Develop Reference

What is the Ideal way to create a Triangle-Coordinates in Google Map API with two location points consisting of latitude & longitude

I have a scenario in my JavaScript application where I have the coordinates of a starting point which consist of Latitude and Longitude, similarly an ending point with it's respective coordinates.
Now I need to search for a location which basically provides with a set of coordinates and find if the recently entered location lies in between the previously mentioned starting point or ending point. However, the location does not need to match exactly within the points of the path of the start and end point. That is even if the location lies around the distance of say 2-3 km from the derived path, it should give a match.
I believe that we can create a triangle by providing three coordinates i.e start-point, end-point and a third point. So once the triangle is formed we can use google.maps.geometry.poly.containsLocation method to find if our searched location is present inside this triangle.
So my question is how can we get a third point to create a triangle which will provide locations that are nearby within 2-3 km from start to end point.
Else is there any alternate approach to deal with my use case?

Use googlemap's geometry library
This function specifically
isLocationOnEdge
Here's an example
0.001 tolerance value would be 100m
var isLocationNear = google.maps.geometry.poly.isLocationOnEdge(
yourLatLng,
new google.maps.Polyline({
path: [
new google.maps.LatLng(point1Lat, point1Long),
new google.maps.LatLng(point2Lat, point2Long),
]
}),
.00001);

Please note that the following answer assumes Plane Geometry where you should be using Spherical Geometry instead. Although this will be fine for less accurate purposes (like approximate distance, etc..)
It seems more of a geometry question than a programming question. A triangle like you mentioned won't be able to cover a straight line path in a uniform way. The situation can be thought of more like a distance between point and a line problem (Refer the given diagram
Here you can just find the distance between point C and line AB which you can check whether it's below 2.5 KMs (I've omitted all the units and conversions for simplicity)
Please note that you will also need to convert the distances from radian to appropriate units that you require using haversine formula, etc. which is not a trivial task (https://www.movable-type.co.uk/scripts/latlong.html).

Growing Google map polygon by certain meters

I have drawn a polygon on google map(with certain lats and lngs) now I would like to grow the outer of this polygon by 40 meters(input from users). How can I achieve this? Thank you
Thank you #geocodezip, I tried buffering polygon data(without using center point of polygon) and got this
distance between black and red lines seems not same.
Thanks for all replies && answers! I found a way to do that without buffering polygon/polylines data. Should have a better way to do this. But I would like to share what I have done in here:
I know two gps points: A and B so I could get bearing between those two points.
Based on this bearing I could get the bearing between A and the point(C) 40 meters away from A because line between A and C is 90 degrees against the line between A and B
With the new bearing and distance(40m) I could get the gps point of C (How to calculate the latlng of a point a certain distance away from another?)
finally do step from 1~3 again with point from B to A
Here is a screenshot of what I have done (I have not finish yet, will upload a final screenshot later.)

Draw lines parallel to the existing ones 40 meters (or whatever the desired buffer is) to either side.
Remove the ones that have either endpoint inside the polygon
Extend the new lines and find their intersections
The intersections define the new "bigger" polygon.

Finally got this problem solved, not perfect I know.XD
Implement lines intersection: from here ( [http://jsfiddle.net/justin_c_rounds/Gd2S2/])

How to do javascript online k-means clustering for many dimensions

I found many examples of javascript online k-means clustering, but all of the are for 2 dimensions.
If I have 56 dimensions (for example), how can I do the clustering?
Bonus question:
Could it be possible, having some new data, to predict some value looking the clusters (like, 76% of belonging to cluster x, so the value should be y)

k-means algorithm should be easy to port to any number of dimensions. It looks like this:
Randomly choose centers of clusters.
For each point check, what is the nearest cluster.
Compute new cluster center by computing avarage from all points.
Repeat until cluster centers don't change.
In 2d, you check the distance between (x1, x2) and (y1, y2) in 2. like this (x1-x2)^2 + (y1-y2)^2 (you don't need to use square root, if you are using distance only to compare it with another distance). In 56 dimensions, you just have 56 components.
In 2d, you compute cluster center by taking avarage of all points. Take the first dimension of all points and take the average avg1, take all the second dimensions avg2 up to 56 and your new cluster center will be (avg1, avg2, avg3 ... avg56).
What is not easy is that it is very expensive. Check out algorithms for dimensionality reduction (feature extraction) like PCA.
Also make sure, that all freatures are normalized. For example - they have ranges between (-100, 100).
If you need more information, check out Machine Learning course at coursera.
Week 8 is all about clustering and its traps.

If I have one lat/lng which I assume is at 0,0 then how do I calculate the x, y coordinates of another lat/lng pair?

I've seen many variations of this question asked but am having trouble relating their answers to my specific need.
I have several sets of 3 lat/lng coordinate pairs. The coordinates in any set are within a few km of eachother.
For each set I would like to convert the coordinates to x/y values so that I can plot them.
I would like to assign 1 of the coordinates to 0,0 and then compute the relative x/y values of the other two coordinates.
This site does what I want but unfortunately doesn't share the algorithm:
http://www.whoi.edu/marine/ndsf/cgi-bin/NDSFutility.cgi?form=0&from=LatLon&to=XY

First some definitions just to be clear
let a be latitude <-pi/2,+pi/2>
let b be longitude <0,+2*pi>
let re,rp be equator and pole radiuses of Earth
a0,b0, a1,b1, a2,b2 are your points in spherical coordinates
and x0,y0, x1,y1, x2,y2 are your wanted cartesian coordinates
convert coordinates to relative to (a0,b0)
Leta assume East is aligned to your X-axis and North pole is aligned to Y-axis
x0=0.0;
y0=0.0;
r1=re*cos(a1)+rp*sin(a1) // actual radius for point 1
r2=re*cos(a2)+rp*sin(a2) // actual radius for point 2
x1=x0+((b1-b0)*r1);
x2=x0+((b2-b0)*r2);
y1=y0+((a1-a0)*re); // here instead of re should be length of ellipse curve from 0 to a1-a0
y2=y0+((a2-a0)*re); // here instead of re should be length of ellipse curve from 0 to a2-a0
if re!=rp then the y1,y2 coordinates will be less accurate
to correct that just replace ((a1-a0)*re) with the propper formula
for ellipse arc->length computation (this one is for circle)
I am too lazy to compute that integral
anyway even this is good enough (earth eccentricity is not that bad)
you also can normalize the angles after substraction
while (a<-pi) a+=2.0*pi;`
while (a>+pi) a-=2.0*pi;`
just to be safe ...

Actually, that's not entirely true. The site does share the algorithm, just not in the way one would expect to.
See http://www.whoi.edu/marine/ndsf/utility/NDSFutility.js .
Hope that helps.

In a spherical condition, given 3 points and their respective distances to a 4th point, how do a find its geolocation? [duplicate]

This question already has answers here:
Closed 10 years ago.
Possible Duplicate:
Trilateration using 3 latitude and longitude points, and 3 distances
This is more of a math question than programming question. Basically, I have P1:(lat1, lon1), P2:(lat2, lon2), P3:(lat3, lon3) and D1, D2, D3, and a 4th unknown point Px:(latx, lonx); also P1, P2, P3 do not lie in the same path of Great Circle, and D1 is the distance between P1 and Px, and D2 is the distance betwwen P2 and Px, etc.
How do I figure out the coordinates of Px?
=edited based on reply=
Thanks a lot!
PS. If you are going to point to any API, I would like it to be in JavaScript.

You have to understand that there will be multiple points that satisfy the mathematical constraint here. Think clearly, if you have two points on a sphere (ignore the geodesic form for now), P1 and P2, and you have another point T1, at distance x from P1 and distance y from P2, then you will have another symmetric (mirrored) point T1' which will satisfy the same distance conditions, on the other side, so to speak.
Even worse: Consider the case of a sphere with a diameter D. Your P1 is at the North Pole, and your P2 is at the South pole. Do you see that the all points on the equator will satisfy your condition?
Apply this to your example: Consider P1 at the North Pole. Consider P2 at the South Pole. Consider distance to Px, i.e. D1 = D2 = (2.pie.r)/4. See the problem? All points on the equator satisfy this, not a single unique point. In fact, for this case, even if D1 != D2, then you have smaller concentric lines (concentric to the equator) whose points satisfy these constraints.
Too many Px's in your case, not one. To come to a singularity point on a spherical surface, the description constraints would be more specific.
Lastly, establishing correctness of the context is important. Should your algorithm support all points that meet the criteria? Or should your criteria be altered such that the algorithm evaluates to a singular point, always. Be careful.
Some links to help you:
Wikipedia: http://en.wikipedia.org/wiki/Spherical_coordinates
SO: Plotting a point on the edge of a sphere
Updates, based on your three point example:
Again, there can be multiple points satisfying your criteria. What if P1, P2, P3 lie on the same arc? See the diagram below. Even with three points, there is no guarantee that there will be a single fourth point satisfying the distance criteria. Even with n points, there is no such guarantee.
In mathematical language, for a set of n random points, and a set of distances from these individual points, the set of resulting points that satisfy the distance criteria MAY have more than one elements.
You may be fooled into thinking: Oh, this guy is always assuming points lying on the same arc. Well, you are not making a special algorithm, are you? Your algo will be a generalized solution, won't it?
You need to guarantee that the points are not on the same arc (in a set of n points, I think at least 1 point cannot be on the same arc).
For keeping source points to a bare minimum : You need to establish traingular relations between points, because then, using ONLY two points, the triangle relation will yield exactly one point.
What triangle? Visualize this: You have two points, and a third unknown point. All distance you mention are SPHERICAL, i.e. curved surface distances. Do you see that there are also flat distances between these points? Can you visualize, that there will be a plane passing through these points, slicing the sphere, right? I say this to emphasize that you do not need to worry about surface curvature (hence 3d steradian angles). You can see the underlying 2d triangle, whose unknown vertex will also be the third point on the sphere surface.
I know this maybe very hard for you to visualize, I'll try making a diagram for this. (Can't find any good online tools!).
Lastly, this will be of significant help: Please read carefully.
Taken from Wikipedia: http://en.wikipedia.org/wiki/Great-circle_distance
The great-circle or orthodromic distance is the shortest distance between any two points on the surface of a sphere measured along a path on the surface of the sphere (as opposed to going through the sphere's interior). Because spherical geometry is different from ordinary Euclidean geometry, the equations for distance take on a different form. The distance between two points in Euclidean space is the length of a straight line from one point to the other. On the sphere, however, there are no straight lines. In non-Euclidean geometry, straight lines are replaced with geodesics. Geodesics on the sphere are the great circles (circles on the sphere whose centers are coincident with the center of the sphere).
Some more updates:
Conversion between long, lats to Cartesian coordinates can be done by the Haversine formula. Google it. See here: Converting from longitude\latitude to Cartesian coordinates
and here: http://en.wikipedia.org/wiki/World_Geodetic_System