提交: adb72c59d2f73bc2989ee7f1f64aa5ae955a9a04 - myprojects-hg-reborn (hg)

Commit MetaInfo

• 修订版: adb72c59d2f73bc2989ee7f1f64aa5ae955a9a04 (tree)
• 时间: 2008-01-07 19:00:56
• 作者: iselllo
• Commiter: iselllo

Log Message

I simply modified the definition of the length I am using in the routine
for numerical derivatives; now it is worked out correctly every time I
define a new r vector where r is the independent variable.

更改概述

• modified: Python-codes/plot_potential.py

差异

diff -r 4c573ce35ef7 -r adb72c59d2f7 Python-codes/plot_potential.py
--- a/Python-codes/plot_potential.py	Sun Jan 06 23:59:46 2008 +0000
+++ b/Python-codes/plot_potential.py	Mon Jan 07 10:00:56 2008 +0000
@@ -314,6 +314,8 @@
 r_cut=1.1
 min_tabul_distance=0.01 #minimum tabulated distance for the force
 r=s.linspace(min_tabul_distance,r_cut,NN)
+#r=s.linspace(0.9,r_cut,NN)
+
 h_max=60.
 sig=0.1
 h_min=-30.
@@ -349,7 +351,9 @@
 my_f_analytic=myforce_modified(h_max,r_core, r_cut,sig,r_min,h_min,r)
 
 my_force=s.zeros(NN)
-length=r_cut-min_tabul_distance
+#length=r_cut-min_tabul_distance
+length=r.max()-r.min()
+
 d.der1(my_pot,my_force,length)
 
 my_force=-my_force #the force is minus the grad of the potential!