Mysql polygon between multiple coordinates - javascript

I have 4 points on the map(latitude, longitude), there can be a more.
And I need to build a MySQL POLYGON with 10 Kilometers wide between these points.
As you see from the picture - I have blue line, and I need a polygon (green outline).
My first thought was to create two separate points for each given point.
For example:
Given 1st point(Bremen)
calculate 5km right and add a point,
calculate 5km left and add a point.
Logic is simple.
BUT the problem is - I don't know what to calculate(right,top,bottom,left), i need some kind of angle, but I am stuck here.
I just need an algorithm, no need for a full code example.
function in javascript i tried:
var meters = 10000 / 2;
var my_lat = 52.51978;
var my_long = 13.388211;
// number of km per degree = ~111km (111.32 in google maps, but range varies
/* between 110.567km at the equator and 111.699km at the poles) */
// 1km in degree = 1 / 111.32km = 0.0089
// 1m in degree = 0.0089 / 1000 = 0.0000089
var coef = meters * 0.0000089;
var new_lat = my_lat + coef;
// pi / 180 = 0.018
var new_long = my_long - coef / Math.cos(my_lat * 0.018);

Related

Draw a perpendicular line through middle of two long/lat points

Hello i'm currently trying to draw a line through two long/lat lines to create a triangle. So far i have manged to draw a line through but the line is not perpendicular and looks skewed. Here is my code:
startPosition = [-0.17640, 51.426700];
endPosition = [0.17640, 51.796700];
triangleSizeY = (endPosition [1] - startPosition[1]) / 6;
/*subtract
end from start to work out direction and also use this divided by 6 to
get distance*/
triangleSize *= -1;
triangleSizeX = (endPosition [0] - startPosition[0]) / 6;
/*subtract
end from start to work out direction and also use this divided by 6 to
get distance*/
triangleSize *= -1;
var cx = (startPosition[0] + endPosition[0]) / 2;
var cy = (startPosition[1] + endPosition[1]) / 2;
var dx = (endPosition[0] - startPosition[0]) / 2;
var dy = (endPosition[1] - startPosition[1]) / 2;
positions[0] = [midPoint[0] + triangleSizeX, midPoint[1] +
triangleSizeY];
positions[1] = [cx - dy, cy + dx];
positions[2] = [cx + dy, cy - dx];
This is what it looks like:
First, lat/lon are angular so you can not do linear type distances. The steps you need to take to solve this:
Compute the distance between the 2 lat/lon pairs you want a perpendicular line from.
Take half the distance computed from the above step to get you the midpoint range.
Calculate the bearing between the 2 lat/lon pairs. (see reference below on computing a bearing from 2 lat/lon's)
With the half distance and bearing, you can compute the lat/lon of the midpoint. This is called computing a lat/lon based on a range and bearing. (See the reference below.)
Now you can go perpendicular from the midpoint by adding/subtracting 90 degrees from the bearing in step 3. Decide on a range you want to compute the new lat/lon from a range/bearing like in step 4.
This site (https://www.movable-type.co.uk/scripts/latlong.html) has the calculations you need to do this. Also, since the distance is relatively small, you can use the Equirectangular approximation over Haversine for distance calculation.

How to find the nearest points to given coordinates

How do i identify if a point with a LAT, LONG coordinate format is near to other point.
Lets Say i want to find all the points near to:
-38.9086621 ,-68.082214
That are at less than 1km of distance. I already know how to find a coordinate in a given quadrant but not how to find a quadrant near a point.
You can use the haversine formular to calculate distances between two given coordinates. The calculated distance is the direct connection between the given coordinates (beeline).
JavaScript example (Source):
var R = 6371e3; // meters
var lat1Radians = lat1.toRadians();
var lat2Radians = lat2.toRadians();
var deltaLat = (lat2-lat1).toRadians();
var deltaLon = (lon2-lon1).toRadians();
var a = Math.sin(deltaLat/2) * Math.sin(deltaLat/2) +
Math.cos(lat1Radians) * Math.cos(lat2Radians) *
Math.sin(deltaLon/2) * Math.sin(deltaLon/2);
var c = 2 * Math.atan2(Math.sqrt(a), Math.sqrt(1-a));
var d = R * c;

Leaflet calculating meters per pixel at zoom level

I am trying to determine a way to calculate the number of meters represented by 1 pixel at a given zoom level and geo centerpoint in Leaflet. Could anyone direct me to the math involved or if there is a way to do this out of the box in leaflet? I am not finding much out there.
You can use the containerPointToLatLng conversion function of L.Map to get the latLngcoordinates for a given pixelcoordinate. If you take one of the first pixel, and one of the next, you can use the distanceTo utility method of L.LatLng to calculate the distance in meters between them. See the following code (assuming map is an instance of L.Map):
var centerLatLng = map.getCenter(); // get map center
var pointC = map.latLngToContainerPoint(centerLatLng); // convert to containerpoint (pixels)
var pointX = [pointC.x + 1, pointC.y]; // add one pixel to x
var pointY = [pointC.x, pointC.y + 1]; // add one pixel to y
// convert containerpoints to latlng's
var latLngC = map.containerPointToLatLng(pointC);
var latLngX = map.containerPointToLatLng(pointX);
var latLngY = map.containerPointToLatLng(pointY);
var distanceX = latLngC.distanceTo(latLngX); // calculate distance between c and x (latitude)
var distanceY = latLngC.distanceTo(latLngY); // calculate distance between c and y (longitude)
That should work, thanks to Jarek PiĆ³rkowski for pointing my mistake before the edit.
You can use this to work out the metres per pixel:
metresPerPixel = 40075016.686 * Math.abs(Math.cos(map.getCenter().lat * Math.PI/180)) / Math.pow(2, map.getZoom()+8);
Take a look at openstreetmap.org page on zoom levels. It gives this formula for calculating the meters per pixel:
The distance represented by one pixel (S) is given by
S=C*cos(y)/2^(z+8) where...
C is the (equatorial) circumference of the Earth
z is the zoom level
y is the latitude of where you're interested in the scale.
Correct me if I am wrong, IMHO, the number of meters per pixel = map height in meters / map height in pixels
function metresPerPixel() {
const southEastPoint = map.getBounds().getSouthEast();
const northEastPoint = map.getBounds().getNorthEast();
const mapHeightInMetres = southEastPoint.distanceTo(northEastPoint);
const mapHeightInPixels = map.getSize().y;
return mapHeightInMetres / mapHeightInPixels;
}

Compute a distance including elevation/terrain data for a KmlLinestring

The Google Earth Desktop Application shows the both the map length and ground length of a line.
In the Google Earth plugin I want to do a similar thing, that is I wish to determine the ground length of a tessellated KmlLineString taking the terrain into account.
Can I do this, and if so, how?
You can certainly get the length pretty easily if you use the earth-api-utility-library. Using that you can do.
var length = (new geo.Path(linestring)).distance();
Granted this method does not take the terrain into account - but there are a number of caveats you should be aware of before trying calculate distances using an elevation gradient.
Firstly any differences between topographic and direct distance are minimal in most cases. Indeed many quality GPS receivers simply don't take any changes in elevation into account when calculating distances.
Secondly ground altitude is one of the most unreliable pieces data. Using a gradient based on elevation to determine distance will often produce greater inaccuracy in distance measurements than using a simple 'as the crow flies' measure.
Bearing that in mind, if you still wanted to do it then one way would be something like the following.
Sample the line string at certain points (say every 10 meters).
Get the ground altitude at each point.
Convert each point to Cartesian coordinates
Calculate the angular distances between each Cartesian point in sequence.
You can improve your precision of this kind of method in two ways, either by increasing the sampling rate (say every meter) or by applying a smoothing procedure to the results.
For a rougher version, you could just loop over the coordinates in the the KmlLinestring itself, rather than resampling at some set distance. You would use the latitude, longitude of the coordinate to get the ground altitude at each point. Then you would construct a Cartesian coordinate from this data (latitude, longitude, elevation => X,Y,Z) and work out the angular distance between it and the next point...and so on.
something like the following idea should work - although it is written here and untested!
var EARTH_RADIUS = 6378135; // approximate in meters
var degreestoRadians = function(degrees) {
return degrees * Math.PI / 180;
}
var arcLength = function(point1 , point2) {
var length = Math.sqrt(Math.pow(point1.X-point2.X, 2)
+ Math.pow(point1.Y-point2.Y, 2)
+ Math.pow(point1.Z-point2.Z, 2));
var angle = 2 * Math.asin(length/2/EARTH_RADIUS);
return EARTH_RADIUS * angle;
}
var sphericalToCartesian = function(latitude, longitude, altitude) {
var phi = degreestoRadians(latitude);
var theta = degreestoRadians(longitude);
var rho = EARTH_RADIUS + altitude;
return {
X: Math.cos(phi) * Math.cos(theta) * rho,
Y: Math.cos(phi) * Math.sin(theta) * rho,
Z: Math.sin(phi) * rho
}
}
var topographicDistance = function(linestring) {
var coordinates = linestring.getCoordinates(); //KmlCoordArray
var last = null;
var distance = 0;
for(var i = 0; i < coordinates.length; i++) {
var coord = coordinates.get(i); //KmlCoord
var lat = coord.getLatitude();
var lng = coord.getLongitude();
var alt = ge.getGlobe().getGroundAltitude(lat, lng);
var latest = sphericalToCartesian(lat, lng, alt);
if(last != null) {
distance += arcLength(last, latest);
}
last = latest;
}
return distance;
}
You would use it like so...
var distance = topographicDistance(yourLinestring);

KinectJS: Algorithm required to determine new X,Y coords after image resize

BACKGROUND:
The app allows users to upload a photo of themselves and then place a pair of glasses over their face to see what it looks like. For the most part, it is working fine. After the user selects the location of the 2 pupils, I auto zoom the image based on the ratio between the distance of the pupils and then already known distance between the center points of the glasses. All is working fine there, but now I need to automatically place the glasses image over the eyes.
I am using KinectJS, but the problem is not with regards to that library or javascript.. it is more of an algorithm requirement
WHAT I HAVE TO WORK WITH:
Distance between pupils (eyes)
Distance between pupils (glasses)
Glasses width
Glasses height
Zoom ratio
SOME CODE:
//.. code before here just zooms the image, etc..
//problem is here (this is wrong, but I need to know what is the right way to calculate this)
var newLeftEyeX = self.leftEyePosition.x * ratio;
var newLeftEyeY = self.leftEyePosition.y * ratio;
//create a blue dot for testing (remove later)
var newEyePosition = new Kinetic.Circle({
radius: 3,
fill: "blue",
stroke: "blue",
strokeWidth: 0,
x: newLeftEyeX,
y: newLeftEyeY
});
self.pointsLayer.add(newEyePosition);
var glassesWidth = glassesImage.getWidth();
var glassesHeight = glassesImage.getHeight();
// this code below works perfect, as I can see the glasses center over the blue dot created above
newGlassesPosition.x = newLeftEyeX - (glassesWidth / 4);
newGlassesPosition.y = newLeftEyeY - (glassesHeight / 2);
NEEDED
A math genius to give me the algorithm to determine where the new left eye position should be AFTER the image has been resized
UPDATE
After researching this for the past 6 hours or so, I think I need to do some sort of "translate transform", but the examples I see only allow setting this by x and y amounts.. whereas I will only know the scale of the underlying image. Here's the example I found (which cannot help me):
http://tutorials.jenkov.com/html5-canvas/transformation.html
and here is something which looks interesting, but it is for Silverlight:
Get element position after transform
Is there perhaps some way to do the same in Html5 and/or KinectJS? Or perhaps I am going down the wrong road here... any ideas people?
UPDATE 2
I tried this:
// if zoomFactor > 1, then picture got bigger, so...
if (zoomFactor > 1) {
// if x = 10 (for example) and if zoomFactor = 2, that means new x should be 5
// current x / zoomFactor => 10 / 2 = 5
newLeftEyeX = self.leftEyePosition.x / zoomFactor;
// same for y
newLeftEyeY = self.leftEyePosition.y / zoomFactor;
}
else {
// else picture got smaller, so...
// if x = 10 (for example) and if zoomFactor = 0.5, that means new x should be 20
// current x * (1 / zoomFactor) => 10 * (1 / 0.5) = 10 * 2 = 20
newLeftEyeX = self.leftEyePosition.x * (1 / zoomFactor);
// same for y
newLeftEyeY = self.leftEyePosition.y * (1 / zoomFactor);
}
that didn't work, so then I tried an implementation of Rody Oldenhuis' suggestion (thanks Rody):
var xFromCenter = self.leftEyePosition.x - self.xCenter;
var yFromCenter = self.leftEyePosition.y - self.yCenter;
var angle = Math.atan2(yFromCenter, xFromCenter);
var length = Math.hypotenuse(xFromCenter, yFromCenter);
var xNew = zoomFactor * length * Math.cos(angle);
var yNew = zoomFactor * length * Math.sin(angle);
newLeftEyeX = xNew + self.xCenter;
newLeftEyeY = yNew + self.yCenter;
However, that is still not working as expected. So, I am not sure what the issue is currently. If anyone has worked with KinectJS before and has an idea of what the issue may be, please let me know.
UPDATE 3
I checked Rody's calculations on paper and they seem fine, so there is obviously something else here messing things up.. I got the coordinates of the left pupil at zoom factors 1 and 2. With those coordinates, maybe someone can figure out what the issue is:
Zoom Factor 1: x = 239, y = 209
Zoom Factor 2: x = 201, y = 133
OK, since it's an algorithmic question, I'm going to keep this generic and only write pseudo code.
I f I understand you correctly, What you want is the following:
Transform all coordinates such that the origin of your coordinate system is at the zoom center (usually, central pixel)
Compute the angle a line drawn from this new origin to a point of interest makes with the positive x-axis. Compute also the length of this line.
The new x and y coordinates after zooming are defined by elongating this line, such that the new line is the zoom factor times the length of the original line.
Transform the newly found x and y coordinates back to a coordinate system that makes sense to the computer (e.g., top left pixel = 0,0)
Repeat for all points of interest.
In pseudo-code (with formulas):
x_center = image_width/2
y_center = image_height/2
x_from_zoom_center = x_from_topleft - x_center
y_from_zoom_center = y_from_topleft - y_center
angle = atan2(y_from_zoom_center, x_from_zoom_center)
length = hypot(x_from_zoom_center, y_from_zoom_center)
x_new = zoom_factor * length * cos(angle)
y_new = zoom_factor * length * sin(angle)
x_new_topleft = x_new + x_center
y_new_topleft = y_new + y_center
Note that this assumes the number of pixels used for length and width stays the same after zooming. Note also that some rounding should take place (keep everything double precision, and only round to int after all calculations)
In the code above, atan2 is the four-quadrant arctangent, available in most programming languages, and hypot is simply sqrt(x*x + y*y), but then computed more carefully (e.g., to avoid overflow etc.), also available in most programing languages.
Is this indeed what you were after?

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