Path: ...!fu-berlin.de!uni-berlin.de!individual.net!not-for-mail From: Jeff Liebermann Newsgroups: rec.bicycles.tech Subject: Re: Biden Drops out on sure defeat. Date: Mon, 22 Jul 2024 18:18:41 -0700 Lines: 68 Message-ID: References: <5q8r9jp7b2l3ggul998o0jo18got6ec9a0@4ax.com> <9ABnO.9374$Bg2c.882@fx01.iad> Mime-Version: 1.0 Content-Type: text/plain; charset=ISO-8859-1 Content-Transfer-Encoding: 8bit X-Trace: individual.net FdBz4/Sj0ScL4PEMHPmqmgGyXP0UBKfWmHr+k1cAM6ZR7v6qG5 Cancel-Lock: sha1:3fOp3PVbV39Ix4l6bUcWyoPzFxU= sha256:PJg+japZVBTM/HB6t1VG8hD95uTRnVEmQowdaJrFHSY= User-Agent: ForteAgent/8.00.32.1272 Bytes: 4120 On Mon, 22 Jul 2024 23:04:37 GMT, Tom Kunich wrote: >Satellite photos are taken from about 20,000 km up. Nope. Last time, you were at 6 miles altitude (9.7km). I guess 20,000 km is closer to the mark. I assume that 20,000 km "up" means 20,000 km "and up" or minimum altitude. The real minimum is more like 167.4 km (104 miles) and up: Your 20,000 km altitude is probably a bad guess or a number picked at random. There's very little in orbit at 20,000 km. "Humanity currently has about 500 operational satellites occupying Earth’s "Clarke Belt". This is about one-third of the Earth’s existing satellites, whereas the rest are at an altitude of 2000 km (1200 mi) or less from the surface – the region known as Low Earth Orbit (LEO)." >Tell us the amount of ground covered by ONE degree from that altitude. Yet again, you're asking me to demonstrate YOUR talking point. It's a simple calculation and you should be able to do it without my help. Do your own calculations and show your work. It would also be helpful if you would specify a more realistic satellite altitude and any slant angle involved. >THAT would require knowing simple algebra or you could pretend that >it is a trig problem. There's no algebra involved. It's 100% trigonometry which you should have learned in the 10th or 11th grade. If you want better accuracy or a large viewing area, spherical geometry will be necessary. For the purposes of this discussion, a flat earth approximation is sufficient. Maybe you should do some reading about diffraction limited angular resolution, which is what limits satellite optical resolution. That's why satellite imagery can't read license plates. >Since EE's are supposed to be educated in partial equation calculus, >tell us what amount of ground that covers which will inform everyone >here what you know about anything and why pictures are taken at >midday and straight down. There are no ordinary differential equations involved. Nothing is changing as the photos are being taken. There are no time or motion dependent variables involved. Similarly, there are no partial differential equations because a partial differential equation is just an ordinary differential equation with more than one variable. >The highest resolution that you can get is about 0.5 meters which >means that SEEING people is difficult at best. 0.5m is correct, however 0.3m is now commercially available: "Satellite data resolution is specified using Ground Sampling Distance (GSD), measured in metres. Satellite imagery of 0.3m GSD means that each pixel covers 30cm x 30cm of the Earth." 0.3m is 11.8 inches. -- Jeff Liebermann jeffl@cruzio.com PO Box 272 http://www.LearnByDestroying.com Ben Lomond CA 95005-0272 Skype: JeffLiebermann AE6KS 831-336-2558