Special cases of thrust constant equations
α = 0 (Hover like orientation)
α = 90 (Propeller or climb)
Hover α = 0 Thrust is aligned with the z-axis.
Consider the general expression for induced velocity
4
3 2 2
vi
V0
vi
V0
vi
+2
sin α +
−1=0
vh
vh
vh
vh
vh
sin α = 0 The induced velocity equation is a biquadratic
4 2 2
vi
V0
vi
+
−1=0
vh
vh
vh
or
2
2 s 4
vi
1 V0
1 V0
=−
+
+1
vh
2 vh
4 vh
or

1
vi
= −
vh
2
V0
vh
2
 12
s 4
1 V0
+
+ 1
4 vh
Climb/propeller α = 90
T = ρA(V0 + vi )(2vi )
T
= V0 vi + vi2
2ρA
vh2 = V0 vi + vi2
2 vi
V0
vi
+
−1=0
vh
vh
vh
But
T
2ρA
= vh for the rotor hovering at the same thrust. Solving for vvhi yields
s 2
vi
1 vi
1 V0
+1
=−
+
4 vh
vh
2 vh
Constant power equations
Variation of vi and T with forward velocity V0 for constant power. In hover the thrust that can be developed
at a given power
Pi,h = Th vh
(A)
With the same power the rotor at an angle of attack α in a freestream velocity V0 produces a T and induced
velocity vi given by the equation
Pi = Pi,h = T (V0 sin α + vi )
(B)
Dividing (B) by (A) we get
Pi
Pi,h
T
=
=1=
Pi,h
Pi,h
Th
or
−1
T
V0 sin α + vi
=
Th
vh
1
V0 sin α + vi
vh
For a hovering rotor
Pi,h = Th vh
1
Th 2
vh =
2ρA
1
Pi,h 3
vh =
2ρA
1
Th 2
Pi,h = Th
2ρA
12
3
1
2
Pi,h = Th
2ρA
3
1
Th2 = Pi,h = (2ρA) 2
2
1
3
Th = Pi,h
(2ρA) 3
From Glauerts’ hypothesis
T = ρAVd (2vi )
Where Vd2 = [V0 cos α)2 + (vi + V0 sin α)2 ]
Dividing by
Th
2ρA
2
T
2ρA
2
= (V02 + 2V0 vi sin α + vi2 )vi2
= vh4 we get
2 4
3
2 2
T
vi
V0
vi
V0
vi
=
+2
sin α +
Th
vh
vh
vh
vh
vh
" 3
2 2 # 2
4
vi
V0
vi
V0
vi
Th
1=
+2
sin α +
vh
vh
vh
vh
vh
T
But
−1
T
V0 sin α + vi
=
Th
vh
Th
V0 sin α + vi
=
T
vh
"
vi
vh
4
+2
V0
vh
vi
vh
3
sin α +
V0
vh
2 vi
vh
2 # V0 sin α + vi
vh
2
−1=0
Solve this equation for given V0 , P , and α. For constant power calculations
1
1. vh = (Pi,h /2ρA) 3
2. For V0 /vh and α solve
" 3
2 2 # 2
4
vi
V0
vi
V0
vi
V0 sin α + vi
+2
sin α +
−1=0
vh
vh
vh
vh
vh
vh
to obtain
3.
T
Th
=
h
vi
vh
from which we get
V0 sin α+vi
vh
i−1
2