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Rotation angle needed to make two dependant vectors point into direction x

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Hi, I'll try to describe the problem im facing as accurate as I can.

I have two vectors, say A and B and a target direction vector T. I want E = A+B to have the same direction as T. A is the vector (R1,0) rotated by an angle alpha and B is the vector (R2, 0) rotated by 5/3alpha.


So we have this:

A = R1 * (cos(alpha), sin(alpha))
B = R2 * (cos(2/3alpha), sin(2/3alpha))
E = A + B
T = (Tx, Ty)

R1 and R2 are known. T is known as well.

I was thinking of something like:

|E| * normalized(T) = E

I can't seem to solve the equation though, I'm not sure if my trigonometry knowledge is not good enough or if its simply not possible...


Can anyone help me out please?


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next time grab ruler, pencil or pen and white paper.then draw the problem or at least draw it in paint, anyway your idea of putting the E to T seems kinda silly but here it is:





or maybe you want to project a vector onto another then this:

//Project A onto B
template <class type> t3dpoint<type>  __fastcall VectorProjectionBDIR(t3dpoint<type> A, t3dpoint<type> B,bool NormalizedB)
t3dpoint<type> i = B;

if ( NormalizedB == false ) i = Normalize( i );
type dotp = Dot(A,i);
return i * dotp;


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Okay seems like I didn't explain the problem properly. Sorry for my poor paint skills...



So I know the length of A and B and I know that the second angle is 2/3s of the first angle (relative to the first). I want to choose alpha, so that E points in the direction of T.

Thanks in advance.

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 everything is clear now, however your (2/3) * alpha is bigger than alpha tongue.png and i am not sure why you draw a pic with angles elow x origin since they should be pointed up liek on this picture:




here is my apporach, find angle of T vector, your assumptions (A = R1 * (cos(alpha), sin(alpha))) show that if t points directly right from coord origin then alpha is 0

when alpha is 90 its pointing straight up when 180 then left and when 270 then down. thats how i like it tongue.png

note i use degrees not radians (for me its more intuitive)



anyway my notebook burned and i am using old one i have no option to compile and test + i had some of my math functions with bugs (not sure if those i show you were bugged, my recent math unit is on other notebook tongue.png)

const long double pild      = 3.1415926535897932384626433832795;

template <class type> type VALIDATE_TO_360(type angle2)
type angle = angle2;

int kat=(int)angle2;


if (angle < 0 )
   angle = angle + (type)(kat+1)*360.0;

if (angle >= 360)
   angle = angle - (type)kat*360.0;

return angle;

long double n2dGetPolarCoordAngleAD(double x,double y)
if ( (x == 0) && (y > 0) )  { 	return 3.1415926535897932384626433832795/2.0;                   }

if ( (x == 0) && (y < 0) )  { 	return 3.0*3.1415926535897932384626433832795/2.0;                 }

if (x == 0) return 0.0; //actually we cant find the angle
long double k; k = (long double)y / (long double)x;
if ( (x > 0) && (y >= 0) ) { 	return atanl(k);        }

if ( (x > 0) && (y < 0) )  { 	return atanl(k)+2.0*3.1415926535897932384626433832795;  	  }

if  (x < 0)                {	return atanl(k)+3.1415926535897932384626433832795;   	  }

//last versions were without .0 just simple 2 division
return 0.0; //actually we cant find the angle

long double GetAngle(vec2 vector)
vec2 origin = vec2(0.0, 0.0);
 long double angle;
 angle = n2dGetPolarCoordAngleAD(
 (long double)vector.x-origin.x,
 (long double)vector.y-origin.y) / (pild*2.0);

	   return VALIDATE_TO_360(360.0*angle); //not sure if there should be * 360.0 ut it seems that it should. since i divide with getpolarcoord by two pi

now actual function
we define the angles of the A and B so B additionally 2/3 of alpha
1 and 2/3 => 3/3 + 2/3 = 5 / 3

float return_my_angle(vec2 vector)
  float angle = float(GetAngle(vector)); //angle between 0-360
//  5/3 * alpha = angle;

return (angle * 3.0) / 5.0;

but it should work


so you now have found your alpha




there could be a problem with floating point precision for function



with if x == 0 or if  y == 0 that needs to be tweaked to use some sort of epsilon  like 0.0001 or even smaller so it will assume small numbers as 0.0

Edited by WiredCat

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Okay thank you first of all, that already helped a little bit, but I'm afraid I dont think this will work. Heres the problem drawn again, this time without mistakes hopefully (sorry about that, it was kinda late for me already).


So the problem is that |A| != |B| and thus your equation of alpha = 3/5*angle won't hold!? Is there any way to extend that relation with the length information of A and B? I couldn't find anything that would help me there..if I had just payed more attention to trigonometry :S



So we know the slope of the target vector, lets call it m.

We can express E as a combination of A and B. This leaves us with

E.x = |A| * cos(alpha) + |B| * cos(2/3alpha)
E.y = |A| * sin(alpha) + |B| * sin(2/3alpha)
m = E.y / E.x

However, I'm not able to solve this..

Edited by .chicken

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