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  • Android: Find coordinates of a certain point X meters from my location moving towards the point I am

    - by Aidan
    Hi Guys, I'm constructing a geolocation based application and I'm trying to figure out a way to make my application realise when a user is facing the direction of the given location (a particular long / lat co-ord). I've done some Googling and checked the SDK but can't really find anything for such a thing. Does anyone know of a way? Example. Point A = Phones current location. Point B = A's orientation in relation to true north + 45 + max distance towards the direction your facing, Point C = A's orientation in relation to true north - 45 + max distance towards the direction your facing. So now you have a triangle constructed. pretty schweet huh? yeah.. I think so.. So now that I have my fancy Triangle I use something called Barycentric Coordinates ( http://en.wikipedia.org/wiki/Barycentric_coordinates_(mathematics) ). This will allow me to test another point and see if it is in the triangle. If it is, it means we're facing it AND it's within the right distance. So it should be displayed on screen. If I'm facing 90 degrees from true north. The distance it travels should be that direction. 90 degrees from true north. It should not be 100 degrees or something from true north! But the problem is I haven't yet figured out how I make the device realise it must go "out" the direction it is facing.

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  • How do I determine when two moving points become visible to each other?

    - by Devin Jeanpierre
    Suppose I have two points, Point1 and Point2. At any given time, these points may be at different positions-- they are not necessarily static. Point1 is located at some position at time t, and its position is defined by the continuous functions x1(t) and y1(t) giving the x and y coordinates at time t. These functions are not differentiable, they are constructed piecewise from line segments. Point2 is the same, with x2(t) and y2(t), each function having the same properties. The obstacles that might prevent visibility are simple (and immobile) polygons. How can I find the boundary points for visibility? i.e. there are two kinds of boundaries: where the points become visible, and become invisible. For a become-visible boundary i, there exists some ?0, such that for any real number a, a ? (i-?, i) , Point1 and Point2 are not visible (i.e. the line segment that connects (x1(a), y1(a)) to (x2(a), y2(x)) crosses some obstacles). For b ? (i, i+?) they are visible. And it is the other way around for becomes-invisible. But can I find such a precise boundary, and if so, how?

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  • Surface Area of a Spheroid in Python

    - by user3678321
    I'm trying to write a function that calculates the surface area of a prolate or oblate spheroid. Here's a link to where I got the formulas (http://en.wikipedia.org/wiki/Prolate_spheroid & http://en.wikipedia.org/wiki/Oblate_spheroid). I think I've written them wrong, but here is my code so far; from math import pi, sqrt, asin, degrees, tanh def checkio(height, width): height = float(height) width = float(width) lst = [] if height == width: r = 0.5 * width surface_area = 4 * pi * r**2 surface_area = round(surface_area, 2) lst.append(surface_area) elif height > width: #If spheroid is prolate a = 0.5 * width b = 0.5 * height e = 1 - a / b surface_area = 2 * pi * a**2 * (1 + b / a * e * degrees(asin**-1(e))) surface_area = round(surface_area, 2) lst.append(surface_area) elif height < width: #If spheroid is oblate a = 0.5 * height b = 0.5 * width e = 1 - b / a surface_area = 2 * pi * a**2 * (1 + 1 - e**2 / e * tanh**-1(e)) surface_area = round(surface_area, 2) lst.append(surface_area, 2) return lst

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  • Find most right and left point of a horizontal circle in 3d Vector environment

    - by Olivier de Jonge
    I'm drawing a 3D pie chart that is rendered with in 3D vectors, projected to 2D vectors and then drawn on a Graphics object. I want to calculate the most left and right point of the circle The method to create a vector, draw and project to a 2d vector are below. Anyone knows the answer? public class Vector3d { public var x:Number; public var y:Number; public var z:Number; //the angle that the 3D is viewed in tele or wide angle. public static var viewDist:Number = 700; function Vector3d(x:Number, y:Number, z:Number){ this.x = x; this.y = y; this.z = z; } public function project2DNew():Vector { var p:Number = getPerspective(); return new Vector(p * x, p * y); } public function getPerspective():Number{ return viewDist / (this.z + viewDist); } }

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  • Determine mouse click for all screen resolutions

    - by Hallik
    I have some simple javascript that determines where a click happens within a browser here: var clickDoc = (document.documentElement != undefined && document.documentElement.clientHeight != 0) ? document.documentElement : document.body; var x = evt.clientX; var y = evt.clientY; var w = clickDoc.clientWidth != undefined ? clickDoc.clientWidth : window.innerWidth; var h = clickDoc.clientHeight != undefined ? clickDoc.clientHeight : window.innerHeight; var scrollx = window.pageXOffset == undefined ? clickDoc.scrollLeft : window.pageXOffset; var scrolly = window.pageYOffset == undefined ? clickDoc.scrollTop : window.pageYOffset; params = '&x=' + (x + scrollx) + '&y=' + (y + scrolly) + '&w=' + w + '&random=' + Date(); All of this data gets stored in a DB. Later I retrieve it and display where all the clicks happened on that page. This works fine if I do all my clicks in one resolution, and then display it back in the same resolution, but this not the case. there can be large amounts of resolutions used. In my test case I was clicking on the screen with a screen resolution of 1260x1080. I retrieved all the data and displayed it in the same resolution. But when I use a different monitor (tried 1024x768 and 1920x1080. The marks shift to the incorrect spot. My question is, if I am storing the width and height of the client, and the x/y position of the click. If 3 different users all with different screen resolutions click on the same word, and a 4th user goes to view where all of those clicks happened, how can I plot the x/y position correctly to show that everyone clicked in the same space, no matter the resolution? If this belongs in a better section, please let me know as well.

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  • resampling a series of points

    - by clamp
    hello, i have an array of points in 3d (imagine the trajectory of a ball) with X samples. now, i want to resample these points so that i have a new array with positions with y samples. y can be bigger or smaller than x but not smaller than 1. there will always be at least 1 sample. how would an algorithm look like to resample the original array into a new one? thanks!

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  • Algorithm to Group All the Cycles Together

    - by Ngu Soon Hui
    I have a lot of cycles ( indicated by numeric values, for example, 1-2-3-4 corresponds to a cycle, with 4 edges, edge 1 is {1:2}, edge 2 is {2:3}, edge 3 is {3,4}, edge 4 is {4,1}, and so on). A cycle is said to be connected to another cycle if they share one and only one edge. For example, let's say I have two cycles 1-2-3-4 and 5-6-7-8, then there are two cycle groups because these two cycles are not connecting to each other. If I have two cycles 1-2-3-4 and 3-4-5-6, then I have only one cycle group because these two cycles share the same edge. What is the most efficient way to find all the cycle groups?

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  • Uniform distance between points

    - by Reonarudo
    Hello, How could I, having a path defined by several points that are not in a uniform distance from each other, redefine along the same path the same number of points but with a uniform distance. I'm trying to do this in Objective-C with NSArrays of CGPoints but so far I haven't had any luck with this. Thank you for any help. EDIT I was wondering if it would help to reduce the number of points, like when detecting if 3 points are collinear we could remove the middle one, but I'm not sure that would help.

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  • Determining if two lines intersect

    - by Faken
    I have two lines that extend to infinity but both have a starting point. They are both described by a starting point and a vector in the direction of the line extending to infinity. I want to find out if the two lines intersect but i don't need to know where they intersect (its part of a collision detection algorithm). Everything i have looked at so far describes finding the intersection point of two lines or line segments. Anyone know a fast algorithm to solve this?

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  • Merging and splitting overlapping rectangles to produce non-overlapping ones

    - by uj
    I am looking for an algorithm as follows: Given a set of possibly overlapping rectangles (All of which are "not rotated", can be uniformly represented as (left,top,right,bottom) tuplets, etc...), it returns a minimal set of (non-rotated) non-overlapping rectangles, that occupy the same area. It seems simple enough at first glance, but prooves to be tricky (at least to be done efficiently). Are there some known methods for this/ideas/pointers? Methods for not necessarily minimal, but heuristicly small, sets, are interesting as well, so are methods that produce any valid output set at all.

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  • Is a point inside or outside a polygon which is on the surface of a globe

    - by richard
    How do I determine if a point is inside or outside a polygon that lies on the the surface of the earth? The inside of the polygon can be determined via the right hand rule, ie. the inside of the polygon is on your right hand side when you walk around the polygon. The polygon may Circle either pole Cross the 180 longitude Cover more than 50% of the globe As the globe is a sphere the normal ray crossing algorithms do not work correctly.

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  • Centre of a circle that intersects two points

    - by Jason
    Given two points in a 2D plane, and a circle of radius r that intersects both of those points, what would be the formula to calculate the centre of that circle? I realise there would two places the circle can be positioned. I would want the circle whose centre is encountered first in a clockwise direction when sweeping the line that joins the two points around one of those points, starting from an arbitrary angle. I guess that is the next stage in my problem, after I find an answer for the first part. I'm hoping the whole calculation can be done without trigonometry for speed. I'm starting with integer coordinates and will end with integer coordinates, if that helps.

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  • Finding whether a point lies inside a rectangle or not

    - by avd
    The rectangle can be oriented in any way...need not be axis aligned. Now I want to find whether a point lies inside the rectangle or not. One method I could think of was to rotate the rectangle and point coordinates to make the rectangle axis aligned and then by simply testing the coordinates of point whether they lies within that of rectangle's or not. The above method requires rotation and hence floating point operations. Is there any other efficient way to do this??

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  • How to draw a filled circle in Java?

    - by Roman
    I have a JPanel with a Grid Layout. In the "cells" of the grid I can put different elements (for example JButtons). There is no problems with that. But now I want to put a filled circle in some of the cells. I also would like to relate an ActionListener with these circles. In more details, if I click the circle it disappears from the current cell and appears in another one. How can I do it in Java? I am using Swing.

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  • Shortest distance between points on a toroidally wrapped (x- and y- wrapping) map?

    - by mstksg
    I have a toroidal-ish Euclidean-ish map. That is the surface is a flat, Euclidean rectangle, but when a point moves to the right boundary, it will appear at the left boundary (at the same y value), given by x_new = x_old % width Basically, points are plotted based on: (x_new, y_new) = ( x_old % width, y_old % height) Think Pac Man -- walking off one edge of the screen will make you appear on the opposite edge. What's the best way to calculate the shortest distance between two points? The typical implementation suggests a large distance for points on opposite corners of the map, when in reality, the real wrapped distance is very close. The best way I can think of is calculating Classical Delta X and Wrapped Delta X, and Classical Delta Y and Wrapped Delta Y, and using the lower of each pair in the Sqrt(x^2+y^2) distance formula. But that would involve many checks, calculations, operations -- some that I feel might be unnecessary. Is there a better way?

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  • Given the lat/lon of 2 close points on earth (<10m), How do I calculate the distance in metres?

    - by Rory
    I have the lat/lon of 2 points on the earth. They are really close together, <10m. Let's assume the earth is flat. How do I calculate the distance between them in metres? I know about tools (PostGIS, etc.) that can do this correctly, however I'm just doing a rough and ready type, and I'm OK with low accuracy. At such small sizes a difference of 1% is only 10cm, which is fine for me. I'm doing this in stock python. I'm OK with a standard Euclidean distance thing.

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  • determine if line segment is inside polygon

    - by dato
    suppose we have simple polygon(without holes) with vertices (v0,v1,....vn) my aim is to determine if for given point p(x,y) any line segment connecting this point and any vertices of polygon is inside polygon or even for given two point p(x0,y0) `p(x1,y1)` line segment connecting these two point is inside polygon? i have searched many sites about this ,but i am still confused,generally i think we have to compare coordinates of vertices and by determing coordinates of which point is less or greater to another point's coordinates,we could determine location of any line segment,but i am not sure how correct is this,please help me

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  • Graphics Question: How do I restrict the mouse cursor to within a circle?

    - by Dan
    I'm playing with XNA. When I click the left mouse button, I record the X,Y co-ordinates. Keeping the mouse button held down, moving the mouse draws a line from this origin to the current mouse position. I've offset this into the middle of the window. Now, what I'd like to do is restrict the mouse cursor to within a circle (with a radius of N, centred on the middle of the screen). Restricting the mouse to a rectangular region is easy enough (by adjusting the origin by the difference of the mouse position and the size of the region), but I haven't a clue on how to start doing it for a circular region. Can anyone explain how to do this? Any advice on where to start would be helpful.

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  • Python point lookup (coordinate binning?)

    - by Rince
    Greetings, I am trying to bin an array of points (x, y) into an array of boxes [(x0, y0), (x1, y0), (x0, y1), (x1, y1)] (tuples are the corner points) So far I have the following routine: def isInside(self, point, x0, x1, y0, y1): pr1 = getProduct(point, (x0, y0), (x1, y0)) if pr1 >= 0: pr2 = getProduct(point, (x1, y0), (x1, y1)) if pr2 >= 0: pr3 = getProduct(point, (x1, y1), (x0, y1)) if pr3 >= 0: pr4 = getProduct(point, (x0, y1), (x0, y0)) if pr4 >= 0: return True return False def getProduct(origin, pointA, pointB): product = (pointA[0] - origin[0])*(pointB[1] - origin[1]) - (pointB[0] - origin[0])*(pointA[1] - origin[1]) return product Is there any better way then point-by-point lookup? Maybe some not-obvious numpy routine? Thank you!

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  • 3D coordinate of 2D point given camera and view plane

    - by Myx
    I wish to generate rays from the camera through the viewing plane. In order to do this, I need my camera position ("eye"), the up, right, and towards vectors (where towards is the vector from the camera in the direction of the object that the camera is looking at) and P, the point on the viewing plane. Once I have these, the ray that's generated is: ray = camera_eye + t*(P-camera_eye); where t is the distance along the ray (assume t = 1 for now). My question is, how do I obtain the 3D coordinates of point P given that it is located at position (i,j) on the viewing plane? Assume that the upper left and lower right corners of the viewing plane are given. NOTE: The viewing plane is not actually a plane in the sense that it doesn't extend infinitely in all directions. Rather, one may think of this plane as a widthxheight image. In the x direction, the range is 0--width and in the y direction the range is 0--height. I wish to find the 3D coordinate of the (i,j)th element, 0

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  • Const unsigned char* to char*

    - by BSchlinker
    So, I have two types at the moment: const unsigned char* unencrypted_data_char; string unencrypted_data; I'm attempting to perform a simple conversion of data from one to the other (string - const unsigned char*) As a result, I have the following: strcpy((unencrypted_data_char),(unencrypted_data.c_str())); However, I'm receiving the error: error C2664: 'strcpy' : cannot convert parameter 1 from 'const unsigned char *' to 'char *' Any advise? I thought using reinterpret_cast would help, but it doesn't seem to make a difference.

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