K.Haroon Site Admin
Joined: 11 Dec 2005 Posts: 1291

Posted: Wed Feb 15, 2012 7:47 am Post subject: Collection of Numerical Problems involved in POF 


The lift coefficient (CL) of an aeroplane in steady horizontal flight is 0.42, increase in angle of attack of 1 degree increases CL by is 0.1. A vertical up gust instantly changes the angle of attack by 3 degrees.
Answer: The load factor will be 1.71
Increase in CL per 1 deg increase in AoA = 0.1
Increase in CL per 3 deg increase in AoA = 0.1 x 3 = 0.3
Change in CL = 0.42 + 0.3 = 0.72
Load factor is the ratio of lift/weight and in level (horizontal) flight it is 1.
Putting it mathematically, for a level flight CL/weight = 1
In the question CL (for the horizontal level flight) is given as 0.42
Thus putting the values in the equation CL/weight = 1:
0.42/weight = 1
or
0.42/0.42 = 1
Change in CL due to gust = 0.72 (as calculated above)
Weight has not changed so it remains 0.42
Putting in the values:
The load factor = 0.72/0.42 = 1.71
If a twin engine airraft with a L/D ratio of 8:1 is in straight and level flight and the engines are each evelopoing 16000 N of thrust, what is the weight of the aircraft? Answer: 256,000 N
Thrust = 32,000 N (@ 16,000/engine)
In a straight and level flight (in equilibrium) all forces are equal.
So thrust = drag
or
Drag = 32,000 N
L/D ratio is 8:1
So lift is 8 times more than drag (8 x 32,000)
Thus lift = 256,000 N
Since Lift = Weight in level flight
Weight = 256,000 N
In straight and level flight at a speed of 1.3 VS, the lift coefficient, expressed as a percentage of its maximum (CLmax), would be 59%.
1/1.3squared
= 0.59 x 100 = 59%
An aircraft whose weight is 237402 N stalls at 132 kt. At a weight of 356103 N it would stall at?
Ans: 162 kts.
356103/237402 = 1.5 (load factor)
Square root of 1.5 = 1.224
Increase in stall speed = 132 x 1.224 = 162
An aircraft with a take off weight of 10,000 kg has a basic stalling speed VS of 240 kts. What is the stalling speed as the aircraft turns on to finals with landing flaps extended (CLmax doubles) at 30 deg AOB at an AUW of 6400 kg?
Reduction in weight will reduce stall speed:
6400/10000 = 0.64
square root of 0.64 = 0.8
0.8 x 240 = 192
Bank will increase stall speed:
1/cos30 = 1.15
square root of 1.15 = 1.07
1.07 x 192 = 206
Landing flaps will reduce the stall speed:
Since CLmax doubles V will be square root of 2 which is 1.41
Use 1.41 to adjust 206
206/1.41 (to get an answer less than 206 as stall speed will be less with flaps).
Answer: 146 kts.
Note: If any factor except V changes, say by a factor K, then to keep things equal V has to change by the square root of K. Likewise. if V changes by a factor of K then some other factor has to change by Ksquared. So, if you double V, and, for example, maintain level flight, you have to reduce CL by a factor of 4. If you double weight and therefore Lift your speed at CLmax will go up by the square root of the weight change.
Compared to a straight wing of the same airfoil section a wing swept at 30 should theoretically have an Mcrit 1.154 times Mcrit for the straight wing, but will, in practice gain Half that increase.
1/cos30 = 1.15
The value of the manoeuvre stability of an aeroplane is 150 N/g. The load factor in straight and level flight is 1. The increase of stick force necessary to achieve the load factor of 2.5 is 225N.
In straight and level balanced flight the aircraft is at 1g and the stick force when trimmed is zero. If the aircraft is now subjected to a positive g manoeuvre, stability of the aircraft resists the movement and control forces (stick force) increases. Since in the question the aircraft has a stick force gradient of 150N per g, the temptation is to multiply the stick force by the amount of g, but as stated in balanced trimmed flight at 1g stick force is zero.
You must always subtract 1g from the g value given. So 2.5 g 1g = 1.5 x 150 = 225N.
Stability increases in a manoevre due to changes in angle of attack on the aircraft, this is called aerodynamic damping. Increased stability increases stick forces.
The manoeuvre stability of a large jet transport aeroplane is 280 N/g. What stick force is required, if the aeroplane is pulled to the limit manoeuvring load factor from a trimmed horizontal straight and steady flight? (cruise configuration)
The limit load factor is 2.5g. Rest of the logic is the same. Thus 1.5 x 280 = 420N which is the answer.
If the maximum pull force acceptable is 50 lbs and the design limit g of the aircraft is 6g, what stick force/g must be achieved to permit the aircraft to be manoeuvred to its design g limit?
Answer: 10 lb/g
6g  1g (level flight) = 5g
50lbs / 5g = 10 lbs/g
If a stick force of 20 lbs is required to pull 4g from the position of trim, the stick force gradient is:
Answer: 6.6 lb/g
4g  1g (level flight) = 3g
20lbs / 3g = 6.6 lbs/g
How does VA (EAS) alter when the aeroplane's mass decreases by 19%?
Once you set Cn at 2.5 for working out Va the answer for Va is Vs1g times sqrt 2.5.
If you reduce aircraft mass you get a new lower Vs1g and therefore a new lower Va.
Because the lift equation has Vsq in it the reduction in Vs1g is proportional to sqrt of the change in mass.
It is a quirk of the mathematics that, for small changes, inside 20%, the sqrt of the proportional change is approximately half the original figure, so a 20%
reduction in mass gives a 10% reduction in Va.
The answer gives figures of 19/10, which is probably the correct answer, not just the approximation.
Source: (http://www.atpforum.eu/showthread.php?t=5465)
An aircraft is flown at 20% below its normal weight. Because of this, VA will be 10% lower.
See the above logic. Even if you calculate by assuming an example it comes around 10%
e.g. If Weight = 10,000, VS1g = 200 and Limit load factor = 2.5
Then VA = 200 x sq root of 2.5 which comes around 316
If weight is 20% below 10,000 which is 8,000 then
VS1g = sq root of 8000/10000 x 200, which comes around 179
Then VA = 179 x sq root of 2.5, which comes around 283
Perecent drop in VA is 316283/283 x 100 = 11.66%
An aircraft has a mass of 60,000 kg and a limiting positive load factor of 2.5. VA is calculated as the EAS at which full positive elevator deflection will give the limiting load factor at the stall, and is 237 kts. If the aircraft mass is reduced to 40,000 kg by fuel burn, what will be the new VA?
Solution:
The formula is VA = VS1g x sq root of limit load factor
New VA = Stall speed at new lower weight of 40,000 x sq root of 2.5
First we need to know the stall speed at 60,000, which we can get from the above formula:
237 = VS x sq root of 2.5
VS = 150
Stall speed at 40,000 will be less than 150
Reduction factor = 40000/60000 = 0.666
sq root of 0.666 = 0.81
150 x 0.81 = 122
So the stall speed at 40,000 will be 122
New VA = 122 x sq root of 2.5
= 193
For a given TAS and bank angle, a heavy aircraft will have the same radius of turn as a lighter one and the same g load.
Radius of turn = TAS squared / g x tan of banke angle
Load factor = 1 / cos of banke angle
Weight is not an issue here.
The lift of an aeroplane of weight W in a constant linear climb with a climb angle (gamma) is approximately:
L = W x cos gamma
For shallow climb angles the following formula can be used: Sin (gamma) = T/W  CD/CL
Sin Gamma = Thrust  Drag / Weight
or
Sin Gamma = Thrust / Weigth  Drag / Weight
For shallow climb angles lift and weight are practically the same. Thus the formula can be written as:
Sin Gamma = Thrust / Weigth  Drag / Lift
or
Sin Gamma = Thrust / Weigth  CD / CL
Given an Aeroplane mass of 50,000 kg, Lift/Drag ratio 10, Thrust per engine 60,000N, assumed g=10 m/s2. For a straight, steady, wings level climb of a twin engine aeroplane, the all engines climb gradient will be:
Formula:
Sin gamma = Thrust / Weight  Drag / Weight (as mentioned above)
Thrust = 60,000 per engine. For 2 engines its 2 x 60,000 = 120,000
Weight = mass x g. i.e. 50,000 x 10 = 500,000
L/D = 10 so Drag = Lift/10
Lift = W x cos gamma
Lift decreases as the angle of climb increases. In this questin we cannot calculate lift since we dont know the climb angle. For small angles of climb (up to about 15) lift is almost equal to weight so we can take the weight value as lift (error is very small).
So Lift = 500,000
Thus Drag = 500,000 / 10 = 50,000
Now putting the values in the formula:
Sin (gamma) = T/W  D/W
Sin (gamma) = 120,000/500,000  50,000/500,000
Sin (gamma) = 0.14
In percentage: Sin (gamma) = 0.14 x 100 = 14%
For calculating angle:
Sin gamma = 0.14
gamma = Inv Sin of 0.14 = 8 degrees.
When an aeroplane performs a straight steady climb with a 20% climb gradient, the load factor is equal to:
20% climb gradient (in a right angled triangle) means:
20 up (opposite side) for 100 horizontal (adjacent side)
opposite/adjacent = tan (angle)
20/100 = tan (angle)
tan angle = 0.2
climb angle = Inv tan of 0.2 = 11.3
So gamma = 11.3
Lift in a climb is:
L = W x cos gamma
So load factor in climb is cos of climb angle
cos of 11.3 = 0.98 (Answer)
With a L/D ratio of 9:1 and flying at 12000 ft the glide range in still air would be 18 nm.
9:1 means 9000 feet horizontal distance covered for 1000 feet loss in height.
If for 1000 feet horizontal distance covered is 9000 feet.
Then for 12000 feet horizontal distance covered is 12 x 9000 feet = 108000 feet.
108000/6076 = 18 NM
An aeroplane is in a steady horizontal turn at a TAS of 194 kt. The turn radius is 1000 m. The bank angle is (assume g = 10 m/s2)?
a) 45 deg
b) 30 deg
c) 50 deg
d) 60 deg
Answer marked as correct in the database is (c)
Whereas correct answer should have been (a)
Given:Aeroplane mass:50,000 kg. Lift/Drag ratio:12 Thrust per engine: 28,000N Assumed g:10 m/s2 For a straight,steady, wings level climb of a four engine aeroplane, the all engines climb gradient will be:
a) 8.5%
b) 8.0%
c) 9.7%
d) 2.9%
Answer marked as correct in the database is (a)
Whereas correct answer should have been 14.07% which is not mentioned in any of the options. 
