( within ~3 millimag ) My present results are 3.68~3.69. If you see any issue with this method or have question, please let me know. Yours truly, Roger
Hi Roger, I guess you are doing photometry with a CCD tho, right? Some of the comp stars you mentioned have magnitudes down to 6.0, and I guess the limited dynamic range of DSLRs (at least those with 12 bit sensors) will make it a challenge to get a good SNR for them without overexposing epsilon Aur ? CS HB
Hi Heinz, It's DSLR ! No problem in fact, we did discuss this with Robin a couple of months ago if you remember. We are not using a single pixel but probably 100 and next at least 10 images. That means, if the saturation level of a 12 bit sensor (12 bit ADC in fact) is 3500 we got at end 3.5 M ADUs, that's far enoughresolution ! More important, the e-capacityis also multiplied by about 1000. Thewell depth of an APS-C 12 Mpixel being about 30.000 e- that means we can accumulate 30 Me-.If wetarget an SNR of about 300 that leaves adynamic rangeof about 300 or 6+ magnitudes. ACCD with aV-Johnson filter is nomuch better, just a factor 2 against a DSLR with its 2 CFAgreens per RGGB Bayer cell. The true issue is the optimisation of the shot conditions. In general I see the CS membersusing a much too high ISOsetting: often 800 ISO !That means the saturation is no more such of the sensor but the ADC range. For my 450D thiswould bea limitation by a factor 7 of the e- capacity ! (or 2 mag dynamicloss, without any significant gain of SNR) Next major factor is thesize of the lens aperture. If the standard Canon 18-55 zoomis used at 55 mm the max aperture diameter is about 10 mm. When I use an old 200 mm tele-lens at F/4 the aperture is 50 mm. This is a factor 25 on number of photons / electrons being accumulated, that's another 3.5 mag! My present shoting conditions are: Canon 450D (CMOS, 14 bits), 100 ISO, 10.4 sec, fl=200 mm at F/4, defocus to about 100 RGGB cells depending sky conditions. I take 24 shots per series. No dark, no bias, but careful vignetting correction, a blue compensation to V-Johnson (instead transform) and the differential extinction correction based on pixel position. The result is currently an SD of about 0.005~0.007, identical for allstars from 3 to 6 mag. That shows the SNR is not the limiting factor but the scintillation (strong at present low heights). SD of stars at mag 7~8 starts to increase due to decreasing SNR. Yours truly, Roger
Could you please further explain what the difference in the ISO setting make in SNR and dynamic range. I indeedis usingISO 800 on my Canon 450D. With this setting I can get confident values of +/-0.02 mag for stars between 3-6.5 and about +/-0.05 for mag 7 stars. I use 10 stacked images and 5 sec and a EF35-80 lens at 50 mm.
Hi, I get this opportunity to make thatpoint clearer ( I hope ! ) ISO setting in DSLR is only an adjustment of theGAIN of the read amplifier. This amplifier get the pixel electron charge as input, makes some amplification and outputs it to the ADC (analog to digital converter) The ADC has a limited range: 4096 ADUs or 16384 ADUs following it's a 12 or 14 bits. It's purely arbitrary units. What istrulyimportant is the number of electrons accumulated from a star at each pixel and in a group of pixels of the imager. For a typical DSLR, APS-C, 12 Mpix, the maximum capacity of each imager pixel (photosite) is about 25000~30000 electrons (the said well depth) that determines the saturation level of a photosite (dynamic limitation). Then: ADUs = Gain x Nelectrons Result is that there are two possible waysforsaturating depending the ISO setting: - At lowest ISO (gain ~0.5, often 100 ISO) the saturation occurs at the photositelevel. Typicaly 28000 e- and for a 0.5 gain: 14000 ADUs. - Now if you push the gain by a factor8 that is 800 ISO the gain of the chain is 4 and the saturation occurs at the max range of the ADC that is 16384ADUs. This is only 16384/4 =4096 electrons at the pixel level. Conclusion: The pixel dynamic range is reduced by a factor 28000/4096 = 6.8 !! Up to a certain point it's possible to compensate it using a larger defocus. Now, where are the noise sources ? Near 100% at the pixel level and the amplifier input stage. - Thefarlargest noise is linked to the electron charge itself: the shot noise = square_root(Ne) - Thepixel read noise (Gaussian) is about ~4e (where Ne is the e- number) - Various "fixed pattern noises" FPN (fixed imager defects... ) are few e- for few secondsexposure. Then theSNRshould bemore or less: SNR = Ne / square_root(Ne+ square(read_noise e-)) Integration shall be done on all pixelscollecting the star light. All these noises are amplifiedby the same factor than the signal. That meansthe SNR expressed in ADUs ismoreor lessindependant of the ISO settings if enough e- are accumulated. In fact an extra read noise from the upper amplifier stagescould appear (depending DSLR type, case of the 450D !) at lower ISO setting, this is due to a non-optimal design of the amplifier gain control. Due to this it could be favorable to increase the ISO settings in case of weak signal when the read noise becomesa dominant factor. A calculation in then necessary to optimize the shotting conditions. Note: for the 450D the electronics chain gain is 0.44 ADU/e- at 100 ISO and the imager saturation occurs about 14800 ADUs at 100 ISO. The imager linearity is excellent (~0.5% ) up to about 13000 ADUs at 100 ISO ( that is about 30000 e- ). Then the gains are proportional to the ISO. The read noiseis10, 10.5 12, 16, 25 ADUs at 100, 200, 400, 800, 1600 ISO. The key to get a low uncertainty (high SNR but also sky scintillation averaging with long enough overall integration time) is to maximize the number of electrons accumulated for a given star. That means a lens aperture as large as possible (its surface - not the F# ! ) and a long enough exposure, next is to use a large enough number of shots. You could probably get a larger aperture at 80 mm focal lengthand next do nothesitate to expose longer including significant star trail. At endthe defocus shall be adjusted to avoid saturation (50~60% level is typical, 50 pixels at least to avoid the CFA sampling effect) I would also recommend to measure individual images and average the results, not stacking. It will allow you to calculate the trueSD (standard deviation) of your endresult instead of a theoretical 1/SNR. It is the right indicator of a good overall process (and sky conditions !). SD about 0.005 is possible with a DSLR on eps AUR after averaging about 12-24 image results depending sky conditions. Yours truly, Roger
Thanks for your excellent explanation. This gives me some new thougts. I think the first I will test is to use a longer exposure time, maybe 8 sec instead of5 and ISO 100instead of ISO 800and just see what effect that give. Doing a couple of series with booth settings should give the answer pretty quick.
Hi Thomask, Not sure ISO 100 in your case is right. I got the info on your lens (it is unknown here) and it seems to me the entrance pupil at 50 mm fl is about 10 mm diameter. With 4 sec exposure this is a ratio downabout 65 timesfrom the case I did discuss with Heinz. Much less photon/electrons per shot. What you could do is to push to 80 mm fl(14 mm aperture) and next push the exposure to 10.4 sec with some star trail, acceptable due to the 80 mm fl. (I use 10.4 s with 200 mm fl) That will give you 5 times more photon/electron per shot. Next you could adjust the defocus to about 200 pixels and adjust the ISO to be about 60% of the ADC range. I have not made the math but I would not be surprised you end up about 400 ISO. Both actionswill improve both the SNR and the scintillation SD. Clear Sky ! Here it's improving after a couple of poor days, but not sure enough for tomorrow morning 4 o'clock ! Yours truly, Roger
I did a lot of testing before I begun a yaer ago with photometry. Different aproches to process and meassur the images and different camera settings. Unfourtunaly I didn't document all the testing I made so I don't remember what result I got with different ISO settings. Since then I have done several 100:s meassurment of epsilon with my current settings and I estimate that the average error is +/-0.02 per image. It is a hard job to do all the testing. I have an other 50 mm lens with bigger aperture but the initial test did not give better values. I want to use the bigger field that 50 mm gives as I have compairsion stars that I want to use a bit from epsilon. So the first thing I will test is lower ISO setting and longer exposure time next time I have the opportunity. It's a cloudy period here right now...
Hi all, Agree with Roger that ISO 800might betoo high. I would recommend you try ISO 200 as well as ISO 100 recommended by Roger. Many image chips seem to have a "natural ISO" of about 200. That is to say, the number of photons to darken ISO 200 film will also just saturate an image chip. To achieve other ISOs the DSLR just amplifies or attenuates the signal from the image chip at the expense of loss of dynamic range and poorer S/N.As Roger pointed out, to achieve ISO 100 a gain of 1/2 or an attenuator is used. In theory this might degrade the S/N,therefore I suggested you also try ISO 200. The rull of thum is most DSLR cameras achieve their best S/N and greatest dynamic range at the lowest ISO setting. The trick is to find the right lens focal length/opening, exposure and acceptable trailing/defocuse to utilize the lowest camera ISO. So why does your Canon 20Dhave aISO 100 setting and my Nikon D100 only goes as low as 200? Marketing I suspect to apeal to photographer's that miss their ASA 100 film. If you want to look at scintaltion error see: Dravins et al 1997 PASP 109,173equation delta Mag scint = (0.09* A 7/4power) / (D2/3power * sqrt(2*t)) Iuse a 50mmf1.4 lens stopped down to f2.8 at 8secexposure at ISO 400 for X=2 airmass and slight defocused so that star radius is about 6 pixels (AIP4WIN default). for A=2 airmass, D= lens opening 1.79cm, and t = 8 seconds delta mag = 0.026, thus average of 10 images I expect about .0026 SD From the formula, we can see, there is less scintilation for lower air mass, wider lense opening, and longer exposure time. Charlie Hofferber
Thanks for the reply....very impressive indeed. Your technique (interpolation based on pixel positions) is interesting. It's kind of like making a synthetic flat field correction, with the flight source outside the atmosphere, isn't it. I have to try this myself next time, also heeding your advice on how to collect more photons. Thanks & Clear Skies HB
Hi Heinz, You are right, this is very similar to a flat-field synthetic correction being directly applied to stars instead the overall image. Yours truly, Roger
Hi Brian! Thanks for checking my data, this gives me the confidence to actually report the value as the fit seems to be not so bad. As for the mysteriously changing sign of the air mass coeff: I guess the air mass correction formula in the Excel sheet will effectively correct any "transmission gradient" that exists in the images and have a orienentation roughly the same as the extinction. So for example if you have a gradient in your images in the alt direction becaiuse of whatever other effect (flat-fielding not optimal, haze, ...) that is stronger than extinction effects and in the other direction, this would lead to the observed effect. Nr of comparison stars: I guess it helps precision a lot to have so many. If the extinction calculation can't be done in Excel itself, another idea would be to have a dedicated Citizen Sky customized airmass web-calulator maybe as a short term Team effort. You would just enter your observation location coordinates, time zone (could be saved in cookies even) and the observation time, push a button and voila, the web app would use the existing (repaired :-) ) code of the AAVSO airmass calulator on the most popular comp stars in the eps aur field (e.g. the ones from your Excel sheet). No need to type in the sky coordinates of those stars over and over again. CS HB