Page Comparison

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Solid angle Omega subtended in an angle A from celestial pole is Omega=2pi*(1-cos(A)). 

So within 2 airmasses we can get to 3/4 of the entire sky. A six-band, annual 3pi survey would require 6*3222 = 19K visits. At 50 seconds per visit, this is essentially 16K minutes ~ 270 hours = 30 nights per year. But of course it's only the marginal investment that should count. 

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Number of fields vs. declination

Cadence, coverage, and passband trades.

Only considering time between astronomical twilight, using skycalc:

  average duration is 9 hours. Obstime=9.2+2*cos(2pi t/yr). 

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Sky Brightness from LSST ETC

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footprint is 26.82 pixels, Gaussian weighted. Units are electrons in 15 sec exposure. Moon is 90 deg from boresight

SDSS cumulative DR1 sky brightness distribution (no bright time imaging...)

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1) additional optical attenuation due to increased atmospheric path length

2) degraded "seeing". 

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Plot of seeing degradation vs. airmass, and polynomial fit:

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SNR scales as source flux in the numerator and (for unresolved objects) as seeing in the denominator. Flux at an airmass "a" is reduced by a factor f(a)=10^(x*(a-1)/2.5) where x is the extinction coefficient listed in the table above. So SNR vs. airmass at fixed exposure time for unresolved point sources scales as SNR(a)~10^(x*(a-1)/2.5)/(0.35+0.72a^2-0.07a).

Slew times. 

Oct 19 2013, CWS. 

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See http://www.gb.nrao.edu/~rcreager/GBTMetrology/140ft/l0058/gbtmemo52/memo52.html for az rates vs. zenith angle. 

Coverage Rate

At 35 seconds per visit and 9.6 square degrees per field, we cover the sky at a rate of 7900 square degrees in an 8 hour night. That means that on average we revisit interval (no weather) is 3 days for 18,000 square degrees.  

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The position of objects on the sky changes in right ascension direction at an angular rate of 15 degrees per hour times cos(declination).  How long does it take the sky to rotate by one field width, as a function of declination? It takes 3.1/15 = 0.2 hours = 12 minutes on the equator, and so t(dec)=cos(dec)*12 minutes elsewhere. So if we were scanning along the meridian we would have return to a given declination at an interval of cos(dec)*12 minutes, to get full coverage at minimum airmass for each declination band. At 50 seconds per field (average) in 12 minutes we would cover 3.1*12*(60/50)=45 degrees of declination.  

One potential approach:

1. determine the rank-ordered priority of all fields on the meridian, or at that night's minimum airmass if they don't transit, in each passband, for different potential values of seeing. 

2. reject the fields that never appear in the top ~1000. These have such low priority we'd never get to them in a single night. 

3. For each parametric value of seeing, compute the sequence of observations that maximizes the merit function, including the slew overhead contribution. 

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Took Gautham's data set and did correction for airmass, make cumulative plot of delta zero point, sorted. 

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Stubbs Notes, May 25 2014. 

Jaimal and I have agreed that weighted sum and a product figure of merit amount to the same thing. So we'll stick with the weighted sum, and compute a Figure of Merit accordingly: , 

FOM=$\Sigma_{programs=1}^M \Sigma_{fields=1}^N w_{program}w_{field} M_{field}$,

where the weights w and merits M are drawn from multiple considerations. We'll tune values of M to range from 0 to 1, where they saturate. Some candidate elements for the merit by field:

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$M_{temporal,1}=(1-e^{(-t/\tau_f)})$, where $t$ is (partial-credit) time elapsed since last observation and $\tau_f$ is field-dependent max unobserved gap.

$M_{temporal,2}=(1/2+(1/\pi)*atan((t-\tau_1)/\tau_2))$

$M_{seeing,WL}=1/2-(1/\pi)*atan((FWHM-FWHM_1)/FWHM_2))$

$M_{uniformity}=1/2-(1/\pi)*atan((depth-mean(depth))/d_2))$

For the atan() function, tau_1 determines the 50% point and tau_2 the slope of the merit function at that point. 

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Here is an example of FWHM-based merit, driving a field higher if seeing is really excellent. This is for FWHM_1= 0.5 and FWHM_2=0.1. Depth uniformity would look the same as this. 

This FOM is computed per field, per passband, for each potential observation. We can also introduce a couple of penalties:

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So we want to compute, for each field and for each observing opportunity, a zenith-angle and sky-brightness adjusted 5 sigma point source magnitude, m5. The signal to noise ratio for a point source scales as 

$SNR=\frac{\Phi}{\sqrt{(FWHM^2*sky}}$

for a fixed integration time and in the sky-dominated regime. Taking the log of both sides, and incorporating the zenith-dependence of FWHM and also passband-dependent extinction A(zenith), and incorporating the attenuation due to clouds AC (in magnitudes), we compute a change in m5 relative to observing at the zenith and under a sky background of m_o magnitudes per square arc sec, 

$dm5=AC+A~sec(z)+0.5~log(FWHM/0.7)+0.5~(m_o-m_{sky})+1.5~log(sec(z))$

This includes the zenith-depedence of FWHM, which scales as airmass^0.6, and extinction in the various bands. The coefficient of the final term comes from the airmass dependence of seeing, a^0.6, and the 2.5 factor for magnitudes, so that 2.5*0.6=1.5. 

Have each science program fill out this table, for each field center. Constrain field weights so they sum to one, for each program. Examples from SN, weak lensing, and static sky are illustrated

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A high value for tau1,2 de-emphasizes that aspect. A low value for FWHM1,2 de-emphasizes seeing. 

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This favors subsequent observations of higher merit, unless urgency is much high at the later time.

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Some references

LSST science book

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