7 ms·
A full-resolution, maximum-size JPEG XL image (1,073,741,823 × 1,073,741,824): Uncompressed: 3.5–7 exabytes Realistically compressed: Tens to hundreds of peta
by m348e912 10mo ago
A full-resolution, maximum-size JPEG XL image (1,073,741,823 × 1,073,741,824):
Uncompressed: 3.5–7 exabytes
Realistically compressed: Tens to hundreds of petabytes
Thats a serious high-res image
- cubefox 10mo agoYes, but unlike AVIF, JPEG XL supports progressive decoding, so you can see the picture in lower quality long before the download has finished. (Ordinary JPEG also supports progressive decoding, but in a much less efficient manner, which means you have to wait longer for previews with lower quality.)
- tyre 10mo agoI don’t think the issue with the exabyte image is progressive decoding, though it would at least get you an image of what is bringing down your machine while you wait for the inevitable!
- flir 10mo agoAn image of earth at very roughly 4cmx4cm resolution? (If I've knocked the zero's off correctly)
- aidenn0 10mo agoEach pixel would represent roughly 16cm^2 using a cylindrical equal-area projection. They would only be square at the equator though (representing less distance E-W and more distance N-S as you move away from the equator). No projection of a sphere on a rectangle can preserve both direction and area.
- flir 10mo agoI admit it, I was applying Cunningham’s Law. Disappointingly(?), you came to the same answer.
- aidenn0 10mo agoI admit I trusted your math; you seem to be off by a factor of 4: You have: 510.1e6km^2/1073741824/1073741824 You want: cm^2 * 4.4244122 / 0.22601872 Strangely enough, units lacks area_earth, so I used the number from https://iere.org/what-is-the-area-of-the-earth/ https://iere.org/what-is-the-area-of-the-earth/
- flir 10mo ago:D I was starting with the length of the equator and assuming a spherical cow^Hplanet.
- aidenn0 10mo agoDid you perhaps use the diameter of the earth rather than the radius? #PiIsWrong https://www.tauday.com/ https://www.tauday.com/ You have: (40075km/tau)^2*4*pi/1073741824/1073741824 You want: cm^2 * 4.434018 / 0.22552908
- mcdonje 10mo ago[flagged]
- yread 10mo agoThe only practical way to work with such large images is if they are tiled and pyramidal anyway
- Akronymus 10mo agowhat does pyramidal mean in this context?
- jjcob 10mo agoI think it means encoded in such a way that you first have low res version, then higher res versions, then even higher res versions etc.
- jjk7 10mo agoTiled at different zoom levels
- scheme271 10mo agoProbably, multiple resolutions of the same thing. E.g. a lower res image of the entire scene and then higher resolution versions of sections. As you zoom in, the higher resolution versions get used so that you can see more detail while limiting memory consumption.
- shadowgovt 10mo agoReplicated at different resolutions depending on your zoom level. One patch at low resolution is backed by four higher-resolution images, each of which is backed by four higher-resolution images, and so on... All on top of an index to fetch the right images for your zoom level and camera position.
- swiftcoder 10mo agoExcept in the case of a format like JPEG, there is no duplication - higher layers are used to "fill in the gaps" in the data from lower layers.
- deleted 10mo ago
- xnorswap 10mo agoAt 600DPI that's over a marathon in each dimension. I do wonder if there are any DOS vectors that need to be considered if such a large image can be defined in relatively small byte space. I was going to work out how many A4 pages that was to print, but google's magic calculator that worked really well has been replaced by Gemini which produces this trash: Number of A4 pages=0.0625 square meters per A4 page * 784 square miles =13,200 A4 pages. No Gemini, you can't equate meters and miles, even if they do both abbreviate to 'm' sometimes.
- fwip 10mo agoWolfram alpha is the better calculator for that sort of thing.
- spider-mario 10mo agoA better Gemini also works. Google Search seems to use the most minimal of Geminis, giving it a bad rep. Prompt: “How many A4 pages would a 1073741823×1073741824 image printed at 600dpi be?” Gemini Pro: “It would require approximately 33.1 billion (33,127,520,230) A4 pages to print that image. To put that into perspective, the image would cover an area of 2,066 square kilometers […]. The Math 1. Image Dimensions: 1,073,741,823 × 1,073,741,824 pixels. 2. Physical Size: At 600 DPI, the image measures roughly 45.45 km wide by 45.45 km tall. 3. A4 Area: A single sheet of A4 paper (210 mm * 297 mm) covers approximately 0.06237 m². 4. Result: 2,066,163,436 m² / 0.06237 m² ≈ 33,127,520,230 pages.” Alternatively, rink (https://rinkcalc.app/ https://rinkcalc.app/) : > (1073741823 / (600/inch))**2 / A4paper approx. 3.312752e10 (dimensionless)
- __patchbit__ 10mo agoGrok 4.1 beta finds the answer: approximately 33.1 billion pages.
- threeducks 10mo ago> I do wonder if there are any DOS vectors that need to be considered if such a large image can be defined in relatively small byte space. You can already DOS with SVG images. Usually, the browser tab crashes before worse things happen. Most sites therefore do not allow SVG uploads, except GitHub for some reason.
- flakes 10mo agoA selfie at that resolution would be some sort of super-resolution microscopy.