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|2 years ago
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|2 years ago
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.