Changeset 3b571ae in sasmodels


Ignore:
Timestamp:
Mar 22, 2017 6:32:33 PM (9 months ago)
Author:
Paul Kienzle <pkienzle@…>
Branches:
master, boltzmann, costrafo411, doc_update, generic_integration_loop, ticket-1043, ticket-786
Children:
61104c8
Parents:
b00a646
Message:

ellipsoid: fix docs so equations are consistent with code; rearrange code for speed and readability

Location:
sasmodels/models
Files:
2 edited

Legend:

Unmodified
Added
Removed
  • sasmodels/models/ellipsoid.c

    r130d4c7 r3b571ae  
    33double Iqxy(double qx, double qy, double sld, double sld_solvent, 
    44    double radius_polar, double radius_equatorial, double theta, double phi); 
    5  
    6 static double 
    7 _ellipsoid_kernel(double q, double radius_polar, double radius_equatorial, double cos_alpha) 
    8 { 
    9     double ratio = radius_polar/radius_equatorial; 
    10     // Using ratio v = Rp/Re, we can expand the following to match the 
    11     // form given in Guinier (1955) 
    12     //     r = Re * sqrt(1 + cos^2(T) (v^2 - 1)) 
    13     //       = Re * sqrt( (1 - cos^2(T)) + v^2 cos^2(T) ) 
    14     //       = Re * sqrt( sin^2(T) + v^2 cos^2(T) ) 
    15     //       = sqrt( Re^2 sin^2(T) + Rp^2 cos^2(T) ) 
    16     // 
    17     // Instead of using pythagoras we could pass in sin and cos; this may be 
    18     // slightly better for 2D which has already computed it, but it introduces 
    19     // an extra sqrt and square for 1-D not required by the current form, so 
    20     // leave it as is. 
    21     const double r = radius_equatorial 
    22                      * sqrt(1.0 + cos_alpha*cos_alpha*(ratio*ratio - 1.0)); 
    23     const double f = sas_3j1x_x(q*r); 
    24  
    25     return f*f; 
    26 } 
    275 
    286double form_volume(double radius_polar, double radius_equatorial) 
     
    3715    double radius_equatorial) 
    3816{ 
     17    // Using ratio v = Rp/Re, we can implement the form given in Guinier (1955) 
     18    //     i(h) = int_0^pi/2 Phi^2(h a sqrt(cos^2 + v^2 sin^2) cos dT 
     19    //          = int_0^pi/2 Phi^2(h a sqrt((1-sin^2) + v^2 sin^2) cos dT 
     20    //          = int_0^pi/2 Phi^2(h a sqrt(1 + sin^2(v^2-1)) cos dT 
     21    // u-substitution of 
     22    //     u = sin, du = cos dT 
     23    //     i(h) = int_0^1 Phi^2(h a sqrt(1 + u^2(v^2-1)) du 
     24    const double v_square_minus_one = square(radius_polar/radius_equatorial) - 1.0; 
     25 
    3926    // translate a point in [-1,1] to a point in [0, 1] 
     27    // const double u = Gauss76Z[i]*(upper-lower)/2 + (upper+lower)/2; 
    4028    const double zm = 0.5; 
    4129    const double zb = 0.5; 
    4230    double total = 0.0; 
    4331    for (int i=0;i<76;i++) { 
    44         //const double cos_alpha = (Gauss76Z[i]*(upper-lower) + upper + lower)/2; 
    45         const double cos_alpha = Gauss76Z[i]*zm + zb; 
    46         total += Gauss76Wt[i] * _ellipsoid_kernel(q, radius_polar, radius_equatorial, cos_alpha); 
     32        const double u = Gauss76Z[i]*zm + zb; 
     33        const double r = radius_equatorial*sqrt(1.0 + u*u*v_square_minus_one); 
     34        const double f = sas_3j1x_x(q*r); 
     35        total += Gauss76Wt[i] * f * f; 
    4736    } 
    4837    // translate dx in [-1,1] to dx in [lower,upper] 
     
    6251    double q, sin_alpha, cos_alpha; 
    6352    ORIENT_SYMMETRIC(qx, qy, theta, phi, q, sin_alpha, cos_alpha); 
    64     const double form = _ellipsoid_kernel(q, radius_polar, radius_equatorial, cos_alpha); 
     53    const double r = sqrt(square(radius_equatorial*sin_alpha) 
     54                          + square(radius_polar*cos_alpha)); 
     55    const double f = sas_3j1x_x(q*r); 
    6556    const double s = (sld - sld_solvent) * form_volume(radius_polar, radius_equatorial); 
    6657 
    67     return 1.0e-4 * form * s * s; 
     58    return 1.0e-4 * square(f * s); 
    6859} 
    6960 
  • sasmodels/models/ellipsoid.py

    r925ad6e r3b571ae  
    1818.. math:: 
    1919 
    20     F(q,\alpha) = \frac{3 \Delta \rho V (\sin[qr(R_p,R_e,\alpha)] 
    21                 - \cos[qr(R_p,R_e,\alpha)])} 
    22                 {[qr(R_p,R_e,\alpha)]^3} 
     20    F(q,\alpha) = \Delta \rho V \frac{3(\sin qr  - qr \cos qr)}{(qr)^3} 
    2321 
    24 and 
     22for 
    2523 
    2624.. math:: 
    2725 
    28     r(R_p,R_e,\alpha) = \left[ R_e^2 \sin^2 \alpha 
    29         + R_p^2 \cos^2 \alpha \right]^{1/2} 
     26    r = \left[ R_e^2 \sin^2 \alpha + R_p^2 \cos^2 \alpha \right]^{1/2} 
    3027 
    3128 
    3229$\alpha$ is the angle between the axis of the ellipsoid and $\vec q$, 
    33 $V = (4/3)\pi R_pR_e^2$ is the volume of the ellipsoid , $R_p$ is the polar radius along the 
    34 rotational axis of the ellipsoid, $R_e$ is the equatorial radius perpendicular 
    35 to the rotational axis of the ellipsoid and $\Delta \rho$ (contrast) is the 
    36 scattering length density difference between the scatterer and the solvent. 
     30$V = (4/3)\pi R_pR_e^2$ is the volume of the ellipsoid, $R_p$ is the polar 
     31radius along the rotational axis of the ellipsoid, $R_e$ is the equatorial 
     32radius perpendicular to the rotational axis of the ellipsoid and 
     33$\Delta \rho$ (contrast) is the scattering length density difference between 
     34the scatterer and the solvent. 
    3735 
    38 For randomly oriented particles: 
     36For randomly oriented particles use the orientational average, 
    3937 
    4038.. math:: 
    4139 
    42    F^2(q)=\int_{0}^{\pi/2}{F^2(q,\alpha)\sin(\alpha)d\alpha} 
     40   \langle F^2(q) \rangle = \int_{0}^{\pi/2}{F^2(q,\alpha)\sin(\alpha)\,d\alpha} 
    4341 
     42 
     43computed via substitution of $u=\sin(\alpha)$, $du=\cos(\alpha)\,d\alpha$ as 
     44 
     45.. math:: 
     46 
     47    \langle F^2(q) \rangle = \int_0^1{F^2(q, u)\,du} 
     48 
     49with 
     50 
     51.. math:: 
     52 
     53    r = R_e \left[ 1 + u^2\left(R_p^2/R_e^2 - 1\right)\right]^{1/2} 
    4454 
    4555To provide easy access to the orientation of the ellipsoid, we define 
     
    4858:ref:`cylinder orientation figure <cylinder-angle-definition>`. 
    4959For the ellipsoid, $\theta$ is the angle between the rotational axis 
    50 and the $z$ -axis. 
     60and the $z$ -axis in the $xz$ plane followed by a rotation by $\phi$ 
     61in the $xy$ plane. 
    5162 
    5263NB: The 2nd virial coefficient of the solid ellipsoid is calculated based 
     
    90101than 500. 
    91102 
     103Model was also tested against the triaxial ellipsoid model with equal major 
     104and minor equatorial radii.  It is also consistent with the cyclinder model 
     105with polar radius equal to length and equatorial radius equal to radius. 
     106 
    92107References 
    93108---------- 
     
    96111*Structure Analysis by Small-Angle X-Ray and Neutron Scattering*, 
    97112Plenum Press, New York, 1987. 
     113 
     114Authorship and Verification 
     115---------------------------- 
     116 
     117* **Author:** NIST IGOR/DANSE **Date:** pre 2010 
     118* **Converted to sasmodels by:** Helen Park **Date:** July 9, 2014 
     119* **Last Modified by:** Paul Kienzle **Date:** March 22, 2017 
    98120""" 
    99121 
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