Search Results

Search found 1603 results on 65 pages for 'coordinate transformation'.

Page 20/65 | < Previous Page | 16 17 18 19 20 21 22 23 24 25 26 27  | Next Page >

  • How to flip the origin.y of an CGRect?

    - by mystify
    I'm trying to draw an UIImageView, but the frame's origin is wrong when I flip the coordinate system for drawing not upside-down. CGRect imgRect = imgView.frame; imgRect.origin.y += 10.0f; CGContextTranslateCTM(context, 0.0f, imgRect.size.height); CGContextScaleCTM(context, 1.0f, -1.0f); CGContextDrawImage(context, imgRect, imgView.image.CGImage); like you can see I move the image down by 10, but instead of going down by 10, it goes UP by 10. That's completely unlogical since I had actually inverted the wrong coordinate system to look exactly like the one in UIView, right? What to do about it?

    Read the article

  • Java :moving ball with angle ?

    - by Meko
    Hi all.I am started to learn game physics and I am trying to move ball with an angle.But it does not change its angle .Java coordinate sydstem is a little different and i think my problem is there.Here my codes this is for calculating x and y speed scale_X= Math.sin(angle); scale_Y=Math.cos(angle); velosity_X=(speed*scale_X); velosity_Y=(speed*scale_Y); This is for moving ball in run() function ball.posX =ball.posX+(int)velosity_X; ball.posY=ball.posY+(int)velosity_Y; I used (int)velosity_X and (int)velosity_Y because in ball class I draw object g.drawOval(posX, posX, width, height); and here g.drawOval requares int.I dont know is it problem or not. Also if I use angle 30 it goes +X and +Y but if I use angle 35 it goes -X and -Y. I didnot figure out how work coordinate system on java.

    Read the article

  • MATLAB: Convert two array to a sparse matrix

    - by CziX
    I'm looking for an a command or trick to convert two arrays to a sparse matrix. The two arrays contain x-values and y-values, which gives a coordinate in the cartesian coordinate system. I want to group the coordinates, which if the value is between some value on the x-axes and the y-axes. % MATLAB x_i = find(x > 0.1 & x < 0.9); y_i = find(y > 0.4 & y < 0.8); %Then I want to find indicies which are located in both x_i and y_i Is there an easy way to this little trick?

    Read the article

  • MKMapView not centered on pin

    - by Val Smith
    Hi all, I have an mkmapview that i'm currently adding pins to, but for some reason when I call [mapView setRegion:[detailItem coordinateRegion] animated:YES]; the pin is off-centered (toward the right side of the screen) on the map. Here is the code for [deailItem coordinateRegion]: - (MKCoordinateRegion)coordinateRegion { MKCoordinateRegion region = { {0.0, 0.0 }, { 0.0, 0.0 } }; region.center = self.coordinate; region.span.longitudeDelta = 0.0075f; region.span.latitudeDelta = 0.0075f; return (region); } I'm setting the coordinateRegion's center to the object's x,y coordinate, so why is it off-center on the map? I feel like there's something I'm missing here... ::Val::

    Read the article

  • OpenLayers Projections.

    - by Jenny
    I can succesfully do: point.transform(new OpenLayers.Projection("EPSG:900913"), new OpenLayers.Projection("EPSG:4326")); To a point that is in the google format (in meters), but when I want to do the reverse: point.transform(new OpenLayers.Projection("EPSG:4326"), new OpenLayers.Projection("EPSG:900913")); to a point that is in 4326 (regular lat/lon format), I am having some issues. Any negative value seems to become NaN (not a number) when I do the transformation. Is there something about the transformation in reverse that I don't understand? Edit: Even worse, when I have no negative values, the coordinates seem off. I am getting the coordinates by drawing a square on the screen, then saving those coordinates to a database and loading them later. I can draw a square near the tip of africa (positive coordinates), and then when it loads it's near the top of africa, in the atlantic ocean. I'm definitely doing something wrong....

    Read the article

  • Erlang list comprehension, traversing two lists and excluding values

    - by ErJab
    I need to generate a set of coordinates in Erlang. Given one coordinate, say (x,y) I need to generate (x-1, y-1), (x-1, y), (x-1, y+1), (x, y-1), (x, y+1), (x+1, y-1), (x+1, y), (x+1, y+1). Basically all surrounding coordinates EXCEPT the middle coordinate (x,y). To generate all the nine coordinates, I do this currently: [{X,Y} || X<-lists:seq(X-1,X+1), Y<-lists:seq(Y-1,Y+1)] But this generates all the values, including (X,Y). How do I exclude (X,Y) from the list using filters in the list comprehension?

    Read the article

  • Zoom on userLocation

    - by Marco
    Hello, how can I zoom the map on my userLocation automatically in my app? I have the following code to zomm in the map but i must zoom on the userLocation and the following code zooms always to africa? MKCoordinateRegion zoomIn = mapView.region; zoomIn.span.latitudeDelta *= 0.2; zoomIn.span.longitudeDelta *= 0.2; zoomIn.center.latitude = mapView.userLocation.location.coordinate.latitude; zoomIn.center.longitude = mapView.userLocation.location.coordinate.longitude; [mapView setRegion:zoomIn animated:YES];

    Read the article

  • Howto project a planar polygon on a plane in 3d-space

    - by sum1stolemyname
    I want to project my Polygon along a vector to a plane in 3d Space. I would preferably use a single transformation matrix to do this, but I don't know how to build a matrix of this kind. Given the plane's parameters (ax+by+cz+d), the world coordinates of my Polygon. As stated in the the headline, all vertices of my polygon lie in another plane. the direction vector along which to project my Polygon (currently the polygon's plane's normal vector) goal -a 4x4 transformation matrix which performs the required projection, or some insight on how to construct one myself

    Read the article

  • How to fix rotations in a Rubik's Cube?

    - by Eindbaas
    I'm trying to create a Rubik's Cube in Flash & Papervision and i'm really stuck here. I'm up to the point where i can rotate any plane of cubes once, but after that...it's messed up because all local coordinate systems are messy. I dont really know where to go from here, can anybody give any advice on what do do? I'm not looking for 'read about transformation matrices', i know i should (and i am doing that), but i'm not really sure what to look for. My idea is that, after each rotation, i should fix each coordinate system of each cube again, but i have no idea how. Any hints on what i want to achieve (in words), and why, are much appreciated. http://dl.dropbox.com/u/250155/rubik/main.html (you can press the K key once ;) )

    Read the article

  • TomTom integration does not work anymore?

    - by viezel
    I did for 6 months back an integration for TomTom navigation with URL Scheme. This worked out great, but then TomTom updated their iPhone app to version 1.10. Since then the integration does not work. Anyone tried the same? My iOS integration looks like this: (Titanium appcelerator) var lat = dbDataArray[chosenRoute]["lat"].toFixed(6); //geo coordinate var lng = dbDataArray[chosenRoute]["lng"].toFixed(6); //geo coordinate //open navigation var url = "tomtomhome://geo:action=navigateto&lat="+lat+"&long="+lng+"&name=name"; //show map //var url = "tomtomhome://geo:action=show&lat="+lat+"&long="+lng+"&name=name"; if (Ti.Platform.canOpenURL(url)) { Ti.Platform.openURL(url); }else { alert('Please install TomTom Navigation app'); } Note: I cannot get TomTom to tell me what has changed.

    Read the article

  • Encode complex number as RGB pixel and back

    - by Vi
    How is it better to encode a complex number into RGB pixel and vice versa? Probably (logarithm of) an absolute value goes to brightness and an argument goes to hue. Desaturated pixes should receive randomized argument in reverse transformation. Something like: 0 - (0,0,0) 1 - (255,0,0) -1 - (0,255,255) 0.5 - (128,0,0) i - (255,255,0) -i - (255,0,255) (0,0,0) - 0 (255,255,255) - e^(i * random) (128,128,128) - 0.5 * e^(i *random) (0,128,128) - -0.5 Are there ready-made formulas for that? Edit: Looks like I just need to convert RGB to HSB and back. Edit 2: Existing RGB - HSV converter fragment: if (hsv.sat == 0) { hsv.hue = 0; // ! return hsv; } I don't want 0. I want random. And not just if hsv.sat==0, but if it is lower that it should be ("should be" means maximum saturation, saturation that is after transformation from complex number).

    Read the article

  • I've got my 2D/3D conversion working perfectly, how to do perspective

    - by user346992
    Although the context of this question is about making a 2d/3d game, the problem i have boils down to some math. Although its a 2.5D world, lets pretend its just 2d for this question. // xa: x-accent, the x coordinate of the projection // mapP: a coordinate on a map which need to be projected // _Dist_ values are constants for the projection, choosing them correctly will result in i.e. an isometric projection xa = mapP.x * xDistX + mapP.y * xDistY; ya = mapP.x * yDistX + mapP.y * yDistY; xDistX and yDistX determine the angle of the x-axis, and xDistY and yDistY determine the angle of the y-axis on the projection (and also the size of the grid, but lets assume this is 1-pixel for simplicity). x-axis-angle = atan(yDistX/xDistX) y-axis-angle = atan(yDistY/yDistY) a "normal" coordinate system like this --------------- x | | | | | y has values like this: xDistX = 1; yDistX = 0; xDistY = 0; YDistY = 1; So every step in x direction will result on the projection to 1 pixel to the right end 0 pixels down. Every step in the y direction of the projection will result in 0 steps to the right and 1 pixel down. When choosing the correct xDistX, yDistX, xDistY, yDistY, you can project any trimetric or dimetric system (which is why i chose this). So far so good, when this is drawn everything turns out okay. If "my system" and mindset are clear, lets move on to perspective. I wanted to add some perspective to this grid so i added some extra's like this: camera = new MapPoint(60, 60); dx = mapP.x - camera.x; // delta x dy = mapP.y - camera.y; // delta y dist = Math.sqrt(dx * dx + dy * dy); // dist is the distance to the camera, Pythagoras etc.. all objects must be in front of the camera fac = 1 - dist / 100; // this formula determines the amount of perspective xa = fac * (mapP.x * xDistX + mapP.y * xDistY) ; ya = fac * (mapP.x * yDistX + mapP.y * yDistY ); Now the real hard part... what if you got a (xa,ya) point on the projection and want to calculate the original point (x,y). For the first case (without perspective) i did find the inverse function, but how can this be done for the formula with the perspective. May math skills are not quite up to the challenge to solve this. ( I vaguely remember from a long time ago mathematica could create inverse function for some special cases... could it solve this problem? Could someone maybe try?)

    Read the article

  • Implementing "Generator" support in a custom language

    - by Roger Alsing
    I've got a bit of fettish for language design and I'm currently playing around with my own hobby language. (http://rogeralsing.com/2010/04/14/playing-with-plastic/) One thing that really makes my mind bleed is "generators" and the "yield" keyword. I know C# uses AST transformation to transform enumerator methods into statemachines. But how does it work in other languages? Is there any way to get generator support in a language w/o AST transformation? e.g. Does languages like Python or Ruby resort to AST transformations to solve this to? (The question is how generators are implemented under the hood in different languages, not how to write a generator in one of them)

    Read the article

  • Rotation towards an object in 3d space

    - by retoucher
    hello, i have two coordinates on a 2d plane in 3d space, and am trying to rotate one coordinate (a vector) to face the other coordinate. my vertical axis is the y-axis, so if both of the coordinates are located flat on the 2d plane, they would both have a y-axis of 0, and their x and z coordinates determine their position length/width-wise on the plane. right now, i'm calculating the angle like so (language agnostic): angle = atan2(z2-z1,x2-x1); and am rotating/translating in space like so: pushMatrix(); rotateY(angle); popMatrix(); this doesn't seem to be working though. are my calculations/process correct?

    Read the article

  • Find top "n" nearby coordinates.

    - by John Hamelink
    I have a coordinate. I want to find the top "n" (n being a variable value) nearest coordinates out of several thousand rows stored on a MySQL database. I also want to be able to define maximum and minimum distances between the coordinate in question and the coordinates in the database. How best am I to go about this? Would it be bonkers to use PHP as I understand the syntax much better than MySQL? If I use a MySQL function, how do I move it between databases if I choose to switch servers? How is it stored? Lastly, what is the most efficient method of getting through all these coordinates accurately - the coordinates are all relatively close to one another? Thanks for your time, John.

    Read the article

  • Can't access annotation property of subclassed uibutton - editted

    - by Tzur Gazit
    Below is my original question. I kept investigating and found out that the type of the button I allocate is of type UIButton instead of the subclassed type CustomButton. the capture below is the allocation of the button and connection to target. I break immediately after the allocation and check the button type (po rightButton at the debugger console). It's turned out tht the type is UIButton instead of CustomButton. CustomButton* rightButton = [CustomButton buttonWithType:UIButtonTypeDetailDisclosure]; [rightButton addTarget:self action:@selector(showDetails:) forControlEvents:UIControlEventTouchUpInside]; I have a mapView to which I add annotations. The pin's callout have a button (rightCalloutAccessoryView). In order to be able to display various information when the button is pushed, i've subclassed uibutton and added a class called "Annotation". @interface CustomButton : UIButton { NSIndexPath *indexPath; Annotation *mAnnotation; } @property (nonatomic, retain) NSIndexPath *indexPath; @property (nonatomic, copy) Annotation *mAnnotation; - (id) setAnnotation2:(Annotation *)annotation; @end Here is "Annotation": @interface Annotation : NSObject <MKAnnotation> { CLLocationCoordinate2D coordinate; NSString *mPhotoID; NSString *mPhotoUrl; NSString *mPhotoName; NSString *mOwner; NSString *mAddress; } @property (nonatomic, assign) CLLocationCoordinate2D coordinate; @property (nonatomic, copy) NSString *mPhotoID; @property (nonatomic, copy) NSString *mPhotoUrl; @property (nonatomic, copy) NSString *mPhotoName; @property (nonatomic, copy) NSString *mOwner; @property (nonatomic, copy) NSString *mAddress; - (id) initWithCoordinates:(CLLocationCoordinate2D)coordinate; - (id) setPhotoId:(NSString *)id url:(NSString *)url owner:(NSString *)owner address:(NSString *)address andName:(NSString *)name; @end I want to set the annotation property of the uibutton at - (MKAnnotationView *)mapView:(MKMapView *)pMapView viewForAnnotation:(id )annotation, in order to refer to it at the button push handler (-(IBAction) showDetails:(id)sender). The problem is that I can't set the annotation property of the button. I get the following message at run time: 2010-04-27 08:15:11.781 HotLocations[487:207] *** -[UIButton setMAnnotation:]: unrecognized selector sent to instance 0x5063400 2010-04-27 08:15:11.781 HotLocations[487:207] *** Terminating app due to uncaught exception 'NSInvalidArgumentException', reason: '*** -[UIButton setMAnnotation:]: unrecognized selector sent to instance 0x5063400' 2010-04-27 08:15:11.781 HotLocations[487:207] Stack: ( 32080987, 2472563977, 32462907, 32032374, 31884994, 55885, 30695992, 30679095, 30662137, 30514190, 30553882, 30481385, 30479684, 30496027, 30588515, 63333386, 31865536, 31861832, 40171029, 40171226, 2846639 ) I appreciate the help. Tzur.

    Read the article

  • Stream/string/bytearray transformations in Python 3

    - by Craig McQueen
    Python 3 cleans up Python's handling of Unicode strings. I assume as part of this effort, the codecs in Python 3 have become more restrictive, according to the Python 3 documentation compared to the Python 2 documentation. For example, codecs that conceptually convert a bytestream to a different form of bytestream have been removed: base64_codec bz2_codec hex_codec And codecs that conceptually convert Unicode to a different form of Unicode have also been removed (in Python 2 it actually went between Unicode and bytestream, but conceptually it's really Unicode to Unicode I reckon): rot_13 My main question is, what is the "right way" in Python 3 to do what these removed codecs used to do? They're not codecs in the strict sense, but "transformations". But the interface and implementation would be very similar to codecs. I don't care about rot_13, but I'm interested to know what would be the "best way" to implement a transformation of line ending styles (Unix line endings vs Windows line endings) which should really be a Unicode-to-Unicode transformation done before encoding to byte stream, especially when UTF-16 is being used, as discussed this other SO question.

    Read the article

  • Color space - RGB and YCbCr question

    - by HardCoder1986
    Hello! I am now trying to understand how JPEG encoding works and everything seems fine except the color transformation part. Before attempting to do a DCT in JPEG algorithm, the image is transformed into YCbCr color space. To me this essentially means that we just (comparing to initial RGB image) take a chunk of color information and dispose it while applying the RGB -> YCbCr transformation. So, our encoding steps look generally like RGB -> YCbCr -> DCT -> Huffman. The decoding means inversing this process. And my question is - why does the image (for example, created and exported to JPEG) remain the same in terms of color, although we have to make inverse YCbCr -> RGB transform. Where does the disposed part of color information comes from or how is it handled?

    Read the article

  • Triangular bounding volumes

    - by Cheery
    I've come up with an alternative for beziers that might be easier to ray-trace, perhaps even though a plain vertex shader. Though there's missing a piece. I need to find the parametric surface equation from the surface normals I have for edge vertices. I also have to know it's peak and valley so I can constraint the depth of my bounding triangle. Image explains the overall idea: I build a bounding-volume from a control triangle. Then apply a function to each parametric coordinate of the triangle (s+t+u=1 where s,t,u = 0) to get the height coordinate for that certain point. Simply put, it produces a procedurally generated height-map for the triangle's surface. I just need to find a function that generates the height-map so I can make it work.

    Read the article

  • Python: Repeat elements in a list comprehension?

    - by User
    I have the following list comprehension which returns a list of coordinate objects for each location. coordinate_list = [Coordinates(location.latitude, location.longitude) for location in locations] This works. Now suppose the location object has a number_of_times member. I want a list comprehension to generate n Coordinate objects where n is the number_of_times for the particular location. So if a location has number_of_times = 5 then the coordinates for that location will be repeated 5 times in the list. (Maybe this is a case for a for-loop but I'm curious if it can be done via list comprehensions)

    Read the article

  • ant - trying to copy to /lib/endorsed, library is not available in windows 7 to the next task

    - by kfox
    On Windows 7 I have an ant target that copies a xalan library into the jdk endorsed directory so that the next xslt transformation task can occur. The first time that the ant target runs, the xslt transformation fails. The second time it runs the jar file is already in the correct place and the xslt tranformation succeeds. The first time that the ant target runs, it looks like the file copied successfully. It feels like a timing issue, but I don't know what I can do to get around it. Here is my copy task: <mkdir dir="${java.home}\lib\endorsed"/> <copy file="${basedir}\xalan.jar" tofile="${java.home}\lib\endorsed\xalan.jar"/> Has anyone seen anything like this before?

    Read the article

  • draw line with php using coordinates from txt file

    - by netmajor
    I have file A2.txt with coordinate x1,y1,x2,y2 in every line like below : 204 13 225 59 225 59 226 84 226 84 219 111 219 111 244 192 244 192 236 209 236 209 254 223 254 223 276 258 276 258 237 337 in my php file i have that code. This code should take every line and draw line with coordinate from line. But something was wrong cause nothing was draw :/: <?php $plik = fopen("A2.txt", 'r') or die("blad otarcia"); while(!feof($plik)) { $l = fgets($plik,20); $k = explode(' ',$l); imageline ( $mapa , $k[0] , $k[1] , $k[2] , $k[3] , $kolor ); } imagejpeg($mapa); imagedestroy($mapa); fclose($plik) ; ?> If I use imagejpeg and imagedestroy in while its only first line draw. What to do to draw every line ?? Please help :)

    Read the article

  • Question about Convolutional neural network.

    - by Nhu Phuong
    I readed few book and acticles about Convolutional neural network, it seem I understand the concept but I don't know how to put it up like in image below: from 28x28 normalized pixel INPUT we get 4 feature map 24x24. but how to get them ? size the INPUT image ? or perform image transformation? but what kind of transformation? or cut up the input image to 4 piece 24x24 by 4 corner? I don't understand the process to me it seem they cut up or resize the image to more smaller at each step. please help thanks.

    Read the article

< Previous Page | 16 17 18 19 20 21 22 23 24 25 26 27  | Next Page >