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Search found 702 results on 29 pages for 'geometry'.

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  • circles and triangles problem

    - by Faken
    Hello everyone, I have an interesting problem here I've been trying to solve for the last little while: I have 3 circles on a 2D xy plane, each with the same known radius. I know the coordinates of each of the three centers (they are arbitrary and can be anywhere). What is the largest triangle that can be drawn such that each vertice of the triangle sits on a separate circle, what are the coordinates of those verticies? I've been looking at this problem for hours and asked a bunch of people but so far only one person has been able to suggest a plausible solution (though i have no way of proving it). The solution that we have come up with involves first creating a triangle about the three circle centers. Next we look at each circle individually and calculate the equation of a line that passes through the circle's center and is perpendicular to the opposite edge. We then calculate two intersection points of the circle. This is then done for the next two circles with a result of 6 points. We iterate over the 8 possible 3 point triangles that these 6 points create (the restriction is that each point of the big triangle must be on a separate circle) and find the maximum size. The results look reasonable (at least when drawn out on paper) and it passes the special case of when the centers of the circles all fall on a straight line (gives a known largest triangle). Unfortunate i have no way of proving this is correct or not. I'm wondering if anyone has encountered a problem similar to this and if so, how did you solve it? Note: I understand that this is mostly a math question and not programming, however it is going to be implemented in code and it must be optimized to run very fast and efficient. In fact, I already have the above solution in code and tested to be working, if you would like to take a look, please let me know, i chose not to post it because its all in vector form and pretty much impossible to figure out exactly what is going on (because it's been condensed to be more efficient). Lastly, yes this is for school work, though it is NOT a homework question/assignment/project. It's part of my graduate thesis (abet a very very small part, but still technically is part of it). Thanks for your help.

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  • Cuboid inside generic polyhedron

    - by DOFHandler
    I am searching for an efficient algorithm to find if a cuboid is completely inside or completely outside or (not-inside and not-outside) a generic (concave or convex) polyhedron. The polyhedron is defined by a list of 3D points and a list of facets. Each facet is defined by the subset of the contour points ordinated such as the right-hand normal points outward the solid. Any suggestion? Thank you

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  • Cone-Line Segment Intersection 2D

    - by nophnoph
    Hi everyone, I would like to know is there any way to determine if a cone is intersecting with a (finite)line segment. The cone actually a circle located at P(x,y) with theta degree field of view and radius r. The simple visualization can be found here (sorry i can't post the picture here) I'm trying to do it in C# but I don't have any idea how to that, so for now this is what I'm doing : Check if the line segment is intersecting with the circle If the line segment is intersecting with the circle then I check every single point in the line segment using a function I found here. But I don't think this is the best way to do it. Does anyone have an idea? For additional info, I need this function to make some kind of simple vision simulator. Thanks in advance :)

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  • Generating a beveled edge for a 2D polygon

    - by Metaphile
    I'm trying to programmatically generate beveled edges for geometric polygons. For example, given an array of 4 vertices defining a square, I want to generate something like this. But computing the vertices of the inner shape is baffling me. Simply creating a copy of the original shape and then scaling it down will not produce the desired result most of the time. My algorithm so far involves analyzing adjacent edges (triples of vertices; e.g., the bottom-left, top-left, and top-right vertices of a square). From there, I need to find the angle between them, and then create a vertex somewhere along that angle, depending on how deep I want the bevel to be. And because I don't have much of a math background, that's where I'm stuck. How do I find that center angle? Or is there a much simpler way of attacking this problem?

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  • Determining if a coordinate is on a line

    - by TGCBraun
    I´m coding a little app that allows the user to draw multiple shapes and then remove or resize them. It´s working perfectly on rectangles and ovals, but I´m having issues with lines. Here´s a method that I wrote to find if the clicked spot on the screen is part of a specific line: public boolean containsLocation(int x, int y) { int m = (getY2() - getY()) / (getX2() - getX()); int b = getY() - (m * getX()); if (y == (m * x) + b) { return true; } return false; I´m using the famous y = mx + b formula and replacing y and x to find if the clicked spot is part of the line. The problem is when I click on the screen to remove the line, it only works if I click on the very fist coordinate (x,y) where the line starts. Nothing happens when I click anywhere else along the line. Can anyone shed a light on what I´m doing wrong? Thanks a lot.

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  • Reverse-projection 2D points into 3D

    - by ehsan baghaki
    Suppose we have a 3d Space with a plane on it with an arbitary equation : ax+by+cz+d=0 now suppose that we pick 3 random points on that plane: (x0,y0,z0) (x1,y1,z1) (x1,y1,z1) now i have a different point of view(camera) for this plane. i mean i have a different camera that will look at this plane from a different point of view. From that camera point of view these points have different locations. for example (x0,y0,z0) will be (x0',y0') and (x1,y1,z1) will be (x1',y1') and (x2,y2,z2) will be (x2',y2') from the new camera point of view. So here is my a little hard question! I want to pick a point for example (X,Y) from the new camera point of view and tell where it will be on that plane. All i know is that 3 points and their locations on 3d space and their projection locations on the new camera view. Do you know the coefficients of the plane-equation and the camera positions (along with the projection), or do you only have the six points? - Nils i know the location of first 3 points. therefore we can calculate the coefficients of the plane. so we know exactly where the plane is from (0,0,0) point of view. and then we have the camera that can only see the points! So the only thing that camera sees is 3 points and also it knows their locations in 3d space (and for sure their locations on 2d camera view plane). and after all i want to look at camera view, pick a point (for example (x1,y1)) and tell where is that point on that plane. (for sure this (X,Y,Z) point should fit on the plane equation). Also i know nothing about the camera location.

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  • Detecting coincident subset of two coincident line segments

    - by Jared Updike
    This question is related to: How do I determine the intersection point of two lines in GDI+? (great explanation of algebra but no code) How do you detect where two line segments intersect? (accepted answer doesn't actually work) But note that an interesting sub-problem is completely glossed over in most solutions which just return null for the coincident case even though there are three sub-cases: coincident but do not overlap touching just points and coincident overlap/coincident line sub-segment For example we could design a C# function like this: public static PointF[] Intersection(PointF a1, PointF a2, PointF b1, PointF b2) where (a1,a2) is one line segment and (b1,b2) is another. This function would need to cover all the weird cases that most implementations or explanations gloss over. In order to account for the weirdness of coincident lines, the function could return an array of PointF's: zero result points (or null) if the lines are parallel or do not intersect (infinite lines intersect but line segments are disjoint, or lines are parallel) one result point (containing the intersection location) if they do intersect or if they are coincident at one point two result points (for the overlapping part of the line segments) if the two lines are coincident

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  • Uniforme 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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  • Open source jigsaw piece generator

    - by codecowboy
    Hi, Does anyone know of a C-based open source class / framework which can generate a random jigsaw puzzle from an image or generate a random puzzle template which could then be applied to an image? The puzzle pieces must have male and female notches/holes. There should be more than one template so that the puzzle does not become too easy. The target system is iOS / Mac. If not, how would you approach this problem? The puzzle pieces should be as close to a real jigsaw piece shape as possible and the system must be dynamic so that the user can use their own photos or download photos. thanks!

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  • WPF, convert Path.DataProperty to Segment objects

    - by user275587
    I was wondering if there was a tool to convert a path data like "M 0 0 l 10 10" to it's equivalent line/curve segment code. Currently I'm using: string pathXaml = "<Path xmlns=\"http://schemas.microsoft.com/winfx/2006/xaml/presentation\" xmlns:x=\"http://schemas.microsoft.com/winfx/2006/xaml\" Data=\"M 0 0 l 10 10\"/>"; Path path = (Path)System.Windows.Markup.XamlReader.Load(pathXaml); It appears to me that calling XamlParser is much slower than explicitly creating the line segments. However converting a lot of paths by hand is very tedious.

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  • Reconstructing simple 3d enviroment(room) from photo

    - by Riz
    I have photo of a room with three walls and floor/ceiling or both. I am trying to reconstruct this room in 3d asking user for minimal input. Right now I use 8 points defined by user, angles of left and right wall(they can be quite different from 90) and one size "InLeftBottom-InRightBottom"(I need to have real size of this room for later use). I have no info about user's camera(I can read EXIF to get FOV and use constant height but this can be only used as additional info). Is this possible to ask user for less info? Maybe it's possible to get wall angles without user interaction? Or maybe I am completly wrong and should use different approach?

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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 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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  • 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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  • 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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  • 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 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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  • 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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  • 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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  • 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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