Forces in the Tip Path Plane
Assumptions: c= constant (untapered rotor)
v = constant (uniform inflow)
Cd = constant
r
   0   tw   1c cos   1s sin 
R
(linear twist)
No cut out, no tip losses.
Effective angle of attack:  eff   
Where,
UP
UT
U T  r  V sin 
U P  V sin  TPP  v  V  cos  RTPP  V  cos
We can now express the lift per unit span L’ as:
1
1
1
L   cC lU 2  caU T2 eff  ca U T2  U T U P 
2
2
2
Substituting the individual terms for UT , , and UP we get:
L 


1
r


ca r  V sin  2  0   tw   1c cos   1s sin    r  V sin  R  V  cos 
2
R




3


r2
r
2 2
2
2
2 r
 2V  tw sin    tw V2 sin 2  
 0  r  2V r 0 sin    0V sin    tw 
R
R
R


1
2 2
2
2


r

cos


2

rV

sin

cos


V

sin

cos




 ca
1c
 1c
 1c
2
 2 2

2
2
3
 r  1s sin   2rV 1s sin   V  1s sin 

  2 R r  V R sin   rV  cos  V 2  cos sin 

TPP

TPP




Notice that L’ has first, second, and third harmonics. Thus, the lift force produced by
each blade will have up to three-per-rev vibrations.
 1 2  R
 

d 

L
dr
Finally, T= b 



2

00
 

We can interchange the order of integration. We can integrate with respect to azimuth
first. Many terms will drop out. For example:
2
2
2
0
0
0
2
 sin d   cosd   sin  cosd  0
2
 sin cosd  0
0
2
2
0
0
2
2
 sin d   cos d  
With these relations, we get:
1
2
3

0 2
r
2 2
2 r


r

V



 V2 tw 
1
0

tw

2
R
R
0 Ld  2 ca 
2
 rV 1s   Rr TPP

2
Next, integrate the above expression on the right hand side with respect to R, multiply by
b. The resulting expression is:
T

1
R3 0 2
R 3  tw 2
R2
R2 
abc  0  2
 V R   tw 2

V R  V1s
  2 TPP

2
3
2
4
4
2
2 

Normalize by dividing T by (R)2(R2). After some very minor algebra, we get:
CT 
a  0 

 
3 2   tw
1   2   1s  TPP 
1    

2 3  2  4
2
2 


While using this expression, the angle 1s must be measured in the tip path plane, and
may be thought of as the angle between the tip path plane and the no-feathering plane,
viewing the helicopter from the pilot’s left side.