What you need is the De Casteljau algorithm. This will allow you to split your curve into whatever segments you’d like. However, since you’re dealing with just cubic curves, I’d like to suggest a slightly easier to use formulation that’ll give you a segment from t0 to t1 where 0 <= t0 <= t1 <= … Read more
The map will not try to get the users location until it’s actually displayed, as far as I know. A good test of that is, when do you get the dialog asking you if it’s OK for the app to get the users location? In my experience that has always come up only after the … Read more
You have to get a reference to you button, for example by calling findViewById(). When you got the reference to the button you can set the x and y value with button.setX() and button.setY(). …. Button myButton = (Button) findViewById(R.id.<id of the button in the layout xml file>); myButton.setX(<x value>); myButton.setY(<y value>); ….
Oki, I have worked this out mathematically myself so please bear with me. If you want to translate an acceleration vector accelerationvalues into an acceleration vector trueacceleration expressed in real world’s coordinates, once you have azimuth,pitch and roll stored in a orientationvalues vector, just do the following: trueacceleration[0] =(float) (accelerometervalues[0]*(Math.cos(orientationvalues[2])*Math.cos(orientationvalues[0])+Math.sin(orientationvalues[2])*Math.sin(orientationvalues[1])*Math.sin(orientationvalues[0])) + accelerometervalues[1]*(Math.cos(orientationvalues[1])*Math.sin(orientationvalues[0])) + accelerometervalues[2]*(-Math.sin(orientationvalues[2])*Math.cos(orientationvalues[0])+Math.cos(orientationvalues[2])*Math.sin(orientationvalues[1])*Math.sin(orientationvalues[0]))); trueacceleration[1] … Read more
The Geocoder API provided in the Android Framework I have previously worked with the Geocoding API which exists in the Android API but it does not work on all devices. In fact, as per my and others experiences, a lot of devices will return null when using the Geocoding API. For that reason, I have … Read more
This post shows how to do this in SQL Server. And here is how to do it in MySQL: SELECT ((ACOS(SIN($lat * PI() / 180) * SIN(lat * PI() / 180) + COS($lat * PI() / 180) * COS(lat * PI() / 180) * COS(($lon – lon) * PI() / 180)) * 180 / PI()) … Read more
In a UIResponder subclass, such as UIView: override func touchesBegan(touches: NSSet, withEvent event: UIEvent) { let touch = touches.anyObject()! as UITouch let location = touch.locationInView(self) } This will return a CGPoint in view coordinates. Updated with Swift 3 syntax override func touchesBegan(touches: NSSet, withEvent event: UIEvent) { let touch = touches.first! let location = touch.location(in: … Read more
Here’s a reusable function you can use. It includes doctests and data validation: def bilinear_interpolation(x, y, points): ”’Interpolate (x,y) from values associated with four points. The four points are a list of four triplets: (x, y, value). The four points can be in any order. They should form a rectangle. >>> bilinear_interpolation(12, 5.5, … [(10, … Read more
Calculate the distance of each point using Math.sqrt( Math.pow(x, 2) + Math.pow(y, 2) ); Take the result that’s the lowest var points = [ {x: 10, y: 20}, {x: 12, y: 18}, {x: 20, y: 30}, {x: 5, y: 40}, {x: 100, y: 2} ]; function d(point) { return Math.pow(point.x, 2) + Math.pow(point.y, 2); } … Read more
You need more information than just a single lat/lon pair to be able to do this. At this stage, the information you have provided is missing two things: how large an area does your image cover (in terms of lat/lon)? Based on what you’ve provided, I don’t know if the image shows an area a … Read more