POWER OFF GLIDING FLIGHT





AA is absolute altitude; GD is glide distance; a = glide angle.
Glide ratio GR = GD / AA
Tan a = AA / GD, so GD / AA = 1/tan a
L = W cos a is the component of weight supported by lift.
D =W sin a is the component of weight that propels the aircraft down the
glide path at a constant glide speed.
 For a given AA, GD is maximum when the glide angle a is minimum
 The small right and large right triangular above are similar (same angles):
this implies that the ratios of any two corresponding sides are equal. Thus:
GR =
GD W cos a
1
L


 . Thus
AA W sin a tan a D
GD = GR (AA) = (L/D) (AA) = (AA) / tan a
AA
W sin a
a
W cos a
a
GD
1
Key Points
 Glide angle a is minimum and glide distance GD is maximum at DMIN,
i.e. when drag is minimum (proven in text).
 Thus best glide speed GS is obtained at the airspeed corresponding to
(L/D)MAX, because this is where D is minimum on the DT curve.
 Best glide airspeed is affected by weight changes according to the
formula V2 = V1 (W2 / W1). However, AOA αBG for best glide is
invariant for fixed configuration.
 A pilot who deviates from best glide speed shortens glide distance,
appreciably if the deviation is large (which results in drag significantly
higher than DMIN).
 AA is often specified in feet and GL in nautical miles (nm) or statute
miles (sm). To use the formula GR = GD/AA or any related formula,
GD and AA must be in the same units.
 A nautical mile has 6076 feet, and a statute mile has 5280 feet. To
convert feet to nm (or sm), divide by 6076 (or 5280).
GD = GR (AA) = (L/D) (AA) = (AA) / tan a.
Example 1:
AA = 40,000 feet; GR = 15.0. Find GD in nm.
GD =GR (AA) = 15.0
40,000 ft
=98.74917709 nm.
ft
6076
nm
Example 2:
Glide angle = 5o. Absolute altitude is 10,000 feet. Find glide distance in statute
miles. Also find GR
GD = GR (AA) = AA / tan a =


 10,000 ft 


 5280 ft 
sm  =21.64782633

tan 5
sm.
GR = 1 / tan a = 1 / tan 5 = 11.43005246 = GD / AA = 21.64782633 sm =


 10,000 ft 


 5280 ft 
sm 

11.43005230 (the difference in the two answers is due to calculator arithmetic
algorithms and imperfect mapping between base two calculations and the base
10 numeric display of results).
2
Example 3:
Can glide 40 nm from 20,000 feet. Find GR.
GR = GD / AA =
40 nm


 20,000 ft 


 6076 ft

nm 

= 12.1520000.
Example 4:
VBG = 240 KEAS at 750,000#. Find best glide speed at 550,000#.
VBG = V2 = V1 (W2 / W1) = 240 (550,000 / 750,000) = 205.5237213 KEAS.
Note: the AOA for VBG is invariant—i.e. the same for both airspeeds.
Example 5:
VBG = 63 KEAS  63 KIAS for light aircraft at 2100#. Find VBG at 2400#.
VBG = V2 = V1 (W2 / W1) = 63 (2400 / 2100) = 67.34983298 KEAS.
A/C approximates the glide performance of the C172.
QUESTION: WHICH GLIDES FURTHER FROM THE SAME ALTITUDE,
A C172 OR A BOEING 767?
 A C172 has a high camber, high lift straight wing which facilitates low
takeoff and landing airspeeds.
 A B767 has a thin, low camber swept wing that requires high airspeeds to
develop adequate lift for takeoff and landing.
 A factor other than lift is relevant in addressing this question. What is it?
 We will understand the answer to this question more thoroughly when we
discuss drag polars in the next lesson.
3