K.Haroon Site Admin
Joined: 11 Dec 2005 Posts: 1311

Posted: Tue Apr 17, 2012 3:44 am Post subject: Questions on Climb Performance 


When comparing Vx to Vy:
a) Vy will always be greater than or equal to Vx < Correct
b) Vy will always be greater than Vx
Climb Performance
A headwind component increasing with altitude, as compared to zero wind condition (assuming IAS is constant):
a) has no effect on rate of climb < Correct
b) improves angle and rate of climb
c) does not have any effect on the angle of flight path during climb
d) decreases angle and rate of climb
With increasing altitude, the rate of climb: "decreases because power available decreases and power required increases"
The rate of climb:
a) Is approximately climb gradient times true airspeed divided by 100 < Correct
The climb gradient is defined as the ratio of:
a) The increase of altitude to horizontal air distance expressed as a percentage < Correct
b) The increase of altitude to distance over ground expressed as a percentage
c) True airspeed to rate of climb
d) Rate of climb to true airspeed
Equations below expresses approximately the unaccelerated percentage climb gradient for small climb angles:
Climb Gradient = [(Thrust  Drag)/Weight] x 100
Assuming that the required lift exists, which forces determine an aeroplane's angle of climb? "Weight, drag and thrust"
For a given aircraft mass, the climb gradient: "decreases with increasing flap angle and increasing temperature"
In unaccelerated climb thrust equals drag plus the downhill component of the gross weight in the flight path direction.
What will happen to VX and VY if the landing gear is extended? "VX and VY decrease"
Increase in the profile drag will shift the drag curve to the left.
If the aircraft mass increases, how does the (i) rate of climb, and (ii) rate of climb speed change?
a) decrease; increase
Rate of Climb = (Power Available  Power Required) / Weight
Climb Performance
Other factors remaining constant, how does increasing altitude affect Vx and Vy: "Both will increase"
Compared to Vx and Vy in clean configuration, Vx and Vy in configuration with flaps extended will be: "Lower"
Profile drag due to flaps shift the drag curve towards left
For a piston engine aircraft the service ceiling corresponds to: "The altitude at which the aircraft is capable of a climb rate of 100 feet per minute"
The absolute ceiling: "is the altitude at which the maximum rate of climb is zero"
On a twin engined piston aircraft with variable pitch propellers, for a given mass and altitude, the minimum drag speed is 125 kt and the holding speed (minimum fuel burn per hour) is 95 kt. The best rate of climb speed will be obtained for a speed:
a) equal to 95 kts < Correct
b) < 95 kts
c) is between 95 and 125 kts
d) equal to 125 kts
The examiner is trying to tell you something by specifying a powerful aircraft with VP propellers. For this aircraft In the lower speed ranges you would expect the THP available to be a nearly flat line, although I admit a constant speed prop would give a flatter trace. If you draw THP available as constant over this range and superimpose the power required curve you will see that the maximum power surplus occurs at minimum power speed, Vimp. You have been given this as 95kt.
My whole argument relies on having THP available as roughly constant at these speeds, which is a special case. It would not necessarily apply in the real world or to a fixed pitch prop. The shape of the THP available curve (assuming constant EHP) depends on propeller efficiency versus TAS and on your ordinary bugsmasher rises rapidly from zero TAS and then falls off again
This speed is now your best rate of climb speed, Vy. Your best angle of climb speed, Vx, is where you have maximum excess thrust over drag and because props give maximum thrust at very low speed to find Vx you would fly as slowly as possible  minimum control speed is usually quoted
Source: (http://www.pprune.org/professionalpilottrainingincludesgroundstudies/410673atplperformance.html#post5623939)
Maximum endurance for a piston engined aeroplane is achieved at:
a) The speed that approximately corresponds to the maximum rate of climb speed < Correct
b) The speed for maximum lift coefficient
c) The speed for minimum drag
d) The speed that corresponds to the speed for maximum climb angle
Max endurance speed for a prop is Vmp. As quoted by the authority, it is the best rate of climb speed (Vy) for a prop.
What is the effect of a head wind component, compared to still air, on the maximum range speed (IAS) and the speed for maximum climb angle respectively? "Maximum range speed increases and maximum climb angle speed stays constant"
Climbing to cruise altitude with a head wind will:
a) decrease ground distance covered to climb < Correct
b) decreased time to climb
Any acceleration in climb, with a constant power setting:
a) decreases the rate of climb and the angle of climb < Correct
b) decreases rate of climb and increases angle of climb
c) improves the climb gradient if the airspeed is below VX
d) improves the rate of climb if the airspeed is below VY
The only way we can accelerate in the climb at a constant power setting is to lower the nose. The initial effect then is for the ROC and angle of climb to decrease. Subsequent effects would be an increase in the angle of climb if the initial speed was below Vx and an increase in ROC if the initial speed was below Vy. Its the initial effect that the examiner is looking for.
Source: (http://www.atpforum.eu/showthread.php?t=3549)
For a given power setting in the climb, if the speed is increased:
a) the ROC will increase if the speed is below Vy < Correct
b) the ROC will decrease if the speed is below Vy
c) the gradient of climb will increase if the speed is above Vx
d) the gradient of climb will decrease if the speed is below Vx
Speed is increased from which point?
From a point below Vy  Speed increase towards Vy  ROC Increases uptil Vy  After crossing Vy the ROC will start decreasing. (option "b" is out).
From a point below Vx  Speed increase towards Vx  Climb gradient increases uptil Vx  After crossing Vx the climb gradient will start decreasing (option "c" and "d" are out).
An aircraft is climbing at VX. If the speed is increased, maintaining the power setting:
a) the climb gradient will decrease and the rate of climb increase < Correct
b) the climb gradient will decrease and the rate of climb decrease
A climb gradient required is 3.3%. For an aircraft maintaining 100 kt true airspeed, no wind, this climb gradient corresponds to a rate of climb of approximately:
CLimb gradient x TAS = Rate of Climb
3.3 x 100 = 330 ft/min
On a segment of the take off flight path an obstacle requires a minimum gradient of climb of 2.6% in order to provide an adequate margin of safe clearance. At a mass of 110000 kg the gradient of climb is 2.8%. For the same power and assuming that the sine of the angle of climb varies inversely with mass, at what maximum mass will the aeroplane be able to achieve the minimum gradient.
a) 118455 kg < Correct
b) 106425 kg
c) 102150 kg
d) 121310 kg
Sine of the climb angle = ThrustDrag /Weight
i.e. Sine of the climb angle varies inversely with mass.
A linear relationship can be used.
Reduction of climb angle (2.6/2.8 = 0.928) will increase the weight limit (2.8/2.6 = 1.077)
110,000 kg x 1.077 = 118,470kg.
Closest option is 118455 Kg.
An aircraft with a mass of 110000 kg is capable of maintaining a grad of 2.6%. With all the atmospheric variables remaining the same, with what mass would it be able to achieve a gradient of 2.4?
a) 119167 kg < Correct
Same as above except that the question does not mention the fact that "sine of the climb angle varies inversely with mass".
110000kg x (2.6/2.4) = 119167kg.
In the climb an aircraft has a thrust to weight ratio of 1:4 and a lift to drag ratio of 12:1. While ignoring the slight difference between lift and weight in the climb, the climb gradient will be:
Answer: 16.7%
Solution:
Sine of Climb angle = Thrust  Drag / Weight
For small angles Adjacent is nearly the same as Hypotenuse and Opposite/Adjacent = Sin relation.
So we get an approximate formula for climb gradient, which is:
Climb Gradient = Thrust  Drag / Weight
Just put in the values.
L/D is 12/1. Considering lift almost equal to weight. Weight = 12 and Drag = 1
Thrust to Weight ration is 1:4
If weight is 12 then Thrust is 12 x 1/4 = 3
Climb Gradient = 31/12 = 1/6 = 16 = 0.166
Climb Gradient % is 0.166 x 100 = 16.66%
Still air gradient of climb: 3.8%, TAS: 100 kts, GS: 130 kts. The wind effective climb gradient is:
a) 2.9% < Correct
b) 4.9%
c) 4.2%
d) 3.8%
The climb gradient in aerodynamic terms is considered in the air mass and is therefore not affected by wind. It is an air gradient. If the gradient is related to ground distance a headwind will increase the flight path climb gradient whereas a tailwind will decrease it. In this case the gradient should be referred to as the Flight Path Angle.
(Refer to figure 032_8.4)Considering a rate of climb diagram (ROC versus TAS) for an aeroplane. Which of the diagrams shows the correct curves for FLAPS DOWN compared to CLEAN configuration?
Figure A is Correct
Considering a rate of climb diagram (ROC versus TAS) for an aeroplane. In which direction does the curve for FLAPS DOWN move when compared to the CLEAN configuration curve?
a) Left and Down < Correct
b) Left and horizontally
c) Right and Up
d) Left and Up
Which statement, in relation to the climb limited take off mass of a jet aeroplane, is correct?
a) The climb limited take off mass decreases with increasing OAT < Correct
b) 50% of a head wind is taken into account when determining the climb limited takeoff mass
c) On high elevation airports equipped with long runways the aeroplane will always be climb limited
d) The climb limited take off mass is determined at the speed for best rate of climb
An aircraft with a climb gradient of 3.3%, flying at an IAS of 85 kts. At a pressure ALT of 8500 ft with a temperature of +15 deg C will have a ROC of:
a) 334 ft/min < Correct
b) 284 ft/min
c) 623 ft/min
d) 1117 ft/min
Convert IAS into TAS
ROC = TAS x Gradient = 100 x 3.3
Multiply IAS of 85 x 3.3 and you get into the trap set at option (b)
A four jet engined aeroplane (mass = 150000 kg) is established on climb with all engines operating. The lifttodrag ratio is 14. Each engine has a thrust of 75000 Newtons. The gradient climb is: (given: g=10m/s?):
Answer: 12.86%
Solution:
Climg gradient = Thrust  Drag / Weight
= (75000 x 4)  Drag / Weight
Weight = m x g = 150,000 x 10 = 150,000,0
L/D = 14 so Drag = L/14
For small angles Lift is almost equal to Weight
So Drag = 150,000,0/14 = 107143
Climb gradient = 300,000  107143 / 150,000,0 = 0.128
Climb gradient % = 0.128 x 100 = 12.8%
If the climb speed schedule is changed from 280/.74 to 290/.74 the new crossover altitude is:
a) lower < Correct
b) higher
c) unchanged
d) only affected by the aeroplane gross mass
The Crossosver Altitude is the altitude at which a specified CAS and Mach value represent the same TAS value.
The curves for constant CAS and constant Mach intersect at this point. Above this altitude the Mach number is used to reference speeds.
CLimb at constant CAS, TAS increases.
CLimb at constant Mach, TAS decreass.
For M0.74 in the climb the cross over altitude will occur at a lower altitude if the IAS/CAS is increased from 280 to 290.
Use this Online Calculator to compute the cross over altitude with M0.74 against 280 and 290 kts CAS and see the results.
In a climb, at a constant IAS/Mach No. 300 kts/0.78 M., what happens at the change over point (29500 ft, ISA)?
a) Find that rate of climb would start to increase < Correct
b) Accelerate from the IAS to the Mach number, and therefore rate of climb will decrease
c) No change in rate of climb since TAS remains constant
d) Find that rate of climb would start to decrease
You climb with a climb speed schedule 300/.78. What do you expect in the crossover altitude 29200 ft (OAT = ISA)?
a) The rate of climb increases since the constant IAS climb is replaced by the constant Mach climb < Correct
b) During the acceleration to the Mach number .78 the rate of climb is approximately zero
c) The rate of climb decreases since climb performance at a constant Mach number is grossly reduced as compared to constant IAS
d) No noticeable effect since the true airspeed at 300 kt IAS and .78 Mach are the same (at ISA temperature TAS = 460 kt)
To change from a constant IAS climb to a constant Mach climb you must pitch up at the crossover altitude to lock on to the Mach no. This gives a temporary increase in the rate of climb, 20 or so seconds later the ROC will start to decrease again because TAS decreases when flying at a constant Mach number.
Source: (http://www.atpforum.eu/showthread.php?t=4320)
A jet aeroplane is climbing with constant IAS. Which operational speed limit is most likely to be reached?
a) The Maximum operating Mach number < Correct
b) The Stalling speed
c) The Minimum control speed air
d) The Mach limit for the Mach trim system
What happens when an aeroplane climbs at a constant Mach number?
a) The lift coefficient increases < Correct
b) The '1.3G' altitude is exceeded, so Mach buffet will start immediately
c) The TAS continues to increase, which may lead to structural problems
d) IAS stays constant so there will be no problems
Lift = 1/2 Rho V(square) x CL x S
1/2 Rho V(square) is the dynamic pressure displayed as IAS.
At a constant IAS climb Rho decreases and V (TAS) increases.
Thus dynamic pressure remains constant and there is no change in CL.
At a constant Mach climb apart from Rho that was already decreasing, TAS starts decreasing as well.
Thus dynamic pressure decreases.
To maintain lift angle of attack needs to be increased, meaning CL increases.
A head wind will:
a) increase the climb flight path angle < Correct
b) increase the angle of climb
A jet aeroplane is climbing at a constant IAS and maximum climb thrust, how will the climb angle/the pitch angle change?
a) Reduce/decrease < Correct
b) Reduce/remain constant
c) Remain constant/decrease
d) Remain constant/become larger
Increase in altitude = decrease in thrust available = pitch down to maintai constant IAS = reduced climb angle
(Refer to figure 032_89)What is the maximum vertical speed of a three engine turbo jet aeroplane with one engine inoperative (N1) and a mass of 75000 kg? Using the following:G = 10 m/s2 1 kt = 100 ft/min SIN (Angle of climb) = (Thrust  Drag)/Weight
Answer: +1267 ft/min
Solution:
Rate of Climb = Gradient x TAS
From the figure:
TAS = 190
Gradient = (ThrustDrag)/Weight
Max rate of climb is with max difference between thrust available and thrust required.
From the figure Thrust available for all engines = 240
With one engine out thrust available = 24080 = 160
Thrust required = 110
Weight = m x g = 750000
gradient = 160,000  110,000 / 750,000 = 0.066
ROC = gradient x TAS = 0.066 x 190 = 12.66
given 1kt = 100 ft/min
ROC = 12.66 x 100 = 1267 ft/min
If the conversion factor is not given then before multiplying TAS with gradient we need to convert TAS from nm/hr to ft/min i.e.
190/60 x 6080 = 19253
19253 x 0.066 = 1270 ft/min
An aircraft is climbing in a standard atmosphere above the tropopause at a constant Mach number:
a) the IAS decreases and TAS remain constant < Correct
b) the IAS and TAS remain constant
c) the IAS decreases and TAS decreases
d) the IAS remains constant and TAS increases
TAS = 39 x M x Square root of Temperature in Kelvin
Above Tropopause, temperature remains constant. So at constant Mach number TAS remains constant.
An aircraft is climbing at a constant IAS, below the Mach limit. As height increases: "drag remains constant, but the climb gradient decreases" 
