How to calculate bearings between two points (in decimal degrees) using Excel
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Having problems formulating the bearing from two points in decimal degrees in excel. I'm using the following formula:
=MOD(ATAN2(SIN(LonB-LonA)*COS(LatB),COS(LatA*SIN(LatB)-SIN(LatA)*COS(LonB-LonA))),6.30318)
Example data
- Lat A (38.6318909290283)
- Lon A (-90.2828979492187)
- Lat B (38.5352759115441)
- Lon B(-89.8448181152343)
The answer I get is 5.024 degrees it should be 105.619 degrees
microsoft-excel
edited Jun 10 '16 at 1:31
Ron Rosenfeld
2,0992611
asked Jun 9 '16 at 22:14
JeffJeff
60129
add a comment |
Having problems formulating the bearing from two points in decimal degrees in excel. I'm using the following formula:
=MOD(ATAN2(SIN(LonB-LonA)*COS(LatB),COS(LatA*SIN(LatB)-SIN(LatA)*COS(LonB-LonA))),6.30318)
Example data
- Lat A (38.6318909290283)
- Lon A (-90.2828979492187)
- Lat B (38.5352759115441)
- Lon B(-89.8448181152343)
The answer I get is 5.024 degrees it should be 105.619 degrees
microsoft-excel
edited Jun 10 '16 at 1:31
Ron Rosenfeld
2,0992611
asked Jun 9 '16 at 22:14
JeffJeff
60129
1
I don't know if the formula is correct or not, but Excel'sSINandCOSfunction expect the angles in Radians, not Degrees
– Ron Rosenfeld
Jun 10 '16 at 1:52
add a comment |
Having problems formulating the bearing from two points in decimal degrees in excel. I'm using the following formula:
=MOD(ATAN2(SIN(LonB-LonA)*COS(LatB),COS(LatA*SIN(LatB)-SIN(LatA)*COS(LonB-LonA))),6.30318)
Example data
- Lat A (38.6318909290283)
- Lon A (-90.2828979492187)
- Lat B (38.5352759115441)
- Lon B(-89.8448181152343)
The answer I get is 5.024 degrees it should be 105.619 degrees
microsoft-excel
edited Jun 10 '16 at 1:31
Ron Rosenfeld
2,0992611
asked Jun 9 '16 at 22:14
JeffJeff
60129
Having problems formulating the bearing from two points in decimal degrees in excel. I'm using the following formula:
=MOD(ATAN2(SIN(LonB-LonA)*COS(LatB),COS(LatA*SIN(LatB)-SIN(LatA)*COS(LonB-LonA))),6.30318)
Example data
- Lat A (38.6318909290283)
- Lon A (-90.2828979492187)
- Lat B (38.5352759115441)
- Lon B(-89.8448181152343)
The answer I get is 5.024 degrees it should be 105.619 degrees
microsoft-excel
microsoft-excel
edited Jun 10 '16 at 1:31
Ron Rosenfeld
2,0992611
asked Jun 9 '16 at 22:14
JeffJeff
60129
edited Jun 10 '16 at 1:31
Ron Rosenfeld
2,0992611
asked Jun 9 '16 at 22:14
JeffJeff
60129
edited Jun 10 '16 at 1:31
Ron Rosenfeld
2,0992611
edited Jun 10 '16 at 1:31
Ron Rosenfeld
2,0992611
edited Jun 10 '16 at 1:31
Ron Rosenfeld
2,0992611
2,0992611
asked Jun 9 '16 at 22:14
JeffJeff
60129
asked Jun 9 '16 at 22:14
JeffJeff
60129
asked Jun 9 '16 at 22:14
JeffJeff
60129
60129
1
I don't know if the formula is correct or not, but Excel'sSINandCOSfunction expect the angles in Radians, not Degrees
– Ron Rosenfeld
Jun 10 '16 at 1:52
add a comment |
1
I don't know if the formula is correct or not, but Excel'sSINandCOSfunction expect the angles in Radians, not Degrees
– Ron Rosenfeld
Jun 10 '16 at 1:52
1
1
I don't know if the formula is correct or not, but Excel's
SIN and COS function expect the angles in Radians, not Degrees– Ron Rosenfeld
Jun 10 '16 at 1:52
I don't know if the formula is correct or not, but Excel's
SIN and COS function expect the angles in Radians, not Degrees– Ron Rosenfeld
Jun 10 '16 at 1:52
add a comment |
2 Answers
2
active
oldest
votes
Looks like you mistyped the formula when entering into excel.
Also, you need to convert degrees to radians, because excel takes input for trigonometric functions as radians. At the end, need to convert the result of ATAN2 back to degrees.
Solution:
=DEGREES(ATAN2(COS(RADIANS(latA))*SIN(RADIANS(latB))-SIN(RADIANS(latA))*COS(RADIANS(latB))*COS(RADIANS(lonB-lonA)),SIN(RADIANS(lonB-lonA))*COS(RADIANS(latB))))
Source for formula: http://www.movable-type.co.uk/scripts/latlong.html
answered Jun 10 '16 at 6:20
JollyMortJollyMort
349110
1
Jolly Thank you. I should have known that. Works and if I add =MOD(<expression>,360) I get the degrees from North.
– Jeff
Jun 10 '16 at 16:36
that's right, glad I could help
– JollyMort
Jun 10 '16 at 16:54
add a comment |
Calculating Azimuth referenced to North (0 degrees) between two x,y locations using UTM Easting and Northing coordinates in metres, I use the following formula:
Azimuth==IF(AND(x>=0,y>=0),90-MOD(360+DEGREES(ATAN2(x,y)),360),450-MOD(360+DEGREES(ATAN2(x,y)),360))
Cell name 'x' = distance (difference) between eastings in metres (target minus source)
Cell name 'y' = distance (difference) between northings in metres (target minus source)
edited Mar 8 at 4:22
Scott
16.2k113990
answered Mar 8 at 3:55
SFR56SFR56
1
add a comment |
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2 Answers
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2 Answers
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Looks like you mistyped the formula when entering into excel.
Also, you need to convert degrees to radians, because excel takes input for trigonometric functions as radians. At the end, need to convert the result of ATAN2 back to degrees.
Solution:
=DEGREES(ATAN2(COS(RADIANS(latA))*SIN(RADIANS(latB))-SIN(RADIANS(latA))*COS(RADIANS(latB))*COS(RADIANS(lonB-lonA)),SIN(RADIANS(lonB-lonA))*COS(RADIANS(latB))))
Source for formula: http://www.movable-type.co.uk/scripts/latlong.html
answered Jun 10 '16 at 6:20
JollyMortJollyMort
349110
1
Jolly Thank you. I should have known that. Works and if I add =MOD(<expression>,360) I get the degrees from North.
– Jeff
Jun 10 '16 at 16:36
that's right, glad I could help
– JollyMort
Jun 10 '16 at 16:54
add a comment |
Looks like you mistyped the formula when entering into excel.
Also, you need to convert degrees to radians, because excel takes input for trigonometric functions as radians. At the end, need to convert the result of ATAN2 back to degrees.
Solution:
=DEGREES(ATAN2(COS(RADIANS(latA))*SIN(RADIANS(latB))-SIN(RADIANS(latA))*COS(RADIANS(latB))*COS(RADIANS(lonB-lonA)),SIN(RADIANS(lonB-lonA))*COS(RADIANS(latB))))
Source for formula: http://www.movable-type.co.uk/scripts/latlong.html
answered Jun 10 '16 at 6:20
JollyMortJollyMort
349110
1
Jolly Thank you. I should have known that. Works and if I add =MOD(<expression>,360) I get the degrees from North.
– Jeff
Jun 10 '16 at 16:36
that's right, glad I could help
– JollyMort
Jun 10 '16 at 16:54
add a comment |
Looks like you mistyped the formula when entering into excel.
Also, you need to convert degrees to radians, because excel takes input for trigonometric functions as radians. At the end, need to convert the result of ATAN2 back to degrees.
Solution:
=DEGREES(ATAN2(COS(RADIANS(latA))*SIN(RADIANS(latB))-SIN(RADIANS(latA))*COS(RADIANS(latB))*COS(RADIANS(lonB-lonA)),SIN(RADIANS(lonB-lonA))*COS(RADIANS(latB))))
Source for formula: http://www.movable-type.co.uk/scripts/latlong.html
answered Jun 10 '16 at 6:20
JollyMortJollyMort
349110
Looks like you mistyped the formula when entering into excel.
Also, you need to convert degrees to radians, because excel takes input for trigonometric functions as radians. At the end, need to convert the result of ATAN2 back to degrees.
Solution:
=DEGREES(ATAN2(COS(RADIANS(latA))*SIN(RADIANS(latB))-SIN(RADIANS(latA))*COS(RADIANS(latB))*COS(RADIANS(lonB-lonA)),SIN(RADIANS(lonB-lonA))*COS(RADIANS(latB))))
Source for formula: http://www.movable-type.co.uk/scripts/latlong.html
answered Jun 10 '16 at 6:20
JollyMortJollyMort
349110
answered Jun 10 '16 at 6:20
JollyMortJollyMort
349110
answered Jun 10 '16 at 6:20
JollyMortJollyMort
349110
answered Jun 10 '16 at 6:20
JollyMortJollyMort
349110
349110
1
Jolly Thank you. I should have known that. Works and if I add =MOD(<expression>,360) I get the degrees from North.
– Jeff
Jun 10 '16 at 16:36
that's right, glad I could help
– JollyMort
Jun 10 '16 at 16:54
add a comment |
1
Jolly Thank you. I should have known that. Works and if I add =MOD(<expression>,360) I get the degrees from North.
– Jeff
Jun 10 '16 at 16:36
that's right, glad I could help
– JollyMort
Jun 10 '16 at 16:54
1
1
Jolly Thank you. I should have known that. Works and if I add =MOD(<expression>,360) I get the degrees from North.
– Jeff
Jun 10 '16 at 16:36
Jolly Thank you. I should have known that. Works and if I add =MOD(<expression>,360) I get the degrees from North.
– Jeff
Jun 10 '16 at 16:36
that's right, glad I could help
– JollyMort
Jun 10 '16 at 16:54
that's right, glad I could help
– JollyMort
Jun 10 '16 at 16:54
add a comment |
Calculating Azimuth referenced to North (0 degrees) between two x,y locations using UTM Easting and Northing coordinates in metres, I use the following formula:
Azimuth==IF(AND(x>=0,y>=0),90-MOD(360+DEGREES(ATAN2(x,y)),360),450-MOD(360+DEGREES(ATAN2(x,y)),360))
Cell name 'x' = distance (difference) between eastings in metres (target minus source)
Cell name 'y' = distance (difference) between northings in metres (target minus source)
edited Mar 8 at 4:22
Scott
16.2k113990
answered Mar 8 at 3:55
SFR56SFR56
1
add a comment |
Calculating Azimuth referenced to North (0 degrees) between two x,y locations using UTM Easting and Northing coordinates in metres, I use the following formula:
Azimuth==IF(AND(x>=0,y>=0),90-MOD(360+DEGREES(ATAN2(x,y)),360),450-MOD(360+DEGREES(ATAN2(x,y)),360))
Cell name 'x' = distance (difference) between eastings in metres (target minus source)
Cell name 'y' = distance (difference) between northings in metres (target minus source)
edited Mar 8 at 4:22
Scott
16.2k113990
answered Mar 8 at 3:55
SFR56SFR56
1
add a comment |
Calculating Azimuth referenced to North (0 degrees) between two x,y locations using UTM Easting and Northing coordinates in metres, I use the following formula:
Azimuth==IF(AND(x>=0,y>=0),90-MOD(360+DEGREES(ATAN2(x,y)),360),450-MOD(360+DEGREES(ATAN2(x,y)),360))
Cell name 'x' = distance (difference) between eastings in metres (target minus source)
Cell name 'y' = distance (difference) between northings in metres (target minus source)
edited Mar 8 at 4:22
Scott
16.2k113990
answered Mar 8 at 3:55
SFR56SFR56
1
Calculating Azimuth referenced to North (0 degrees) between two x,y locations using UTM Easting and Northing coordinates in metres, I use the following formula:
Azimuth==IF(AND(x>=0,y>=0),90-MOD(360+DEGREES(ATAN2(x,y)),360),450-MOD(360+DEGREES(ATAN2(x,y)),360))
Cell name 'x' = distance (difference) between eastings in metres (target minus source)
Cell name 'y' = distance (difference) between northings in metres (target minus source)
edited Mar 8 at 4:22
Scott
16.2k113990
answered Mar 8 at 3:55
SFR56SFR56
1
edited Mar 8 at 4:22
Scott
16.2k113990
edited Mar 8 at 4:22
Scott
16.2k113990
edited Mar 8 at 4:22
Scott
16.2k113990
16.2k113990
answered Mar 8 at 3:55
SFR56SFR56
1
answered Mar 8 at 3:55
SFR56SFR56
1
answered Mar 8 at 3:55
SFR56SFR56
1
1
add a comment |
add a comment |
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1
I don't know if the formula is correct or not, but Excel's
SINandCOSfunction expect the angles in Radians, not Degrees– Ron Rosenfeld
Jun 10 '16 at 1:52