MIT2_019S11_SK4 - 2.019 Design of Ocean Systems Lecture 8...

Info iconThis preview shows pages 1–5. Sign up to view the full content.

View Full Document Right Arrow Icon
2.019 Design of Ocean Systems Lecture 8 Seakeeping (IV) March 4, 2011
Background image of page 1

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full DocumentRight Arrow Icon
General Response of A Floating Body in Regular Ambient Waves ζ 3 ( t )= <{ ¯ i ω t ζ 3 e } ζ 6 e Incident wave: z ζ 6 ( t <{ ¯ i ω t } x Φ tt + g Φ z =0 η ( t Φ t /g ζ 5 ( t <{ ¯ ζ 5 e } η I = a cos( ω t kx ) y ζ 2 ( t <{ ζ ¯ 2 e i i ω ω t t } ζ 1 e ~n ζ 1 ( t <{ ζ ¯ 4 e i i ω ω t t } ζ 4 ( t <{ ¯ } 6 X Equation of motion: [( M j` + A ) ζ ¨ ` + B ζ ˙ ` + C ζ ` ]= F ¯ Ej e i ω t ( j =1 ,..., 6) (1) ` =1 6 X [ ω 2 ( M + A )+ i ω B + C ] ζ ¯ j = F ( j 6) ` =1 M :6 × 6 elements of the egeneralized mass matrix A ,B × 6 elements of added mass and wave damping matrices C × 6 elements of hydrostatic restoring matrix F : 6 elements of the excitation vector ¯ Transfer function or Response Amplitude Operator (RAO): H j ( ω ζ j a ( ω ) ( j 6)
Background image of page 2
Numerical Method for Potential-Flow Problems Uniform free stream : Φ = Ux u = U,v =0 ,w 2D point source : u r Φ = m 2 π ln x 2 + z 2 = m 2 π ln r m u r = 2 π r 2D point source plus point sink : Φ = m 2 π ln p ( x + s ) 2 + z 2 p m ( x s ) 2 + z 2 ln 2 π source sink
Background image of page 3

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full DocumentRight Arrow Icon
2D doublet or dipole : source + sink, as s 0 while keeping 2 ms = µ .
Background image of page 4
Image of page 5
This is the end of the preview. Sign up to access the rest of the document.

Page1 / 12

MIT2_019S11_SK4 - 2.019 Design of Ocean Systems Lecture 8...

This preview shows document pages 1 - 5. Sign up to view the full document.

View Full Document Right Arrow Icon
Ask a homework question - tutors are online