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PONAŠANJE BRODA NA TALASIMA____________________________Ustvari, krak b predstavlja jedinu nepoznatu ujednačini...Sledib =2( aη+ gϕ) h−j Caζ+ g ϕDa bi se teret valjao zajedno s brodom, krak bmora biti unitar gabarita teretab ≤ lU protivnom, doći će do preturanja teretaUl Uslov da ne dolazi ido preturanja tereta t može senapisati u obliku2ha ⋅ + gh⋅ϕ−la ⋅ − j ⋅ϕ ≤ glηZbir četiri harmonijske funkcije...ζTo je opet harmonijska a funkcije iste frekvencije...amplitude A prUslov da ne dolazi do preturanja tereta, izraženpreko ove amplitude glasi≤ glCAprPošli smo od pretpostavke da se teret,iako je slobodno oslonjen na palubu,kreće zajedno s brodom...Ovakva kruta veza tereta i broda ćepostojati ukoliko je istovremenozadovoljena oba uslovaAkl< μ gtApr≤glU protivnom, doći će do pomeranjatereta u odnosu na brod...Pojava pomeranja tereta može imativeoma opasne posledice, i predstavljajedan od čestih uzroka havarija, pa iprevrtanja brodova koji plove na olujnimtalasima...PREDAVANJA 2009. 181
PONAŠANJE BRODA NA TALASIMA____________________________Na osnovu prethodne analize pomeranjatereta na regularnim talasima, moguće jeodrediti verovatnoću proklizavanja ipreturanja na realnim, neregularnimtalasima...Potrebno je odrediti prenosne funkcijeuvedenih amplituda A kl i A pr ...A AklprPkl= , Ppr=A ATada jeTT2S = P ( ω ) ⋅S( ω )kl kl p T p2S = P ( ω ) ⋅S( ω )pr pr p T p2⎡ ( μ g)⎤tV( A > μ g) = expkl t ⎢−⎥⎣ 2σ( kl)o ⎦2⎡ ( gl)⎤V ( A > gl) = exp−pr⎢⎥⎣ 2σ( pr)o ⎦Primer, verovatnoća preturanjaslobodno oslonjenog kontejnera unajvišem redu broda dužine 110 m∞σ ( kl) = ∫oS kl( ω p)dωp∫0∞σ ( pr) = ∫ S ( ω ) dωo pr p p0PREDAVANJA 2009. 182
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