5.18e7 fields which would take 1.6 million years Guess the answer to Patrick's original question (How's your math?) is bad. Bill B On Mar 29, 2005, at 9:25 PM, William Biesele wrote:

The hudf was a million second exposure. It would take 6 billion years to complete a sky survey at this resolution.

Bill B.

p.s. Don't know how many mac users there are on this list, but I found a good calculator program called Longhand which is great for this kind of stuff. http://longhand.pansophists.net/

6 years = 189345600 seconds 1 square degree = 60^4 arc seconds = 12960000 square arc minutes 360^2 square degrees of sky = 129600 square degrees of sky = 1679616000000 square arc minutes of sky 9 square arc minutes per field = 186624000000 fields The answer to the original question

One year is 3.15576e7 seconds

On Mar 29, 2005, at 8:03 PM, Patrick Wiggins wrote:

...

Given that the Hubble Ultra Deep Field measures 3 arc minutes on a side how many HUDFs would it take to cover the entire sky?

Thanks,

Patrick

_______________________________________________ Utah-Astronomy mailing list Utah-Astronomy@mailman.xmission.com http://mailman.xmission.com/cgi-bin/mailman/listinfo/utah-astronomy Visit the Photo Gallery: http://www.utahastronomy.com

_______________________________________________ Utah-Astronomy mailing list Utah-Astronomy@mailman.xmission.com http://mailman.xmission.com/cgi-bin/mailman/listinfo/utah-astronomy Visit the Photo Gallery: http://www.utahastronomy.com

O.K. So let's take a look at the math of the question. (The mathematical answer is 16.5 HUDF images to cover the whole sky. Details follow) Given - Each HUDF image covers 3 arc minutes on a side. We need to figure out the surface area of an imaginary sphere such that a plane sliced out of the sphere passing through the exact center of the sphere would describe a circle with a circumference of 360 degrees. We need to find out the radius of the sphere so we can calculate it's surface area. The radius of the circle and the sphere would be the same and be equal to 360 degrees divided by 2pi. So the radius of this circle and the sphere would be ~57.3 having a unit of degrees, (which does not make sense in this context, but we need the units later on.) The surface area of a sphere is 4pi*(radius squared). (Too bad I cannot do symbols in this email). Since the radius is ~57.3 degrees, the surface area of a sphere in this example would be 4pi*(~57.3 degrees)^2 or 41,253 square degrees. Since there are 3600 square arc minutes in a square degree and 9 square arc minutes in a HUDF image, there would be 400 HUDF (3600 / 9) in a square degree and about 16.5 million (400 * 41,253) HUDF images required to image the entire sky. Now since the HUDF imager is most likely slightly larger than a perfect 3 arc minutes on a side, Patrick's answer of the whole sky being 12.7 million times as large as a HUDF image seems very much in line to me. (Working the equation backwards, the HUDF would need to be 3.4 arc minutes on a side for 12.7 million images to cover the whole sky... Anyone know the exact size of a HUDF image?) John Zeigler

Oops! I should have specified 16.5 million images instead of just saying 16.5 at the beginning. John Zeigler O.K. So let's take a look at the math of the question. (The mathematical answer is 16.5 HUDF images to cover the whole sky. Details follow) Given - Each HUDF image covers 3 arc minutes on a side. We need to figure out the surface area of an imaginary sphere such that a plane sliced out of the sphere passing through the exact center of the sphere would describe a circle with a circumference of 360 degrees. We need to find out the radius of the sphere so we can calculate it's surface area. The radius of the circle and the sphere would be the same and be equal to 360 degrees divided by 2pi. So the radius of this circle and the sphere would be ~57.3 having a unit of degrees, (which does not make sense in this context, but we need the units later on.) The surface area of a sphere is 4pi*(radius squared). (Too bad I cannot do symbols in this email). Since the radius is ~57.3 degrees, the surface area of a sphere in this example would be 4pi*(~57.3 degrees)^2 or 41,253 square degrees. Since there are 3600 square arc minutes in a square degree and 9 square arc minutes in a HUDF image, there would be 400 HUDF (3600 / 9) in a square degree and about 16.5 million (400 * 41,253) HUDF images required to image the entire sky. Now since the HUDF imager is most likely slightly larger than a perfect 3 arc minutes on a side, Patrick's answer of the whole sky being 12.7 million times as large as a HUDF image seems very much in line to me. (Working the equation backwards, the HUDF would need to be 3.4 arc minutes on a side for 12.7 million images to cover the whole sky... Anyone know the exact size of a HUDF image?) John Zeigler

We are cleaning up the shop and I have an industrial 2hp Leeson DC motor I picked up for a large mirror grinding machine project that has now been canceled. The motor has a right angle industrial head and pneumatic clutch. This is a true industrial motor rated for the full 2hp output running 24 hours a day. The output is 0-60 rpm. The only catch - you need to pick it up by this Saturday April 2nd. This motor would be perfect to move an observatory dome or a roll off roof. This motor is FREE for the taking. Just come and get it. If there are no takers, it will end up in the landfill as I need to get it out of the shop before this weekend. John Zeigler

Yes, but in that 6 billion years, how many more would form? -- Joe

The hudf was a million second exposure. It would take 6 billion years to complete a sky survey at this resolution.

Bill B.

p.s. Don't know how many mac users there are on this list, but I found a good calculator program called Longhand which is great for this kind of stuff. http://longhand.pansophists.net/

6 years = 189345600 seconds 1 square degree = 60^4 arc seconds = 12960000 square arc minutes 360^2 square degrees of sky = 129600 square degrees of sky = 1679616000000 square arc minutes of sky 9 square arc minutes per field = 186624000000 fields The answer to the original question

One year is 3.15576e7 seconds

On Mar 29, 2005, at 8:03 PM, Patrick Wiggins wrote:

...

Given that the Hubble Ultra Deep Field measures 3 arc minutes on a side how many HUDFs would it take to cover the entire sky?

Thanks,

Patrick

_______________________________________________ Utah-Astronomy mailing list Utah-Astronomy@mailman.xmission.com http://mailman.xmission.com/cgi-bin/mailman/listinfo/utah-astronomy Visit the Photo Gallery: http://www.utahastronomy.com

_______________________________________________ Utah-Astronomy mailing list Utah-Astronomy@mailman.xmission.com http://mailman.xmission.com/cgi-bin/mailman/listinfo/utah-astronomy Visit the Photo Gallery: http://www.utahastronomy.com