I have the following setting:
I know P1, P2 and the angle alpha, now how do i calculate the coordinates of P3? (Note: P3 is on the same circle with origin P1 and radius P1P2)
The blue lines indicate the coordinate system
+4
A:
If you rotate the vector P1->P2 by alpha about P1 you'll get the vector P1->P3. Then knowing P1 you can get P3.
The basic equation for rotation about the origin is:
[ cos(α) -sin(α) ] [x]
[ sin(α) cos(α) ] [y]
You might have to change the signs with your coordinate system, but I always have to it by trial and error as I can never remember!
Don't forget - as S.C. Madsen says sin and cos expect angles to be in radians not degrees.
Wikipeida's article on rotation has more information.
ChrisF
2010-10-28 15:31:09
Thanks. However I cant solve the problem. P1 is (150/210). P2 is (150/110) and alpha is 30. But if i calculated this and draw the point it isn't on the line.
Roflcoptr
2010-10-28 16:21:35
My solution for P3 is P3(135/111). Maybe this is wrong :D
Roflcoptr
2010-10-28 16:28:29
The problem is that you need to translate P1 to the origin before the rotation and then translate back to its original place.
ysap
2010-10-28 21:40:58
+1
A:
Let r be the distance from P1 to P2. Then P3 lies r*sin(α) in the negative x-direction from P1, and r*cos(α) in the negative y-direction from P1. For more details, see Wikipedia on trigonometry. Hence P3 has coordinates P1 - (r*sin(α), r*cos(α)).
Sidenote: It's a shame SO doesn't have LaTeX support, like MO.
gspr
2010-10-28 15:31:48
+1
A:
The angle has to be in radians when invoking sin and cos, you state the angle alpha is 30, so it seems to be in degrees. Other than that I think 'gspr' and 'ChrisF' have given excellent advice on how to solve this.
S.C. Madsen
2010-10-28 17:01:24
Thanks! this solved the problem! Now i have to find out which answer to accept :D all helped!
Roflcoptr
2010-10-28 17:19:45
I'm glad you solved your problem. May I suggest you select Fischer's answer as the solution, since he has the lowest reputation of all the people who helped out.
S.C. Madsen
2010-10-28 17:32:29
+7
A:
The formula stated above from Wikipedia is usable here to rotate the vector P1->P2 (V12).
V12 = [0, -100]
When rotated (beware of α is -30 degrees in your drawing) the vector P1->P3 becomes
x' = V12(x)*cos(α) - V12(y)*sin(α) = 0*cos(-30) - (-100)*sin(-30) = -50
y' = V12(x)*sin(α) + V12(y)*cos(α) = 0*sin(-30) + (-100)*cos(-30) = -86.6
When translated with the point P1 the coordinates for P3 becomes
[x, y] = [-50+150, -86.6+210] = [100, 123.4]
Fischer
2010-10-28 17:10:06
+2
A:
Complex_To_Vector(Vector_To_Complex(P_2 - P_1) * ei*alpha) + P_1.
(Just for fun -- not a serious suggestion)
Josephine
2010-10-28 19:03:46
Well, thank you, but I'm just worried the level of the answer compared to the level of the question is too high -- I'm concerned it may be more mystifying than helpful/educational to the original author. But don't let me stop you from upvoting! :)
Josephine
2010-10-28 23:01:32