# NavList:

## A Community Devoted to the Preservation and Practice of Celestial Navigation and Other Methods of Traditional Wayfinding

Message:αβγ
Message:abc
 Add Images & Files Posting Code: Name: Email:
Re: SD topocentric formula
From: David Iwancio
Date: 2020 Sep 11, 14:01 -0700

Eric:

Trying to wrap my head around that formula took an awful lot of effort for a "rigorous" model that doesn't pay lip service to the true shape of the earth.

Anyway, all terms are multiplied by 1 earth radii to produce distances measured in earth radii.

The formula produces the tangent of the observed semidiameter by constructing a right triangle between the observer, the selenocenter and the radius from the selenocenter to the point of tangency of the line of sight.  The semidiameter angle's tangent is the radius of the moon divided by the length of the line of sight from the observer to the point of tangency on the moon's limb.

The terms under the radical in the denominator are from the Pythagorean theorem for a right triangle with the distance between the centers of the earth and the moon are the hypotenuse.  This distance is equal to csc(HP), also known as 1/sin(HP).  Deriving that expression can be easier to visualize if you remember that the geocentric horizontal paralax of the moon is equal to the selenocentric semidiameter of the earth.

The sum of the second and third terms under the radical, sin(h) ± k, produces the length of a line from the center of the earth to a particular point inside of the figure of the earth, where a line parallel to the line of sight, produced from the selenocenter, intercepts the earth's radius at a right angle.  It helped me to visualize this point as the corner of a rectangle, where the line between the centers of the earth and moon is a diagonal.

The entire radical itself produces the distance from the point of LoS tangency on the moon to the subterranean point.  The last term in the denominator, outside of the radical, is the distance between the observer and that subterranean point (the radius from the geocenter fo the observer is treated as the hypotenuse).  Removing it leaves the distance from the observer to the point of LoS tangency.

The geometry of the whole thing checks out for a spherical earth, but (in my amateur opinion) the effort spent on determining the length of the subterranean line, which is going to be very small compared to the earth-moon distance (the length of the diagonal is almost the same as the length of the longer side in the aforementioned rectangle), is misplaced when you're already ignoring the small deflection of the observer's vertical on a non-spherical earth.

Browse Files

Drop Files

### Join NavList

 Name: (please, no nicknames or handles) Email:
 Do you want to receive all group messages by email? Yes No
You can also join by posting. Your first on-topic post automatically makes you a member.

### Posting Code

Enter the email address associated with your NavList messages. Your posting code will be emailed to you immediately.
 Email:

### Email Settings

 Posting Code:

### Custom Index

 Subject: Author: Start date: (yyyymm dd) End date: (yyyymm dd)