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Up: Analysis of the Solutions Previous: Southern Outliers

Northern Outliers

In progress...


  
Figure: (a) Pooled root variance plot for Southern solution. Sources with large variance ( $\sigma > 0.1$) are marked and analyzed in Appendix A. (b) The histogram of the root variance for sources with <m>South < 12.5. (c) Same as (b), but for the all the sources.
\begin{figure} \centering \mbox{\epsfxsize=5.0in \epsfbox{Ppool_south.ps}} \end{figure}


  
Figure 2: Same as Figure 1, but for the Northern solution.
\begin{figure} \centering \mbox{\epsfxsize=5.0in \epsfbox{Ppool_north.ps}}\end{figure}


  
Figure: Average Northern magnitude vs average Southern magnitude. The slope of the straight line is $0.998\pm 0.001$.
\begin{figure} \centering \mbox{\epsfxsize=2.9in \epsfbox{Pmagmag.ps}}\end{figure}


  
Figure: Residuals (the difference between Northern and Southern magnitudes) as the function of Southern magnitude. Each point represents a field star. The slope of the straight line is $(-1.80\pm 1.10)\times 10^{-3}$.
\begin{figure} \centering \mbox{\epsfxsize=2.9in \epsfbox{Pmagres.ps}}\end{figure}


  
Figure 5: Residuals as the function of the R.A. coordinate. Each data point represents a field star. Clearly seen are the 10 calibration fields used in the analysis. Large triangles show the mean residuals in each field.
\begin{figure} \centering \mbox{\epsfxsize=5.0in \epsfbox{Prares.ps}}\end{figure}


  
Figure 6: Residuals as the function of the relative cross-scan position (offset from the center of the scan.
\begin{figure} \centering \mbox{\epsfxsize=5.0in \epsfbox{Pdrares.ps}}\end{figure}


  
Figure: The same as Figure 4, but for the central 1/3 of the scans. The best-fit straight line has the slope of $(0.95\pm 1.63) \times 10^{-3}$.
\begin{figure} \centering \mbox{\epsfxsize=5.0in \epsfbox{Pmagres3.ps}}\end{figure}


  
Figure 8: The same as Figure 5, but for the central 1/3 of the scans.
\begin{figure} \centering \mbox{\epsfxsize=5.0in \epsfbox{Prares3.ps}}\end{figure}


  
Figure: Residuals (full scan width) as the function of J-K color. The straight line fit produces the slope of $-0.001\pm 0.006$.
\begin{figure} \centering \mbox{\epsfxsize=5.0in \epsfbox{Pcoldif.ps}}\end{figure}


  
Figure: Source colors as the function of the cross-scan position. The correlation coefficient is r = 0.039 and the $3\sigma $ confidence interval is [-0.055, 0.132].
\begin{figure} \centering \mbox{\epsfxsize=5.0in \epsfbox{Pdracol.ps}}\end{figure}

next up previous
Up: Analysis of the Solutions Previous: Southern Outliers
Martin Weinberg
1998-10-26