# Arctg table!

This topic is 3093 days old which is more than the 365 day threshold we allow for new replies. Please post a new topic.

## Recommended Posts

Anyone could past here, a harcoded Arctg table as array of 360 entries? Would apreciate your help! Thanks.

##### Share on other sites
Do you mean a table with the tangent of each angle that is an integer number of degrees?

##### Share on other sites
DIY:

-- lua5.1local f = assert( io.open( "invtan.txt", "w" ) )for i=0, 360 do	f:write( 1 / math.tan( i * math.pi / 180 ) .. ' ' )end

Output.

EDIT:
Did you want atan, tan or 1 / tan?

[Edited by - _fastcall on August 3, 2009 3:11:01 PM]

##### Share on other sites
surely you could make that list yourself via a program? :P

shouldnt be too hard, esp not for someone who knows asm (;

##### Share on other sites
Quote:
 Original post by alvaroDo you mean a table with the tangent of each angle that is an integer number of degrees?

I mean the Arc tangent values for the selected angle entry, so ArcTg[0] will store the ArcTg value of angle 0, and ArcTg[359] will store the ArcTg value of angle 259.

See, I have a circle(x0,y0), a point in the circle(x,y).
What I want to do is to detrmine the angle with this formula:

Angle= ArcTg (y-y0/x-x0)

so with the lookup table I can simply do this to get the angle:

float Angle= ArcTgArray[(y-y0/x-x0)];

Thanks.

##### Share on other sites
But that slope will not always be in [0,360]... You can't have a table that covers all the reals. You could have a table for values between 0 and 1 (in however many steps you want) and reduce all the other cases to this one. Could that be roughly what you want?

##### Share on other sites
const int step = 361;double arctan[step];// Create a table that uses values from -1 to 1for ( int i = 0; i < step; ++i )	arctan = std::atan( (double)(i*2 - step) / (double)step );// Then:double angle = arctan[ std::min( step-1, std::max( 0, (int)( (test * step + step)/2 ) ) ) ];

You should be using atan2( y-y0, x-x0 ) anyways ...

##### Share on other sites
If you don't have atan2, you can use something like this:
double atan2(double y, double x) {  double alpha=3.141592653589793238462643383279;  double gx=-1.0, gy=0.0, result=-alpha;  static double cosine_and_sine[32][2]={    {-1,0},    {0,1},    {0.70710678118654757274,0.70710678118654746172},    {0.92387953251128673848,0.38268343236508978178},    {0.98078528040323043058,0.19509032201612824808},    {0.99518472667219692873,0.098017140329560603629},    {0.99879545620517240501,0.049067674327418014935},    {0.99969881869620424997,0.024541228522912288124},    {0.99992470183914450299,0.012271538285719925387},    {0.99998117528260110909,0.0061358846491544752691},    {0.99999529380957619118,0.0030679567629659761432},    {0.99999882345170187925,0.0015339801862847655001},    {0.99999970586288222663,0.0007669903187427044855},    {0.99999992646571789212,0.00038349518757139556321},    {0.99999998161642933425,0.00019174759731070329153},    {0.99999999540410733356,9.5873799095977344669e-05},    {0.99999999885102686115,4.793689960306688131e-05},    {0.99999999971275665978,2.3968449808418219318e-05},    {0.99999999992818922046,1.1984224905069705298e-05},    {0.9999999999820472496,5.9921124526424275272e-06},    {0.9999999999955118124,2.9960562263346608352e-06},    {0.99999999999887800861,1.4980281131690111427e-06},    {0.99999999999971944664,7.490140565847157414e-07},    {0.9999999999999298339,3.7450702829238412872e-07},    {0.99999999999998245848,1.8725351414619534661e-07},    {0.99999999999999567013,9.3626757073098083589e-08},    {0.99999999999999888978,4.6813378536549088116e-08},    {0.99999999999999977796,2.3406689268274550676e-08},    {0.99999999999999988898,1.1703344634137276992e-08},    {1,5.8516723170686384961e-09},    {1,2.925836158534319248e-09},    {1,1.462918079267159624e-09}  };  for(int i=0;i<32;++i) {    double c=cosine_and_sine[0], s=cosine_and_sine[1];    double nx=gx*c-gy*s, ny=gy*c+gx*s;    if(x*ny<y*nx) {      result += alpha;      gx=nx;      gy=ny;    }    alpha*=.5;  }  return result;}

##### Share on other sites
@alvaro: I can have only 360 array's entry, and then check for the nearest array value to the value i got from calculating x-x0/y-y0.

@_fastcall: Could you please provide me with the output as you did in your first post? My environement compiler does not have a math library (no cos, sin... functions) and I cannot install other compilers.

Thanks.

##### Share on other sites
Program:
#include <iterator>#include <fstream>#include <iomanip>#include <cmath>int main(){	const int step = 361;	double arctan[step];	// Note:  Range is from -1 to 1 inclusive.	for ( int i = 0; i < step; ++i )		arctan = std::atan( (double)(i*2 - step) / (double)(step-1) );	std::ofstream fs( "out.txt" );	fs.precision( 16 );	std::copy( arctan, arctan + step, std::ostream_iterator<double>( fs, " " ) );	fs.flush();}

The output:
Output.

You can use the binary interpreters that come with lua or python to generate your tables for you, as shown in my first post.

##### Share on other sites
1. Your use of the abbreviation "ArcTg" for "arctangent" is unusual. The standard abbreviation is "atan" or occasionally "arctan;" you will see either this or "tan-1" in math texts.

2. You say you have no math libraries? You can compute your table using the series expansion,

which I have taken from the Wikipedia article. Note that the return value is in radians.

[Edited by - Emergent on August 3, 2009 6:00:36 PM]

##### Share on other sites
What alvaro is saying is that you have tangent and arctangent mixed up. Arctangent takes in the ratio of sides on the triangle, which could be anything, and then returns the angle, which is between 0 and 360. It is regular tangent that takes the angle as input.
Tangent:[0,360]->[-inf,inf]
Arctangent:[-inf,inf]->[-90,90]

So it sounds like you want tangent(x) for x from 0 to 360, right?

##### Share on other sites
Quote:
 Original post by Maze MasterWhat alvaro is saying is that you have tangent and arctangent mixed up. Arctangent takes in the ratio of sides on the triangle, which could be anything, and then returns the angle, which is between 0 and 360. It is regular tangent that takes the angle as input.Tangent:[0,360]->[-inf,inf]Arctangent:[-inf,inf]->[-90,90]So it sounds like you want tangent(x) for x from 0 to 360, right?

Actually, I want something to return the angle from the diference between
the point in the circle and the center of the circle. I believe this fromula applies in this case:

Angle= arctangen (PointY-CenterY/PointX-CenterX)

Cheers.

##### Share on other sites
Quote:
 Original post by asmcoder@alvaro: I can have only 360 array's entry, and then check for the nearest array value to the value i got from calculating x-x0/y-y0.

You mean you will search for the value in the table? Then you need a table of tangent, not arc-tangent, as we have been saying from the beginning.

Quote:
 [...]My environement compiler does not have a math library (no cos, sin... functions) and I cannot install other compilers.

Can you use the code I posted above?

Thanks.

##### Share on other sites
You are welcome. By the way, if you want the answer in degrees (which I don't recommend), simply change the line double alpha=3.141592653589793238462643383279;' to double alpha=180.0;'.

##### Share on other sites
I've done that using this formula: Degree=Radian*180/Pi
Btw, why don't you recommend getting degree answers?

Thanks a lot.

##### Share on other sites
Quote:
 Original post by asmcoderI've done that using this formula: Degree=Radian*180/PiBtw, why don't you recommend getting degree answers?

1) It's a more natural choice. Dividing the circle in 360 degrees is a completely arbitrary decision, although it's a familiar one. The length of a circle measured in radii is 2*pi, so it would be natural to measure angles by specifying the length of the arc determined by the angle, measured in radii, and that's what radians are.
2) It makes derivatives simple. The derivative of sin(x) is cos(x) and the derivative of cos(x) is -sin(x), if you measure angles in radians.
3) Libraries (except for OpenGL) and hardware use radians.
4) exp(i*x) = cos(x) + i*sin(x), as long as x is in radians (if this doesn't make sense, I won't try to explain it).

I've been meaning to write a little article advocating the use of the most natural representation in several types of variables. In short, it would say:
* Prefer vectors to angles.
* Prefer radians to degrees (if you have to use angles at all, which you probably don't).
* Start counting from 0.
* Use ranges of the form [x,y), which include the bottom element and don't include the top element.
* Use a 24-hour format for time, instead of am/pm.
* Use fractions, not percentages.

If you follow those simple pieces of advice, your code will have fewer special cases and many formulas will be simpler.

##### Share on other sites
Cheers for the explanations dude!

Thanks.

##### Share on other sites
Quote:

Quote:
 I've been meaning to write a little article advocating the use of the most natural representation in several types of variables. In short, it would say

;)