Performance.pdf

We will talk about
static and dynamic
performance.

We will answer questions such as:
How fast?
How high?
How far?
How long can an aircraft fly?

Coverage Airplane Performance
Static Performance
(zero acceleration)
Dynamic Performance
(finite acceleration)
Thrust Required
Thrust Available
Maximum
Velocity
Power Required
Power Available
Maximum Velocity
Rate of Climb
Takeoff
Landing
Equations of Motions
V-n Diagram
Turning Flight
Range and Endurance
Time to Climb
Maximum Altitude
Gliding Flight
Service Ceiling
Absolute Ceiling
Drag Polar

A Prerequisite
Performance analysis hinges
on knowledge of the airplane
drag polar.

What is a drag polar?
It is a term coined by Eiffel.
The same monsieur of Eiffel tower fame.
The same guy who designed Quiapo bridge
(a.k.a. Quezon bridge)
It is a graph or an equation that accounts for all types of drag
in an airplane and how it relates to lift.

Quezon bridge: FEATI’s vantage point
Gustav’s Tower

What types of drag are included then? Let’s
enumerate.
Bullets would come in handy for
this sort of job.

What are the different types of drag?
▪ Skin friction drag
▪ Pressure drag
▪ Profile drag
▪ Interference drag
▪ Parasite drag
▪ Induced drag
▪ Zero-lift drag
▪ Drag due to lift
▪ Wave drag

Drag Types
Skin-friction drag. Drag due to frictional shear
stress integrated over the surface.

Pressure drag due to flow separation (form drag):
The drag due to the pressure imbalance in the
drag direction caused by separated flow.
Drag Types

Profile drag. The sum of skin friction drag and
form drag. (The term profile drag is usually used
in conjunction with two-dimensional airfoils; it is
sometimes called section drag.)
Drag Types

Interference drag. An additional pressure drag
caused by the mutual interaction of the flow
fields around each component of the airplane.
The total drag of the combined body is usually
greater than that of the sum of its individual
parts; the difference is the interference drag.
Drag Types

Parasite drag. The term used for the profile drag
for a complete airplane. It is that portion of the
total drag associated with skin friction and
pressure drag due to flow separation, integrated
over the complete airplane surface. It includes
interference drag.
Drag Types

Induced drag. A pressure drag due to the
pressure imbalance in the drag direction
caused by the induced flow (downwash)
associated with the vortices created at the tips
of finite wings.
Drag Types

Zero-lift drag. (Usually used in conjunction
with a complete airplane configuration.) The
parasite drag that exists when the airplane is
at its zero-lift angle of attack, that is, when the
lift of the airplane is zero.
Drag Types

Drag due to lift. (Usually used in conjunction with a
complete airplane.) That portion of the total airplane
drag measured above the zero-lift drag. It consists of
the change in parasite drag when the airplane is at an
angle of attack different from the zero-lift angle, plus
the induced drag from the wings and other lifting
components of the airplane.
Drag Types

Wave drag. The pressure drag associated with
transonic and supersonic flow (or shock waves,
hence the name). It can be expressed as the
sum the zero-lift wave drag and wave drag
due to lift.
Drag Types

Total Drag
Skin Friction Drag Pressure Drag
Induced Drag Wave Drag
Note : Profile Drag = Skin Friction Drag + Form Drag
Total Drag
Form Drag (Drag due to flow separation)
drag
induced
drag
profile
drag
total +
=

Drag Polar
eAR
π
C
C
C L
d
D
2
+
=
eAR
C
C
C L
D
D

2
0
, +
=
A wing or airfoil have its own drag polar
An airplane have its own drag polar
profile drag
zero-lift drag

Drag Polar
eAR
C
C
C L
e
D
D

2
, +
=
parasite drag coefficient
-profile drag of wing
-friction and pressure drag of:
tail surfaces
fuselage
engine nacelles
landing gear
other components exposed to the flow
-a function of angle of attack
lift span
efficiency factor
induced drag coefficient

Drag Polar
2
0
,
, L
D
e
D rC
C
C +
=
2
0
, )
1
( L
D
D C
eAR
r
C
C

+
+
=
eAR
π
C
C
C L
e
D
D
2
, +
=
eAR
π
C
C
C L
D
D
2
0
, +
=

Drag polar of a complete airplane
i
D
D
L
D
D C
C
eAR
C
C
C ,
0
,
2
0
, +
=
+
=

induced drag
coefficient
parasite drag
coefficient at
zero lift
Oswald’s
efficiency
factor

Drag polar of a complete airplane
eAR
C
C
C L
D
D

2
0
, +
=

Drag Polar
eAR
π
C
C
C
C drag
L
L
D
D
2
min
,
)
( min
−
+
=

For level, unaccelerated flight,
If thrust line is aligned with flight path,
Equations of Motion

If thrust line is
aligned with flight
path,
Level, unaccelerated flight
Equations of Motion

Thrust Required
Thrust required for steady level flight at
given speed.

D
TR =
NOTE:
Thrust Required is a
function of velocity.
It has two
components.
It has a minimum.
Thrust Required for Level, Unaccelerated Flight
at a given velocity

D
TR =
)
(
2
1
)
(
2
1
2
1
2
2
2
2
eAR
C
C
S
V
C
C
S
V
SC
V
D
T L
D
D
D
D
R o
i
o



 +
=
+
=
=
= 





)
)
2
/
1
(
(
2
1
2
2
2
eAR
S
V
L
C
S
V
T o
D
R











+
= 



)
1
)(
2
1
(
2
1
2
2
2
eAR
S
V
W
C
S
V
T o
D
R






 +
=

)
1
)(
2
1
(
2
1
2
2
2
eAR
S
V
W
C
S
V
T o
D
R






 +
=
zero-lift
thrust required
lift-induced
thrust required

Applying a first and a second derivative test to this function
will confirm the existence of a minimum. This minimum will
exist at velocity,
2
/
1
1
2
min
,








=
 S
W
eAR
C
V
o
R
D
T


)
(
)
1
)(
2
1
(
2
1
2
2
2




 =
+
= V
f
eAR
S
V
W
C
S
V
T o
D
R




2
/
1
1
2
min
,








=
 S
W
eAR
C
V
o
R
D
T



Thrust Required: Alternative Approach
D
TR =
L
D
R
C
C
W
T
W
L
=
=
Since
D
L
R
C
C
W
T
/
=
Since we have already established the existence of a minimum
thrust required, this equation implies the existence of a
maximum lift-to-drag ratio.

Indeed there is a
maximum L/D ratio
exhibited by every
aircraft.
You will see how this
ratio is an indicator of
performance
(aerodynamic efficiency)
of an aircraft.

D
L
W
C
C
W
T
D
L
R
/
/
=
=

Different
points on TR
curve
correspond to
different
angles of
attack.








+
=
=
=
=
=





eAR
C
C
S
q
SC
q
D
SC
q
SC
V
W
L
L
D
D
L
L


2
0
,
2
2
1
At a:
Large q∞
Small CL and a
D large
At b:
Small q∞
Large CL (or CL
2) and a to support W
D large

Thrust Required Computation
TR is thrust required
to fly at a given
velocity in level,
unaccelerated flight
1. Select a flight speed,V∞ and calculate CL.
S
V
W
CL
2
2
1


=

eAR
C
C
C L
D
D

2
0
, +
=
2. Calculate CD.
3. Calculate CL/CD and calculateTR.






=
D
L
R
C
C
W
T

CP-1: A light, single-engine, propeller-driven, private airplane,
approximately modelled after the Cessna Skylane, having the following
characteristics:
Wingspan = 35.8 ft
Wing area = 174 ft2
Normal gross weight = 2950 lb
Fuel capacity: 65 gal of aviation gasoline
Power plant: one-piston engine, 230 hp (SL)
Specific fuel consumption= 0.45 lb/(hp)(h)
Parasite drag coefficient CD,o = 0.025
Oswald efficiency factor, e = 0.8
Propeller efficiency = 0.8
Example

Example
357
.
0
)
174
(
)
200
)(
002377
.
0
(
2
1
2950
2
1 2
2
=
=
=

 S
V
W
CL

AtV=200 ft/s = 136.4 mi/h
37
.
7
174
)
8
.
35
( 2
2
=
=
=
S
b
AR
0319
.
0
)
37
.
7
)(
8
.
0
(
)
357
.
0
(
025
.
0
2
2
0
, =
+
=
+
=

eAR
C
C
C L
D
D
lb
263
2
.
11
2950
/
=
=
=
D
L
W
TR
2
.
11
0319
.
0
357
.
0
=
=
=
D
L
C
C
D
L
1
2
3

Example
At other velocities…
1 2 3

CJ-1: A jet-powered executive aircraft, approximately modelled after
the Cessna Citation 3, having the following characteristics:
Wingspan = 53.3 ft
Wing area = 318 ft2
Normal gross weight = 19,815 lb
Fuel capacity: 1119 gal of kerosene
Power plant: two turbofan engines of 3650-lb thrust each at sea level
Specific fuel consumption = 0.6 lb of fuel/(lb thrust)(h)
Oswald efficiency factor e = 0.81
Example

Example
210
.
0
)
318
(
)
500
)(
002377
.
0
(
2
1
19815
2
1 2
2
=
=
=

 S
V
W
CL

AtV=500 ft/s = 341 mi/h
93
.
8
318
)
3
.
53
( 2
2
=
=
=
S
b
AR
022
.
0
)
93
.
8
)(
81
.
0
(
)
21
.
0
(
02
.
0
2
2
0
, =
+
=
+
=

eAR
C
C
C L
D
D
lb
2075
55
.
9
19815
/
=
=
=
D
L
W
TR
55
.
9
022
.
0
21
.
0
=
=
=
D
L
C
C
D
L
1
2
3

How do we compute for (L/D)max?
At TRmin we found (by differentiating TR with
respect to V and equating to zero),
2
/
1
1
2
min
,








=
 S
W
eAR
C
V
o
R
D
T


i
D
L
D C
eAR
C
C ,
2
0
, =
=

From this formula for V at TRmin, the following
relationship (which has already been revealed in
the graph) can be derived:
0
,
0
,
0
, 4
/
2
/
/ D
D
D
D
L C
eAR
C
eAR
C
C
C 
 =
=
Thus,
and this is a maximum
because an (L/D)max is simultaneous with a TRmin.

i
D
L
D C
eAR
C
C ,
2
0
, =
=

How do we compute for (L/D)max?
0
,
max
4
/ D
D
L C
eAR
C
C 
=






At TRmin
Thus,

How do we compute for TRmin?
You can substitute
Or you can substitute
0
,
max 4
/
)
/
( D
D
L C
eAR
C
C 
=
)
1
)(
2
1
(
2
1
2
2
2
eAR
S
V
W
C
S
V
T o
D
R






 +
=






=
D
L
R
C
C
W
T
to to
2
/
1
1
2
min
,








=
 S
W
eAR
C
V
o
R
D
T



Effects of compressibility onTR

Effects of altitude onTR
)
1
)(
2
1
(
2
1
2
2
2
eAR
S
V
W
C
S
V
T o
D
R


 +
=
Note that the minimum thrust required is independent of altitude.
2
/
1
1
2
min
,








=
S
W
eAR
C
V
o
R
D
Lower
T


2
/
1
1
2
min
,








=
S
W
eAR
C
V
o
R
D
Higher
T


max
min
,






=
D
L
R
C
C
W
T

Propeller-Piston Engine
Jet Engine
Thrust Available

Maximum Velocity: Graphical
The intersection of the
TA and TR curve gives
Vmax at a certain
altitude.

Example
Calculate the maximum velocity for the sample jet plane.
Vmax = 975 ft/s
= 665 mi/h
Intersection of TR
curve and maximum
TA defines maximum
flight speed of airplane.

Example
Some remarks. Computation of TR
curve assumed constant CD,o
At this speed, drag
divergence effects are
significant, and adds to
the CD,o

Maximum Velocity: Analytical








+
=
=
= 

eAR
C
C
S
q
SC
q
T
D L
D
D

2
0
,
S
q
W
CL

=
eAR
S
q
W
SC
q
eAR
S
q
W
C
S
q
T D
D

 


 +
=








+
=
2
0
,
2
2
2
0
,
0
2
0
,
2
=
+
− 

eAR
S
W
T
q
SC
q D

Steady, level flight:T = D
Steady, level flight: L =W
Substitute into
drag equation
Turn this equation into a
quadratic
equation (by multiplying by q∞)
and rearranging.

Maximum Velocity: Analytical
2
1
0
,
0
,
2
max
max
max
4














−












+












=
 D
D
A
A
C
eAR
C
W
T
S
W
S
W
W
T
V


Solving the quadratic equation and setting thrust, T, to
maximum available thrust, TA,max results in,

Maximum Velocity: Design Considerations
2
1
0
,
0
,
2
max
max
max
4














−












+












=
 D
D
A
A
C
eAR
C
W
T
S
W
S
W
W
T
V


• TA,max does not appear alone, but only in ratio: (TA/W)max
• S does not appear alone, but only in ratio: (W/S)
• Vmax does not depend on thrust alone or weight alone, but rather on
ratios
• (TA/W)max: maximum thrust-to-weight ratio
• W/S: wing loading

Maximum Velocity: Design Considerations
• Vmax also depends on density (altitude), CD,0, eAR
• Increase Vmax by
• Increase maximum thrust-to-weight ratio, (TA/W)max
• Increasing wing loading, (W/S)
• Decreasing zero-lift drag coefficient, CD,0
2
1
0
,
0
,
2
max
max
max
4














−












+












=
 D
D
A
A
C
eAR
C
W
T
S
W
S
W
W
T
V



Example
CalculateVmax for the CP-1.
2
lb/ft
95
.
16
174
2950
=
=
S
W
Wingspan = 35.8 ft
Wing area = 174 ft2
Power plant: one-piston engine, 230 hp (SL)
3
-
2
0
,
10
x
4066
.
5
]
174
/
)
8
.
35
)[(
8
.
0
(
)
025
.
0
(
4
4
=
=

eAR
CD
3
5
0
,
slug/ft
10
x
9425
.
5
)
025
.
0
(
002377
.
0
−

=
=
D
C

2
1
0
,
0
,
2
max
max
max
4














−












+












=
 D
D
A
A
C
eAR
C
W
T
S
W
S
W
W
T
V



Example
lb)/s
(ft
10
x
012
.
1
)
550
)(
230
(
8
.
0 5

=
=
=
=
 P
P
V
T A
A 
( )
max
max
V
P
TA

=
For maxTA and PA,V∞ =Vmax
max
max
max
305
.
34
1
V
V
W
P
W
TA
=
=





 
?
max
=






W
TA
2
1
0
,
0
,
2
max
max
max
4














−












+












=
 D
D
A
A
C
eAR
C
W
T
S
W
S
W
W
T
V



Example
2
/
1
3
2
max
max
max
max
max 10
x
4066
.
5
305
.
34
305
.
34
97
.
558








−








+








= −
V
V
V
Solve this by trial and error.

Jets Engines are usually rated inThrust
Thrust is a Force with units (N = kg m/s2)
For example, the PW4000-112 is rated at 98,000 lb of thrust
Piston-Driven Engines are usually rated in terms of Power
Power is a precise term and can be expressed as:
Energy /Time with units (kg m2/s2) / s = kg m2/s3 = Watts
Note that Energy is expressed in Joules = kg m2/s2
Force *Velocity with units (kg m/s2) * (m/s) = kg m2/s3 =Watts
Usually rated in terms of horsepower (1 hp = 550 ft lb/s = 746W)
Why is there a need for a new parameter?

PR vs. V∞ curve qualitatively
resembles TR vs. V∞ curve.
Power Required
PR = TRV∞

Power Required
NOTE:
Power Required is
a function of
velocity.
It has two
components.
It has a minimum.

Power Required
)
1
)(
2
1
(
2
1
2
2
2
eAR
S
V
W
C
S
V
T o
D
R






 +
=
)
1
)(
2
1
(
2
1 2
3
eAR
S
V
W
C
S
V
P o
D
R




 +
=
zero-lift power required lift-induced power required

= V
T
P R
R

V
zero-lift PR ~V3
lift-induced PR ~ 1/V

Power Required, Minimum
)
(
)
1
)(
2
1
(
2
1 2
3


 =
+
= V
f
eAR
S
V
W
C
S
V
P o
D
R



Get f’(V∞).
Equate to zero.
Solve forV∞ in f’(V∞)=0 to getVPR,min.
Substitute V∞ in f(V∞) to get PR,min.
The results are…

Power Required, Minimum
2
1
0
,
,
3
1
2
min
, 







=


S
W
eAR
C
V
D
PR


The results are…
At PRmin ,
i
D
D C
C =
0
,
3
and

Power Required
2
1
0
,
,
3
1
2
min
, 







=


S
W
eAR
C
V
D
PR


i
D
D C
C =
0
,
3
2
/
1
1
2
min
,








=
 S
W
eAR
C
V
o
R
D
T


min
,
min
,
4
1
3
1
R
R T
P V
V 





=


 =
= V
C
C
W
V
T
P
D
L
R
R
Power Required: Alternative Approach
L
SC
V
W
L
2
2
1


=
= 
L
SC
W
V

 =

2
L
D
L
R
R
SC
W
C
C
W
V
T
P

 =
=

2
C
1
2
2
/
3
L
3
2
3
D
L
D
R
C
SC
C
W
P a

= x

C
1
2
max
D
2
3
L
3
min
,








=
 C
S
W
PR

2
3
2
3
L
D
R
SC
C
W
P

=

Power Required: Alternative Approach

Example

= V
T
P R
R
Calculate the power required curve for (a) the CP-1 at sea level and
(b) the CJ-1 at an altitude of 22,000 ft.

Example

= V
T
P R
R
3
slug/ft
001184
.
0
=


At an altitude of 22,000 ft
Thrust required is re-computed using this density.

How do we compute for (L3/2/D)max?
eAR
C
C
C L
D
D i

2
0
,
3 =
=
( ) 4
3
3
1
0
,
0
,
4
3
0
,
max
2
3
3
4
1
4
3








=
=








D
D
D
D
L
C
eAR
C
eAR
C
C
C 

eAR
C
C D
L 
0
,
3
=

i
D
D C
C =
0
,
3
How do we compute for (L3/2/D)max?
At PRmin
Thus,
4
3
3
1
0
,
max
2
3
3
4
1








=








D
D
L
C
eAR
C
C 

Locating (L/D)max in the PR curve

How do we compute for PR,min?
You can substitute
Or you can substitute
to to
2
1
0
,
,
3
1
2
min
, 







=


S
W
eAR
C
V
D
PR


)
1
)(
2
1
(
2
1 2
3
eAR
S
V
W
C
S
V
P o
D
R




 +
=
4
3
3
1
0
,
max
2
3
3
4
1








=








D
D
L
C
eAR
C
C 
C
1
2
D
2
3
L
3








=
 C
S
ρ
W
PR

Effects of altitude on PR
2
1
0
0
,
,
2
1
0
0








=








=




R
ALT
R
ALT
P
P
V
V
C
1
2
D
2
3
L
3








=
 C
S
W
PR

2
1
0
,
,
3
1
2
min
, 







=


S
W
eAR
C
V
D
PR



Effects of altitude on PR
2
1
0
0
,
,
2
1
0
0








=








=




R
ALT
R
ALT
P
P
V
V

SUMMARY
thrust required
power required
)
1
)(
2
1
(
2
1 2
3
eAR
S
V
W
C
S
V
P o
D
R




 +
=
C
1
2
D
2
3
L
3








=
 C
S
W
PR

)
1
)(
2
1
(
2
1
2
2
2
eAR
S
V
W
C
S
V
T o
D
R






 +
=
D
L
R
C
C
W
T
/
=

SUMMARY
At minimum thrust required At minimum power required
i
D
D C
C ,
0
, =
0
,
max
4
/ D
D
L C
eAR
C
C 
=






i
D
D C
C =
0
,
3
2
1
0
,
,
3
1
2
min
, 







=


S
W
eAR
C
V
D
PR


2
/
1
1
2
min
,








=
 S
W
eAR
C
V
o
R
D
T


( ) 4
3
3
1
0
,
0
,
4
3
0
,
max
2
3
3
4
1
4
3








=
=








D
D
D
D
L
C
eAR
C
eAR
C
C
C 


Power Available VS Thrust Available

Effects of Altitude on Maximum Velocity

Minimum Velocity
Sometimes
minimum or
stall velocity
is dictated by
powerplant
considerations.
It is true,
Chuck Norris’
legendary
kick can also
cause a stall,
but…

Rate of Climb


cos
sin
W
L
W
D
T
=
+
=

Rate of Climb

sin


 +
= WV
DV
TV

sin
W
D
T +
=

sin
/ 
=V
C
R

sin



=
−
V
W
DV
TV
W
DV
TV
C
R 
 −
=
/

Rate of Climb
W
DV
TV
C
R 
 −
=
/
Power Available ~ Power Required
(for small Ѳ)

sin
W
D
T +
=

 −
= DV
TV
power
excess
W
C
R
power
excess
/ =

Rate of Climb
W
C
R
power
excess
maximum
)
/
( max =

Example
Calculate the rate of climb vs velocity at sea level for (a) the CP-1
and (b) the CJ-1.
ft/min
1398
ft/s
3
.
23
2950
32600
-
10120
2950
P
P
power
excess
)
/
( R
A
=
=
=
−
=
=
W
C
R
AtV = 150 ft/s PR = 32,600 ft-lb/s and PA = 10,120 ft-lb/s. Hence,

Example
ft/min
7914
ft/s
132
19815
1884
-
6636
550
19815
P
P
power
excess
)
/
( R
A
=
=
=
−
=
=
W
C
R
AtV = 500 ft/s PR = 1884 hp and PA = 6636 hp. Hence,

R/Cmax: Analytical
( ) ( )2
max
2
max /
/
3
1
1
W
T
D
L
Z +
+
=
( ) ( )
( ) ( ) 





−
−














=
 Z
D
L
W
T
Z
W
T
C
Z
S
W
C
R
D
2
max
2
max
2
/
3
max
2
/
1
0
,
max
/
/
2
3
6
1
3
/
/

( )
( ) 2
/
3
max
0
,
max
max
/
1
/
8776
.
0
/
D
L
C
S
W
W
P
C
R
D

−






=


For a piston-propeller aircraft:
For a jet aircraft:
Where:

Service Ceiling
min
/
100
)
/
( max ft
C
R =

Example
Calculate the absolute and service ceilings for (a) the CP-1 and (b) the CJ-
1.
W
C
R
power
excess
maximum
)
/
( max =

Example
Calculate the absolute and service ceilings for (a) the CP-1 and (b) the CJ-
1. (a) the CP-1 (b) the CJ-1
service ceilings = 25,000 ft
absolute ceilings = 27,000 ft
service ceilings = 48,000 ft
absolute ceilings = 49,000 ft

Time to Climb
dt
dh
C
R =
/
C
R
dh
dt
/
=

=
2
1
h
h
dt
t

=
2
1
/
h
h
C
R
dh
t

Time to Climb: Graphical

=
2
1
/
h
h
C
R
dh
t

Time to Climb
b
mx
y +
=
0
max
0
max,
0
)
/
(
)
/
(
H
C
R
C
R
H
H +
−
=
0
H
0
max,
)
/
( C
R

=
H
C
R
dh
t
0 max
)
/
(
)
(
)
/
(
)
/
( 0
0
0
max,
max H
H
H
C
R
C
R −
=
 −
=
H
H
H
dh
C
R
H
t
0 0
0
max,
0
)
/
(








−
=
H
H
H
C
R
H
t
0
0
0
max,
0
ln
)
/
(
Altitude,
H
Maximum Rate of Climb, (R/C)max

Gliding Flight
D
L
L
D
1
tan
cos
sin
=
=





cos
sin
0
W
L
W
D
T
=
=
=

Gliding Flight
D
L
1
tan =

D
L
1
tan 1
−
=

max
1
min
1
tan






= −
D
L


Gliding Flight
D
L
1
tan 1
−
=

D
L
h
h
R =
=

tan
max
max )
(
tan D
L
h
h
R =
=


Gliding Flight
To maximize range, glide at smallest  (at (L/D)max )
A modern sailplane may have a glide ratio as high as 60:1
So  = tan-1(1/60) ~ 1°


Example
Calculate the minimum glide angle and the maximum range measured
along the ground covered by the CP-1 and the CJ-1 in a power-off glide
that starts at an altitude of 10,000 ft.
10,000 ft

CP-1: A light, single-engine, propeller-driven, private airplane,
approximately modelled after the Cessna Skylane, having the following
characteristics:
Example
Aspect Ratio = 7.37

Example
along the ground covered by the CP-1 in a power-off glide that starts at
an altitude of 10,000 ft.
( )

=
=
= −
2
.
4
61
.
13
1
1
tan
max
1
min
D
L

ft
136,000
)
61
.
13
(
10000
)
( max
max
=
=
=
D
L
h
R
10,000 ft

CJ-1: A jet-powered executive aircraft, approximately modelled after
the Cessna Citation 3, having the following characteristics:
Example
Aspect Ratio = 8.93
( )
9
.
16
)
02
.
0
(
4
/
)
93
.
8
)(
81
.
0
(
4
/ 0
,
max
=
=
=

 D
C
eAR
D
L

Example
along the ground covered by the CJ-1 in a power-off glide that starts at
an altitude of 10,000 ft.
( )

=
=
= −
39
.
3
9
.
16
1
1
tan
max
1
min
D
L

ft
169,000
)
9
.
136
(
10000
)
( max
max
=
=
=
D
L
h
R
10,000 ft

Example
For the CP-1, calculate the equilibrium glide velocities at altitudes of
10,000 ft and 2,000 ft, each corresponding to the minimum glide angle.
L
SC
V
W
L
2
2
1
cos 

=
= 

S
W
C
V
L

 =


cos
2
i
D
L
D C
eAR
C
C ,
2
0
, =
=

CL corresponding to
(L/D)max
At
(L/D)max
eAR
C
C D
L 
0
,
=
634
.
0
)
37
.
7
)(
8
.
0
(
)
025
.
0
(
=
=
L
L
C
C 

Example
For the CP-1, calculate the equilibrium glide velocities at altitudes of
10,000 ft and 2,000 ft, each corresponding to the minimum glide angle.
S
W
C
V
L

 =


cos
2
2
lb/ft
95
.
16
174
2950
=
=
S
W

= 2
.
4
min

)
634
.
0
(
0017556
.
0
)
95
.
16
)(
2
.
4
cos
2
( 
=

V
ft
10,000
h
at
ft/s
3
.
174 =
=

V
)
634
.
0
(
0022409
.
0
)
95
.
16
)(
2
.
4
cos
2
( 
=

V
ft
2,000
h
at
ft/s
3
.
154 =
=

V

f
W
W
W +
= 1
dt
dW
dt
dW f
=
Weight Equation
W –Weight of the airplane at any instant during flight.
W0 – Gross weight of the airplane, including everything: full fuel
load, payload, crew, structures, etc.
Wf – Weight of fuel: this is an instantaneous value, and it
changes as fuel is consumed during flight.
W1 –Weight of the airplane when the fuel tanks are empty.
f
W
W 
 =

( )( )
hour
BHP
fuel
of
lb
SFC =
SFC VS TSFC
( )( )
hour
thrust
of
lb
fuel
of
lb
TSFC =
P
dt
dW
P
W
c
f
f
−
=
−
=

T
dt
dW
T
W
c
f
f
t −
=
−
=

pr
t
V
c
c


=

Range: Piston-Propeller
( )( )

=
=
V
mile
(HP)
fuel
of
lb
hour
HP
fuel
of
lb
SFC
( )
( )
R
T
SFC)
(
V
HP
SFC
mile
fuel
of
lb



To cover longest distance use minimum pounds of fuel per mile.
To cover longest distance fly at minimum thrust required.

dt
V
ds
dt
ds
V 
 =

=
T
c
dW
dt
T
dt
dW
c
t
f
f
t −
=

−
=








−
= 
T
c
dW
V
ds
t
f
f
W
W 
 =








−
= 
T
c
dW
V
ds
t
W
W
T
c
dW
V
ds
t








−
= 
W
dW
D
L
c
V
W
L
D
c
dW
V
ds
t
t

 −
=








−
=

W
dW
D
L
c
V
ds
t

−
=

 
−
=
=
1
0
0
W
W t
R
W
dW
D
L
c
V
ds
R

 
−
=
=
1
0
0
W
W
t
R
W
dW
D
L
c
V
ds
R
Assumptions made: level,
unaccelerated flight with
constantTSFC and L/D.
1
0
ln
W
W
D
L
c
V
R
t

=
BREGUET RANGE
EQUATION
1
0
ln
W
W
D
L
c
R
pr

=
pr
t
V
c
c


=

1
0
ln
W
W
D
L
c
R
pr

=
To maximize range:
Fly at largest propeller efficiency
Lowest possible SFC
Highest ratio of W0 toW1 (fly with the largest fuel weight)
Fly at maximum L/D (minimumTR)
propulsion
aerodynamics
structures
and materials

Example
Estimate the maximum range for the CP-1.
1
0
max
max ln
W
W
D
L
c
R
pr






=


Example
Estimate the maximum range for the CP-1.
( ) 61
.
13
4
/ 0
,
max =
= D
C
eAR
D
L 
1
-
7
ft
10
x
27
.
2
s
3600
h
1
lb/s
-
ft
550
hp
1
(hp)(h)
lb
45
.
0 −
=
=
c
lb
367
)
64
.
5
(
65 =
=
f
W
Since aviation gasoline weighs 5.64 lb/gal,
lb
2583
367
2950
1 =
−
=
W
mi
1207
ft
10
x
38
.
6
2583
2950
ln
)
62
.
13
(
10
x
27
.
2
8
.
0
ln 6
7
1
0
max
max =
=






=






= −
W
W
D
L
c
R
pr


Range: Jet Aircraft
( ) 

V
T
TSFC A
)
(
mile
fuel
of
lb
To cover longest distance use minimum pounds of fuel per mile.
To cover longest distance fly at maximum L1/2/D.
( )( ) ( )

=
=
V
miles
thrust
of
lb
fuel
of
lb
hour
thrust
of
lb
fuel
of
lb
TSFC
D
L
D
L
R
C
C
C
SC
W
S
V
T
2
1
1
2
2
1

=


 


Range: Jet Aircraft
 
−
=
1
0
W
W t W
dW
D
L
c
V
R
L
SC
W
V 
 = 
2
 
−
=
1
0
2
1
2
1
2
W
W t
D
L
W
dW
c
C
C
S
R

)
(
2
2 2
1
1
2
1
0
2
1
W
W
C
C
S
c
R
D
L
t
−
=


Assumptions made: level,
unaccelerated flight with
constantTSFC and L1/2/D.

Range: Jet Aircraft
To maximize range:
Fly at minimumTSFC
Maximum fuel weight
Maximum L1/2/D
Fly at high altitudes (low density)
)
(
2
2 2
1
1
2
1
0
2
1
W
W
C
C
S
c
R
D
L
t
−
=



How is (L1/2/D)max computed?
)
(
2
0
,
2
/
1
2
/
1
L
L
D
L
D
L
C
f
KC
C
C
C
C
=
+
=
( )
( )
0
)
2
(
)
2
/
1
(
)
(
' 2
2
0
,
2
/
1
2
/
1
2
0
,
=
+
−
+
=
−
L
D
L
L
L
L
D
L
KC
C
KC
C
C
KC
C
C
f
( ) 0
)
2
(
)
2
/
1
(
2
/
1
2
/
1
2
0
, =
−
+
−
L
L
L
L
D KC
C
C
KC
C
i
D
L
D C
KC
C ,
2
0
, 3
3 =
=
πeAR
K /
1
Where =

How is (L1/2/D)max computed?
K
C
C
KC
C D
L
L
D 3
3 0
,
2
0
, =

=
0
,
0
,
0
, )
3
/
4
(
)
3
/
1
( D
D
D
D C
C
C
C =
+
=
0
,
,
,
0
, )
3
/
1
(
3 D
i
D
i
D
D C
C
C
C =

=
( ) 4
/
1
3
0
,
0
,
2
/
1
0
,
max
2
/
1
)
(
1
256
27
)
3
/
4
(
3








=
=








D
D
D
D
L
C
K
C
K
C
C
C

Summary
4
/
1
3
0
,
max
2
/
1
)
(
1
256
27








=








D
D
L
C
K
C
C
( ) )
4
/(
1 0
,
max D
D
L KC
C
C =
4
3
3
1
0
,
max
2
3
3
4
1








=








D
D
L
KC
C
C
πeAR
K /
1
Where =
i
D
D C
C ,
0
, 3
=
i
D
D C
C ,
0
, =
i
D
D C
C ,
0
,
3 =

Example
Estimate the maximum range for the CJ-1.
)
(
2
2 2
1
1
2
1
0
max
2
1
max W
W
C
C
S
c
R
D
L
t
−








=


Normal gross weight = 19,815 lb
Fuel capacity: 1119 gal of kerosene
Specific fuel consumption = 0.6 lb of fuel/(lb thrust)(h)

Example
1
-
4
s
10
x
667
.
1
s
3600
h
1
(lb)(h)
lb
6
.
0 −
=
=
t
c
lb
7463
)
67
.
6
(
1119 =
=
f
W
Since kerosene weighs 6.67 lb/gal,
lb
12352
7463
19815
1 =
−
=
W
4
.
23
)
02
.
0
(
)
93
.
8
)(
81
.
0
(
256
27
)
(
1
256
27
4
/
1
3
4
/
1
3
0
,
max
2
/
1
=








=








=







 
D
D
L
C
K
C
C

Example
)
(
2
2 2
1
1
2
1
0
max
2
1
max W
W
C
C
S
c
R
D
L
t
−








=


)
12352
19815
)(
4
.
23
(
)
318
(
001184
.
0
2
10
x
667
.
1
2 2
1
2
1
4
max −
= −
R
miles
3630
ft
10
x
2
.
19 6
max =
=
R

Endurance: Piston-Propeller
( )( )
hour
HP
fuel
of
lb
SFC =
To stay in the air for the longest time,
fly at minimum pounds of fuel per hour.
For maximum endurance, fly at minimum power required.
( )
)
(SFC)(P
hour
fuel
of
lb
R
a


/

= DV
P
cP
dW
dt
P
dt
dW
c −
=

−
=
1
 
 
=
=
=
0
1
0
1
0
W
W
W
W
E
DV
dW
c
cP
dW
dt
E

L
SC
W
V 
 = 
2
 
=
0
1
2
3
2
W
W
L
D
L
W
dW
SC
C
C
c
E


W
dW
DV
L
c
E
W
W
 
=
0
1

( ) ( )
2
1
0
2
1
1
2
1
2
3
2
−
−
 −
= W
W
S
C
C
c
E
D
L


Assumptions made: level, unaccelerated
flight with constant SFC, η and L3/2/D.

( ) ( )
2
1
0
2
1
1
2
1
2
3
2
−
−
 −
= W
W
S
C
C
c
E
D
L


To maximize endurance, fly at:
Largest propeller efficiency, η
Lowest possible SFC
Largest fuel weight
Fly at maximum CL
3/2/CD
Flight at sea level

Example
Estimate the maximum endurance for the CP-1.
81
.
12
3
4
1
4
3
3
1
0
,
max
2
3
=








=








D
D
L
C
eAR
C
C 
( ) ( )
2
1
0
2
1
1
2
1
max
2
3
max 2
−
−
 −








= W
W
S
C
C
c
E
D
L


  





−
= − 2
/
1
2
/
1
2
1
7
2950
1
2583
1
)
174
)(
002377
.
0
(
2
)
81
.
12
(
10
x
7
.
2
8
.
0
E
h
4
.
14
s
10
x
19
.
5 4
=
=
E

Endurance: Jet Aircraft
( )( )
hour
thrust
of
lb
fuel
of
lb
TSFC =
To stay in the air for the longest time,
fly at minimum pounds of fuel per hour.
For maximum endurance, fly at minimum thrust required.
( )
)
(TSFC)(T
)
(TSFC)(T
hour
fuel
of
lb
R
A a
a

A
t
A
t
T
c
dW
dt
T
dt
dW
c −
=

−
=
1

 −
=
=
1
0
0
W
W A
t
E
T
c
dW
dt
E

−
=
1
0
1
W
W t W
dW
D
L
c
E
1
0
ln
1
W
W
C
C
c
E
D
L
t
=
Assumptions made:
level, unaccelerated
flight with constant
TSFC and L/D.

1
0
ln
1
W
W
C
C
c
E
D
L
t
=
To maximize endurance, fly at:
MinimumTSFC
Maximum fuel weight
Maximum L/D

Example
Estimate the maximum endurance for the CJ-1.
h
3
.
13
s
10
x
79
.
4 4
=
=
E
1
0
max
max ln
1
W
W
C
C
c
E
D
L
t








=
12352
19815
ln
)
9
.
16
(
10
x
667
.
1
1
4
max −
=
E

Coverage Airplane Performance
Static Performance
(zero acceleration)
Dynamic Performance
Thrust Required
Thrust Available
Maximum
Velocity
Power Required
Power Available
Maximum Velocity
Rate of Climb
Takeoff
Landing
V-n Diagram
Turning Flight
Range and Endurance
Time to Climb
Maximum Altitude
Gliding Flight
Service Ceiling
Absolute Ceiling

Ground Roll (Liftoff Distance)
Preliminary (purely kinematic) considerations
dt
dV
m
ma
F =
=
dt
m
F
dV =
t
m
F
dt
m
F
dV
V
t
V
=
=
= 
 '
'
0
0
tdt
m
F
Vdt
ds =
=
2
'
'
'
2
0
0
t
m
F
dt
t
m
F
ds
s
t
s
=
=
= 

F
m
V
F
Vm
m
F
s
2
2
1 2
2
=






=

Ground Roll (Liftoff Distance)
Rolling resistance
mr = 0.02 relatively smooth paved surface
mr = 0.10 grass field
( )
dt
dV
m
L
W
D
T
R
D
T
F r =
−
−
−
=
−
−
= m
Forces in an aircraft during takeoff ground roll

Coefficient of Rolling Friction

Ground Roll
F
m
V
s
2
2
=
Is the assumption of a
constant force
reasonable?
( )
L
W
D
T
F r −
−
−
= m

Ground Roll
L
SC
V
L
2
2
1


= 
constant force
reasonable?








+
= 

eAR
C
C
S
V
D L
D



2
2
0
2
1
( )
( )2
2
16
1
16
b
h
b
h
+
=


Ground Effect
( )
( )2
2
16
1
16
b
h
b
h
+
=

Reduction of induced drag
by a factor Φ≤1.








+
= 

eAR
C
C
S
V
D L
D



2
2
0
2
1

Ground Roll
constant force
reasonable?
T is approximately constant
(especially for a jet)
The difference between the
drag and friction combined
and the thrust is also
approximately constant
( ) constant?
=
−
−
−
= L
W
D
T
F r
m

Ground Roll
AssumeT is constant.
Assume an average value
ofT-[D+μR(W-L)].
( ) ave
r
eff L
W
D
T
F ]
[ −
−
−
= m
Shevell suggests computing
this average atV=0.7VLO.
eff
LO
LO
F
g
W
V
s
2
)
(
2
=

Ground Roll
max
,
2
2
.
1
2
.
1
L
stall
LO
SC
W
V
V

=
=

}
)]
(
[
{
44
.
1
max
,
2
ave
R
L
LO
L
W
D
T
SC
g
W
s
−
+
−
=
 m

T
SC
g
W
s
L
LO
max
,
2
44
.
1

=


Ground Roll
}
)]
(
[
{
44
.
1
max
,
2
ave
R
L
LO
L
W
D
T
SC
g
W
s
−
+
−
=
 m
 T
SC
g
W
s
L
LO
max
,
2
44
.
1

=

Lift-off distance:
Is very sensitive to weight; varies asW2
Depends on ambient density
May be decreased by:
Increasing wing area, S
Increasing CL,max
Increasing thrust,T

Example
Estimate the liftoff distance for the CJ-1 at sea level. Assume a paved
runway; hence, μr = 0.02. Also, during the ground roll, the angle of
attack of the airplane is restricted by the requirement that the tail not
drag the ground; therefore, assume that CL,max during ground roll is
limited to 1.0. Also, when the airplane is on the ground, the wings are 6
ft above the ground.
( )
( )
764
.
0
16
1
16
2
2
=
+
=
b
h
b
h


Example
ft/s
230
)
0
.
1
)(
318
(
002377
.
0
)
19815
(
2
2
.
1
2
2
.
1
2
.
1
max
,
=
=
=
=
 L
stall
LO
SC
W
V
V

ft/s
3
.
160
7
.
0 =
LO
V
lb
9712
)
0
.
1
)(
318
(
)
3
.
160
)(
002377
.
0
)(
2
/
1
(
2
1 2
2
=
=
= 
 L
SC
V
L 
lb
7
.
520
)
93
.
8
)(
81
.
0
(
0
.
1
764
.
0
02
.
0
)
318
(
)
3
.
160
)(
002377
.
0
(
2
1
2
1
2
2
2
2
0
=






+
=








+
= 





eAR
C
C
S
V
D L
D

Example
}
)]
(
[
{
44
.
1
max
,
2
ave
R
L
LO
L
W
D
T
SC
g
W
s
−
+
−
=
 m

)]}
9712
19815
)(
02
.
0
(
7
.
520
[
7300
){
0
.
1
)(
318
)(
002377
.
0
(
2
.
32
)
19815
(
44
.
1 2
−
+
−
=
LO
s
ft
3532
=
LO
s

Total Takeoff Distance
Total takeoff distance as per FAR
35 ft (jet-powered civilian transport)
50 ft (all other airplanes)
ground roll

Balanced Field Length
A + B
Distance up toV1
Additional distance travelled =
Distance required to clear an obstacle
= Distance required for a full stop

Distance to clear obstacle

sin
R
sa =
Where,
g
V
R stall
2
)
(
96
.
6
=
)
1
(
cos 1
R
h
−
= −

h is the obstacle height.
Analysis is based on pull up maneuver

Landing Roll
 
=
0
0
'
'
L
s
t
dt
t
m
F
ds
2
2
t
m
F
sL −
=
F
m
V
sL
2
2
−
=
Can we assume a
constant landing
force just as we did
in takeoff
performance?

Landing Roll
( )
dt
dV
m
L
W
D
R
D
F r =
−
+
−
=
+
−
= ]
[
)
( m
( )
dt
dV
m
L
W
D
T
R
D
T
F r =
−
−
−
=
−
−
= m
0 0

Landing Roll
( )
dt
dV
m
L
W
D
F r =
−
+
−
= ]
[ m
Assume a constant
effective force,
( ) ave
r
eff L
W
D
F ]
[ −
+
−
= m
Compute this average by
evaluating the quantity at
0.7VT , where VT is the
touchdown velocity.

Landing Roll
F
m
V
sL
2
2
−
=
T
V
R
T
L
L
W
D
g
W
V
s
7
.
0
2
)]
(
[
2
)
/
(
−
+
−
=
m
max
,
2
3
.
1
3
.
1
L
stall
T
SC
W
V
V

=
=

T
V
R
L
L
L
W
D
SC
g
W
s
7
.
0
max
,
2
)]
(
[
69
.
1
−
+
−
=
 m

μR = 0.4 for paved surface

Landing Roll
T
V
R
R
L
L
L
W
D
T
SC
g
W
s
7
.
0
max
,
2
)]
(
[
69
.
1
−
+
+
−
=
 m

with reverse thrust
0
with spoilers

Example
Estimate the landing ground roll distance at sea level for the CJ-1. No
thrust reversal is used; however, spoilers are employed such that L = 0.The
spoilers increase the zero-lift, drag coefficient by 10 percent.The fuel
tanks are essentially empty, so neglect the weight of any fuel carried by
the airplane.The maximum lift coefficient, with flaps fully employed at
touchdown, is 2.5.
ft/s
6
.
148
)
5
.
2
)(
318
(
002377
.
0
)
12353
(
2
3
.
1
2
3
.
1
3
.
1
max
,
=
=
=
=
 L
stall
T
SC
W
V
V

ft/s
104
7
.
0 =
T
V
022
.
0
)
02
.
0
(
1
.
0
02
.
0
0
, =
+
=
D
C

Example
0
0 =

= L
CL
lb
9
.
89
)
022
.
0
)(
318
(
)
104
)(
002377
.
90
2
1
2
1 2
2
0
=
=
= 
 D
SC
V
D 
T
V
R
L
L
W
D
SC
g
W
s
7
.
0
max
,
2
)
(
69
.
1
m
 +
−
=

ft
842
)]
12352
(
4
.
0
9
.
89
)[
5
.
2
)(
318
)(
002377
.
0
(
2
.
32
)
12353
(
69
.
1 2
=
+
−
=
L
s



Approach Distance

cos
W
L =

sin
W
T
D +
=
W
T
D
L
W
T
W
D
−

−
=
1
sin

cos
R
R
hf −
=
g
V
R
f
2
.
0
2
=

tan
50 f
a
h
s
−
=
from pull up maneuver analysis

Flare Distance

sin
R
sf =


W
L =

cos
Load Factor
Turn Radius
Turn Rate
Level Turn

 −
=
=

V
n
g
R
V
dt
d 1
2


2
2
W
L
Fr −
=
W
L
n 
1
2
−
= n
W
Fr
R
V
m
Fr
2

=
1
2
2
−
= 
n
g
V
R

Constraints on n and V∞
At any given velocity the maximum possible load factor for a
sustained level turn is constrained by the maximum thrust
available.
2
/
1
0
,
2
max
2
max
/
2
1
)
/
(
2
1
















−






= 



S
W
C
V
W
T
S
W
K
V
n D


eAR
K

1
=

2
/
1
0
,
2
max
2
max
/
2
1
)
/
(
2
1
















−






= 



S
W
C
V
W
T
S
W
K
V
n D


max
max 





=
W
T
D
L
n
n is also constrained by
CLmax
S
W
C
V
n L
/
2
1 max
,
2
max 

= 
max
max
1
cos
n
=


V∞ is constrained by stall.
max
,
2
L
stall
C
n
S
W
ρ
V

=
n is also constrained by regulation. Example:
category)
(utility
4
.
4
=
n

Minimum Turn Radius
Minimum R occurs at the right combination of n and V∞.
)
/
(
)
/
(
4
)
( min
W
T
S
W
K
V R

 =

2
0
,
)
/
(
4
2
min
W
T
KC
n D
R −
=
2
0
,
min
)
/
/(
4
1
)
/
(
)
/
(
4
W
T
KC
W
T
g
S
W
K
R
D
−
=


1
2
2
−
= 
n
g
V
R

Maximum Turn Rate
Maximum ω occurs at the right combination of n and V∞.
4
/
1
0
,
2
/
1
)
/
(
2
)
( max 













=


D
C
K
S
W
V


2
/
1
0
,
1
/
min








−
=
D
KC
W
T
n
















−
= 

2
/
1
0
,
max
2
/
/ K
C
K
W
T
S
W
q D



−
=
V
n
g 1
2


Pull-Up Maneuver
( )
1
2
−
= 
n
g
V
R

cos
2
W
L
R
V
m −
=

( )

−
=
V
n
g 1

W
L
R
V
m −
=

2

Pull-Down Maneuver
( )
1
2
+
= 
n
g
V
R
( )

+
=
V
n
g 1

W
L
R
V
m +
=

2

For large load factors
gn
V
R
2

=

=
V
gn

R for level turn, pull-up and pull down
ω for level turn, pull-up and pull down

For large load factors
S
W
gC
R
L max
,
min
2

=

)
/
(
2
max
max
,
max
S
W
n
C
g L

=


Minimum R for level turn, pull-up and pull down
Maximum ω for level turn, pull-up and pull down

S
W
C
V
n
W
SC
V
W
L
n
L
L
max
,
2
max
2
2
1
2
1




=
=
=


V-n Diagram

Topics Discussed Airplane Performance
Static Performance
(zero acceleration)
Dynamic Performance
Thrust Required
Thrust Available
Maximum
Velocity
Power Required
Power Available
Maximum Velocity
Rate of Climb
Takeoff
Landing
V-n Diagram
Turning Flight
Range and Endurance
Time to Climb
Maximum Altitude
Gliding Flight
Service Ceiling
Absolute Ceiling

• John D. Anderson. Introduction to Flight
• John D. Anderson, Airplane Performance and Design
References

Performance.pdf

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