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Beating the dead horse of CN
Before I get started on my questions, I assure you, I have read all the threads here and on other sites regarding Celestial Navigation and have learned a ton so far. My questions are for the serious sextant users.
As a light scout in the Army, Land Navigation is my bread and butter. When I train my soldiers, I enforce the compass, map and protractor as primary means of navigation and to use the GPS to confirm their plots. I teach hasty, primitive means of finding cardinal directions and I am a big believer in knowing how to navigate using the most primitive of resources before I rely on GPS. When I attend ASA 101-108 courses after I retire, I want to go in with a fairly good grasp of CN.
Recently I have purchased an Astra IIIB sextant, nautical almanac (altitude correction tables) and a few books on how to use them. Since I am landlocked for the next year, I can only practice land navigation (artificial horizon) with the sextant. So far, best accuracy is on the order of 1 mile.
So, getting to the point of this post, being a simple minded knuckle dragger, I will start with my issues in finding latitude by double meridian altitude, using an artificial horizon.
To paraphrase some of what I have read and understood, "A meridian altitude of the sun's lower limb ('noon shot'), when an artificial horizon is used, this type of observation may be called a double meridian altitude, because the angle measured by the sextant is twice what it would have been if a sea horizon had been used."
This basically comes down to a ton of math.... "keep in mind, by "height of the sun" I mean height of the sun above the horizon." J. Gottfred.
"Height of sun times 2 (artificial horizon), less index correction gives corrected height of the sun. Divide by 2 to get height of the sun, less refraction of the air correction gives height of sun corrected for refraction. Adding semi-diameter of sun for current date gives height of the sun as observed. Angle between the zenith and horizon, less the height of sun, gives zenith distance.
Subtract south declination of sun for final latitude." J. Gottfred.
Am I just slow, or doing this totally wrong; is it supposed to take several hours to find latitude this way?
Is finding latitude by double altitudes, using an upper limb observation of the sun, which takes about the same amount of time to take readings, an easier or better method?
I will get into finding latitude by double altitudes and longitude from GT, local time, magnetic variation in another post, my brain is kinda fried at the moment. I look forward to your replies to these 2 questions.