SME
Flight Mechanics, Stability & Control · Problem type
Bank angle for a coordinated turn
Use this for “What bank angle gives a standard-rate turn of 3°/s at 60 m/s, and what load factor does it pull?”
Diagram
Up
and
down:
L
cos
φ
=
W,
so
n
=
L
W
=
1
cos 17.76
°
=
1.05
Level:
L
sin
φ
=
mV²
R
,
the
centripetal
force
R
=
V
2
g tan φ
=
60
2
9.81 × tan 17.76°
R = 1,146 m
ω
=
V
R
=
0.05236
rad/s
=
3°/s
This is an example. Type your own values to start a new problem.
tan(
φ
°)
=
V (m/s)
×
ω (°/s)
9.81
n
=
1
cos(
°)
Formulas
tan(
φ
)
=
V
×
ω
9.81
tan(17.76°)
=
60
×
0.05236
9.81
n
=
1
cos(
φ
)
1.05
=
1
cos(17.76°)
Formulas are worked in rad/s. The step-by-step shows the conversions.
Assumptions
•
A level, coordinated turn: no sideslip, and L cos φ = W holds the height.
•
g = 9.81 m/s²; ω is used in rad/s.
•
The load factor n = L ÷ W = 1 ÷ cos φ.
Step-by-step
WE KNOW
V = 60 m/s, ω = 3 °/s
FIND
bank angle (φ), load factor (n)
Step 1 · Convert to the formula’s units
The formulas don’t work directly in the units you chose, so convert first (to rad/s).
ω = 3 °/s = 0.05236 rad/s (1 rad/s = 57.2958 °/s)
Step 2 · Find bank angle (φ)
Use tan(φ) = V × ω ÷ 9.81
Divide
L
sin
φ
=
mVω
by
L
cos
φ
=
mg,
then
take
the
inverse
tangent.
φ
=
tan⁻¹(
V
×
ω
9.81
)
φ
=
tan⁻¹(60
×
0.05236
9.81
)
φ
=
17.76°
Step 3 · Find load factor (n)
Use n = 1 ÷ cos(φ)
Up
and
down,
L
cos
φ
=
W,
so
L
W
=
1
cos
φ
.
n
=
1
cos(17.76°)
n
=
1
0.9523
n
=
1.05
Check
Put the numbers back in. Both sides should match.
tan(17.76°)
=
60
×
0.05236
9.81
✓
1.05
=
1
cos(17.76°)
✓
Topic check
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