# DNV vertical accelerations formula

Discussion in 'Boat Design' started by jcamilleri, Mar 23, 2013.

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### jcamilleriJunior Member

Is the formula given by the DNV for the average 1/10th or 1/100th highest accelerations.
The formula is:

a_CG=(k_h×g)/1650 (H_s/B_WL2 +0.084)(50-β_CG ) (V/√(3.28L_OA ))^2 ((L_OA×B_WL2^2)/∆)

I computed the vertical accelerations using both the Savitsky and Brown formula and this formula. For a speed of 40 knots i got 13.88g (Savitsky and Brown) and 13.2g (DNV). However Savitsk and Koelbel in their paper Seakeeping of hard chine planing hulls mention something that the formula given by the DNV is for the average 1/100th accelerations.

Any thoughts?

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### Mr EfficiencySenior Member

My thought is 13g will put you in very serious trouble. 3 is not too bad, 5 about the limit of sane human tolerance. But the Solo lemon drink man can probably stand much more.

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### jcamilleriJunior Member

Are you talking about the 1/10th highest accelerations? my average acceleration for a top speed of 40 knots is 5.33g. I attached the excel sheet i developed. Please have a look.

Thanks

#### Attached Files:

• ###### Seakeeping Assessment - Feasibilty Study.xlsx
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4. Joined: Oct 2008
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Does it matter what it is or is not based upon?

If you are designing to DNV, you must use their formula, unless you have a large database with supporting evidence of previous vessels using said database to act as justification for not using it.

The key aspect of DNV rules, to obtain the acg, are: the speed, the Hs and the deadrise angle.

What values of these are you using?

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### jcamilleriJunior Member

Speed - 40 knots
Hs - 3.21ft

The reason why i want to know if its 1/10th or 1/100th is because i want to convert this average value to RMS acceleration using a formula developed by Savitsky and Koelbel:
(1⁄3 octave RMS)_MAX=n_(1/10)/6

Then i want to use the boundary limit curves of the ISO 2631 standard (attached) to check whether the accelerations experienced are within the limits.

#### Attached Files:

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• ###### Fatigue Reduced Proficiency Limits - Vertical Motions.png
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### Mr EfficiencySenior Member

Deadrise angle 12.9 degrees at COG + length 6.38m + beam 1.9m + 1-metre waves+ speed 40 knots = unbearable. IMO.

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### jcamilleriJunior Member

Mr Efficiency, i am aware of that. I intend to change those parameters when "optimizing" the design. This is just a feasibility study. But i want to know if the method i am using for assessing the seakeeping performance is good.

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### Mr EfficiencySenior Member

No good if the crew are bashed senseless. What does, say, 22 degrees do to alter the g's ?

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### jcamilleriJunior Member

The deadrise angle at CG has to be at least 35 degrees to lower the acceleration levels within required limits

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### Mr EfficiencySenior Member

What are the limits ?

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### jcamilleriJunior Member

As for limits i am using the ISO motion sickness criteria (figures given in an earlier post). But i am not sure that this is sufficient. When i calculated the centre frequency the maximum value is 0.74 Hz. Based on the ISO plots that means that the crew can be subject to only seasickness. And with those g values i am sure there is more than that. I cant seem to find a sufficient criteria suitable for high speed vessels. there is one given by Nordsfolk:
0.65g RMS value at the bow
0.275g RMS value at bridge

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You're missing the point.

Those formulae are for calculating the vertical accelerations, all based upon the principal dimension of the boat.

However all accelerations are influenced by the wavelength and wave period and heading. Since a 1.0m (as you have it) of say a nominal 6 second period has a frequency of encounter of 0.53. Yet that same wave, if it is a nominal 12 second period the encounter frequency is 0.07 and if the period is a mere 4seconds then the encounter frequency is 1.07....thus prediciting motions when using accelerators needs the correct application. It is not cart blanch.

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### jcamilleriJunior Member

Ad hoc i am accounting for that by using the empirical formula
f_c=f_m+(2πVf_m^2)/g
where
f_m=1/(2.77√(H_s )) - f_m is the significant wave frequency which is related to the significant wave period

No?

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### AlikSenior Member

Usually 1/100 is used for structural design.

Regarding Savitsky-Brown formula and its modifications (used in Rules of classification societies), on should be careful. We found that S-B formula overestimates acceleration levels for non-prismatic hulls sometimes by 40%. Our recommendation is to use deadrise on Stn.4 (40% from bow) as it is made in ISO12215-5, for monohulls.

Another problem in S-B formula is trim angle that seriously effects the results.

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There is half your problem, it's empirical!

I'm sorry but you still don't get it. In the formula you posted above, the only input is Hs, which relates to a wave. That is it. Nothing about the wave length, nothing about the period.

And then in my other post about indicating how the frequency of encounter changes with differing wave length, again you're missing the point.

If you're in a Hs=1.0m, but the wave length is 10m, what would be the vertical acceleration at 40knots? Then taking the same conditions except that the Hs=1.0 now has a wave length of 1000m. What is the vertical acceleration?

The formula does not account for wave length or period on the calculated accelerations. All you're doing is plugging in numbers into a formula hoping to get an answer you want!

You can only get these values by model testing, since the response of your hull in a given sea way is unique to that hull and the sea spectrum in which you have tested to obtain your RAOs. Since the vessels natural period of pitch/heave/roll influence the motions.

Where is any of that in the DNV/Savitsky formula...it isn't!

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