This is quick and dirty. Most of this was adapted from the following website:
http://www.movable-type.co.uk/scripts/latlong.html
I haven't tested this too much but it seemed to work after initial testing. It will return a list of lat, long coordinate pairs along a line at a specified interval. I wrote it in python since I don't know php very well.
from math import cos, sin, atan2, sqrt, radians, degrees, asin, modf
def getPathLength(lat1,lng1,lat2,lng2):
'''calculates the distance between two lat, long coordinate pairs'''
R = 6371000 # radius of earth in m
lat1rads = radians(lat1)
lat2rads = radians(lat2)
deltaLat = radians((lat2-lat1))
deltaLng = radians((lng2-lng1))
a = sin(deltaLat/2) * sin(deltaLat/2) + cos(lat1rads) *
cos(lat2rads) * sin(deltaLng/2) * sin(deltaLng/2)
c = 2 * atan2(sqrt(a), sqrt(1-a))
d = R * c
return d
def getDestinationLatLong(lat,lng,azimuth,distance):
'''returns the lat an long of destination point
given the start lat, long, aziuth, and distance'''
R = 6378.1 #Radius of the Earth in km
brng = radians(azimuth) #Bearing is degrees converted to radians.
d = distance/1000 #Distance m converted to km
lat1 = radians(lat) #Current dd lat point converted to radians
lon1 = radians(lng) #Current dd long point converted to radians
lat2 = asin( sin(lat1) * cos(d/R) + cos(lat1)* sin(d/R)* cos(brng))
lon2 = lon1 + atan2(sin(brng) * sin(d/R)* cos(lat1),
cos(d/R)- sin(lat1)* sin(lat2))
#convert back to degrees
lat2 = degrees(lat2)
lon2 = degrees(lon2)
return[lat2, lon2]
def main(interval,azimuth,lat1,lng1,lat2,lng2):
'''returns every coordinate pair inbetween two coordinate
pairs given the desired interval'''
coords = []
d = getPathLength(lat1,lng1,lat2,lng2)
remainder, dist = modf((d / interval))
counter = 1.0
coords.append([lat1,lng1])
for distance in xrange(1,int(dist)):
c = getDestinationLatLong(lat1,lng1,azimuth,counter)
counter += 1.0
coords.append(c)
counter +=1
coords.append([lat2,lng2])
return coords
if __name__ == "__main__":
#point interval in meters
interval = 1
#direction of line in degrees
azimuth = 279.91
#start point
lat1 = 41.97809
lng1 = -91.65573
#end point
lat2 = 41.97834
lng2 = -91.65767
coords = main(interval,azimuth,lat1,lng1,lat2,lng2)
print coords
I hope this helps. sorry if not. :)
Ok, I added the ability for it to calculate the azimuth for you. I tested with the same location you were using and it turned out perfectly.http://www.darrinward.com/lat-long/?id=705140
import math
def getPathLength(lat1,lng1,lat2,lng2):
'''calculates the distance between two lat, long coordinate pairs'''
R = 6371000 # radius of earth in m
lat1rads = math.radians(lat1)
lat2rads = math.radians(lat2)
deltaLat = math.radians((lat2-lat1))
deltaLng = math.radians((lng2-lng1))
a = math.sin(deltaLat/2) * math.sin(deltaLat/2) + math.cos(lat1rads) * math.cos(lat2rads) * math.sin(deltaLng/2) * math.sin(deltaLng/2)
c = 2 * math.atan2(math.sqrt(a), math.sqrt(1-a))
d = R * c
return d
def getDestinationLatLong(lat,lng,azimuth,distance):
'''returns the lat an long of destination point
given the start lat, long, aziuth, and distance'''
R = 6378.1 #Radius of the Earth in km
brng = math.radians(azimuth) #Bearing is degrees converted to radians.
d = distance/1000 #Distance m converted to km
lat1 = math.radians(lat) #Current dd lat point converted to radians
lon1 = math.radians(lng) #Current dd long point converted to radians
lat2 = math.asin(math.sin(lat1) * math.cos(d/R) + math.cos(lat1)* math.sin(d/R)* math.cos(brng))
lon2 = lon1 + math.atan2(math.sin(brng) * math.sin(d/R)* math.cos(lat1), math.cos(d/R)- math.sin(lat1)* math.sin(lat2))
#convert back to degrees
lat2 = math.degrees(lat2)
lon2 = math.degrees(lon2)
return[lat2, lon2]
def calculateBearing(lat1,lng1,lat2,lng2):
'''calculates the azimuth in degrees from start point to end point'''
startLat = math.radians(lat1)
startLong = math.radians(lng1)
endLat = math.radians(lat2)
endLong = math.radians(lng2)
dLong = endLong - startLong
dPhi = math.log(math.tan(endLat/2.0+math.pi/4.0)/math.tan(startLat/2.0+math.pi/4.0))
if abs(dLong) > math.pi:
if dLong > 0.0:
dLong = -(2.0 * math.pi - dLong)
else:
dLong = (2.0 * math.pi + dLong)
bearing = (math.degrees(math.atan2(dLong, dPhi)) + 360.0) % 360.0;
return bearing
def main(interval,azimuth,lat1,lng1,lat2,lng2):
'''returns every coordinate pair inbetween two coordinate
pairs given the desired interval'''
d = getPathLength(lat1,lng1,lat2,lng2)
remainder, dist = math.modf((d / interval))
counter = float(interval)
coords = []
coords.append([lat1,lng1])
for distance in xrange(0,int(dist)):
coord = getDestinationLatLong(lat1,lng1,azimuth,counter)
counter = counter + float(interval)
coords.append(coord)
coords.append([lat2,lng2])
return coords
if __name__ == "__main__":
#point interval in meters
interval = 1.0
#direction of line in degrees
#start point
lat1 = 43.97076
lng1 = 12.72543
#end point
lat2 = 43.969730
lng2 = 12.728294
azimuth = calculateBearing(lat1,lng1,lat2,lng2)
print azimuth
coords = main(interval,azimuth,lat1,lng1,lat2,lng2)
print coords
I think I have fixed the bugs from the original and it should work (including change the interval). Let me know how it turns out.
Best Answer
Assuming you know AB, AC, and BC, then the answer is yes. It requires trigonometry. The following assumes a flat plane and that points A anb B are not on the same line of latitude or longitude. To get the lat/long of C you need to figure out the ΔX and ΔY from a known point (let's use A).
You're going to need the value of the angle θ, which is 90-a-λ
Let ΔXB and ΔYB be the lat/long difference from A to B (both of which have known coordinates). From basic trigonometry we know that sin(λ) = ΔYB/AB. Therefore λ = arcsin(ΔYB/AB).
By the Law of Cosines cos(a) = (AB²+AC²-BC²)/(2*AB*AC), which means a = arccos((AB²+AC²-BC²)/(2×AB×AC))
Therefore θ = 90 - arccos((AB²+AC²-BC²)/(2×AB×AC)) - arcsin(ΔYB/AB)
With θ calculated, we're back to basic trigonometry again: sin(θ)= ΔX/AC and cos(θ) = ΔY/AC. Therefore ΔX = sin(θ)×AC and ΔY = cos(θ)×AC. Just add those values to the coordinates of A, and you have your answer.
TLDR version: