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GPxz

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  1. Quote:Original post by SpaceDude Quote:Original post by GPxz Dont worry, the problem is my hardware, i am using a really crappy onboard graphic display, and the drivers of it werent installed, i installed But DirectX still dont work OpenGL works, but the colors all screwed up, looks like a problem with the textures only, cause the menus colors are right, which led me to suspect its a problem with the graphic displayer Ok, yes OpenGL does tend to work better with lower end systems. I have however also noticed a problem with the colours being messed up. I will try to fix that in the next version. Direct3D is the preferred renderer though, it is generally faster and less likely to have graphical display problems. I constantly get an error after some time of playing
  2. Quote:Original post by SpaceDude Quote:Original post by GPxz By the screenshots it remember me BaboViolent2 I dled it First, it requested the MSVCP71.DLL, which i had to download separetely. Just to warn you, so next time include this file Now its crashing when i try to run it :( In OpenGL it crashes before anything appear, on DirectX it gives me this error "OGRE EXCEPTION(3:RenderingAPIException): Failed to create Direct3D9 Device: Not available in D3D9RenderWindow::createD3DResources at \c++\ogrenew\RenderSystems\Direct3D9\src\OgreD3DRenderWindow.cpp (line 573)" Maybe its missing the DirectX dll, i am downloading DirectX9 and i will try again LATER: Yeah, not working at all, its a shame, i really wanted to give a try :( Yes the game style is similar to that of BaboViolent2. Thanks for the heads up on the MSVCP71.DLL missing file. Its odd that I haven't had any complaints from anybody else about this though. I can't think what could be the problem if you have the latest DX9 drivers and your gfx card drivers are up to do. Unless you have a very old machine? Dont worry, the problem is my hardware, i am using a really crappy onboard graphic display, and the drivers of it werent installed, i installed But DirectX still dont work OpenGL works, but the colors all screwed up, looks like a problem with the textures only, cause the menus colors are right, which led me to suspect its a problem with the graphic displayer
  3. By the screenshots it remember me BaboViolent2 I dled it First, it requested the MSVCP71.DLL, which i had to download separetely. Just to warn you, so next time include this file Now its crashing when i try to run it :( In OpenGL it crashes before anything appear, on DirectX it gives me this error "OGRE EXCEPTION(3:RenderingAPIException): Failed to create Direct3D9 Device: Not available in D3D9RenderWindow::createD3DResources at \c++\ogrenew\RenderSystems\Direct3D9\src\OgreD3DRenderWindow.cpp (line 573)" Maybe its missing the DirectX dll, i am downloading DirectX9 and i will try again LATER: Yeah, not working at all, its a shame, i really wanted to give a try :( [Edited by - GPxz on November 22, 2007 1:47:59 PM]
  4. Quote:Original post by TheAdmiral Quote:Original post by GPxz Knowing the x and y, you can retrieve the Z using the plane equation You can determine the Z from the X and Y on a heightmap or plane system in a constant-time operation. But if the terrain is made up of arbitrary bounding meshes, then you need a per-object, per-primitive interpolated method, which is far slower in general. The reason I asked which layout you're using is that each one requires a very different approach (for optimal performance). I only listed the most common, but there are several ways to represent your terrain. There is a lot to learn, and it's probably not worthwhile trying to understand it all at once. I recommend you pick a single, simple representation and get it working. Once you're confident with local heightmaps, for example, you can move on to the next representation. In particular, get a good understanding of heightmaps (if that's your representation of choice) and learn to render them successfully. Once that's done, you'll have a much easier time with the corresponding physics. By the way, I apologise if you felt my original reply was a little terse [wink]. Dont know what you mean exactly with bounding meshes, but i will try the Z incognita plane equation method and i think it will be enough, since i am using triangles. I have been thinking about it for a time, and it looks pretty robust for dealing with not only heightmap, but for any planar polygons I have rendered heightmaps a long time ago, so i dont have problems with this part
  5. Umm, giving some more read there seems to be a more faster and elegant way Knowing the x and y, you can retrieve the Z using the plane equation Any help would be cool, i still feel a little unsecure about this
  6. Cool, such a common thing like terrain walking, and information on it is poorly documented After reading some random comments texts around google, i think i got it, If you are on the top of a triangle vertex, just set to the triangle vertex height, if youre inside the triangle, interpolate between the triangle vertexs height to get the correct height on that position
  7. Quote:Original post by TheAdmiral 2D or 3D? If 3D, Heightmap, procedural, arbitrary meshes or something else? If 2D, bitmap-mask, block-based or what? Sorry for dont being specific, its for 3D, it can be for any 3d, but lets say, heightmap
  8. How to make caracther correctly walk in irregular terrains (like heightmaps)? I have one idea how to do it, but i'd like to hear how ppl usually do it if its like the way i am wondering [Edited by - GPxz on November 21, 2007 12:49:03 PM]
  9. Quote:Original post by Zipster The direction of movement of pA is always perpendicular to the radial vector (A-B). Remember that the torque at any point in time is the sum of all forces acting on the point (including that caused by gravity), crossed with the radial vector. The point can't move in anything but the θ-hat direction. This is assuming the wire is a rigid constraint of course. I think i understood But whats the influence of torque/angular momentum on it, is it a force happening too? Cause in the Wikipedia article http://en.wikipedia.org/wiki/Angular_momentum It shows angular momentum as the cross product of the radial vector and linear momentum as you said, and in the picture its a direction vector up, this would explain why the gyroscope remains upright, cause it would anulate gravity, since its opposite to it, is it? Quote:Original post by Zipster If the point pB is suddenly removed, then you no longer have a centripetal force keeping the point in radial motion. It will only be under the influence of gravity, and as such behave like a falling particle with some initial velocity, depending on when pB was removed. You mean if i remove pB, pA speed will be completely null? This is not what happen in real life, it seems it follow the radial vector
  10. Hello, I have a "point" A, which lets say, connected by a "wire", to another "point" B which is immovable. If i apply a force on pA, unless the force is enough to move point B or break the wire, it will be transformed into rotational movement. So i will have a rotational movement, which circle radius is the size of the wire. My doubts is, If i have a force (gravity) which is constantly attracting everything in a certain vector direction. When pA is rotating, what will be is vector direction in that moment, so i can reduce gravity from it? I mean if i pause (pause, not stop) for a second, what will be pA vector direction in that moment, it will be the linear momentum? And if when pA is rotating, i remove pB, what direction will it go? It will be the vector point-pivot? PS: The wire is not elastic! [Edited by - GPxz on October 31, 2007 12:03:40 PM]
  11. Problem is, the dot product is not something so obvious for our brain as the math of a straight line for example, its something that you will have to think a little more if you want to understand it If i know how it works? Not completely, but i am just getting used to accept it and use it for its purposes, as i go deeper in math that is more complex for me. And so, you will have too, if you want to continue in your ways of math, until one day you will be inspired and enlightned enough to learn it. Otherwise, look for lesser elegants methods, for example, take the atan2() of the two points and subtract them I hope someday i can fully understand everyone of those stuffs By the way, there is a good proof that a.b=|a||b|cos(theta) on the Wikipedia article
  12. Quote:Original post by nsmadsen Sephyx- A solid portfolio is WAY more important than a degree, every single time. Every company I've worked for (both large and small) has been more concerned about WHAT I can do rather than where or what I studied. Now the important question for you to ask is: do you have to get the degree to be able to have a solid, professional demo? That depends on you. Some people can learn everything on their own and don't have to study at college or trade schools. Others completely fall apart in that environment and have to attend some formal school. Only you know what category you fall into. Best of luck! Nathan A self-taught with portfolio, but no degree, has good chances of getting a job in the game industry?
  13. Quote:Original post by Alrecenk If light bounces off multiple(we'll say 3 for simplicity) surfaces and the color of those surfaces is a set of 3 component vectors S1-S3, and angles and distances are not considered, then the final light will be S1*S2*S3*light (note that that times is a per component times, not a dot or cross product). In reality there is a lot more to it though. http://en.wikipedia.org/wiki/Rendering_equation If no light hits a pixel it will be black, but generally, in direct illumination, light that doesn't reach the eye is not calculated. Light does loose strength every time it reflects since some of the energy is absorbed. However, if you're only absorbing it multiplicatively, and none of your surfaces are completely black then the light will never fade out completely. It'll just become too small to be worth calculating. Ok, and how angles and distance affect it?
  14. Quote:Original post by Flatus 1) segment and circle The ellipse can be obtained from a circle by stretching it along the direction of either axis; the same trivially invertible stretching transforms line segments to line segments. I never thought of that! I will give it a try. Thanks! Good idea
  15. Quote:Original post by ToohrVyk Create a constant vector called 'ones' equal to (1,1,1). Then, multiply it by the scalar before substraction: v2 = v1 - 0.5 * ones; Interesting, i had the same doubt sometime ago