The flight speed of the so-called "Slow-Fly" planes (the real ones, not those Foam stuffs equipped with Speed 400 motors and 500 mAh cells, or more, and which monopolise your gymnasium for hours or cut your plane in slices unscrupulously...) can be measured quite easily with just a rule and a stopwatch. Interesting aerodynamical data can then be determined.

A first flight test with the motor will give you an idea on the cruise speed.

Then, remove the propeller and replace it with an lead of the same weight in order to keep the balance and throw your model over you head with an initial speed similar to the cruise speed and control it so it will glide straight ahead with constant speed and slope. Ask a friend to time the flight duration and measure the distance between the launching and the landing points. Do that several times and take the mean values. In this way, flight duration t (in seconds), distance d and altitude loss h (in meters) can be determined.

Put the propeller again, fully charge the battery and fly level and straight (as much as possible) your model at cruise speed until the battery is discharged. Note the total flight duration T (in seconds).

We can now easily calculate:

- The ground
speed Vx = d / t in metre / second

- The sink rate
Vz = h / t en metre / second

- The glide
ratio f = d / h,

From this data, we can determined:

- The minimum power Pu (in Watts) needed for a level flight. This power is equivalent to the power needed to compensate (per time unit) for the loss of potential energy Ep of the plane (Ep = m * g * h). So, we have :

Pu = m * g* h / t or Pu = m * g * Vz where m is the model's mass (in kilogrammes) and g the gravitational acceleration (9,81).

- The thrust force Fp (in Newton) for a level flight (at constant speed). As a matter of fact, the energy of this force is W = Fp * d that is to say per time unit, W / t = Fp * d / t = Fp * Vx = Pu = m * g * Vz or:

Fp = m * g * Vz / Vx or Fp = m * g * h / d or Fp = m * g / f

- The overall efficiency of the model (Needed power / Consumed Power) :

R = Pu / Pe = (m * g * Vz) / (U * C / T) where U is the battery's voltage in Volt and C the battery's capacity in Ampere-second.

In the case of the "Magicien d'Oz" (data given by J.M. Piednoir), we have:

- Flight
duration T with 3 x 50 mAh cells : 14 minutes or 840 seconds,

- Total masse :
31 grams,

- Glide duration
over 13 metres and from an altitude of 1,8 metres : 5,5 seconds.

Therefore we have:

- Ground
speed : Vx = 13 / 5.5 = 2,36 m/s (or 8,5 km/h),

- Sink rate : Vz
= 1.8 / 5.5 = 0,327 m/s

- Glide ratio :
f = 13 / 1,8 = 7,22

- Level thrust :
Fp = 31 / 7, 22 = 4,29 grams

- Required power
: Pu = 0,030 * 9,81 * 1,8 / 5,5 = 0,0995 W

- Electrical
power : Pe = 3,45 * 0,050 * 60 * 60 / (14 * 60) = 0,77 W

- Overall
efficiency : R = 0,0995 / 0,77 = 12,9 %

Assuming the motor efficiency is 45 %, the efficiency of the frame plus the propeller is 12,9 / 0,45 = 29 %. If we suppose the propeller efficiency is 55 %, then the efficiency of the frame only is 29 / 0,55 = 52 %.

We have considered the glide and the level flight. Let's go a little bit further and study the climbing with a climb angle A.

On a level flight (cruise), you can consider the thrust force Fp we have seen above is used to compensate for the drag. During climbing, the required force F will have to compensate for not only the drag but also an additional force which is in fact the projection of the plane's weight on the flight line, that is to say m * g * sin (A). And therefore:

F = Fp + ( m * g * sin (A) ) or F = ( m * g / f ) + ( m * g * sin (A) ) and then:

sin (A) = ( F / P ) - ( 1 / f )

Still in the case of the "Magicien d'Oz", the maximum thrust ( at full throttle ) given by the propeller is 15 gramme-force and the maximum climb angle is defined by sin (A) = ( 15 / 31 ) - ( 1 / 7,22 ) = 0,345 and therefore : A = 20 degrees.