Don't rely on any of this to design your airplane. It may all be wrong. These are just some of the exercises I went through in piddling with mine. Who knows if it will really fly ?????
Daisy
Mae Real Bipe Data -Formula's
First
I need the Control Lift number:
Control
Lift coefficient estimate: 391 (fixed standard) /speed squared
x aircraft estimated average weight / sq. ft of wing ........... or (391/
80 x 80) x (1000 lbs. / 173) = (391/6400) x (1000/ 173 = .061 x 5.78 =
CL of .35.
If I
need to lift 1200 lb., then my C.L. comes out at .42
Next I need to know where to set the angle of incidence for the wings:
Control Lift Weight (angle) = .0011887 x C.L. x Wing Sq. Ft. x feet per second squared.
C.L. is found in the NACA Airfoil book, right side of the page, bottom graph
Feet per second is figured at anticipated cruise of 80-85 mph. i.e. - 80 x 5,280 feet = 422,400 to 448,800 and this number divided by 3600 seconds in an hour.
The C.L. Weight (Angle) answer is the amount of weight your wings will fly at the specified C.L. angle. You then go back and re-figure the C.L. angle to get it where it will fly your specified or calculated aircraft weight.
For example: Our airfoil is a NACA 2415 where I have determined the section Lift Coefficient to be between (.4) and (.45). The wing to be 173 square feet. If I assume a cruise speed of 80 mph (or traveling 80 miles in one hour) then in the course of an hour I will travel 80 times 5,280 (feet in a mile) or 422,400 feet. There are 3600 seconds in an hour (i.e.-60 seconds per minute x 60 minutes). So if I divide 422,400 feet by 3600 I will see that I will be traveling 117.33 feet per second. If I square 117.33 I get 13,766. The formula now looks like this:
.0011887
x .35 x 173 x 13,766 = 990.71 lbs. to
.0011887
x .4 x 173 x 13766 = 1,132 lbs.
Since
I may need to lift upwards of 1200 lb. I can see that I need to increase
my C.L. to about .43 and rework the formula.
.0011887
x .43 x 173 x 13766 = 1217 lbs.
Knowing
this (i.e.- C.L. between .4 and .43) I can now go to my lift curve in the
Theory of Wing Sections book for the NACA 2415, left page (pg. 480) and
see that if I need a C.L. of .4 to .43, then I need to set wing incidence
at about 2.5 degrees.
Estimating Stall Speed - On the 2415 plot I see that at Rn of around 3 million and an angle of attack of about 14 degrees, the wing will stall. Looking to the left of that graph, I see that corresponds to a CL of about 1.4. So now I need to figure out at what speed will I be traveling when my CL of 1.4 arrives. To solve for speed I use this formula:
Take the
square root of ...........normal flying weight x 390 (a constant)
divided
by the wing area in sq. ft times CL of 1.4. This is what it looks
like:
1000 x
390 divided by 173 x 1.4 = 390,000/242.2 = 1610.23 and the square root
of
1610.23
is 40.12 mph, which should be the approximate stall speed. Adding
15%
to that
we should have a landing speed of about 46 mph.
Fully
loaded at 1200 lbs. my stall speed increases to 44 mph and a landing speed
of 51 mph.
NOSE
MOMENT is the distance between the balance
point of the wing(s) and the front of the cowl. And is a percentage of
the overall fuselage length. I am using 29% initially.
Tail
Moment is the distance
aft of the balance point and also a percentage of the total fuselage length.
Reynolds
number (Rn) = Aircraft
cruise speed in feet per second x wing (avg. or mean aerodynamic) cord
in square feet x 6380. To get speed in feet per second multiply your
speed in mph times 1.466. For the bipe I used 80 mph cruise and an avg.
cord of 3.86 Sq.ft. (one wing). Had I used the average cord of both
wings I would use 62.6 inches or 5.21 sq. ft. So in the first
example (80 x 1.466) x 3.86 x 6380 = 117.28 x 3.86 x 6380= 2,888,231.
If I used the combination cord, allowing for the stagger of both wings
I would come up with (80 x 1.466) x 5.21 x 6380 = 3,898,363 Rn. So
now I conclude I will be operating at an Rn of around 3 million.
Where to place the Landing Gear: Pazmany suggests if I draw a vertical line up from the touch point of the tire, with the aircraft level and then project another line from that same point upwards through the center of gravity, that the angular difference should be between 15 and 25 degrees. Trouble is I don't exactly know where the center of gravity is. I know where the balance point is, but not the C.G. I estimate the upper wing will weigh more than the landing gear. So if I draw a line from each wings 25% of cord point I am going to estimate the C.G.. will fall somewhere in this area. So I just placed the landing gear where it was on the model, then drew the lines and find that my angular difference is 19 degrees if I have guessed the C.G. in the right place. If not, then my assumption is that the angular difference would increase rather than decrease and that is good.
How much fuel in that imaginary tank? First I need to know cubic inches of the tank. Height times width times length gets that but what about the cylindrical top part of the tank? My granddaughter suggests I need to find the volume of the complete cylinder by using pi R squared. 3.14 x the radius squared, then to get that into cubic inches I would multiply that answer times the length of the cylinder. At which point I would then know the volume of a complete cylinder. I am only having the top half of the tank rounded so I need to divide that answer by 1/2. After that answer comes up I need to add it to the cubic inches found in the rectangular part of the tank. Lets say I had 1800 cubic inches in the half cylinder and 1800 cubic inches in the rectangular part of the tank for a total of 3600 cubic inches. Looking in the weights and measures part of the dictionary I see there are 231 cubic inches in a gallon, so if I divide 3600 by 231 I come up with a 15.5 gallon tank. Mine will come in around 11-12 gallons I think. Changed the design of the tank several times and ended up with a 15 gallon job.
Figuring
the Neutral Point Total wing area = 172 Sq. Ft
Total
Horizontal area = 22.8 Sq. Ft.
Total
Vertical area = 10.1 Sq. Ft.
Distance
between quarter chord points = 10.25 Ft.
Formula:
Hoiz.
area x distance divided by Wing area x wing cord
22.8 x
10.25 / 172 x 3.86 = 233.7 / 663.9 = .35 or 35% of the wing cord
Therefore
Neutral Point = .35 x 3.86 = 1.35 ft or 16.2 inches aft of L.E. therefore
the balance point should be 10 to 15% of cord added to this dimension for
stability.
However
haven't decided if the wing cord should be increased 18 or 19 inches due
to the stagger between the wings. Will update later and will check
the model for current, accurate balance point and work backwards for the
Neutral Point.
Summary:
I have found that anytime you balance beyond 33% you are getting into an
unstable condition. For rough estimates I would use that normally
and just make sure I was balancing ahead of that point. In the case
of this airplane I wanted to at least attempt a better number and then
add a margin of safety. So the more advanced formula suggests a neutral
point of 36.5% and the rougher formula suggests 35% I have to assume these
are good numbers and anything 15% of chord forward of those points will
be a good start.
The Mean
Aerodynamic chord is 46.5" ( 3.875 ft.) and 15% of that is about 7 inches.
So my starting point will be 46.5 x .355 = 16.5 inches aft of the
leading edge less 7 inches or 9.5 inches aft of the leading edge of each
wing at the true C.G.. 9.5 divided by 46.5 = 20.4 % which
would be 5% forward of optimum. (see below for my attempt at the real formula.)
...................................................................................................
Horizontal
Tail Volume Coefficient and Vertical Tail Vol. Coefficient is figured about
the same way as the Neutral Point.
For Horizontal
TVC you take the area of the horizontal times the distance between quarter
chord points and divide this by the wing area times the wing chord. - ie:
22.8 x 10.2 / 172 x 3.86 = .351
.351 then
is the Horizontal T.V.C.
Vertical Tail Vol. Coefficient is a little different, but not if you think of viewing the craft in a 90 degree turn 9( ie- wings vertical). Here is that formula:
Tail Area
x distance / wing area x wing span or:
10.1 x 10.25
divided by 172 x 23 = .026
.026 is
the Vertical Tail Vol. Coefficient
...............................................................................................................
Estimating weight - Gross, Empty and Useful Load
Estimated Empty Wt. = 760
2 people
= 340
Fuel ( 15
gal x 6 lb) = 90
Oil
= 7.5 lbs
Baggage
= 40 lbs.
Added Wt.
=477.5 lbs
Using a constant of .38 for similar airplanes, divide 477.5 by .38 = 1257 lbs. for gross weight. A few pounds heavy for the new Sport Pilot category.
1257 minus
760 = Useful Load of 497 lbs.
.................................................................................................................
Wing loading
at normal 1100 lbs = 1100 divided by 172 sq. ft. = 6.39
Wing loading
at gross of 1257 lbs = 7.3
.................................................................................................................
Power Loading
= 1100 lbs. divided by 85 hp = 12.94 lbs.
...............................................................................................................
Estimating
Fabric Needs - To estimate the
material
need " (top wing span X 2) + (bottom wing span x2)+ (tail span X2)+(Rudder
height) +(Fuse length X 3) X 1.2 = total
...........................................................................................................
Cowl Air
Inlet Opening should equal .35 x rated horse power. Therefore .35
x 85 = 29.75 sq. inches or roughly 5"x6"
.............................................................................................................
Geometric
Pitch
RPM X Pitch
Inches X .000947 = Speed (MPH)
This approximates
the aircraft forward speed.
2400 cruise
RPM x 44 pitch x .000947 = 100 mph for 42 pitch = 95
2300 cruise
RPM x 44 pitch x .000947 = 95 = for 42 pitch = 91
2200 cruise
RPM x 44 pitch x .000937 = 91 mph or for 42 pitch= 87
2200
" " x 38 "
" " = 78 mph
fuel consumption
about 45 lbs. per hour at full throttle or 7 g.p.hr.
Gasoline
weighs 6.15 lbs per gal.
C-85 max
RPM = 2575
Since a
bipe isn't the most drag resistant airframe these conclusions have to be
very optimistic. Am guessing I need a 72 x 40 prop to cruise 82 mph
at 2200 rpm. Less pitch will give faster take off.
..............................................................
Fuel Consumption
Fuel consumption,
leaned, is about .44 pounds of fuel per hour per horsepower. For the C-85
this gives, .44 X 85 hp /6 lb. per gal = 6.23...gallons per hour.
Fuel available at climb attitude needs to be 150% of full power usage which in this case is 45.5 pounds per hour (according to the engine manual) at full throttle or 7.6 gph ..... multiply that times 1.5 I come up with 11.4 gal per hour needed to be available........ so if I put in a gallon and set it at climb attitude, I figure it needs to drain that gallon in less than 5.5 minutes. (11.4 gph/60 minutes =.19 gal per minute x 5.5 minutes = 1.04 gallons.)
Some useful
conversions
1 mile/hour
(mph) = 0.8689762 knots
1 mile/hour
(mph) = 1.4666667 foot/second
1 feet =
0.3048 meter
1 inch =
2.54 centimeter
1 meter
= 3.2808399 feet
1 centimeter
= 0.3937008 inch
Next is my attempt at the real deal Neutral Point formula
and it came out at 36.5% vs 35% .
Needed for
formula:
Ho = aerodynamic
center of lift. Got this from Theory of Wing sections on a NACA 2415
@ Rn of 3million which suggests the a.c. is 24.1% for that particular airfoil.
Slope of the wing lift curve (a) estimated at .11 (same as the Clark Y) because the 2415 is similar in shape and I had the number for the Clark Y.
Slope of the tail lift curve.(a-1) Tail is flat plate so estimated .095 for that.
Da =
Here is
the formula to solve for Da :
Slope
of the wing lift (.11)
.11_
.11__ Da = .082
1 + ( 18.25
)
= 1.334
(aspect ratio of wing=6) x .11 ) = 1+(.334
)
De =Same
formula as above :
Slope
of the horiz. Lift curve = .095
.095___
.095___ De = .060
1 + ( 18.25
)
=
(aspect ratio of horiz. =3) x .11 ) =
1 + .577
1.577
Lc is the
Ratio of both lift curves & the symbol I will use. Lc is obtained by
dividing the slope of the wing lift number Da into the slope of the horizontal
De)
Therefore
initial Lc = .73
.06___
= .73
.082
However
Lc is then
determined by estimating De/Da to compensate for the fact, that because
of the downwash, the change of angle at the tail will be less than the
wing. So I used this suggested rough formula:
55 (a
constant for a biplane & 35 for monoplane) x Da
= 55 x .082 = 4.51 = .75
Aspect ratio
of the wing
6
6
Lc = .75
Ns = horizontal
stab efficiency assumed at .6 for a normal (not T-tail) stab
Vs = Horiz.
Stab vol. Coefficient. This is figured by dividing the distance between
the quarter chord points of the wing vs. the horizontal x the
area of the horizontal / by
The wing
area times the wing chord.
Stab
area x distance
= 22.8 x 10.38___ = 236.66
= .35
Wing area
x wing chord
124.58 x 5.41
673.97
NOTE: Since I am dealing with a biplane with staggered wings I decided to just use the total area of the wings as seen from the top view which is 124.58. (Treating both wings as one.) Had I used both wings their area is 171 sq. ft.
Now I have
all of the numbers needed for this formula:
Hn (Neutral
Point) = Ho + ( Ns x Vs x (a-1/a) x Lc)
Hn = .241
+ (.6 x .35 x .73 x .75) = .341 + (.115) = .365 or 36.5% of average chord
of one wing.