Chapter XIV: Useful Notes, Tables, Formulæ, Etc

§ 1. COMPARATIVE VELOCITIES.

Miles per hr. Feet per sec. Metres per sec.
10 = 14·7 = 4·470
15 = 22 = 6·705
20 = 29·4 = 8·940
25 = 36·7 = 11·176
30 = 44 = 13·411
35 = 51·3 = 15·646

§ 2. A metre = 39·37079 inches.

_In order to convert_:--
Metres into inches multiply by 39·37
" feet " 3·28
" yards " 1·09
" miles " 0·0006214
Miles per hour into ft. per min. multiply by 88·0
" min. " sec. " 88·0
" hr. into kilometres per hr. " 1·6093
" " metres per sec. " 0·44702
Pounds into grammes multiply by 453·593
" kilogrammes " 0·4536

§ 8. Total surface of a cylinder = circumference of base × height + 2 area of base.

Area of a circle = square of diameter × 0·7854.

Area of a circle = square of rad. × 3·14159.

Area of an ellipse = product of axes × 0·7854.

Circumference of a circle = diameter × 3·14159.

Solidity of a cylinder = height × area of base.

Area of a circular ring = sum of diameters × difference of diameters × 0·7854.

For the area of a sector of a circle the rule is:--As 360 : number of degrees in the angle of the sector :: area of the sector : area of circle.

To find the area of a segment less than a semicircle:--Find the area of the sector which has the same arc, and subtract the area of the triangle formed by the radii and the chord.

The areas of corresponding figures are as the squares of corresponding lengths.

§ 4. 1 mile = 1·609 kilometres.
1 kilometre = 1093 yards.
1 oz. = 28·35 grammes.
1 lb. = 453·59 "
1 lb. = 0·453 kilogrammes.
28 lb. = 12·7 "
112 lb. = 50·8 "
2240 lb. = 1016 "
1 kilogram = 2·2046 lb.
1 gram = 0·0022 lb.
1 sq. in. = 645 sq. millimetres.
1 sq. ft. = 0·0929 sq. metres.
1 sq. yard = 0·836 "
1 sq. metre = 10·764 sq. ft.

§ 5. One atmosphere = 14·7 lb. per sq. in. = 2116 lb. per sq. ft. = 760 millimetres of mercury.

A column of water 2·3 ft. high corresponds to a pressure of 1 lb. per sq. in.

1 H.P. = 33,000 ft.-lb. per min. = 746 watts.

Volts × amperes = watts.

{pi} = 3·1416. _g_ = 32·182 ft. per sec. at London.

§ 6. TABLE OF EQUIVALENT INCLINATIONS.

Rise. Angle in Degs.
1 in 30 1·91
1 " 25 2·29
1 " 20 2·87
1 " 18 3·18
1 " 16 3·58
1 " 14 4·09
1 " 12 4·78
1 " 10 5·73
1 " 9 6·38
1 " 8 7·18
1 " 7 8·22
1 " 6 9·6
1 " 5 11·53
1 " 4 14·48
1 " 3 19·45
1 " 2 30·00
1 " {square root}2 45·00

§ 7. TABLE OF SKIN FRICTION.

Per sq. ft. for various speeds and surface lengths.

-----------------+-------------+-------------+-------------+------------
Velocity of Wind | 1 ft. Plane | 2 ft. Plane | 4 ft. Plane | 8 ft. Plane
-----------------+-------------+-------------+-------------+------------
10 | ·00112 | ·00105 | ·00101 | ·000967
15 | ·00237 | ·00226 | ·00215 | ·00205
20 | ·00402 | ·00384 | ·00365 | ·00349
25 | ·00606 | ·00579 | ·00551 | ·00527
30 | ·00850 | ·00810 | ·00772 | ·00736
35 | ·01130 | ·0108 | ·0103 | ·0098
-----------------+-------------+-------------+-------------+------------

This table is based on Dr. Zahm's experiments and the equation

_f_ = 0·00000778_l_^{-0·07}_v_^{1·85}

Where _f_ = skin friction per sq. ft.; _l_ = length of surface; _v_ = velocity in feet per second.

In a biplane model the head resistance is probably from twelve to fourteen times the skin friction; in a racing monoplane from six to eight times.

§ 8. TABLE I.--(METALS).

--------------+------------+-----------------+-------------
Material | Specific | Elasticity E[A] | Tenacity
| Gravity | | per sq. in.
--------------+------------+-----------------+-------------
Magnesium | 1·74 | | {22,000-
| | | {32,000
Magnalium[B] | 2·4-2·57 | 10·2 |
Aluminium- } | | |
Copper[C]} | 2·82 | | 54,773
Aluminium | 2·6 | 11·1 | 26,535
Iron | 7·7 (about)| 29 | 54,000
Steel | 7·8 (about)| 32 | 100,000
Brass | 7·8-8·4 | 15 | 17,500
Copper | 8·8 | 36 | 33,000
Mild Steel | 7·8 | 30 | 60,000
| | |
--------------+------------+-----------------+-------------
[A] E in millions of lb. per sq. in.
[B] Magnalium is an alloy of magnesium and aluminium.
[C] Aluminium 94 per cent., copper 6 per cent. (the best
percentage), a 6 per cent. alloy thereby doubles the
tenacity of pure aluminium with but 5 per cent.
increase of density.
--------------+------------+-----------------+-------------

§ 9. TABLE II.--WIND PRESSURES.

_p_ = _kv²_.

_k_ coefficient (mean value taken) ·003 (miles per hour) = 0·0016 ft. per second. _p_ = pressure in lb. per sq. ft. _v_ = velocity of wind.

Miles per hr. Ft. per sec. Lb. per sq. ft.
10 14·7 0·300
12 17·6 0·432
14 20·5 0·588
16 23·5 0·768
18 26·4 0·972
20 29·35 1·200
25 36·7 1·875
30 43·9 2·700
35 51·3 3·675

§ 10. Representing normal pressure on a plane surface by 1; pressure on a rod (round section) is 0·6; on a symmetrical elliptic cross section (axes 2:1) is 0·2 (approx.). Similar shape, but axes 6:1, and edges sharpened (_see_ ch. ii., § 5), is only 0·05, or 1/20, and for the body of minimum resistance (_see_ ch. ii., § 4) about 1/24.

§ 11. TABLE III.--LIFT AND DRIFT.

On a well shaped aerocurve or correctly designed cambered surface. Aspect ratio 4·5.

Inclination. Ratio Lift to Drift.
0° 19:1
2·87° 15:1
3·58° 16:1
4·09° 14:1
4·78° 12:1
5·73° 9·6:1
7·18° 7·9:1

Wind velocity 40 miles per hour. (The above deduced from some experiments of Sir Hiram Maxim.)

At a velocity of 30 miles an hour a good aerocurve should lift 21 oz. to 24 oz. per sq. ft.

§ 12. TABLE IV.--LIFT AND DRIFT.

On a plane aerofoil.

N = P(2 sin {alpha}/1 + sin² {alpha})

Inclination. Ratio Lift to Drift.
1° 58·3:1
2° 29·2:1
3° 19·3:1
4° 14·3:1
5° 11·4:1
6° 9·5:1
7° 8·0:1
8° 7·0:1
9° 6·3:1
10° 5·7:1

P = 2_kd_ AV² sin {alpha}.

A useful formula for a single plane surface. P = pressure supporting the plane in pounds per square foot, _k_ a constant = 0·003 in miles per hour, _d_ = the density of the air.

A = the area of the plane, V relative velocity of translation through the air, and {alpha} the angle of flight.

Transposing we have

AV² = P/(2_kd_ sin {alpha})

If P and {alpha} are constants; then AV² = a constant or area is inversely as velocity squared. Increase of velocity meaning diminished supporting surface (_and so far as supporting surface goes_), diminished resistance and skin friction. It must be remembered, however, that while the work of sustentation diminishes with the speed, the work of penetration varies as the cube of the speed.

§ 13. TABLE V.--TIMBER.

Column Headings:

A. Material
B. Specific Gravity
C. Weight per Cub. Ft. in Lb.
D. Strength per Sq. In. in Lb.
E. Ultimate Breaking Load (Lb.) span 1' x 1" x 1"
F. Relative Resilience in Bending
G. Modulus of Elasticity in millions of Lb. per Sq. In. for Bending
H. Relative Value. Bending Strength compared with Weight

---------------+-----+-------+-------------+-------+-----+-----+----
A |B | C | D |E |F |G | H
---------------+-----+-------+-------------+-------+-----+-----+----
Ash | ·79 | 43-52 |14,000-17,000| 622 |4·69 |1·55 |13·0
Bamboo | | 25[A]| 6300[53] | |3·07 |3·20 |
Beech | ·69 | 43 |10,000-12,000| 850 | |1·65 |19·8
Birch | ·71 | 45 | 15,000 | 550 | |3·28 |12·2
Box |1·28 | 80 |20,000-23,000| 815 | | |10·2
Cork | ·24 | 15 | | | | |
Fir (Norway | | | | | | |
Spruce) | ·51 | 32 | 9,000-11,000| 450 |3·01 |1·70 |14·0
American | | | | | | |
Hickory | | 49 | 11,000 | 800 |3·47 |2·40 |16·3
Honduras | | | | | | |
Mahogany | ·56 | 35 | 20,000 | 750 |3·40 |1·60 |21·4
Maple | ·68 | 44 | 10,600 | 750 | | |17·0
American White | | | | | | |
Pine | ·42 | 25 | 11,800 | 450 |2·37 |1·39 |18·0
Lombardy Poplar| | 24 | 7,000 | 550 |2·89 | 0·77|22·9
American Yellow| | | | | | |
Poplar | | 44 | 10,000 | |3·63 |1·40 |
Satinwood | ·96 | 60 | |1,033 | | |17·2
Spruce | ·50 | 31 | 12,400 | 450 | | |14·5
Tubular Ash, | | | | | | |
_t_ = 1/8 _d_ | | 47 | | |3·50 |1·55 |
---------------+-----+-------+-------------+-------+-----+-----+----

_t_ = thickness: _d_ = diameter.

[A] Given elsewhere as 55 and 22,500 (_t_ = 1/3_d_), evidently
regarded as solid.

§ 14.--=Formula connecting the Weight Lifted in Pounds per Square Foot and the Velocity.=--The empirical formula

W = (V²C)/_g_

Where W = weight lifted in lb. per sq. ft.
V = velocity in ft. per sec.
C = a constant = 0·025.
_g_ = 32·2, or 32 approx.

may be used for a thoroughly efficient model. This gives (approximately)

1 lb. per sq. ft. lift at 25 miles an hour.
21 oz. " " 30 "
6 oz. " " 15 "
4 oz. " " 12 "
2·7 oz. " " 10 "

Remember the results work out in feet per second. To convert (approximately) into miles per hour multiply by 2/3.

§ 15. =Formula connecting Models of Similar Design, but Different Weights.=

D {proportional to} {square root}W.

or in models of _similar design_ the distances flown are proportional to the square roots of the weights. (Derived from data obtained from Clarke's flyers.)

For models from 1 oz. to 24-30 oz. the formula appears to hold very well. For heavier models it appears to give the heavier model rather too great a distance.

Since this was deduced a 1 oz. Clarke model of somewhat similar design but longer rubber motor has flown 750 ft. at least; it is true the design is not, strictly speaking, similar, but not too much reliance must be placed on the above. The record for a 1 oz. model to date is over 300 yards (with the wind, of course), say 750 ft. in calm air.

§ 16. =Power and Speed.=--The following formula, given by Mr. L. Blin Desbleds, between these is--

W/W{0} = (3_v{0}_)/(4_v_) + ¼(_v_/_v{0}_)³.

Where _v{0}_ = speed of minimum power
W{0} = work done at speed _v{0}_.
W = work done at speed _v_.

Making _v_ = 2_v{0}_, i.e. doubling the speed of minimum power, and substituting, we have finally

W = (2-3/8)W{0}

i.e. the speed of an aeroplane can be doubled by using a power 2-3/8 times as great as the original one. The "speed of minimum power" being the speed at which the aeroplane must travel for the minimum expenditure of power.

§ 17. The thrust of the propeller has evidently to balance the

Aerodynamic resistance = R
The head resistance (including skin friction) = S

Now according to Renard's theorem, the power absorbed by R + S is a minimum when

S = R/3.

Having built a model, then, in which the total resistance

= (4/3)R.

This is the thrust which the propeller should be designed to give. Now supposing the propeller's efficiency to be 80 per cent., then P--the minimum propulsion power

= (4/3)R × 100/80 × 100/75 × _v_.

Where 25 per cent. is the slip of the screw, _v_ the velocity of the aeroplane.

§ 18. =To determine experimentally the Static Thrust of a Propeller.=--Useful for models intended to raise themselves from the ground under their own power, and for helicopters.

The easiest way to do this is as follows: Mount the propeller on the shaft of an electric motor, of sufficient power to give the propeller 1000 to 1500 revolutions per minute; a suitable accumulator or other source of electric energy will be required, a speedometer or speed counter, also a voltmeter and ammeter.

Place the motor in a pair of scales or on a suitable spring balance (the former is preferable), the axis of the motor vertical, with the propeller attached. Rotate the propeller so that the air current is driven _upwards_. When the correct speed (as indicated by the speed counter) has been attained, notice the difference in the readings if a spring balance be used, or, if a pair of scales, place weights in the scale pan until the downward thrust of the propeller is exactly balanced. This gives you the thrust in ounces or pounds.

Note carefully the voltage and amperage, supposing it is 8 volts and 10 amperes = 80 watts.

Remove the propeller and note the volts and amperes consumed to run the motor alone, i.e. to excite itself, and overcome friction and air resistance; suppose this to be 8 volts and 2 amperes = 16; the increased load when the propeller is on is therefore

80 - 16 = 64 watts.

All this increased power is not, however, expended on the propeller.

The lost power in the motor increases as C²R.

R = resistance of armature and C = current. If we deduct 10 per cent. for this then the propeller is actually driven by 56 watts.

Now 746 watts = 1 h.p.

{therefore} 56/746 = 1/13 h.p. approx.

at the observed number of revolutions per minute.

§ 19. N.B.--The h.p. required to drive a propeller varies as the cube of the revolutions.

_Proof._--Double the speed of the screw, then it strikes the air twice as hard; it also strikes twice as much air, and the motor has to go twice as fast to do it.

§ 20. To compare one model with another the formula

Weight × velocity (in ft. per sec.)/horse-power

is sometimes useful.

§ 21. =A Horse-power= is 33,000 lb. raised one foot in one minute, or 550 lb. one foot in one second.

A clockwork spring raised 1 lb. through 4½ ft. in 3 seconds. What is its h.p.?

1 lb. through 4½ ft. in 3 seconds
is 1 lb. " 90 ft. " 1 minute.

{therefore} Work done is 90 ft.-lb.
= 90/33000 = 0·002727 h.p.

The weight of the spring was 6¾ oz. (this is taken from an actual experiment), i.e. this motor develops power at the rate of 0·002727 h.p. for 3½ seconds only.

§ 22. =To Ascertain the H.P. of a Rubber Motor.= Supposing a propeller wound up to 250 turns to run down in 15 seconds, i.e. at a mean speed of 1200 revolutions per minute or 20 per second. Suppose the mean thrust to be 2 oz., and let the pitch of the propeller be 1 foot. Then the number of foot-pounds of energy developed

= (2 oz. × 1200 revols. × 1 ft. (pitch)) / 16 oz.

= 150 ft.-lb. per minute.

But the rubber motor runs down in 15 seconds.

{therefore} Energy really developed is

= (150 × 15) / 60 = 37·5 ft.-lb.

The motor develops power at rate of 150/33000 = 0·004545 h.p., but for 15 seconds only.

§ 23. =Foot-pounds of Energy in a Given Weight of Rubber= (experimental determination of).

Length of rubber 36 yds.
Weight " 2-7/16 oz.
Number of turns = 200.

12 oz. were raised 19 ft. in 5 seconds.
i.e. ¾ lb. was raised 19 × 12 ft. in 1 minute.
i.e. 1 lb. was raised 19 × 3 × 3 ft. in 1 minute.
= 171 ft. in 1 minute.

i.e. 171 ft.-lb. of energy per minute. But actual time was 5 seconds.

{therefore} Actual energy developed by 2-7/16 oz. of rubber of 36 yards, i.e. 36 strands 1 yard each at 200 turns is

= 171/12 ft.-lb.

= 14¼ ft.-lb.

This allows nothing for friction or turning the axle on which the cord was wound. Ball bearings were used; but the rubber was not new and twenty turns were still unwound at the end of the experiment. Now allowing for friction, etc. being the same as on an actual model, we can take ¾ of a ft.-lb. for the unwound amount and estimate the total energy as 15 ft.-lb. as a minimum. The energy actually developed being at the rate of 0·0055 h.p., or 1/200 of a h.p. if supposed uniform.

§ 24. The actual energy derivable from 1 lb. weight of rubber is stated to be 300 ft.-lb. On this basis 2-7/16 oz. should be capable of giving 45·7 ft.-lb. of energy, i.e. three times the amount given above. Now the motor-rubber not lubricated was only given 200 turns--lubricated 400 could have been given it, 600 probably before rupture--and the energy then derivable would certainly have been approximating to 45 ft.-lb., i.e. 36·25. Now on the basis of 300 ft.-lb. per lb. a weight of ½ oz. (the amount of rubber carried in "one-ouncers") gives 9 ft.-lb. of energy. Now assuming the gliding angle (including weight of propellers) to be 1 in 8; a perfectly efficient model should be capable of flying eight times as great a distance in a horizontal direction as the energy in the rubber motor would lift it vertically. Now 9 ft.-lb. of energy will lift 1 oz. 154 ft. Therefore theoretically it will drive it a distance (in yards) of

(8 × 154)/3 = 410·6 yards.

Now the greatest distance that a 1 oz. model has flown in perfectly calm air (which never exists) is not known. Flying with the wind 500 yards is claimed. Admitting this what allowance shall we make for the wind; supposing we deduct half this, viz. 250 yards. Then, on this assumption, the efficiency of this "one ouncer" works out (in perfectly still air) at 61 per cent.

The gliding angle assumption of 1 in 8 is rather a high one, possibly too high; all the writer desires to show is the method of working out.

Mr. T.W.K. Clarke informs me that in his one-ouncers the gliding angle is about 1 in 5.

§ 25. =To Test Different Motors or Different Powers of the Same Kind of Motor.=--Test them on the same machine, and do not use different motors or different powers on different machines.

§ 26. =Efficiency of a Model.=--The efficiency of a model depends on the weight carried per h.p.

§ 27. =Efficiency of Design.=--The efficiency of some particular design depends on the amount of supporting surface necessary at a given speed.

§ 28. =Naphtha Engines=, that is, engines made on the principle of the steam engine, but which use a light spirit of petrol or similar agent in their generator instead of water with the same amount of heat, will develop twice as much energy as in the case of the ordinary steam engine.

§ 29.=Petrol Motors.=

Horse-power. No. of Cylinders. Weight.
¼ Single 4½ lb.
½ to ¾ " 6½ "
1½ Double 9 "

§ 30. =The Horse-power of Model Petrol Motors.=--Formula for rating of the above.

(R.P.M. = revolutions per minute.)

H.P. = ((Bore)² × stroke × no. of cylinders × R.P.M.)/12,000

If the right-hand side of the equation gives a less h.p. than that stated for some particular motor, then it follows that the h.p. of the motor has been over-estimated.

§ 30A. =Relation between Static Thrust of Propeller and Total Weight of Model.=--The thrust should be approx. = ¼ of the weight.

§ 31. =How to find the Height of an Inaccessible Object by Means of Three Observations taken on the Ground (supposed flat) in the same Straight Line.=--Let A, C, B be the angular elevations of the object D, as seen from these points, taken in the same straight line. Let the distances B C, C A and A B be _a_, _b_, _c_ respectively. And let required height P D = _h_; then by trigonometry we have (see Fig. 56)

_h²_ = _abc_/(_a_ cot²A - _c_ cot²C + _b_ cot²B).

§ 32. =Formula= for calculating the I.H.P. (indicated horse-power) of a single-cylinder double-acting steam-engine.

Indicated h.p. means the h.p. actually exerted by the steam in the cylinder without taking into account engine friction. Brake h.p. or effective h.p. is the actual h.p. delivered by the crank shaft of the engine.

I.H.P. = (2 × S × R × A × P)/33,000.

Where S = stroke in feet.
R = revolutions per minute.
A = area of piston in inches.
P = mean pressure in lb. exerted per sq. in. on the piston.

The only difficulty is the mean effective pressure; this can be found approximately by the following rule and accompanying table.

TABLE VI.

---------+----------+---------+----------+---------+---------
Cut-off | Constant | Cut-off | Constant | Cut-off | Constant
---------+----------+---------+----------+---------+---------
1/6 | ·566 | 3/8 | ·771 | 2/3 | ·917
1/5 | ·603 | ·4 | ·789 | ·7 | ·926
1/4 | ·659 | 1/2 | ·847 | 3/4 | ·937
·3 | ·708 | ·6 | ·895 | ·8 | ·944
1/3 | ·743 | 5/8 | ·904 | 7/8 | ·951
---------+----------+---------+----------+---------+---------

Rule.--"Add 14·7 to gauge pressure of boiler, this giving 'absolute steam pressure,' multiply this sum by the number opposite the fraction representing the point of cut-off in the cylinder in accompanying table. Subtract 17 from the product and multiply the remainder by 0·9. The result will be very nearly the M.E.P." (R.M. de Vignier.)

FOOTNOTE:

[53] Given elsewhere as 55 and 22,500 (_t_ = 1/3 _d_), evidently regarded as solid.