View Angle CalculationCalculating View Angle?Calculating View Angle?Calculate field of view / angle of a...
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View Angle Calculation
Calculating View Angle?Calculating View Angle?Calculate field of view / angle of a point to a shapefileTrasformation from ECEF to ENURotate symbol base on polyline angleMeasuring Angle of two Points from North in ArcGIS 10.3 C# SDKHow to measure angle in QGISRotate feature by angle (QGIS)tool to output xy from an input xy, distant, and angleAdvanced Digitizing Panel angle unitsSolar angle of Incident calculation confusion!
In Calculating View Angle? the accepted answer contains the following calculations:
Given two points (x,y,z) and (x',y',z') in an earth-centered coordinate system, the vector from the first to the second is (dx,dy,dz) = (x'-x, y'-y, z'-z)
And then provides an example:
The XYZ coordinates of the airplanes are (x,y,z) = (1285410, -4797210, 3994830) and (x',y',z') = (1202990, -4824940, 3999870), respectively (in the ITRF00 datum, which uses the GRS80 ellipsoid). The pilot's view vector therefore is (dx,dy,dz) = (-82404.5, -27735.3, 5034.56).
What am I missing such that:
1202990 - 1285410 = -82404.5 (vice -82420)
-4824940 - -4797210 = -27735.3 (vice -27730)
3999870 - 3994830 = 5034.56 (vice 5040) ?
coordinates angles
New contributor
RudolfSchmidt is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
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In Calculating View Angle? the accepted answer contains the following calculations:
Given two points (x,y,z) and (x',y',z') in an earth-centered coordinate system, the vector from the first to the second is (dx,dy,dz) = (x'-x, y'-y, z'-z)
And then provides an example:
The XYZ coordinates of the airplanes are (x,y,z) = (1285410, -4797210, 3994830) and (x',y',z') = (1202990, -4824940, 3999870), respectively (in the ITRF00 datum, which uses the GRS80 ellipsoid). The pilot's view vector therefore is (dx,dy,dz) = (-82404.5, -27735.3, 5034.56).
What am I missing such that:
1202990 - 1285410 = -82404.5 (vice -82420)
-4824940 - -4797210 = -27735.3 (vice -27730)
3999870 - 3994830 = 5034.56 (vice 5040) ?
coordinates angles
New contributor
RudolfSchmidt is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
add a comment |
In Calculating View Angle? the accepted answer contains the following calculations:
Given two points (x,y,z) and (x',y',z') in an earth-centered coordinate system, the vector from the first to the second is (dx,dy,dz) = (x'-x, y'-y, z'-z)
And then provides an example:
The XYZ coordinates of the airplanes are (x,y,z) = (1285410, -4797210, 3994830) and (x',y',z') = (1202990, -4824940, 3999870), respectively (in the ITRF00 datum, which uses the GRS80 ellipsoid). The pilot's view vector therefore is (dx,dy,dz) = (-82404.5, -27735.3, 5034.56).
What am I missing such that:
1202990 - 1285410 = -82404.5 (vice -82420)
-4824940 - -4797210 = -27735.3 (vice -27730)
3999870 - 3994830 = 5034.56 (vice 5040) ?
coordinates angles
New contributor
RudolfSchmidt is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
In Calculating View Angle? the accepted answer contains the following calculations:
Given two points (x,y,z) and (x',y',z') in an earth-centered coordinate system, the vector from the first to the second is (dx,dy,dz) = (x'-x, y'-y, z'-z)
And then provides an example:
The XYZ coordinates of the airplanes are (x,y,z) = (1285410, -4797210, 3994830) and (x',y',z') = (1202990, -4824940, 3999870), respectively (in the ITRF00 datum, which uses the GRS80 ellipsoid). The pilot's view vector therefore is (dx,dy,dz) = (-82404.5, -27735.3, 5034.56).
What am I missing such that:
1202990 - 1285410 = -82404.5 (vice -82420)
-4824940 - -4797210 = -27735.3 (vice -27730)
3999870 - 3994830 = 5034.56 (vice 5040) ?
coordinates angles
coordinates angles
New contributor
RudolfSchmidt is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
New contributor
RudolfSchmidt is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
edited 2 hours ago
PolyGeo♦
53.6k1780240
53.6k1780240
New contributor
RudolfSchmidt is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
asked 4 hours ago
RudolfSchmidtRudolfSchmidt
83
83
New contributor
RudolfSchmidt is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
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New contributor
RudolfSchmidt is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
RudolfSchmidt is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
add a comment |
add a comment |
1 Answer
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The ECEF coordinates that @whuber provided appear to have been rounded to the nearest 10 meters. Looking at the revisions to the answer reveals that originally, the reference points that he used were (N39°, W75°, 4000m) and (N39°, W76°, 12000m), translating to ECEF coordinates:
(1285408.203, -4797208.722, 3994834.304)
(1202993.662, -4824944.042, 3999868.867)
The vectors are then:
dX = -82414.541
dY = -27735.320
dZ = 5034.563
So they all match except dX which looks like a typo.
EDIT
After checking @whuber 's answer more carefully, I noticed that there is a slight issue with the methodology. The vector that is used is from the center of the Earth to the reference point, but that vector is not the appropriate "up" direction to base the calculation on. On the ellipsoid, the normal does not pass through the center of the Earth, except for the Equator or the Poles:

That difference amounts to around 0.2° at most.
To get a slightly better result, I would suggest translating and rotating the frame to topocentric (ENU), as mentioned in this Wikipedia article. A previous version of this article used the wrong calculations, but it has been fixed.
Also, you can take a look at this answer on Math that states the whole formulas for the calculation of the distance and angles.
Thanks for taking the time to share your knowledge! I hope to be able to do the same.
– RudolfSchmidt
2 hours ago
You're welcome. Also, you can have a look at my answer on Math StackExchange, which describes the full formulas as well.
– FSimardGIS
2 hours ago
add a comment |
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1 Answer
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1 Answer
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oldest
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The ECEF coordinates that @whuber provided appear to have been rounded to the nearest 10 meters. Looking at the revisions to the answer reveals that originally, the reference points that he used were (N39°, W75°, 4000m) and (N39°, W76°, 12000m), translating to ECEF coordinates:
(1285408.203, -4797208.722, 3994834.304)
(1202993.662, -4824944.042, 3999868.867)
The vectors are then:
dX = -82414.541
dY = -27735.320
dZ = 5034.563
So they all match except dX which looks like a typo.
EDIT
After checking @whuber 's answer more carefully, I noticed that there is a slight issue with the methodology. The vector that is used is from the center of the Earth to the reference point, but that vector is not the appropriate "up" direction to base the calculation on. On the ellipsoid, the normal does not pass through the center of the Earth, except for the Equator or the Poles:

That difference amounts to around 0.2° at most.
To get a slightly better result, I would suggest translating and rotating the frame to topocentric (ENU), as mentioned in this Wikipedia article. A previous version of this article used the wrong calculations, but it has been fixed.
Also, you can take a look at this answer on Math that states the whole formulas for the calculation of the distance and angles.
Thanks for taking the time to share your knowledge! I hope to be able to do the same.
– RudolfSchmidt
2 hours ago
You're welcome. Also, you can have a look at my answer on Math StackExchange, which describes the full formulas as well.
– FSimardGIS
2 hours ago
add a comment |
The ECEF coordinates that @whuber provided appear to have been rounded to the nearest 10 meters. Looking at the revisions to the answer reveals that originally, the reference points that he used were (N39°, W75°, 4000m) and (N39°, W76°, 12000m), translating to ECEF coordinates:
(1285408.203, -4797208.722, 3994834.304)
(1202993.662, -4824944.042, 3999868.867)
The vectors are then:
dX = -82414.541
dY = -27735.320
dZ = 5034.563
So they all match except dX which looks like a typo.
EDIT
After checking @whuber 's answer more carefully, I noticed that there is a slight issue with the methodology. The vector that is used is from the center of the Earth to the reference point, but that vector is not the appropriate "up" direction to base the calculation on. On the ellipsoid, the normal does not pass through the center of the Earth, except for the Equator or the Poles:

That difference amounts to around 0.2° at most.
To get a slightly better result, I would suggest translating and rotating the frame to topocentric (ENU), as mentioned in this Wikipedia article. A previous version of this article used the wrong calculations, but it has been fixed.
Also, you can take a look at this answer on Math that states the whole formulas for the calculation of the distance and angles.
Thanks for taking the time to share your knowledge! I hope to be able to do the same.
– RudolfSchmidt
2 hours ago
You're welcome. Also, you can have a look at my answer on Math StackExchange, which describes the full formulas as well.
– FSimardGIS
2 hours ago
add a comment |
The ECEF coordinates that @whuber provided appear to have been rounded to the nearest 10 meters. Looking at the revisions to the answer reveals that originally, the reference points that he used were (N39°, W75°, 4000m) and (N39°, W76°, 12000m), translating to ECEF coordinates:
(1285408.203, -4797208.722, 3994834.304)
(1202993.662, -4824944.042, 3999868.867)
The vectors are then:
dX = -82414.541
dY = -27735.320
dZ = 5034.563
So they all match except dX which looks like a typo.
EDIT
After checking @whuber 's answer more carefully, I noticed that there is a slight issue with the methodology. The vector that is used is from the center of the Earth to the reference point, but that vector is not the appropriate "up" direction to base the calculation on. On the ellipsoid, the normal does not pass through the center of the Earth, except for the Equator or the Poles:

That difference amounts to around 0.2° at most.
To get a slightly better result, I would suggest translating and rotating the frame to topocentric (ENU), as mentioned in this Wikipedia article. A previous version of this article used the wrong calculations, but it has been fixed.
Also, you can take a look at this answer on Math that states the whole formulas for the calculation of the distance and angles.
The ECEF coordinates that @whuber provided appear to have been rounded to the nearest 10 meters. Looking at the revisions to the answer reveals that originally, the reference points that he used were (N39°, W75°, 4000m) and (N39°, W76°, 12000m), translating to ECEF coordinates:
(1285408.203, -4797208.722, 3994834.304)
(1202993.662, -4824944.042, 3999868.867)
The vectors are then:
dX = -82414.541
dY = -27735.320
dZ = 5034.563
So they all match except dX which looks like a typo.
EDIT
After checking @whuber 's answer more carefully, I noticed that there is a slight issue with the methodology. The vector that is used is from the center of the Earth to the reference point, but that vector is not the appropriate "up" direction to base the calculation on. On the ellipsoid, the normal does not pass through the center of the Earth, except for the Equator or the Poles:

That difference amounts to around 0.2° at most.
To get a slightly better result, I would suggest translating and rotating the frame to topocentric (ENU), as mentioned in this Wikipedia article. A previous version of this article used the wrong calculations, but it has been fixed.
Also, you can take a look at this answer on Math that states the whole formulas for the calculation of the distance and angles.
edited 16 mins ago
answered 3 hours ago
FSimardGISFSimardGIS
1,38129
1,38129
Thanks for taking the time to share your knowledge! I hope to be able to do the same.
– RudolfSchmidt
2 hours ago
You're welcome. Also, you can have a look at my answer on Math StackExchange, which describes the full formulas as well.
– FSimardGIS
2 hours ago
add a comment |
Thanks for taking the time to share your knowledge! I hope to be able to do the same.
– RudolfSchmidt
2 hours ago
You're welcome. Also, you can have a look at my answer on Math StackExchange, which describes the full formulas as well.
– FSimardGIS
2 hours ago
Thanks for taking the time to share your knowledge! I hope to be able to do the same.
– RudolfSchmidt
2 hours ago
Thanks for taking the time to share your knowledge! I hope to be able to do the same.
– RudolfSchmidt
2 hours ago
You're welcome. Also, you can have a look at my answer on Math StackExchange, which describes the full formulas as well.
– FSimardGIS
2 hours ago
You're welcome. Also, you can have a look at my answer on Math StackExchange, which describes the full formulas as well.
– FSimardGIS
2 hours ago
add a comment |
RudolfSchmidt is a new contributor. Be nice, and check out our Code of Conduct.
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RudolfSchmidt is a new contributor. Be nice, and check out our Code of Conduct.
RudolfSchmidt is a new contributor. Be nice, and check out our Code of Conduct.
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