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Original file line number | Diff line number | Diff line change |
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#include <stdio.h> | ||
#include <math.h> | ||
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# define M_PI 3.14159265358979323846 /* pi */ | ||
#ifndef M_PI | ||
#define M_PI 3.14159265358979323846 /* pi */ | ||
#endif | ||
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void newton_array(const int n_elements, | ||
const double manom[], | ||
const double ecc[], | ||
const double tol, | ||
const int max_iter, | ||
double eanom[]){ | ||
/* | ||
Vectorized C++ Newton-Raphson solver for eccentric anomaly. | ||
const double manom[], | ||
const double ecc[], | ||
const double tol, | ||
const int max_iter, | ||
double eanom[]){ | ||
/* | ||
Vectorized C Newton-Raphson solver for eccentric anomaly. | ||
Args: | ||
manom (double[]): array of mean anomalies | ||
ecc (double[]): array of eccentricities | ||
eanom0 (double[]): array of first guess for eccentric anomaly, same shape as manom (optional) | ||
Return: | ||
None: eanom is changed by reference | ||
None: eanom (double[]): is changed by reference | ||
Written: Devin Cody, 2018 | ||
*/ | ||
register int i; | ||
for (i = 0; i < n_elements; i ++){ | ||
double diff; | ||
int niter = 0; | ||
int half_max = max_iter/2.0; // divide max_iter by 2 using bit shift | ||
// Let's do one iteration to start with | ||
eanom[i] -= (eanom[i] - (ecc[i] * sin(eanom[i])) - manom[i]) / (1.0 - (ecc[i] * cos(eanom[i]))); | ||
diff = (eanom[i] - (ecc[i] * sin(eanom[i])) - manom[i]) / (1.0 - (ecc[i] * cos(eanom[i]))); | ||
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while ((fabs(diff) > tol) && (niter <= max_iter)){ | ||
eanom[i] -= diff; | ||
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// If it hasn't converged after half the iterations are done, try starting from pi | ||
if (niter == half_max) { | ||
eanom[i] = M_PI; | ||
} | ||
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diff = (eanom[i] - (ecc[i] * sin(eanom[i])) - manom[i]) / (1.0 - (ecc[i] * cos(eanom[i]))); | ||
niter += 1; | ||
} | ||
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// If it has not converged, set eccentricity to -1 to signal that it needs to be | ||
// solved using the analytical version. Note this behavior is a bit different from the | ||
// numpy implementation | ||
if (niter >= max_iter){ | ||
printf("%f %f %f %f >= %d iter\n", manom[i], eanom[i], diff, ecc[i], max_iter); | ||
eanom[i] = -1; | ||
} | ||
} | ||
*/ | ||
int i; | ||
for (i = 0; i < n_elements; i ++){ | ||
double diff; | ||
int niter = 0; | ||
int half_max = max_iter/2.0; // divide max_iter by 2 using bit shift | ||
// Let's do one iteration to start with | ||
eanom[i] -= (eanom[i] - (ecc[i] * sin(eanom[i])) - manom[i]) / (1.0 - (ecc[i] * cos(eanom[i]))); | ||
diff = (eanom[i] - (ecc[i] * sin(eanom[i])) - manom[i]) / (1.0 - (ecc[i] * cos(eanom[i]))); | ||
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while ((fabs(diff) > tol) && (niter <= max_iter)){ | ||
eanom[i] -= diff; | ||
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// If it hasn't converged after half the iterations are done, try starting from pi | ||
if (niter == half_max) { | ||
eanom[i] = M_PI; | ||
} | ||
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diff = (eanom[i] - (ecc[i] * sin(eanom[i])) - manom[i]) / (1.0 - (ecc[i] * cos(eanom[i]))); | ||
niter += 1; | ||
} | ||
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// If it has not converged, set eccentricity to -1 to signal that it needs to be | ||
// solved using the analytical version. Note this behavior is a bit different from the | ||
// numpy implementation | ||
if (niter >= max_iter){ | ||
printf("%f %f %f %f >= %d iter\n", manom[i], eanom[i], diff, ecc[i], max_iter); | ||
eanom[i] = -1; | ||
} | ||
} | ||
} | ||
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void mikkola_array(const int n_elements, const double manom[], const double ecc[], double eanom[]){ | ||
/* | ||
Vectorized C Analtyical Mikkola solver for the eccentric anomaly. | ||
Adapted from IDL routine keplereq.pro by Rob De Rosa http://www.lpl.arizona.edu/~bjackson/idl_code/keplereq.pro | ||
Args: | ||
manom (double[]): mean anomaly, must be between 0 and pi. | ||
ecc (double[]): eccentricity | ||
eanom0 (double[]): array for eccentric anomaly | ||
Return: | ||
None: eanom (double[]): is changed by reference | ||
Written: Devin Cody, 2019 | ||
*/ | ||
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int i; | ||
double alpha, beta, aux, z, s0, s1, se0, ce0; | ||
double f, f1, f2, f3, f4, u1, u2, u3; | ||
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for (i = 0; i < n_elements; i++){ | ||
alpha = (1.0 - ecc[i]) / ((4.0 * ecc[i]) + 0.5); | ||
beta = (0.5 * manom[i]) / ((4.0 * ecc[i]) + 0.5); | ||
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aux = sqrt(beta*beta + alpha*alpha*alpha); | ||
z = pow(fabs(beta + aux), (1.0/3.0)); | ||
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s0 = z - (alpha/z); | ||
s1 = s0 - (0.078*(pow(s0, 5))) / (1.0 + ecc[i]); | ||
eanom[i] = manom[i] + (ecc[i] * (3.0*s1 - 4.0*(s1*s1*s1))); | ||
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se0=sin(eanom[i]); | ||
ce0=cos(eanom[i]); | ||
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f = eanom[i]-ecc[i]*se0-manom[i]; | ||
f1 = 1.0-ecc[i]*ce0; | ||
f2 = ecc[i]*se0; | ||
f3 = ecc[i]*ce0; | ||
f4 = -f2; | ||
u1 = -f/f1; | ||
u2 = -f/(f1+0.5*f2*u1); | ||
u3 = -f/(f1+0.5*f2*u2+(1.0/6.0)*f3*u2*u2); | ||
eanom[i] += -f/(f1+0.5*f2*u3+(1.0/6.0)*f3*u3*u3+(1.0/24.0)*f4*(u3*u3*u3)); | ||
} | ||
} | ||
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int main(void){ | ||
// Test functions with a small program | ||
// Test functions with a small program | ||
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// Define variables for newton array method | ||
double m[] = {.5, 1, 1.5}; | ||
double ecc[] = {.25, .75, .83}; | ||
double tol = 1e-9; | ||
int mi = 100; | ||
double eanom[] = {0, 0, 0}; | ||
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// test newton_array | ||
// Answer should be: [ 0.65161852, 1.73936894, 2.18046524]) | ||
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newton_array(3, m, ecc, tol, mi, eanom); | ||
int i; | ||
for (i = 0; i < 3; i++){ | ||
printf("eanom[%d] = %f\n", i, eanom[i]); | ||
eanom[i] = 0; | ||
} | ||
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//Define variables for newton array method | ||
double m[] = {.5, 1, 1.5}; | ||
double ecc[] = {.25, .75, .83}; | ||
double tol = 1e-9; | ||
int mi = 100; | ||
double eanom[] = {0, 0, 0}; | ||
// test mikkola_array | ||
// Answer should be: [ 0.65161852, 1.73936894, 2.18046524]) | ||
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//test newton_array | ||
newton_array(3, m, ecc, tol, mi, eanom); | ||
//Answer should be: [ 0.65161852, 1.73936894, 2.18046524]) | ||
mikkola_array(3, m, ecc, eanom); | ||
for (i = 0; i < 3; i++){ | ||
printf("eanom[%d] = %f\n", i, eanom[i]); | ||
} | ||
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return 0; | ||
return 0; | ||
} |
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