CHAPTER XVI. HEAD RESISTANCE CALCULATIONS.

*Effect of Resistance*. Resistance to the forward motion of an aeroplane

can be divided into two classes, (1) The resistance or drag due to the

lift of the wings, and (2) The useless or "Parasitic" resistance due to

the body, chassis and other structural parts of the machine. The total

resistance is the sum of the wing drag and the parasitic resistance.

Since every pound of resistance calls for a definite amount of power, it

is of the greatest importance to reduce this loss to the lowest possible

amount. The adoption of an efficient wing section means little if there

is a high resistance body and a tangle of useless struts and wires

exposed to the air stream. The resistance has a much greater effect on

the power than the weight.

*Weight and Resistance*. We have seen that the average modern wing

section will lift about 16 times the value of the horizontal drag, that

is, an addition of 16 pounds will be equal to 1 pound of head

resistance. If, by unnecessary resistance, we should increase the drag

by 10 pounds, we might as well gain the benefit of 10 x 16 = 160 pounds

of useful load. The higher the lift-drag efficiency of the wing, the

greater will be the proportional loss by parasitic resistance.

*Gliding Angle*. The gliding angle, or the inclination of the path of

descent when the machine is operating without power, is determined by

the weight and the total head resistance. With a constant weight the

angle is greatest when the resistance is highest. Aside from

considerations of power, the gliding angle is of the greatest importance

from the standpoint of safety. The less the resistance, and the flatter

the angle of descent, the greater the landing radius.

Numerically this angle can be expressed by: Glide = W/R, where W = the

weight of the aeroplane, and R = total resistance. Thus if the weight is

2500 pounds and the head resistance is 500 pounds, the rate of glide

will be: 2500/500 = 5. This means that the machine will travel forward 5

feet for every foot that it falls vertically. If the resistance could be

decreased to 100 pounds, the rate of glide would be extended to 2500/100

= 25, or the aeroplane would travel 25 feet horizontally for every foot

of descent. This will give an idea as to the value of low resistance.

*Resistance and Speed*. The parasitic resistance of a body in uniform

air varies as the square of the velocity at ordinary flight speeds.

Comparing speeds of 40 and 100 miles per hour, the ratio will be as 40°

is to 100° = 1600: 10,000 = 6.25, that is, the resistance at 100 miles

per hour will be 6.25 times as great as at 40 miles per hour.

The above remarks apply only to bodies making constant angle with the

air stream. Wings and lifting surfaces make varying angles at different

speeds and hence do not show the same rate of increase. In carrying a

constant load, the angle of the aeroplane wing is decreased as the speed

increases and up to a certain point the resistance actually decreases

with an increase in the speed. The wing resistance is greatest at

extremely low speeds and at very high speeds. As the total resistance is

made up of the sum of the wing and parasitic resistance at the different

speeds, it does not vary according to any fixed law. The only true

knowledge of the conditions existing through the range of flight speeds

is obtained by drawing a curve in which the sums of the drag and head

resistance are taken at intervals.

*Resistance and Power*. The power consumed in overcoming parasitic

resistance increases at a higher rate than the resistance, or as the

cube of the speed. Thus if the speed is increased from 40 to 100 miles

per hour, the power will be increased 15.63 times. This can be shown by

the following: Let V = velocity in miles per hour, H = Horsepower, K=

Resistance coefficient of a body, A = Total area of presentation, and R

= resistance in pounds. Then H = RV/375. Since R = KAV², then H = KAV² x

V/375 = KAV³/375.

*Resistance and Altitude*. The resistance decreases with a reduction in

the density of the air at constant speed. In practice, the resistance of

an aeroplane is not in direct proportion to a decrease in the density as

the speed must be increased at high altitudes in order to obtain the

lift. The following example given by Capt. Green will show the actual

relations.

Taking an altitude of 10,000 feet above sea level where the density is

0.74 of that at sea level, the resistance at equal speeds will be

practically in proportion to the densities. In order to gain

sustentation at the higher altitude, the speed must be increased, and

hence the true resistance will be far from that calculated by the

relative densities. Assume a sea level speed of 100 ft./sec., a weight

of 3000 pounds, a lift-drag ratio of L/D = 15, and a body resistance of

40 pounds at sea level.

Because of the change in density at 10,000 feet, the flying speed will

be increased from 100 feet per second to 350 feet per second in order to

obtain sustentation. With sea level density this increase in speed (3.5

times) would increase the body resistance 3.5 x 3.5 = 12.25 times,

making the total resistance 12.25 x 40 = 490 pounds. Since the density

at the higher altitude is only 0.74 of that at sea level, this will be

reduced by 0.26, or 0.26 x 490 = 364 pounds. Thus, the final practical

result is that the sea level resistance of the body (40 pounds) is

increased 9.1 times because of the speed increase necessary for

sustentation. Since the wing angle and hence the liftdrag ratio would

remain constant under both conditions, the wing drag would be constant

at both altitudes, or 3000/15 = 200 pounds. The total sea level

resistance at 100 feet per second is 200 + 40 = 240 pounds, while the

total resistance at 10,000 feet becomes 364 + 200 = 564 pounds.

The speed varies as the square root of the change in density percentage.

If V = velocity at sea level, v = velocity at a higher level, and d =

percentage of the sea level density at the higher altitude, then v =

V/√D. When the velocity at the high altitude is thus determined, the

resistance can be easily obtained by the method given in Capt. Green’s

article. The following table gives the percentage of densities referred

to sea level density.

───────────────────────────────────────────────────────────────────

Altitude Feet Density Percent Altitude Feet Density Percent

───────────────────────────────────────────────────────────────────

Sea-level 1.00 7,500 0.78

───────────────────────────────────────────────────────────────────

1,000 .97 10,000 .74

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2,000 .95 12,500 .66

───────────────────────────────────────────────────────────────────

3,000 .91 15,000 .61

───────────────────────────────────────────────────────────────────

5,000 .85 20,000 .52

───────────────────────────────────────────────────────────────────

If the velocity at sea level is 100 miles per hour, the velocity at

20,000 feet will be 100/0.72 = 139 miles per hour, where 0.72 is the

square root of the density percentage, or the square root of 0.52 = .72

at 20,000 feet.

*Total Parasitic Resistance*. Aside from the drag of the wings, the

resistance of the structural parts, body, tail and chassis depends upon

the size and type of aeroplane. A speed scout has less resistance than a

larger machine because of the small amount of exposed bracing, although

the relative resistance of the body is much greater. The type of engine

also has a great influence on the parasitic resistance. The following

gives the approximate distribution of a modern fighting aeroplane:

─────────────────────────────────────────────────────────────────

Body 62 percent

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Landing gear 16 "

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Tail, fin, rudder 7 "

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Struts, wires, etc. 15 "

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The body resistance is by far the greatest item. A great part of the

body resistance can be attributed to the motor cooling system, since in

either case it is diverted from the true streamline form in order to

accommodate the radiator, or the rotary motor cowl. The body resistance

is also influenced by the necessity of accommodating a given cargo or

passenger-carrying capacity, and by the distance of the tail surfaces

from the wings. A body is not a streamline form when its length greatly

exceeds 6 diameters.

*Calculation of Total Resistance*. The nearest approach that we can make

to the actual head resistance by means of a formula is to adopt an

expression in the form of R = KV² where K is a factor depending upon the

size and type of machine. The true method would be to go over the planes

and sum up the individual resistance of all the exposed parts. The parts

lying in the propeller slip stream should be increased by the increased

velocity of the slip stream. The parasitic resistance of biplanes

weighing about 1800 pounds will average about, R = 0.036V² where V =

velocity in miles per hour. Biplanes averaging 2500 pounds give R =

0.048V². Machines of the training or 2-seater type weigh from 1800 to

2500 pounds, and have an average head resistance distribution as

follows:

─────────────────────────────────────────────────────────────────

Body, radiators, shields 35.5 percent.

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Tail surface and bracing 14.9 "

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Landing gear 17.2 "

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Interplane struts, wires and fittings 23.6 "

─────────────────────────────────────────────────────────────────

Ailerons, aileron bracing, etc. 8.8 "

─────────────────────────────────────────────────────────────────

The averages in the above table differ greatly from the values given for

the high speed fighting machine, principally because of the large

control surfaces used in training machines, and the difference in the

size of the motors.

With the wing drag being equal to D = Kx AV², and the total parasitic

resistance equal to R = KV², the total resistance can be expressed by Rt

= KxAV² + KV², where K = coefficient of parasitic resistance for

different types and sizes of machines. The value of K for training

machines will average 0.036, for machines weighing about 2500 pounds K =

0.048. Scouts and small machines will be safe at K = 0.028. The wing

drag coefficient Kx varies with the angle of incidence and hence with

the speed. For example, we will assume that the wing drag (Kx) of a

scout biplane at 100 miles per hour is 0.00015, that the area is 200

square feet, and that the parasitic resistance coefficient is K = 0.028.

The total resistance becomes: R = (0.00015 x 200 x 100 x 100) + 0.028 x

100 x 100 = 300 + 280 = 580 pounds. The formula in this case would be R

= KxAV² + 0.028V².

*Strut Resistance*. The struts are of as nearly streamline form as

possible. In practice the resistance must be compromised with strength,

and for this reason the struts having the least resistance are not

always applicable to the practical aeroplane. From the best results

published by the N. P. L. the resistance was about 12.8 pounds per 100

feet strut at 60 miles per hour. The width of the strut is 1 inch. A

rectangular strut under the same conditions gave a resistance of 104.4

pounds per 100 feet. A safe value would be 25 pounds per 100 feet at 60

miles per hour. If a wider strut is used, the resistance must be

increased in proportion. With a greater speed, the resistance must be

increased in proportion to the squares of the velocity. When the struts

are inclined with the wind, the resistance is much decreased, and this

is one advantage of a heavy stagger in a biplane.

The "Fineness ratio" or the ratio of the width to the depth of the

section has a great effect on the resistance. With the depth equal to

twice the width measured across the stream, a certain strut section gave

a resistance of 24.8 pounds per 100 feet, while with a ratio of 3.5 the

resistance was reduced to 11.4 pounds per hundred feet. Beyond this

ratio the change is not as great, for with a ratio of 4.6 the resistance

only dropped to 11.2 pounds.

*Radiator Resistance*. For the exact calculation of the radiator

resistance it is first necessary to know the motor power and the fuel

consumption since the radiator area, and hence the resistance, depends

upon the size of the motor and the amount of heat transmitted to the

jacket water. An aeronautic motor may be considered to lose as much

through the water jackets as is developed in useful power, so that on

this basis we should allow about 1.6 square feet of radiation surface

per horsepower. This figure is arrived at by J. C. Hunsaker and assumes

that the wind speed is 50 miles per hour (73 feet per second). The most

severe cooling condition is met with in climbing at low speed, and it is

here assumed that 50 miles per hour will represent the lowest speed that

would be maintained for any length of time with the motor full out. For

a racing aeroplane that will not climb for any length of time, one-half

of the surface given above will be sufficient, and if the radiator is

placed in the propeller slip stream it can be made relatively still

smaller as the increased propeller slip at rapid rates of climb

partially offsets the additional heating.

In the above calculations, Hunsaker does not take any particular type of

radiator into consideration, merely assuming a smooth cooling surface.

The Rome–Turney Company states that they allow 1.08 square feet of

cooling surface per horsepower for honeycomb radiators, and 0.85 square

feet for the helical tube type. The surface referred to means the actual

surface measured all over the tubes and cells, and does not refer to the

front area nor the exterior dimensions of the radiator. While a radiator

may be made 25 percent smaller when placed in the slipstream, the

resistance is increased by about 25 per cent, with a very small saving

in weight, hence the total saving is small, if any. Side mounted

radiators have a lower cooling effect per square foot than those placed

in any other position, owing to the fact that the air must pass through

a greater length of tube than where the broad side faces the wind.

In the radiator section tested by Hunsaker, there were about 64 square

feet of cooling surface per square foot of front face area, but for

absolute assurance on this point one should determine the ratio for the

particular type of radiator that is to be used. The Auto Radiator

Manufacturing Corporation, makers of the "Flexo" copper core radiators,

have published some field tests made under practical conditions and for

different types and methods of mounting. The four classes of radiators

described are: (1) Front Type, in which the radiator is mounted in the

end of the fuselage; (2) Side Type, mounted on the sides of the body;

(3) Overhead Type, mounted above the fuselage and near the top plane;

(4) Over-Engine Type, placed above and connected directly to the motor,

as in the Standard H-3.

The following table gives the effectiveness of the different mountings

in terms of the frontal area required per horsepower and the cooling

surface, the area being in square inches (Front face area of radiator).

Area in wind of type (3) is half the calculated frontal area since one

core lies behind the other: Taking the value of the Rome–Turney

honeycomb radiator as 60 square feet of cooling surface per horsepower,

the frontal area per horsepower will be 0.0169 square feet, assuming

that the radiator is approximately 6 inches thick. This amounts to 2.43

square inches of frontal area per horsepower.

*Example*. Find the approximate frontal area of a Rome-Turney type

honeycomb radiator used with a motor giving 100 brake-horsepower. Find

Resistance at 50 miles per hour (73 feet per second).

─────────────────────────────────────────────────────────────────

Class of Mounting Square Inches Cooling Surface Per H.P

Per H.P. Square Inch

─────────────────────────────────────────────────────────────────

Front Type 4.00 117.00

─────────────────────────────────────────────────────────────────

Side Type 7.20 104.00

─────────────────────────────────────────────────────────────────

Overhead Type 2.70 112.00

─────────────────────────────────────────────────────────────────

Over-Engine Type 5.00 121.00

─────────────────────────────────────────────────────────────────

*Solution*. Area = A = 0.0169 HP = 0.0169 x 100 = 1.69 square feet. The

honeycomb portion of surface for a square radiator of the above area

will measure 16.2" x 16.2". Allowing a 1-inch water passage or frame all

around the core, the side of the completed square radiator will measure

16.2" + 1" = 18.2". The diameter of a circular radiator core of the same

1.69 x 144 area will be 17.4 inches, since D = 1.69x144/0.7854. Adding

the water passage, the overall diameter becomes 17.4 + 1 + 1 = 19.4

inches. The round honeycomb front radiator used on the 100 horsepower

Curtiss Baby Scout measures 20 inches. Hunaker found the resistance of a

honeycomb radiator to be R=0.000814 AV², there being 4 honeycomb cells

per square inch. A = area of radiator in square feet, and V = velocity

in feet per second. Adopting, for example, a speed of 73 feet per

second, and an area equivalent to a 19.4-inch diameter circular radiator

as above, the total resistance becomes:

R = 0.000814 AV² = 0.000814 x 3.1x (73 x 73) = 13.32 pounds, at 50 M. P.

H. where A = 3.1 square feet.

*Resistance of Chassis*. Disc wheels (Enclosed spokes) have a resistance

of about one-half that of open-wire wheels. The N. P. L. and Eiffel have

agreed that the resistance of a wheel approximating 26" x 4" has a

resistance of 1.7 pounds at 60 miles per hour (Disc type). For any other

speed, the wheel resistance will be R = 1.7 V²/3600, where V = speed in

miles per hour. We must also take into consideration the axle, chassis

struts, wiring, shock absorbers, etc. The itemization of the chassis

resistance, as given by the N.P.L. for the B.E.-2 biplane is as follows

(60 miles per hour):

─────────────────────────────────────────────────────────────────

Wheels 2(a)1.75 pounds 3.5 pounds

─────────────────────────────────────────────────────────────────

Axle 2.0 "

─────────────────────────────────────────────────────────────────

Chassis struts and connections 1.1 "

─────────────────────────────────────────────────────────────────

Total chassis resistance(3)60 MPH. 6.6 pounds

─────────────────────────────────────────────────────────────────

At any other speed, the resistance for the complete chassis can be given

by the formula R = 6.6V²/3600. This allowance will be ample, as the B.

E.-2 is an old type and is equipped with skids.

*Interplane Resistance*. The interplane struts and wires are difficult

to estimate by an approximate formula, the only exact way being to

figure up each item separately from a preliminary drawing. The

resistance varies with the form of the strut or wire section, the

length, and the thickness. The fact that some of the struts lie in the

propeller slipstream, and some outside of it, makes the calculation

doubly difficult. The only recourse that we have at present is to

analyze the conditions on the B. E.-2. With struts approximating true

streamline form, a great percentage of the total resistance is skin

friction, and as before explained, this item varies at a lesser rate

than the square of the speed.

According to a number of experiments on full size biplanes averaging

1900 pounds, it has been found that the interplane resistance (Struts,

wires and fittings) amounts to about 24 per cent of the total parasitic

head resistance of the entire machine, the drag of wings not being

included. The maximum observed gave 29 per cent and the minimum 15 per

cent. The resistance of the interplane bracing of speed scouts will be

considerably less in proportion, as there are fewer exposed struts and

cables on this type, the resistance probably averaging 15 per cent of

the total head resistance. Based on these figures the resistance of the

interplane bracing can be expressed by the following formula, in which I

= resistance of interplane bracing in pounds, and V = translational

speed in miles per hour:

I = 0.009 V² (For two-place biplanes weighing 1900 pounds).

I = 0.0054V² (For biplane speed scouts or racing type biplanes).

*Strut Resistance*. The above estimate includes wiring, strut fittings,

etc., complete, and also takes the effect of the slipstream into

consideration. A more accurate estimate can be made on the basis of

strut length. To obtain this unit value we have recourse to the B. E.-2

tests. The translational speed in 60 miles per hour (88 feet per second)

and the slipstream is taken at 25 feet per second. This gives a total

velocity in the slipstream of 113 feet per second. The struts are 1%

inches wide, and vary in length from 3’ 0" to 6’ 0". In the slipstream

the increased velocity increases the resistance of the items by 64 per

cent.

Total running length = 110’–0". Total resistance = 10.81 pounds. The

resistance per foot = 10.81/110 = 0.099 pounds.

*Resistance of Wire and Cable*. In this estimate we will take the

resistance given in the B. E.-2 tests, since values are given in the

slipstream as well as for the outer portions. In the translational

stream there is 240’ 0" of cable, 70’ 0" of No. 12 solid wire, and 52

turnbuckles, the total giving a resistance of 38.10 pounds. In the

slipstream there is 50’ 0" of cable and 30’ 0" of solid wire with a

resistance of 11.00 pounds. The total wire and cable resistance for the

wings is therefore 49.10 pounds. The resistance of the wire and cable

combined is 0.127 pounds per running foot.

*Summary of Interplane Resistance*. The total interplane resistance

includes the struts, wires, cables and turnbuckles, a portion of which

are in the slipstream. Since the total head resistance of the entire

machine (B.E.-2) is 140 pounds at 60 M. P. H., and the interplane

resistance = 10.81 + 49.10= 59.91 pounds, the relation of the interplane

resistance to the total resistance is 43 per cent. This is much higher

than the average (24 per cent), but the B.E.-2 is an old type of machine

and the number of struts and wires were much greater than with modern

aeroplanes.

*Control Surface Resistance*. The resistance of the control surfaces is

a variable quantity, since so much depends upon the arrangement and

form. Another variation occurring among machines of the same make and

type is due to the various angles of the surfaces during flight, or at

least during the time that they are used in correcting the attitude of

the machine. With the elevator flaps or ailerons depressed to their

fullest extent, the drag is many times that with the surfaces in

"neutral," and as a general thing the controls are depressed at the time

when the power demand is the greatest—that is, on landing, flying slow,

or in "getting off."

Ailerons "in neutral" can be considered as being an integral part of the

wings when they are hinged to the wing spar. In the older types of

Curtiss machines the ailerons were hinged midway between the planes and

the resistance was always in existence, whether the ailerons were in

neutral or not. Wing warping, in general, can be assumed as in the case

where the wings and ailerons are combined. With ailerons built into the

wings, the resistance of the ailerons, and their wires and fittings, can

be taken as being about 4 per cent of the total head resistance. With

the aileron located between the two wings, the resistance may run as

high as 20 per cent of the total.

Like the ailerons, the elevator surfaces and rudder are variable in

attitude and therefore give a varying resistance. In neutral attitude

the complete tail, consisting of the rudder, stabilizer, elevator, fin

and bracing, will average about 15 per cent of the total resistance, it

being understood that a non-lifting stabilizer is fitted. With lifting

tails the resistance will be increased in proportion to the load carried

by the stabilizer. In regard to the tail resistance it should be noted

that these surfaces are in the slipstream and are calculated

accordingly, although the velocity of the slipstream is somewhat reduced

at the point where it encounters the tail surfaces. The total tail

resistance of the B. E.-2 is given as 3.3 pounds.

*Resistance of Seaplane Floats*. The usual type of seaplane with double

floats may be considered as having about 12 per cent higher resistance

than a similar land machine. Some forms of floats have less resistance

than others, owing to their better streamline form, but the above figure

will be on the safe side for the average pontoon. Basing our formula on

a 12 per cent increase on the total head resistance, the formula for the

floats and bracing will become: Rt = 0.00436V² where R1 = resistance of

floats and fittings.

*Body Resistance*. This item is probably the most difficult of any to

compute, owing to the great variety of forms, the difference in the

engine mounting, and the disposition of the fittings and connections.

The resistance of the pilot’s and passenger’s heads, wind shields, and

propeller arrangement all tend to increase the difficulty of obtaining a

correct value. Aeroplanes with rotary air-cooled motors, or with large

front radiators have a higher resistance than those arranged with other

types of motors or radiator arrangements. Probably the item having the

greatest influence on the resistance of the fuselage is the ratio of the

length to the depth, or the "fineness ratio." In tractor monoplanes and

biplanes, of the single propeller type, the body is in the slipstream,

and compensation must be made for this factor.

If it were not for the motor and radiator, the tractor fuselage could be

made in true dirigible streamline form, and would therefore present less

resistance than the present forms of "practical" bodies. The necessity

of placing the tail surfaces at a fixed distance from the wings also

involves the use of a body that is longer in proportion than a true

streamline form, and this factor alone introduces an excessive head

resistance. The ideal ratio of depth to length would seem to range from

1 to 5.5 or 1 to 6. The fineness ratio of the average two-seat tractor

is considerably greater than this, ranging from 1 to 7.5 or 8.5. A

single-seat machine of the speed-scout type can be made much shorter and

has more nearly the ideal proportions.

The only possible way of disposing of this problem is to compare the

results of wind tunnel tests made on different types of bodies, and even

with this data at hand a liberal allowance should be made because of the

influence of the connections and other accessories. Eiffel, the N. P.

L., and the Massachusetts Institute of Technology have made a number of

experiments with scale models of existing aeroplane bodies. It is from

these tests that we must estimate our body resistance, hence a table of

the results is attached, the approximate outlines being shown by the

figures.

As in calculating the resistance of other parts, the resistance of the

body can be expressed by R = KxAV², where Kx = coefficient of the body

form, A = Cross-sectional area of body in square feet (Area of

presentation), and V = velocity in miles per hour. The area A is

obtained by multiplying the body depth by the width. The "area of

presentation" of a body 2’ 6" wide and 3’ 0" deep will be 2.5 x 3 = 7.5

square feet.

The experimental data does not give a very ready comparison between the

different types, as the bodies not only vary in shape and size, but are

also shown with different equipment. Some have tail planes and some have

not; two are shown with the heads of the pilot and passenger projecting

above the fuselage, while the remainder have either a simple cock-pit

opening or are entirely closed. The presence of the propeller in two

cases may have a great deal to do with raising the value of the

experimental results. The propeller was stationary during the tests, but

it was noted that the resistance was considerably less when the

propeller was allowed to run as a windmill, driving the motor. This

latter condition would correspond to the resistance in gliding with the

motor cut off. In all cases, except the Deperdussin, the bodies are

covered with fabric, and the sagging of the cloth in flight will

probably result in higher resistance than would be indicated by the

solid wood or metal model used in the tests. The pusher type bodies give

less resistance than the tractors, but the additional resistance of the

outriggers and tail bracing will probably bring the total far above the

tractor body.

In the accompanying body chart are shown 7 representative bodies: (a)

Deperdussin Monocoque Monoplane Body, a single-seater; (b) N. P. L.-5

Tractor Biplane Body, single-seater; (c) B. F.-36 Dirigible Form,

without propeller or cock-pit openings; (d) B. E.-3. Two-Place Tractor

Body, with passenger and pilot; (e) Curtiss JN Type Tractor Body, with

passengers, chassis and tail; (f) Farman Pusher type, with motor,

propeller and exposed passengers; (g) N. P. L. Pusher Body, bare. Body

(a) was tested with a 1/5 scale model at a wind tunnel speed of 28

meters per second, the resistance of the model being 0.377 kilograms

(0.83 pounds). Body (d) in model form was 1/16 scale and was tested at

20.5 miles per hour, at which speed the resistance was 0.0165 pounds.

Model (e) was 1/12 scale and was tested at 30 miles per hour. These

varying test speeds, it will be seen, do not allow of a very accurate

means of comparison. The resistance of model (e) was 0.1365 pounds at

the specified wind-tunnel air speed.

The speeds given in the above table are simply translational speeds, and

are not corrected for slipstream velocity. With a slipstream of 25 per

cent, increase the body resistance by 40 per cent. It would be safe to

add an additional 10 per cent to make up for projecting fittings, baggy

fabric, and scale variations.

Since a body of approximately streamline form has a considerable

percentage of skin friction, scale corrections for size and velocity are

even of more importance than with wing sections. No wind-tunnel

experiments can determine the resistance exactly because of the

uncertainty of the scale factor. The resistance as given in the table is

also affected by the proximity of the wing and tail surfaces, and by

projections emanating from the motor compartment. It will be noted that

the dirigible form B.F.-36 is markedly better than any of the others,

being almost of perfect streamline form. The nearest approximation to

the ideal form is N.P.L.-5, which has easy curves, low resistance, and

is fairly symmetrical about the center line. Because of their small

size, the pusher bodies or "nacelles" have a small total resistance, but

the value of Kx is high.

[Illustration: Chart Showing Forms of 7 Typical Aeroplane Fuselage]

*Problem*. Find the resistance of a Curtiss Tractor Type JN body with a

breadth of 2’ 6" and a depth of 3’ 3", the speed being 90 miles per

hour. The slipstream is assumed to be 25 per cent, with an additional 10

per cent for added fabric loss, etc.

[Illustration: Typical Stream Line Strut Construction.]

*Solution*. The cross-sectional area = 2’ 6" x 3’ 3" = A = 8.13 square

feet. The velocity of translation is 90 M. P. H., or V² = 8100. The

value of the resistance coefficient is taken from the table, Ko-0.00273.

The total resistance R = KxAV² = 0.00273 x 8.13 x 8100 = 178.2 pounds.

Since a slipstream of 25 per cent increases the resistance by 40 per

cent, the resistance in the slipstream is 1782 x 1.4 = 249.48 pounds.

The addition of the 10 per cent for extra friction makes the total

resistance = 249.48 x 1.1 = 274.43 pounds. The resistance of this body,

used with "twin" motors, would be 178.2 x 1.1 = 196.02, but as a tractor

with the body in the slipstream, the resistance would be equal to 274.43

pounds as calculated above.

Aeroplane Construction and Operation · 第 16 / 21 章