Propulsion And Manoeuvring Systems 

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

We learned alot from this.thanks

Great presentation indeed...

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

What beautiful presentation!

Excelente produto.

Thank you so much my ser 🌹

Thanks for the video....................................... & THUMBS UP :D

Nice video congrats

Thank you

Thanks you.

Zabardast

Nice video

Please more in next 😘

WE SHOULD INVEST HEAVILY ON JET PROPULSION TECHNOLOGY

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

Well read

motori, propulzioni tutto da buttare via nella spazzatura e scafi pure...tutto completamente posso innovare ma costo carissimo....Saluti

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?

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.

Great video... Sucess

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.

Good motor

I did heavy equipment service engineering can I be able to fix ships

bang, mau tanya. untuk part number filter berapa ya.terima kasih

Admin dmet se hai💃💃

The spelling of manoeuvring broke my brain.

What is the value of k?

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

Add eco surface drives

Every heard of electric propulsion ?

What these metal bumps on the rudder and nozzle are for (13:33) ? Thank you .

Why is he reading what is on the chart? I can read. Tell me something else.

thank you