
Rotational Transformation 
ngiacomelli
Member #5,114
October 2004

I have a triangle that I wish to rotate to represent the player. I want the tip of the triangle (+) to represent their current angle and the direction in which they would travel (if they were to apply thrust). + / \ / \ /_____\ The movement code isn't my problem here, though. It's actually drawing the triangle. I want to do this using primitives (so no drawing to a bitmap, then rotating). I came up with some messy code here as a quick test: float t_x, t_y, t_x2, t_y2, t_x3, t_y3; t_x = 100 * fixtof(fcos(angle))  100 * fixtof(fsin(angle)); t_y = 100 * fixtof(fcos(angle)) + 100 * fixtof(fsin(angle)); t_x2 = 200 * fixtof(fcos(angle))  200 * fixtof(fsin(angle)); t_y2 = 200 * fixtof(fcos(angle)) + 200 * fixtof(fsin(angle)); t_x3 = 0 * fixtof(fcos(angle))  200 * fixtof(fsin(angle)); t_y3 = 200 * fixtof(fcos(angle)) + 0 * fixtof(fsin(angle)); Which works to slowly rotate the triangle when the angle is adjusted. But I really need the pivot point to be in the triangles centre (I should think). So how do you do this effectively?

Pavel Hilser
Member #5,788
April 2005

I did rotation around a point, if this is what you want, i did it this way x  1 \ \ \ \ 2 rotating point 1 to 2  adding 45degrees, rotating around x get x<>1 lengt //get the length (to point 0,0) delka=sqrt(((f*5)*(f*5))+((g*5)*(g*5))); //get the angle curr_angle = (atan2(g*5, f*5)*180)/M_PI; //new angle curr_angle+=angle; //new position mymodel[f+5][g+5].square_x = delka * cos((float)(M_PI*((float)curr_angle))/180); mymodel[f+5][g+5].square_y = (delka * sin((float)(M_PI*((float)curr_angle))/180))/2; (it's taken from my new project, where points are rotated around zero) anyway, if you have fixed triangle sides, and its always triangle, you can use the pivot as a center of a circle and the two poinst could be easi calculated with the last point (youw kno both angles and both length (cause its a triangle)) _____________________________________ 
Ceagon Xylas
Member #5,495
February 2005

Angles, rotations, and circles sure make for some 'leet' looking code. 
GullRaDriel
Member #3,861
September 2003

it's taken from his project, so you can imagine that: //get the length (to point 0,0) delka=sqrt(((f*5)*(f*5))+((g*5)*(g*5))); //get the angle curr_angle = (atan2(g*5, f*5)*180)/M_PI; //new angle curr_angle+=angle; //new position square_x = delka * cos((float)(M_PI*((float)curr_angle))/180); square_y = (delka * sin((float)(M_PI*((float)curr_angle))/180))/2; is what you need. EDIT: I'll test this and update my post to make it doing the trick. LAST EDIT: WORKING EXAMPLE:
"Code is like shit  it only smells if it is not yours" 
Paul Pridham
Member #250
April 2000

Turtle is a nice and simple abstraction for Asteroidslike vector drawing with relative movement and rotations.  
gillius
Member #119
April 2000

If you are wanting to draw an equilateral triangle with the pivot point in its center then it is easy as the three points are at theta, theta + 120deg, and theta + 240deg, where theta is the direction you are pointing. Then given theta you do sin and cos and multiply by one half of the height of the triangle (since the point is in the center, you want use use "radius" of the triangle).
Gillius 
GullRaDriel
Member #3,861
September 2003

With all this code posted and all those ideas , can you give your mean of the whole, Nial ? "Code is like shit  it only smells if it is not yours" 
