Questions to principles of flight

[keith williams]

1 An aeroplane in straight and level flight at 200 kt is subjected to a sudden disturbance in speed. Assuming the angle of attack remains constant initially and the load factor reaches a value of 1.21:“Answer: the speed will have increased by 20 kt.



Load factor = lift/weight so in straight and level flight lift = weight and load factor = 1
Increasing load factor to 1.21 means that lift has multiplied by a factor of 1.21.


Lift = CL1/2RhoVsquared

So lift is proportional to the square of the speed

In the case of a sudden gust the increase in lift is proportional to the square of the increase in speed.

So to increase lift by a factor of 1.21 we must increase sped by a factor of the square root of 1.21.

The square root of 1.21 is 1.1. So the speed in this question has increased by 0.1 which is 10%.
The initial speed was 200 knots so the increase is 20 knots.


2) „Approximately how long does it take to fly a complete circle during a horizontal steady coordinated turn with a bank angle of 45° and a TAS of 288 kt?“
Answer: 95 s


Radius of turn = Vsquared / g TanAOB

Circumference of turn = 2Pi x radius of trun
So Circumference of turn = 2 Pi V squared / g TanAOB

Time to turn = Circumference of turn / Velocity
So Time to turn = 2 Pi V / g Tan AOB ……………Equation 1

We must ensure that both the Velocity and g are in the same units. If we use g in metres per second squared we must convert the knots into metres squared per second squared by multiplying by 0.515.

AOB is 45 degrees and TAN AOB = 1G = 9.81 metres per second squared.

TAS = 288 knots

Inserting the above data into equation 1 gives

Time to turn = ( 2 Pi x (288 x 0.515) ) / (9.81 x 1) = 95 seconds.




3) An aeroplane in straight and level flight at a speed of 2 VS. If, at this speed, a vertical gust causes a load factor of 3, the load factor n caused by the same gust at a speed of 1.2 VS would be:“
Answer: not greater than 1.44, because the aeroplane is stalled with a higher load facto rat 1.2 VS I have no idea how to figure it out.


Stall speed at any given weight is proportional to the square root of the load factor.

Rearranging this equation the load factor at which an aircraft stalls at any given weight is proportional to the square of the speed.

So for example at VS the load factor at which the aircraft stalls will be in the stalls will be 1 squared which is 1.

And at 1.1VS the load factor at which an aircraft stalls will be 1.1 squared which is 1.21.

In this question the speed is 1.2VS so the load factor at which it stall will be 1.2 squared which is 1.44. This means that the load factor in the gust cannot be greater than 1.44 because that is the load factor at which it stalled.