Changes between Version 16 and Version 17 of u/johannjc

Timestamp:

04/13/12 13:45:18 (13 years ago)

Author:

Jonathan

Comment:

—

u/johannjc

@@ -63 +63 @@
    Determine the path's distance of closest approach to [[latex($\vec{x}$)]] and call that [[latex($\vec{x}'(t')$)]]
   
-   Estimate the change in initial velocity needed [[latex($\vec{v'}(t_r,x_p(t_r) = \vec{v'}(t_r,x_p(t_r))+\frac{x-x'(t')}{t'-t_r}$)]]
+   Estimate the change in initial velocity needed [[latex($\vec{v'}(t_r,x_p(t_r) = \vec{v}(t_r,x_p(t_r))+\frac{x-x'(t')}{t'-t_r}$)]]
 
+   And then solve for the unit vector that gives that direction
+    
+   [[latex($ \hat{n} \times \vec{v}' = \frac{-1}{v_w}\vec{V}_p(t_r) \times \vec{v}' $)]]
 
-
-
-
+   
+   Cycle until [[latex($x' \approx x$)]]
 
   Update the retarded time using the new distance and wind speed
 
-  [[latex($t_r=t-\frac{d}{v}$)]]
+  [[latex($t_r=t-(t'-t_r)$)]]
 
 END DO
@@ -106 +108 @@
  * Then when distributing the level 1 work load - we have to check that nobody will be waiting for the level 1 synchronizations as well as the level 0 synchronization.  So how long before any 1 processor is idle after level 1 step?  It's level 1 workload + remaining level 0 work load.  So max level 1 workload assigned = min(level 1 workload + remaining level 0 workload)
 
-