Sprite Rotation Math Formula for Screen Width and Height - c

Question

I am programming an asteroids type game in C, and I have a sprite sheet of 36 sprites that is the ship rotating. I would like to know a math formula for figuring out how to move the ship in the direction of the sprite I have chosen from the sprite sheet. Note that I am incrementing by 10 degrees (hence 36 sprites for 360 degrees).
For example, my screen is 320 pixels wide by 256 pixels high.
If I select sprite image 10 (which is 90 degrees (the ship is facing right)), how can I calculate (using some sort of formula) the X and Y coordinates to move the ship in? I know 90 degrees is an easy one, by imagine if it were 30 degrees. There is a certain value for X and a certain value for Y. Since the screen in wider in width that height, the X speed would be higher than the Y speed.
Hope that makes sense.
Many thanks.

Answer 1

There are two easy approaches: you can build a table of [x,y] distances for each of the 36 angles, or you can do the math "on the fly".
The advantage of calculating the distances immediately is that you can easily increase the accuracy later on, if you decide you want more than 36 angles (and don't mind the sprite is off by a couple of degrees). Also, since you will be working with floats anyway, you can do all of your calculations with a far greater accuracy. Your speed could be as low as 0.01 pixel per second, and if you store your position as floats as well, you'd see your sprite move a tiny bit every few minutes.
Pre-calculating a table is easy and fast, though. Run this program to create the arrays xmove and ymove. Then, for an angle a, you can set xpos += ((speed*xmove[a])>>8) and ypos += ((speed*ymove[a])>>8).
The table stores sin and cos times 256, as integers. The values need to be multiplied by some large factor because they always fall inside the floating point range -1..1; storing them as their original floating point value is possible but unnecessary (it would only re-introduce floating point calculations in what can be reasonably approximated with pure integers, in your case). Now since the values are "premultiplied" by 256, you need to divide the speed*move calculation again by that number -- shifting right by 8 bits is all it takes. (There is a small rounding issue here; if it bothers you, add 128 before the right-shift.)
You can use a larger accuracy by using a multiplier of 1024 or higher, but again, more accuracy is probably entirely invisible for your purposes. ('1024' instead of '1000' because you can still efficiently use bit-shifting with that number.)
I believe that nowadays any modern screen has nigh-on square pixels, so unless you want it as some sort of special effect, speed in the y direction should be the same as x-speed. However, it's simple to add. Instead of dividing by 256, you'd use something like ypos += ((speed*ymove[angle])/341); -- this is (4*256/3), so the vertical speed is 75% of the horizontal speed.
A final possible refinement: you can also store your xpos,ypos as pre-multiplied by 256! Then you would not shift right the new coordinates, but immediately add the correct value. Only when displaying the actual sprite, you'd divide the coordinates by 256. That way your ship will not move by "entire pixels" only, but way more smooth. If your speed is variable, you can store it with higher accuracy the same way (remember to scale down correctly, because it'd make your 'virtual' speed is 256*256 higher than your 'screen' speed).
The table created below assumes #0 is "straight up", #9 (not 10!) is "right", #18 is down and #27 is "left", where positive y points downwards.
By the way: the size of your ship doesn't really matter ... You probably don't want it to "jump" distances equal to its own size.
#include <stdio.h>
#include <math.h>
#ifndef M_PI
#define M_PI 3.14159265358979323846
#endif
int main (void)
{
int i, angle;
printf ("int xmove[36] = {\n");
for (i=0; i<36; i++)
{
angle = 10*i;
// x distance: sin
printf ("\t%d,", (int)(round(256*sin(angle * M_PI/180))));
printf ("\t\tangle: %d\n", angle);
}
printf ("};\n");
printf ("\n");
printf ("int ymove[36] = {\n");
for (i=0; i<36; i++)
{
angle = 10*i;
// y distance: cos
printf ("\t%d,", (int)(round(-256*cos(angle * M_PI/180))));
printf ("\t\tangle: %d\n", angle);
}
printf ("};\n");
return 0;
}