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Re: Computaion of table of offsets.
From: Gary LaPook
Date: 2007 Oct 30, 00:54 -0700
Gary LaPook writes:

Thank you for your explanation and I thought it would be something like that. I see where the numbers come from, 60 NM in a degree and 360 degrees in the circumference of the earth for a total distance of 21,600 minutes or NM. Then you divide this by 2 pi to find the radius of the earth 3437.7467...NM which Bowditch rounds to 3438 NM. ( I have always remembered it as being 3440 NM but that is the mean radius based on the Clarke and other spheroids.)

Based on the formula the difference between the radius of the circle of position (a small circle) computed by this formula and the radius computed by the formula of 60 times the zenith distance (ZD) will not exceed 1 NM for any altitude above 84º 36' so it doesn't matter which formula you use for high altitude shots that you want to plot as circles around the GP (geographical position.)

But I still have a question, as the table you provided shows, the radius for a circle of position derived from an altitude of 10º is 324.94 degrees, which multiplied by 60 NM per degree results in the circle of position having a radius of  19, 496.43076...NM which seems to be an impossibility since the radius of the earth is only 3,437.7467...NM  meaning the radius derived by the formula is more than 16,058.8406 NM too large. In fact, for any altitude less than 45º the radius determined by this formula will exceed the radius of the earth.

But this doesn't take into account that the distance from the GP is on the curved surface of the earth. Using 60 times ZD you find a radius of 4,800 NM which makes sense because it is 5,400 NM (60 times 90) for a zero degree altitude so an altitude of 10 º (80º ZD) should produce a smaller radius. Comparing the 19,496.43076...NM radius derived from the formula with the 4,800 NM radius derived from 60 times ZD still results in a radius that is more than 14,696.43076 NM too large.

And looking at the case of a zero degree altitude, the radius as computed by 60 times ZD is 5,400 NM, as previously noted, but the radius as computed by the Bowditch formula is infinite. Using an altitude of one minute as approaching the zero degree altitude case, you end up with 5,399 NM using the 60 times ZD formula and 11,818,102.53....NM radius using the Bowditch formula, more than 11,812,270.53 NM larger and more than 11,814,664.78 NM larger than the radius of the earth.

So, what is going on with this formula?

gl

Andres Ruiz wrote: Considere this figure: Earth’s normal section to the plane of the CoP.

Tan(90-H) = R(CoP)/R(Earth)

A spherical Earth hás R(Earth) = 60*360/2PI = 21600/2PI

Then R(CoP) = R(Earth)/ Tan(H)

We can discern between the radius of the CoP in his plane, and the angular distance that is the radius of the CoP on the surface of a sphere: R = 90-H

Gary, In this table you can see that for high altitude CoP are near the same

 H zd = 90-H 360/(2*PI)/TAN(H) 10 80 324.94 20 70 157.42 30 60 99.24 40 50 68.28 50 40 48.08 60 30 33.08 70 20 20.85 71 19 19.73 72 18 18.62 73 17 17.52 74 16 16.43 75 15 15.35 76 14 14.29 77 13 13.23 78 12 12.18 79 11 11.14 80 10 10.10 81 9 9.07 82 8 8.05 83 7 7.04 84 6 6.02 85 5 5.01 86 4 4.01 87 3 3.00 88 2 2.00 89 1 1.00 90 0 0.00

Andrés Ruiz

http://www.geocities.com/andresruizgonzalez

-----Mensaje original-----
De: NavList@fer3.com [mailto:NavList@fer3.com] En nombre de glapook{at}PACBELL.NET
Enviado el: lunes, 29 de octubre de 2007 10:01
Para: NavList
Asunto: [NavList 3691] Computaion of table of offsets.

In the discusssion of the St. Hilaire method the subject of the table

of offsets used to approximate the curved LOP came up. This is table 4

in Bowditch (1975 ed), table 19 in the online Bowditch and is also

printed in each volume of H.O. 229. Bowditch gives the formulas for

this calculation and one of the formulas gives the radius of the

circle of postion. The formula for this is: R = 3438' cot altitude.

Ruiz gives a similar formula: R = (60*180/pi) cot altitude which

reduces to the other formula given in Bowditch.

My question is why do you use this formula? Why isn't the radius of

the circle of position simply  60 times the zenith distance? This is

the formula we have always used for plotting high altitude circles of

positions around the GP.  Should we use the Bowditch formula for this

purpose also.

gl

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