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Ttelmah
Joined: 11 Mar 2010
Posts: 19515
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 Posted: Sat Aug 03, 2013 4:32 am |
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No guarantees on this, and you need to work out what is causing your problems first, but this is an alternative to using atan for this problem.
It is a generic function, that accepts an X and Y co-ordinate, and returns the corresponding angle in degrees.
I may well have got the quadrant solutions wrong, but a quick test suggests it looks right. The accuracy is better than half a degree, and it is _fast_. Takes about 70uSec on a 40Mhz PIC, while atan takes at least ten times longer. It is also smaller.... It only gives positive angles, from 0 to 360 as it's result.
| 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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Best Wishes |
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dorinm
Joined: 07 Jan 2006
Posts: 38
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| Posted: Sat Aug 03, 2013 5:41 pm |
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| excellent! |
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RF_Developer
Joined: 07 Feb 2011
Posts: 839
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| Posted: Mon Aug 05, 2013 2:42 am |
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| The function atan2(y, x) is intended for just this sort of problem, but if the approximation works accurately enough then use it. |
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Ttelmah
Joined: 11 Mar 2010
Posts: 19515
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| Posted: Mon Aug 05, 2013 4:08 am |
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The approximation is more accurate than the chip. It's about half the size in ROM.
The maximum error, for the entire range of the chip output (-2047 to 2048), is under 0.1 degrees.
As a comment though, using atan, could be the reason for the original problems. atan2, is specifically written to 'cope' with the special cases where X=0, while a division, feeding atan, will return a maths error, which unless trapped, will result in garbage results. The approximation also handles this case.
Best Wishes |
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