Good to know that the smoothed polygon does not pass through the
vertices, but that doesn't fully answer my question.
What I need to know is if the smoothed polygon mapping is symmetric or
not. I'm asking this because I run simulations on a thin-flat-plate
(created with the smoothed polygon mapping) in a uniform wind, and i
found that even at angle of attach = 0, the lift force and moment are
not zero, and the flat plate reaches equilibrium at an angle different
from 0.
Also looking closer to the mapping i see that the radial lines above
and belove the plate seems to be a bit shifted, and i'm guessing that
this "broken symmetry" can cause an unbalance in the moment evaluation
and a rotation...
Thanks,
Alessandro
------------------------------------------------------------------------
Date: Fri, 16 Dec 2011 12:52:12 -0800
From: henshaw@xxxxxxxx
To: overture@xxxxxxxxxxxxx
Subject: [overture] Re: smoothed polygon
Hi Alessandro,
The smoothed polygon does not actually pass through the vertices.
I suggest that you start the polygon on the middle of the bottom edge
and maybe add another vertex on the middle of the top edge, even though
the polygon is flat there (or to be even more symmetric add a few
vertices along the
bottom and top edges.
...Bill
Alessandro Orchini wrote:
Thanks Bill, that was really helpful.
Actually I've another question about the smoothed-polygon mapping.
I usually create them with a routine similar to the following one:
***********************************
smoothedPolygon
vertices
5
0. -0.025
1.0 -0.025
1.0 0.025
0. 0.025
0. -0.025
n-dist
fixed normal distance
-1.0
lines
250,150
boundary conditions
-1, -1, 1, 0
n-stretch
2. 10. 0.00001
exit
***********************************
and I think that the leading edge should be in the origin. Is that
so, or the smooth function makes the mapping a bit "longer" and it
is somehow shifted?
I do appreciate your help,
Alessandro
