Again. For about a week now. Damn…
すべての我々は星で作られています。
Buy me a coffee?
Writing this blog is a labour of love, and one which I evidently need a little encouragement towards on occasion. If you like what I write, the best way to show your gratitude would be to fuel my caffeine addiction and buy me a coffee to say thank you. All coffee donations are gratefully accepted!
To be honest, that’s better than the spam of three to five updates per day, though only for a while. Maybe when the writers rush starts again, you might think about building up a buffer?
On a not-so-completely unrelated note, could you tell me about the resolution of the Hubble (or any hypothetical or actual) telescope? I’ve heard of figures of 0.1 arcseconds being thrown around, but nothing that’d put FOV angle and imaging matrix resolution into perspective.
Spam? Ouch. I’m going to go with the benefit of the doubt here and hope that comment wasn’t intended to come across quite the way it sounded. Not particularly helpful when dealing with a crisis of confidence.
Resolution… Well technically speaking…
Due to diffraction the resolution of any telescope is wholly dependent upon wavelength. Assuming we’re talking about spatial resolution here, the resolution of any telescope is given by the equation;
R = λ/D
where R is resolution in radians, λ is the observed wavelength of light, and D is the diameter of the telescope’s objective. This gives the highest intrinsic resolution possible for any given telescope – and this is always better at shorter wavelengths. This is nearly impossible to achieve for ground based telescopes, due to atmospheric effects. Spaceborne telescopes like Hubble, on the other hand, are free from atmospheric limitations and so the optics can be assumed to be perfect. Secondarily, the pixel size of the detector needs to be considered. Under some circumstances, this can be the limiting factor in a telescope’s resolving power.
For example, in the ultraviolet at 300 nm, the Hubble Space Telescope’s angular resolving power works out to be 0.026 arcseconds. The Wide Field Camera 3 (WFC3) instrument, however, has a pixel size of 0.04 arcseconds. An estimate of the final resolution can be given by the equation;
Rfinal = √(R² + p²)
where p is pixel size. This gives a final estimate of the angular resolution in images taken with WFC3 at 300 nm of 0.048 arcseconds, so anything with an apparent angular diameter of less than this will appear as a point source. Final image quality will still be affected by signal-to-noise ratio, so either a bright source or a long exposure time will be needed to capture a good quality image.
It was definitely well-meaning. The thing is, you’ve been so productive in the past that on some days, I couldn’t keep up with your articles, though I definitely want to. Five articles a day is a lot of reading after a full workday. That’s why I recommended the buffer. And that’s why I said spam; not in the meaning of unsolicited, unwanted mail, but as being overwhelming. My excuses for giving the impression of the wrong implication.
Ok, fair point. I do usually try and queue things so I don’t post too much at once. It really depends how I’m feeling, I guess. Either way – no problem. I’ll try and curb my overzealousness. :)
Also: Oh, look, there’s a new article on Supernova Condensate. Let’s see if I can put numbers into those formulas after a little nap.
Heh… Overly technical and somewhat ineloquent, but I’m glad you liked it all the same!