Page 465 - eProceeding for IRSTC2017 and RESPeX2017

M. Dwi Trisna / JOURNAL ONLINE JARINGAN COT POLIPD

Geometry of balade from inside:
q. Central radius of the blade curvature
2
R = R  R 2  / 1 2 (mm) (17)
2 3
r. Central angle of the blades radius

R
 = arc.Sin 3
R (18)

R 2  R  R 2 
 = arc.Cos  1 3 
1
 R . 2 1 R . 
 = .   1
2
s. Length of bowstring (AB)

2
AB = R  R  . 2 R 2 .R 1 .Cos  2  (mm). (19)
2
1
2
t. Angle of blades curvature
 .2 R 2  AB 
 = arc.Cos  3 2 
 2R 3  (20)

u. Jari-jari pusat titik berat sudu
0
  1 2 .   180     1 (21)
1
2
2
R  R  R  . 2 R .R 3 .Cos  .
2
3
4
Geometry of blades from outside
v. Length of C, (mm) (22)
2
C  R  R  2 R R COS(    )
2
2
2
2
1
1
1
w. Value of  : (23)

 sin    
 R
  arcsin  2 1 2 
 c 


x. Value of  :
o
 = 180 – ( 1 +  2 +  ) (24)
y. Value of  :
o
 =  1 +  2 – (180 - 2 ) (25)
z. Value of d (mm):

 R sin    

d   2 1 2  (26)
 2sin 180   


aa. Angle of blades curvatutre:
o
 = 180 – ( 1 +  ) (27)
464 | V O L 1 0 - I R S T C 2 0 1 7 & R E S P E X 2 0 1 7