# NavList:

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

**Re: Hints2Travellers 1883: graphical LD clearing?**

**From:**Lars Bergman

**Date:**2022 May 12, 07:13 -0700

I think Frank is right. Prolong the line S''S' until it crosses the prolonged line AB. Call this point F. Let the radius of the circle equal unity. Call the crossing closest to the Moon sign M. Call the crossing closest to the Mars sign N.

From the text one can figure out that angle BAS is equal to the lunar distance LD, angle BAx is equal to 90°-ALT_{m}, and angle SAS'' is equal to 90°-ALT_{s}. The angle at F is equal to 90°-LD. Then it follows that angle MES' is equal to LD. The length AN is equal to sin(ALT_{s}) and the length AM is equal to sin(ALT_{m}). And AF equals sin(ALT_{s}) / cos(LD). Then in the triangle MFE you have tan(LD) = MF / ME = (AF - AM) / ME, hence ME = (sin(ALT_{s}) / cos(LD) - sin(ALT_{m})) / tan(LD) or

ME = sin(ALT_{s}) / sin(LD) - sin(ALT_{m}) / tan(LD) = (P_{2} - P_{1}) / HP. "Multiply the line of correction by the horizontal parallax" yields ME·HP = P_{2} - P_{1} and as the correction in this case should be subtractive, we have LD_{c} = LD + P_{1} - P_{2} , identical to Frank's formula. Regarding the constant 62 (or 53) I have no explanation.

Lars