Just curious . . . but if anyone knows exactly what percentage of the sky from our vantage point has the HST imaged? I would assume not even close to 50%, but that would be my guess. There appears to be more sky to be imaged . . . and more discoveries to be made!

Many thanks for the input,
-Kevin

Not even close to 50%! There are about 40,000 square degrees in the sky, and the biggest Hubble field of view is about 3.5 x 3.5 arcminutes (there are 60 arcminutes per degree). If I did my math correctly, it would take 12 million images to cover the sky with that camera, and most of the cameras have a (much) smaller field of view.

Also, a lot of images are of the same fields over and over again. I cannot say how much of the sky it's covered, but I would guess it's far less than even 1%. Surprising, eh?

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aren't the deep sky images mostly from areas that are devoid of stars, so that they don't obscure the far galaxies? I imagine any of the 8 or 9 thousand naked-eye visible stars would look pretty bright in one of those Hubble images

As an upper limit - The Hubble archive contains something like 700,000 exposures, as of the 15th-anniversary press release. If those were all of different pieces of sky (which is not true by some factor of at least 5, from multiple exposures and multiple filters) and all used the widest-field camera ACS (again not true by a large factor, especially since ACS was installed only in 2000), the area covered would be 700,000 x (200 arcsec)^2 or 2000 square degrees (and I'm stunned it's that high). The sky contains about 42,000 square degrees, so the lesson is that HST could have looked at 5% of the sky if its goal had been such a survey. The actual number is smaller by (my estimate) aboput 20 times, allowing for spectroscopy and other cameras plus multiple observations of the same field. That would make maybe 100 square degrees observed. (You get a similar number if you ask how much of the sky could have been covered by pointing one of the cameras at a different field for one orbital viewing period each since launch). For Northern Hemisphere orientation, the bowl of the Big Dipper subtends about half that. There was one program (I think the COSMOS survey) which got images of a single 2-square-degree area of sky to map large-scale structure and galaxy statistics. They used a single filter with ACS for galaxy structures, plus a 17-filter ground-based survey and spectroscopy for redshifts.

I just saw an archival image of Sirius and Sirius B with WFPC2. Sure is bright. You can see diffraction spikes from the tangent points of the first couple of Airy rings (which helps get the coordinates of Sirius itself to refine the relative orbits).

In retrospect, I am extremely delighted that there will soon be a Hubble servicing mission. I look forward to all the glorious pictures and observations to come!

Again, many thanks Bad Astro as well as the others!

-Kevin

That would be cool. I'll look that image up.

I wrote a sophisticated piece of software that used the diffraction spikes to get the centroids of bright stars so that we could look for low mass companions and disks around young stars. I calculated that it was good to about 1/50th of a STIS pixel, or 0.002 arcseconds. :-) Best code I ever wrote.

A a quick start - this is one of the images in that sequence. WFPC2, F1042M to get nice big centroidable Airy rings and diffraction spikes. (Note to BABB: the reason I was grabbing that image is the same one I've been a bit quiet here lately and will be quieter for the next couple of weeks. Book deadline and 120 pages left to get in final form...)

1/50 of a pixel - cool. A tenth is easy, but doing much better takes really understanding the instrument and optical train.

Actually, it was all post-processing. I took a series of slices across the diffraction spikes, assumed a gaussian profile, found the max, which then was the center of brightness of the spike in the X-direction. I then did a least squares fit to all the centers I found for every pixel row, which I assumed defined the best linear fit for the spike. Then I solved for the intersection of the two spikes. Voila! The center of the PSF.

Trivial, really. 8)

Yeah - it's all trivial when you know how. The clever bit is spotting the solution when no one else has beaten you to it, then working it through. =D>

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