Quote:
Originally Posted by sspoldi
one forward Euler integration, one backward Euler integration
|
V
n+1 = V
n + A
n∙dt;
X
n+1 = X
n + V
n+1∙dt;
Maybe provides some insight: the above is algebraically equivalent to
V
n+1 = V
n + A
n∙dt;
X
n+1 = X
n + dt∙(V
n+V
n+1)/2 + ½∙A
n∙dt
2