Comparison of photometry of one star in the M51 field

A quick comparison of photometry values from the USNO B1.0 and that available on

Given the following image:

If you take: the 14.9/17.5 star at about 1:30 from the core of M51

Now look at the USNO B1.0 mags for star ID=1372-0290320 (see below for url)
'average'  the B1.0 R and B mags:

B=17.385 R=15.195 then the differences are B~=0.1 and R~=0.3mag
which is in reasonable agreement with the 'accuracy' statement here:
see: Section 5. Photometric Calibration
for the B1.0 and plates containing 'faint photometry standards' and the target star.

The kicker with this star is it's I mag 13.87 which kinda makes sense if you look at how much brighter in R than B the star is.

See this long url for the full B1.0 data on the star::,_DEJ2000&-source=I/284/out&USNO-B1.0===1372-0290320

The star ID is the last field in the url.
The BIC/RIC on the page is identified to be a '1' meaning the standard star and the target star are on the same plate.

Here is a small Excel 97 spreadsheet that uses the Blue/Red/Infared mags from the USNO B1.0 and
calculates the 'approximate' Vmagnitude:
Click here to download Excel spreadsheet

It uses calculated flux for the magnitudes in B1.0 and interpolates to the Vmag based upon an exponential function fit.

I have found quite a few statements of filter vs wavelenght versus Jansky's. The spreadsheet above allows
you to change them and your changes will be used in the calculations if you do not like the ones from the referenced url below

Reference for math and constants below:

Flter        Wavelength        Flux(Jy)
U                  360                    1880
B                  440                    4400
V                  550                    3880
R                  700                    3010
I                    880                   2430

Magnitudes cannot be added to get the total flux from two or more objects. First you must convert to flux then add the flux then convert back to magnitude.
B=-2.5log(fB / 4400Jy) = -2.5log(fB ) + 2.5 log(4400Jy)
V=-2.5log(fV / 3880Jy) = -2.5log(fV ) + 2.5 log(3880Jy)
R=-2.5log(fR / 3010Jy) = -2.5log(fR ) + 2.5 log(3010Jy)
I= -2.5log(fI / 2430Jy) = -2.5log(fI ) + 2.5 log(2430Jy)

Or Flux = Jy*10**(m/(-2.5))