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  • How to know coordinates in a real image from a scaled image.

    - by ocell
    Hi folks, First of all thanks for your time reading my question :-) I have an original image (w': 2124, h': 3204) and the same image scaled (w: 512, h: 768). The ratio for width is 4.14 (rw) and the ratio for height is 4.17 (rh). I'm trying to know the coordinates (x', y') in the original image when I receive the coordinates in the scaled image (x, y). I'm using the formula: x' = x * rw and y' = y * rh. But when I'm painting a line, or a rectangle always appears a shift that is incremented when x or y is higher. Please anybody knows how to transform coordinates without loosing of accuracy. Thanks in advance! Oscar.

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  • How to get coordinates when iPhone is locked/sleeping?

    - by lionfly
    For normal situations, the didUpdateToLocation (of CLLocationManager) does not work any more when the lock/sleep button is pushed, and your app cannot get further newLocation coordinates during the locked period. However, some apps CAN track during the locked period, and they will show the correct route when you unlocked and see the Mapview. Obviously there are ways to get the coordinates during the locked period. How to get the coordinates when the iPhone is locked/sleeping? Who has hints and further info, please post. Thanks.

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  • shader coding: calculate screen coordinates of fragment

    - by Jay
    Good morning, I'm new to shader coding and trying to implement some visual effects code in shaders using billboards. (Yes, I couldn't have picked anything harder to start with, but I'm lucky that way) Setup: I have rendered the full screen z depth to an array of floats in a previous pass. In the fragment shader I need the scene depth where the rendered fragment is displayed (to see if it's occluded). I can use tex2d() to get the depth value if I have the screen coordinates of the point being rendered in the fragment shader. Question: In the fragment shader how do you calculate the screen coordinates of the pixel (in the range 0-1.0)? Is the position passed to the fragment shader a pixel offset? If so, I guess it would be: float2( position.x / screen-width, position.y / screen-height ) Thanks for any help/

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  • Finding Z given X & Y coordinates on terrain?

    - by mrky
    I need to know what the most efficient way of finding Z given X & Y coordinates on terrain. My terrain is set up as a grid, each grid block consisting of two triangles, which may be flipped in any direction. I want to move game objects smoothly along the floor of the terrain without "stepping." I'm currently using the following method with unexpected results: double mapClass::getZ(double x, double y) { int vertexIndex = ((floor(y))*width*2)+((floor(x))*2); vec3ray ray = {glm::vec3(x, y, 2), glm::vec3(x, y, 0)}; vec3triangle tri1 = { glmFrom(vertices[vertexIndex].v1), glmFrom(vertices[vertexIndex].v2), glmFrom(vertices[vertexIndex].v3) }; vec3triangle tri2 = { glmFrom(vertices[vertexIndex+1].v1), glmFrom(vertices[vertexIndex+1].v2), glmFrom(vertices[vertexIndex+1].v3) }; glm::vec3 intersect; if (!intersectRayTriangle(tri1, ray, intersect)) { intersectRayTriangle(tri2, ray, intersect); } return intersect.z; } intersectRayTriangle() and glmFrom() are as follows: bool intersectRayTriangle(vec3triangle tri, vec3ray ray, glm::vec3 &worldIntersect) { glm::vec3 barycentricIntersect; if (glm::intersectLineTriangle(ray.origin, ray.direction, tri.p0, tri.p1, tri.p2, barycentricIntersect)) { // Convert barycentric to world coordinates double u, v, w; u = barycentricIntersect.x; v = barycentricIntersect.y; w = 1 - (u+v); worldIntersect.x = (u * tri.p0.x + v * tri.p1.x + w * tri.p2.x); worldIntersect.y = (u * tri.p0.y + v * tri.p1.y + w * tri.p2.y); worldIntersect.z = (u * tri.p0.z + v * tri.p1.z + w * tri.p2.z); return true; } else { return false; } } glm::vec3 glmFrom(s_point3f point) { return glm::vec3(point.x, point.y, point.z); } My convenience structures are defined as: struct s_point3f { GLfloat x, y, z; }; struct s_triangle3f { s_point3f v1, v2, v3; }; struct vec3ray { glm::vec3 origin, direction; }; struct vec3triangle { glm::vec3 p0, p1, p2; }; vertices is defined as: std::vector<s_triangle3f> vertices; Basically, I'm trying to get the intersect of a ray (which is positioned at the x, and y coordinates specified facing pointing downwards toward the terrain) and one of the two triangles on the grid. getZ() rarely returns anything but 0. Other times, the numbers it generates seem to be completely off. Am I taking the wrong approach? Can anyone see a problem with my code? Any help or critique is appreciated!

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  • 2D camera perspective projection from 3D coordinates -- HOW?

    - by Jack
    I am developing a camera for a 2D game with a top-down view that has depth. It's almost a 3D camera. Basically, every object has a Z even though it is in 2D, and similarly to parallax layers their position, scale and rotation speed vary based on their Z. I guess this would be a perspective projection. But I am having trouble converting the objects' 3D coordinates into the 2D space of the screen so that everything has correct perspective and scale. I never learned matrices though I did dig the topic a bit today. I tried without using matrices thanks to this article but every attempt gave awkward results. I'm using ActionScript 3 and Flash 11+ (Starling), where the screen coordinates work like this: Left-handed coordinates system illustration I can explain further what I did if you want to help me sort out what's wrong, or you can directly tell me how you would do it properly. In case you prefer the former, read on. These are images showing the formulas I used: upload.wikimedia.org/math/1/c/8/1c89722619b756d05adb4ea38ee6f62b.png upload.wikimedia.org/math/d/4/0/d4069770c68cb8f1aa4b5cfc57e81bc3.png (Sorry new users can't post images, but both are from the wikipedia article linked above, section "Perspective projection". That's where you'll find what all variables mean, too) The long formula is greatly simplified because I believe a normal top-down 2D camera has no X/Y/Z rotation values (correct ?). Then it becomes d = a - c. Still, I can't get it to work. Maybe you could explain what numbers I should put in a(xyz), c(xyz), theta(xyz), and particularly, e(xyz) ? I don't quite get how e is different than c in my case. c.z is also an issue to me. If the Z of the camera's target object is 0, should the camera's Z be something like -600 ? ( = focal length of 600) Whatever I do, it's wrong. I only got it to work when I used arbitrary calculations that "looked" right, like most cameras with parallax layers seem to do, but that's fake! ;) If I want objects to travel between Z layers I might as well do it right. :) Thanks a lot for your help!

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  • Kinect joint coordinates and XNA animation

    - by Sweta Dwivedi
    I have written a program to record the x,y,z coordinated of the Hand joint and I want to animate my models 2D or 3D according to these coordinates. . .However the output of the x,y,z coordinates are fluctuating from -0 to 1 but not more than that.. So i assume I will need to multiply them back with the screen width and height, however it still doesnt seem to animate according to the original x,y,z points Any transformations I might be missing out? while ((line = r.ReadLine()) != null) { string[] temp = line.Split(','); int x = (int) float.Parse(temp[0]))* maxWidth); int y = (int) float.Parse(temp[1])) * maxHeight); }

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  • Converting world space coordinate to screen space coordinate and getting incorrect range of values

    - by user1423893
    I'm attempting to convert from world space coordinates to screen space coordinates. I have the following code to transform my object position Vector3 screenSpacePoint = Vector3.Transform(object.WorldPosition, camera.ViewProjectionMatrix); The value does not appear to be in screen space coordinates and is not limited to a [-1, 1] range. What step have I missed out in the conversion process? EDIT: Projection Matrix Perspective(game.GraphicsDevice.Viewport.AspectRatio, nearClipPlaneZ, farClipPlaneZ); private void Perspective(float aspect_Ratio, float z_NearClipPlane, float z_FarClipPlane) { nearClipPlaneZ = z_NearClipPlane; farClipPlaneZ = z_FarClipPlane; float yZoom = 1f / (float)Math.Tan(fov * 0.5f); float xZoom = yZoom / aspect_Ratio; matrix_Projection.M11 = xZoom; matrix_Projection.M12 = 0f; matrix_Projection.M13 = 0f; matrix_Projection.M14 = 0f; matrix_Projection.M21 = 0f; matrix_Projection.M22 = yZoom; matrix_Projection.M23 = 0f; matrix_Projection.M24 = 0f; matrix_Projection.M31 = 0f; matrix_Projection.M32 = 0f; matrix_Projection.M33 = z_FarClipPlane / (nearClipPlaneZ - farClipPlaneZ); matrix_Projection.M34 = -1f; matrix_Projection.M41 = 0f; matrix_Projection.M42 = 0f; matrix_Projection.M43 = (nearClipPlaneZ * farClipPlaneZ) / (nearClipPlaneZ - farClipPlaneZ); matrix_Projection.M44 = 0f; } View Matrix // Make our view matrix Matrix.CreateFromQuaternion(ref orientation, out matrix_View); matrix_View.M41 = -Vector3.Dot(Right, position); matrix_View.M42 = -Vector3.Dot(Up, position); matrix_View.M43 = Vector3.Dot(Forward, position); matrix_View.M44 = 1f; // Create the combined view-projection matrix Matrix.Multiply(ref matrix_View, ref matrix_Projection, out matrix_ViewProj); // Update the bounding frustum boundingFrustum.SetMatrix(matrix_ViewProj);

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  • Converting to and from local and world 3D coordinate spaces?

    - by James Bedford
    Hey guys, I've been following a guide I found here (http://knol.google.com/k/matrices-for-3d-applications-view-transformation) on constructing a matrix that will allow me to 3D coordinates to an object's local coordinate space, and back again. I've tried to implement these two matrices using my object's look, side, up and location vectors and it seems to be working for the first three coordinates. I'm a little confused as to what I should expect for the w coordinate. Here are couple of examples from the print outs I've made of the matricies that are constructed. I'm passing a test vector of [9, 8, 14, 1] each time to see if I can convert both ways: Basic example: localize matrix: Matrix: 0.000000 -0.000000 1.000000 0.000000 0.000000 1.000000 0.000000 0.000000 1.000000 0.000000 0.000000 0.000000 5.237297 -45.530716 11.021271 1.000000 globalize matrix: Matrix: 0.000000 0.000000 1.000000 0.000000 -0.000000 1.000000 0.000000 0.000000 1.000000 0.000000 0.000000 0.000000 -11.021271 -45.530716 -5.237297 1.000000 test: Vector4f(9.000000, 8.000000, 14.000000, 1.000000) localTest: Vector4f(14.000000, 8.000000, 9.000000, -161.812256) worldTest: Vector4f(9.000000, 8.000000, 14.000000, -727.491455) More complicated example: localize matrix: Matrix: 0.052504 -0.000689 -0.998258 0.000000 0.052431 0.998260 0.002068 0.000000 0.997241 -0.052486 0.052486 0.000000 58.806095 2.979346 -39.396252 1.000000 globalize matrix: Matrix: 0.052504 0.052431 0.997241 0.000000 -0.000689 0.998260 -0.052486 0.000000 -0.998258 0.002068 0.052486 0.000000 -42.413120 5.975957 -56.419727 1.000000 test: Vector4f(9.000000, 8.000000, 14.000000, 1.000000) localTest: Vector4f(-13.508600, 8.486917, 9.290090, 2.542114) worldTest: Vector4f(9.000190, 7.993863, 13.990230, 102.057129) As you can see in the more complicated example, the coordinates after converting both ways loose some precision, but this isn't a problem. I'm just wondering how I should deal with the last (w) coordinate? Should I just set it to 1 after performing the matrix multiplication, or does it look like I've done something wrong? Thanks in advance for your help!

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  • CS3, Illustrator - Where Do X/Y Coordinates Measure from?

    - by nicorellius
    I have a project where there are several PDF files. I'm using Illustrator to make these. It seems the point/line of origin is inconsistent from image to image (file to file). Where is the point of origin, by default, in CS3 Illustrator? It would be nice if, while I was positioning images, I could just say, "OK, x coordinate is 5.5 inches in this document, so it is 5.5 in that one." But it seems this is not the case. Anyone know how Illustrator sets these parameters?

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  • GlassFish Server 3.1.2.2 Maven Coordinates

    - by arungupta
    GlassFish Server 3.1.2.2 was released a few weeks ago. This micro release fixed a couple of important bugfixes - one in JAX-WS (JAX-WS-1059) and another one related to JK listener with Apache + mod_ajp_proxy (GLASSFISH-18446). This release is already integrated in NetBeans 7.2 and you can download separately from here. Maven coordinates for this build are now also described on the download page. The following fragment in your pom.xml will allow you to invoke embedded-glassfish:run, embedded-glassfish:deploy, and other similar commands. <dependency>    <groupId>org.glassfish.embedded</groupId>     <artifactId>maven-embedded-glassfish-plugin</artifactId>     <version>3.1.2.2</version> </dependency> GlassFish Embedded Server Guide provide more details about setup etc. Similarly full platform or Web profile implementation of GlassFish can be included as a single JAR using <dependency>    <groupId>org.glassfish.main.extras</groupId>    <artifactId>glassfish-embedded-web</artifactId>    <version>3.1.2.2</version></dependency> Of course, you need to replace "glassfish-embedded-web" with "glassfish-embedded-all" to get complete platform.  The download page provide more details different bundles and complete maven coordinates. Or you can get started with the simple zip bundle as well.

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  • Isometric Screen View to World View

    - by Sleepy Rhino
    I am having trouble working out the math to transform the screen coordinates to the Grid coordinates. The code below is how far I have got but it is totally wrong any help or resources to fix this issue would be great, had a complete mind block with this for some reason. private Point ScreenToIso(int mouseX, int mouseY) { int offsetX = WorldBuilder.STARTX; int offsetY = WorldBuilder.STARTY; Vector2 startV = new Vector2(offsetX, offsetY); int mapX = offsetX - mouseX; int mapY = offsetY - mouseY + (WorldBuilder.tileHeight / 2); mapY = -1 * (mapY / WorldBuilder.tileHeight); mapX = (mapX / WorldBuilder.tileHeight) + mapY; return new Point(mapX, mapY); }

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  • How do you calculate UVW coordinates?

    - by Jenko
    I'm working on a 3d engine and I'm calculating UVT coordinates, where U and V represent pixels on the texture measured in 0-1, and T is: T = perspective / Z But I'm trying to use this perspective-correct triangle rasteriser, which requires a W, per vertex. How do I calculate the W for each vertex for the drawPerspectiveTexturedPolygon() function? Hint: The code comments refer to W as the "homogenous coordinate" ... does that mean anything?

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  • Calculating 2D (screen) coordinates from 3D positions in XNA 4.0

    - by NDraskovic
    I have a program that draws some items to the scene by loading their positions from a file. Now I want to place a Ray on the same location where the items are drawn. So my question is how can I calculate the position of the ray (it's 2D components) by using 3D coordinates of each particular item? The items don't move anywhere, so once they are placed they stay until the end of the programs execution. Thanks.

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  • 2D mouse coordinates from 3d object projection

    - by user17753
    Not entirely certain of the nomenclature here -- basically, after placing a model in world coordinates and setting up a 3D camera to look at it the model has been projected onto the screen in a 2D fashion. What I'd like to do is determine if the mouse is inside the projected view of the model. Is there a way to "unproject" in the XNA framework? Or what is this process called as, so that I can better search for it?

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  • MATLAB image corner coordinates & referncing to cell arrays

    - by James
    Hi, I am having some problems comparing the elements in different cell arrays. The context of this problem is that I am using the bwboundaries function in MATLAB to trace the outline of an image. The image is of a structural cross section and I am trying to find if there is continuity throughout the section (i.e. there is only one outline produced by the bwboundaries command). Having done this and found where the is more than one section traced (i.e. it is not continuous), I have used the cornermetric command to find the corners of each section. The code I have is: %% Define the structural section as a binary matrix (Image is an I-section with the web broken) bw(20:40,50:150) = 1; bw(160:180,50:150) = 1; bw(20:60,95:105) = 1; bw(140:180,95:105) = 1; Trace = bw; [B] = bwboundaries(Trace,'noholes'); %Traces the outer boundary of each section L = length(B); % Finds number of boundaries if L > 1 disp('Multiple boundaries') % States whether more than one boundary found end %% Obtain perimeter coordinates for k=1:length(B) %For all the boundaries perim = B{k}; %Obtains perimeter coordinates (as a 2D matrix) from the cell array end %% Find the corner positions C = cornermetric(bw); Areacorners = find(C == max(max(C))) % Finds the corner coordinates of each boundary [rowindexcorners,colindexcorners] = ind2sub(size(Newgeometry),Areacorners) % Convert corner coordinate indexes into subcripts, to give x & y coordinates (i.e. the same format as B gives) %% Put these corner coordinates into a cell array Cornerscellarray = cell(length(rowindexcorners),1); % Initialises cell array of zeros for i =1:numel(rowindexcorners) Cornerscellarray(i) = {[rowindexcorners(i) colindexcorners(i)]}; %Assigns the corner indicies into the cell array %This is done so the cell arrays can be compared end for k=1:length(B) %For all the boundaries found perim = B{k}; %Obtains coordinates for each perimeter Z = perim; % Initialise the matrix containing the perimeter corners Sectioncellmatrix = cell(length(rowindexcorners),1); for i =1:length(perim) Sectioncellmatrix(i) = {[perim(i,1) perim(i,2)]}; end for i = 1:length(perim) if Sectioncellmatrix(i) ~= Cornerscellarray Sectioncellmatrix(i) = []; %Gets rid of the elements that are not corners, but keeps them associated with the relevent section end end end This creates an error in the last for loop. Is there a way I can check whether each cell of the array (containing an x and y coordinate) is equal to any pair of coordinates in cornercellarray? I know it is possible with matrices to compare whether a certain element matches any of the elements in another matrix. I want to be able to do the same here, but for the pair of coordinates within the cell array. The reason I don't just use the cornercellarray cell array itself, is because this lists all the corner coordinates and does not associate them with a specific traced boundary.

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  • Interpreting Inkscape SVG path coordinates for HTML map

    - by tovare
    I needed some coordinates for a HTML MAP and tried to use inkskape by opening the image and just draw a path with my polygon coordinates. My document properties are set to 256 x 256 pixels and units: px When opening the svg file i get coordinates which are not immediately apparent. <path style="fill:none;stroke:#000000;stroke-width:1px;stroke-linecap:butt; stroke-linejoin:miter;stroke-opacity:1" d="m 23.864407,126.91525 3.254237, 44.47458 35.79661, 44.47458 71.593216, 19.52542 71.59322, -37.9661 22.77967, -72.67797 L 218.0339, 64 192,49.898305 l -32.54237, 8.677966 -18.44068, -35.79661 1.08474, -17.3559322 -71.593215,0 L 45.559322,34.711864 35. 79661,57.491525 5.4237288, 74.847458 6.5084746,101.9661 23.864407,126.91525 z" id="path2840" /> How can I get coordinates I can use ? The original image The SVG file from inkscape Link to SVG Progress: I tried a tool called InkscapeMap which looks promising and simple, but unfortunately it looks like it didn't work with this particular svn file. Solved! Saving the file as a Plain SVG solved the problem and InkscapeMap worked perfectly. (Btw. saving as an optimized svg caused a parsing error) Update 13.11 Using inkscapeMap 0.6 and Inkscape 0.48 i needed to uncheck relative coordinates in SVG output preferences. Also if you get a C error message, hunt down the polygon with a C in it, and redraw the polygon using the XML editor in inkscape. Update 25.11.2011 I modified the source to improve parsing. http://tovare.com/articles/createhtmlimagemapsusinginkscape/

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  • Generate random coordinates from area outside of a rectangle?

    - by Rockmaninoff
    Hi all, I'm working on a simple tutorial, and I'd like to randomly generate the positions of the red and green boxes in the accompanying images anywhere inside the dark gray area, but not in the white area. Are there any particularly elegant algorithms to do this? There are some hackish ideas I have that are really simple (continue to generate while the coordinates are not outside the inside rectangle, etc.), but I was wondering if anyone had come up with some neat solutions. Thanks for any help!

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  • Isometric - precise screen coordinates to isometric

    - by Rawrz
    I'm trying to translate mouse coords to precise isometric coords (I can already find the tile the mouse is over, but I want it to be more precise). I've tried several different methods but I seem to keep falling short. For drawing I use: batch.draw( texture, (y * tileWidth / 2) + (x * tileWidth / 2), (x * tileHeight / 2) - (y * tileHeight / 2)) This is what I currently use for figuring out a tile position: float xt = x + camPosition.x - (ScreenWidth/2) ; float yt = (ScreenHeight) - y + camPosition.y - (ScreenHeight/2); int tileY = Math.round((((xt) / tileWidth) - ((yt) / tileHeight))); int tileX = Math.round((((xt) / tileWidth) + ((yt) / tileHeight))- 1); I'm just wondering how I could update these to allow for more precise coordinates, instead of tile only. EDIT: Following what ccxvii said below, and removing the -1 from tileX, the object follows my mouse just like I had wanted. Just going to re-examine the math and figure out if that change will result in other messes =o

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  • Raycasting mouse coordinates to rotated object?

    - by SPL
    I am trying to cast a ray from my mouse to a plane at a specified position with a known width and length and height. I know that you can use the NDC (Normalized Device Coordinates) to cast ray but I don't know how can I detect if the ray actually hit the plane and when it did. The plane is translated -100 on the Y and rotated 60 on the X then translated again -100. Can anyone please give me a good tutorial on this? For a complete noob! I am almost new to matrix and vector transformations.

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  • Creating a curved mesh on inside of sphere based on texture image coordinates

    - by user5025
    In Blender, I have created a sphere with a panoramic texture on the inside. I have also manually created a plane mesh (curved to match the size of the sphere) that sits on the inside wall where I can draw a different texture. This is great, but I really want to reduce the manual labor, and do some of this work in a script -- like having a variable for the panoramic image, and coordinates of the area in the photograph that I want to replace with a new mesh. The hardest part of doing this is going to be creating a curved mesh in code that can sit on the inside wall of a sphere. Can anyone point me in the right direction?

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  • Android how to get opengl 3D coordinates in ontouch event

    - by Sandy
    I created a cube in opengl and it rotates in ontouch event. To to this I created a CustomSurfaceView as follows public class CustomSurfaceView extends GLSurfaceView { @Override public boolean onTouchEvent(MotionEvent e) { float x = e.getX() float y = e.getY(); } } Here x and y are screen coordinates. How can I get 3D coordinated from this? I have already looked gluProject and NeHe. But I dont know how to implement this in my project, it shows that there is no GLdouble,GLfloat class.

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  • How to set sprite source coordinates?

    - by ChaosDev
    I am creating own sprite drawer with DX11 on C++. Works fine but I dont know how to apply source rectangle to texture coordinates of rendering surface(for animation sprite sheets) //source = (0,0,32,64); //RECT D3DXVECTOR2 t0 = D3DXVECTOR2( 1.0f, 0.0f); D3DXVECTOR2 t1 = D3DXVECTOR2( 1.0f, 1.0f); D3DXVECTOR2 t2 = D3DXVECTOR2( 0.0f, 1.0f); D3DXVECTOR2 t3 = D3DXVECTOR2( 0.0f, 1.0f); D3DXVECTOR2 t4 = D3DXVECTOR2( 0.0f, 0.0f); D3DXVECTOR2 t5 = D3DXVECTOR2( 1.0f, 0.0f); VertexPositionColorTexture vertices[] = { { D3DXVECTOR3( dest.left+dest.right, dest.top, z),D3DXVECTOR4(1,1,1,1), t0}, { D3DXVECTOR3( dest.left+dest.right, dest.top+dest.bottom, z),D3DXVECTOR4(1,1,1,1), t1}, { D3DXVECTOR3( dest.left, dest.top+dest.bottom, z),D3DXVECTOR4(1,1,1,1), t2}, { D3DXVECTOR3( dest.left, dest.top+dest.bottom, z),D3DXVECTOR4(1,1,1,1), t3}, { D3DXVECTOR3( dest.left , dest.top, z),D3DXVECTOR4(1,1,1,1), t4}, { D3DXVECTOR3( dest.left+dest.right, dest.top, z),D3DXVECTOR4(1,1,1,1), t5}, };

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