POF Q23 - Level Flight & the Climb Calculations

Question 1

A twin engine aircraft flies a straight, steady, wings level, all engines operating climb with a mass of 40250kg. The aircraft’s thurst per engine is 67500N. The aircraft has a total Drag of 35000N.

Calculate the aircraft’s Angle of Climb.
(Please give answer to 2 decimal places. Answer is assumed in º )

Answer 1

Thurst = 67500 x 2 = 135000N

Weight = 40250 x 9.81 = 394,852.5N

sin𝛾 = 𝑇−𝐷𝑊
sin𝛾 = 135000−35000/394852.5
sin𝛾 = 0.253259…..
𝛾 = sin-1 (0.253259….)

𝛾 = 14.67º

Question 2

A four engine aircraft is flying in a straight, steady and wings level climb and has lost thrust from one engine. The aircraft has a mass of 500,000kg and the thrust per engine is 356,810N. The total drag of the aircraft is 82143N. Assuming that 1g = 10m/s2, calculate the aircraft’s Angle of Climb.

Question 2 Answer

a.15.6º

b.3.14º

c.11.4º

d.33.3º

Answer 2

c.11.4º

Thrust = 356810 x 3 = 1,070,430N

Weight = 500,000 x 10 = 5,000,000N

sin𝛾 = 𝑇−𝐷𝑊
sin𝛾 = 1070430−82143/5000000
sin𝛾 = 0.1976….
𝛾 = sin-1 (0.1976….)

𝛾 = 11.4º

Question 3

A Boeing 737 of mass 62,000kg is flying in a steady, wings level climb. Its Lift/Drag ratio is 15.5. The total thrust produced by the two engines at climb power is 122kN. Assuming 1g = 10m/s2, calculate the aircraft’s Angle of Climb.

(Please give answer to 1 decimal place. Answer is assumed in º )

Answer 3

Thrust = 122,000N
Lift = 62,000 x 10 = 620,000N

Drag = 620,000 ÷ 15.5 = 40,000N

sin𝛾 = 𝑇−𝐷𝑊
sin𝛾 = 122000−40000620000
sin𝛾 = 0.132258…..
𝛾 = sin-1 (0.132258….)

𝛾 = 7.6º

Question 4

A DA40 has taken off with a mass of 1870kg. It is climbing at an angle of 9º. It currently has 1700N of Drag. Assuming 1g = 10m/s2, what is the thrust produced?

(Please give answer to 2 decimal places. Answer is assumed in N )

Answer 4

T = Wsin𝛾 + D
T = (18700 x sin(9)) + 1700

T = 4625.32N

Question 5

A DA42 is in a level climb at 8.4º AoC. It is producing 9200N of thrust and has a total drag of 6400N. What is the aircraft’s mass?

(Please give answer to the nearest whole number. Answer is assumed in kg )

Answer 5

W = 9200−6400𝑠𝑖𝑛(8.4)
W = 19167.18N

19167.18 ÷ 9.81 = 1953.84kg

Question 6

An aircraft is in a steady wings level climb at an angle of 10.4º. The total thrust being produced is 8700N and the total drag being produced is 5270N. Assuming that 1g = 10m/s2, calculate the aircraft’s Weight.

a.32981N

b.1850kg

c.19,000N

d.1940kg

Answer 6

c.19,000N

W = (T-D) ÷ sin𝛾
W = (8700-5270) ÷ sin(10.4)
W = 19,000N

Question 7

A twin engine aircraft is in a steady wings level climb. If one engine fails, the thrust (I) and the Drag (II) and the excess thrust (III)

a.(I) reduces by more than 50% (II) increases by 50% (III) reduces by more than 50%

b.(I) reduces by 50% (II) slightly increases (III) reduces by more than 50%

c.(I) reduces by 50% (II) remains constant (III) reduces by 50%

d.(I) reduces by more than 50% (II)remains constant (III) reduces by 50%

Answer 7

b.(I) reduces by 50% (II) slightly increases (III) reduces by more than 50%

Question 8

A Boeing 737 of mass 69,000kg is flying in a steady, wings level climb. Its Lift/Drag ratio is 15.5. The total thrust produced by the two engines at climb power is 190kN. Assuming 1g = 10m/s2, calculate the aircraft’s Angle of Climb.

(Please give answer to 2 decimal place. Answer is assumed in º )

Answer 8

Thrust = 190,000N

Weight = 69,000 x 10 = 690,000N

Drag = 690,000 ÷ 15.5 = 44,516N

sin𝛾 = 𝑇−𝐷𝑊
sin𝛾 = 190000−44516/690000
sin𝛾 = 0.2108…..
𝛾 = sin-1 (0.2108….)

𝛾 = 12.17º

Question 9

A three engine aircraft that weighs 69 tonnes is in a steady wings level climb and one engine fails. The aircraft now has a climb gradient of 8.06%. The thrust per engine is 80,000N. What is the total drag of the aircraft?

a.105,443N

b.185,443N

c.184,386N

d.104,386N

Answer 9

a.105,443N

The angle of climb formula is sin𝛾 = 𝑇−𝐷𝑊
D = T - Wsin𝛾

At small angles of climb sin𝛾 ≈ tan𝛾
We can therefore say that the Gradient ÷ 100 = 0.0806 = tan𝛾 ≈ sin𝛾
sin𝛾 = 0.0806

W = 69,000 x 9.81 = 676,890N

Thurst = 80,000 x 2 = 160,000N

D = T - Wsin𝛾
D = 160,000 - (676890 x 0.0806)
D = 105,443N

Question 10

An Airbus A320 takes off with a mass of 69,500kg. The thrust per engine in the climb is 111kN. The total drag produced in the climb is 77.5kN. Assuming that the total drag is not affected, what would be the reduction in the climb gradient if one engine fails?
Assume 1g = 10m/s2

a.4.9%

b.20.8%

c.12%

d.15.9%

Answer 10

d.15.9%

The angle of climb formula is sin𝛾 = 𝑇−𝐷/𝑊

Thurst = 111000 x 2 = 222000N
Weight = 69,500 x 10 = 695,000N

sin𝛾 = 𝑇−𝐷/𝑊
sin𝛾 = 222000−77500/695000
sin𝛾 = 0.2079

At small angles of climb sin𝛾 ≈ tan𝛾
We can therefore say that the climb gradient is ≈ sin𝛾 x 100.
*Climb gradient = 0.2079 x 100 = 20.8%

Next we will work out the one engine operating climb gradient. We will do the same as above except only use the thrust value from one engine

sin𝛾 = 𝑇−𝐷/𝑊
sin𝛾 = 111000−77500/695000
sin𝛾 = 0.0482

At small angles of attack sin𝛾 ≈ tan𝛾
We can therefore say that the climb gradient is ≈ sin𝛾 x 100.
*Climb gradient = 0.0482 x 100 = 4.9%

The question asks for the reduction in climb gradient. So finally we must take one away from the other to find the difference in gradients.
20.8% - 4.9% = 15.9 %