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    Re: Correcting for the movement of an observer: a plausible explanation?
    From: Tony Oz
    Date: 2019 Dec 26, 02:12 -0800


    I want to compare the intercept fix with a result of the 2-body direct computation. The latter method requires the knowledge of altitude values related to the same moment in time. To obtain such knowledge the MoO ans MoB corrections are available.

    Recently Frank gave an explanation:

    To picture the effect of the MOO table, imagine a vessel crossing the Earth's surface under that celestial sphere. Stars off the bow climb. They rise at a rate of one minute of arc for every nautical miles that we travel. Stars astern sink at the same rate. What happens to the rest of the celestial sphere? If you imagine an axis running abeam, from exactly port to exactly starboard, the celestial sphere rotates about that axis. Stars to port and starboard rotate around the fixed points on the horizon which are exactly perpendicular to our track over the ocean. The changes in altitude, Δh ("delta h"), are given by

    Δh = d · cos(C – Zn)

    where d is the distance made good which may be computed from the elapsed time in hours multiplied by the speed in knots (adjusted for currents if possible). The difference between C, the true course of the vessel, and Zn, the true azimuth of the star, gives the angle off the bow in degrees. You don't need to use the table if you don't have access to a copy. You can work this up easily on a calculator. You do have to think a little about signs, but it's pretty obvious in practice. Suppose I'm in latitude 41.6° N, and I observe the altitude of the star Vega when its azimuth is 310° (which I get from the main table while planning the round of sights) and my course is 090°. I'm travelling at 10 knots. Five minutes and 45 seconds later I observe the altitude of Pollux. Rather than plotting and advancing the Vega LOP, I can bring my altitude of Vega forward. How much? Distance made good is 0.96 nautical miles, and the cosine of course minus star's azimuth is -0.77 so the change in altitude is just about -0.7 minutes of arc. Our vessel is moving (mostly) away from Vega, so the sign is right as it stands. To synchronize the earlier Vega sight with the later Pollux sight, we subtract 0.7 minutes from the observed altitude of Vega. Not a big correction, but easy to do. But we're not done. Time for MOB.

    During the 5.75 minutes between the Pollux and the Vega sights, the Earth was turning, so we also need to adjust the altitude of Vega for the daily rotation of the celestial sphere. The sphere is turning "under its own power", if you like, on an axis that is parallel to the Earth's axis. It's like a spinning umbrella above our heads with the handle pointed nearly at Polaris. It makes one turn in a sidereal day (23h56m). The rate of change of change of altitude is directly proportional to the elapsed time in minutes, T, the cosine of our latitude, and the sine of the true azimuth of the star. The maximum rising rate for stars for observers on the equator with azimuths due East or due West (Zn 90° or 270°) is 15.04’ per minute of time. The rate of change of altitude, in full, is:

    Δh = 15.04ʹ · T · cos(Lat) · sin(Zn)

    This is tabulated in MOB tables, and it's also easy to work up on a calculator. For the Vega sight above, T is 5.75 minutes (as before) and Lat is 41.6° while Zn is 310°. Putting it all together, the change in altitude is -49.5'. If we were at rest, we would synchronize our Vega sight to the time of our Pollux sight, by lowering it, reducing the observed altitude, by 49.5 minutes of arc. Combining this with the MOO correction, the net amount is a reduction in altitude of 50.2 minutes of arc. The Vega and Pollux sights are not synchronized. They have the same GHA Aries and thus the same LHA Aries, and we do not need to advance the earlier LOP. Where the two synchronized lines of position cross, we mark the fix and that's it. By the way, it's normal for surface navigation cases that the MOO correction will be something like two orders of magnitude smaller than MOB. That's because the speed of rotation of points on the Earth's equator (relative to points on the axis of rotation) is 900 knots. Surface vessel speeds will usually be 50 to 100 times smaller. If you're flying in a supersonic aircraft, you can make MOO and MOB cancel out. The stars will stand still.

    Calculations and tables like these should not be used for time intervals greater than 15 to 30 minutes though naturally it all depends on how much accuracy you want in the end.

    In my attempt to compare the fixes I used an example from a booklet on CN (if needed I can qoute it here). I took the intercept-obtained coordinates as the reference. I notice that the directly computed coordinates are not perfectly equal to the reference ones, - they are close but differ by ~0.7~1.5 nm.

    To understand the reason for that very noticeable difference I calculated the altitude of the body (which was corrected by MoO and MoB) as it should be seen if taken simultaneously and from the same place as the second body. The MoO/MoB-corrected values is rather different from thus calculated.

    Looking into this I got the impression that it is for the value of the azimuth that was used in the formulas. If I use an "amended" azimuth - not the one that was at the first sight, but a value closer to the calculated at the (fictitious) simultaneous sight - the MoO and the MoB corrections give much better approximation of the altitude.

    So, my (temporary) "conclusion" is - this is an iterative process, one has to re-calculate azimuths and re-apply MoO+MoB corrections if more accurate 2-body direct fix is necessary.

    Please comment.

    Warm regards,
    60°N 30°E

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