[C] a circle in 2D/3D - return a pairs o points

[heterodon7]

hello,

can anyone give me a clue or simple code on task:

for example we have in 2D an equation fo circle:
(x - 3)^2 + (y - 4)^2 = 25.

now the program must return for example a 40 pairs of points (x,y)
which belong to this cirlce. and these points must be taken evenly
form the circle because i need them for planning a trajectory in CNC.

 

[Dave Vandervies]

hello,

can anyone give me a clue or simple code on task:

I hope not. DYODH.

for example we have in 2D an equation fo circle:
(x - 3)^2 + (y - 4)^2 = 25.

now the program must return for example a 40 pairs of points (x,y)
which belong to this cirlce. and these points must be taken evenly
form the circle because i need them for planning a trajectory in CNC.

If you'd found somewhere appropriate to ask the question, instead of
asking a bunch of C programmers a question that has nothing to do with C,
you might have heard something about parametric curves.


dave

 

[per9000]

hello,

can anyone give me a clue or simple code on task:

Why not solve the problem without code first? Do not let the eager to
code destroy your project - I have done many a mistake because of too
much eager to code.

for example we have in 2D an equation fo circle:
(x - 3)^2 + (y - 4)^2 = 25.

Thus a circle with center-point in (3,4) and a radius of 5 (since 5^2
=25), see more at f.x. http://en.wikipedia.org/wiki/Circle

now the program must return for example a 40 pairs of points (x,y)
which belong to this circle. and these points must be taken evenly
form the circle because i need them for planning a trajectory in CNC.

Now I'd recommend you to divide 2*pi by 40 and read up on the Unit
Circle: http://en.wikipedia.org/wiki/Unit_Circle

Now you've practically solved the problem, good luck.

HTH,
Per

--

Per Erik Strandberg
home: www.pererikstrandberg.se
work: www.incf.org
also: www.spongswedencare.se

 

[Army1987]

hello,

can anyone give me a clue or simple code on task:

for example we have in 2D an equation fo circle:
(x - 3)^2 + (y - 4)^2 = 25.

now the program must return for example a 40 pairs of points (x,y)
which belong to this cirlce. and these points must be taken evenly
form the circle because i need them for planning a trajectory in CNC.

You can return a struct with two elements, e.g. struct point { double x,
y; }
To compute pi, you can use 4.0 * atan(1.0), or you can #define PI 3.1415926
with enough digits.

Each point is point.x = 3 + 5*cos(i * 2*pi / 40); point.y = 4 + 5*sin(i *
2*pi / 40);

Use a for loop, and you're done.

 

[Martin Ambuhl]

hello,

can anyone give me a clue or simple code on task:

for example we have in 2D an equation fo circle:
(x - 3)^2 + (y - 4)^2 = 25.

now the program must return for example a 40 pairs of points (x,y)
which belong to this cirlce. and these points must be taken evenly
form the circle because i need them for planning a trajectory in CNC.

#include <stdio.h>
#include <stdlib.h>
#include <time.h>
#include <math.h>

struct point
{
double x;
double y;
};

struct point circle40(struct point c, double r, size_t which);
inline double sqr(double x)
{
return x * x;
}
inline double radius(struct point c, struct point p)
{
return (sqrt(sqr(p.x - c.x) + sqr(p.y - c.y)));
}

int main(void)
{
int i;
struct point this, center = { 3, 4 };
printf("Here are ten points chosen from the 40\n"
"nearly equally spaced ones on a circle centered\n"
"at (3,4) with a radius 5.\n\n"
"%6s%13s%18s\n", "x", "y", "radius");
srand(time(0));
for (i = 0; i < 10; i++) {
this = circle40(center, 5, rand() * 40. / (1. + RAND_MAX));
printf("%12f %12f %12f\n",
this.x, this.y, radius(center, this));
}
}


struct point circle40(struct point c, double r, size_t which)
{
static struct point edge[40] = {
{1, 0},
{0.987688340595138, 0.156434465040231},
{0.951056516295154, 0.309016994374947},
{0.891006524188368, 0.453990499739547},
{0.809016994374947, 0.587785252292473},
{0.707106781186548, 0.707106781186547},
{0.587785252292473, 0.809016994374947},
{0.453990499739547, 0.891006524188368},
{0.309016994374947, 0.951056516295154},
{0.156434465040231, 0.987688340595138},
{6.12303176911189e-17, 1},
{-0.156434465040231, 0.987688340595138},
{-0.309016994374947, 0.951056516295154},
{-0.453990499739547, 0.891006524188368},
{-0.587785252292473, 0.809016994374947},
{-0.707106781186547, 0.707106781186548},
{-0.809016994374947, 0.587785252292473},
{-0.891006524188368, 0.453990499739547},
{-0.951056516295154, 0.309016994374948},
{-0.987688340595138, 0.156434465040231},
{-1, 1.22460635382238e-16},
{-0.987688340595138, -0.156434465040231},
{-0.951056516295154, -0.309016994374947},
{-0.891006524188368, -0.453990499739547},
{-0.809016994374948, -0.587785252292473},
{-0.707106781186548, -0.707106781186547},
{-0.587785252292473, -0.809016994374947},
{-0.453990499739547, -0.891006524188368},
{-0.309016994374948, -0.951056516295154},
{-0.156434465040231, -0.987688340595138},
{-1.83690953073357e-16, -1},
{0.156434465040231, -0.987688340595138},
{0.309016994374947, -0.951056516295154},
{0.453990499739547, -0.891006524188368},
{0.587785252292473, -0.809016994374948},
{0.707106781186547, -0.707106781186548},
{0.809016994374947, -0.587785252292473},
{0.891006524188368, -0.453990499739547},
{0.951056516295154, -0.309016994374948},
{0.987688340595138, -0.156434465040231}
};
struct point retval;
retval.x = edge[which].x * r + c.x;
retval.y = edge[which].y * r + c.y;
return retval;
}


[output from one run]
Here are ten points chosen from the 40
nearly equally spaced ones on a circle centered
at (3,4) with a radius 5.

x y radius
-1.938442 4.782172 5.000000
7.755283 2.454915 5.000000
6.535534 0.464466 5.000000
1.454915 8.755283 5.000000
5.938926 8.045085 5.000000
0.061074 -0.045085 5.000000
6.535534 0.464466 5.000000
-1.455033 6.269952 5.000000
0.730048 8.455033 5.000000
7.455033 1.730048 5.000000

 

[Army1987]

#include <stdio.h>
#include <stdlib.h>
#include <time.h>
#include <math.h>

struct point
{
double x;
double y;
};

struct point circle40(struct point c, double r, size_t which);
inline double sqr(double x)
{
return x * x;
}
inline double radius(struct point c, struct point p)
{
return (sqrt(sqr(p.x - c.x) + sqr(p.y - c.y)));
}

If you're using C99, as the use of "inline" suggests, you have the
hypot() function. BTW, IMO it would make more sense to call it
distance() than radius().

int main(void)
{
int i;
struct point this, center = { 3, 4 };
printf("Here are ten points chosen from the 40\n"
"nearly equally spaced ones on a circle centered\n"
"at (3,4) with a radius 5.\n\n"
"%6s%13s%18s\n", "x", "y", "radius");
srand(time(0));
for (i = 0; i < 10; i++) {
this = circle40(center, 5, rand() * 40. / (1. + RAND_MAX));
printf("%12f %12f %12f\n",
this.x, this.y, radius(center, this));
}
}

The OP asked for 40 evenly spaced points on the circle, not 10
(pseudo)randomly choosen (and not necessarily distinct) vertices of
the regular 40gon. So you should've used
for (i = 0; i < 40; i++) {
this = circle40(center, 5, i);
/* etc... */

struct point circle40(struct point c, double r, size_t which)
{
static struct point edge[40] = {
{1, 0},
{0.987688340595138, 0.156434465040231},
{0.951056516295154, 0.309016994374947},
{0.891006524188368, 0.453990499739547},
{0.809016994374947, 0.587785252292473},
{0.707106781186548, 0.707106781186547},
{0.587785252292473, 0.809016994374947},
{0.453990499739547, 0.891006524188368},
{0.309016994374947, 0.951056516295154},
{0.156434465040231, 0.987688340595138},
{6.12303176911189e-17, 1},

If you are indeed doing something like that rather than simplily
use sin() and cos(), at least use 'exact' values. For example,
cos(pi/2) is exactly 0.

struct point circlepoint(struct point c, double r, double angle)
{
struct point result;
result.x = c.x + r * cos(angle);
result.y = c.y + r * sin(angle);
return result;
}

 

[Martin Ambuhl]

If you're using C99, as the use of "inline" suggests, you have the
hypot() function. BTW, IMO it would make more sense to call it
distance() than radius().

Hey! If you want to do this lazy person's homework, go ahead. There is
nothing wrong with the code as posted. And making it C89 compilable
requires nothing more than adding
#define inline
while your "suggestion" requires writing the hypot() function.
When you do this turkey's homework, you can call the function any damn
thing you want.

The OP asked for 40 evenly spaced points on the circle, not 10
(pseudo)randomly choosen (and not necessarily distinct) vertices of
the regular 40gon. So you should've used
for (i = 0; i < 40; i++) {
this = circle40(center, 5, i);
/* etc... */

Did you bother to look at the function circle40? circle40 provides 40
evenly spaced points on a circle. There is no reason why my example of
use should be subject to your asinine criticism.

struct point circle40(struct point c, double r, size_t which)
{
static struct point edge[40] = {
{1, 0},
{0.987688340595138, 0.156434465040231},
{0.951056516295154, 0.309016994374947},
{0.891006524188368, 0.453990499739547},
{0.809016994374947, 0.587785252292473},
{0.707106781186548, 0.707106781186547},
{0.587785252292473, 0.809016994374947},
{0.453990499739547, 0.891006524188368},
{0.309016994374947, 0.951056516295154},
{0.156434465040231, 0.987688340595138},
{6.12303176911189e-17, 1},

If you are indeed doing something like that rather than simplily
use sin() and cos(), at least use 'exact' values. For example,
cos(pi/2) is exactly 0.

struct point circlepoint(struct point c, double r, double angle)
{
struct point result;
result.x = c.x + r * cos(angle);
result.y = c.y + r * sin(angle);
return result;
}

 

[Army1987]

Hey! If you want to do this lazy person's homework, go ahead. There is
nothing wrong with the code as posted. And making it C89 compilable
requires nothing more than adding
#define inline
while your "suggestion" requires writing the hypot() function.

#define hypot(x, y) (sqrt((x)*(x) + (y)*(y)))

When you do this turkey's homework, you can call the function any damn
thing you want.



Did you bother to look at the function circle40? circle40 provides 40
evenly spaced points on a circle. There is no reason why my example of
use should be subject to your asinine criticism.

Yes, I did.
The OP asked for a *program* which provided 40 points. And your
main() function provides 10 points. Also, he said "for example 40",
so why did you hard-code that number in the function rather than
use a parameter?Why do you use size_t? Details about how the function is
implemented shouldn't affect the interface. The function returns
a point at a distance r from c, at an angle of n/40 of a full
circle starting from the easternmost point counterclockwise. What
has that purpose to do with bytes or members of arrays, except the
fact that you implement it the way you do?

BTW, I would replace edge[which] with edge[which % 40]. Ever heard
about the least astonishment principle? If someone calls
circlepoint(origin, 1.0, 43) or circlepoint(origin, 1.0, -4) they
are far more likely to expect to get the third or 36th point than
to expect demons to fly out of their nose.
(Also, I'd rename edge to vertex...)

{
static struct point edge[40] = {
{1, 0},
{0.987688340595138, 0.156434465040231},
{0.951056516295154, 0.309016994374947},
{0.891006524188368, 0.453990499739547},
{0.809016994374947, 0.587785252292473},
{0.707106781186548, 0.707106781186547},
{0.587785252292473, 0.809016994374947},
{0.453990499739547, 0.891006524188368},
{0.309016994374947, 0.951056516295154},
{0.156434465040231, 0.987688340595138},
{6.12303176911189e-17, 1},

If you are indeed doing something like that rather than simplily
use sin() and cos(), at least use 'exact' values. For example,
cos(pi/2) is exactly 0.

struct point circlepoint(struct point c, double r, double angle)
{
struct point result;
result.x = c.x + r * cos(angle);
result.y = c.y + r * sin(angle);
return result;
}