rudder size

hello..i'm a student of marine technology.recently my lecturer assigned me to find the rudder size of 78 ft tugboat and also how much torque can be applied to steer the boat.can anyone tell me the answer?

 

You are supposed to figure it out for yourself.

 

ask for help, not answer...

Hey,

I think you might get a better response if you ask for help rather than asking for the answer outright... There are some rules of thumb for rudder area, perhaps in PNA or ship design and construction... I've heard 1-2% of lateral area for typical merhant ships. This might be very different for tugs as they've got a lot of power and relatively little lateral area.

As far as lift or turning moment goes, pick a section and do the calcs... Abbot n' Von Doenhoff might be a good place to start...

-Dave

 

thanks! it just i'm not so good in english words...i will remember that

 

Since tugs tend to have slow turning wheels, a good rule of thumb is:
The square root of the length at waterline equals the TOTAL surface area (surface area of both sides) of the rudder. So, if the waterline length of the tug is 78' then the square root is 8.83 square feet for the total surface area. One side therefore has a surface area of 4.415 (round to 4.5) square feet. The rudder should extend both above and below the propeller at least a couple of inches, so the diameter of the propeller plus 4" will be the height of the rudder and, by dividing the area by that you will find the length of the rudder. Try to fit 1/3 of the surface area forward of the rudder post to give a "balanced" rudder.