SME
Flight Mechanics, Stability & Control · Problem type
Rate of climb from excess thrust
Use this for “A 50,000 N airplane at 80 m/s has 6000 N of thrust and 3553 N of drag. Find its rate of climb and climb angle.”
Diagram
Square to the path: L = W cos γ = 50,000 × cos 2.81° = 49,940 N
Along the path: T = D + W sin γ = 3,553 + 2,447 = 6,000 N
This is an example. Type your own values to start a new problem.
RC (m/s)
=
V (m/s)
×
(
T (N)
−
D (N)
)
W (N)
sin(
γ
°)
=
T (N)
−
D (N)
W (N)
Formulas
RC
=
V
×
T
−
D
W
3.915
=
80
×
6,000 − 3,553
50,000
sin(
γ
)
=
T
−
D
W
sin(2.805°)
=
6,000 − 3,553
50,000
Assumptions
•
A steady climb at angle γ: along the path T = D + W sin γ, and square to it L = W cos γ.
•
The rate of climb RC is the vertical part of the airspeed, V sin γ.
•
The excess power (T − D)V lifts the weight; a negative RC is a descent.
Step-by-step
WE KNOW
W = 50,000 N, T = 6,000 N, D = 3,553 N, V = 80 m/s
FIND
rate of climb (RC), climb angle (γ)
Step 1 · Find rate of climb (RC)
Use RC = V × (T − D) ÷ W
The
excess
power
(
T
−
D
)
V
,
over
the
weight
it
lifts.
RC
=
80
×
6,000 − 3,553
50,000
RC
=
80
×
2,447
50,000
RC
=
195,760
50,000
RC = 3.915 m/s
Step 2 · Find climb angle (γ)
Use sin(γ) = (T − D) ÷ W
Along
the
path
the
excess
thrust
T
−
D
holds
up
W
sin
γ
.
γ
=
sin⁻¹(
T
−
D
W
)
γ
=
sin⁻¹(
6,000 − 3,553
50,000
)
γ
=
2.805°
Check
Put the numbers back in. Both sides should match.
3.915
=
80
×
6,000 − 3,553
50,000
✓
sin(2.805°)
=
6,000 − 3,553
50,000
✓
Topic check
See what you know: 5 questions on this topic
New questions each time. Pass to mark it complete.
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