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I need to calculate distance from subscriber position to Position B in the image. I have the GPS co-ordinate of the subscriber and the "B" position. How can I calculate the distance?
Simple case: Express lat and long values in decimal form and use the standard geometry distance formula if subscriber is less than 100 miles from position B. distance = sqrt((lat1-lat2)^2 - (long1-long2)^2).
More general case: Use the haversine formulas using a great circle to calculate distances from points on a sphere for more accurate measurements if position B might be a continent or two away from the subscriber. Let's call the subscriber position A and say and say he is at lat[a], long[a] and the fixed point B is at lat[b], long[b]. Let r represent the radius of the earth (about 3961 miles).
distance = 2*r*arcsin(sqrt(sin^2((lat[b]-lat[a])/2) + cos(lat[a])*cos(lat[b])*sin^2((long[b]-long[a])/2)))
If you specify r in miles, your answer will come out in miles. If you use kilometers use 6373 for a good number for the earth's radius, and of course the answer will come out in kilometers.
Exact case: The haversine formula will not provide a perfect answer because the earth is not a perfect sphere. Even apart from the mountains and the canyons, the earth has a larger radius at the equator than it does at the poles. The radius at the equator is the equator is about 3963 miles, and at the poles it is about 3950 miles. So you really need to devise your own lookup table (or borrow one from google maps) if you are measuring distances halfway around the globe and you have to be exact.
The haversine formula will be accurate to less than half of a percentage point. In 1000 miles your answer will be accurate to within 5 miles.
Haversine formula: https://en.wikipedia.org/wiki/Haversine_formula
Radius of the earth: https://en.wikipedia.org/wiki/Earth_radius
Related
I'm trying to decide whether it makes cpu processing time sense to use the more complex Haversine formula instead of the faster Pythagorean's formula but while there seems to be a pretty unanimous answer on the lines of: "you can use Pythagora's formula for acceptable results on small distances but haversine is better", I can not find even a vague definition on what "small distances" mean.
This page, linked in the top answer to the very popular question Calculate distance between two latitude-longitude points? claims:
If performance is an issue and accuracy less important, for small distances Pythagoras’ theorem can be used on an equirectangular projection:*
Accuracy is somewhat complex: along meridians there are no errors, otherwise they depend on distance, bearing, and latitude, but are small enough for many purposes*
the asterisc even says "Anyone care to quantify them?"
But this answer claims that the error is about 0.1% at 1000km (but it doesn't cite any reference, just personal observations) and that for 4km (even assuming the % doesn't shrink due to way smaller distance) it would mean under 4m of error which for public acces GPS is around the open-space best gps accuracy.
Now, I don't know what the average Joe thinks of when they say "small distances" but for me, 4km is definitely not a small distance (- I'm thinking more of tens of meters), so I would be grateful if someone can link or calculate a table of errors just like the one in this answer of Measuring accuracy of latitude and longitude? but I assume the errors would be higher near the poles so maybe choose 3 representative lattitudes (5*, 45* and 85*?) and calculate the error with respect to the decimal degree place.
Of course, I would also be happy with an answer that gives an exact meaning to "small distances".
Yes ... at 10 meters and up to 1km meters you're going to be very accurate using plain old Pythagoras Theorem. It's really ridiculous nobody talks about this, especially considering how much computational power you save.
Proof:
Take the top of the earth, since it will be a worst case, the top 90 miles longitude, so that it's a circle with the longitudinal lines intersecting in the middle.
Note above that as you zoom in to an area as small as 1km, just 50 miles from the poles, what originally looked like a trapezoid with curved top and bottom borders, essentially looks like a nearly perfect rectangle. In other words we can assume rectilinearity at 1km, and especially at a mere 10M.
Now, its true of course that the longitude degrees are much shorter near the poles than at the equator. For example any slack-jawed yokel can see that the rectangles made by the latitude and longitude lines grow taller, the aspect ratio increasing, as you get closer to the poles. In fact the relationship of the longitude distance is simply what it would be at the equator multiplied by the cosine of the latitude of anywhere along the path. ie. in the image above where "L" (longitude distance) and "l" (latitude distance) are both the same degrees it is:
LATcm = Latitude at *any* point along the path (because it's tiny compared to the earth)
L = l * cos(LATcm)
Thus, we can for 1km or less (even near the poles) calculate the distance very accurately using Pythagoras Theorem like so:
Where: latitude1, longitude1 = polar coordinates of the start point
and: latitude2, longitude2 = polar coordinates of the end point
distance = sqrt((latitude2-latitude1)^2 + ((longitude2-longitude1)*cos(latitude1))^2) * 111,139*60
Where 111,139*60 (above) is the number of meters within one degree at the equator,
because we have to convert the result from equator degrees to meters.
A neat thing about this is that GPS systems usually take measurements at about 10m or less, which means you can get very accurate over very large distances by summing up the results from this equation. As accurate as Haversine formula. The super-tiny errors don't magnify as you sum up the total because they are a percentage that remains the same as they are added up.
Reality is however that the Haversine formula (which is very accurate) isn't difficult, but relatively speaking Haversine will consume your processor at least 3 times more, and up to 31x more computational intensive according to this guy: https://blog.mapbox.com/fast-geodesic-approximations-with-cheap-ruler-106f229ad016.
For me this formula did come useful to me when I was using a system (Google sheets) that couldn't give me the significant digits that are necessary to do the haversine formula.
I am Using Haversine Formula to find the great circle distance in my work
But I want to know The most Exact formula for small distance like 10 meters only
Continue using the haversine formula. The [spherical] law of cosines formula is known to be inaccurate at short distances. See https://en.wikipedia.org/wiki/Great-circle_distance ... also this SO question: MySQL WordPress Query Returning a Distance of Zero for Some Records
There are a couple of alternative options.
One is to convert points from geodetic to Cartesian system of coordinates and then use Euclidean distance in space. Relative error due to curvature is about 10^-9 for distances below 1 km and 10^-3 for distances below 1000 km.
Another option is to project locations on plane that is tangent to surface of the Earth and then calculate Euclidean distance on the plane. But this solution will not work near poles and additional care should be taken at 180th meridian. A careful implementation for ellipsoid datum can use just one computation if sine and one computation of square root aside of arithmetic operations, and can potentially be faster than Haversine formula for spherical datum, at the same time being much more accurate.
I have GPS coordinates provided as degrees latitude, longitude and would like to offset them by a distance and an angle.E.g.: What are the new coordinates if I offset 45.12345, 7.34567 by 22km along bearing 104 degrees ?Thanks
For most applications one of these two formulas are sufficient:
"Lat/lon given radial and distance"
The second one is slower, but makes less problems in special situations (see docu on that page).
Read the introduction on that page, and make sure that lat/lon are converted to radians before and back to degrees after having the result.
Make sure that your system uses atan2(y,x) (which is usually the case) and not atan2(x,y) which is the case in Excell.
The link in the previous answer no longer works, here is the link using the way back machine:
https://web.archive.org/web/20161209044600/http://williams.best.vwh.net/avform.htm
The formula is:
A point {lat,lon} is a distance d out on the tc radial from point 1 if:
lat=asin(sin(lat1)*cos(d)+cos(lat1)*sin(d)*cos(tc))
IF (cos(lat)=0)
lon=lon1 // endpoint a pole
ELSE
lon=mod(lon1-asin(sin(tc)*sin(d)/cos(lat))+pi,2*pi)-pi
ENDIF
This algorithm is limited to distances such that dlon <pi/2, i.e those that extend around less than one quarter of the circumference of the earth in longitude. A completely general, but more complicated algorithm is necessary if greater distances are allowed:
lat =asin(sin(lat1)*cos(d)+cos(lat1)*sin(d)*cos(tc))
dlon=atan2(sin(tc)*sin(d)*cos(lat1),cos(d)-sin(lat1)*sin(lat))
lon=mod( lon1-dlon +pi,2*pi )-pi
Greetings,
I have two coordinates:
(52.4412396, -6.563223)
and
(52.8912397, -6.683669)
The delta is:
(-0.4499999, 0.120446)
The distance moved is:
sqrt((-0.4499999)^2+(0.120446)^2)
=.465840261
How do I convert this to meters?!
I hope someone can help.
Many thanks in advance,
You have mistakenly done the sum of squares on spherical coordinates. Each difference has to be converted to its longitudinal and latitudinal distance before getting the hypotenuse. While latitude converts directly to distance, (each degree is equal to 60 nautical miles) the longitude will only do that at the equator) That means that you have to multiply the above by the cosine of the latitude. Then you can move on to a simple hypotenuse calculation before converting to meters.
as topic, the Coordinates value (Latitude and Longitude) is known , these Coordinates will compose as polygonal area , my question is how to calculate the area of the polygonal that is base the geography ?
thanks for your help .
First you would need to know whether the curvature of the surface would be significant. If it is a relatively small then you can get a good approximation by projecting the coordinates onto a plane.
Determine units of measure per degree of latitude (eg. meters per degree)
Determine units of meature per degree of longitude at a given latitude (the conversion factor varies as you go North or South)
Convert latitude and longitude pairs to (x,y) pairs in the plane
Use an algorithm to compute area of a polygon. See StackOverflow 451425 or Paul Bourke
If you are calculating a large area then spherical techniques must be used.
If I understand your question correctly - triangulation should help you. Basically you break the polygonal to triangles in such a way that they don't overlap and sum their areas.