UFC 4-159-03
3 October 2005
Figure 4-10. Example of Transverse Current Drag Coefficients
3.20
Data taken from tests conducted at the US
3.00
Naval Academy at scales 1/24.75 and 1/80.
Some data taken at 5 and 6 knots is not
2.80
included. (Kreibel, 1992)
2.60
2.40
2.20
χ = 14.89
2.00
1.80
1.60
FFG-7
1.40
= 0.78
Cm
= 124.36 m
LwL
1.20
B
= 11.58 m
T
= 4.389 m
1.00
D
= 3590 long ton (LT)
3
= 3590 LT * 0.9904 m /LT
V
0.80
3
= 3555.7 m
0.60
2
Am = 0.78 *B *T = 39.64 m
2
Model data points
χ = LwL *Am/(B*V) = 14.89
0.40
C0 = 0.8489
C1 = 3.2
0.20
K =2
0.00
0.00
0.10
0.20
0.30
0.40
0.50
0.60
0.70
0.80
0.90
1.00
T/d
88
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