Undetected laser

Hi, I am doing a game on plane shooting, but its seems that whenever I try to shoot the enemy, my laser have not effect on the enemy. Can anyone advise me on this? I suspect that it is due to the fact that my sphere size of my laser beam is too small, but even when I enlarge it, the problem exist. Hope to hear from you guys soon. Thanks.

Please rephrase your question in terms of OpenGL rendering. It’s very abstract right now and pretty meaningless w.r.t. graphics functionality. It sounds like you’re using software on top of OpenGL and not OpenGL directly and therefore your question seems a bit off topic. Also what do you mean by effect, do you mean your laser is not visible on the plane? Please be very specific.

Erm okay… I am making a plane shooting game, using OpenGl. I have created a plane which is suppose to shoot laser to attack an object, let say it’s a tanker. But the problem is that when the laser reach the tanker, it just went through it, without collision detecting the Tanker, and thus delete the Tanker away. What I hope to know is that why is it that even when I have set the collision detection of the laser, the tanker is still not “dead” after my laser hit it? I am quite new to programming games, so I hope that you will bear with me. Thanks.

So, does your card have an ARB_laser_tanker_collision_detection extension or how does that work?


Please be more specific about the way you detect the collision and how you handle it.

Kyo83 OpenGL doesnt do collision detection, you have to do that yourself or use a library for it.

Its mostlikly an issue with the library you are using, perhaps your “LASER” is so fast that its already behind your “TANKER” when the library does the checks or something like this.

The problem is most likely that the tanker is laser-proof.

hey kyo83, you should post questions unrelated to OpenGL (general math, collision, physics, etc) in the “math and algorithms” forum. there, i’m sure you’ll recieve more than just friendly jibes :slight_smile:

Okay thanks. Will do.