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    Re: It's Moon-landing Monday
    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
    
    
    
    
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