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Author Topic: "Mare Australe" thruster has way too less power  (Read 13555 times)


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Re: "Mare Australe" thruster has way too less power
« Reply #25 on: December 14, 2017, 01:49:16 »

We are allowed to disregard resistances because the only thing they do is braking/damping the movements, and are negligible at small speeds, Newton's first law which is easy to check, no wind, no stream of course.
As you seem to dislike computing physics, I tried a best guess of the 110 m barge pivoting freely.
Loa 110 m, Lpp 105 m, B 11.45 m, thruster 5 m astern from the waterline or 10 m from the bow, draught 3m and block coefficient 0.9 -> Mass M=3246 mt, reckoned yaw mass moment of inertia Iyaw= (0.3 Lpp)^2 x M as 3220843 (mt.m²). Take a hm hm Vetus thruster of 140 kN (700kW), at 47.5 m from the C/(MoM.
The static pivot point is at a distance of Iyaw/(47.5 x M)=20.89 m astern from the CoM (26.61 m from the stern).
The lateral acceleration of the CoM is 140 kN/3246000 kg or 4.313 cm/s², the rotational acceleration of the thruster about the CoM is 140 x 4.75/Iyaw = 2.065 mrad/s² and about the static pivot point (63.39 m instead of 47.5 m) 5.32 deg/min per second.
50 seconds after start, the CoM lateral speed is 4 kt and the rate of turn about the pivot point around 266 deg/min, and a bow thruster circle radius 63.39m, 294.3 m/min i.e. 9.53 kts. Also, the ship turned 148 degrees around its pivot point.
Check, 4.90 m/s and 140kN for the thruster gives 686 kW, almost its maximum power, even having disregarded water resistances and added masses, unlike what the editor achieves.
So, indeed, 360°/min cannot be reached for such a 110 m fully loaded barge, but I don't intend to check it for different loadings and to go farther to disprove the topic starter.

Conclusion, you are right, even as the speed doesn't seem incredible to me, and certainly not to the point that you could charge the user of cheating, or of being a young snotneus. What else?



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Re: "Mare Australe" thruster has way too less power
« Reply #26 on: December 14, 2017, 10:00:59 »

hey luc ! like said no intention to argue this , more to share views and opinions on this old thread .

nevertheless in my humble opinion you forget one thing in your computations : you imagine whatkind of drag a 3 meter draught sidewall of the entire bowsection would create when you try to plow it with 9.5 knots thru water basicly sideways and whatkind of thrust would be needed to overcome this ?

in fact a bowthruster is of course not for spinning circles but to dock / manouver against the wind / current etc .

i,m not saying somebody cheats but wanted just to suggest to take a deep breath before starting editing when next time an user describing himself as the captain of the real PoR registers here to tell you she should be able to fly


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