Geographic location determination

In summary, the conversation was about determining the geographic location of an observer by measuring the elevation of two stars at a certain time. The script provided for Octave was used, but there were some issues with its accuracy. The expert suggests a different approach of finding the equivalent latitudes and longitudes of the stars and using the spherical law of cosines to find the observer's location. This method may require inputting the curvature of the Earth for greater precision.
  • #1
alseth
15
0
Hello guys. I was trying to determine the geographic location of the observer by measuring the elevation of two distinct stars at a certain time. Well, I am not very skilled in programming and I found this script for Octave and tried to use it.

My data:
Star A: Arcturus(α Boo)
Right ascension: 14h 15 m 39.7s
Declination: +19° 10' 56"
Time: 19:31:11 UT
Elevation: (100-89.9)*0.9= 9.09°
Elevation after the refraction correction: 9.0°
JD1 = 2455097.31332

Star B: Mirfak (α Per)
Right ascension: 03h 24m 19.4s
Declination: +49° 51′ 40″
Time: 20:05:29 UT
Elevation: (100-61.5)*0.9= 34.65°
Elevation after the refraction correction: 34.62°
JD2= 2455097.33714

script:
# 1 rad
rad=180/pi
# 90degrees-elevation
r1 = 81
r2 = 55.38
# coordinates of the first star
# Arcturus, a1 = a1 - ts
a1 = 213.92 - 15*(19+(38/60)+(20.2/3600))
d1 = 19.18
# coordinates of the second star
# Mirfak
a2 = 51.081 - 15*(20+(12/60)+(43.9/3600))
d2 = 49.86
# their distance
rad*acos(sin(d1/rad)*sin(d2/rad) +
cos(d1/rad)*cos(d2/rad)*cos((a2-a1)/rad))
# first estimate of the geographical location
a = 16
d = 50
for i = 1:3
t1=sin(d1/rad)*sin(d/rad)+cos(d1/rad)\
*cos(d/rad)*cos((a-a1)/rad)
t2=sin(d2/rad)*sin(d/rad)+cos(d2/rad)\
*cos(d/rad)*cos((a-a2)/rad)
p = [ rad*acos(t1) - r1,
rad*acos(t2) - r2]
sqrt(p(1)**2+p(2)**2)
# susbstitution
u = -1/sqrt(1 - t1**2)
v = -1/sqrt(1 - t2**2)
# derivation matrix
m = [ -u*sin((a-a1)/rad)*cos(d1/rad)*cos(d/rad),\
u*(sin(d1/rad)*cos(d/rad) - \
cos(d1/rad)*sin(d/rad)*cos((a-a1)/rad));\
-v*sin((a-a2)/rad)*cos(d2/rad)*cos(d/rad),\
v*(sin(d2/rad)*cos(d/rad) - \
cos(d2/rad)*sin(d/rad)*cos((a-a2)/rad))]
# inverse matrix
[im,c] = inv(m)
# linear equations
x = -(im*p)
# and addition to the initial estimates
a = a + x(1)
d = d + x(2)
endfor

ts is apparently the sidereal time a and r1, r2 are 90 degrees minus elevation

The observer should be around 16.6 degrees east and 49.2 degrees north but I am getting very imprecise values. Even some parts of the script do not seem to e doing anything. Is there some error there or maybe is there a more elegant way to do the computation?
Thanks for any help.
 
Astronomy news on Phys.org
  • #2
I'm not an astrophysicist and don't know why matrices are being used in the program, but here's what I would do.

First, find the equivalent latitudes and longitudes of the two stars. By "equivalent latitude and longitude", I mean the position that a line going from Earth's center to the star would pass through. Then, calculate what the second star's longitude would have been at the time the first star was observed. This should be trivial: Earth takes 23 hours, 56 minutes to rotate once.

If you imagine drawing a circle of radius 90-alt with the first star at the center, and doing the same for the second, the observer's location would be at one of the intersections of the circles. Written out explicitly:

cos theta1=sin(A)sin(o)+cos(A)cos(o)cos(lonA-lonO)
cos theta2=sin(A2)sin(o)+cos(A2)cos(o)cos(lonA2-lonO)

The two theta's are equal to 90-alt; A, lonA, A2, lonA2 are the stars' positions while o,lonO is the observer's position. These equations are just the spherical law of cosines for the angular distance between Earth's center and the two stars. I don't know if it's possible to solve these equations analytically for o and lonO, but if it's not, naively trying out every integer for o and every integer for lonO, picking the most accurate combination, then reducing the step size to 0.1 degrees will quickly give you an answer.
 
  • #3
Remember you are taking measurements from a curved surface. Inputing curvature should considerably improve the precision - note, ideasrule has outlined how to do this.
 

Related to Geographic location determination

1. What is geographic location determination?

Geographic location determination is the process of determining the exact location of a specific point on the Earth's surface. This can be done using various methods such as GPS, cellular triangulation, or satellite imagery.

2. Why is geographic location determination important?

Geographic location determination is important for a variety of reasons. It allows us to accurately navigate and map our surroundings, track the movement of objects or individuals, and make informed decisions based on location-specific data.

3. What are some common techniques used for geographic location determination?

Some common techniques used for geographic location determination include GPS, Wifi positioning, cellular triangulation, and satellite imagery. Each method has its own advantages and limitations, and the most appropriate technique will depend on the specific application.

4. How accurate is geographic location determination?

The accuracy of geographic location determination can vary depending on the method used and the environmental conditions. GPS, for example, can provide location accuracy within a few meters, while cellular triangulation may only be accurate within a few hundred meters.

5. What are some potential challenges or limitations of geographic location determination?

One potential challenge of geographic location determination is the reliance on technology and infrastructure. In remote or underdeveloped areas, accurate location determination may be difficult or even impossible. Additionally, factors such as weather, terrain, and signal interference can also impact the accuracy of location determination methods.

Similar threads

  • Introductory Physics Homework Help
Replies
11
Views
799
Replies
28
Views
2K
Replies
25
Views
2K
  • Precalculus Mathematics Homework Help
Replies
17
Views
13K
  • Special and General Relativity
Replies
11
Views
372
  • Precalculus Mathematics Homework Help
Replies
16
Views
3K
  • Advanced Physics Homework Help
Replies
2
Views
3K
Replies
2
Views
2K
  • Engineering and Comp Sci Homework Help
Replies
1
Views
2K
  • Differential Equations
Replies
1
Views
1K
Back
Top