# ISO 6185-4 maximum maneouvring speed

Discussion in 'Class Societies' started by meren, Dec 4, 2015.

1. Joined: Sep 2005
Posts: 51
Likes: 7, Points: 0, Legacy Rep: 83
Location: Northern shore of The Baltic Sea

### merenJunior Member

I read this standard recently and I don't think I have got it right? There are lower limits for speed 1,1 Froude and Power, if both exceeded You shall proceed to Avoidance line test or Quick turn test. But limit for Power

P = BH*LH*m^0,333

P = Power [kW]
BH = Beam of RIB [m]
LH = Hull length [m]
m = mass in test drive condition [m]

Example 1:
LH = 8 m
BH = 2,6 m
m = 3500 kg
--> P = 315 kW = 428 hp

Example 2:
LH = 11,5 m
BH = 3,7 m
m = 4500 kg
--> P = 700 kW = 952 hp

Can this really be right? You shall test only boats that insanely powered?

Last edited: Dec 8, 2015
2. Joined: Sep 2005
Posts: 51
Likes: 7, Points: 0, Legacy Rep: 83
Location: Northern shore of The Baltic Sea

### merenJunior Member

I just wonder that Speed Froude number 1,1 doesn't add up with the power equation at all. Crafts powered according to the equation LH*BH*m^0,333 have speeds way over 55 knots at minimum (froude say 3,3...4,5). Power requirement does not follow LH*BH relation so could there be an error? Maybe it was meant to be (LH*BH)^(2/3)*m^0,333 ?

In that case we would have more realistic power limits giving speeds not So far away from 1,1 Froude:

Example 1A:
LH = 8 m
BH = 2,6 m
m = 3500 kg
--> P = 116 kW = 158 hp

If an example boat 1 would do 20 knots it's about 1,1 froude. Hull design would have gone seriously wrong if it would require power more than 315 kW to do that speed...if we are talking RIBs, not Tugs.

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