[time-nuts] tube GPS receivers

[Bob Camp]

Hi

Well if the chip rate is at or above 1 MHz, a wavelength is 300 meters or less. A 1 KM error is probably a bit to large.

Bob

On Jun 22, 2013, at 7:52 PM, Jim Lux <jimlux at earthlink.net> wrote:

> On 6/22/13 4:38 PM, Magnus Danielson wrote:
> 
>>> 
>>> electromechanical.. like omega receivers. rotary transformers can do
>>> very high quality trig functions, but do you actually need trig
>>> functions assuming you're just solving for X,Y,Z,T.
>> 
>> Oh yes. Check IS-GPS-200F, clause 20.3.3.4.3 User Algorithm for
>> Ephemeris Determination, found on page 113 and forward. The Table 20-IV
>> contains the actual formulas. The Kepler's Equation for Eccentric
>> Anomaly is a bit annoying, since it is not in closed form, so one way or
>> another of approximation iteration is needed.
>> 
>> Quite a bit of trigonometry goes on just to have each tracked satellites
>> current position estimated, such that the pseudo-range value taken for
>> the bird can be diffed out with the position. That process becomes
>> trivial if the position is known and only time is needed, given that we
>> cranked out the birds X, Y, Z and T position, which requires trigonometry.
> 
> Yes, but that trig can be done VERY slowly, since the cycle time is 12 hours, which is why a resolver/rotary transformer approach seems viable.
> 
> (rather, than, say, integrating the satellite state vector)
> 
> 
>> 
>>> Are you allowed to externally supply the almanac, in the form of a
>>> electromechanical system. The satellites are in circular orbits and
>>> fairly stable, and with multiple satellites in the same plane.
>> 
>> You could naturally cheat in several interesting ways, but you need
>> fairly accurate X, Y and Z values for the birds at any given time.
> 
> 
> How accurate??   Resolvers are good to about 16 bit accuracy, so I guess 1 part in 60,000.  if the orbit circumference is 163 Mm, then a resolver can determine the position to a few km.
> However, I don't know that that is good enough.  If you need to know to 1 chip at C/A code rates, 1 microsecond, that's a pretty small fraction of one 12 hour rev of 43200 seconds. But maybe not.
> 
> 
> 
>> 
>>> You'd only need trig to convert X,Y,Z into lat/lon, and for us timenuts
>>> types, do you really need lat/lon? In fact, do you even need to solve
>>> for earth centered coordinates? Why not work in inertial space (whether
>>> your receiver happens to be moving in a circle at 1 rev/24 hrs or flying
>>> in a plane at something else is sort of immaterial)
>> 
>> Once you come to having a X, Y, Z and T, the remaining trig operations
>> is trivial to what you already have done, so you might as well do them.
>> 
>>> I envision something with a common shaft running at 1 rev/12 hours that
>>> drives N rotors (one for each satellite). there's a small motor that
>>> sets the offset of the rotor relative to the shaft to account for small
>>> movements along the orbit plane. That, plus some other transformers
>>> would give you X,Y, and Z for each satellite.
>> 
>> You have a sick mind. What worse is, I understood what you actually meant!
>> 
>>> Actually, how bad would your time estimate be if you just assumed
>>> perfect circular orbits with no higher order corrections?
>> 
>> Grabbing a modern set of data, doing the calculations with and without
>> the proper values would tell you. I would not be surprised if it where
>> way over the km off. On the other hand, the precision we talk about in
>> general already throws us off sufficiently, so who cares.
>> 
>> One should realize that we talk about tens of Mm numbers in pseudo-range
>> distances.
>> 
> 
> So I think you probably can't get a position fix within 10km, but hey, what a beast it would be.
> 
> 
> 
> 
>> Cheers,
>> Magnus
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