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  • Image rotation algorithm

    - by Stefano Driussi
    I'm looking for an algorithm that rotates an image by some degrees (input). public Image rotateImage(Image image, int degrees) (Image instances could be replaced with int[] containing each pixel RGB values, My problem is that i need to implement it for a JavaME MIDP 2.0 project so i must use code runnable on JVM prior to version 1.5 Can anyone help me out with this ? EDIT: I forgot to mention that i don't have SVG APIs available and that i need a method to rotate by arbitrary degree other than 90 - 180- 270 Also, no java.awt.* packages are available on MIDP 2.0

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  • Need help programming with Mclauren series and Taylor series!

    - by user352258
    Ok so here's what i have so far: #include <stdio.h> #include <math.h> //#define PI 3.14159 int factorial(int n){ if(n <= 1) return(1); else return(n * factorial(n-1)); } void McLaurin(float pi){ int factorial(int); float x = 42*pi/180; int i, val=0, sign; for(i=1, sign=-1; i<11; i+=2){ sign *= -1; // alternate sign of cos(0) which is 1 val += (sign*(pow(x, i)) / factorial(i)); } printf("\nMcLaurin of 42 = %d\n", val); } void Taylor(float pi){ int factorial(int); float x; int i; float val=0.00, sign; float a = pi/3; printf("Enter x in degrees:\n"); scanf("%f", &x); x=x*pi/180.0; printf("%f",x); for(i=0, sign=-1.0; i<2; i++){ if(i%2==1) sign *= -1.0; // alternate sign of cos(0) which is 1 printf("%f",sign); if(i%2==1) val += (sign*sin(a)*(pow(x-a, i)) / factorial(i)); else val += (sign*cos(a)*(pow(x-a, i)) / factorial(i)); printf("%d",factorial(i)); } printf("\nTaylor of sin(%g degrees) = %d\n", (x*180.0)/pi, val); } main(){ float pi=3.14159; void McLaurin(float); void Taylor(float); McLaurin(pi); Taylor(pi); } and here's the output: McLaurin of 42 = 0 Enter x in degrees: 42 0.733038-1.00000011.0000001 Taylor of sin(42 degrees) = -1073741824 I suspect the reason for these outrageous numbers goes with the fact that I mixed up my floats and ints? But i just cant figure it out...!! Maybe its a math thing, but its never been a strength of mine let alone program with calculus. Also the Mclaurin fails, how does it equal zero? WTF! Please help correct my noobish code. I am still a beginner...

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  • How can I flip/rotate a PrintDocument in .NET ?

    - by Simon_Weaver
    I have a document that I want to be flipped / rotated 180 degrees when printed. (This is due to the orientation of label stock in the printer). There is a property PrintDocument.PrinterSettings.LandscapeAngle but it is read only. I think this property is influenced by the printer driver and therefore not 'settable'. Is there a nice way i can rotate the print by 180 degrees without having to do anything too nasty?

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  • gps orientation in android

    - by klaus-vlad
    Hi I know that the function Location.getBearing() returns the bearing if any when in move public void onLocationChanged(Location lastLocation) { int bearing=lastLocation.getBearing() } ,so now bearing might be , 170 degrees..but, I'd like to know if there is anything in android that will give me the direction of the orientation (ex for 170 degrees , the direction is : SSE south, south east)

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  • How can I "flip" an image using PHP?

    - by learner
    Here's what I've tried: $image = "images/20100609124341Chrysanthemum.jpg"; $degrees = 40; // Content type header('Content-type: image/jpeg'); // Load $source = imagecreatefromjpeg($filename); // Rotate $rotate = imagerotate($source, $degrees, 0); // Output imagejpeg($rotate); ...But I get no output. Can anyone tell me what's wrong with this?

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  • sidewaysfire and twosided

    - by hanno
    I try two use sidewaysfigure from the rotating package in the twosided memoir class. The resulting figures look correct in the pdf that is generated, with the page rotated by 90 degrees. However, when I print the document (on linux, using CUPS), some of the pages with a sidewaysfigure are upside down (rotated by 180 degreeS).

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  • How to find vector for the quaternion from X Y Z rotations

    - by can poyrazoglu
    I am creating a very simple project on OpenGL and I'm stuck with rotations. I am trying to rotate an object indepentdently in all 3 axes: X, Y, and Z. I've had sleepless nights due to the "gimbal lock" problem after rotating about one axis. I've then learned that quaternions would solve my problem. I've researched about quaternions and implementd it, but I havent't been able to convert my rotations to quaternions. For example, if I want to rotate around Z axis 90 degrees, I just create the {0,0,1} vector for my quaternion and rotate it around that axis 90 degrees using the code here: http://iphonedevelopment.blogspot.com/2009/06/opengl-es-from-ground-up-part-7_04.html (the most complicated matrix towards the bottom) That's ok for one vector, but, say, I first want to rotate 90 degrees around Z, then 90 degrees around X (just as an example). What vector do I need to pass in? How do I calculate that vector. I am not good with matrices and trigonometry (I know the basics and the general rules, but I'm just not a whiz) but I need to get this done. There are LOTS of tutorials about quaternions, but I seem to understand none (or they don't answer my question). I just need to learn to construct the vector for rotations around more than one axis combined. UPDATE: I've found this nice page about quaternions and decided to implement them this way: http://www.cprogramming.com/tutorial/3d/quaternions.html Here is my code for quaternion multiplication: void cube::quatmul(float* q1, float* q2, float* resultRef){ float w = q1[0]*q2[0] - q1[1]*q2[1] - q1[2]*q2[2] - q1[3]*q2[3]; float x = q1[0]*q2[1] + q1[1]*q2[0] + q1[2]*q2[3] - q1[3]*q2[2]; float y = q1[0]*q2[2] - q1[1]*q2[3] + q1[2]*q2[0] + q1[3]*q2[1]; float z = q1[0]*q2[3] + q1[1]*q2[2] - q1[2]*q2[1] + q1[3]*q2[0]; resultRef[0] = w; resultRef[1] = x; resultRef[2] = y; resultRef[3] = z; } Here is my code for applying a quaternion to my modelview matrix (I have a tmodelview variable that is my target modelview matrix): void cube::applyquat(){ float& x = quaternion[1]; float& y = quaternion[2]; float& z = quaternion[3]; float& w = quaternion[0]; float magnitude = sqrtf(w * w + x * x + y * y + z * z); if(magnitude == 0){ x = 1; w = y = z = 0; }else if(magnitude != 1){ x /= magnitude; y /= magnitude; z /= magnitude; w /= magnitude; } tmodelview[0] = 1 - (2 * y * y) - (2 * z * z); tmodelview[1] = 2 * x * y + 2 * w * z; tmodelview[2] = 2 * x * z - 2 * w * y; tmodelview[3] = 0; tmodelview[4] = 2 * x * y - 2 * w * z; tmodelview[5] = 1 - (2 * x * x) - (2 * z * z); tmodelview[6] = 2 * y * z - 2 * w * x; tmodelview[7] = 0; tmodelview[8] = 2 * x * z + 2 * w * y; tmodelview[9] = 2 * y * z + 2 * w * x; tmodelview[10] = 1 - (2 * x * x) - (2 * y * y); tmodelview[11] = 0; glMatrixMode(GL_MODELVIEW); glPushMatrix(); glLoadMatrixf(tmodelview); glMultMatrixf(modelview); glGetFloatv(GL_MODELVIEW_MATRIX, tmodelview); glPopMatrix(); } And my code for rotation (that I call externally), where quaternion is a class variable of the cube: void cube::rotatex(int angle){ float quat[4]; float ang = angle * PI / 180.0; quat[0] = cosf(ang / 2); quat[1] = sinf(ang/2); quat[2] = 0; quat[3] = 0; quatmul(quat, quaternion, quaternion); applyquat(); } void cube::rotatey(int angle){ float quat[4]; float ang = angle * PI / 180.0; quat[0] = cosf(ang / 2); quat[1] = 0; quat[2] = sinf(ang/2); quat[3] = 0; quatmul(quat, quaternion, quaternion); applyquat(); } void cube::rotatez(int angle){ float quat[4]; float ang = angle * PI / 180.0; quat[0] = cosf(ang / 2); quat[1] = 0; quat[2] = 0; quat[3] = sinf(ang/2); quatmul(quat, quaternion, quaternion); applyquat(); } I call, say rotatex, for 10-11 times for rotating only 1 degree, but my cube gets rotated almost 90 degrees after 10-11 times of 1 degree, which doesn't make sense. Also, after calling rotation functions in different axes, My cube gets skewed, gets 2 dimensional, and disappears (a column in modelview matrix becomes all zeros) irreversibly, which obviously shouldn't be happening with a correct implementation of the quaternions.

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  • Quaternion dfference + time --> angular velocity (gyroscope in physics library)

    - by AndrewK
    I am using Bullet Physic library to program some function, where I have difference between orientation from gyroscope given in quaternion and orientation of my object, and time between each frame in milisecond. All I want is set the orientation from my gyroscope to orientation of my object in 3D space. But all I can do is set angular velocity to my object. I have orientation difference and time, and from that I calculate vector of angular velocity [Wx,Wy,Wz] from that formula: W(t) = 2 * dq(t)/dt * conj(q(t)) My code is: btQuaternion diffQuater = gyroQuater - boxQuater; btQuaternion conjBoxQuater = gyroQuater.inverse(); btQuaternion velQuater = ((diffQuater * 2.0f) / d_time) * conjBoxQuater; And everything works well, till I get: 1 rotating around Y axis, angle about 60 degrees, then I have these values in 2 critical frames: x: -0.013220 y: -0.038050 z: -0.021979 w: -0.074250 - diffQuater x: 0.120094 y: 0.818967 z: 0.156797 w: -0.538782 - gyroQuater x: 0.133313 y: 0.857016 z: 0.178776 w: -0.464531 - boxQuater x: 0.207781 y: 0.290452 z: 0.245594 - diffQuater -> euler angles x: 3.153619 y: -66.947929 z: 175.936615 - gyroQuater -> euler angles x: 4.290697 y: -57.553043 z: 173.320053 - boxQuater -> euler angles x: 0.138128 y: 2.823307 z: 1.025552 w: 0.131360 - velQuater d_time: 0.058000 x: 0.211020 y: 1.595124 z: 0.303650 w: -1.143846 - diffQuater x: 0.089518 y: 0.771939 z: 0.144527 w: -0.612543 - gyroQuater x: -0.121502 y: -0.823185 z: -0.159123 w: 0.531303 - boxQuater x: nan y: nan z: nan - diffQuater -> euler angles x: 2.985240 y: -76.304405 z: -170.555054 - gyroQuater -> euler angles x: 3.269681 y: -65.977966 z: 175.639420 - boxQuater -> euler angles x: -0.730262 y: -2.882153 z: -1.294721 w: 63.325996 - velQuater d_time: 0.063000 2 rotating around X axis, angle about 120 degrees, then I have these values in 2 critical frames: x: -0.013045 y: -0.004186 z: -0.005667 w: -0.022482 - diffQuater x: -0.848030 y: -0.187985 z: 0.114400 w: 0.482099 - gyroQuater x: -0.834985 y: -0.183799 z: 0.120067 w: 0.504580 - boxQuater x: 0.036336 y: 0.002312 z: 0.020859 - diffQuater -> euler angles x: -113.129463 y: 0.731925 z: 25.415056 - gyroQuater -> euler angles x: -110.232368 y: 0.860897 z: 25.350458 - boxQuater -> euler angles x: -0.865820 y: -0.456086 z: 0.034084 w: 0.013184 - velQuater d_time: 0.055000 x: -1.721662 y: -0.387898 z: 0.229844 w: 0.910235 - diffQuater x: -0.874310 y: -0.200132 z: 0.115142 w: 0.426933 - gyroQuater x: 0.847352 y: 0.187766 z: -0.114703 w: -0.483302 - boxQuater x: -144.402298 y: 4.891629 z: 71.309158 - diffQuater -> euler angles x: -119.515343 y: 1.745076 z: 26.646086 - gyroQuater -> euler angles x: -112.974533 y: 0.738675 z: 25.411509 - boxQuater -> euler angles x: 2.086195 y: 0.676526 z: -0.424351 w: 70.104248 - velQuater d_time: 0.057000 2 rotating around Z axis, angle about 120 degrees, then I have these values in 2 critical frames: x: -0.000736 y: 0.002812 z: -0.004692 w: -0.008181 - diffQuater x: -0.003829 y: 0.012045 z: -0.868035 w: 0.496343 - gyroQuater x: -0.003093 y: 0.009232 z: -0.863343 w: 0.504524 - boxQuater x: -0.000822 y: -0.003032 z: 0.004162 - diffQuater -> euler angles x: -1.415189 y: 0.304210 z: -120.481873 - gyroQuater -> euler angles x: -1.091881 y: 0.227784 z: -119.399445 - boxQuater -> euler angles x: 0.159042 y: 0.169228 z: -0.754599 w: 0.003900 - velQuater d_time: 0.025000 x: -0.007598 y: 0.024074 z: -1.749412 w: 0.968588 - diffQuater x: -0.003769 y: 0.012030 z: -0.881377 w: 0.472245 - gyroQuater x: 0.003829 y: -0.012045 z: 0.868035 w: -0.496343 - boxQuater x: -5.645197 y: 1.148993 z: -146.507187 - diffQuater -> euler angles x: -1.418294 y: 0.270319 z: -123.638245 - gyroQuater -> euler angles x: -1.415183 y: 0.304208 z: -120.481873 - boxQuater -> euler angles x: 0.017498 y: -0.013332 z: 2.040073 w: 148.120056 - velQuater d_time: 0.027000 The problem is the most visible in diffQuater - euler angles vector. Can someone tell me why it is like that? and how to solve that problem? All suggestions are welcome.

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  • How would you solve this graph theory handshake problem in python?

    - by Zachary Burt
    I graduated college last year with a degree in Psychology, but I also took a lot of math for fun. I recently got the book "Introductory Graph Theory" by Gary Chartrand to brush up on my math and have some fun. Here is an exercise from the book that I'm finding particularly befuddling: Suppose you and your husband attended a party with three other married couples. Several handshakes took place. No one shook hands with himself (or herself) or with his (or her) spouse, and no one shook hands with the same person more than once. After all the handshaking was completed, suppose you asked each person, including your husband, how many hands he or she had shaken. Each person gave a different answer. a) How many hands did you shake? b) How many hands did your husband shake? Now, I've been reasoning about this for a while, and trying to draw sample graphs that could illustrate a solution, but I'm coming up empty-handed. My logic is this: there are 8 different vertices in the graph, and 7 of them have different degrees. The values for the degrees must therefore be 0, 1, 2, 3, 4, 5, 6, and x. The # of degrees for one married couple is (0, 6). Since all graphs have an even number of odd vertices, x must be either 5, 3, or 1. What's your solution to this problem? And, if you could solve it in python, how would you do it? (python is fun.) Cheers.

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  • How would you solve this graph theory handshake problem in python?

    - by Zachary Burt
    I graduated college last year with a degree in Psychology, but I also took a lot of math for fun. I recently got the book "Introductory Graph Theory" by Gary Chartrand to brush up on my math and have some fun. Here is an exercise from the book that I'm finding particularly befuddling: Suppose you and your husband attended a party with three other married couples. Several handshakes took place. No one shook hands with himself (or herself) or with his (or her) spouse, and no one shook hands with the same person more than once. After all the handshaking was completed, suppose you asked each person, including your husband, how many hands he or she had shaken. Each person gave a different answer. a) How many hands did you shake? b) How many hands did your husband shake? Now, I've been reasoning about this for a while, and trying to draw sample graphs that could illustrate a solution, but I'm coming up empty-handed. My logic is this: there are 8 different vertices in the graph, and 7 of them have different degrees. The values for the degrees must therefore be 0, 1, 2, 3, 4, 5, 6, and x. The # of degrees for one married couple is (0, 6). Since all graphs have an even number of odd vertices, x must be either 5, 3, or 1. What's your solution to this problem? And, if you could solve it in python, how would you do it? (python is fun.) Cheers.

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  • Spatial Rotation in Gmod Expression2.

    - by Fascia
    I'm using expression2 to program behavior in Garry's mod (http://wiki.garrysmod.com/?title=Wire_Expression2) Okay so, to set the precedent. In Gmod I have a block and I am at a complete loss of how to get it to rotate around the 3 up, down and right vectors (Which are local. ie; if I pitch it 45 degrees the forward vector is 0.707, 0.707, 0). Essentially, From the 3 vectors I'd like to be able to get local Pitch/Roll/Yaw. By Local Pitch Roll Yaw I mean that they are completely independent of one another allowing true 3d rotation. So for example; if I place my craft so its nose is parallel to the floor the X,Y,Z would be 0,0,0. If I turn it parallel to the floor (World and Local Yaw) 90 degrees it's now 0, 0, 90. If I then pitch it (World Roll, Local Pitch) it 180 degrees it's now 180, 0, 90. I've already explored quaternions however I don't believe I should post my code here as I think I was re-inventing the wheel. I know I didn't explain that well but I believe the problem is pretty generic. Any help anyone could offer is greatly appreciated. Oh, I'd like to avoid gimblelock too. Essentially calculating the rotation around each of the crafts up/forward/right vectors using the up/forward/right vectors. To simply the question a generic implementation rather than one specific to Gmod is absolutely fine.

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  • calculating min and max of 2-D array in c

    - by m2010
    This program to calculate sum,min and max of array elements Max value is the problem, it is always not true. void main(void) { int degree[3][2]; int min_max[][]; int Max=min_max[0][0]; int Min=min_max[0][0]; int i,j; int sum=0; clrscr(); for(i=0;i<3;i++) { for(j=0;j<2;j++) { printf("\n enter degree of student no. %d in subject %d:",i+1,j+1); scanf("%d",&degree[i][j]); } } for(i=0;i<3;i++) { for(j=0;j<2;j++) { printf("\n Student no. %d degree in subject no. %d is %d",i+1,j+1,degree[i][j]); } } for(i=0;i<3;i++) { sum=0; for(j=0;j<2;j++) { sum+=degree[i][j]; } printf("\n sum of degrees of student no. %d is %d",i+1,sum); min_max[i][j]=sum; if(min_max[i][j] <Min) { Min=min_max[i][j]; } else if(min_max[i][j]>Max) { Max=min_max[i][j]; } } printf("\nThe minimum sum of degrees of student no. %d is %d",i,Min); printf("\nThe maximum sum of degrees of student no. %d is %d",i,Max); getch(); }

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  • determining if value is in range with 0=360 degree problem.

    - by Raven
    Hi, I am making a piece of code for DirectX app. It's meaning is to not show faces that are not visible. Normaly it would be just using Z-buffer, but I'm making many moves and rotations of mesh, so I would like to not do them and save computing power. I will describe this on cube. You are looking from the front so you see just one face and you don't need to rotate the 5 that left. If you would have one side of cube from 100*100 meshes, it would be great to not have to turn around 50k meshes that you really don't need. So I have stored X,Y,Z rotation of camera(the Z rotation I'm not using), and also X,Y,Z rotation of faces. In this cube simplified I would see faces that makes this statement true: cRot //camera rotation in degrees oRot //face rotation in degrees if(oRot.x > cRot.x-90 && oRot.x < cRot.x+90 && oRot.y > cRot.y-90 && oRot.y < cRot.y+90) But there comes a problem. If I will rotate arround, the camera can get to value 330 for exapmple. In this state, I would see front and right side of cube. Right side have rotation 270 so that's allright in IF statement. Problem is with 0 rotation of front face, which is also 360 degrees. So my question is how to make this statement to work, because when I use modulo, it will be failing for that right side and in this way it won't work for 0=360.

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  • Manually changing keyboard orientation for a view that's on top of a camera view

    - by XKR
    I'm basically trying to reproduce the core functionality of the "At Once" app. I have a camera view and another view with a text view on it. I add both views to the window. All is well so far. [window addSubview:imagePicker.view]; [window addSubview:textViewController.view]; I understand that the UIImagePickerController does not support autorotation, so I handle it manually by watching UIDeviceOrientationDidChangeNotifications and applying the necessary transforms to the textViewController.view. Now, the problem here is the keyboard. If I do nothing, it just stays in portrait mode. I can get it to rotate by adding the following code to the notification handler. [[UIApplication sharedApplication] setStatusBarOrientation:interfaceOrientation]; [textView resignFirstResponder]; [textView becomeFirstResponder]; However, the following simple test produces weird behavior. Start the app in portrait mode. Rotate the device 90 degrees clockwise. Rotate the device 90 degrees counterclockwise (back to the initial position). Rotate the device 90 degrees clockwise. After step 4, instead of the landscape-mode keyboard, the portrait-style keyboard is shown, skewed to fit in the landscape keyboard frame. Perhaps my approach is wrong from the start. I was wondering if anyone has been able to reliably make the keyboard change its orientation in response to setStatusBarOrientation.

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  • How to use MySQL geospatial extensions with spherical geometries

    - by Joshua
    Hi Everyone, I would like to store thousands of latitude/longitude points in a MySQL db. I was successful at setting up the tables and adding the data using the geospatial extensions where the column 'coord' is a Point(lat, lng). Problem: I want to quickly find the 'N' closest entries to latitude 'X' degrees and longitude 'Y' degrees. Since the Distance() function has not yet been implemented, I used GLength() function to calculate the distance between (X,Y) and each of the entries, sorting by ascending distance, and limiting to 'N' results. The problem is that this is not calculating shortest distance with spherical geometry. Which means if Y = 179.9 degrees, the list of closest entries will only include longitudes of starting at 179.9 and decreasing even though closer entries exist with longitudes increasing from -179.9. How does one typically handle the discontinuity in longitude when working with spherical geometries in databases? There has to be an easy solution to this, but I must just be searching for the wrong thing because I have not found anything helpful. Should I just forget the GLength() function and create my own function for calculating angular separation? If I do this, will it still be fast and take advantage of the geospatial extensions? Thanks! josh UPDATE: This is exactly what I am describing above. However, it is only for SQL Server. Apparently SQL Server has a Geometry and Geography datatypes. The geography does exactly what I need. Is there something similar in MySQL?

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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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  • Oracle CRM in the UK- Gartner CRM Summit 2010

    - by divya.malik
    We are now headed to the UK to co-sponsor and participate in the Gartner Customer Relationship Management Summit 2010 on the 16th and 17th of March in London. Oracle CRM Vice President Mark Woollen will be presenting on Tuesday, 16 March 2010 from 15:20-15:50 on                                                                                                                                          CRM is dead, long live CRM?  Everyone is saying the world has changed and with it a new set of acronyms/buzzwords/vendors etc have appeared. What does this really mean for CRM software? Is it Dead or Alive? Listen to Mark’s view from Oracle and its customers.                  Location- Westbourne 2, Level –1. Also stop by the Oracle booth at the demogrounds.  The event looks promising with some great content from the Gartner analysts and from what the Gartner folks just told me, the event is oversold. And the weather in London town? As expected…slight showers on Monday with a high of 49 degrees F and partly cloudy on Tuesday, with a high of 50 degrees F.

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  • What's the maximum safe temperature for a HD Radeon 6870?

    - by Adrian Grigore
    I'm running a passively cooled HD Radeon 6870 in my PC. While using 3D Acceleration, the temperature climbs up to 95 degrees Celsius according to SpeedFan. It seems a bit hot, but on the other hand I've seen other GPUs being specified to run up to 120 Degrees Celsius. The system is very stable, but Battlefield 3 crashes every few hours or so. On the other hand it might be the game's fault and not related to the GPU temperature at all. Does anyone know where I can find some manufacturer specs on the maximum allowed temperature for this GPU? Thanks, Adrian

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  • glm quaternion camera rotating on wrong axis

    - by Jarrett
    I'm trying to get my camera implemented with a glm::quat used to store the rotation. However, whenever I do circles with the mouse, the camera rotates along the axis I am viewing (i.e. I think it's called the target axis). For example, if I rotated the mouse in a clockwise fashion, the camera rotates clockwise around the axis. I initialize my quaternion like so: void Camera::initialize() { orientationQuaternion_ = glm::quat(); orientationQuaternion_ = glm::normalize(orientationQuaternion_); } I rotate like so: void Camera::rotate(const glm::detail::float32& degrees, const glm::vec3& axis) { orientationQuaternion_ = orientationQuaternion_ * glm::normalize(glm::angleAxis(degrees, axis)); } and I set the viewMatrix like so: void Camera::render() { glm::quat temp = glm::conjugate(orientationQuaternion_); viewMatrix_ = glm::mat4_cast(temp); viewMatrix_ = glm::translate(viewMatrix_, glm::vec3(-pos_.x, -pos_.y, -pos_.z)); } The only axis' I actually try to rotate are the X and Y axis (i.e. (1,0,0) and (0,1,0)). Anyone have any idea why I see my camera rotating around the target axis?

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  • Finding direction of travel in a world with wrapped edges

    - by crazy
    I need to find the shortest distance direction from one point in my 2D world to another point where the edges are wrapped (like asteroids etc). I know how to find the shortest distance but am struggling to find which direction it's in. The shortest distance is given by: int rows = MapY; int cols = MapX; int d1 = abs(S.Y - T.Y); int d2 = abs(S.X - T.X); int dr = min(d1, rows-d1); int dc = min(d2, cols-d2); double dist = sqrt((double)(dr*dr + dc*dc)); Example of the world : : T : :--------------:--------- : : : S : : : : : : T : : : :--------------: In the diagram the edges are shown with : and -. I've shown a wrapped repeat of the world at the top right too. I want to find the direction in degrees from S to T. So the shortest distance is to the top right repeat of T. but how do I calculate the direction in degreed from S to the repeated T in the top right? I know the positions of both S and T but I suppose I need to find the position of the repeated T however there more than 1. The worlds coordinates system starts at 0,0 at the top left and 0 degrees for the direction could start at West. It seems like this shouldn’t be too hard but I haven’t been able to work out a solution. I hope somone can help? Any websites would be appreciated.

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  • Is learning how to use C (or C++) a requirement in order to be a good (excellent) programmer?

    - by blueberryfields
    When I first started to learn how to program, real programmers could write assembly in their sleep. Any serious schooling in computer science would include a hefty bit of training and practice in programming using assembly. That has since changed, to the point where I see Computer Science degrees with assembly, if included at all, is relegated to one assignment, and one chapter, for a total of two weeks' work out of 4 years' schooling. C/C++ programming seems to have followed a similar path. I'm no longer surprised to interview university graduates who have not spent more than two weeks programming in C++, and have only read of C in a book somewhere. While the most serious CS degrees still seem to include significant time learning and using one or both of the languages, the trend is clearly towards less enforced C/C++ in school. It's clearly possible to make a career producing good work without ever reading or writing a single line of C or C++ code. Given all of that, is learning the two languages worth the effort? Are they at all required to excel? (beyond the obvious, non-language specific advice, such as "a good selection of languages is probably important for a comprehensive education", and "it's probably a good idea to keep trying out and learning new languages throughout a programmers' career, just to stretch the gray cells")

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  • Will an online degree get you a job that requires "CS or equivalent 4-year degree"? [on hold]

    - by qel
    I'm a nerdy slacker type who didn't get my life together till I was 30. I've had a real job for a couple years doing C#/SQL. I've gotten several raises, but I'm making less than most developers, and the atmosphere is ... not positive. Looking for a new job, I think my applications get thrown out because I don't have a degree. And I want to finish a Bachelor's just to feel like less of a loser. I have a lot of college credits from 1996-2003 and a low GPA, so I don't know if that's worth much. An online degree looks like a good option, but I just don't know what I should be looking at for online schools because they all look like fake degrees. If they had programs equivalent to a real Comp Sci degree, I don't think they would have weird sounding names like they do. University of Phoenix has a B.S./Information Technology-Software Engineering. DeVry has a B.S./Computer Engineering Technology program. But that's not CS, and most other things I see have even more fake-sounding names. Are these useless degrees? Some people say DeVry and UoP are acceptable, some people say they're a joke. I have enough experience now, though, that maybe all I'm missing is being able to check the box that I have a 4-year degree. Harvard Extension seems like a real degree, even if it isn't a real Harvard degree, but I'd have to live there at least 3 months, which kinda defeats the purpose of an online degree fitting around work.

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  • Client Side Prediction for a Look Vector

    - by Mike Sawayda
    So I am making a first person networked shooter. I am working on client-side prediction where I am predicting player position and look vectors client-side based on input messages received from the server. Right now I am only worried about the look vectors though. I am receiving the correct look vector from the server about 20 times per second and I am checking that against the look vector that I have client side. I want to interpolate the clients look vector towards the correct one that is server side over a period of time. Therefore no matter how far you are away from the servers look vector you will interpolate to it over the same amount of time. Ex. if you were 10 degrees off it would take the same amount of time as if you were 2 degrees off to be correctly lined up with the server copy. My code looks something like this but the problem is that the amount that you are changing the clients copy gets infinitesimally small so you will actually never reach the servers copy. This is because I am always calculating the difference and only moving by a percentage of that every frame. Does anyone have any suggestions on how to interpolate towards the servers copy correctly? if(rotationDiffY > ClientSideAttributes::minRotation) { if(serverRotY > clientRotY) { playerObjects[i]->collisionObject->rotation.y += (rotationDiffY * deltaTime); } else { playerObjects[i]->collisionObject->rotation.y -= (rotationDiffY deltaTime); } }

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