So for example, I have two pairs of coordinates: (103.82 W, 32.024 N) and (104.2 W, 32.587 N). I'm trying to figure out how to calculate the distance between these two points and then represent that distance in DMS form. When I use the distance formula, I obtain a decimal number that does not accurately represent my desired number when I convert it to DMS format. I'm supposed to use the distance formula, so I'm not sure what I'm missing here or what I could be doing wrong. I don't believe I need to use the haversine formula for this.
Any help would be greatly appreciated!
The Rosetta Code web site has working code examples in various computer languages (Java, JavaScript, Python, and many others) to compute the distance between two lat/lon coordinates using the Haversine formula.
You can verify your distance results using this online distance calculator for comparison.
Given a coordinate in decimal degrees you can convert that number into the DMS components or vice versa. The formula has been answered in a related answer and there are online converters to convert between DMS and decimal degrees.
Related
I am trying to figure out a way to convert the result of presto geo spatial function ST_DISTANCE to meters.
If I run the this example query:
SELECT ST_Distance(ST_Point(0.3476, 32.5825),ST_Point(0.0512, 32.4637))
The result I get from Presto is: 0.3193217812802629. The actual distance between these two places is 40,000m.
The presto documentation states that ST_DISTANCE: Returns the 2-dimensional cartesian minimum distance (based on spatial ref) between two geometries in projected units.
What I can understand about spatial ref is at links such as these:
http://webhelp.esri.com/arcgiSDEsktop/9.3/index.cfm?TopicName=Defining_a_spatial_reference
Which leads me to believe I need to figure you what spatial-ref Presto is using.
If I check the prest docs here:
https://github.com/prestodb/presto/blob/master/presto-geospatial/src/main/java/com/facebook/presto/geospatial/GeoFunctions.java
I can guess that is using the ESRI libraries so I assume the ESRI spatial ref? But that is where I get a bit lost as to where to proceed?
Thank you for your help..
I would recommend using Presto’s great_circle_distance() function instead of ST_Distance(). It will interpret your coordinates as WGS84 (aka EPSG:4326), and find the distance between them in kilometres by treating the shape of the earth as a sphere.
ST_Distance() would be appropriate if the coordinate system being used was already projected into a system that used metres or miles or some other unit, but there's no trivial way to do that in Presto.
From looking at the docs, it appears that presto supports a geometry type but not a geography type. That means it's not working with Latitude and Longitude, which is what I assume you're supplying as those point parameters. It's just an arbitrary 2D grid and so the resulting units are in whatever units you supplied as input.
The distance, in meters, between two points which are both approximately 32.5 meters "up" from the origin and about 0.5 meters "left" from the origin (how presto will have interpreted your points) is, indeed, 0.3193217812802629, the value that was returned to you.
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
I create a polygon where each x/y point is WGS84 format
lat/long values.
The polygons are good approximations to circles and sectors of
radius R (each circumference/arc point is a projected lat/long
value of distance R from a centre/apex coordinate - which I have
verified is correct by computing the Haversine distance between
the edge and reference points and getting a value of R back) .
I use GEOSSetSRID(4326) to indicate the coords are WGS84 format.
GEOSGetSRID() confirms the SRID is set.
Use of GEOSArea then gives a value not even remotely close to
the expected value.
I do not see what else I can programmatically do.
If I set the points in cartesian format, and then set the SRID to
4326, will GEOS implicitly convert the polygon points to WGS84 ??
Is the basic GEOS C API incapable of doing the above ??
Dos SRID have no meaning to the API at all ??
Any info/pointers to correct usage/solutions would be much appreciated.
TIA.
The distance that is given is something like degrees between the two points. In actuality, the GEOS API (at least the C++ interface) is units agnostic; the units it gives the distance in is based on whatever you passed in.
In general, multiplying the result you get by 111000 gives you a fairly accurate measurement in meters. For area, you have to do 111000^2.
I'm trying to figure out how to calculate a min/max lat/long bound on the specific given range of a gps coordinate.
for example: gps coord 37.42935699924869,-122.16962099075317 range .2 miles
I'm looking at the point + range + bearing in the http://www.movable-type.co.uk/scripts/latlong.html site but im not sure if this is exactly what i want.
This gives 4 unique lat/long pairs and I want/need a max/min lat and a max/min long.
Calculate the distance between the (constant) central point and the point you want to test. (This page should give you the distance (in meters)).
If (distance < 0.2) then ...
Well, given a point and a distance, you will get a circle.
You're looking for two points, which will essentially describe a square (two opposite corners). The two points you're looking for won't even be on the circle. I'm not exactly sure why you want this, but I don't think there is an answer to your question.
Perhaps you could tell us what you're trying to accomplish.
EDIT: Added image to illustrate. The orange line is the distance from the centre (e.g. 0.2 miles)
alt text http://img155.imageshack.us/img155/1315/diagramp.png
After your clarification, here is a less elegant answer that might give you what you want. Well, you want the inverse of a really complicated function. I'm afraid my math skills aren't up to the task, but it should be doable.
A less elegant solution is to find it by trial and error. Essentially, keep longitude the same and vary latitude. Using the right algorithm, you should be able to find one that is very close to the distance you want. This will give you a point on the circle (one of four that is also on the square).
Then keep latitude the same and vary longitude. This will give you a second point on the square (on the middle of one of the sides), from there you can find the 4 corners of the square.
This will slow, depending on how often you have to do it, that might or might not matter.