The Theory and Practice of Model Aeroplaning Part 17

4 oz. " " 12 "

27 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_) + (_v_/_v{0}_).

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 subst.i.tuting, 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 acc.u.mulator 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 CR.

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 4 ft. in 3 seconds. What is its h.p.?

1 lb. through 4 ft. in 3 seconds is 1 lb. " 90 ft. " 1 minute.

{therefore} Work done is 90 ft.-lb.

= 90/33000 = 0002727 h.p.

The weight of the spring was 6 oz. (this is taken from an actual experiment), i.e. this motor develops power at the rate of 0002727 h.p. for 3 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.