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I'm currently working on a game project and need to render a point in front of the current players vision, the game is written in a custom c++ engine. I have the current position (x,y,z) and the current rotation (pitch,yaw,roll). I need to extend the point forward along the known angle at a set distance.

edit: What I Used As A Solution (Its slightly off but that's ok for me)

Vec3 LocalPos = {0,0,0};
Vec3 CurrentLocalAngle = {0,0,0};
float len   = 0.1f;
float pitch = CurrentLocalAngle.x * (M_PI / 180);
float yaw   = CurrentLocalAngle.y * (M_PI / 180);
float sp = sinf(pitch);
float cp = cosf(pitch);
float sy = sinf(yaw);
float cy = cosf(yaw);
Vec3 dir = { cp * cy, cp * sy, -sp };
LocalPos = { LocalPos.x + dir.x * len, LocalPos.y + dir.y * len,LocalPos.z + dir.z * len };
Sebastian
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Pinecone
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2 Answers2

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You can get the forward vector of the player from matrix column 3 if it is column based, then you multiply its normal by the distance you want then add the result to the player position you will get the point you need.

Mohamed Moussa
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Convert the angle to a directional vector or just get the "forward vector" from the player if it's available in the engine you're using (it should be the same thing).

Directional vectors are normalized by nature (they have distance = 1), so you can just multiply them by the desired distance to get the desired offset. Multiply this vector by the distance you want the point to be relative to the reference point (the player's camera vector I presume), and then you just add one to the other to get the point in the world where this point belongs.

Havenard
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