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  • Best way to collect and store data daily?

    - by mktb
    I have a bunch of statistics: # of users, # of families, ratio user/family, etc. I'd like to store these daily so I can view this data historically. However, I'm looking for the most effective way to store this data. Should I run a cron job that writes to the database DATE: today USERS: 123 FAMILIES: 456 RATIO: 7.89 or whatever? (or should I write multiple rows like DATE: today DATATYPE: users VALUE: 123?) Or is there another option I can use that is more efficient or more effective? Thanks!

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  • based map follow layer wms tiled

    - by user32263
    hye all, im new in openlayer, just want to ask, i want to develop one system i already get layer from geoserver using wms services; this is the code: var hilirPerak = new OpenLayers.Layer.WMS("hilirPerak", "http://localhost:8080/geoserver/hilirPerak/wms", { workspace: 'hilirPerak', layers: 'hilirPerak:lot', styles:'', format: 'image/png', tiled: true, transitionEffect: 'resize', units: 'degrees' }, { maxResolution: 1000, singleTile: false, ratio:1, isBaseLayer:false, tilesOrigin : map.maxExtent.left + ',' + map.maxExtent.bottom, buffer: 0, visibility:true, maxExtent: OpenLayers.Bounds.fromString("3.873635,4.074956,101.14974"), //ratio: 1.9, //buffer: 0, //displayOutsideMaxExtent: true, //isBaseLayer: true, yx : {'EPSG:4326' : true}, displayOutsideMaxExtent: true, transparent:true } ); its show on system, for based map im using google, but the problem is, when i tick check layer, the based map follow size layer. if the layer point at certain place, based map follow layer bounds. please help me.. Thank you

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  • optimized grid for rectangular items

    - by peterchen
    I have N rectangular items with an aspect ratio Aitem (X:Y). I have a rectangular display area with an aspect ratio Aview The items should be arranged in a table-like layout (i.e. r rows, c columns). what is the ideal grid rows x columns, so that individual items are largest? (rows * colums = N, of course - i.e. there may be "unused" grid places). A simple algorithm could iterate over rows = 1..N, calculate the required number of columns, and keep the row/column pair with the largest items. I wonder if there's a non-iterative algorithm, though (e.g. for Aitem = Aview = 1, rows / cols can be approximated by sqrt(N)).

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  • BCB: how to get the (approximate) width of a character in a given TFont?

    - by mawg
    It's a TMemo, not that that should make any difference. Googling suggests that I can use Canvas->TextWidth() but those are Delphi examples and BCB doesn't seem to offer this property. I really want something analogous to memo->Font->Height for width. I realize that not all fonts are fixed width, so a good estimate will do. All that I need is to take the width of a TMemo in pixels and make a reasonable guess at how many characters of the current font it will hold. Of course, if I really want to be lazy, I can just google for the average height/width ratio, since height is known. Remember, an approximation is good enough for me if it is tricky to get exact. http://www.plainlanguagenetwork.org/type/utbo211.htm says, " A width to height ratio of 3:5 (0.6) is recommended for most applications"

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  • Howto plot two cumulative frequency graph together

    - by neversaint
    I have data that looks like this: #val Freq1 Freq2 0.000 178 202 0.001 4611 5300 0.002 99 112 0.003 26 30 0.004 17 20 0.005 15 20 0.006 11 14 0.007 11 13 0.008 13 13 ...many more lines.. Full data can be found here: http://dpaste.com/173536/plain/ What I intend to do is to have a cumulative graph with "val" as x-axis with "Freq1" & "Freq2" as y-axis, plot together in 1 graph. I have this code. But it creates two plots instead of 1. dat <- read.table("stat.txt",header=F); val<-dat$V1 freq1<-dat$V2 freq2<-dat$V3 valf1<-rep(val,freq1) valf2<-rep(val,freq2) valfreq1table<- table(valf1) valfreq2table<- table(valf2) cumfreq1=c(0,cumsum(valfreq1table)) cumfreq2=c(0,cumsum(valfreq2table)) plot(cumfreq1, ylab="CumFreq",xlab="Loglik Ratio") lines(cumfreq1) plot(cumfreq2, ylab="CumFreq",xlab="Loglik Ratio") lines(cumfreq2) What's the right way to approach this?

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  • Calculate percentage of up votes

    - by MakuraYami
    I have searched this site and Google and even though the idea is pretty simple I can't figure it out. I need to (like seen on YouTube) calculate the % of up-votes based on the amount up-votes and down-votes. I have two vars, $upvotes and $downvotes now i need to calculate $ratio For example $upvotes = 3; $downvotes = 1; The ratio here needs to be 75 (%) If you have $upvotes = 0; $downvotes = 100; It needs to be 0 (%) How do I calculate the percentage (in PHP)?

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  • How to label a cuboid?

    - by usha
    Hi this is how my 3dcuboid looks, I have attached the complete code. I want to label this cuboid using different names across sides, how is this possible using opengl on android? public class MyGLRenderer implements Renderer { Context context; Cuboid rect; private float mCubeRotation; // private static float angleCube = 0; // Rotational angle in degree for cube (NEW) // private static float speedCube = -1.5f; // Rotational speed for cube (NEW) public MyGLRenderer(Context context) { rect = new Cuboid(); this.context = context; } public void onDrawFrame(GL10 gl) { // TODO Auto-generated method stub gl.glClear(GL10.GL_COLOR_BUFFER_BIT | GL10.GL_DEPTH_BUFFER_BIT); gl.glLoadIdentity(); // Reset the model-view matrix gl.glTranslatef(0.2f, 0.0f, -8.0f); // Translate right and into the screen gl.glScalef(0.8f, 0.8f, 0.8f); // Scale down (NEW) gl.glRotatef(mCubeRotation, 1.0f, 1.0f, 1.0f); // gl.glRotatef(angleCube, 1.0f, 1.0f, 1.0f); // rotate about the axis (1,1,1) (NEW) rect.draw(gl); mCubeRotation -= 0.15f; //angleCube += speedCube; } public void onSurfaceChanged(GL10 gl, int width, int height) { // TODO Auto-generated method stub if (height == 0) height = 1; // To prevent divide by zero float aspect = (float)width / height; // Set the viewport (display area) to cover the entire window gl.glViewport(0, 0, width, height); // Setup perspective projection, with aspect ratio matches viewport gl.glMatrixMode(GL10.GL_PROJECTION); // Select projection matrix gl.glLoadIdentity(); // Reset projection matrix // Use perspective projection GLU.gluPerspective(gl, 45, aspect, 0.1f, 100.f); gl.glMatrixMode(GL10.GL_MODELVIEW); // Select model-view matrix gl.glLoadIdentity(); // Reset } public void onSurfaceCreated(GL10 gl, EGLConfig config) { // TODO Auto-generated method stub gl.glClearColor(0.0f, 0.0f, 0.0f, 1.0f); // Set color's clear-value to black gl.glClearDepthf(1.0f); // Set depth's clear-value to farthest gl.glEnable(GL10.GL_DEPTH_TEST); // Enables depth-buffer for hidden surface removal gl.glDepthFunc(GL10.GL_LEQUAL); // The type of depth testing to do gl.glHint(GL10.GL_PERSPECTIVE_CORRECTION_HINT, GL10.GL_NICEST); // nice perspective view gl.glShadeModel(GL10.GL_SMOOTH); // Enable smooth shading of color gl.glDisable(GL10.GL_DITHER); // Disable dithering for better performance }} public class Cuboid{ private FloatBuffer mVertexBuffer; private FloatBuffer mColorBuffer; private ByteBuffer mIndexBuffer; private float vertices[] = { //width,height,depth -2.5f, -1.0f, -1.0f, 1.0f, -1.0f, -1.0f, 1.0f, 1.0f, -1.0f, -2.5f, 1.0f, -1.0f, -2.5f, -1.0f, 1.0f, 1.0f, -1.0f, 1.0f, 1.0f, 1.0f, 1.0f, -2.5f, 1.0f, 1.0f }; private float colors[] = { // R,G,B,A COLOR 0.0f, 1.0f, 0.0f, 1.0f, 0.0f, 1.0f, 0.0f, 1.0f, 1.0f, 0.5f, 0.0f, 1.0f, 1.0f, 0.5f, 0.0f, 1.0f, 1.0f, 0.0f, 0.0f, 1.0f, 1.0f, 0.0f, 0.0f, 1.0f, 0.0f, 0.0f, 1.0f, 1.0f, 1.0f, 0.0f, 1.0f, 1.0f }; private byte indices[] = { // VERTEX 0,1,2,3,4,5,6,7 REPRESENTATION FOR FACES 0, 4, 5, 0, 5, 1, 1, 5, 6, 1, 6, 2, 2, 6, 7, 2, 7, 3, 3, 7, 4, 3, 4, 0, 4, 7, 6, 4, 6, 5, 3, 0, 1, 3, 1, 2 }; public Cuboid() { ByteBuffer byteBuf = ByteBuffer.allocateDirect(vertices.length * 4); byteBuf.order(ByteOrder.nativeOrder()); mVertexBuffer = byteBuf.asFloatBuffer(); mVertexBuffer.put(vertices); mVertexBuffer.position(0); byteBuf = ByteBuffer.allocateDirect(colors.length * 4); byteBuf.order(ByteOrder.nativeOrder()); mColorBuffer = byteBuf.asFloatBuffer(); mColorBuffer.put(colors); mColorBuffer.position(0); mIndexBuffer = ByteBuffer.allocateDirect(indices.length); mIndexBuffer.put(indices); mIndexBuffer.position(0); } public void draw(GL10 gl) { gl.glFrontFace(GL10.GL_CW); gl.glVertexPointer(3, GL10.GL_FLOAT, 0, mVertexBuffer); gl.glColorPointer(4, GL10.GL_FLOAT, 0, mColorBuffer); gl.glEnableClientState(GL10.GL_VERTEX_ARRAY); gl.glEnableClientState(GL10.GL_COLOR_ARRAY); gl.glDrawElements(GL10.GL_TRIANGLES, 36, GL10.GL_UNSIGNED_BYTE, mIndexBuffer); gl.glDisableClientState(GL10.GL_VERTEX_ARRAY); gl.glDisableClientState(GL10.GL_COLOR_ARRAY); } } public class Draw3drect extends Activity { private GLSurfaceView glView; // Use GLSurfaceView // Call back when the activity is started, to initialize the view @Override protected void onCreate(Bundle savedInstanceState) { super.onCreate(savedInstanceState); glView = new GLSurfaceView(this); // Allocate a GLSurfaceView glView.setRenderer(new MyGLRenderer(this)); // Use a custom renderer this.setContentView(glView); // This activity sets to GLSurfaceView } // Call back when the activity is going into the background @Override protected void onPause() { super.onPause(); glView.onPause(); } // Call back after onPause() @Override protected void onResume() { super.onResume(); glView.onResume(); } }

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  • how to label a cuboid using open gl?

    - by usha
    hi this is how my 3dcuboid looks ,i have attached complete code , i want to label this cuboid using different name across sides how is it possible using opengl in android...plz help me out public class MyGLRenderer implements Renderer { Context context; Cuboid rect; private float mCubeRotation; // private static float angleCube = 0; // Rotational angle in degree for cube (NEW) // private static float speedCube = -1.5f; // Rotational speed for cube (NEW) public MyGLRenderer(Context context) { rect = new Cuboid(); this.context = context; } public void onDrawFrame(GL10 gl) { // TODO Auto-generated method stub gl.glClear(GL10.GL_COLOR_BUFFER_BIT | GL10.GL_DEPTH_BUFFER_BIT); gl.glLoadIdentity(); // Reset the model-view matrix gl.glTranslatef(0.2f, 0.0f, -8.0f); // Translate right and into the screen gl.glScalef(0.8f, 0.8f, 0.8f); // Scale down (NEW) gl.glRotatef(mCubeRotation, 1.0f, 1.0f, 1.0f); // gl.glRotatef(angleCube, 1.0f, 1.0f, 1.0f); // rotate about the axis (1,1,1) (NEW) rect.draw(gl); mCubeRotation -= 0.15f; //angleCube += speedCube; } public void onSurfaceChanged(GL10 gl, int width, int height) { // TODO Auto-generated method stub if (height == 0) height = 1; // To prevent divide by zero float aspect = (float)width / height; // Set the viewport (display area) to cover the entire window gl.glViewport(0, 0, width, height); // Setup perspective projection, with aspect ratio matches viewport gl.glMatrixMode(GL10.GL_PROJECTION); // Select projection matrix gl.glLoadIdentity(); // Reset projection matrix // Use perspective projection GLU.gluPerspective(gl, 45, aspect, 0.1f, 100.f); gl.glMatrixMode(GL10.GL_MODELVIEW); // Select model-view matrix gl.glLoadIdentity(); // Reset } public void onSurfaceCreated(GL10 gl, EGLConfig config) { // TODO Auto-generated method stub gl.glClearColor(0.0f, 0.0f, 0.0f, 1.0f); // Set color's clear-value to black gl.glClearDepthf(1.0f); // Set depth's clear-value to farthest gl.glEnable(GL10.GL_DEPTH_TEST); // Enables depth-buffer for hidden surface removal gl.glDepthFunc(GL10.GL_LEQUAL); // The type of depth testing to do gl.glHint(GL10.GL_PERSPECTIVE_CORRECTION_HINT, GL10.GL_NICEST); // nice perspective view gl.glShadeModel(GL10.GL_SMOOTH); // Enable smooth shading of color gl.glDisable(GL10.GL_DITHER); // Disable dithering for better performance }} public class Cuboid{ private FloatBuffer mVertexBuffer; private FloatBuffer mColorBuffer; private ByteBuffer mIndexBuffer; private float vertices[] = { //width,height,depth -2.5f, -1.0f, -1.0f, 1.0f, -1.0f, -1.0f, 1.0f, 1.0f, -1.0f, -2.5f, 1.0f, -1.0f, -2.5f, -1.0f, 1.0f, 1.0f, -1.0f, 1.0f, 1.0f, 1.0f, 1.0f, -2.5f, 1.0f, 1.0f }; private float colors[] = { // R,G,B,A COLOR 0.0f, 1.0f, 0.0f, 1.0f, 0.0f, 1.0f, 0.0f, 1.0f, 1.0f, 0.5f, 0.0f, 1.0f, 1.0f, 0.5f, 0.0f, 1.0f, 1.0f, 0.0f, 0.0f, 1.0f, 1.0f, 0.0f, 0.0f, 1.0f, 0.0f, 0.0f, 1.0f, 1.0f, 1.0f, 0.0f, 1.0f, 1.0f }; private byte indices[] = { // VERTEX 0,1,2,3,4,5,6,7 REPRESENTATION FOR FACES 0, 4, 5, 0, 5, 1, 1, 5, 6, 1, 6, 2, 2, 6, 7, 2, 7, 3, 3, 7, 4, 3, 4, 0, 4, 7, 6, 4, 6, 5, 3, 0, 1, 3, 1, 2 }; public Cuboid() { ByteBuffer byteBuf = ByteBuffer.allocateDirect(vertices.length * 4); byteBuf.order(ByteOrder.nativeOrder()); mVertexBuffer = byteBuf.asFloatBuffer(); mVertexBuffer.put(vertices); mVertexBuffer.position(0); byteBuf = ByteBuffer.allocateDirect(colors.length * 4); byteBuf.order(ByteOrder.nativeOrder()); mColorBuffer = byteBuf.asFloatBuffer(); mColorBuffer.put(colors); mColorBuffer.position(0); mIndexBuffer = ByteBuffer.allocateDirect(indices.length); mIndexBuffer.put(indices); mIndexBuffer.position(0); } public void draw(GL10 gl) { gl.glFrontFace(GL10.GL_CW); gl.glVertexPointer(3, GL10.GL_FLOAT, 0, mVertexBuffer); gl.glColorPointer(4, GL10.GL_FLOAT, 0, mColorBuffer); gl.glEnableClientState(GL10.GL_VERTEX_ARRAY); gl.glEnableClientState(GL10.GL_COLOR_ARRAY); gl.glDrawElements(GL10.GL_TRIANGLES, 36, GL10.GL_UNSIGNED_BYTE, mIndexBuffer); gl.glDisableClientState(GL10.GL_VERTEX_ARRAY); gl.glDisableClientState(GL10.GL_COLOR_ARRAY); } } public class Draw3drect extends Activity { private GLSurfaceView glView; // Use GLSurfaceView // Call back when the activity is started, to initialize the view @Override protected void onCreate(Bundle savedInstanceState) { super.onCreate(savedInstanceState); glView = new GLSurfaceView(this); // Allocate a GLSurfaceView glView.setRenderer(new MyGLRenderer(this)); // Use a custom renderer this.setContentView(glView); // This activity sets to GLSurfaceView } // Call back when the activity is going into the background @Override protected void onPause() { super.onPause(); glView.onPause(); } // Call back after onPause() @Override protected void onResume() { super.onResume(); glView.onResume(); } }

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  • In 3D camera math, calculate what Z depth is pixel unity for a given FOV

    - by badweasel
    I am working in iOS and OpenGL ES 2.0. Through trial and error I've figured out a frustum to where at a specific z depth pixels drawn are 1 to 1 with my source textures. So 1 pixel in my texture is 1 pixel on the screen. For 2d games this is good. Of course it means that I also factor in things like the size of the quad and the size of the texture. For example if my sprite is a quad 32x32 pixels. The quad size is 3.2 units wide and tall. And the texcoords are 32 / the size of the texture wide and tall. Then the frustum is: matrixFrustum(-(float)backingWidth/frustumScale,(float)backingWidth/frustumScale, -(float)backingHeight/frustumScale, (float)backingHeight/frustumScale, 40, 1000, mProjection); Where frustumScale is 800 for a retina screen. Then at a distance of 800 from camera the sprite is pixel for pixel the same as photoshop. For 3d games sometimes I still want to be able to do this. But depending on the scene I sometimes need the FOV to be different things. I'm looking for a way to figure out what Z depth will achieve this same pixel unity for a given FOV. For this my mProjection is set using: matrixPerspective(cameraFOV, near, far, (float)backingWidth / (float)backingHeight, mProjection); With testing I found that at an FOV of 45.0 a Z of 38.5 is very close to pixel unity. And at an FOV of 30.0 a Z of 59.5 is about right. But how can I calculate a value that is spot on? Here's my matrixPerspecitve code: void matrixPerspective(float angle, float near, float far, float aspect, mat4 m) { //float size = near * tanf(angle / 360.0 * M_PI); float size = near * tanf(degreesToRadians(angle) / 2.0); float left = -size, right = size, bottom = -size / aspect, top = size / aspect; // Unused values in perspective formula. m[1] = m[2] = m[3] = m[4] = 0; m[6] = m[7] = m[12] = m[13] = m[15] = 0; // Perspective formula. m[0] = 2 * near / (right - left); m[5] = 2 * near / (top - bottom); m[8] = (right + left) / (right - left); m[9] = (top + bottom) / (top - bottom); m[10] = -(far + near) / (far - near); m[11] = -1; m[14] = -(2 * far * near) / (far - near); } And my mView is set using: lookAtMatrix(cameraPos, camLookAt, camUpVector, mView); * UPDATE * I'm going to leave this here in case anyone has a different solution, can explain how they do it, or why this works. This is what I figured out. In my system I use a 10th scale unit to pixels on non-retina displays and a 20th scale on retina displays. The iPhone is 640 pixels wide on retina and 320 pixels wide on non-retina (obsolete). So if I want something to be the full screen width I divide by 20 to get the OpenGL unit width. Then divide that by 2 to get the left and right unit position. Something 32 units wide centered on the screen goes from -16 to +16. Believe it or not I have an excel spreadsheet do all this math for me and output all the vertex data for my sprite sheet. It's an arbitrary thing I made up to do .1 units = 1 non-retina pixel or 2 retina pixels. I could have made it .01 units = 2 pixels and someday I might switch to that. But for now it's the other. So the width of the screen in units is 32.0, and that means the left most pixel is at -16.0 and the right most is at 16.0. After messing a bit I figured out that if I take the [0] value of an identity modelViewProjection matrix and multiply it by 16 I get the depth required to get 1:1 pixels. I don't know why. I don't know if the 16 is related to the screen size or just a lucky guess. But I did a test where I placed a sprite at that calculated depth and varied the FOV through all the valid values and the object stays steady on screen with 1:1 pixels. So now I'm just calculating the unityDepth that way. If someone gives me a better answer I'll checkmark it.

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  • More Fun With Math

    - by PointsToShare
    More Fun with Math   The runaway student – three different ways of solving one problem Here is a problem I read in a Russian site: A student is running away. He is moving at 1 mph. Pursuing him are a lion, a tiger and his math teacher. The lion is 40 miles behind and moving at 6 mph. The tiger is 28 miles behind and moving at 4 mph. His math teacher is 30 miles behind and moving at 5 mph. Who will catch him first? Analysis Obviously we have a set of three problems. They are all basically the same, but the details are different. The problems are of the same class. Here is a little excursion into computer science. One of the things we strive to do is to create solutions for classes of problems rather than individual problems. In your daily routine, you call it re-usability. Not all classes of problems have such solutions. If a class has a general (re-usable) solution, it is called computable. Otherwise it is unsolvable. Within unsolvable classes, we may still solve individual (some but not all) problems, albeit with different approaches to each. Luckily the vast majority of our daily problems are computable, and the 3 problems of our runaway student belong to a computable class. So, let’s solve for the catch-up time by the math teacher, after all she is the most frightening. She might even make the poor runaway solve this very problem – perish the thought! Method 1 – numerical analysis. At 30 miles and 5 mph, it’ll take her 6 hours to come to where the student was to begin with. But by then the student has advanced by 6 miles. 6 miles require 6/5 hours, but by then the student advanced by another 6/5 of a mile as well. And so on and so forth. So what are we to do? One way is to write code and iterate it until we have solved it. But this is an infinite process so we’ll end up with an infinite loop. So what to do? We’ll use the principles of numerical analysis. Any calculator – your computer included – has a limited number of digits. A double floating point number is good for about 14 digits. Nothing can be computed at a greater accuracy than that. This means that we will not iterate ad infinidum, but rather to the point where 2 consecutive iterations yield the same result. When we do financial computations, we don’t even have to go that far. We stop at the 10th of a penny.  It behooves us here to stop at a 10th of a second (100 milliseconds) and this will how we will avoid an infinite loop. Interestingly this alludes to the Zeno paradoxes of motion – in particular “Achilles and the Tortoise”. Zeno says exactly the same. To catch the tortoise, Achilles must always first come to where the tortoise was, but the tortoise keeps moving – hence Achilles will never catch the tortoise and our math teacher (or lion, or tiger) will never catch the student, or the policeman the thief. Here is my resolution to the paradox. The distance and time in each step are smaller and smaller, so the student will be caught. The only thing that is infinite is the iterative solution. The race is a convergent geometric process so the steps are diminishing, but each step in the solution takes the same amount of effort and time so with an infinite number of steps, we’ll spend an eternity solving it.  This BTW is an original thought that I have never seen before. But I digress. Let’s simply write the code to solve the problem. To make sure that it runs everywhere, I’ll do it in JavaScript. function LongCatchUpTime(D, PV, FV) // D is Distance; PV is Pursuers Velocity; FV is Fugitive’ Velocity {     var t = 0;     var T = 0;     var d = parseFloat(D);     var pv = parseFloat (PV);     var fv = parseFloat (FV);     t = d / pv;     while (t > 0.000001) //a 10th of a second is 1/36,000 of an hour, I used 1/100,000     {         T = T + t;         d = t * fv;         t = d / pv;     }     return T;     } By and large, the higher the Pursuer’s velocity relative to the fugitive, the faster the calculation. Solving this with the 10th of a second limit yields: 7.499999232000001 Method 2 – Geometric Series. Each step in the iteration above is smaller than the next. As you saw, we stopped iterating when the last step was small enough, small enough not to really matter.  When we have a sequence of numbers in which the ratio of each number to its predecessor is fixed we call the sequence geometric. When we are looking at the sum of sequence, we call the sequence of sums series.  Now let’s look at our student and teacher. The teacher runs 5 times faster than the student, so with each iteration the distance between them shrinks to a fifth of what it was before. This is a fixed ratio so we deal with a geometric series.  We normally designate this ratio as q and when q is less than 1 (0 < q < 1) the sum of  + … +  is  – 1) / (q – 1). When q is less than 1, it is easier to use ) / (1 - q). Now, the steps are 6 hours then 6/5 hours then 6/5*5 and so on, so q = 1/5. And the whole series is multiplied by 6. Also because q is less than 1 , 1/  diminishes to 0. So the sum is just  / (1 - q). or 1/ (1 – 1/5) = 1 / (4/5) = 5/4. This times 6 yields 7.5 hours. We can now continue with some algebra and take it back to a simpler formula. This is arduous and I am not going to do it here. Instead let’s do some simpler algebra. Method 3 – Simple Algebra. If the time to capture the fugitive is T and the fugitive travels at 1 mph, then by the time the pursuer catches him he travelled additional T miles. Time is distance divided by speed, so…. (D + T)/V = T  thus D + T = VT  and D = VT – T = (V – 1)T  and T = D/(V – 1) This “strangely” coincides with the solution we just got from the geometric sequence. This is simpler ad faster. Here is the corresponding code. function ShortCatchUpTime(D, PV, FV) {     var d = parseFloat(D);     var pv = parseFloat (PV);     var fv = parseFloat (FV);     return d / (pv - fv); } The code above, for both the iterative solution and the algebraic solution are actually for a larger class of problems.  In our original problem the student’s velocity (speed) is 1 mph. In the code it may be anything as long as it is less than the pursuer’s velocity. As long as PV > FV, the pursuer will catch up. Here is the really general formula: T = D / (PV – FV) Finally, let’s run the program for each of the pursuers.  It could not be worse. I know he’d rather be eaten alive than suffering through yet another math lesson. See the code run? Select  “Catch Up Time” in www.mgsltns.com/games.htm The host is running on Unix, so the link is case sensitive. That’s All Folks

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  • % new visitor vs. % returning visitor

    - by Torben Gundtofte-Bruun
    I'm not sure how to interpret the results in Google Analytics. I understand that some metrics should be high, and some should be low. But this one I don't get: % new visitor vs. % returning visitor: It's good that users are returning, but surely it's also good to get new, fresh visitors. How do I evaluate this %-vs-% ratio? The higher the better: visits unique visitors pageviews pages per visit avg. visit duration The lower the better: bounce rate drop-offs

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  • problem in array of shooter sprites which contain different colour bubbles

    - by prakash s
    everyone i am developing bubble shooter game in cocos2d I have placed shooter array which contain different color bubbles like this 00000000 it is 8 bubbles array if i tap the screen, first bubbles should move for shooting the target .png .And if i again tap the screen again 2nd position bubble should move for shooting the target.png bubbles,how it will possible for me because i have already created the array of target which contain different color bubbles, here i write the code : - (void)ccTouchesEnded:(NSSet *)touches withEvent:(UIEvent *)event { // Choose one of the touches to work with UITouch *touch = [touches anyObject]; CGPoint location = [touch locationInView:[touch view]]; location = [[CCDirector sharedDirector] convertToGL:location]; // Set up initial location of projectile CGSize winSize = [[CCDirector sharedDirector] winSize]; NSMutableArray * movableSprites = [[NSMutableArray alloc] init]; NSArray *images = [NSArray arrayWithObjects:@"1.png", @"2.png", @"3.png", @"4.png",@"5.png",@"6.png",@"7.png", @"8.png", nil]; for(int i = 0; i < images.count; ++i) { int index = (arc4random() % 8)+1; NSString *image = [NSString stringWithFormat:@"%d.png", index]; CCSprite*projectile = [CCSprite spriteWithFile:image]; //CCSprite *projectile = [CCSprite spriteWithFile:@"3.png" rect:CGRectMake(0, 0,256,256)]; [self addChild:projectile]; [movableSprites addObject:projectile]; float offsetFraction = ((float)(i+1))/(images.count+1); //projectile.position = ccp(20, winSize.height/2); //projectile.position = ccp(18,0 ); //projectile.position = ccp(350*offsetFraction, 20.0f); projectile.position = ccp(10/offsetFraction, 20.0f); // projectile.position = ccp(projectile.position.x,projectile.position.y); // Determine offset of location to projectile int offX = location.x - projectile.position.x; int offY = location.y - projectile.position.y; // Bail out if we are shooting down or backwards if (offX <= 0) return; // Ok to add now - we've double checked position //[self addChild:projectile]; // Determine where we wish to shoot the projectile to int realX = winSize.width + (projectile.contentSize.width/2); float ratio = (float) offY / (float) offX; int realY = (realX * ratio) + projectile.position.y; CGPoint realDest = ccp(realX, realY); // Determine the length of how far we're shooting int offRealX = realX - projectile.position.x; int offRealY = realY - projectile.position.y; float length = sqrtf((offRealX*offRealX)+(offRealY*offRealY)); float velocity = 480/1; // 480pixels/1sec float realMoveDuration = length/velocity; // Move projectile to actual endpoint [projectile runAction:[CCSequence actions: [CCMoveTo actionWithDuration:realMoveDuration position:realDest], [CCCallFuncN actionWithTarget:self selector:@selector(spriteMoveFinished:)], nil]]; // Add to projectiles array projectile.tag = 1; [_projectiles addObject:projectile]; } }

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  • rotate sprite and shooting bullets from the end of a cannon

    - by Alberto
    Hi all i have a problem in my Andengine code, I need , when I touch the screen, shoot a bullet from the cannon (in the same direction of the cannon) The cannon rotates perfectly but when I touch the screen the bullet is not created at the end of the turret This is my code: private void shootProjectile(final float pX, final float pY){ int offX = (int) (pX-canon.getSceneCenterCoordinates()[0]); int offY = (int) (pY-canon.getSceneCenterCoordinates()[1]); if (offX <= 0) return ; if(offY>=0) return; double X=canon.getX()+canon.getWidth()*0,5; double Y=canon.getY()+canon.getHeight()*0,5 ; final Sprite projectile; projectile = new Sprite( (float) X, (float) Y, mProjectileTextureRegion,this.getVertexBufferObjectManager() ); mMainScene.attachChild(projectile); int realX = (int) (mCamera.getWidth()+ projectile.getWidth()/2.0f); float ratio = (float) offY / (float) offX; int realY = (int) ((realX*ratio) + projectile.getY()); int offRealX = (int) (realX- projectile.getX()); int offRealY = (int) (realY- projectile.getY()); float length = (float) Math.sqrt((offRealX*offRealX)+(offRealY*offRealY)); float velocity = (float) 480.0f/1.0f; float realMoveDuration = length/velocity; MoveModifier modifier = new MoveModifier(realMoveDuration,projectile.getX(), realX, projectile.getY(), realY); projectile.registerEntityModifier(modifier); } @Override public boolean onSceneTouchEvent(Scene pScene, TouchEvent pSceneTouchEvent) { if (pSceneTouchEvent.getAction() == TouchEvent.ACTION_MOVE){ double dx = pSceneTouchEvent.getX() - canon.getSceneCenterCoordinates()[0]; double dy = pSceneTouchEvent.getY() - canon.getSceneCenterCoordinates()[1]; double Radius = Math.atan2(dy,dx); double Angle = Radius * 180 / Math.PI; canon.setRotation((float)Angle); return true; } else if (pSceneTouchEvent.getAction() == TouchEvent.ACTION_DOWN){ final float touchX = pSceneTouchEvent.getX(); final float touchY = pSceneTouchEvent.getY(); double dx = pSceneTouchEvent.getX() - canon.getSceneCenterCoordinates()[0]; double dy = pSceneTouchEvent.getY() - canon.getSceneCenterCoordinates()[1]; double Radius = Math.atan2(dy,dx); double Angle = Radius * 180 / Math.PI; canon.setRotation((float)Angle); shootProjectile(touchX, touchY); } return false; } Anyone know how to calculate the coordinates (X,Y) of the end of the barrel to draw the bullet?

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  • Test your internet connection - Emtel Mobile Internet

    After yesterday's report on Emtel Fixed Broadband (I'm still wondering where the 'fixed' part is), I did the same tests on Emtel Mobile Internet. For this I'm using the Huawei E169G HSDPA USB stick, connected to the same machine. Actually, this is my fail-safe internet connection and the system automatically switches between them if a problem, let's say timeout, etc. has been detected on the main line. For better comparison I used exactly the same servers on Speedtest.net. The results Following are the results of Rose Hill (hosted by Emtel) and respectively Frankfurt, Germany (hosted by Vodafone DE): Speedtest.net result of 31.05.2013 between Flic en Flac and Rose Hill, Mauritius (Emtel - Mobile Internet) Speedtest.net result of 31.05.2013 between Flic en Flac and Frankfurt, Germany (Emtel - Mobile Internet) As you might easily see, there is a big difference in speed between national and international connections. More interestingly are the results related to the download and upload ratio. I'm not sure whether connections over Emtel Mobile Internet are asymmetric or symmetric like the Fixed Broadband. Might be interesting to find out. The first test result actually might give us a clue that the connection could be asymmetric with a ratio of 3:1 but again I'm not sure. I'll find out and post an update on this. It depends on network coverage Later today I was on tour with my tablet, a Samsung Galaxy Tab 10.1 (model GT-P7500) running on Android 4.0.4 (Ice Cream Sandwich), and did some more tests using the Speedtest.net app. The results are actually as expected and in areas with better network coverage you will get better results after all. At least, as long as you stay inside the national networks. For anything abroad, it doesn't really matter. But see for yourselves: Speedtest.net result of 31.05.2013 between Cascavelle and servers in Rose Hill, Mauritius (Emtel - Mobile Internet), Port Louis, Mauritius and Kuala Lumpur, Malaysia It's rather shocking and frustrating to see how the speed on international destinations goes down. And the full capability of the tablet's integrated modem (HSDPA: 21 Mbps; HSUPA: 5.76 Mbps) isn't used, too. I guess, this demands more tests in other areas of the island, like Ebene, Pailles or Port Louis. I'll keep you updated... The question remains: Alternatives? After the publication of the test results on Fixed Broadband I had some exchange with others on Facebook. Sadly, it seems that there are really no alternatives to what Emtel is offering at the moment. There are the various internet packages by Mauritius Telecom feat. Orange, like ADSL, MyT and Mobile Internet, and there is Bharat Telecom with their Bees offer which is currently limited to Ebene and parts of Quatre Bornes.

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  • How to resolve concurrent ramp collisions in 2d platformer?

    - by Shaun Inman
    A bit about the physics engine: Bodies are all rectangles. Bodies are sorted at the beginning of every update loop based on the body-in-motion's horizontal and vertical velocity (to avoid sticky walls/floors). Solid bodies are resolved by testing the body-in-motion's new X with the old Y and adjusting if necessary before testing the new X with the new Y, again adjusting if necessary. Works great. Ramps (rectangles with a flag set indicating bottom-left, bottom-right, etc) are resolved by calculating the ratio of penetration along the x-axis and setting a new Y accordingly (with some checks to make sure the body-in-motion isn't attacking from the tall or flat side, in which case the ramp is treated as a normal rectangle). This also works great. Side-by-side ramps, eg. \/ and /\, work fine but things get jittery and unpredictable when a top-down ramp is directly above a bottom-up ramp, eg. < or > or when a bottom-up ramp runs right up to the ceiling/top-down ramp runs right down to the floor. I've been able to lock it down somewhat by detecting whether the body-in-motion hadFloor when also colliding with a top-down ramp or hadCeiling when also colliding with a bottom-up ramp then resolving by calculating the ratio of penetration along the y-axis and setting the new X accordingly (the opposite of the normal behavior). But as soon as the body-in-motion jumps the hasFloor flag becomes false, the first ramp resolution pushes the body into collision with the second ramp and collision resolution becomes jittery again for a few frames. I'm sure I'm making this more complicated than it needs to be. Can anyone recommend a good resource that outlines the best way to address this problem? (Please don't recommend I use something like Box2d or Chipmunk. Also, "redesign your levels" isn't an answer; the body-in-motion may at times be riding another body-in-motion, eg. a platform, that pushes it into a ramp so I'd like to be able to resolve this properly.) Thanks!

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  • How do I crop a video clip?

    - by Lekensteyn
    I've recorded a video using recordmydesktop, but have unfortunately chosen for capturing the whole screen (1600*896) instead of a small part with geometric 900*690. How do I crop this video? Preferably in the editor I'm using, Kdenlive, to minimize quality loss. I've tried the Crop, Pan & Zoom and Scale0tilt effects, modifying the pixel ratio, movie size inside project settings options without success. A step-by-step guide would be preferred or at least some hints.

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  • Organization &amp; Architecture UNISA Studies &ndash; Chap 4

    - by MarkPearl
    Learning Outcomes Explain the characteristics of memory systems Describe the memory hierarchy Discuss cache memory principles Discuss issues relevant to cache design Describe the cache organization of the Pentium Computer Memory Systems There are key characteristics of memory… Location – internal or external Capacity – expressed in terms of bytes Unit of Transfer – the number of bits read out of or written into memory at a time Access Method – sequential, direct, random or associative From a users perspective the two most important characteristics of memory are… Capacity Performance – access time, memory cycle time, transfer rate The trade off for memory happens along three axis… Faster access time, greater cost per bit Greater capacity, smaller cost per bit Greater capacity, slower access time This leads to people using a tiered approach in their use of memory   As one goes down the hierarchy, the following occurs… Decreasing cost per bit Increasing capacity Increasing access time Decreasing frequency of access of the memory by the processor The use of two levels of memory to reduce average access time works in principle, but only if conditions 1 to 4 apply. A variety of technologies exist that allow us to accomplish this. Thus it is possible to organize data across the hierarchy such that the percentage of accesses to each successively lower level is substantially less than that of the level above. A portion of main memory can be used as a buffer to hold data temporarily that is to be read out to disk. This is sometimes referred to as a disk cache and improves performance in two ways… Disk writes are clustered. Instead of many small transfers of data, we have a few large transfers of data. This improves disk performance and minimizes processor involvement. Some data designed for write-out may be referenced by a program before the next dump to disk. In that case the data is retrieved rapidly from the software cache rather than slowly from disk. Cache Memory Principles Cache memory is substantially faster than main memory. A caching system works as follows.. When a processor attempts to read a word of memory, a check is made to see if this in in cache memory… If it is, the data is supplied, If it is not in the cache, a block of main memory, consisting of a fixed number of words is loaded to the cache. Because of the phenomenon of locality of references, when a block of data is fetched into the cache, it is likely that there will be future references to that same memory location or to other words in the block. Elements of Cache Design While there are a large number of cache implementations, there are a few basic design elements that serve to classify and differentiate cache architectures… Cache Addresses Cache Size Mapping Function Replacement Algorithm Write Policy Line Size Number of Caches Cache Addresses Almost all non-embedded processors support virtual memory. Virtual memory in essence allows a program to address memory from a logical point of view without needing to worry about the amount of physical memory available. When virtual addresses are used the designer may choose to place the cache between the MMU (memory management unit) and the processor or between the MMU and main memory. The disadvantage of virtual memory is that most virtual memory systems supply each application with the same virtual memory address space (each application sees virtual memory starting at memory address 0), which means the cache memory must be completely flushed with each application context switch or extra bits must be added to each line of the cache to identify which virtual address space the address refers to. Cache Size We would like the size of the cache to be small enough so that the overall average cost per bit is close to that of main memory alone and large enough so that the overall average access time is close to that of the cache alone. Also, larger caches are slightly slower than smaller ones. Mapping Function Because there are fewer cache lines than main memory blocks, an algorithm is needed for mapping main memory blocks into cache lines. The choice of mapping function dictates how the cache is organized. Three techniques can be used… Direct – simplest technique, maps each block of main memory into only one possible cache line Associative – Each main memory block to be loaded into any line of the cache Set Associative – exhibits the strengths of both the direct and associative approaches while reducing their disadvantages For detailed explanations of each approach – read the text book (page 148 – 154) Replacement Algorithm For associative and set associating mapping a replacement algorithm is needed to determine which of the existing blocks in the cache must be replaced by a new block. There are four common approaches… LRU (Least recently used) FIFO (First in first out) LFU (Least frequently used) Random selection Write Policy When a block resident in the cache is to be replaced, there are two cases to consider If no writes to that block have happened in the cache – discard it If a write has occurred, a process needs to be initiated where the changes in the cache are propagated back to the main memory. There are several approaches to achieve this including… Write Through – all writes to the cache are done to the main memory as well at the point of the change Write Back – when a block is replaced, all dirty bits are written back to main memory The problem is complicated when we have multiple caches, there are techniques to accommodate for this but I have not summarized them. Line Size When a block of data is retrieved and placed in the cache, not only the desired word but also some number of adjacent words are retrieved. As the block size increases from very small to larger sizes, the hit ratio will at first increase because of the principle of locality, which states that the data in the vicinity of a referenced word are likely to be referenced in the near future. As the block size increases, more useful data are brought into cache. The hit ratio will begin to decrease as the block becomes even bigger and the probability of using the newly fetched information becomes less than the probability of using the newly fetched information that has to be replaced. Two specific effects come into play… Larger blocks reduce the number of blocks that fit into a cache. Because each block fetch overwrites older cache contents, a small number of blocks results in data being overwritten shortly after they are fetched. As a block becomes larger, each additional word is farther from the requested word and therefore less likely to be needed in the near future. The relationship between block size and hit ratio is complex, and no set approach is judged to be the best in all circumstances.   Pentium 4 and ARM cache organizations The processor core consists of four major components: Fetch/decode unit – fetches program instruction in order from the L2 cache, decodes these into a series of micro-operations, and stores the results in the L2 instruction cache Out-of-order execution logic – Schedules execution of the micro-operations subject to data dependencies and resource availability – thus micro-operations may be scheduled for execution in a different order than they were fetched from the instruction stream. As time permits, this unit schedules speculative execution of micro-operations that may be required in the future Execution units – These units execute micro-operations, fetching the required data from the L1 data cache and temporarily storing results in registers Memory subsystem – This unit includes the L2 and L3 caches and the system bus, which is used to access main memory when the L1 and L2 caches have a cache miss and to access the system I/O resources

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  • Beyond the Great Wall

    This highway is traversed everyday by roughly 338 million Chinese Internet users. With the largest population in the world of 1.3 billion, the increase of Chinese Internet users in the next years would undoubtedly be viably incremental. Reaching out to an established and growing target market of that size and potential at a relatively lower cost of advertising makes for a lucrative ratio.

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  • Top Notch Results With the Best SEO Techniques

    Search engine optimization is one method in which you would be able to attract higher number of traffic to your website. You can boost up your profit ratio with search engines. With this aspect you can perform a flourishing internet marketing. It is also very crucial that when you are forming the SEO campaign you include the factors which offer higher profits. You can be successful with your search engine optimization only if you make use of the fundamental points.

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  • CSS hack for Google Chrome and Safari

    - by Renso
    When wanting to hack css in an external stylesheet just for Google Chrome and Safari. Here is an example of where I override the margin-top for Chrome and Safari.Normal:#AccountMaintenanceWrapper #callDetailsPreviewWrapper{    border: none;    padding: 0px;    width: 209px;    position: fixed;    margin-top: 84px;    z-index: 1;}Google Chrome and Safari:@media screen and (-webkit-min-device-pixel-ratio:0){    #AccountMaintenanceWrapper #callDetailsPreviewWrapper    {        margin-top: 12px;    }}

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  • If I project a sphere in 3D will it be a circle?

    - by yuumei
    Assuming I have infinite vertices to represent the sphere, if I project the sphere from any position/scale in 3D to 2D, will it be a circle? I know it will not be a circle on the screen, because of scaling and different resolutions. But do field of view and aspect ratio effect the results? Edit: Sorry yes, I am talking about perspective projection. Seems the answer is no then, perspective will distort the sphere. Thanks!

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  • How I Use Emotional Keywords and Why it Works!

    A very important part of our marketing work is finding good keywords. The main criteria when choosing a keyword (apart from the good ratio between searches and results) is that we want the people to type that keyword to be in the end of their purchase cycle, looking to "take action" now and not just to "find some info".

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