Incidence Is Not Incidental By Noel J. Becar, EAA 725 San Francisco - East Bay Chapter 20 ITH MANY homebuilders, setting the angle of inciW dence is something of a mystery, so the decision made in reference to it is often, "let's ignore it and maybe
airplane at its cruising speed. This can be solved by applying the basic lift equation:
it will go away." Moreover, I find practically no information in reference to this topic in current textbooks and no article seems to have appeared in SPORT AVIATION defining the parameters that should be considered. So let's take a good analytical look at some factors entering into the solution, for optimum performance, of the question, "At what angle should we mount the wing?" We might well begin this clambake by asking, "At what angle of incidence should a wing be placed on any airplane?" The most cogent answer to this question is, "At that angle which allows the wing to develop the precise amount of lift necessary to support the actual
Where: CL = Lift coefficient as given on airfoil data charts
gross weight of the airplane at some desired velocity."
You will note we said, "the actual gross weight," etc., because the estimated gross, or the maximum gross, or
any other gross weight may not be what the wing is lifting during most of its useful life. The weight which it will be lifting, however, is a weight which must be realistically estimated in order to obtain the maximum benefit from the applied angle of incidence. Remember this, and we'll come back to it later. The desired velocity, in the case of a Goodyear racer, would be the maximum speed for which it's designed, whereas, with a crop-duster it would logically be the speed at which the dusting operation is performed.
CL = 391/(V)2 x W/S V = Desired velocity, or speed in mph W = Actual gross weight in Ibs.
S = Wing area in sq. ft. In the form given, the multiplication factor of 391 in the lift equation solves for the CL value at sea level only. As most cruising would be done at altitude and not sea level, the following table gives substitute values for use at various altitudes: Altitude
Numerical factor in lift equation
in teet
Sea level 1000 2000 3000 4000
391 403
6000 7000 8000 10000
415 427
440
NACA 4412
Therefore, the POLLIWOG'S engines will be mounted
so that their thrust lines will be negative 3 deg. to the datum. This will result in a positive angle of 3 deg. for
MARCH 1964
Numerical factor in lift equation
454 468 482 498 530
EXAMPLE: Solve for the C, value required where single-place operation with 20 Ibs. of baggage is desired for a
This might be only 20 or 30 mph above stalling speed.
10
in teet 5000
However, in the case of most homebuilts, the desired velocity at which the airplane would be expected to be flown most of the time would, no doubt, be its cruising speed. This, then, is our starting point. Our next consideration is to obtain the most efficient performance at this velocity. To accomplish this, we must have the attitude of the fuselage parallel to the path of level flight or at any given angle relative to the flight path which would allow the fuselage to offer the least drag. If the fuselage has its top longerons level in the position of minimum drag (and the majority of homebuilts do), then this would be our reference point or datum. In most cases this will also coincide with the thrust line of the engine. Some fuselages and hulls, and many have been tested by the NACA in numerous wind tunnel tests, present an attitude of minimum drag when they are placed at an angle of 1 or 2 deg. positive to the airflow measured relative to a datum, as previously mentioned. Others have to be placed at a negative angle of 1 or 2 deg. to obtain optimum results. As an example, my little amphibian, the POLLIWOG, records minimum drag when its keel is at an angle of from 2 to 4 deg., positive to the relative airflow. This has been verified by extensive wind tunnel tests by the NACA on planing-tail hulls.
the hull with the thrust lines level. Now that we have the foregoing condition recognized, we need to know what the lift coefficient value should be in order to support the gross weight of the
Altitude
-0.4C
majority of the flight time in a homebuilt designed to carry a maximum of two people and 60 Ibs. of baggage at a gross weight of 1550 Ibs. The above mentioned single-place operation for a majority of the time illustrates what we meant by consideration of actual gross weight as mentioned earlier in this article. In this case, ths actual design gross weight for cruising would be 1550 —
(170+40) = 1340 Ibs. Using this value for W and assuming a realistic cruising speed of 115 mph, with 130 sq. ft. of wing area, we will solve for C, at an expected normal cross-country cruising altitude of 5000 ft. The above table gives a factor of 454 for this altitude, hence: C, ^ 4547(115)2 x 1340/130 = .354
Our hypothetical airplane must therefore develop a lift
coefficient of .354 in order to exactly balance its desired gross weight of 1340 Ibs. at 115 mph and at 50CO ft. altitude.
We shall also assume this airplane is to use an NACA
4412 airfoil section in a rectangular wing without aerodynamic twist, or washout, and with the same airfoil used from root to tip. In this case, we refer to an aerodynamic characteristics chart for the NACA 4412, reproduced herein as FIG. 1. A word needs to be said here regarding which C,, curve to refer to, as you will note curves are
given for various Reynolds Numbers as listed under "R" at the top of FIG. 1. As has often been noted before in these pages, the curve identified as "standard roughness" is closer to the results which can be expected from amateur construction than any other curve. Of course, if you are fabricating with sheet metal, using flush rivets and enough
internal support for the skin so that practically no deformation, ripples or waves, or protuberances exist, you may find the tests covered by the upper three symbols on the chart, under "R", more realistic. Which of these
to use will naturally depend upon the length of your
wing chord and the cruising speed; for Reynolds Number is computed as: R = 6380 x c x V Where: c = Mean aerodynamic chord, (MAC) in feet V = Velocity in feet per second, (1.466 x mph)
ACA 652-415
COMPUTATION OF ANGLE OF INCIDENCE (Before correction for Aspect Ratio ) Using different airfoils at root and tip
of wing and a washout of three de-
grees of incidence between root & tip
C|_ curve for M A C
If we assume a MAC length for our example of 57 in., or
4.75 ft. and a cruise speed of 115 mph, or 168.6 ft. per sec., (1.466 x 115) we can substitute and obtain the R
value for cruise:
R = 6380 x 4.75 x 168.6 = 5,109,423
This value is so close to 6.0 x 1C6, (or 6,000,000) that the curve using the square as an identifying symbol should be the one to use. After locating the intersection of the C, value of .354 with the selected C, coefficient curve, the angle of attack can be read off by dropping a vertical line down
to the abscissa, (or horizontal) values, given in degrees
of angle of attack which, in this example, reads off as: —0.4, or 4/10's of a degree negative. This ends the problem except for a correction of Aspect Ratio, which we will go into at the end of the next problem. Quite frequently, the problem is not so simple, for some designers see fit to use one airfoil at the wing
root and a different one at the tip with a constantly
varying airfoil section between the two, and/or taper in planform. To really complicate the problem, some measure of aerodynamic twist, or washout of incidence is often applied between the wing root and tip. This latter
feature, of course, is to give reliable lateral or rolling control, so that the ailerons will be effective after the root section stalls. Wipe that frown off your face, fellows, because we've
developed a method to lick this problem, too, and it
really is not difficult to apply. The explanation is more time-consuming and difficult to put over than the actual operation of a solution by yourself. After considerable
research and delving into all the factors concerned, we have originated a graphical solution as covered in FIG. 3. The best way to explain the process is to originate a
typical example and then go through the solution, step by step. For this second hypothetical design, we shall apply
the same values for V, W and S used in the first example, but we will use an NACA 65L,-415 airfoil at the wing root with a 72 in. chord and an NACA 65,-412 airfoil at the wing tip with a 42 in. chord. The semi-span or wing panel will be 144 in. from root to tip. In addition, we will consider 3 deg. of washout applied from root to tip to insure good lateral control. Problem: At what angle of incidence should the root chord be placed at the fuselage? Step 1: Determine the length of the mean aerodynamic chord and its distance out from the fuselage. Various methods of obtaining this information arc available but the simplest is probably the graphical method covered in FIG. 2, which is self-explanatory. From this graph we find that the MAC length is 58 in. and that it is located 65.5 in. out from the root chord, or 45.5 percent of the 144 in. semi-span, (65.5/144 = .455).
1'
Step 2: On the left side of a sheet of graph paper duplicate the straight section of the C, curve for the root airfoil, (the 65.,-415 in this case) covering C, values from 0 to about onehalf the maximum, as shown in FIG. 3. On the right side of the sheet do the same for the tip section, (ths 65,-412 in this example) spacing
IUCA - 65| -142
(Continued on next page)
Lines connecting points on ROOT C[_curve to point on TIP C(_ci
three degrees lower in angle of attack ANGLE OF ATTACK
- MiC» - 652-U1S
-1"
(f
SPORT
AVIATION
11
Graphic Computation of Mean Aerodynamic Chord 144"
L.B.
T.K.
FIGURE 2
INCIDENCE IS NOT INCIDENTAL . . . (Continued from preceding page)
the two curves at any distance that can be conveniently figured in percentage, (10 in. is a good dimension for this purpose). Be careful to lay out the angle of attack as accurately as possible below the curves at the bottom of the sheet. Now, connect the Ctj curve for the root section at each increment of 1 deg. for angle of attack intersecting the curve, using horizontally down-slanting lines terminating at intersections of the tip section C r curve with the vertical lines associated with angles of attack 3 deg. less in value. EXAMPLE: Draw a line from the 0 deg. angle of attack intersection at the root curve to the —3 deg. angle of attack intersection on the tip curve. Of course, if your design has a washout of only 2 deg. in place of the 3 deg. used in this example, you would connect from 0 deg. on the root curve to —2 deg. at the tip curve, etc. Step 3: Measure the length of the bottom slant line from its intersection with the root curve to its intersection with the tip curve. Measure how far from the root curve the MAC curve should be, based on its percentage distance from the root, as obtained in FIG. 2, (in this case, 45.5%). EXAMPLE: If the distance between the 0 deg. intersection on the root curve and the —3 deg. intersection on the tip curve should be exactly 10 in., we would measure out 4.55 in. from the root curve intersection and establish a mark on the connecting slant line. Then go through the same operation at the top slant line,
(in this illustration, the slant line from the 4 deg. intersect of the root curve to the 1 deg. intersect of the tip curve). Now, connect the points you have established on the upper and lower slant lines and this will represent the C, curve of the MAC. Step 4: Now, all we have to do is to locate our C r value of .354 where it intersects the MAC curve, then travel back, upward and parallel to the slant lines, (as indicated by the arrowed dash lines on FIG. 3) until we intersect the root CL curve, whereupon we drop vertically from this intersection to the bottom of the chart and read off the angle of attack in degrees, (which, in this case is 1.9 deg.). As a matter of academic interest, the tip section angle would be 3 deg. less, or — 1.1 deg. The above explanation has taken about twice as long as it takes to set up the whole chart and perform the operation as described. As the airfoil section characteristics are plotted from two-dimensional tests, (infinite span, as the model airfoil extends from wall to wall in the wind tunnel) we must now correct for Aspect Ratio which involves the following simple formula:
