Paul Lamar wrote:
Technically speaking frontal area is rarely around the front of the
airplane. It is defined as the maximum cross-sectional area of the
aircraft. In the case of many pushers it occurs about 2/3-rds of the way
towards the back.
If you park the airplane at night in front of a brick wall
and you shine a spot light on if from a distance located on the
longitudinally axis a shadow will appear on the brick wall. If you
count the bricks that are in the shadow and add up their total
area then that is the frontal area of the aircraft.
Russell Kent responded:
Actually, for that to work the light rays must be parallel, i.e. the "spotlight" would need
to be very,
very far away. Like as far away as the Sun. :-)
But you can still make it work by multiplying the counted brick area by the correction
factor:
1 / (( Dsub2 / Dsub1 ) ^ 2)
where Dsub1 is the distance from the spotlight to the "location" of the maximum
cross-sectional area,
and Dsub2 is the distance from the spotlight to the brick wall. The "location" of the
maximum
cross-sectional area would be the average of all the spots where the plane's surface
transitions from
lighted to dark. Generally it's about the same spot as the Cg.
Paul Lamar responded:
Run some of these numbers Russell for the airplane parked right
in front of the brick wall and the spot 100 feet away and get back
to us :)
Well, assuming the "100 feet" (above) is the distance from spotlight to wall, and the plane's
"location" of
maximum cross-sectional area is 10 feet in front of the wall (plane's rudder at the wall), the
area of the
shadowed bricks will be 23% larger than the true maximum cross-sectional area, i.e. correction
factor is:
1 / (( 100 / 90 ) ^ 2) = 0.81.
Do you consider 23% to be an acceptable error factor, Paul? :-)
Russell Kent
Wow! That much? Are you sure about that equation?
I would think it would also be a function of the actual frontal area.
For example if the the frontal area were disk shaped and one foot in diameter
and it was 10 feet in front of the wall the ray angle would be
6/1080 or .0055 or .3 degree. If it were one inch in diameter the ray angle
would be a lot less or .5/1080 or .00046 or .022 degree.
If it were zero diameter the area of the shadow would be the same as
the area of the non-existent disk. If it were 10 feet in diameter?
If it were 100 feet in diameter? 1000 feet in diameter?
Obviously it is also a function of the shape. If you have a long
skinny shape such as a wing the wing span shadow would obviously
be a lot longer than the thickness variation.
Paul Lamar
Bob White wrote:
There is an assumption that the light source is a point source. Then it's just the ratio of two
similar triangles. (Rotate the 2D triangle to get the 3D equilivent.) One with a base at the
frontal area of the plane, and one at the wall. If the light source is not very small compared to
the object, it won't work. There is also an error introduced by a long object because the
distance from the light source is not the same at each point. The light source must be "far"
compared to the size of the object to minimize this. A 20 foot long object 1 ft high (wing) would
appear be about 4-5% thicker at the end than near the fuselage with the light source 90 ft. away.
(This is the same reason your torus has a thicker shadow than the sphere.)
Bob White
--
http://www.bobwhite-abq.com
The AirCraft Rotary Engine NewsLetter. Powered by Linux.
ACRE NL web site. http://home.earthlink.net/~rotaryeng/
Copyright 1998-2002 All world wide rights reserved.