27 karma · joined November 27, 2017
According to Wikipedia, the order is: Sun, L1, Earth, L2. So L2 is always in the night sky.
I don't see issues with focus. 1.5M km is effectively infinity. Indeed for my camera on earth, 10m resolves as pretty much as well as infinity, f stops notwithstanding.
JWST is about 20m by 14m in size. And it's out of the earth + moon shadow because of the nature of its orbit. It has a sun shield, and radio dish in the shield.
The resolving power of 0.05 arcseconds at 1,500,000 km apogee is indeed 350 m (ish), which is much bigger than the JWST. The situation is worse at perigee. An image of JWST would take up less than 1% of Hubble 'pixel'. So there may be a faint blob, depending on the albedo of the JWST. (I don't know how Hubble's the photodetector arrays' resolution compares to the maximum resolving power of the lens.)
However, Hubble's guidance systems have permitted astronomy at 0.0003 arcseconds, which comes out at a little over 2 m at JWST's range. So after a lot of computation, stacked images and fine-guidance shifting, presumably there could be an image of around 10 x 7 pixels (or whatever they are after the computation).
And that's a result!
But that makes little sense - we're measuring the number of simultaneous calls. It gets reflected in buffers, circuits, sockets, channels allocated, etc.
It would be better to consider 'call starts / second' as the rate at which calls are initiated, and the average 'call.seconds / call start', which when multiplies gives the more sensible unit 'call'.
Or simpler: just measure the average length in seconds. So we have so many 'calls / second' initiated, and the average length is so many 'seconds', which when multiplied again gives a unit of 'call'.