Good point Chuck. It’s only an approximation anyway. The answer to the surface brightness question goes into another formula. The formula I plug the surface brightness into only gives an approximation of how long an exposure is needed, so it doesn’t have to be exact. Since my camera sees in rectangles I figure it will be good enough to get an idea. The math for computing the area in a rectangle is easier than using methods of calculus to figure the exact answer only then to be used as an approximation. Depending on the answer, I would use the information to plan a set of integrations that would equal or exceed the estimate anyway. I would like to know why my math was so far off, though. I know you are and have been real busy for months, so don’t worry about figuring that out. I would like to bring my grandkids to SPOC the night you are there, but I don’t know if I can work it out yet. Jim --- On Mon, 3/30/09, Chuck Hards <chuck.hards@gmail.com> wrote: From: Chuck Hards <chuck.hards@gmail.com> Subject: Re: [Utah-astronomy] Surface Brightness To: "Utah Astronomy" <utah-astronomy@mailman.xmission.com> Date: Monday, March 30, 2009, 1:20 PM Jim, I haven't the time to run all the numbers right now, but I can tell you now that M1 isn't a rectangle. On Mon, Mar 30, 2009 at 9:31 AM, Jim Gibson <jimgibson00@yahoo.com> wrote:
I Need help from some of you math wiz-bangs. I looked up on a chart of surface brightness that M1 has a surface brightness of 35 * 10 to the minus 6; or .000035. I found a formula for surface brightness ( http://en.wikipedia.org/wiki/Surface_brightness) that states S = m + 2.5 * log10A where m is point-magnitude and A is the area in square arcseconds. I went on my planetarium software and measured M1 to be approximately 440” X 260” arcseconds or 114,400 square arcseconds.
To further complicate maters I used an Excel spread sheet to do the math for me so I could use it for different objects. Excel is telling me that Log10(114,400) = 5.058, and plugging back into the formula 2.5 * 5.058 = 12.6 and + 8.4 = 21. 21 is a loooong ways from .000035. I figure the problem has to be in the Log10 area of my math. Or, they are saying to different things from which I am confused.
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