Author

Topic: Calculating Lat/Lon from the Sun historical

Captain K

posted May 03, 2004 05:08 PM
A few friends and I are on a five stage treasure hunt (all for fun) in Northern Virginia. This puzzle hunt was created by an engineer, history professor and scientist from NASA. It has taken us a few weeks worth of work to get to this point in the puzzle. This is all for fun, there are no monetary rewards, just bragging rights.
Where we have to use celestial navigation to find the final stage. We have an azimuth, elevation, and a date and time from the 19th century. The stages have been interwoven with the civil war history of the area. The 5th and final stage is located somewhere in the woods in Virginia. If I had to guess it is located in a park that has civil war history associated with it. The final stage is a Tupperware container, so we need the lat/long to be pretty accurate. If we can get within 75 feet, we can do a brute force physical search. This is the four pieces of date that I have: Date – Sunday, May 10, 1863 (This is the day that Stonewall Jackson died)
Time – 3:15 PM (This is the time that Stonewall Jackson died)
Azimuth  257.05005
Elevation  43.02465
I need to calculate where on earth (As discussed above, I would think it is somewhere in Virginia) that at the above date and time where on earth (lat/lon) can you get the above azimuth / elevation. It would be really cool if you could show me / explain to me how to calculate the answer, show me the steps and formulas use. Thanks for your help. It is really appreciated. Can't wait to hear from all of you.


David Burch

posted May 03, 2004 07:04 PM
We need a bit more detail... and even with that there are going to be more quesitons.
This is the four pieces of date that I have:
Date – Sunday, May 10, 1863 (This is the day that Stonewall Jackson died)
== that is OK. Calendar is same as now.
Time – 3:15 PM (This is the time that Stonewall Jackson died) === what time is this, GMT or some zone time. if not GMT what zone? if some historical time, then we need to clarify very precisely what is meant. also on the precision level of the following data, we must assume this is 3:15:00 exactly!
Azimuth  257.05005
=== is this meant to be the true bearing to the sun? if so, is this is decimal degrees?
Elevation  43.02465
=== Is this the "elevation" of the sun, ie its height above the visbible, true horizon in degrees? or does it mean 43° 02' 46.5"  a convention sometimes used.
=== if we know these things, we can compute a *theoretical* location, however it must be stressed that this is all theoretical, as neither of these values can be measured to the precision indicated nor what you need to establish a location to such a precision using cel nav.
david
From: Starpath, Seattle, WA


Captain K

posted May 04, 2004 05:33 AM
Thanks for your swift reply, here is the additional data that you requested.
The time is 3:15 PM local time in Virginia, which I believe is 5 hours off of GMT (correct me if I am incorrect). For the sake of this problem you can assume 3:15:00 PM (no seconds).
Azimuth  257.05005 is the true bearing of the sun in decimal degrees.
Elevation  43.02465 is this the elevation of the sun in decimal degrees.
I hope that this helps. Thanks.


David Burch

posted May 04, 2004 05:05 PM
we cannot correct you as we have no way of knowing. we can guess if you give an approximate lat lon for the anticipated location and expain if local time means daylight savings time is in effect.
note if you give us a street address we can convert that to a lat lon using one of our online resources.
in the meantime, i will look into the computations to see if we are at all close, using Lynchburg 37° 20' N, 79° 12'W, time = 15:15:00 ZD = +5 (they are likely on +4 in the summer), on Sunday, May 10, 1863.... also assuming that since these are all calculations, not observations, that the elevation is to the center of the sun.
Aanswer: Using lynchburg, as above, with ZD+5, then at the time and date given, Zn = 256.4° and Hc = 45° 38.7', so it appears this is roughly consistent.
We need the time issue resolved, ie ZD+4 or +5, we can likely solve that ourselves, but it would be quicker to know.
=======
note i am underway with our onboard nav course leaving Friday 7th and not back till may 19th. might be able to work on this some during that time, but i might ask that any of our students or online participants who have a starpilot or other means of precise compuations might join in and work out the solution.
it will have to be an iterative procedure: gha and dec of sun can be computed for historical date, then maybe check with NAO data online to confirm. starpilot will do 1863, i do not know which other sources will.
my guess would be start with a GCDR computation from sun's dec and GHA using 90Hc for distance and Zn180 for course, then from that lat lon, iterate to home in on answer. the problem is, most packaged math routines will not carry the precision given in the problem data, so that has to be resolved somehow, or do the computations manually.
subtlties: the person who generated the problem most likely used a spherical earth, since cel nav relies on that, but if you solve the GC part with a GPS then they use ellipsoidal compuations and differences could be a dozen miles or so... won't affect the final iterated answer, but will get you closer to start with spherical. we have an article online to show how to fool the GPS into spherical computations. typical standard PC programs, however, still use spherical earth.
i have email underway, but not much free time.
david
From: Starpath, Seattle, WA


David Burch

posted May 04, 2004 06:00 PM
A few more thoughts that ultimately have to be considered.
imagine how the problem might have been made. one possibilitymaybe the only possibility. Person buries the treasure,
(1) then somehow establishes the lat lon of the location.
this is either done from a detailed chart or map, which may or may not be right, depending on several factors.... or they use a GPS. to the precision we care about here, this takes much care and knowledge of GPS. first they have to have the right chart datum selected. Next they have to use a modern unit and let it average out the data. this way they could get a position to within say ±20 meters. if they had good conections to real surveyors with expensive equipment, they could do much better. Or if they had access to dgps in the region (again surveyor connections) they could do very accurately, submeter in the extreme.
2) once they have a lat lon, they need to do the cel nav computations... if they just randomly punch this into any commercial or freeware program, they could end up with vastly different data because of the almanac computations from 1863, and the required handling of the precision of each step of the computation  but since they specify the data so precisely, they are likely using some academic computation... and we have to assume that they are using the best historical almanac data.
so both of these points have to answered in an affirmative, bestanswer solution, or we are still on a goose chase on the level of finding a tupper ware box, or even the Molokai Lighthouse in the middle of the night.
then we have to look to what the inherent accuracy of the historical almanac computation might be.... standard thinking would limit this to 0.1' or some 600 feet, but for the sun, one might be able to do better. this has to be researched and i am hoping someone will jump in with the answer.
From: Starpath, Seattle, WA


Luis Soltero

posted May 04, 2004 09:14 PM
This is really a straight forward sight reduction in reverse. To solve this problem you need (a) an accurate date and time (for the computation of GHA and DEC for the sun and distance between center of the earth and the sun) (b) height of eye to compute dip correction (assume 0 if not given) (c) Observed Altitude (to center of body) (d) Observed Azimuth (in True)
Step 1  compute the ephemeris information for the sun at given date.
Use Mica or some other astronomy program to compute an accurate position for the sun. You need the RA (right of ascension), DEC, and distance from the center of the earth. RA can easily be converted to GHA (this is left as an exercise for the reader). This needs to be as accurate as you can get it. The Nautical Almanac will not cut it. The error in GHA for NA is .3' which is not good enough for this problem.
Step 2
The next step in solving this problem is to compute Hc from the observed altitude. There are two components that come into play here. Refraction, parallax and dip. The dip angle is assumed to be 0 since a height of eye was not given. Refraction and parallax can be computed using the following equations.
Parallax = cos h (0.147'/d) where h is the observed altitude of the body and d is the distance between the center of the sun and the center of the earth in au. This computed value should be about 0.1'.
Refraction = 12.242373(z  arc sin (0.998606 sin(0.996761z)))  0.039826z where z is the zenith distance (90  h).
Now you compute Ho using Ho = h  Refraction + Parallax
Step 3
Now that you have Ho, Z, GHA, DEC you can solve for Lat and Lon using the sight reduction equations in reverse.
You have 2 unknowns (Lat and Lon) and therefore need two equations
Cos Z = (sin d  sin L sin h)/(cos L cos h) where d is declination, L is lat, h is Ho and Z is the Azimuth.
solve for Lat.
now use
sin h = sin L sin d + cos L cos d cos LHA
substitute L from above and solve for LHA use GHA from Step 1 to compute Long.
You are done.
Luis Soltero, Ph.D., MCS Author of StarPilot


