About the geodesic coordinates, and their conversion into cartesian ones
This is more a sort of "let me know if I'm right" question, rather than a real question, but I thought that if is there a place where I can find true and solid trustful answers then it's here.
Suppose someone gives me those coordinates:
$$109^{circ} 39' N$$
$$30^{circ} 10' W$$
If I want to localize this place over the Earth surface, via cartesian coordinates (or better: if I simply want to convert them into cartesian), here is what I understood to be the process.
Please, notice that this is what I understood so if it's wrong you have to correct me because I did not find any book or notes where this has been explained clearly.
First step
Transform the coordinates I got into "true angles" that is: latitude $theta$ and longitude $phi$ according to the DMS transformation:
$$theta = 109 + frac{39}{60} = 109.65$$
$$phi = 30 + frac{10}{60} = 39.16$$
Second Step
Using the polar coordinates to find the space vector, that is:
$$x = Rsinthetacosphi$$
$$y = Rsinthetasinphi$$
$$z = Rcostheta$$
Where $R = $ Earth radius.
Finally I will obtain the space vector $mathbf{r} = (x, y, z)$ for the place in question.
Final Question
Is this all right? Are there any errors or information I shall take into account? Any remark?
Thank you all.
analytic-geometry coordinate-systems polar-coordinates spherical-coordinates geodesy
edited Nov 26 at 0:45
Henry
98k475160
asked Mar 8 '17 at 14:04
Von Neumann
16.3k72543
add a comment |
This is more a sort of "let me know if I'm right" question, rather than a real question, but I thought that if is there a place where I can find true and solid trustful answers then it's here.
Suppose someone gives me those coordinates:
$$109^{circ} 39' N$$
$$30^{circ} 10' W$$
If I want to localize this place over the Earth surface, via cartesian coordinates (or better: if I simply want to convert them into cartesian), here is what I understood to be the process.
Please, notice that this is what I understood so if it's wrong you have to correct me because I did not find any book or notes where this has been explained clearly.
First step
Transform the coordinates I got into "true angles" that is: latitude $theta$ and longitude $phi$ according to the DMS transformation:
$$theta = 109 + frac{39}{60} = 109.65$$
$$phi = 30 + frac{10}{60} = 39.16$$
Second Step
Using the polar coordinates to find the space vector, that is:
$$x = Rsinthetacosphi$$
$$y = Rsinthetasinphi$$
$$z = Rcostheta$$
Where $R = $ Earth radius.
Finally I will obtain the space vector $mathbf{r} = (x, y, z)$ for the place in question.
Final Question
Is this all right? Are there any errors or information I shall take into account? Any remark?
Thank you all.
analytic-geometry coordinate-systems polar-coordinates spherical-coordinates geodesy
edited Nov 26 at 0:45
Henry
98k475160
asked Mar 8 '17 at 14:04
Von Neumann
16.3k72543
add a comment |
This is more a sort of "let me know if I'm right" question, rather than a real question, but I thought that if is there a place where I can find true and solid trustful answers then it's here.
Suppose someone gives me those coordinates:
$$109^{circ} 39' N$$
$$30^{circ} 10' W$$
If I want to localize this place over the Earth surface, via cartesian coordinates (or better: if I simply want to convert them into cartesian), here is what I understood to be the process.
Please, notice that this is what I understood so if it's wrong you have to correct me because I did not find any book or notes where this has been explained clearly.
First step
Transform the coordinates I got into "true angles" that is: latitude $theta$ and longitude $phi$ according to the DMS transformation:
$$theta = 109 + frac{39}{60} = 109.65$$
$$phi = 30 + frac{10}{60} = 39.16$$
Second Step
Using the polar coordinates to find the space vector, that is:
$$x = Rsinthetacosphi$$
$$y = Rsinthetasinphi$$
$$z = Rcostheta$$
Where $R = $ Earth radius.
Finally I will obtain the space vector $mathbf{r} = (x, y, z)$ for the place in question.
Final Question
Is this all right? Are there any errors or information I shall take into account? Any remark?
Thank you all.
analytic-geometry coordinate-systems polar-coordinates spherical-coordinates geodesy
edited Nov 26 at 0:45
Henry
98k475160
asked Mar 8 '17 at 14:04
Von Neumann
16.3k72543
This is more a sort of "let me know if I'm right" question, rather than a real question, but I thought that if is there a place where I can find true and solid trustful answers then it's here.
Suppose someone gives me those coordinates:
$$109^{circ} 39' N$$
$$30^{circ} 10' W$$
If I want to localize this place over the Earth surface, via cartesian coordinates (or better: if I simply want to convert them into cartesian), here is what I understood to be the process.
Please, notice that this is what I understood so if it's wrong you have to correct me because I did not find any book or notes where this has been explained clearly.
First step
Transform the coordinates I got into "true angles" that is: latitude $theta$ and longitude $phi$ according to the DMS transformation:
$$theta = 109 + frac{39}{60} = 109.65$$
$$phi = 30 + frac{10}{60} = 39.16$$
Second Step
Using the polar coordinates to find the space vector, that is:
$$x = Rsinthetacosphi$$
$$y = Rsinthetasinphi$$
$$z = Rcostheta$$
Where $R = $ Earth radius.
Finally I will obtain the space vector $mathbf{r} = (x, y, z)$ for the place in question.
Final Question
Is this all right? Are there any errors or information I shall take into account? Any remark?
Thank you all.
analytic-geometry coordinate-systems polar-coordinates spherical-coordinates geodesy
analytic-geometry coordinate-systems polar-coordinates spherical-coordinates geodesy
edited Nov 26 at 0:45
Henry
98k475160
asked Mar 8 '17 at 14:04
Von Neumann
16.3k72543
edited Nov 26 at 0:45
Henry
98k475160
asked Mar 8 '17 at 14:04
Von Neumann
16.3k72543
edited Nov 26 at 0:45
Henry
98k475160
edited Nov 26 at 0:45
Henry
98k475160
edited Nov 26 at 0:45
Henry
98k475160
98k475160
asked Mar 8 '17 at 14:04
Von Neumann
16.3k72543
asked Mar 8 '17 at 14:04
Von Neumann
16.3k72543
asked Mar 8 '17 at 14:04
Von Neumann
16.3k72543
16.3k72543
add a comment |
add a comment |
2 Answers
2
active
oldest
votes
if your $z-$axis travels from south to north pole, then $z = R sin(lattitude, angle)$ where North is assumed positive and south is assumed negative. (Center of earth is System of coordinates center). Consequently $x = R cos(lattitude, angle) cos(longitude, angle)$ and $y = R cos(lattitude, angle) sin(longitude, angle)$ for longitude angle : East is positive and West is negative limit is [-180,180[
answered Mar 8 '17 at 14:18
Hmath
1625
add a comment |
You have a BIG mistake, which is the limit of the lattitude are $90S$ which is the South Pole to $90N$ which is the North Pole. You are here mentioning a lattitude which is 109N. This is outside the range. I assume your angles should be $109W$ and $30N$ which is somewhere Texas....
Fix your coordinates then, we can verify your transformation .
edited Feb 6 at 8:41
Von Neumann
16.3k72543
answered Mar 8 '17 at 14:11
Hmath
1625
Oh, so you mean I shall have something like $109$ for $phi$ and $39$ for $theta$?
– Von Neumann
Mar 8 '17 at 14:12
add a comment |
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2 Answers
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active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
if your $z-$axis travels from south to north pole, then $z = R sin(lattitude, angle)$ where North is assumed positive and south is assumed negative. (Center of earth is System of coordinates center). Consequently $x = R cos(lattitude, angle) cos(longitude, angle)$ and $y = R cos(lattitude, angle) sin(longitude, angle)$ for longitude angle : East is positive and West is negative limit is [-180,180[
answered Mar 8 '17 at 14:18
Hmath
1625
add a comment |
if your $z-$axis travels from south to north pole, then $z = R sin(lattitude, angle)$ where North is assumed positive and south is assumed negative. (Center of earth is System of coordinates center). Consequently $x = R cos(lattitude, angle) cos(longitude, angle)$ and $y = R cos(lattitude, angle) sin(longitude, angle)$ for longitude angle : East is positive and West is negative limit is [-180,180[
answered Mar 8 '17 at 14:18
Hmath
1625
add a comment |
if your $z-$axis travels from south to north pole, then $z = R sin(lattitude, angle)$ where North is assumed positive and south is assumed negative. (Center of earth is System of coordinates center). Consequently $x = R cos(lattitude, angle) cos(longitude, angle)$ and $y = R cos(lattitude, angle) sin(longitude, angle)$ for longitude angle : East is positive and West is negative limit is [-180,180[
answered Mar 8 '17 at 14:18
Hmath
1625
if your $z-$axis travels from south to north pole, then $z = R sin(lattitude, angle)$ where North is assumed positive and south is assumed negative. (Center of earth is System of coordinates center). Consequently $x = R cos(lattitude, angle) cos(longitude, angle)$ and $y = R cos(lattitude, angle) sin(longitude, angle)$ for longitude angle : East is positive and West is negative limit is [-180,180[
answered Mar 8 '17 at 14:18
Hmath
1625
answered Mar 8 '17 at 14:18
Hmath
1625
answered Mar 8 '17 at 14:18
Hmath
1625
answered Mar 8 '17 at 14:18
Hmath
1625
1625
add a comment |
add a comment |
You have a BIG mistake, which is the limit of the lattitude are $90S$ which is the South Pole to $90N$ which is the North Pole. You are here mentioning a lattitude which is 109N. This is outside the range. I assume your angles should be $109W$ and $30N$ which is somewhere Texas....
Fix your coordinates then, we can verify your transformation .
edited Feb 6 at 8:41
Von Neumann
16.3k72543
answered Mar 8 '17 at 14:11
Hmath
1625
Oh, so you mean I shall have something like $109$ for $phi$ and $39$ for $theta$?
– Von Neumann
Mar 8 '17 at 14:12
add a comment |
You have a BIG mistake, which is the limit of the lattitude are $90S$ which is the South Pole to $90N$ which is the North Pole. You are here mentioning a lattitude which is 109N. This is outside the range. I assume your angles should be $109W$ and $30N$ which is somewhere Texas....
Fix your coordinates then, we can verify your transformation .
edited Feb 6 at 8:41
Von Neumann
16.3k72543
answered Mar 8 '17 at 14:11
Hmath
1625
Oh, so you mean I shall have something like $109$ for $phi$ and $39$ for $theta$?
– Von Neumann
Mar 8 '17 at 14:12
add a comment |
You have a BIG mistake, which is the limit of the lattitude are $90S$ which is the South Pole to $90N$ which is the North Pole. You are here mentioning a lattitude which is 109N. This is outside the range. I assume your angles should be $109W$ and $30N$ which is somewhere Texas....
Fix your coordinates then, we can verify your transformation .
edited Feb 6 at 8:41
Von Neumann
16.3k72543
answered Mar 8 '17 at 14:11
Hmath
1625
You have a BIG mistake, which is the limit of the lattitude are $90S$ which is the South Pole to $90N$ which is the North Pole. You are here mentioning a lattitude which is 109N. This is outside the range. I assume your angles should be $109W$ and $30N$ which is somewhere Texas....
Fix your coordinates then, we can verify your transformation .
edited Feb 6 at 8:41
Von Neumann
16.3k72543
answered Mar 8 '17 at 14:11
Hmath
1625
edited Feb 6 at 8:41
Von Neumann
16.3k72543
edited Feb 6 at 8:41
Von Neumann
16.3k72543
edited Feb 6 at 8:41
Von Neumann
16.3k72543
16.3k72543
answered Mar 8 '17 at 14:11
Hmath
1625
answered Mar 8 '17 at 14:11
Hmath
1625
answered Mar 8 '17 at 14:11
Hmath
1625
1625
Oh, so you mean I shall have something like $109$ for $phi$ and $39$ for $theta$?
– Von Neumann
Mar 8 '17 at 14:12
add a comment |
Oh, so you mean I shall have something like $109$ for $phi$ and $39$ for $theta$?
– Von Neumann
Mar 8 '17 at 14:12
Oh, so you mean I shall have something like $109$ for $phi$ and $39$ for $theta$?
– Von Neumann
Mar 8 '17 at 14:12
Oh, so you mean I shall have something like $109$ for $phi$ and $39$ for $theta$?
– Von Neumann
Mar 8 '17 at 14:12
add a comment |
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