Darn if this isn't an interesting discussion. Data from the NASA site
pasted below my uninformed rant.
Well - if you took a 55 gallon drum and glued it to the front of a packing
crate with the same length as the drum, and the same height as the drum
diameter, the picture from the side would be a big rectangle. But we all
know that the barrel would go sideways through the air easier than the
packing crate - so I agree, the 'actual' shape around the longitudinal axis
will have a big effect. (Kind of explains why belly tanks are rumored to be
difficult to stabilize - from the side both ends look like they have an
equal chance - and, in this case, the body is symmetrical around the long
axis (circular)) I can also see that huge complexity is added for an
'irregular' shaped body and as soon as you start moving and your rearview
side mirror (or whatever) starts to stir up the air, turbulence is going to
change the airflow, different at different speeds. (Bottom line - for
anything other than a simple symmetric shape calculations are going to get
really hard really fast - and a simple deterministic method would move
farther from the truth?) Also, from a side view, can one practically
streamline the 'sideways' body enough to overcome the raw effect of the
(side) frontal area?
I'm still left with a question as to how valuable this might be for a more
regularly shaped vehicle - streamliners look to be fairly symmetric around
their long axis. As for hanging a model out the window of your moving car -
I think (?) it would still be important to weight the model for the same
weight distribution as the real one.
------ From the NASA website
Calculating cp
You can calculate the center of pressure. But, in general, this is a
complicated procedure requiring the use of calculus. The aerodynamic forces
are the result of pressure variations around the surface of the rocket. In
general, you would have to determine the integral of the pressure times the
unit normal, times the area, times the distance from a reference line. Then
divide by the integral of the pressure times the unit normal, times the
area. (Lot's of work!) For a model rocket, there are some simplifying
assumptions that we can use to make this task much easier. Model rockets are
fairly symmetric about the axis of the rocket. This allows us to reduce the
full three dimensional problem to a simple, two dimensional cut through the
axis of the rocket. For model rockets, the magnitude of the pressure
variation is quite small. If we assume that the pressure is nearly constant,
finding the average location of the pressure times the area distribution
reduces to finding just the average location of the projected area
distribution.
Simplified Calculation of cp
This figure shows a simplified version of the calculation procedure that you
can use. We assume that we already know the projected area and location,
relative to some reference location, of each of the major parts of the
rocket: the nose, body tube, and fins. The projected area of the rocket is
the sum of the projected area of the components. Since the center of
pressure (cp) is an average location of the projected area, we can say that
the area of the whole rocket times the location of the center of pressure is
equal to the sum of the projected area of each component times the distance
of that component from the reference location. The "location" of each
component is the distance of each component's center of pressure from the
reference line. So you have to be able to calculate or determine the center
of pressure of each of the components. For example, the projected area of
the body tube is a rectangle. The center of pressure is on the axis, half
way between the end planes.
Mechanically determining cp
For a model rocket, there is a simple mechanical way to determine the center
of pressure for each component or for the entire rocket. Make a two
dimensional tracing of the shape of the component (or rocket) on a piece of
cardboard and cut out the shape. Hang the cut out shape by a string, and
determine the point at which it balances. (Just like balancing a pencil with
a string!) The point at which the component (or rocket) is balanced is the
center of pressure. You obviously could not use this procedure for a very
large rocket like the Space Shuttle. But it works quite well for a model.
http://www.grc.nasa.gov/WWW/K-12/airplane/rktcp.html
Thanks,
Jim.
/// unsubscribe/change address requests to majordomo@autox.team.net or try
/// http://www.team.net/mailman/listinfo
/// Archives at http://www.team.net/archive/land-speed
/// what is needed. It isn't that difficult, folks.
|