Elphaba Posted March 10 Share Posted March 10 (edited) I need to know the position in 3D space of where an aircraft will be at a specific time, assuming no changes in speed or direction. The getPosition() returns 4 vec 3's. The first is p - which is the position in 3d space, the next 3 are x, y and z; each vec 3's that are the nomalised vectors for the translation in each axis right? So all I need to do is multiply the x, y and z by a magnitude (in metres?) and ADD them to the p.x, p.y, p.z - to get the new position... right? Or have I misunderstood something? Edited March 10 by Elphaba Link to comment Share on other sites More sharing options...
Exorcet Posted March 10 Share Posted March 10 What about using get velocity? You should be able to multiply the values for it by (desired time - current time) and then add to the position components. I haven't tried to do something like what you're attempting, but that seems like it should work. Awaiting: DCS F-15C Win 10 i5-9600KF 4.6 GHz 64 GB RAM RTX2080Ti 11GB -- Win 7 64 i5-6600K 3.6 GHz 32 GB RAM GTX970 4GB -- A-10C, F-5E, Su-27, F-15C, F-14B, F-16C missions in User Files Link to comment Share on other sites More sharing options...
Elphaba Posted March 10 Author Share Posted March 10 16 minutes ago, Exorcet said: What about using get velocity? You should be able to multiply the values for it by (desired time - current time) and then add to the position components. I haven't tried to do something like what you're attempting, but that seems like it should work. Oh, okay, let me look into that - thanks! Link to comment Share on other sites More sharing options...
Solution cfrag Posted March 10 Solution Share Posted March 10 (edited) 3 hours ago, Elphaba said: So all I need to do is multiply the x, y and z by a magnitude (in metres?) and ADD them to the p.x, p.y, p.z - to get the new position... right? Yes. Since you say that the direction that the aircraft moving in is normalized, it means that the magnitude of the direction is one. If you multiply that vector by the speed (itself a scalar value in m/s) you have the velocity vector for that aircraft. It is moving in that direction by that amount every second. So, to see where that plane will be in 7 seconds, multiply the velocity vector by 7, and add the resulting vector to the starting point. The result is the point where (no changes in speed or direction assumed) the plane will be in 7 seconds. Edited March 10 by cfrag 1 Link to comment Share on other sites More sharing options...
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