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Antoine Couette wrote:
> Feb 12th, 2011
> Height of Eye : 6.40m/21ft , Outside Temp : 9°C/48.2°F , Pressure : 1011 
> mb/hPa / 29.85 In.Hg
> All heights corrected for only Instrument error (Refraction, SD and 
> Parallax need to be performed)
> 1 - For the first of observations our bottom course/speed were 255° True 
> / 12.0 kts
> I first shot 3 heights of the SunLL, averaged as follows : watch-time 
> t1=00h08m54.6s , h1=7°46'.7

Estimated lat +47.000° lon -2.000° at the above watch time.

Estimated clock correction +16 h.

UT1-UTC = -.17 s.

Dip correction for 6.4 m = -4.5'.

> 2 - At watch-time t=00h12m our bottom course/speed became 184° True / 
> 16.2 kts

I will use mid-latitude sailing to compute the position of each 
observation. Since the latitude differences and the departures are 
independent of latitude and longitude, they need be calculated only once.

 From the Sun observation to first course and speed change, DLat1 = 
-.0027°, Dep1 = -.0099°.

> I then shot 4 heights of the MoonLL, averaged as follows : 
> t2=01h26m57.9s , h2=61°02'.9, and

To the above observation, DLat2 = -.3365°, Dep2 = -.0235°.

> I then shot 3 heights of Jupiter, averaged as follows : t3=01h42m01.8s , 
> h3=28°28'8

DLat3 = -.0676, Dep3 = -.0047°.

> 3 - At watch time t=01h47m our bottom course and speed became 135° True/ 
> 22.0 kts

Dlat4 = -.0223°, Dep4 = -.0016°.

> I then took 3 Jupiter-Moon Near Distances averaged as follows 
> t4=01h50m21.4s , d4=66°12'9

DLat5 = -.0145°, Dep5 = +.0145°.

Based on all the above, first approximation to the observation positions:

+47.000° -2.000° Sun
+46.661° -2.049° Moon
+46.593° -2.056° Jupiter
+46.556° -2.037° lunar distance

Set of linear equations for the observations:

-3.175° = -.562 ∆n - .827 ∆e - .141 ∆t (Sun)
  2.657° = -.624 ∆n + .782 ∆e + .134 ∆t (Moon)
-3.054° = -.628 ∆n - .778 ∆e - .134 ∆t (Jupiter)
   .148° = -.007 ∆n - .016 ∆e + .006 ∆t (lunar distance)

The method for forming these equations is basically what I have
described in other postings, but there are some differences, so I'll
describe it from scratch.

These equations represent the four observations. Each value on the left 
is the residual, i.e., the observed angle minus the computed angle 
(computed on the basis of the estimated time and position). For altitude 
observations, the residual is the same thing as the "altitude intercept" 
in celestial navigation. The coefficients (e.g., -.562, -.827, and -.141 
in the first equation) describe the response of the computed value to 
changes in northing, easting, and time. The variables ∆n, ∆e, and ∆t are 
the unknown corrections to northing, easting, and clock correction.

For example, the first equation says the observed Sun altitude was
3.175° less than the altitude computed for the estimated time and
position. It also says the computed altitude changes -.562° for each 
degree the observer moves north, -.827° for each degree east, and -.141° 
per minute of time. The first two values are simply the cosine and sine 
of the body's azimuth. The last is the sine of azimuth, times the cosine 
of latitude, and divided by 4. Also, the second value is in terms of a 
great circle degree, not a degree of longitude.

If you solve a set of linear equations by hand, it's soon obvious that 
only the numbers are important. The numbers are what you manipulate, but 
the = signs and variables don't change from step to step. So a common 
notation is to show the numbers only, in two matrices. (In fact, that's 
how the equations are entered into my calculator for solution.) Then the 
equations become:

|-3.175 |  | -.562 -.827 -.141 |
| 2.657 |  | -.624  .782  .134 |
|-3.054 |  | -.628 -.778 -.134 |
|  .148 |  | -.007 -.016  .006 |

For an overall measure of the agreement between the observed and 
computed angles, we can use the root of the mean squared error. That is, 
square each residual (the values at the left), sum the squares, divide 
by the number of values (4) to get the mean, and take the square root. 
According to the theory of least squares, the most likely solution to 
the problem is the one that brings this value to a minimum.

In this case, the initial root mean squared error is 2.57°.

My calculator says the least squares solution to the equations is ∆n = 
+.31°, ∆e = -.45°, ∆t = +23.9 m. Remember, ∆e is easting, so it must be 
divided by the cosine of latitude to obtain the longitude correction.

Then I apply those corrections to the initial position and recompute the 
positions of the observations:

+47.310° -2.664° Sun
+46.971° -2.713° Moon
+46.903° -2.720° Jupiter
+46.866° -2.701° lunar

I apply the 23.9 minute correction to the clock correction, and compute 
a new set of expected angles at each observation. Again, I subtract 
those from the observed angles to form the matrix of residuals on the 
left. The matrix of "partial derivatives" on the right is computed as 
before. Though the residuals are much smaller, the partial derivatives 
are close to the old values. That's because these are a function of the 
observing geometry, which hasn't changed very much because the initial 
guesses of time and position were fairly good.

Nevertheless, the matrix does change with any correction to the 
observer's position or clock correction. That is, linear equations don't 
precisely describe the behavior of the computed angles. But the 
approximation is good enough that a few iterations are enough to 
converge on the optimum solution.

|  .045 |  | -.500 -.866 -.147 |
|  .258 |  | -.729  .684  .117 |
|  .064 |  | -.558 -.830 -.142 |
| -.003 |  | -.006 -.017  .006 |

The root of the mean squared residual has decreased from 2.57° to
.135°.

The solution of this second set of equations gives new corrections to 
latitude, easting, and clock correction of -.261°, .154°, and -.324 
minutes. With these, plus the first set of corrections, the result is a 
third set of positions and equations:

+47.049° -2.438°
+46.710° -2.487°
+46.642° -2.493°
+46.605° -2.475°

|  .000 |  | -.498 -.867 -.148 |
| -.005 |  | -.728  .685  .117 |
|  .002 |  | -.554 -.832 -.143 |
|  .000 |  | -.006 -.016  .006 |

The root mean squared residual decreased from .135° to .003°. Again 
solve the above equations and get corrections .004° north, 0 east, and 
-.018 minute. Apply all corrections. Final UTC, latitude, longitude, and 
residual:

16:32:28.1 +47.053° (47 03.2) -2.441° (2 26.5) -.001° Sun
17:50:31.4 +46.714° (46 42.8) -2.490° (2 29.4)  .000° Moon
18:05:35.3 +46.646° (46 38.8) -2.496° (2 29.8)  .001° Jupiter
18:13:55.0 +46.609° (46 36.5) -2.478° (2 28.7)  .000° lunar

The small residuals do not imply a result accurate to .001°! They only 
say "the curve fits the points" to that accuracy. But all the points 
have errors, and the solution can be quite sensitive to them. E.g., 
error in the lunar distance appears in longitude, magnified 30x. And 
it's possible for the computer (me) to make a blunder but still get low 
residuals.

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