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Ttelmah
Joined: 11 Mar 2010
Posts: 19529
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| Fast trigonometric approximations |
 Posted: Fri Jun 09, 2017 2:50 pm |
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Some time ago, I posted a much faster atan2 in the general forum.
Decided to post it here together with another fast set of code for sin and cosine:
The accuracy is better than half a degree, and it is _fast_. Takes about 70uSec on a 40Mhz PIC, while atan2 takes at least ten times longer. It is also smaller.... It only gives positive angles, from 0 to 360 as it's result. This works in degrees but is easy to convert to radians if required.
| Code: |
//Processor defines here....
#include <math.h>
float angle(float X, float Y)
{
//routine to give a fast solution for angle, from X/Y co-ordinates - result in degrees
float AX,AY,ival,oval,aival;
int8 quad;
AX=fabs(X);
AY=fabs(Y);
//Now the approximation used works for tan from -1 to 1, so we have to keep the
//values inside this range and adjust the input/output.
//Four 'quadrants' are decoded -1 to 1, (315 to 45 degrees), then 45 to 135,
//135 to 225, 225 to 315
If (X >= 0) //Right hand half of the circle
{
If (AY > X)
{
If (Y < 0)
{
quad = 4;
ival = -X / Y;
}
Else
{
quad = 2;
ival = X / -Y;
}
}
Else
{
If (AY > X)
{
quad = 4;
ival = -Y / X;
}
else
{
quad = 1;
ival = Y / X;
}
}
}
else
{
//Now the others
If (Y > AX)
{
quad = 2;
ival = X / -Y;
}
Else
{
If (AY > AX)
{
quad = 4;
ival = -X / Y;
}
Else
{
quad = 3;
ival = -Y / -X;
}
}
}
//A lot of lines of code, but small and quick really.....
//Now the solution
//Now approximation for atan from -1 to +1, giving an answer in degrees.
aival = fAbs(ival);
oval = 45 * ival - ival * (aival - 1) * (14.02 + 3.79 * aival);
//Now solve back to the final result
If (quad != 1)
{
If (quad == 2)
oval = oval + 90;
Else
{
If (quad == 3)
oval = oval + 180;
Else
oval = oval + 270;
}
}
if (oval<0)
oval+=360;
return oval;
}
//Demo program using pairs of numbers from the array to test
void main()
{
const signed int16 source[] = {0,300,600,1000,0,-300,-600,-1000};
int8 ctr,ctr2=0;
signed int16 X,Y;
while (TRUE)
{
for (ctr=0;ctr<8;ctr++)
{
//Now loop through the array, using pairs from two counters as X/Y
X=source[ctr];
Y=source[ctr2];
printf("X %ld, Y %ld, angle %5.1f\n\r", X,Y,angle(X,Y));
}
if (ctr2<7)
ctr2++;
else
ctr2=0;
}
}
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Now I have added this code for sin/cosine. This is in radians for +PI to -PI input range.
| Code: |
//A fast float approximation to sin over the range +/-PI
#define PI2 (PI*PI)
#define INV4 (4/PI)
#define INV4SQ (-4/(PI2))
#define P 0.225
float fast_sin(float x)
{
float y;
y = (INV4 * x) + (INV4SQ * x * fabs(x));
return P * (y * fabs(y) - y) + y;
}
float fast_cos(float x)
{ //This is done by shifting the quadrants
x += PIDIV2;
if(x > PI) // Original x > pi/2
{
x -= PI2; // Wrap: cos(x) = cos(x - 2 pi)
}
return fast_sin(x);
}
void main(void)
{
//Test the fast sin algorithm for angles from 0 to PI in steps of PI/128
float an, res, sres,;
for (an=0.0;an<(PI);an+=(PI/128))
{
//res1=fast_sin1(an);
res=fast_sin(an);
sres=sin(an);
printf ("AN=%5.3f sinfast=%5.3f sin=%5.3f\n",an,res,sres);
}
while(TRUE)
delay_cycles(1);
}
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I don't show a test for the cos, but this works exactly the same. Slightly slower. |
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tinley
Joined: 09 May 2006
Posts: 67
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| Great code thanks |
| Posted: Tue Jan 07, 2020 7:03 am |
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Thanks Ttelmah. This code is exactly what I was looking for. You mention that the quadrants may not be correct. They don't match CCS atan2 function and I think you were right to suspect. I substituted the following:
result = (atan2( input_value_X , input_value_Y ) * 57.2958 ); // 180/pi = 57.2958
With:
result = angle(input_value_X, input_value_Y);
And I had to swap the X and Y to get it to work.
But it does work very well and quickly! Thank you! |
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tinley
Joined: 09 May 2006
Posts: 67
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| Fast COS |
| Posted: Tue Jan 07, 2020 9:07 am |
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As it happens, I can also use your fast COS in the same project. But this did have a couple of bugs which are fixed in the following:
| Code: |
//A fast float approximation to sin over the range +/-PI
#define PI2 (PI*PI)
#define INV4 (4/PI)
#define INV4SQ (-4/(PI2))
#define P 0.225
#define PIDIV2 (PI/2)
#define PIX2 (PI*2)
float fast_sin(float x)
{
float y;
y = (INV4 * x) + (INV4SQ * x * fabs(x));
return P * (y * fabs(y) - y) + y;
}
float fast_cos(float x)
{ //This is done by shifting the quadrants
x += PIDIV2;
if(x > PI) // Original x > pi/2
{
x -= PIX2; // Wrap: cos(x) = cos(x - 2 pi)
}
return fast_sin(x);
}
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But still save me heaps of time trying to save time! Thank you! |
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Ttelmah
Joined: 11 Mar 2010
Posts: 19529
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| Posted: Mon Jan 27, 2020 1:16 pm |
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Well done.
Glad you got it sorted.  |
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