I am trying to code with JavaScript a function to calculate the surface area of a sphere. The radius using a NumericUpDown.
I have tried to use what I did in a different assignment and it is not working like the other did. I am new at this and have not had any help.
function calcSurface() {
var radius = document.getElementByID("radius").value;
radius = math.abs(radius);
Surface = 4*Math.PI*Math.pow(radius);
}
Math.abs() needs a capital M, and getElementById() needs a lower-case d. Math.pow takes two arguments, the base and the exponent, while you're only supplying one: the radius (which is the base).
Try this:
function calcSurface() {
var radius = document.getElementById("radius").value;
radius = Math.abs(radius);
var surface= 4 * Math.PI * Math.pow(radius, 2);
}
You can use this function
function calculateSurface() {
let radius = document.getElementById("radius").value;
radius = Math.abs(radius);
return 4 * Math.PI * Math.pow(radius, 2);
}
And if you have some div in your html like
<div id='result'></div>
You can do this to get the result in your div
document.getElementById('result').innerText = calculateSurface();
I'm trying to use sin on a variable and I saw you can convert it by using / (Math.PI / 180) but that seems contradictory if you want to convert it to degrees.. How do I properly convert and use sin in degree form? (On an iPhone calculator, for example it returns ~.707 from an input of 45, while this returns ~.806).
function click25() {
if (vi === 0) {
reactant = Math.sin(reactant / (Math.PI / 180))
}
}
You need to multiply the value in degree by (pi/180) to convert into the equivalent value in radians
var reactant = 45;
var vi = 0;
function click25() {
if (vi === 0) {
reactant = Math.sin(reactant * (Math.PI / 180))
}
console.log(reactant);
}
click25();
I wish to create the compass / arrow exactly like the one we see in AroundMe Mobile App that point exactly to a pin to the map accordingly with my mobile position and update the arrow when I move the phone.
I'm getting crazy to understand exactly how to do that and I can not find any guide or tutorial that explain a bit it.
What I found online is a bearing function and I created a directive around it:
app.directive('arrow', function () {
function bearing(lat1, lng1, lat2, lng2) {
var dLon = (lng2 - lng1);
var y = Math.sin(dLon) * Math.cos(lat2);
var x = Math.cos(lat1) * Math.sin(lat2) - Math.sin(lat1) * Math.cos(lat2) * Math.cos(dLon);
var rad = Math.atan2(y, x);
var brng = toDeg(rad);
return (brng + 360) % 360;
}
function toRad(deg) {
return deg * Math.PI / 180;
}
function toDeg(rad) {
return rad * 180 / Math.PI;
}
return {
restrict: 'E',
link: function (scope, element, attrs) {
var arrowAngle = bearing(scope.user.position.lat, scope.user.position.lng, attrs.lat, attrs.lng);
element.parent().css('transform', 'rotate(' + arrowAngle + 'deg)');
}
};
});
It seems to update the arrow in a direction but unfortunately it is not the right direction because it is not calculated using also the mobile magneticHeading position.
So I added the ngCordova plugin for Device Orientation to get the magneticHeading and now I don't know exactly how to use it and where in the bearing function.
$cordovaDeviceOrientation.getCurrentHeading().then(function(result) {
var magneticHeading = result.magneticHeading;
var arrowAngle = bearing(scope.user.position.lat, scope.user.position.lng, attrs.lat, attrs.lng, magneticHeading);
element.parent().css('transform', 'rotate(' + arrowAngle + 'deg)');
});
I tried to add it in the return statement:
return (brng - heading) % 360;
or:
return (heading - ((brng + 360) % 360));
Implementing this code with a watcher I see the arrow moving but not in the exact position... For example from my position and the pin the arrow should point to N and it is pointing to E.
Always looking online I can not find any tutorial / question to find the bearing between a lat/lng point and a magnetingHeading.
Maybe I'm close to the solution but I can not go ahead alone.
I also tried to search for a mathematical formulas but even there is a huge pain to understand and implement it.
I hope you can help.
It's hard to give a plain answer to this question because a lot depends on the actual graphical representation. For instance, in what direction do you point where rotate(0deg).
I can explain the formula you've found, which might help you to clear the issue yourself. The hard part is the following:
var dLon = (lng2 - lng1);
var y = Math.sin(dLon) * Math.cos(lat2);
var x = Math.cos(lat1) * Math.sin(lat2) - Math.sin(lat1) * Math.cos(lat2) * Math.cos(dLon);
var rad = Math.atan2(y, x);
What you see here is Haversines formula (https://en.wikipedia.org/wiki/Haversine_formula). Normally one could suffice with two sets of coordinates and calculate the angle between them. When working with latitude and longitude this will not work because the earth is not a flat surface. The first three lines are Haversine and the result (x and y) are coordinates on the unit circle (https://en.wikipedia.org/wiki/Unit_circle)
The next step is to calculate the angle from the point on this unit circle to the center. We can use the arctangant for this. Javascript has a helper function atan2 which simplifies this process. The result is simple the angle of your point towards the center of the circle. In other words, your position towards your point of interest. The result is in radians and needs to be converted to degrees. (https://en.wikipedia.org/wiki/Atan2)
A simplified version with a flat coordinate system would look like this:
var deltaX = poi.x - you.x;
var deltaY = you.y - poi.y;
var rotation = toDeg(Math.atan2(deltaX, deltaY));
bearingElement.css('transform', 'rotate(' + rotation + 'deg)');
Where poi is the Point of Interest and you is your position.
To compensate for your own rotation, you need to substract your own rotation. In the above sample the result would become:
var deltaX = poi.x - you.x;
var deltaY = you.y - poi.y;
var rotation = toDeg(Math.atan2(deltaX, deltaY));
rotation -= you.rotation;
bearingElement.css('transform', 'rotate(' + rotation + 'deg)');
I've made a Plunckr in which you can see a simple flat coordinate system. You can move and rotate you and the point of interest. The arrow inside 'you' will always point towards the poi, even if you rotate 'you'. This is because we compensate for our own rotation.
https://plnkr.co/edit/OJBGWsdcWp3nAkPk4lpC?p=preview
Note in the Plunckr that the 'zero'-position is always to the north. Check your app to see your 'zero'-position. Adjust it accordingly and your script will work.
Hope this helps :-)
I'm working on some coordinates function to my canvas in HTML5, and I want to make a function which can move an object by degrees.
My dream is to make a function which works like this:
box.x=10;
box.y=10;
// now the box has the coordinates (10,10)
moveTheBoxWithThisAmountOfDistance=10;
degreesToMoveTheBox=90;
box.moveByDegrees(moveTheBoxWithThisAmountOfDistance,degreesToMoveTheBox);
// now the box.x would be 20 and box.y wouldn't be changed because
// we only move it to the right (directional 90 degrees)
I hope this makes any sense!
So my question is:
How does the mathematical expression look like when I have to turn a degree into to coordinates?
You use sin and cos to convert an angle and a distance into coordinates:
function moveByDegrees(distance, angle) {
var rad = angle * Math.pi / 180;
this.x += Math.cos(rad) * distance;
this.y += Math.sin(rad) * distance;
}
That's it: http://en.wikipedia.org/wiki/Rotation_matrix , all the math is described :)
Also beware that if you need multiple sequential rotations (i.e. you do continuous animation), it's better to recompute x' and y' from initial ones and not just previous. Not doing so will result in rounding errors accumulation, and after some thousand rotations the result will become too rough.
I have this image. It's a map of the UK (not including Southern Ireland):
I have successfully managed to get a latitude and longitude and plot it onto this map by taking the leftmost longitude and rightmost longitude of the UK and using them to work out where to put the point on the map.
This is the code (for use in Processing.js but could be used as js or anything):
// Size of the map
int width = 538;
int height = 811;
// X and Y boundaries
float westLong = -8.166667;
float eastLong = 1.762833;
float northLat = 58.666667;
float southLat = 49.95;
void drawPoint(float latitude, float longitude){
fill(#000000);
x = width * ((westLong-longitude)/(westLong-eastLong));
y = (height * ((northLat-latitude)/(northLat-southLat)));
console.log(x + ", " + y);
ellipseMode(RADIUS);
ellipse(x, y, 2, 2);
}
However, I haven't been able to implement a Mercator projection on these values. The plots are reasonably accurate but they are not good enough and this projection would solve it.
I can't figure out how to do it. All the examples I find are explaining how to do it for the whole world. This is a good resource of examples explaining how to implement the projection but I haven't been able to get it to work.
Another resource is the Extreme points of the United Kingdom where I got the latitude and longitude values of the bounding box around the UK. They are also here:
northLat = 58.666667;
northLong = -3.366667;
eastLat = 52.481167;
eastLong = 1.762833;
southLat = 49.95;
southLong = -5.2;
westLat = 54.45;
westLong = -8.166667;
If anyone could help me with this, I would greatly appreciate it!
Thanks
I wrote a function which does exactly what you were looking for. I know it's a bit late, but maybe there are some other people interested in.
You need a map which is a mercator projection and you need to know the lat / lon positions of your map.
You get great customized mercator maps with perfect matching lat / lon positions from TileMill which is a free software from MapBox!
I'm using this script and tested it with some google earth positions. It worked perfect on a pixel level. Actually I didnt test this on different or larger maps. I hope it helps you!
Raphael ;)
<?php
$mapWidth = 1500;
$mapHeight = 1577;
$mapLonLeft = 9.8;
$mapLonRight = 10.2;
$mapLonDelta = $mapLonRight - $mapLonLeft;
$mapLatBottom = 53.45;
$mapLatBottomDegree = $mapLatBottom * M_PI / 180;
function convertGeoToPixel($lat, $lon)
{
global $mapWidth, $mapHeight, $mapLonLeft, $mapLonDelta, $mapLatBottom, $mapLatBottomDegree;
$x = ($lon - $mapLonLeft) * ($mapWidth / $mapLonDelta);
$lat = $lat * M_PI / 180;
$worldMapWidth = (($mapWidth / $mapLonDelta) * 360) / (2 * M_PI);
$mapOffsetY = ($worldMapWidth / 2 * log((1 + sin($mapLatBottomDegree)) / (1 - sin($mapLatBottomDegree))));
$y = $mapHeight - (($worldMapWidth / 2 * log((1 + sin($lat)) / (1 - sin($lat)))) - $mapOffsetY);
return array($x, $y);
}
$position = convertGeoToPixel(53.7, 9.95);
echo "x: ".$position[0]." / ".$position[1];
?>
Here is the image I created with TileMill and which I used in this example:
In addition to what Raphael Wichmann has posted (Thanks, by the way!),
here is the reverse function, in actionscript :
function convertPixelToGeo(tx:Number, ty:Number):Point
{
/* called worldMapWidth in Raphael's Code, but I think that's the radius since it's the map width or circumference divided by 2*PI */
var worldMapRadius:Number = mapWidth / mapLonDelta * 360/(2 * Math.PI);
var mapOffsetY:Number = ( worldMapRadius / 2 * Math.log( (1 + Math.sin(mapLatBottomRadian) ) / (1 - Math.sin(mapLatBottomRadian)) ));
var equatorY:Number = mapHeight + mapOffsetY;
var a:Number = (equatorY-ty)/worldMapRadius;
var lat:Number = 180/Math.PI * (2 * Math.atan(Math.exp(a)) - Math.PI/2);
var long:Number = mapLonLeft+tx/mapWidth*mapLonDelta;
return new Point(lat,long);
}
Here's another Javascript implementation. This is a simplification of #Rob Willet's solution above. Instead of requiring computed values as parameters to the function, it only requires essential values and computes everything from them:
function convertGeoToPixel(latitude, longitude,
mapWidth, // in pixels
mapHeight, // in pixels
mapLngLeft, // in degrees. the longitude of the left side of the map (i.e. the longitude of whatever is depicted on the left-most part of the map image)
mapLngRight, // in degrees. the longitude of the right side of the map
mapLatBottom) // in degrees. the latitude of the bottom of the map
{
const mapLatBottomRad = mapLatBottom * Math.PI / 180
const latitudeRad = latitude * Math.PI / 180
const mapLngDelta = (mapLngRight - mapLngLeft)
const worldMapWidth = ((mapWidth / mapLngDelta) * 360) / (2 * Math.PI)
const mapOffsetY = (worldMapWidth / 2 * Math.log((1 + Math.sin(mapLatBottomRad)) / (1 - Math.sin(mapLatBottomRad))))
const x = (longitude - mapLngLeft) * (mapWidth / mapLngDelta)
const y = mapHeight - ((worldMapWidth / 2 * Math.log((1 + Math.sin(latitudeRad)) / (1 - Math.sin(latitudeRad)))) - mapOffsetY)
return {x, y} // the pixel x,y value of this point on the map image
}
I've converted the PHP code provided by Raphael to JavaScript and can confirm it worked and this code works myself. All credit to Raphael.
/*
var mapWidth = 1500;
var mapHeight = 1577;
var mapLonLeft = 9.8;
var mapLonRight = 10.2;
var mapLonDelta = mapLonRight - mapLonLeft;
var mapLatBottom = 53.45;
var mapLatBottomDegree = mapLatBottom * Math.PI / 180;
*/
function convertGeoToPixel(latitude, longitude ,
mapWidth , // in pixels
mapHeight , // in pixels
mapLonLeft , // in degrees
mapLonDelta , // in degrees (mapLonRight - mapLonLeft);
mapLatBottom , // in degrees
mapLatBottomDegree) // in Radians
{
var x = (longitude - mapLonLeft) * (mapWidth / mapLonDelta);
latitude = latitude * Math.PI / 180;
var worldMapWidth = ((mapWidth / mapLonDelta) * 360) / (2 * Math.PI);
var mapOffsetY = (worldMapWidth / 2 * Math.log((1 + Math.sin(mapLatBottomDegree)) / (1 - Math.sin(mapLatBottomDegree))));
var y = mapHeight - ((worldMapWidth / 2 * Math.log((1 + Math.sin(latitude)) / (1 - Math.sin(latitude)))) - mapOffsetY);
return { "x": x , "y": y};
}
I think it's worthwhile to keep in mind that not all flat maps are Mercator projections. Without knowing more about that map in particular, it's hard to be sure. You may find that most maps of a small area of the world are more likely to be a conical type projection, where the area of interest on the map is "flatter" than would be on a global Mercator projection. This is especially more important the further you get away from the equator (and the UK is far enough away for it to matter).
You may be able to get "close enough" using the calculations you're trying, but for best accuracy you may want to either use a map with a well-defined projection, or create your own map.
I know the question was asked a while ago, but the Proj4JS library is ideal for transforming between different map projections in JavaScript.
UK maps tend to use the OSGB's National Grid which is based on a Transverse Mercator projection. Ie. like a conventional Mercator but turned 90 degrees, so that the "equator" becomes a meridian.
#Xarinko Actionscript snippet in Javascript (with some testing values)
var mapWidth = 1500;
var mapHeight = 1577;
var mapLonLeft = 9.8;
var mapLonRight = 10.2;
var mapLonDelta = mapLonRight - mapLonLeft;
var mapLatBottom = 53.45;
var mapLatBottomRadian = mapLatBottom * Math.PI / 180;
function convertPixelToGeo(tx, ty)
{
/* called worldMapWidth in Raphael's Code, but I think that's the radius since it's the map width or circumference divided by 2*PI */
var worldMapRadius = mapWidth / mapLonDelta * 360/(2 * Math.PI);
var mapOffsetY = ( worldMapRadius / 2 * Math.log( (1 + Math.sin(mapLatBottomRadian) ) / (1 - Math.sin(mapLatBottomRadian)) ));
var equatorY = mapHeight + mapOffsetY;
var a = (equatorY-ty)/worldMapRadius;
var lat = 180/Math.PI * (2 * Math.atan(Math.exp(a)) - Math.PI/2);
var long = mapLonLeft+tx/mapWidth*mapLonDelta;
return [lat,long];
}
convertPixelToGeo(241,444)
C# implementation:
private Point ConvertGeoToPixel(
double latitude, double longitude, // The coordinate to translate
int imageWidth, int imageHeight, // The dimensions of the target space (in pixels)
double mapLonLeft, double mapLonRight, double mapLatBottom // The bounds of the target space (in geo coordinates)
) {
double mapLatBottomRad = mapLatBottom * Math.PI / 180;
double latitudeRad = latitude * Math.PI / 180;
double mapLonDelta = mapLonRight - mapLonLeft;
double worldMapWidth = (imageWidth / mapLonDelta * 360) / (2 * Math.PI);
double mapOffsetY = worldMapWidth / 2 * Math.Log((1 + Math.Sin(mapLatBottomRad)) / (1 - Math.Sin(mapLatBottomRad)));
double x = (longitude - mapLonLeft) * (imageWidth / mapLonDelta);
double y = imageHeight - ((worldMapWidth / 2 * Math.Log((1 + Math.Sin(latitudeRad)) / (1 - Math.Sin(latitudeRad)))) - mapOffsetY);
return new Point()
{
X = Convert.ToInt32(x),
Y = Convert.ToInt32(y)
};
}
If you want to avoid some of the messier aspects of lat/lng projections intrinsic to Proj4JS, you can use D3, which offers many baked-in projections and renders beautifully. Here's an interactive example of several flavors of Azimuthal projections. I prefer Albers for USA maps.
If D3 is not an end-user option -- say, you need to support IE 7/8 -- you can render in D3 and then snag the xy coordinates from the resultant SVG file that D3 generates. You can then render those xy coordinates in Raphael.
This function works great for me because I want to define the mapHeight based on the map I want to plot. I'm generating PDF maps. All I need to do is pass in the map's max Lat , min Lon and it returns the pixels size for the map as [height,width].
convertGeoToPixel(maxlatitude, maxlongitude)
One note in the final step where $y is set, do not subtract the calculation from the mapHeight if your coordinate system 'xy' starts at the bottom/left , like with PDFs, this will invert the map.
$y = (($worldMapWidth / 2 * log((1 + sin($lat)) / (1 - sin($lat)))) - $mapOffsetY);