If GPS had existed in the past, how long would it have taken to compute position?

Question

I have been reading a book called Fun with GPS which mentions differences in the models being used (in 2005 and earlier) and it suddenly occurred to me that the mention of computing power varying might lead to an interesting experiment - which we might do by reference. Lets suppose the GPS satellites have been up there since 1960 and a receiver exists that can capture the time clocks from a good number, perhaps 6 or 8 to compute the position at that moment. But instead of computing the location internally, suppose it were passed off to a then good and common computer. Say an IBM 7090 Mainframe in 1960 and a Digital Minicomputer in offices in 1968 or so, and an Apple IIe (floating point math) in 1979 and an IBM PC (8086) in 1982 or AT (20286) and some more recent computers with faster processors of specific speeds - Pentium, Dual Pentium, etc.
How many minutes or hours would it take?
All this needs is connection from one to the next - Garmon current model (or 76 in 2004-5) to common computer, then common to common of earlier date?
Anybody got information?

I have long said that early personal computers were such that one was equal to computing power of all the computers in the world in a certain year, like 1960 when I started college and MIT had just gotten their first 7090 which they had to share along with their (3 I think) 709's. Of course, today, a single terabyte drive has more storage than was available on all the hard drive platters of a major city in the late 60's and probably matched a huge stack of tapes.

Answer 1

The computation for position from GPS is pretty simple trig. Wouldn't have taken more than a few dozen milliseconds on a Apple II+ even though done at 32-bits. Math packs for 32-bit and higher had to run in 64kbytes of 8-bit RAM, clock speed was slightly over 1MHz.



A better example for such a benchmark would be something like generating a frame of Call Of Duty. Similar sort of math, but must be applied hundreds of thousands of times per frame at 60 frames per second. And operating in a 3-D virtual world. Just guessing, I'd say each frame would take a couple of hours to do on a 1980 Apple IF you built a 4GB memory expansion for it, with the drivers to use it. Of course, that memory expansion board would have been built out of 64k x 1-bit dynamic RAM chips of the day, so the board would have required, lessee, 4 gig is 32 bits, 64k is 16 bits, so 64K of chips times 8 bits equals 512,000 chips. Pretty big board... each chip was about 0.30" x 0.90", call it 0.5" x 1.0" for traces, doesn't that work out to a board about 42 by 42 FEET? Naw, let's make it practical and call it 1764 boards, 1-foot square.



EDIT - Hmmm, those RAM chips were power hogs compared to the RAM of today, something like a hundred mills each at 5V? (Good thing those 64kb chips only needed a 5V supply, the previous generation of 16k chips needed +5, -5, and +12!). So those 512k of chips would have required 50,000 amps at 5 volts, or 250kW of energy.



I guess things HAVE improved over the last few decades...



PS - The math for GPS positioning would fit in a spreadsheet of Visicalc maybe 20 x 20 cells or smaller. With the optional 8087 co-processor in an original IBM PC, results would update virtually instantly.

Answer 2

you are talking adding 2 or three clock signals..



even on a computer that is 1 megahz like that in a apple 2 would take microseconds..



it would take LESS than 1 minute to calculate the location based on those computers..

remember they used a computer that only had something like 32 k bits of ram in it to get to the moon I am sure they couldnt wait 10 minutes to figure out where they were ...



most of the computing power we use now is wasted on the graphics not the actual position

Answer 3

I don't think it would take more than 10 minutes or so, the math is not that complicated, and even the tiny processor in a GPS receiver does this in a few seconds.