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Search found 1333 results on 54 pages for 'geometry shader'.

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• Generate a polygon from line.

  - by VOX

  I want to draw a line with thickness in j2me. This can easily be achieved in desktop java by setting Pen width as thickness value. However in j2me, Pen class does not support width. My idea is to generate a polygon from the line I have, that resembles the line with thickness i want to draw. In picture, on the left is what I have, a line with points. On the right is what I want, a polygon that when filled, a line with thickness. Could anyone know how to generate a polygon from line? http://www.freeimagehosting.net/uploads/140e43c2d2.gif

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• How to know the line joining two points?

  - by dafero

  I have two points and I want to know the line which is joining them. I don't want to draw the line. I want to create a matrix with all the points which formed the line. In the future, I want to solve if two points belong or not to a shape. And this is the first part.

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• Determining if two rays intersect

  - by Faken

  I have two rays on a 2D plane 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 ray extending to infinity. I want to find out if the two rays 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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• How would I create this background effect?

  - by William

  What would you call the effect applied to the backgrounds in the Giygas fight of Earthbound, and the battle backgrounds in Mother 3? This is what I'm talking about. http://www.youtube.com/watch?v=tcaErqaoWek http://www.youtube.com/watch?v=ubVnmeTRqhg Now anyone know how I could go about this without using animated images, or using openGL?

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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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• 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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• 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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• 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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• 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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• 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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• 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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• QT Question (general math really): How can I tell if an Ellipse and a Line have an intersection?

  - by asdsa

  The ellipse is actually a circle. I know all the coordinates of both things, I just need to know whether or not a line goes through an ellipse. Is this built in anywhere? SPecifically I am using QGraphicsEllipseIem and QLine but i can convert them to any other type thanks

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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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• Fast way to compute the minimal distance of two sets of k-dimensional vectors

  - by vrld

  I two sets of k-dimensional vectors, where k is around 500 and the number of vectors is usually smaller. I want to compute the (arbitrarily defined) minimal distance between the two sets. A naive approach would be this: (loop for a in set1 for b in set2 minimizing (distance a b)) However, this requires O(n² * distance) computations. Is there a faster way of doing this?

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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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• changing the intensity of lighten/darken on bitmaps using PorterDuffXfermode in the Android Paint class

  - by user1116836

  Ok my orignal question has changed. How do i change the intensity of how something like this is effected? DayToNight.setXfermode(new PorterDuffXfermode(Mode.DST_IN)); in my dream world it would have worked like this DayToNight.setXfermode(new PorterDuffXfermode(Mode.DST_IN(10))); the 10 being a level of intensity. An example would be if I had a flickering candle, when the candle burns bright I want the bitmaps I am drawing to the screen to retain their origanol color and brightness, when it flickers I want the bitmaps to be almost blacked out, and I want to darken the Bitmaps as the light dims. I have equations, timers and all that figured out, just not how to actually apply it to change the color/brightness. Maybe burning the images is what im looking for? I just want to change the lightness lol. I feel like using paint.setShader might be a solution, but the information in this area is pretty limited from what i have been able to find. Any help would be appreciated. edit: to be crystal clear, i am looking for a way to lighten/darken bitmaps as I draw them to the canvas

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• I'm looking for a blend mode that gives 'realistic' paint colors. (Subtractive)

  - by almosnow

  I've been looking for a blend mode to (well ...) blend two RGB pixels in order to build colors in the samw way that a painter builds them (i.e: subtractive). Here are quick examples of the type of results that I'm expecting: CYAN + MAGENTA = BLUE CYAN + YELLOW = GREEN MAGENTA + YELLOW = RED RED + YELLOW = ORANGE RED + BLUE = PURPLE YELLOW + BLUE = GREEN I'm looking for a formula, like: dest_red = first_red + second_red; dest_green = first_green + second_green; dest_blue = first_blue + second_blue; I've tried with the commonly used 'multiply' formula but it doesn't work; I've tried with custom made formulas but I'm still not able to 'crack' how it should work. And I know already a lot of color theory so please refrain from answers like: Check this link: http://the_difference_betweeen_additive_and_subtractive_lightning.html

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