SME
Flight Mechanics, Stability & Control
Dynamic stability
Use this for “An airplane cruises at 100 m/s with L/D = 12. Find its phugoid period, damping ratio and time to half amplitude.”
Chart
A free decay inside the envelope ±exp(−ζωₙt): ζ = 0.05893, ωₙ = 0.1387 rad/s
The
envelope
halves
every
t
1
2
=
ln 2
ζωₙ
=
84.79
s.
This is an example. Type your own values to start a new problem.
ωₚₕ (rad/s)
=
2
×
9.81
V (m/s)
ζ
=
1
2
×
T
Period
calculated
t½
Time to half amplitude
calculated
Formulas
ωₚₕ
=
√
2
×
9.81
V
0.1387
=
√
2
×
9.81
100
T
=
2π
ωₚₕ
45.29
=
2π
0.1387
ζ
=
1
√
2
× (L/D)
0.05893
=
1
√
2
× 12
t
1
2
=
ln(2)
ζ
× ωₚₕ
84.79
=
ln(2)
0.05893 × 0.1387
Assumptions
•
Lanchester’s approximation: constant angle of attack; the airplane trades speed for height.
•
Light damping, so the damped and natural periods are about equal.
•
g = 9.81 m/s².
Step-by-step
WE KNOW
V = 100 m/s, L/D = 12
FIND
phugoid frequency (ωₚₕ), period (T), damping ratio (ζ), time to half amplitude (t½)
Step 1 · Find phugoid frequency (ωₚₕ)
Use ωₚₕ = √2 × 9.81 ÷ V
Lanchester: the slower the airplane flies, the faster it swings.
ωₚₕ
=
√
2
×
9.81
100
ωₚₕ
=
1.4142
×
9.81
100
ωₚₕ
=
13.873
100
ωₚₕ = 0.13873 rad/s
Step 2 · Find period (T)
Use T = 2π ÷ ωₚₕ
One full swing is 2π radians.
T
=
2π
0.13873
T
=
45.29
s
Step 3 · Find damping ratio (ζ)
Use ζ = 1 ÷ (√2 × (L/D))
Drag damps the swing, so a cleaner airplane (higher L/D) is damped less.
ζ
=
1
√
2
× 12
ζ
=
1
1.414 × 12
ζ
=
1
16.968
ζ
=
0.058926
Step 4 · Find time to half amplitude (t½)
Use t½ = ln(2) ÷ (ζ × ωₚₕ)
The
envelope
e
−ζωₚₕ t
halves
when
ζωₚₕ
t
=
ln
2.
t
1
2
=
ln(2)
0.058926 × 0.13873
t
1
2
=
ln(2)
0.008175
t
1
2
=
0.6931
0.008175
t
1
2
=
84.79
s
Check
Put the numbers back in. Both sides should match.
0.13873
=
√
2
×
9.81
100
✓
45.29
=
2π
0.13873
✓
0.058926
=
1
√
2
× 12
✓
84.79
=
ln(2)
0.058926 × 0.13873
✓
Topic check
See what you know: 5 questions on this topic
New questions each time. Pass to mark it complete.
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