ATPL theory questions

[keith williams]

1. Whilst maintaining straight & level flight with a lift coefficient of CL=1, what will be the new value of CL after the speed is increased by 41%:

a) 0.25
b) 0.50 (correct answer)
c) 0.60
d) 0.30

Lift = Cl 1/2Rho V squared S

Where V is the TAS.

If speed increase by 41% it becomes 1.41 of its initial value.

So V squared becomes approximate 2 times its initial value.

The question states that SL flight is to be maintained, so the lift must not change.

So to maintain level flight the Cl must become ½ of its initial value.


2. Airplane flying at 100 kt in straight & level flight is subject to a disturbance that suddenly increases the speed by 20 kt. Assuming AoA remains constant, load factor will initially be:

a) 1.41
b) 1.44 (correct answer)
c) 1.30
d) 1.04

Lift = Cl 1/2Rho V squared S

Where V is the TAS.

Cl is proportional to AoA, so if AoA remains unchanged then Cl remains unchanged.

The 20 kt increase from 100 kts increases speed to 120 kts. This is 1.2 times its initial value.
So V squared becomes 1.2 squared = 1.44 times its initial value.

So the new lift = 1.44 times its initial value.

If the aircraft was initially in SL flight the lift must have been equal to the weight and the load factor was Lift / weight = 1

So if we use 1 to represent the initial lift we can use 1.44 to represent the lift in the gust.

So in the gust the load factor = 1.44 / 1 = 1.44


3. In straight & level flight at a speed of 1.3 VS, the lift coefficient, expressed as a percentage of of its maximum CLMAX, would be:

a) 169%
b) 130%
c) 59% (correct answer)
d) 77%

In SL flight at any speed above Vs
Lift = Cl 1/2Rho V squared S

At the stall in SL flight
Lift = ClMax 1/2Rho Vs squared S

Assuming weight is unchanged in the two equations above we can say that the first equation is equal to the second.

If we take out the common factors of Lift and 1/2Rho S we leave the following:

Cl V squared = ClMax Vs squared

Rearranging this gives us

Cl = ClMax Vs squared / V squared)

If we use 1 to represent Vs we can use 1.3 to represent 1.3Vs. Putting these values into the equation gives us
Cl = ClMax x (1 squared / 1.3 squared)

Cl = ClMax x (1 / 1.69)

Cl = 0.59 17 ClMax.

This is approximately 59% of Cl Max

For this type of question just remember the following

Cl at any speed = ClMax x ( Vs squared /Speed chosen squared )