calculate position a fixed offset left or right of avatar

[Judy Hynes]

I imagine this problem has been discussed here before but I couldn't seem to hit on the right search terms to use.  So I apologize in advance if it has been covered.

I have a script where I need to figure out the position vector (in global coordinates) of a position a fixed distance to the left or right of the avatar based on its current rotation.   So for example consider this code:

rotation rot = llGetRot();
vector start = llGetPos();
vector end = start + <100.0,0.0,0.0>*rot;

This will give me a start position where avatar is and end position that's 100 meters straight ahead of where the avatar is facing.  It's like a pole sticking out from avatar's chest.

Now let's say I want the same thing that's 1 meter to the left.  That is, I want to know the start and end positions of a "pole" that's parallel to the original one and shifted over 1 meter to the avatar's left.  How can I calculate the start and end points?  Sorry if the answer is obvious but doing vector and rotation math always makes my head hurt.

 

[Quistess Alpha]

"shifting to the left" is just adding to the y-coordinate, like forward is adding to the x coordinate or going up is adding to the z coordinate.

3 useful functions for things like this are llRot2Fwd, llRot2Left and llRot2Up. (although they're less efficient than just multiplying a unit-vector by the rotation)

rotation MyRot = llGetRot();
vector MyPosition = llGetPos();
vector InFrontOfMe1 =  MyPosition + <1, 0, 0>*MyRot;
vector ToMyLeft1 = MyPosition + <0,1,0>*MyRot;
vector AboveMe1 = MyPosition + <0,0,1>*MyRot;

// the start and end posiitons of a rod extending from your groin to 10m infront of you:
list forward_rod = [MyPosition,10*InFrontOfMe];
// the start and end positions of a rod shift-coppied 1 unit to the left:
list left_rod = [MyPosition+ToMyLeft, 10*InFrontOfMe+ToMyLeft];

 

[Judy Hynes]

Thanks for the reply.  This makes more sense now.  One follow-up on the example you gave.  You wrote:

list left_rod = [MyPosition+ToMyLeft, 10*InFrontOfMe+ToMyLeft];

But ToMyLeft already has MyPosition added to it.  So if I put your example all in one line it looks like this:

list left_rod = [MyPosition+MyPosition+<0,1,0>*MyRot, 10*(MyPosition+<1,0,0>*MyRot)+MyPosition+<0,1,0>*MyRot];

Is this really correct that the starting point has MyPosition added twice?  It feels like ToMyLeft has already done that addition.

[Quistess Alpha]

Oh, Yeah your right, MyPosition should only be in there once in the total sum.  Corrected:

rotation MyRot = llGetRot();
vector MyPosition = llGetPos();
vector FWD = <1, 0, 0>*MyRot; // = llRot2Fwd(MyRot);
vector LFT = <0,1,0>*MyRot;   // = llRot2Left(MyRot);
vector UP =  <0,0,1>*MyRot;   // = llRot2Up(MyRot);

// the start and end posiitons of a rod extending from your groin to 10m infront of you:
list forward_rod = [MyPosition,MyPosition+10*FWD];
// the start and end positions of a rod shift-coppied 5 unit to the left:
list left_rod = [MyPosition+5*LFT, MyPosition+10*FWD+5*LFT];

is closer to how I would usually do something like this.

If you think of looking down at it from above, and you draw a little Cartesian coordinate plane with (0,0) where your object is, and (1,0) one unit in front of your object (assuming your object is an x-forward object) you just have to add scaled versions of your local FWD and LFT vectors to get to any point of interest. (2,3)  is just MyPos+2*FWD+3*LFT.

You can also do this sort of thing directly:

list left_rod = [MyPosition+<0,5,0>*MyRot, MyPosition+<10,5,0>*MyRot];

would be equivalent; multiplying by your rotation changes from local coordinates to global offsets.

Edited July 30, 2021 by Quistess Alpha

[Judy Hynes]

Yup, it all makes much more sense now.  Having success with this technique now.  Thanks for the help!