PLEASE somebody look at these equations
and tell me where I am going wrong. There must be a mistake in here
somewhere.
*********************************************************************
CLS
Weight = 1900 'pounds
statthrust = 2000 'pounds
distance = 0 'feet
FlatPlate = 4 'square feet
PropEff = .8
HP = 300 '1 HP = 550 pound foot per second
FPS = .001
distance = .001
PRINT "Time Drag HP NetThrust G's FPS MPH Feet"
PRINT
FOR time = .1 TO 4.6 STEP .1
PRINT USING "###.#"; time;
Drag = FlatPlate * MPH ^ 2 * .0026
PRINT USING "#######.##"; Drag;
PRINT USING " #####"; HP;
Thrust = (HP * 550 * PropEff) / FPS
IF Thrust > 2000 THEN Thrust = 2000 'lower limit for thrust
NetThrust = Thrust - Drag
PRINT USING "#########"; NetThrust;
Gs = NetThrust / Weight
PRINT USING " ##.##"; Gs;
velincFPS = Gs * 32 * .1
VEL = VEL + velincFPS
PRINT USING "#########"; VEL;
MPH = 60 / 88 * VEL
PRINT USING "######"; MPH;
distance = (VEL * time) + distance
PRINT USING "########"; distance
NEXT
You can only climb straight up for a few seconds so you better
start pitching over right away or its going to hang on the
prop. This program assumes you are going to pitch over right
away. If you don't the velocity will drop to near zero and the
the airplane will stop climbing vertically. It also assumes the prop
pitch is adjusted for maximum thrust at all times.
Top speed will be limited by the rotor tips going supersonic.
Paul Lamar ...No rotor no motor.
Paul,
The way I read your data, at takeoff, your are making 6.66 lbs of thrust
per
hp. At 11,000 ft you are making 9.18 lbs of thrust per hp. I think that
portion of your formula is inverted, as the other attached chart shows the
prop loosing thrust with airspeed (and air density), not gaining.
The other thing is the velocity appears to be gaining at a steady rate and
it should be decaying in proportion to the vertical g force.
As a reality check, look up the gross weight of a 300 hp helicopter and
what
it's vertical performance is.
Larry
Thanks Larry. I'll check it over.
--
Paul Lamar ...No rotor no motor.
Thanks Larry for your help in finding those errors and thinking about it.
I lost focus there for a minute. This is not altitude. It is distance
traveled
through an arc over.
I added altitude compensation to the HP by mistake. I removed it in this
version.
I am assuming the aircraft is going to pitch over once it reaches
flying speed around 70 MPH and at around 1000 feet so this is mostly a
crude
approximation of the vertical and horizontal acceleration up to four
seconds.
No need for altitude effect on HP below 1000 feet.
This model needs to be refined to add the effects of gravity and perhaps
induced
drag due to lift. Gravity hardly affects the acceleration in horizontal
mode and induced drag due to lift hardly affects it in vertical mode.
I will probably assume the flight path is an arc consisting of one
quarter of a circle and calculate the various forces based on
that assumption. This is new to me but I am sure it was new to the
Osprey designers as well. I notice the Osprey starts moving forward
almost immediately and within a few hundred feet of altitude.
I'll refine the model more later.
Here is a chart showing rotor diameter tip speeds using the
6.24 PSRU ratio. We are going to be limited to about a 12 to 14 foot
rotor due to the forward velocity vector adding to these tip
velocity numbers approaching Mach 1. That means the blade area must be
large
enough to absorb the HP available and provide the thrust required.
I am sure Perry is working on this as we speak.
The Osprey is affected by this same phenomena but it has the advantages
of twice the disk area. The name of the game in vertical take off
is a low rotor disk loading.
The engine and rotor RPM will probably be forced to slow due to tip mach
effects
at WOT as the airplane pitches over and builds forward speed with a
consequent
loss in available Horse Power. Fortunately we don't need all that HP in
forward
flight anyway nor do we want to run the engine at 8500 RPM continuously.
We are very fortunate indeed to have a very low cost engine with such
an exceptional power to weight ratio as the Mazda p-port Wankel
for this application.
Paul Lamar ...No rotor no motor.
Paul,
Suppose that you have been on site
http://www.vstol.org
Colin Smith
Yes but there are no free technical papers on there as far as I could find.
This is a bit hard to access but there are free NASA and NACA technical
papers on here. Just be careful of how you cut and past this into
your browser.
http://naca.larc.nasa.gov/search.jsp?R=846225&id=9&qs=Ntt%3DVertical%252B
takeoff%26Ntk%3Dall%26Ntx%3Dmode%2520matchall%26N%3D0%26Ns%3DHarvestDate%257c1
Paul Lamar ...No rotor no motor.
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