Q1: "Why is the stalling speed lower with the engine producing thrust than without?"

When the wings are near the stalling angle of attack ( about 15 degrees) in level flight, the aircraft has a considerable nose-up attitude and the line of thrust is inclined upward. Thus with the engine producing thrust part of that thrust is acting vertically upward and the aircraft needs less lift to balance weight, so the airspeed at the stall will be lower.

A second effect comes from the slipstream which both increases air velocity over the inner wing and reduces the relative angle of attack at the inner wing.

All these engine effects cause an appreciable lowering of stalling speed.

 Q2: "Why does the nose of the aircraft drop at the stall?"

As the angle of attack of the aircraft's wings moves toward the stalling angle, the centre of pressure of the aerofoil moves forward giving a nose-up turning moment - rather than the normal nose-down turning moment of the aerofoil at normal angles of attack. [The horizontal stabilizer normally provides a built-in nose-up turning moment balancing that of the wings.]

As the stalling angle is reached the aerofoil centre of pressure moves rapidly back giving a large nose-down turning moment and the nose of the aircraft will drop. If the downwash from the wing is affecting the tailplane up to the stalling angle and ceases to affect it near the stalling angle or is replaced by the turbulent flow from the wing, a very large and fast acting, nose-down moment may be produced.

 Q3: "What are seven reasons for taking off into wind?"

(a) The ground [roll] speed for take-off is lower.
(b) It is easier to keep straight because of the aircraft's increased directional stability.
(c) The take-off distance is shorter.
(d) The into wind climb out will be steeper and provide better obstacle clearance.
(e) The vertical wind profile is such that the wind velocity changes encountered during the climb are most likely to be an increase in wind speed coming from ahead of the aircraft, thus providing a momentary increase in lift.
(f) If the engine should fail after take-off the aircraft can readily land into wind thus reducing impact force, because the ground speed is reduced quite significantly at ultralight velocities. And remember that [impact] energy increases with the square of the velocity.
(g) It is safer to conform to an accepted traffic pattern.

 Q4: "If an aircraft of all-up mass 400 kg is flying straight and level at 6500 feet (2000 m) where air density is 1.0 kg/m³ and the pilot - Teraya Miller of Top Air Services, who takes pride in her accurate flying - is increasing speed from 60 to 80 knots TAS, what lift is being generated (in kg or newtons) as the velocity passes through 69 knots (35 m/sec)?

If an aircraft is maintaining constant altitude and constant heading, even though it is accelerating then the lift generated at any speed or altitude must be equal to the aircraft's 'normal' weight, which is mass x 1g = 400 kg or about 4000 newtons.

Incidently, as the engine power is increased Teraya must decrease the angle of attack otherwise the additional thrust provided will result in a climb. Let's surmise that the decrease in aoa between 60 and 69 knots is 2° then the inclination of the engine thrust line must be reduced by 2°. Because of the inclination of the thrust line some (the vertical component of the thrust vector) vertical force is generated by thrust. So in this instance the inclination being reduced by 2° reduces the vertical component of thrust by 2/60 (remember the 1 in 60 rule) of the total thrust. We could guess that the aircraft lift/drag ratio is 8:1 at normal cruise and total thrust is then about 500 newtons, If the engine power is increased by say 20% to accelerate then thrust is 600 newtons so 2/60 is 20 newtons - completely insignificant in the total lift of 4000 newtons but more than enough for the aircraft to climb if Teraya doesn't reduce the aoa.

 Q5: "What are five advantages provided by using flaps for landing?"

(a) A lower safe approach speed.
(b) The nose down attitude provides a better view of the landing area.
(c) The steeper approach path provides better obstacle clearance and can be controlled at will.
(d) The float after rounding out is shorter.
(e) The ground roll is shorter.

 Q6: "What is an acceleration – or high speed – stall and how can it be brought about?"

The stall occurs at a particular angle of attack not a particular speed. The speed below which the stall will occur depends on the load factor. If the aircraft reaches the critical aoa under a load higher than 1g the stalling speed will be higher than the normal 1g stall speed – Vs1 – at that mass. This latter stall is called an acceleration stall, usually more pronounced than a normal stall, and the speed at which an acceleration stall occurs is proportional to the square root of the load factor. Remember that the load factor increases in a turn. In a 60° banked level turn the load factor, at 2g, is twice normal, therefore the stall will occur at a speed which is 1.41 times Vs1.

Similarly if you sustain 3.8g (the usual minimum design load factor) when recovering from a dive an accelerated stall will occur if airspeed drops below 1.95 times Vs1. The possibility of accelerated stall during dive recovery is enhanced by the aircraft's tendency to follow the original flight path (to sink) thereby further increasing aoa, possibly past the critical aoa.

 Q7: "You are parked on an airstrip (elevation 800 feet where the air temperature is 40°C) and about to start your aircraft. You set 1013 hPa on the altimeter sub-scale and the altimeter then reads 1000 feet pressure altitude. What's the density altitude and how do you check whether you have sufficient take-off distance available to clear those average height trees at the end of the paddock?"

You can estimate density altitude by ascertaining the difference between the ISA standard temperature for a pressure altitude and the actual temperature. The ISA standard sea-level temperature is 15°C with a temperature lapse rate is 2°C per 1000 feet. Thus the standard temperature for 1000 feet pressure altitude is 13°C and the difference between actual and standard is 40 - 13 = 27°C.

Density altitude is roughly 120 feet greater than pressure altitude for each 1°C that the temperature exceeds ISA for that level so that the density altitude on this occasion is 27 x 120 + 1000 feet = 4240 feet - approximately.

There is no way that you can make a quick check of the take-off distance required under abnormal conditions except with a TODR/Weight/Altitude chart which should be supplied by the aircraft/kit manufacturer for each airframe/engine/propeller combination.

 Q8: "You are a prudent ultralight pilot planning a two hour flight to an airfield due east of you on a day with light northerly winds at levels up to 10,000 feet and clear skies. The terrain is open country at an elevation of about 800 - 1000 feet all the way with scattered minor hills. What altitude would you cruise at?

An aircraft operating under Visual Flight Rules, and below 5000 feet amsl ( area QNH), may cruise at any safe altitude. However a prudent ultralight pilot undertaking a two hour flight would choose a hemispherical VFR cruising level whenever practicable. For an aircraft heading east the VFR cruising levels are 1500 and 3500 feet. The ultralight flight could be operated at a level above 5000 feet if safety considerations dictated so. Then the pilot could choose one of the three mandatory easterly cruising levels below 10000 feet – 5500, 7500 or 9500 feet.

However this flight over open terrain in clear conditions would not warrant an ultralight intrusion above 5000 feet thus the only practicable cruising level available to a prudent pilot is 3500 feet.

 Q9: "For an ultralight making a level turn, what speed and what bank angle would provide the smallest radius of turn and what speed and bank angle would provide the fastest rate of turn?"

The slowest possible speed and the steepest possible bank angle will provide both the smallest radius and the fastest rate of turn. However there are several limitations:-

(a) When the steepest bank angle and slowest speed is applied the necessary centripetal force for the turn is provided by the extra lift gained by increasing the angle of attack ( or C_L) to a very high value. Also due to the lower velocity a larger portion of the total lift is provided by C_L rather than V². Consequently the induced drag will increase substantially - requiring increased thrust power and there will be a bank angle beyond which the engine will not be able to supply sufficient thrust to maintain the required lift, and thus height in the turn.

(b) All aircraft that are not certificated under the utility or aerobatic categories are limited to bank angles not exceeding 60°. A bank angle of 75° in a level turn would induce a 3.8g load factor – the load limit for a normal category certification. Similarly a level turn bank angle of 77° would induce the 4.4g load limit for an utility category aircraft.

(c) The stall speed increases with bank angle, or more correctly with load factor, (see question 6) thus the lowest possible flight speed increases as bank in a level turn increases.

(d)Turns at high bank angles, near the accelerated stall speed, with maximum power applied, leaves the aircraft with nothing in reserve. Any mishandling or turbulence may result in a violent wing and nose drop with substantial loss of height.

 Q10: "When you are flying alone your aircraft's all up weight is 360 kg. How much more induced drag will be produced at your normal cruise speed if you fly with a 90 kg passenger?

Induced drag is directly proportional to the aircraft's weight squared. The weight with the passenger is increased 1.25 times and 1.25 squared is 1.56, thus 56% more induced drag will be produced at the same airspeed with the passenger on board. The increase in induced drag results from the higher angle of attack needed to produce the increased lift at the same cruise speed but higher weight.

 Q11: "Name 5 types of drag that affect an ultralight in flight."

(a) Induced drag resulting from the production of lift.

(b) Parasite drag which is the collective name for all drag components produced by air resistance. These components are:-

(c) Pressure drag, skin friction and interference drag between structural members. (When these components are applied to a wing the collective term is profile drag.)

(d) The engine cooling drag is sometimes specified separately from pressure drag and interference drag.

 Q12: "You are an eagle-eyed ultralight pilot departed Mangalore for an airfield near Wagga Wagga, distance about 125 nm, on a bright, clear day with unlimited visibility and maintaining 4500 feet on your north-east track. The basic ground elevation is about 800 feet all the way. The Rock is a very distinctive sentinel outcrop of elevation 1800 feet near your destination. At what distance might you expect to have the whole top 500 feet of it in view? And what have you done wrong?"

At about 84 nm from The Rock. See the faqs page re line of sight distance. And you should have been maintaining 3,500 feet, the correct VFR cruising level for a north-east track.

 Q13: "Ultralight pilots are required to use two of the four primary air navigation systems – name them."

Pilotage and dead reckoning – see the faq.

 Q14: "Why does a propeller continue to rotate - windmill - following engine shut-down in flight and does a windmilling propeller create more or less aerodynamic drag than a stationary propeller?"

The angle of attack of a fixed pitch propeller, and thus its thrust (lift), depends on the forward speed of the aircraft and the rotational velocity. Following a non catastrophic engine failure the pilot tends to lower the nose so that forward airspeed is maintained while at the same time the rotational velocity of the engine/propeller is winding down. As the forward velocity remains more or less unchanged while the rotational velocity is decreasing the angle of attack must be continually decreasing and at some particular rpm the angle of attack will become negative to the point where the thrust force reverses and the propeller autorotates, driving the engine. This acts as greatly increased aerodynamic drag which seriously affects the L/D ratio and thus glide angles. The drag (reverse thrust) is much greater than that of a stationary propeller, also the engine rotation may cause additional mechanical problems if oil supply is affected.

If the forward speed is increased windmilling will increase, if forward speed is decreased windmilling will decrease, thus the windmilling may be stopped by temporarily reducing airspeed so that the reverse thrust is removed.

In the diagram the upper figure shows the forces associated with a section of a propeller blade operating normally. The lower figure shows the forces and the negative angle of attack (aoa) associated with the propeller now windmilling at the same forward velocity.

 Q15: "You may have come across the term "flat plate area" in aviation magazines. What does it mean?

Equivalent flat plate area, or just flat plate area - FPA, is a term used when comparing the aerodynamic cleanness of aircraft. It is the coefficient of parasite drag multiplied by the wing area in square feet. It does not represent the frontal area of an aircraft. A typical fixed undercarriage general aviation aircraft would have a FPA around 6 square feet. The lowest known FPA for a manned powered aeroplane is that of an ultralight (holder of the world speed for aircraft under 300 kg) - 0.88 square feet. Refer groundschool module 4.

 Q16: "An ultralight at mtow is climbing out from an airfield, elevation near sea level, on a cold but windless morning. Its airspeed is 60 knots (6,000 feet/minute) and the rate of climb is a constant 600 feet/minute. Approximately how much more additional lift is required to maintain the climb, over that required for level flight at 60 knots at the same weight?

No additional lift is required. The force required for climb is not lift but thrust power, in fact the lift in the 60 knot climb will be slightly less than the lift at 60 knots level flight. Refer forces in a climb.

 Q17: "What is meant by the terms 'tail volume', 'longitudinal dihedral' and 'decalage'?"

Tail volume is the length of the moment arm of the stabilizer multiplied by the surface area of the stabilizer. Longitudinal dihedral is the difference in the angles of incidence of the mainplanes and the horizontal stabilizer with reference to the 'fuselage reference line'. Decalage is the difference between the angles of incidence of the upper and lower mainplanes of a biplane. Further information.

 Q18: "What's a 'chandelle'?"

A chandelle is a 180° maximum rate co-ordinated climbing turn which trades kinetic energy for potential energy i.e. altitude. It is designed to demonstrate the pilot's ability to coax maximum altitude in shortest time from airspeed energy, to smoothly substitute C_L for V² and to control roll-out and recovery at very low airspeed.

The manoeuvre is started at an airspeed say two times Vs1 and is ended at Vs1. Any other airspeed will do but 2 × Vs1 is a good standard with which to start – with all these manoeuvres nothing’s written in stone! The sequence is as follows-
    • Start at a height around 3,000 feet agl, aircraft trimmed for cruise, increase airspeed to entry speed.
    • Roll in about 25° bank then add back-pressure (which should be increasing throughout the turn) and full power but maintain the same bank, thus pulling the aircraft into a maximum rate climbing turn. At the 90° turn point the airspeed should be at Vx. The back pressure (i.e. angle of attack) must be continually increasing to compensate for the decreasing velocity (thus maintaining constant lift) otherwise the aircraft will tend to sink. As the outside wing will generate more lift than the inner in a climbing turn and also the inner wing will have a lower aoa the bank angle will naturally tend to increase, so to maintain a constant angle you must hold off bank during the first 90° of the turn.
    • After passing the 90° point start decreasing bank as the airspeed further decays until, just as the airspeed reaches Vs1, the aircraft arrives at the 180° point and the wings become level. Then relax enough back-pressure to fly out of the stall. After recovering note the altitude gain from the initial cruise level.
    • Then try other bank angles in subsequent chandelles to ascertain which angle provides highest altitude gain at the recovery point. Also try higher and lower entry speeds but remember the object, once having established the optimum bank angle and airspeed, is to demonstrate how precise is the arrival at top of climb and how smooth the transition during the turn. It is a simple drill but not so easy to do accurately.

 Q19: "What is the area of a wing having a span of 10 metres and an aspect ratio of 8?"

Aspect ratio is span squared divided by wing area, thus wing area = span² / aspect ratio
e.g. wing area = 10 × 10 / 8 = 12.5 m²

 Q20: "If, in nil wind conditions, an ultralight has a ground roll of 100 metres before reaching the normal liftoff speed of 40 knots what would be the takeoff ground roll into a headwind of 15 knots?"

The takeoff ground roll = the nil wind ground roll × [ (unstick speed – wind speed) / unstick speed ]²
thus the takeoff ground roll = 100 × [ (40 – 15) / 40 ]²
        = 100 × 0.625² = 100 × 0.39 = 39 metres.

 Q21: "Why does rudder deflection cause an aircraft to bank and turn?"

Rudder deflection initiates yaw and the outside wing - i.e. opposite to the direction of yaw - will have a higher velocity than the inner wing and thus provide more lift than the inner. Hence a rolling moment leading to a turn. Refer directional stability.

 Q22: "A Thruster and a Sapphire are conducting full 360° turns around a point on the ground, the turns are at constant height, speed and radius from the ground point. The radius of the Thruster's turn is 300 metres and that of the Sapphire is 500 metres. There is nil wind and both aircraft complete a full 360° turn in exactly 60 seconds. Which aircraft maintains the higher angle of bank?"

The Sapphire. In a level turn the tangent of the bank angle = V² /gr where 'V' is the airspeed in metres/second, 'g' is the acceleration of gravity ( near enough to 10 metres/sec² ) and 'r' is the radius of the turn. The circumference of a circle is 2πr thus the Thruster travels 1884 metres in 60 seconds so V is 31.4 m/sec, V² is 986 and V² /gr is 986/(10 × 300) = 0.33. The angle where the tangent is 0.33 is about 19°. The Sapphire travels 3140 metres in 60 seconds so V is 52.3 m/sec, V² is 2737 and V² /gr is 2737/(10 × 500) = 0.55. The angle where the tangent is 0.33 is about 29°.

For further information refer to centripetal force in the groundschool manoeuvring module.

 Q23: "Can you legally carry baggage or freight on the otherwise empty passenger seat of an ultralight?"

Yes. But it may not exceed 77 kg and shall be restrained so that the cargo and the means of restraint can't interfere with the operation of aircraft controls. If the duplicate control column is designed for easy removal it shall be removed. CAO 20.16.2. para.6 refers, also check the groundschool section on ballasting.

 Q24: "What is the difference between aileron drag and adverse yaw?"

Aileron drag is the cause and adverse yaw is an effect. See the groundschool section on ailerons

 Q25: "What serviceable instruments are mandatory for flight in an ultralight?"

Just the magnetic compass and an airspeed indicator. If there is no panel clock the pilot should have a watch. The same conditions apply to all private VFR flights in aircraft under 5,700 kg.

 Q26: "What does the term 'last light' signify?"

Last light is the end of evening civil twilight and the official end of daylight for aerial navigation. The regulations state that all VFR [i.e. all ultralight ] navigation should be planned to be completed 10 minutes before last light, however pilots should be aware that an overcast sky will cause actual last light to occur well before the official event.

Evening civil twilight is the period from sunset until the centre of the sun’s disc is 6° below the horizon. If the sky is clear it is usually practicable to carry out normal outdoor activities without artificial light. Nautical twilight ends when the sun is 12° below the horizon. The horizon is still clearly defined, weather permitting, and the brighter stars are visible thus providing good conditions for position fixing using the early navigation instruments - hence nautical. Astronomical twilight ends when the sun is 18° below the horizon and all traces of sun illumination in the sky are gone. The duration of twilight is geometrically dependent on latitude, season and the observers elevation. First light is the beginning of morning civil twilight. Further information on twilight effects can be found in the meteorology section. Daily 'End of Daylight' graphs for latitudes 0° to 45° South can be found in AIP GEN 2.7.

 Q27: "What is the minimum forward flight visibility conditions in which an ultralight may legally operate?"

Ultralights may only operate in Visual Meteorological Conditions [VMC] and the minimum average range of visibility forward from the cockpit in VMC is 5,000 metres. If the visibility is less than 5,000 metres then Instrument Meteorological Conditions [IMC] exist. 'Visibility' means the ability to see and identify prominent objects. The problem of course is that there may not be any prominent identifiable objects when flying over featureless areas and, secondly, few people are adept at judging distance from the cockpit.

 Q28: "What do the letters 'QNH' and 'QFE' mean?"

The letters in themselves have no literal significance. The three letter Q-codes are remnants of an extensive coding system from the days of wireless-telegraphy. There were some 200 three letter Q-codes each representing a sentence, a phrase or a question, for instance QRM "I am being interfered with" [!]. Some 30 or so of the Q-codes are still used by amateur radio / morse code enthusiasts and four still survive in aviation meteorology.

QFE: the barometric pressure at the station location or aerodrome elevation datum point. If QFE is set on the altimeter baro-setting scale while parked at an airfield, the instrument should read close to zero altitude – if the local pressure is close to the ISA standard for that elevation. However the use of QFE is deprecated.

QNH: the msl pressure derived from the barometric pressure at the station location by calculating the weight of an imaginary air column, extending from the location to sea level, assuming the temperature at the location is the ISA temperature for that elevation, the temperature lapse rate is ISA and the air is dry throughout the column. If QNH is set on the altimeter baro-setting scale at an airfield, an acceptable instrument should read within 100 feet of the official airfield elevation. QNH is generally not available for airfields from which ultralights operate so the area QNH should be ascertained and used for the altimeter setting.

QFF: the msl pressure derived from the barometric pressure at the station location by calculating the weight of an imaginary air column, extending from the location to sea level, assuming the temperature and relative humidity at the location are the long term monthly mean, the temperature lapse rate is ISA and the relative humidity lapse rate is zero. This is the Australian Bureau of Meteorology method - QFF calculations differ among meteorological organisations. QFF is the location value plotted on surface synoptic charts and is closer to reality than QNH.

QNE:is the ISA Standard Pressure altimeter setting of 1013.2 hPa. The term QNE is now rarely encountered but if you set 1013.2 on the altimeter baro-setting scale while parked the altimeter will indicate the current ISA pressure altitude of the airfield – which is the first step in calculating density altitude.

For further information refer to altitude definitions.

 Q29: "If the stall speed [Vs1] of a two seat ultralight, at maximum gross weight of 540 kg, is 36 knots what will be the stall speed if no passenger is carried and the gross weight is reduced to 455 kg?"

Stall speed varies with wing loading. A reasonably accurate rule of thumb is that the percentage reduction in Vs1 is one half the percentage reduction in gross weight. So if the aircraft weight is reduced by 16% [85 kg] then Vs1 will be reduced by 8% [3 knots] to 33 knots. For further information refer 'Effect of weight'.

 Q30: "You are at an airfield [elevation 2700 feet and situated in flat terrain] and the base of an extensive layer of stratocumulus has been reported as 4000 feet but visibility is at least 10 km. Can you legally take off in an ultralight?"

Ultralight operations may only be conducted in Visual Meteorological Conditions and flight below 500 feet agl is forbidden except when taking off or descending to land. The visual meteorological conditions applicable below 10,000 feet amsl, and thus the VMC for ultralight operations [take-off, en route and landing] are:-

• Visibility: 5,000 metres
• Horizontal cloud clearance: 1500 metres
• Vertical cloud clearance: 1000 feet

Copyright © 2001 John Brandon

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