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