4 ms·

Does anyone have an estimation if the amount of pixels if james web would make a full 360 degree image?

pvg 3y ago
It probably depends a lot on the instruments and processing used. This deep field image from a while back https://webbtelescope.org/contents/media/images/2022/035/01G7DCWB7137MYJ05CSH1Q5Z1Z https://webbtelescope.org/contents/media/images/2022/035/01G... Is about 4.5k pixels(as published)/0.04 degrees (2.4 arcmin) a side so 360 degrees of that would be an image with 40 million pixel sides unless I fatfingered something in calculator.
GavinB 3y ago
[flagged]
- krapp 3y ago
  Why would you do some "quick napkin math" with, of all things, a system notorious for its inability to do precise math? You could have used a calculator. You could have literally written on a napkin. Why choose the worst possible tool for such a simple task?
  - GavinB 3y ago
    If you're going to yell at someone, at least do your own calculation so you can prove they're wrong first, please? Total Sky Area=4π×(2π360)^2×60^2 = 148,510,660 square arcminutes (you can also look this up easily) NIRCam has 9.7 square arcminutes FOV (Yes, I double checked this) NIRCam resolution is 2048 × 2048 x 2 modules for 2.4–5.0 µm (there are lots of different cameras and modules you could look at to make this estimate, but for a sense of magnitude this will work. If you looked at 0.6–2.3 µm that would be roughly 4x as many pixels for the 8 modules, I think? Again, napkin math for fun.) Total FoVs Needed = 148,510,660.8 / 9.7 ≈ 15,310,377 FoVs Total Pixels = (2048×2048×2)×15,310,377 ≈ 128,432,754,096,706 pixels LLMs used to be quite awful at math. Now they're solving quite complicated problems with sophisticated plugins. This is a simple and straightforward calculation, just getting the FOV and its resolution and then projecting that onto a sphere. And, yes, I probably should have posted the full calculation, it's definitely more interesting to see it all written out.
    - krapp 3y ago
      What I'm complaining about is LLMs becoming so normalized that people are using them for even basic intellectual tasks and trusting them implicitly, and apparently (judging from the downvotes) being defensive about the premise that they should even consider doing otherwise. I've seen people post results from ChatGPT with confidence only to realize they get another result if they try again a few more times (of course, they never bothered.) It's a statistical model, in which case the math can never be trusted entirely, or else it's just making an API call to a normal math library, in which case what even is the point? Either way, there are already better tools for the job. Needing "sophisticated plugins" to do basic math accurately should send people running away screaming from these damned things. I'll concede they have utility but we don't need to abandon our minds, hearts and wills to them entirely.
      - GavinB 3y ago
        I got downvotes for not showing the calculation, fair enough, my mistake. I understand the frustration there. And yeah, you've got to check ChatGPT's work and ask for clarification to understand what it did and why. That said, I disagree with the assertions made in your post. ChatGPT is a great tool for saving time in doing simple research and calculations, as long as you double check it. It wasn't a year ago, but it is now. I know, I'm scared by it, too!
wolverine876 3y ago
There are a few issues that cross my mind ... First I assume you mean a '360 x 360 degree' image - meaning you don't want a circle, but a sphere. [0] Second, much of your image would not be useful, obstructed by the Sun, Moon, Earth, and all that direct and reflected sunlight. I suppose JWST, being in L2, follows the Earth's orbit around the Sun. Over a year, does L2 allow a clear view of the entire celestial sphere? Third, the number of pixels would seem to depend very much on the chosen depth of focus. I don't know what the zoom levels are, but theoretically it could snap a nice, sharp wide-angle shot of the entire unobstructed sky. (I don't think wide angle is really JWST's mission - more high zoom to spots over 13 billion light years away, but I also don't know much about its optics.) [0] What is the proper mathematical expression instead of the awkward '360x360 degrees'? what do astronomers and airplane pilots use? I can't think of one atm, but there must be one? A spherical coordinate system for direction, analogous to degrees of a circle?
- maxnoe 3y ago
  The "area" in a spherical coordinate system is cold "solid angle" with unit steradian. The whole sky has 4pi steradian. https://en.m.wikipedia.org/wiki/Solid_angle https://en.m.wikipedia.org/wiki/Solid_angle
  - wolverine876 3y ago
    Thanks!

pvg 3y ago

It probably depends a lot on the instruments and processing used. This deep field image from a while back https://webbtelescope.org/contents/media/images/2022/035/01G7DCWB7137MYJ05CSH1Q5Z1Z https://webbtelescope.org/contents/media/images/2022/035/01G... Is about 4.5k pixels(as published)/0.04 degrees (2.4 arcmin) a side so 360 degrees of that would be an image with 40 million pixel sides unless I fatfingered something in calculator.

GavinB 3y ago

[flagged]

krapp 3y ago
Why would you do some "quick napkin math" with, of all things, a system notorious for its inability to do precise math? You could have used a calculator. You could have literally written on a napkin. Why choose the worst possible tool for such a simple task?
- GavinB 3y ago
  If you're going to yell at someone, at least do your own calculation so you can prove they're wrong first, please? Total Sky Area=4π×(2π360)^2×60^2 = 148,510,660 square arcminutes (you can also look this up easily) NIRCam has 9.7 square arcminutes FOV (Yes, I double checked this) NIRCam resolution is 2048 × 2048 x 2 modules for 2.4–5.0 µm (there are lots of different cameras and modules you could look at to make this estimate, but for a sense of magnitude this will work. If you looked at 0.6–2.3 µm that would be roughly 4x as many pixels for the 8 modules, I think? Again, napkin math for fun.) Total FoVs Needed = 148,510,660.8 / 9.7 ≈ 15,310,377 FoVs Total Pixels = (2048×2048×2)×15,310,377 ≈ 128,432,754,096,706 pixels LLMs used to be quite awful at math. Now they're solving quite complicated problems with sophisticated plugins. This is a simple and straightforward calculation, just getting the FOV and its resolution and then projecting that onto a sphere. And, yes, I probably should have posted the full calculation, it's definitely more interesting to see it all written out.
  - krapp 3y ago
    What I'm complaining about is LLMs becoming so normalized that people are using them for even basic intellectual tasks and trusting them implicitly, and apparently (judging from the downvotes) being defensive about the premise that they should even consider doing otherwise. I've seen people post results from ChatGPT with confidence only to realize they get another result if they try again a few more times (of course, they never bothered.) It's a statistical model, in which case the math can never be trusted entirely, or else it's just making an API call to a normal math library, in which case what even is the point? Either way, there are already better tools for the job. Needing "sophisticated plugins" to do basic math accurately should send people running away screaming from these damned things. I'll concede they have utility but we don't need to abandon our minds, hearts and wills to them entirely.
    - GavinB 3y ago
      I got downvotes for not showing the calculation, fair enough, my mistake. I understand the frustration there. And yeah, you've got to check ChatGPT's work and ask for clarification to understand what it did and why. That said, I disagree with the assertions made in your post. ChatGPT is a great tool for saving time in doing simple research and calculations, as long as you double check it. It wasn't a year ago, but it is now. I know, I'm scared by it, too!

wolverine876 3y ago

There are a few issues that cross my mind ... First I assume you mean a '360 x 360 degree' image - meaning you don't want a circle, but a sphere. [0] Second, much of your image would not be useful, obstructed by the Sun, Moon, Earth, and all that direct and reflected sunlight. I suppose JWST, being in L2, follows the Earth's orbit around the Sun. Over a year, does L2 allow a clear view of the entire celestial sphere? Third, the number of pixels would seem to depend very much on the chosen depth of focus. I don't know what the zoom levels are, but theoretically it could snap a nice, sharp wide-angle shot of the entire unobstructed sky. (I don't think wide angle is really JWST's mission - more high zoom to spots over 13 billion light years away, but I also don't know much about its optics.) [0] What is the proper mathematical expression instead of the awkward '360x360 degrees'? what do astronomers and airplane pilots use? I can't think of one atm, but there must be one? A spherical coordinate system for direction, analogous to degrees of a circle?

maxnoe 3y ago
The "area" in a spherical coordinate system is cold "solid angle" with unit steradian. The whole sky has 4pi steradian. https://en.m.wikipedia.org/wiki/Solid_angle https://en.m.wikipedia.org/wiki/Solid_angle
- wolverine876 3y ago
  Thanks!