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[Kofthefens]

I have been programming a strategy game, and I have run into a problem. I made this diagram:

Spoiler:  Diagram (click to show/hide)

So how can I find V_x, V_y, W_x, and W_y?

In case you're wondering, I will use this for flanking (in the AI). An enemy warrior will be at point T, and the 2 flanking warriors will go to V and W. They don't approach from opposite sides so that they can be close enough to lend each other aid.

I hope I explained this well enough, and thanks in advance for the help.

[Virex]

Given your description, you can't find it. There are 3 degrees of freedom (the location of 1 vertex or the center, and a third one to describe the orientation of the triangle) and you only fix 2 of them (via the location).It's possible to give a solution as function of the third degree of freedom, but no absolute values can be given.

[Kofthefens]

(the location of 1 vertex or the center, and a third one to describe the orientation of the triangle) and you only fix 2 of them (via the location). It's possible to give a solution as function of the third degree of freedom, but no absolute values can be given.

The angle the triangle is rotated at is a given. In this example it's 0 degrees, as line TV is horizontal.

[zombie urist]

(V_x,V_y) = (30cos(Θ), 30sin(Θ))
(W_x,W_y) = (30cos(Θ+2ⲡ/3), 30sin(Θ+2ⲡ/3)

[Lectorog]

V_x will be cos(angle)*30 ; V_y will similarly be sin(angle)*30 .
W_x will be sin(30-angle)*30 ; W_y will be cos(30-angle)*30 .
I did this all without paper, so I may not be correct. I think it is, though.
This is in degrees, too. I think zombie urist's is in radians, but I can't tell, given the pi symbol isn't appearing for me.
For the Ws I operated off of the angle between TW and the vertical; zombie urist's approach is different, but it might work.

[Virex]

Oh, then it's simple. There are two options, the linear algebra approach and the single-dimensional algebra approach.

Assuming the angle is given by T in degrees and defined as the counterclockwise direction from the X-axis.

The linear algebra approach is pretty simple. We just take V' and W' as the known vectors of an unrotated triangle, relative to T' = (0,0) and then apply a rotation matrix R to them, finally transforming them back to world space by adding T.

V' = (30,0)*R
W' = (15,30)/sqrt(45)*R
R = [cos T, -sin T; sin T, cos T]
V = V'+T
W = W'+T

The single-dimensional algebra approach is slightly less straight-forward. It can be obtained by decomposing V and W into an X and an Y section and realizing they can be calculated using trigonometry. It yields the following expressions:

V = T + (30*cos T, 30*sin T)
W = T + (30*cos T+60, 30*sin T+60)

[Kofthefens]

Thanks! Now to let the thread slowly slide into oblivion.

[GalenEvil]

I'm going to go ahead and post the calculations I did, in easily typed words instead of symbols, so that anyone else that searches can go ahead and find it

Given a normalized direction vector for object T at position (Tx, Ty):

T's normalized direction vector will be denoted as dvT with x-component being dvTx and y-component dvTy. Same format for objects V and W.

also, inside of the cos and sin functions the values are in degrees not radians, so you would have to convert those over to radians.

V = (d*(cos(arccos(dvTx)+120), d*(sin(arcsine(dvTy)+120))

W = (d*(cos(arccos(dvTx)-120), d*(sin(arcsine(dvTy)-120))

(d) is the distance away from T that V and W have to be placed.

Hope that isn't horrible math or too much of an eyesore. Not sure how to make the nice looking symbols >.>