NavList:
A Community Devoted to the Preservation and Practice of Celestial Navigation and Other Methods of Traditional Wayfinding
Re: It's Moon-landing Monday
From: Frank Reed
Date: 2009 Jul 21, 00:54 -0700
From: Frank Reed
Date: 2009 Jul 21, 00:54 -0700
Brad, you wrote: "The distance (900 miles) is an easy one. Use your sextant to measure the diameter of the moon, and since you know the true diameter, you can calculate the distance to the moons center, using simple triangles similar to HP." Yes, that would work. The Moon would loom awfully large from 900 miles above the surface. Could you measure it through the small windows typical on spacecraft? You could do something similar with a couple of small, prominent craters. Presumably you have detailed lunar topography (it's 2029 after all), so you could ask your software to compute the exact angles between some craters beneath your trajectory for the correct altitude. Any other ways to navigate visually?? You added: "As a check, you can also perform the same calculation using the diameter of the earth." That wouldn't be accurate enough, would it? The Earth would be about two degrees across seen from the Moon. Suppose I can measure its angular diameter to +/-0.25 minutes of arc. That's about one five-hundredth of the diameter and thus corresponds to a similar proportional error in the distance from the Earth or roughly 500 miles. That is, if I measure the Earth's apparent diameter as two degrees +/-0.25' then the distance away is 240,000 miles +/-500 miles (off the top of my head --somebody check my math) "In terms of the RA and declination, use the earth as a nadir point and determine where you are relative to the star patterns shown. Nominally, in celestial navigation, we use the zenith, but because it is easy to look down at the earth and see the star patterns, look at your nadir. Since the star patterns will not shift in parallax for an orbit within the earth-moon system, this will provide a very reasonable RA and declination." Indeed. You measure a couple of angles between stars and the Earth at known GMT. We could call them "terran distances" instead of "lunar distances". Then those together place you on a "ray of position" extending from the center of the Earth. I will mention (again --can't resist) that you can do the same thing on the surface of the Earth to fix your position by measuring lunar distances at known GMT. It's space navigation on the ground. You concluded: "Finally, since we can assume you are "out of earth orbit" for 6 hours, you are generally pointed in the correct direction anyway. As a result, the time to fire the rocket is most dependent on the distance and not so much on the RA and declination." I was thinking of spacecraft orientation. If your rocket isn't pointed in the right direction when you fire, even by a few degrees, you will not get where you want to go. In fact, the only real practical use of star sights on the Apollo missions was to "align the platform" of the inertial navigation system which amounts to using star sights as a 3d astro-compass. Without an INS and with thrusters on manual, there's no way you could depend on the pointing direction of a manned spacecraft for more than a few hours. Light pressure differences on various parts of the spacecraft, slight outgassing from thrusters and other components, and astronauts shifting around inside would all change the spacecraft's orientation substantially. The Apollo spacecraft had an auto-pilot system tied to the INS which fired thrusters automatically to maintain orientation (or planned rolls). You can see a dramatization of it trying to do its job after the explosion during the movie "Apollo 13". -FER --~--~---------~--~----~------------~-------~--~----~ NavList message boards: www.fer3.com/arc Or post by email to: NavList@fer3.com To , email NavList-@fer3.com -~----------~----~----~----~------~----~------~--~---