Propulsion And Manoeuvring Systems
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- เผยแพร่เมื่อ 10 ธ.ค. 2016
- This video will give you a general overview of the most common propulsion and manoeuvring systems used to day.
Manoeuvring Characteristics.
The traditional rudder and propeller system.
Motor ship - Low speed diesel engines.
Motor ship - Medium and high speed diesel engines.
Turbines.
Controllable pitch propeller (CPP).
Ducted propellers.
The conventional rudder profile.
Rudder balance.
Rudder designs.
Other rudder designs.
Operational aspects.
Twin schilling rudders.
Turning diagram for twin schilling rudder.
Conventional single screw arrangement.
Accurate and predictable turning manoeuvres.
Turning with constant rate of turn.
Calculation of rate of turn (ROT).
Turning with fixed turn radius.
What is ship handling?
• What is ship handling?
Forces Involved During Manoeuvring and Sailing of Ships
• Forces Involved During...
Manoeuvring Characteristics And Interaction
• Manoeuvring Characteri...
Marine Useful Manoeuvres
• Marine Useful Manoeuvres
Propulsion And Manoeuvring Systems
• Propulsion And Manoeuv...
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An excellent video - so much great info for the cadet mariner about to sit for their company exams.
cheers mate, saved me an hour of research
*Watched in 1.5speed
I know of a kitchen rudders which we used to use in the British Navy. Well, it is the same principles used on the reversals of jet engines on planes so there is no need to describe it. One could steer going backwards with it.
Highly educative
Thanks for your work
You Welcome Sir.
Thanks for the video....................................... & THUMBS UP :D
Nice video congrats
What beautiful presentation!
Thanks
We learned alot from this.thanks
Great presentation indeed...
Nice video
Excelente produto.
Thanks Sir.
Thank you
Thanks for your Comment.
Thank you so much my ser 🌹
So nice of you
thank you
You Welcome. Like, share and comment.
Marine Online I saw that some videos have links to further explanations. these videos take part in some online training? I'm looking for btm, brm and naval architecture courses. Thanks. @marine online
Thanks you.
Thank you.
Good motor
Zabardast
Great video... Sucess
Thanks.
Please more in next 😘
Can u please tell me what engine capacity including diesl requirement is needed to drive a boat between long distance islands, say up to 2, 000 kms? Does the size of the engine matter, taking into account the need for lower diesl costs etc?
googel long range trawler
What these metal bumps on the rudder and nozzle are for (13:33) ? Thank you .
Probably cathodic electrolysis protection (zincs). Protecting the bronze propeller.
Zinc sacrificial anodes. Reduces galvanic corrosion due to dissimilar metals.
motori, propulzioni tutto da buttare via nella spazzatura e scafi pure...tutto completamente posso innovare ma costo carissimo....Saluti
Can anybody explain me how you can have a drive shaft driving a propeller without having water flooding the boat? Is it the same system in surface ships and submarines? Thank you.
Vasco Carvalho ...now I ‘m a fly boy myself but lesser water pressure on the surface ship .. might be a factor ... where the sub is in much higher pressure ...
That's a complex seals system which includes several seals to be assembled together, spring to push it for mating with polished surface,... which you will not see in any commercial catalogue...
One method is to use a grease box where the hull meets the water and the shaft goes through. It is literally a box filled with grease. The grease allows the prop shaft to rotate but keeps water out.
I did heavy equipment service engineering can I be able to fix ships
What is the value of k?
Paul Angelos Filippakis. Two over pie ÷ E over I × R.
Paul Angelos Filippakis й
Well read
Add eco surface drives
Admin dmet se hai💃💃
WE SHOULD INVEST HEAVILY ON JET PROPULSION TECHNOLOGY
The math at 19:04 makes no sense whatsoever.
In the text presentation of the problem the speed of the ship (V) is stated as 20 knots (nautical miles per hour or 0.67 nm/min) and the turn radius (R) is 0.5 nautical miles. If k=180° ÷ π = 57.3° then in the stated formula k÷60 can be rounded to 1° (though the unit of measure for 60 is undefined), thus (V ÷ R) • (k÷60) = (0.67 nm/min ÷ 0.5 nm) • 1° = 1.3°/min.
But the author fills in the blanks with values not stated in the text of the problem. He enters a speed (V) of 3 (no units of measure) and how the author arrived at this value after stating in the text that (V) = 20 knots is not revealed & enters a turn radius (R) of 22 (Again, there are no units of measure and again the author does not reveal how he arrived at this value when the text states the turn radius (R) = 0.5 nautical miles) thus arriving at a value of 3 ÷ 22 = 0.1°/min or 6°/hour (rounded down from the more accurate 0.14°/min or 8.4°/hour - an underestimate of 29%) thus arriving at a calculated 15 hours to make a 90° turn ( or 10 hours 47 minutes if one doesn't round down his result).
It's entirely possible I've missed something as I am neither a navigator nor a ship handler. If so then I'd be happy to be corrected.
Thanks.
bang, mau tanya. untuk part number filter berapa ya.terima kasih
untuk part number hyd filter sir?
Only what is not true is that a ships engine is not economical at high rpms the ships engine is laid out so that at normal sailing speed she is most efficient and uses as little feul as is possible if you for example use 30% of power it will be less efficient
Every heard of electric propulsion ?
The spelling of manoeuvring broke my brain.
Why is he reading what is on the chart? I can read. Tell me something else.