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  • Scheme procedure problem

    - by Zun
    I defined the Scheme procedure to return another procedure with 2 parameters : (define (smooth f) (?(x dx)(/ (+ (f (- x dx)) (f x) (f (+ x dx))) 3.0))) if i run this procedure with sin procedure with 2 arguments 10 and 0.0001 then it is ok ((smooth sin) 10 0.0001) ==> -0.544021109075966 if i run this procedure recursively, then it has error ((smooth (smooth sin)) 10 0.0001) ==> procedure expects 2 arguments, given 1: #<promise:temp6> So can anyone tell me where is my problem? Thank you in advance !!! PS:this is apart of exercise 1.44 in SICP

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  • Can anyone explain this pience of snippet?

    - by karthick6891
    Can anyone explain the following code,forget the sin and cosine parts..Is it trying to build a space for the object objectsInScene = new Array(); for (var i=space; i<180; i+=space) { for (var angle=0; angle<360; angle+=space) { var object = {}; var x = Math.sin(radian*i)*radius; object.x = Math.cos(angle*radian)*x; object.y = Math.cos(radian*i)*radius; object.z = Math.sin(angle*radian)*x; objectsInScene.push(object); } }

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  • converting string to int in C++

    - by xbonez
    I am trying to convert a string I read in from a file to an int value so I can store it in an integer variable. This is what my code looks like: ifstream sin; sin.open("movie_output.txt"); string line; getline(sin,line); myMovie.setYear(atoi(line)); Over here, setYear is a mutator in the Movie class (myMovie is an object of Movie class) that looks like this: void Movie::setYear(unsigned int year) { year_ = year; } When I run the code, I get the following error: error C2664: 'atoi' : cannot convert parameter 1 from 'std::string' to 'const char *' 1> No user-defined-conversion operator available that can perform this conversion, or the operator cannot be called

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  • How to implement line of sight restriction in actionscript?

    - by Michiel Standaert
    I have a problem with a game i am programming. I am making some sort of security game and i would like to have some visual line of sight. The problem is that i can't restrict my line of sight so my cops can't see through the walls. Below you find the design, in which they can look through windows, but not walls. Further below you find an illustration of what my problem is exactly. this is what it looks like now. As you can see, the cops can see through walls. This is the map i would want to use to restrict the line of sight. So the way i am programming the line of sight now is just by calculating some points and drawing the sight accordingly, as shown below. Note that i also check for a hittest using bitmapdata to check whether or not my player has been spotted by any of the cops. private function setSight(e:Event=null):Boolean { g = copCanvas.graphics; g.clear(); for each(var cop:Cop in copCanvas.getChildren()) { var _angle:Number = cop.angle; var _radians:Number = (_angle * Math.PI) / 180; var _radius:Number = 50; var _x1:Number = cop.x + (cop.width/2); var _y1:Number = cop.y + (cop.height/2); var _baseX:Number = _x1 + (Math.cos(_radians) * _radius); var _baseY:Number = _y1 - (Math.sin(_radians) * _radius); var _x2:Number = _baseX + (25 * Math.sin(_radians)); var _y2:Number = _baseY + (25 * Math.cos(_radians)); var _x3:Number = _baseX - (25 * Math.sin(_radians)); var _y3:Number = _baseY - (25 * Math.cos(_radians)); g.beginFill(0xff0000, 0.3); g.moveTo(_x1, _y1); g.lineTo(_x2, _y2); g.lineTo(_x3, _y3); g.endFill(); } var _cops:BitmapData = new BitmapData(width, height, true, 0); _cops.draw(copCanvas); var _bmpd:BitmapData = new BitmapData(10, 10, true, 0); _bmpd.draw(me); if(_cops.hitTest(new Point(0, 0), 10, _bmpd, new Point(me.x, me.y), 255)) { gameover.alpha = 1; setTimeout(function():void{ gameover.alpha = 0; }, 5000); stop(); return true; } return false; } So now my question is: Is there someone who knows how to restrict the view so that the cops can't look through the walls? Thanks a lot in advance. ps: i have already looked at this tutorial by emanuele feronato, but i can't use the code to restric the visual line of sight.

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  • Que es Virtualbox?

    - by [email protected]
    VIRTUALBOX (Open Source para virtualización)Las herrramientas de virtualización se han puesto de moda de manera casi exponencial durante los últimos años. Una de ellas es VirtualBox, esta corre sobre diferentes sistemas operativos y para los desarrolladores e ingenieros en computo que desean realizar pruebas de productos sin afectar sus maquinas, esta es una de las mejores alternativas. Adicionalmente hay algo que tiene VirtualBox que aun es mejor, es un program Open Source, eso significa que la podras usar en tu maquina teniendo un costo de $0.Esta herramienta realiza basicamente la misma funcion que Vmware Workstation, e incluso puede ejecutar la maquinas virtuales creadas con vmware workstation sin necesidad de realizar conversiones de alguna manera.La ultima version disponible al dia de hoy es la 3.1.6 y este puede ser descargado facilmente.  Solo visita uno de estos sites."Get  the lastest VirtualBox version at here"http://www.virtualbox.org/wiki/Downloadso en la pagina de Oracle"Get  the lastest VirtualBox version at here"http://dlc.sun.com/virtualbox/vboxdownload.html

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  • Taking fixed direction on hemisphere and project to normal (openGL)

    - by Maik Xhani
    I am trying to perform sampling using hemisphere around a surface normal. I want to experiment with fixed directions (and maybe jitter slightly between frames). So I have those directions: vec3 sampleDirections[6] = {vec3(0.0f, 1.0f, 0.0f), vec3(0.0f, 0.5f, 0.866025f), vec3(0.823639f, 0.5f, 0.267617f), vec3(0.509037f, 0.5f, -0.700629f), vec3(-0.509037f, 0.5f, -0.700629), vec3(-0.823639f, 0.5f, 0.267617f)}; now I want the first direction to be projected on the normal and the others accordingly. I tried these 2 codes, both failing. This is what I used for random sampling (it doesn't seem to work well, the samples seem to be biased towards a certain direction) and I just used one of the fixed directions instead of s (here is the code of the random sample, when i used it with the fixed direction i didn't use theta and phi). vec3 CosWeightedRandomHemisphereDirection( vec3 n, float rand1, float rand2 ) float theta = acos(sqrt(1.0f-rand1)); float phi = 6.283185f * rand2; vec3 s = vec3(sin(theta) * cos(phi), sin(theta) * sin(phi), cos(theta)); vec3 v = normalize(cross(n,vec3(0.0072, 1.0, 0.0034))); vec3 u = cross(v, n); u = s.x*u; v = s.y*v; vec3 w = s.z*n; vec3 direction = u+v+w; return normalize(direction); } ** EDIT ** This is the new code vec3 FixedHemisphereDirection( vec3 n, vec3 sampleDir) { vec3 x; vec3 z; if(abs(n.x) < abs(n.y)){ if(abs(n.x) < abs(n.z)){ x = vec3(1.0f,0.0f,0.0f); }else{ x = vec3(0.0f,0.0f,1.0f); } }else{ if(abs(n.y) < abs(n.z)){ x = vec3(0.0f,1.0f,0.0f); }else{ x = vec3(0.0f,0.0f,1.0f); } } z = normalize(cross(x,n)); x = cross(n,z); mat3 M = mat3( x.x, n.x, z.x, x.y, n.y, z.y, x.z, n.z, z.z); return M*sampleDir; } So if my n = (0,0,1); and my sampleDir = (0,1,0); shouldn't the M*sampleDir be (0,0,1)? Cause that is what I was expecting.

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  • Circular Bullet Spread not Even

    - by SoulBeaver
    I'm creating a bullet shooter much in the style of Touhou. Right now I want to have a very simple circular shot being fired from the enemy. See this picture: As you can see, the spacing is very uneven, which isn't very good if you want to survive. The code I'm using is this: private function shoot() : void { const BULLETS_PER_WAVE : int = 72; var interval : Number = BULLETS_PER_WAVE / 360; for (var i : int = 0; i < BULLETS_PER_WAVE; ++i { var xSpeed : Number = GameConstants.BULLET_NORMAL_SPEED_X * Math.sin(i * interval); var ySpeed : Number = GameConstants.BULLET_NORMAL_SPEED_Y * Math.cos(i * interval); BulletFactory.createNormalBullet(bulletColor_, alice_.center, xSpeed, ySpeed); } canShoot_ = false; cooldownTimer_.start(); } I imagine my mistake is in the sin, cos functions, but I'm not entirely sure what's wrong.

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  • 2D Car Simulation with Throttle Linear Physics

    - by James
    I'm trying to make a simulation game for an automatic cruise control system. The system simulates a car on varying inclinations and throttle speeds. I've coded up to the car physics but these do note make sense. The dynamics of the simulation are specified as follows: a = V' - V T = (k1)V + ?(k2) + ma V' = (1 - (k1 / m) V) + T - ( k2 / m) * ? Where T = throttle position k1 = viscous friction V = speed V' = next speed ? = angle of incline k2 = m g sin ? a = acceleration m = mass Notice that the angle of incline in the equation is not chopped up by sin or cos. Even the equation for acceleration isn't right. Can anyone correct them or am I misinterpreting the physics?

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  • How much C/C++ knowledge is needed for Objective-C/iPhone development?

    - by BFree
    First, a little background. I'm a .Net developer (C#) and have over 5 years experience in both web development and desktop applications. I've been wanting to look into iPhone development for some time now, but for one reason or another always got side tracked. I finally have a potential project on the horizon, and I'm now going full steam ahead learning this stuff. My question is this: I haven't done any C/C++ programming since my schooling days, I've been living in managed land ever since. How much knowledge if any is needed to be successful as an iOS developer? Obviously memory management is something that I'll have to be conscious about (although with iOS 5 there seems to be something called ARC which should make my life easier), but what else? I'm not just talking about the C API (for example, in order to get the sin of a number, I call the sin() function), that's what Google is for. I'm talking about fundamental C/C++ idioms that the average C# developer is unaware of.

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  • Geolocation SQL query not finding exact location

    - by Iridium52
    I have been testing my geolocation query for some time now and I haven't found any issues with it until now. I am trying to search for all cities within a given radius, often times I'm searching for cities surrounding a city using that city's coords, but recently I tried searching around a city and found that the city itself was not returned. I have these cities as an excerpt in my database: city latitude longitude Saint-Mathieu 45.316708 -73.516253 Saint-Édouard 45.233374 -73.516254 Saint-Michel 45.233374 -73.566256 Saint-Rémi 45.266708 -73.616257 But when I run my query around the city of Saint-Rémi, with the following query... SELECT tblcity.city, tblcity.latitude, tblcity.longitude, truncate((degrees(acos( sin(radians(tblcity.latitude)) * sin(radians(45.266708)) + cos(radians(tblcity.latitude)) * cos(radians(45.266708)) * cos(radians(tblcity.longitude - -73.616257) ) ) ) * 69.09*1.6),1) as distance FROM tblcity HAVING distance < 10 ORDER BY distance desc I get these results: city latitude longitude distance Saint-Mathieu 45.316708 -73.516253 9.5 Saint-Édouard 45.233374 -73.516254 8.6 Saint-Michel 45.233374 -73.566256 5.3 The town of Saint-Rémi is missing from the search. So I tried a modified query hoping to get a better result: SELECT tblcity.city, tblcity.latitude, tblcity.longitude, truncate(( 6371 * acos( cos( radians( 45.266708 ) ) * cos( radians( tblcity.latitude ) ) * cos( radians( tblcity.longitude ) - radians( -73.616257 ) ) + sin( radians( 45.266708 ) ) * sin( radians( tblcity.latitude ) ) ) ),1) AS distance FROM tblcity HAVING distance < 10 ORDER BY distance desc But I get the same result... However, if I modify Saint-Rémi's coords slighly by changing the last digit of the lat or long by 1, both queries will return Saint-Rémi. Also, if I center the query on any of the other cities above, the searched city is returned in the results. Can anyone shed some light on what may be causing my queries above to not display the searched city of Saint-Rémi? I have added a sample of the table (with extra fields removed) below. I'm using MySQL 5.0.45, thanks in advance. CREATE TABLE `tblcity` ( `IDCity` int(1) NOT NULL auto_increment, `City` varchar(155) NOT NULL default '', `Latitude` decimal(9,6) NOT NULL default '0.000000', `Longitude` decimal(9,6) NOT NULL default '0.000000', PRIMARY KEY (`IDCity`) ) ENGINE=MyISAM AUTO_INCREMENT=52743 DEFAULT CHARSET=latin1 AUTO_INCREMENT=52743; INSERT INTO `tblcity` (`city`, `latitude`, `longitude`) VALUES ('Saint-Mathieu', 45.316708, -73.516253), ('Saint-Édouard', 45.233374, -73.516254), ('Saint-Michel', 45.233374, -73.566256), ('Saint-Rémi', 45.266708, -73.616257);

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  • Simple haskell string manage

    - by paurullan
    Theres is a little problem I want to solve with Haskell: let substitute a function that change all of the wildcards in a string for one concrete parameter. The function has de signature of: subs :: String -> String -> String -> String -- example: -- subs 'x' "x^3 + x + sin(x)" "6.2" will generate -- "6.2^3 + 6.2 + sin(6.2)"

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  • pandas: complex filter on rows of DataFrame

    - by duckworthd
    I would like to filter rows by a function of each row, e.g. def f(row): return sin(row['velocity'])/np.prod(['masses']) > 5 df = pandas.DataFrame(...) filtered = df[apply_to_all_rows(df, f)] Or for another more complex, contrived example, def g(row): if row['col1'].method1() == 1: val = row['col1'].method2() / row['col1'].method3(row['col3'], row['col4']) else: val = row['col2'].method5(row['col6']) return np.sin(val) df = pandas.DataFrame(...) filtered = df[apply_to_all_rows(df, g)] How can I do so?

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  • Base de Datos Oracle, su mejor opción para reducir costos de IT

    - by Ivan Hassig
    Por Victoria Cadavid Sr. Sales Cosultant Oracle Direct Uno de los principales desafíos en la administración de centros de datos es la reducción de costos de operación. A medida que las compañías crecen y los proveedores de tecnología ofrecen soluciones cada vez más robustas, conservar el equilibrio entre desempeño, soporte al negocio y gestión del Costo Total de Propiedad es un desafío cada vez mayor para los Gerentes de Tecnología y para los Administradores de Centros de Datos. Las estrategias más comunes para conseguir reducción en los costos de administración de Centros de Datos y en la gestión de Tecnología de una organización en general, se enfocan en la mejora del desempeño de las aplicaciones, reducción del costo de administración y adquisición de hardware, reducción de los costos de almacenamiento, aumento de la productividad en la administración de las Bases de Datos y mejora en la atención de requerimientos y prestación de servicios de mesa de ayuda, sin embargo, las estrategias de reducción de costos deben contemplar también la reducción de costos asociados a pérdida y robo de información, cumplimiento regulatorio, generación de valor y continuidad del negocio, que comúnmente se conciben como iniciativas aisladas que no siempre se adelantan con el ánimo de apoyar la reducción de costos. Una iniciativa integral de reducción de costos de TI, debe contemplar cada uno de los factores que  generan costo y pueden ser optimizados. En este artículo queremos abordar la reducción de costos de tecnología a partir de la adopción del que según los expertos es el motor de Base de Datos # del mercado.Durante años, la base de datos Oracle ha sido reconocida por su velocidad, confiabilidad, seguridad y capacidad para soportar cargas de datos tanto de aplicaciones altamente transaccionales, como de Bodegas de datos e incluso análisis de Big Data , ofreciendo alto desempeño y facilidades de administración, sin embrago, cuando pensamos en proyectos de reducción de costos de IT, además de la capacidad para soportar aplicaciones (incluso aplicaciones altamente transaccionales) con alto desempeño, pensamos en procesos de automatización, optimización de recursos, consolidación, virtualización e incluso alternativas más cómodas de licenciamiento. La Base de Datos Oracle está diseñada para proveer todas las capacidades que un área de tecnología necesita para reducir costos, adaptándose a los diferentes escenarios de negocio y a las capacidades y características de cada organización.Es así, como además del motor de Base de Datos, Oracle ofrece una serie de soluciones para optimizar la administración de la información a través de mecanismos de optimización del uso del storage, continuidad del Negocio, consolidación de infraestructura, seguridad y administración automática, que propenden por un mejor uso de los recursos de tecnología, ofrecen opciones avanzadas de configuración y direccionan la reducción de los tiempos de las tareas operativas más comunes. Una de las opciones de la base de datos que se pueden provechar para reducir costos de hardware es Oracle Real Application Clusters. Esta solución de clustering permite que varios servidores (incluso servidores de bajo costo) trabajen en conjunto para soportar Grids o Nubes Privadas de Bases de Datos, proporcionando los beneficios de la consolidación de infraestructura, los esquemas de alta disponibilidad, rápido desempeño y escalabilidad por demanda, haciendo que el aprovisionamiento, el mantenimiento de las bases de datos y la adición de nuevos nodos se lleve e cabo de una forma más rápida y con menos riesgo, además de apalancar las inversiones en servidores de menor costo. Otra de las soluciones que promueven la reducción de costos de Tecnología es Oracle In-Memory Database Cache que permite almacenar y procesar datos en la memoria de las aplicaciones, permitiendo el máximo aprovechamiento de los recursos de procesamiento de la capa media, lo que cobra mucho valor en escenarios de alta transaccionalidad. De este modo se saca el mayor provecho de los recursos de procesamiento evitando crecimiento innecesario en recursos de hardware. Otra de las formas de evitar inversiones innecesarias en hardware, aprovechando los recursos existentes, incluso en escenarios de alto crecimiento de los volúmenes de información es la compresión de los datos. Oracle Advanced Compression permite comprimir hasta 4 veces los diferentes tipos de datos, mejorando la capacidad de almacenamiento, sin comprometer el desempeño de las aplicaciones. Desde el lado del almacenamiento también se pueden conseguir reducciones importantes de los costos de IT. En este escenario, la tecnología propia de la base de Datos Oracle ofrece capacidades de Administración Automática del Almacenamiento que no solo permiten una distribución óptima de los datos en los discos físicos para garantizar el máximo desempeño, sino que facilitan el aprovisionamiento y la remoción de discos defectuosos y ofrecen balanceo y mirroring, garantizando el uso máximo de cada uno de los dispositivos y la disponibilidad de los datos. Otra de las soluciones que facilitan la administración del almacenamiento es Oracle Partitioning, una opción de la Base de Datos que permite dividir grandes tablas en estructuras más pequeñas. Esta aproximación facilita la administración del ciclo de vida de la información y permite por ejemplo, separar los datos históricos (que generalmente se convierten en información de solo lectura y no tienen un alto volumen de consulta) y enviarlos a un almacenamiento de bajo costos, conservando la data activa en dispositivos de almacenamiento más ágiles. Adicionalmente, Oracle Partitioning facilita la administración de las bases de datos que tienen un gran volumen de registros y mejora el desempeño de la base de datos gracias a la posibilidad de optimizar las consultas haciendo uso únicamente de las particiones relevantes de una tabla o índice en el proceso de búsqueda. Otros factores adicionales, que pueden generar costos innecesarios a los departamentos de Tecnología son: La pérdida, corrupción o robo de datos y la falta de disponibilidad de las aplicaciones para dar soporte al negocio. Para evitar este tipo de situaciones que pueden acarrear multas y pérdida de negocios y de dinero, Oracle ofrece soluciones que permiten proteger y auditar la base de datos, recuperar la información en caso de corrupción o ejecución de acciones que comprometan la integridad de la información y soluciones que permitan garantizar que la información de las aplicaciones tenga una disponibilidad de 7x24. Ya hablamos de los beneficios de Oracle RAC, para facilitar los procesos de Consolidación y mejorar el desempeño de las aplicaciones, sin embrago esta solución, es sumamente útil en escenarios dónde las organizaciones de quieren garantizar una alta disponibilidad de la información, ante fallo de los servidores o en eventos de desconexión planeada para realizar labores de mantenimiento. Además de Oracle RAC, existen soluciones como Oracle Data Guard y Active Data Guard que permiten replicar de forma automática las bases de datos hacia un centro de datos de contingencia, permitiendo una recuperación inmediata ante eventos que deshabiliten por completo un centro de datos. Además de lo anterior, Active Data Guard, permite aprovechar la base de datos de contingencia para realizar labores de consulta, mejorando el desempeño de las aplicaciones. Desde el punto de vista de mejora en la seguridad, Oracle cuenta con soluciones como Advanced security que permite encriptar los datos y los canales a través de los cueles se comparte la información, Total Recall, que permite visualizar los cambios realizados a la base de datos en un momento determinado del tiempo, para evitar pérdida y corrupción de datos, Database Vault que permite restringir el acceso de los usuarios privilegiados a información confidencial, Audit Vault, que permite verificar quién hizo qué y cuándo dentro de las bases de datos de una organización y Oracle Data Masking que permite enmascarar los datos para garantizar la protección de la información sensible y el cumplimiento de las políticas y normas relacionadas con protección de información confidencial, por ejemplo, mientras las aplicaciones pasan del ambiente de desarrollo al ambiente de producción. Como mencionamos en un comienzo, las iniciativas de reducción de costos de tecnología deben apalancarse en estrategias que contemplen los diferentes factores que puedan generar sobre costos, los factores de riesgo que puedan acarrear costos no previsto, el aprovechamiento de los recursos actuales, para evitar inversiones innecesarias y los factores de optimización que permitan el máximo aprovechamiento de las inversiones actuales. Como vimos, todas estas iniciativas pueden ser abordadas haciendo uso de la tecnología de Oracle a nivel de Base de Datos, lo más importante es detectar los puntos críticos a nivel de riesgo, diagnosticar las proporción en que están siendo aprovechados los recursos actuales y definir las prioridades de la organización y del área de IT, para así dar inicio a todas aquellas iniciativas que de forma gradual, van a evitar sobrecostos e inversiones innecesarias, proporcionando un mayor apoyo al negocio y un impacto significativo en la productividad de la organización. Más información http://www.oracle.com/lad/products/database/index.html?ssSourceSiteId=otnes 1Fuente: Market Share: All Software Markets, Worldwide 2011 by Colleen Graham, Joanne Correia, David Coyle, Fabrizio Biscotti, Matthew Cheung, Ruggero Contu, Yanna Dharmasthira, Tom Eid, Chad Eschinger, Bianca Granetto, Hai Hong Swinehart, Sharon Mertz, Chris Pang, Asheesh Raina, Dan Sommer, Bhavish Sood, Marianne D'Aquila, Laurie Wurster and Jie Zhang. - March 29, 2012 2Big Data: Información recopilada desde fuentes no tradicionales como blogs, redes sociales, email, sensores, fotografías, grabaciones en video, etc. que normalmente se encuentran de forma no estructurada y en un gran volumen

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  • Applications: The Mathematics of Movement, Part 2

    - by TechTwaddle
    In part 1 of this series we saw how we can make the marble move towards the click point, with a fixed speed. In this post we’ll see, first, how to get rid of Atan2(), sine() and cosine() in our calculations, and, second, reducing the speed of the marble as it approaches the destination, so it looks like the marble is easing into it’s final position. As I mentioned in one of the previous posts, this is achieved by making the speed of the marble a function of the distance between the marble and the destination point. Getting rid of Atan2(), sine() and cosine() Ok, to be fair we are not exactly getting rid of these trigonometric functions, rather, replacing one form with another. So instead of writing sin(?), we write y/length. You see the point. So instead of using the trig functions as below, double x = destX - marble1.x; double y = destY - marble1.y; //distance between destination and current position, before updating marble position distanceSqrd = x * x + y * y; double angle = Math.Atan2(y, x); //Cos and Sin give us the unit vector, 6 is the value we use to magnify the unit vector along the same direction incrX = speed * Math.Cos(angle); incrY = speed * Math.Sin(angle); marble1.x += incrX; marble1.y += incrY; we use the following, double x = destX - marble1.x; double y = destY - marble1.y; //distance between destination and marble (before updating marble position) lengthSqrd = x * x + y * y; length = Math.Sqrt(lengthSqrd); //unit vector along the same direction as vector(x, y) unitX = x / length; unitY = y / length; //update marble position incrX = speed * unitX; incrY = speed * unitY; marble1.x += incrX; marble1.y += incrY; so we replaced cos(?) with x/length and sin(?) with y/length. The result is the same.   Adding oomph to the way it moves In the last post we had the speed of the marble fixed at 6, double speed = 6; to make the marble decelerate as it moves, we have to keep updating the speed of the marble in every frame such that the speed is calculated as a function of the length. So we may have, speed = length/12; ‘length’ keeps decreasing as the marble moves and so does speed. The Form1_MouseUp() function remains the same as before, here is the UpdatePosition() method, private void UpdatePosition() {     double incrX = 0, incrY = 0;     double lengthSqrd = 0, length = 0, lengthSqrdNew = 0;     double unitX = 0, unitY = 0;     double speed = 0;     double x = destX - marble1.x;     double y = destY - marble1.y;     //distance between destination and marble (before updating marble position)     lengthSqrd = x * x + y * y;     length = Math.Sqrt(lengthSqrd);     //unit vector along the same direction as vector(x, y)     unitX = x / length;     unitY = y / length;     //speed as a function of length     speed = length / 12;     //update marble position     incrX = speed * unitX;     incrY = speed * unitY;     marble1.x += incrX;     marble1.y += incrY;     //check for bounds     if ((int)marble1.x < MinX + marbleWidth / 2)     {         marble1.x = MinX + marbleWidth / 2;     }     else if ((int)marble1.x > (MaxX - marbleWidth / 2))     {         marble1.x = MaxX - marbleWidth / 2;     }     if ((int)marble1.y < MinY + marbleHeight / 2)     {         marble1.y = MinY + marbleHeight / 2;     }     else if ((int)marble1.y > (MaxY - marbleHeight / 2))     {         marble1.y = MaxY - marbleHeight / 2;     }     //distance between destination and marble (after updating marble position)     x = destX - (marble1.x);     y = destY - (marble1.y);     lengthSqrdNew = x * x + y * y;     /*      * End Condition:      * 1. If there is not much difference between lengthSqrd and lengthSqrdNew      * 2. If the marble has moved more than or equal to a distance of totLenToTravel (see Form1_MouseUp)      */     x = startPosX - marble1.x;     y = startPosY - marble1.y;     double totLenTraveledSqrd = x * x + y * y;     if ((int)totLenTraveledSqrd >= (int)totLenToTravelSqrd)     {         System.Console.WriteLine("Stopping because Total Len has been traveled");         timer1.Enabled = false;     }     else if (Math.Abs((int)lengthSqrd - (int)lengthSqrdNew) < 4)     {         System.Console.WriteLine("Stopping because no change in Old and New");         timer1.Enabled = false;     } } A point to note here is that, in this implementation, the marble never stops because it travelled a distance of totLenToTravelSqrd (first if condition). This happens because speed is a function of the length. During the final few frames length becomes very small and so does speed; and so the amount by which the marble shifts is quite small, and the second if condition always hits true first. I’ll end this series with a third post. In part 3 we will cover two things, one, when the user clicks, the marble keeps moving in that direction, rebounding off the screen edges and keeps moving forever. Two, when the user clicks on the screen, the marble moves towards it, with it’s speed reducing by every frame. It doesn’t come to a halt when the destination point is reached, instead, it continues to move, rebounds off the screen edges and slowly comes to halt. The amount of time that the marble keeps moving depends on how far the user clicks from the marble. I had mentioned this second situation here. Finally, here’s a video of this program running,

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  • Rotate triangle so that its tip points in the direction of the point on the screen that we last touched

    - by Sid
    OpenGL ES - Android. Hello all, I am unable to rotate the triangle accordingly in such a way that its tip always points to my finger. What i did : Constructed a triangle in by GL.GL_TRIANGLES. Added touch events to it. I can rotate the triangle along my Z-axis successfully. Even made the vector class for it. What i need : Each time when I touch the screen, I want to rotate the triangle to face the touch point. Need some help. Here's what i implemented. I wonder that where i am going wrong? My code : public class Graphic2DTriangle { private FloatBuffer vertexBuffer; private ByteBuffer indexBuffer; private float[] vertices = { -1.0f,-1.0f, 0.0f, 2.0f, 0.0f, 0.0f, -1.0f, 1.0f, 0.0f }; private byte[] indices = { 0, 1, 2 }; public Graphic2DTriangle() { ByteBuffer vbb = ByteBuffer.allocateDirect(vertices.length * 4); vbb.order(ByteOrder.nativeOrder()); // Use native byte order vertexBuffer = vbb.asFloatBuffer(); // Convert byte buffer to float vertexBuffer.put(vertices); // Copy data into buffer vertexBuffer.position(0); // Rewind // Setup index-array buffer. Indices in byte. indexBuffer = ByteBuffer.allocateDirect(indices.length); indexBuffer.put(indices); indexBuffer.position(0); } public void draw(GL10 gl) { gl.glEnableClientState(GL10.GL_VERTEX_ARRAY); gl.glVertexPointer(3, GL10.GL_FLOAT, 0, vertexBuffer); gl.glDrawElements(GL10.GL_TRIANGLES, indices.length, GL10.GL_UNSIGNED_BYTE, indexBuffer); gl.glDisableClientState(GL10.GL_VERTEX_ARRAY); } } My SurfaceView class where i've done some Touch Events. public class BallThrowGLSurfaceView extends GLSurfaceView{ MySquareRender _renderObj; View _viewObj; float oldX,oldY,dX,dY; final float TOUCH_SCALE_FACTOR = 0.6f; Vector2 touchPos = new Vector2(); float angle=0; public BallThrowGLSurfaceView(Context context) { super(context); // TODO Auto-generated constructor stub _renderObj = new MySquareRender(context); this.setRenderer(_renderObj); this.setRenderMode(RENDERMODE_WHEN_DIRTY); } @Override public boolean onTouchEvent(MotionEvent event) { // TODO Auto-generated method stub touchPos.x = event.getX(); touchPos.y = event.getY(); Log.i("Co-ord", touchPos.x+"hh"+touchPos.y); switch(event.getAction()){ case MotionEvent.ACTION_MOVE : dX = touchPos.x - oldX; dY = touchPos.y - oldY; if(touchPos.y > getHeight()/2){ dX = dX*-1; } if(touchPos.x < getWidth()/2){ dY = dY*-1; } _renderObj.mAngle += (dX+dY) * TOUCH_SCALE_FACTOR; requestRender(); Log.i("AngleCo-ord", _renderObj.mAngle +"hh"); } oldX = touchPos.x; oldY = touchPos.y; Log.i("OldCo-ord", oldX+" hh "+oldY); return true; } } Last but not the least. My vector2 class. public class Vector2 { public static float TO_RADIANS = (1 / 180.0f) * (float) Math.PI; public static float TO_DEGREES = (1 / (float) Math.PI) * 180; public float x, y; public Vector2() { } public Vector2(float x, float y) { this.x = x; this.y = y; } public Vector2(Vector2 other) { this.x = other.x; this.y = other.y; } public Vector2 cpy() { return new Vector2(x, y); } public Vector2 set(float x, float y) { this.x = x; this.y = y; return this; } public Vector2 set(Vector2 other) { this.x = other.x; this.y = other.y; return this; } public Vector2 add(float x, float y) { this.x += x; this.y += y; return this; } public Vector2 add(Vector2 other) { this.x += other.x; this.y += other.y; return this; } public Vector2 sub(float x, float y) { this.x -= x; this.y -= y; return this; } public Vector2 sub(Vector2 other) { this.x -= other.x; this.y -= other.y; return this; } public Vector2 mul(float scalar) { this.x *= scalar; this.y *= scalar; return this; } public float len() { return FloatMath.sqrt(x * x + y * y); } public Vector2 nor() { float len = len(); if (len != 0) { this.x /= len; this.y /= len; } return this; } public float angle() { float angle = (float) Math.atan2(y, x) * TO_DEGREES; if (angle < 0) angle += 360; return angle; } public Vector2 rotate(float angle) { float rad = angle * TO_RADIANS; float cos = FloatMath.cos(rad); float sin = FloatMath.sin(rad); float newX = this.x * cos - this.y * sin; float newY = this.x * sin + this.y * cos; this.x = newX; this.y = newY; return this; } public float dist(Vector2 other) { float distX = this.x - other.x; float distY = this.y - other.y; return FloatMath.sqrt(distX * distX + distY * distY); } public float dist(float x, float y) { float distX = this.x - x; float distY = this.y - y; return FloatMath.sqrt(distX * distX + distY * distY); } public float distSquared(Vector2 other) { float distX = this.x - other.x; float distY = this.y - other.y; return distX * distX + distY * distY; } public float distSquared(float x, float y) { float distX = this.x - x; float distY = this.y - y; return distX * distX + distY * distY; } } PS : i am able to handle the touch events. I can rotate the triangle with the touch of my finger. But i want that ONE VERTEX of the triangle should point at my finger position respective of the position of my finger.

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  • Drawing Quadratic Bezier circles with a given radius: how to determine control points

    - by Casey
    Just to clarify; the code below works, but I don't understand where the formula for the variable "controlRadius" comes from. I wrote this function by dissecting an example I found elsewhere, but I can't find any explanation and the original code comments were not able to be translated. Thanks in advance //returns an array of quadratic Bezier segments public static function generateCircularQuadraticBezierSegments(radius:Number, numControlPoints:uint, centerX:Number, centerY:Number):Array { var segments:Array = []; var arcLength:Number = 2 * Math.PI / numControlPoints; var controlRadius:Number; var segment:QuadraticBezierSegment; for (var i:int = 0; i < numControlPoints; i++) { var startX:Number = centerX + radius * Math.cos(arcLength * i); var startY:Number = centerY + radius * Math.sin(arcLength * i); //control radius formula //where does it come from, why does it work? controlRadius = radius / Math.cos(arcLength * .5); //the control point is plotted halfway between the arcLength and uses the control radius var controlX:Number = centerX + controlRadius * Math.cos(arcLength * (i + 1) - arcLength * .5); var controlY:Number = centerY + controlRadius * Math.sin(arcLength * (i + 1) - arcLength * .5); var endX:Number = centerX + radius * Math.cos(arcLength * (i + 1)); var endY:Number = centerY + radius * Math.sin(arcLength * (i + 1)); segment = new QuadraticBezierSegment(new Point(startX, startY), new Point(controlX, controlY), new Point(endX, endY)); segments.push(segment); } return segments; }

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  • MySQL 1064 error, works in command line and phpMyAdmin; not in app

    - by hundleyj
    Here is my query: select * from (select *, 3956 * 2 * ASIN(SQRT(POWER(SIN(RADIANS(45.5200077 - lat)/ 2), 2) + COS(RADIANS(45.5200077)) * COS(RADIANS(lat)) * POWER(SIN(RADIANS(-122.6942014 - lng)/2),2))) AS distance from stops order by distance, route asc) as p group by route, dir order by distance asc limit 10 This works fine at the command line and in PHPMyAdmin. I'm using Dbslayer to connect to MySQL via my JavaScript backend, and the request is returning a 1064 error. Here is the encoded DBSlayer request string: http://localhost:9090/db?{%22SQL%22:%22select%20*%20from%20%28select%20*,%203956%20*%202%20*%20ASIN%28SQRT%28POWER%28SIN%28RADIANS%2845.5200077%20-%20lat%29/%202%29,%202%29%20+%20COS%28RADIANS%2845.5200077%29%29%20*%20COS%28RADIANS%28lat%29%29%20*%20POWER%28SIN%28RADIANS%28-122.6942014%20-%20lng%29/2%29,2%29%29%29%20AS%20distance%20from%20%60stops%60%20order%20by%20%60distance%60,%20%60route%60%20asc%29%20as%20p%20group%20by%20%60route%60,%20%60dir%60%20order%20by%20%60distance%60%20asc%20limit%2010%22} And the response: {"MYSQL_ERRNO" : 1064 , "MYSQL_ERROR" : "You have an error in your SQL syntax; check the manual that corresponds to your MySQL server version for the right syntax to use near '(RADIANS(45.5200077)) * COS(RADIANS(lat)) * POWER(SIN(RADIANS(-122.6942014 - lng' at line 1" , "SERVER" : "trimet"} Thanks!

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  • Matlab fft function

    - by CTZStef
    The code below is from the Matlab 2011a help about fft function. I think there is a problem here : why do they multiply t(1:50) by Fs, and then say it's time in millisecond ? Certainly, it happens to be true in this very particular case, but change the value of Fs to, say, 2000, and it won't work anymore, obviously because of this factor of 2. Right ? Quite misleading, isn't it ? What do I miss ? Fs = 1000; % Sampling frequency T = 1/Fs; % Sample time L = 1000; % Length of signal t = (0:L-1)*T; % Time vector % Sum of a 50 Hz sinusoid and a 120 Hz sinusoid x = 0.7*sin(2*pi*50*t) + sin(2*pi*120*t); y = x + 2*randn(size(t)); % Sinusoids plus noise plot(Fs*t(1:50),y(1:50)) title('Signal Corrupted with Zero-Mean Random Noise') xlabel('time (milliseconds)') Clearer with this : fs = 2000; % Sampling frequency T = 1 / fs; % Sample time L = 1000; % Length of signal t2 = (0:L-1)*T; % Time vector f = 50; % signal frequency s2 = sin(2*pi*f*t2); figure, plot(fs*t2(1:50),s2(1:50)); % NOT good figure, plot(t2(1:50),s2(1:50)); % good

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  • Cannot change font size /type in plots

    - by Sameet Nabar
    I recently had to re-install my operating system (Ubuntu). The only thing I did differently is that I installed Matlab on a separate partition, not the main Ubuntu partition. After re-installing, the fonts in my plots are no longer configurable. For example, if I ask the title font to be bold, it doesn't happen. I ran the sample code below on my computer and then on my colleague's computer and the 2 results are attached. This cannot be a problem with the code; rather in the settings of Matlab. Could somebody please tell me what settings I need to change? Thanks in advance for your help. Regards, Sameet. x1=-pi:.1:pi; x2=-pi:pi/10:pi; y1=sin(x1); y2=tan(sin(x2)) - sin(tan(x2)); [AX,H1,H2]=plotyy(x1,y1,x2,y2); xlabel ('Time (hh:mm)'); ylabel (AX(1), 'Plot1'); ylabel (AX(2), 'Plot2'); axes(AX(2)) set(H1,'linestyle','none','marker','.'); set(H2,'linestyle','none','marker','.'); title('Plot Title','FontWeight','bold'); set(gcf, 'Visible', 'off'); [legh, objh] = legend([H1 H2],'Plot1', 'Plot2','location','Best'); set(legend,'FontSize',8); print -dpng Trial.png; Bad image: http://imageshack.us/photo/my-images/708/trial1u.png/ Good image: http://imageshack.us/photo/my-images/87/trial2.png/

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  • LWJGL SlickUtil Texture Binding

    - by Matthew Dockerty
    I am making a 3D game using LWJGL and I have a texture class with static variables so that I only need to load textures once, even if I need to use them more than once. I am using Slick Util for this. When I bind a texture it works fine, but then when I try to render something else after I have rendered the model with the texture, the texture is still being bound. How do I unbind the texture and set the rendermode to the one that was in use before any textures were bound? Some of my code is below. The problem I am having is the player texture is being used in the box drawn around the player after it the model has been rendered. Model.java public class Model { public List<Vector3f> vertices = new ArrayList<Vector3f>(); public List<Vector3f> normals = new ArrayList<Vector3f>(); public ArrayList<Vector2f> textureCoords = new ArrayList<Vector2f>(); public List<Face> faces = new ArrayList<Face>(); public static Model TREE; public static Model PLAYER; public static void loadModels() { try { TREE = OBJLoader.loadModel(new File("assets/model/tree_pine_0.obj")); PLAYER = OBJLoader.loadModel(new File("assets/model/player.obj")); } catch (Exception e) { e.printStackTrace(); } } public void render(Vector3f position, Vector3f scale, Vector3f rotation, Texture texture, float shinyness) { glPushMatrix(); { texture.bind(); glColor3f(1, 1, 1); glTranslatef(position.x, position.y, position.z); glScalef(scale.x, scale.y, scale.z); glRotatef(rotation.x, 1, 0, 0); glRotatef(rotation.y, 0, 1, 0); glRotatef(rotation.z, 0, 0, 1); glMaterialf(GL_FRONT, GL_SHININESS, shinyness); glBegin(GL_TRIANGLES); { for (Face face : faces) { Vector2f t1 = textureCoords.get((int) face.textureCoords.x - 1); glTexCoord2f(t1.x, t1.y); Vector3f n1 = normals.get((int) face.normal.x - 1); glNormal3f(n1.x, n1.y, n1.z); Vector3f v1 = vertices.get((int) face.vertex.x - 1); glVertex3f(v1.x, v1.y, v1.z); Vector2f t2 = textureCoords.get((int) face.textureCoords.y - 1); glTexCoord2f(t2.x, t2.y); Vector3f n2 = normals.get((int) face.normal.y - 1); glNormal3f(n2.x, n2.y, n2.z); Vector3f v2 = vertices.get((int) face.vertex.y - 1); glVertex3f(v2.x, v2.y, v2.z); Vector2f t3 = textureCoords.get((int) face.textureCoords.z - 1); glTexCoord2f(t3.x, t3.y); Vector3f n3 = normals.get((int) face.normal.z - 1); glNormal3f(n3.x, n3.y, n3.z); Vector3f v3 = vertices.get((int) face.vertex.z - 1); glVertex3f(v3.x, v3.y, v3.z); } texture.release(); } glEnd(); } glPopMatrix(); } } Textures.java public class Textures { public static Texture FLOOR; public static Texture PLAYER; public static Texture SKYBOX_TOP; public static Texture SKYBOX_BOTTOM; public static Texture SKYBOX_FRONT; public static Texture SKYBOX_BACK; public static Texture SKYBOX_LEFT; public static Texture SKYBOX_RIGHT; public static void loadTextures() { try { FLOOR = TextureLoader.getTexture("PNG", new FileInputStream(new File("assets/model/floor.png"))); FLOOR.setTextureFilter(GL11.GL_NEAREST); PLAYER = TextureLoader.getTexture("PNG", new FileInputStream(new File("assets/model/tree_pine_0.png"))); PLAYER.setTextureFilter(GL11.GL_NEAREST); SKYBOX_TOP = TextureLoader.getTexture("PNG", new FileInputStream(new File("assets/textures/skybox_top.png"))); SKYBOX_TOP.setTextureFilter(GL11.GL_NEAREST); SKYBOX_BOTTOM = TextureLoader.getTexture("PNG", new FileInputStream(new File("assets/textures/skybox_bottom.png"))); SKYBOX_BOTTOM.setTextureFilter(GL11.GL_NEAREST); SKYBOX_FRONT = TextureLoader.getTexture("PNG", new FileInputStream(new File("assets/textures/skybox_front.png"))); SKYBOX_FRONT.setTextureFilter(GL11.GL_NEAREST); SKYBOX_BACK = TextureLoader.getTexture("PNG", new FileInputStream(new File("assets/textures/skybox_back.png"))); SKYBOX_BACK.setTextureFilter(GL11.GL_NEAREST); SKYBOX_LEFT = TextureLoader.getTexture("PNG", new FileInputStream(new File("assets/textures/skybox_left.png"))); SKYBOX_LEFT.setTextureFilter(GL11.GL_NEAREST); SKYBOX_RIGHT = TextureLoader.getTexture("PNG", new FileInputStream(new File("assets/textures/skybox_right.png"))); SKYBOX_RIGHT.setTextureFilter(GL11.GL_NEAREST); } catch (Exception e) { e.printStackTrace(); } } } Player.java public class Player { private Vector3f position; private float yaw; private float moveSpeed; public Player(float x, float y, float z, float yaw, float moveSpeed) { this.position = new Vector3f(x, y, z); this.yaw = yaw; this.moveSpeed = moveSpeed; } public void update() { if (Keyboard.isKeyDown(Keyboard.KEY_W)) walkForward(moveSpeed); if (Keyboard.isKeyDown(Keyboard.KEY_S)) walkBackwards(moveSpeed); if (Keyboard.isKeyDown(Keyboard.KEY_A)) strafeLeft(moveSpeed); if (Keyboard.isKeyDown(Keyboard.KEY_D)) strafeRight(moveSpeed); if (Mouse.isButtonDown(0)) yaw += Mouse.getDX(); LowPolyRPG.getInstance().getCamera().setPosition(-position.x, -position.y, -position.z); LowPolyRPG.getInstance().getCamera().setYaw(yaw); } public void walkForward(float distance) { position.setX(position.getX() + distance * (float) Math.sin(Math.toRadians(yaw))); position.setZ(position.getZ() - distance * (float) Math.cos(Math.toRadians(yaw))); } public void walkBackwards(float distance) { position.setX(position.getX() - distance * (float) Math.sin(Math.toRadians(yaw))); position.setZ(position.getZ() + distance * (float) Math.cos(Math.toRadians(yaw))); } public void strafeLeft(float distance) { position.setX(position.getX() + distance * (float) Math.sin(Math.toRadians(yaw - 90))); position.setZ(position.getZ() - distance * (float) Math.cos(Math.toRadians(yaw - 90))); } public void strafeRight(float distance) { position.setX(position.getX() + distance * (float) Math.sin(Math.toRadians(yaw + 90))); position.setZ(position.getZ() - distance * (float) Math.cos(Math.toRadians(yaw + 90))); } public void render() { Model.PLAYER.render(new Vector3f(position.x, position.y + 12, position.z), new Vector3f(3, 3, 3), new Vector3f(0, -yaw + 90, 0), Textures.PLAYER, 128); GL11.glPushMatrix(); GL11.glTranslatef(position.getX(), position.getY(), position.getZ()); GL11.glRotatef(-yaw, 0, 1, 0); GL11.glScalef(5.8f, 21, 2.2f); GL11.glDisable(GL11.GL_LIGHTING); GL11.glLineWidth(3); GL11.glBegin(GL11.GL_LINE_STRIP); GL11.glColor3f(1, 1, 1); glVertex3f(1f, 0f, -1f); glVertex3f(-1f, 0f, -1f); glVertex3f(-1f, 1f, -1f); glVertex3f(1f, 1f, -1f); glVertex3f(-1f, 0f, 1f); glVertex3f(1f, 0f, 1f); glVertex3f(1f, 1f, 1f); glVertex3f(-1f, 1f, 1f); glVertex3f(1f, 1f, -1f); glVertex3f(-1f, 1f, -1f); glVertex3f(-1f, 1f, 1f); glVertex3f(1f, 1f, 1f); glVertex3f(1f, 0f, 1f); glVertex3f(-1f, 0f, 1f); glVertex3f(-1f, 0f, -1f); glVertex3f(1f, 0f, -1f); glVertex3f(1f, 0f, 1f); glVertex3f(1f, 0f, -1f); glVertex3f(1f, 1f, -1f); glVertex3f(1f, 1f, 1f); glVertex3f(-1f, 0f, -1f); glVertex3f(-1f, 0f, 1f); glVertex3f(-1f, 1f, 1f); glVertex3f(-1f, 1f, -1f); GL11.glEnd(); GL11.glEnable(GL11.GL_LIGHTING); GL11.glPopMatrix(); } public Vector3f getPosition() { return new Vector3f(-position.x, -position.y, -position.z); } public float getX() { return position.getX(); } public float getY() { return position.getY(); } public float getZ() { return position.getZ(); } public void setPosition(Vector3f position) { this.position = position; } public void setPosition(float x, float y, float z) { this.position.setX(x); this.position.setY(y); this.position.setZ(z); } } Thanks for the help.

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  • Combine 3D objects in XNA 4

    - by Christoph
    Currently I am writing on my thesis for university, the theme I am working on is 3D Visualization of hierarchical structures using cone trees. I want to do is to draw a cone and arrange a number of spheres at the bottom of the cone. The spheres should be arranged according to the radius and the number of spheres correctly. As you can imagine I need a lot of these cone/sphere combinations. First Attempt I was able to find some tutorials that helped with drawing cones and spheres. Cone public Cone(GraphicsDevice device, float height, int tessellation, string name, List<Sphere> children) { //prepare children and calculate the children spacing and radius of the cone if (children == null || children.Count == 0) { throw new ArgumentNullException("children"); } this.Height = height; this.Name = name; this.Children = children; //create the cone if (tessellation < 3) { throw new ArgumentOutOfRangeException("tessellation"); } //Create a ring of triangels around the outside of the cones bottom for (int i = 0; i < tessellation; i++) { Vector3 normal = this.GetCircleVector(i, tessellation); // add the vertices for the top of the cone base.AddVertex(Vector3.Up * height, normal); //add the bottom circle base.AddVertex(normal * this.radius + Vector3.Down * height, normal); //Add indices base.AddIndex(i * 2); base.AddIndex(i * 2 + 1); base.AddIndex((i * 2 + 2) % (tessellation * 2)); base.AddIndex(i * 2 + 1); base.AddIndex((i * 2 + 3) % (tessellation * 2)); base.AddIndex((i * 2 + 2) % (tessellation * 2)); } //create flate triangle to seal the bottom this.CreateCap(tessellation, height, this.Radius, Vector3.Down); base.InitializePrimitive(device); } Sphere public void Initialize(GraphicsDevice device, Vector3 qi) { int verticalSegments = this.Tesselation; int horizontalSegments = this.Tesselation * 2; //single vertex on the bottom base.AddVertex((qi * this.Radius) + this.lowering, Vector3.Down); for (int i = 0; i < verticalSegments; i++) { float latitude = ((i + 1) * MathHelper.Pi / verticalSegments) - MathHelper.PiOver2; float dy = (float)Math.Sin(latitude); float dxz = (float)Math.Cos(latitude); //Create a singe ring of latitudes for (int j = 0; j < horizontalSegments; j++) { float longitude = j * MathHelper.TwoPi / horizontalSegments; float dx = (float)Math.Cos(longitude) * dxz; float dz = (float)Math.Sin(longitude) * dxz; Vector3 normal = new Vector3(dx, dy, dz); base.AddVertex(normal * this.Radius, normal); } } // Finish with a single vertex at the top of the sphere. AddVertex((qi * this.Radius) + this.lowering, Vector3.Up); // Create a fan connecting the bottom vertex to the bottom latitude ring. for (int i = 0; i < horizontalSegments; i++) { AddIndex(0); AddIndex(1 + (i + 1) % horizontalSegments); AddIndex(1 + i); } // Fill the sphere body with triangles joining each pair of latitude rings. for (int i = 0; i < verticalSegments - 2; i++) { for (int j = 0; j < horizontalSegments; j++) { int nextI = i + 1; int nextJ = (j + 1) % horizontalSegments; base.AddIndex(1 + i * horizontalSegments + j); base.AddIndex(1 + i * horizontalSegments + nextJ); base.AddIndex(1 + nextI * horizontalSegments + j); base.AddIndex(1 + i * horizontalSegments + nextJ); base.AddIndex(1 + nextI * horizontalSegments + nextJ); base.AddIndex(1 + nextI * horizontalSegments + j); } } // Create a fan connecting the top vertex to the top latitude ring. for (int i = 0; i < horizontalSegments; i++) { base.AddIndex(CurrentVertex - 1); base.AddIndex(CurrentVertex - 2 - (i + 1) % horizontalSegments); base.AddIndex(CurrentVertex - 2 - i); } base.InitializePrimitive(device); } The tricky part now is to arrange the spheres at the bottom of the cone. I tried is to draw just the cone and then draw the spheres. I need a lot of these cones, so it would be pretty hard to calculate all the positions correctly. Second Attempt So the second try was to generate a object that builds all vertices of the cone and all of the spheres at once. So I was hoping to render a cone with all its spheres arranged correctly. After a short debug I found out that the cone is created and the first sphere, when it turn of the second sphere I am running into an OutOfBoundsException of ushort.MaxValue. Cone and Spheres public ConeWithSpheres(GraphicsDevice device, float height, float coneDiameter, float sphereDiameter, int coneTessellation, int sphereTessellation, int numberOfSpheres) { if (coneTessellation < 3) { throw new ArgumentException(string.Format("{0} is to small for the tessellation of the cone. The number must be greater or equal to 3", coneTessellation)); } if (sphereTessellation < 3) { throw new ArgumentException(string.Format("{0} is to small for the tessellation of the sphere. The number must be greater or equal to 3", sphereTessellation)); } //set properties this.Height = height; this.ConeDiameter = coneDiameter; this.SphereDiameter = sphereDiameter; this.NumberOfChildren = numberOfSpheres; //end set properties //generate the cone this.GenerateCone(device, coneTessellation); //generate the spheres //vector that defines the Y position of the sphere on the cones bottom Vector3 lowering = new Vector3(0, 0.888f, 0); this.GenerateSpheres(device, sphereTessellation, numberOfSpheres, lowering); } // ------ GENERATE CONE ------ private void GenerateCone(GraphicsDevice device, int coneTessellation) { int doubleTessellation = coneTessellation * 2; //Create a ring of triangels around the outside of the cones bottom for (int index = 0; index < coneTessellation; index++) { Vector3 normal = this.GetCircleVector(index, coneTessellation); //add the vertices for the top of the cone base.AddVertex(Vector3.Up * this.Height, normal); //add the bottom of the cone base.AddVertex(normal * this.ConeRadius + Vector3.Down * this.Height, normal); //add indices base.AddIndex(index * 2); base.AddIndex(index * 2 + 1); base.AddIndex((index * 2 + 2) % doubleTessellation); base.AddIndex(index * 2 + 1); base.AddIndex((index * 2 + 3) % doubleTessellation); base.AddIndex((index * 2 + 2) % doubleTessellation); } //create flate triangle to seal the bottom this.CreateCap(coneTessellation, this.Height, this.ConeRadius, Vector3.Down); base.InitializePrimitive(device); } // ------ GENERATE SPHERES ------ private void GenerateSpheres(GraphicsDevice device, int sphereTessellation, int numberOfSpheres, Vector3 lowering) { int verticalSegments = sphereTessellation; int horizontalSegments = sphereTessellation * 2; for (int childCount = 1; childCount < numberOfSpheres; childCount++) { //single vertex at the bottom of the sphere base.AddVertex((this.GetCircleVector(childCount, this.NumberOfChildren) * this.SphereRadius) + lowering, Vector3.Down); for (int verticalSegmentsCount = 0; verticalSegmentsCount < verticalSegments; verticalSegmentsCount++) { float latitude = ((verticalSegmentsCount + 1) * MathHelper.Pi / verticalSegments) - MathHelper.PiOver2; float dy = (float)Math.Sin(latitude); float dxz = (float)Math.Cos(latitude); //create a single ring of latitudes for (int horizontalSegmentsCount = 0; horizontalSegmentsCount < horizontalSegments; horizontalSegmentsCount++) { float longitude = horizontalSegmentsCount * MathHelper.TwoPi / horizontalSegments; float dx = (float)Math.Cos(longitude) * dxz; float dz = (float)Math.Sin(longitude) * dxz; Vector3 normal = new Vector3(dx, dy, dz); base.AddVertex((normal * this.SphereRadius) + lowering, normal); } } //finish with a single vertex at the top of the sphere base.AddVertex((this.GetCircleVector(childCount, this.NumberOfChildren) * this.SphereRadius) + lowering, Vector3.Up); //create a fan connecting the bottom vertex to the bottom latitude ring for (int i = 0; i < horizontalSegments; i++) { base.AddIndex(0); base.AddIndex(1 + (i + 1) % horizontalSegments); base.AddIndex(1 + i); } //Fill the sphere body with triangles joining each pair of latitude rings for (int i = 0; i < verticalSegments - 2; i++) { for (int j = 0; j < horizontalSegments; j++) { int nextI = i + 1; int nextJ = (j + 1) % horizontalSegments; base.AddIndex(1 + i * horizontalSegments + j); base.AddIndex(1 + i * horizontalSegments + nextJ); base.AddIndex(1 + nextI * horizontalSegments + j); base.AddIndex(1 + i * horizontalSegments + nextJ); base.AddIndex(1 + nextI * horizontalSegments + nextJ); base.AddIndex(1 + nextI * horizontalSegments + j); } } //create a fan connecting the top vertiex to the top latitude for (int i = 0; i < horizontalSegments; i++) { base.AddIndex(this.CurrentVertex - 1); base.AddIndex(this.CurrentVertex - 2 - (i + 1) % horizontalSegments); base.AddIndex(this.CurrentVertex - 2 - i); } base.InitializePrimitive(device); } } Any ideas how I could fix this?

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  • Descubre en una mañana todo lo que Oracle puede hacer por ti

    - by Noelia Gomez
    v\:* {behavior:url(#default#VML);} o\:* {behavior:url(#default#VML);} w\:* {behavior:url(#default#VML);} .shape {behavior:url(#default#VML);} En la actualidad, la tecnología está cambiando el mundo de una forma sin precedentes. La convergencia de novedades como la informática en la nube, los dispositivos móviles, las redes sociales, el Big Data y el «Internet de las cosas» está impulsando la innovación y revolucionando los antiguos modelos de negocio. ¿Cómo lograrán las empresas adaptarse a los cambios con rapidez sin poner en peligro el funcionamiento de la actividad comercial? Oracle siempre se ha puesto este reto y por ello queremos presentar en exclusiva para nuestros clientes las mayores novedades de nuestra gama de soluciones, el próximo 5 de Noviembre en el Oracle Day. En la parte de aplicaciones hablaremos de la oportunidad significativa de conseguir una posición de liderazgo en CX, ya que ofrecer una experiencia excelente está directamente vinculado con un aumento de las ventas. Cuanto más relevante y constante sea la experiencia de sus clientes, más probable es que compren. Disfrute de una experiencia única en este evento interactivo, donde podrá participar en debates con directivos de Oracle, ver vídeos y conocer experiencias de clientes, ampliar su red de contactos, asistir a demostraciones prácticas de productos, y un largo etcétera. Para más información acceda aquí. Normal 0 false false false EN-US X-NONE X-NONE MicrosoftInternetExplorer4 /* Style Definitions */ table.MsoNormalTable {mso-style-name:"Table Normal"; mso-tstyle-rowband-size:0; mso-tstyle-colband-size:0; mso-style-noshow:yes; mso-style-priority:99; mso-style-qformat:yes; mso-style-parent:""; mso-padding-alt:0cm 5.4pt 0cm 5.4pt; mso-para-margin:0cm; mso-para-margin-bottom:.0001pt; mso-pagination:widow-orphan; font-family:"Calibri","sans-serif"; mso-ascii- mso-ascii-theme-font:minor-latin; mso-fareast-font-family:"Times New Roman"; mso-fareast-theme-font:minor-fareast; mso-hansi- mso-hansi-theme-font:minor-latin; mso-bidi-font-family:"Times New Roman"; mso-bidi-theme-font:minor-bidi;}

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  • Resultant Vector Algorithm for 2D Collisions

    - by John
    I am making a Pong based game where a puck hits a paddle and bounces off. Both the puck and the paddles are Circles. I came up with an algorithm to calculate the resultant vector of the puck once it meets a paddle. The game seems to function correctly but I'm not entirely sure my algorithm is correct. Here are my variables for the algorithm: Given: velocity = the magnitude of the initial velocity of the puck before the collision x = the x coordinate of the puck y = the y coordinate of the puck moveX = the horizontal speed of the puck moveY = the vertical speed of the puck otherX = the x coordinate of the paddle otherY = the y coordinate of the paddle piece.horizontalMomentum = the horizontal speed of the paddle before it hits the puck piece.verticalMomentum = the vertical speed of the paddle before it hits the puck slope = the direction, in radians, of the puck's velocity distX = the horizontal distance between the center of the puck and the center of the paddle distY = the vertical distance between the center of the puck and the center of the paddle Algorithm solves for: impactAngle = the angle, in radians, of the angle of impact. newSpeedX = the speed of the resultant vector in the X direction newSpeedY = the speed of the resultant vector in the Y direction Here is the code for my algorithm: int otherX = piece.x; int otherY = piece.y; double velocity = Math.sqrt((moveX * moveX) + (moveY * moveY)); double slope = Math.atan(moveX / moveY); int distX = x - otherX; int distY = y - otherY; double impactAngle = Math.atan(distX / distY); double newAngle = impactAngle + slope; int newSpeedX = (int)(velocity * Math.sin(newAngle)) + piece.horizontalMomentum; int newSpeedY = (int)(velocity * Math.cos(newAngle)) + piece.verticalMomentum; for those who are not program savvy here is it simplified: velocity = v(moveX² + moveY²) slope = arctan(moveX / moveY) distX = x - otherX distY = y - otherY impactAngle = arctan(distX / distY) newAngle = impactAngle + slope newSpeedX = velocity * sin(newAngle) + piece.horizontalMomentum newSpeedY = velocity * cos(newAngle) + piece.verticalMomentum My Question: Is this algorithm correct? Is there an easier/simpler way to do what I'm trying to do?

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  • Membership Provider Parte 1

    - by Jason Ulloa
    Asp.net ha sido una de las tecnologías creadas por Microsoft de mas rápido crecimiento por la facilidad para los desarrolladores de crear sitios web. Una de las partes de mayor importancia que tiene asp.net es el contar con el Membership Provider o proveedor de Membrecía, que permite la creación, manejo y mantenimiento de un sistema completo de control y autenticación de usuarios. Para dar inicio a la serie de post que escribiré sobre que es Membeship y cuáles son las funcionalidades principales daremos unas definiciones. Tal como se menciono anteriormente con el membership provider podemos crear un sistema de control de usuarios completos, entre las funcionalidades principales podemos encontrar: * Creación de usuarios * Almacenamiento de información en base de datos * Autenticación, bloqueos y seguimiento Otras de las ventajas que cabe resaltar, es que, algunos de los controles de asp.net ya traen "naturalmente" en sus funciones la implementación del membership provider, tal como el control "Login" o los controles de estado de usuario, lo cual nos permite que con solo arrastrarlos al diseñador estén funcionando. Membership provider es poderoso, pero su funcionalidad y seguridad se ven aumentadas cuando se integra con otros proveedores de asp.net como lo son RoleProvider y Profile Provider (estos los discutiremos en otros post). En la siguiente figura, podemos ver como se distribuyen algunoS provider creados por Microsoft Antes de iniciar con la implementación de membership debes conocer cosas básicas como el espacio de nombres al que pertenece, el cual es: System.Web.Security que se encuentra dentro del ensamblado System.Web. Algo que debe tomarse en cuenta, es que, para poder utilizar cualquiera de los miembros de la clase, debemos hacer la referencia respectiva. Por defecto, el membership provider está diseñado para trabajar directamente con SQL Server, de ahí que su nombre completo seria SQL Membership Provider. Sin embargo, debido a su gran flexibilidad podemos extenderlo a cualquier base de datos o bien modificarlo para adapatarlo a nuestras necesidades. En los siguientes posts, discutiremos como crear un proveedor personalizado utilizando Entity Framework, separando las capas de acceso y datos y cuáles son las principales funciones que podemos aplicar. En palabras básicas y sin entrar muy hondo en el tema, hemos descrito el objetivo del Membership Provider, para todos los que desean ampliar pueden hacerlo en: http://msdn.microsoft.com/es-es/library/system.web.security.membership%28v=vs.100%29.aspx

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