# Comparing Righting Moments

Discussion in 'Hydrodynamics and Aerodynamics' started by rwatson, Apr 26, 2018.

1. Joined: Sep 2011
Posts: 6,947
Likes: 542, Points: 123, Legacy Rep: 300
Location: Spain

### TANSLSenior Member

Righting moment = Ship's displacement times righting arm
That is, as always, a moment is equal to a force multiplied by a distance.
The two curves what they are indicating is the value of the GZ (distance) for several heel angles (see picture)

#### Attached Files:

• ###### GZ.jpg
File size:
22.4 KB
Views:
232
Last edited: Apr 27, 2018
2. Joined: Aug 2007
Posts: 5,880
Likes: 312, Points: 83, Legacy Rep: 1749
Location: Tasmania,Australia

### rwatsonSenior Member

Thanks Tansl.

The diagram is helpful, but my problem was deciphering what the typical "righting moment graph" was trying to represent.

For some reason, I always pictures the number in the vertical left column to be a "force", not just a distance.

My question "could two righting moment graphs for two boats be similar, but the weight of the two boats mean that the heaviest would be harder to incline" is still awaiting confirmation.

To be clearer, if two boats both had a righting moment of .04 metres at 10 degrees, but one boat was twice as heavy - would the heavier boat be harder to push over ?

3. Joined: Apr 2015
Posts: 483
Likes: 106, Points: 43, Legacy Rep: 37
Location: Berlin, Germany

### HeimfriedSenior Member

No, it isn't, if I understand the question right.

Assuming two boats with similar righting moments and equal lateral overwater areas, the wind will heel both to a similar angle. The wind at a certain velocity is causing a corresponding heeling moment and the boat will heel towards an angle for which its own righting moment has an equal amount but an opposit direction to the heeling moment.

(The heeling moment of the wind is considered as wind force times height of the centroid of the lateral overwater area. Wind force is density of air (kg/m³) times squared velocity of the wind (m/s)² times lateral overwater area (m²) times form factor (dimensionless). Areas and heights are taken from upright boat in calm water.)

4. Joined: Apr 2015
Posts: 483
Likes: 106, Points: 43, Legacy Rep: 37
Location: Berlin, Germany

### HeimfriedSenior Member

If we replace the "righting moment" in your question with "righting arm", then the answer is yes. Equal righting arm times the double of the weight force gives the double of the righting moment.

5. Joined: Aug 2007
Posts: 5,880
Likes: 312, Points: 83, Legacy Rep: 1749
Location: Tasmania,Australia

### rwatsonSenior Member

Ah - thanks Heimfried.

It was the definition of the term "righting moment" that I should have been asking about.

In the interim, I have been watching a few videos on the topic, and getting my head around the whole process.

The interesting thing to me, is that the GZ value alters significantly IF the weight of the boat is added below the Centre of Gravity.

So as I understand it, to achieve the righting moment of two boats with the same GZ value but different weights , the weight of the heavier boat would have to be placed so as not to change the Centre of Gravity ( more passengers for example), or the GZ would change for that trip.

Many thanks for everyone's attempts to explain the principles. I look at the stability graphs with new appreciation now.

Following from that, the graph on the left would result in vastly better righting moments than the one on the right.

From your initial conversion explanation -
So I would understand the 27.5" (maximum left diagram) is 27.5 inch* 0,0254 m/inch = 0.6985 m

I
f each boat weighed 2 tonne, the Maximum Righting Moment (edit) on the left boat would be 2000 Kilos x .69 metres of force

But the Right Boat would only have 2000 kilos x .25 metres of force - or approx 1/3rd of the Maximum Righting Moment.

Edit - no, perhaps that's NOT the right conclusion.
Is the lever action of the increased GZ linear in overall force ?

Edit x 2 - Its all good. It is a linear comparison, not Geometric. Thanks to an Engineer in the family.

Last edited: Apr 27, 2018
6. Joined: Sep 2011
Posts: 6,947
Likes: 542, Points: 123, Legacy Rep: 300
Location: Spain

### TANSLSenior Member

When you add or remove a weight, not only does the position of the center of gravity change but the value of the KN also changes so you must calculate a new curve of the GZ values. The previous one does not work.
I would like to know how the boat corresponding to the curve on the right is because I do not understand how this curve can be generated.

7. Joined: Aug 2007
Posts: 5,880
Likes: 312, Points: 83, Legacy Rep: 1749
Location: Tasmania,Australia

### rwatsonSenior Member

That certainly makers sense Tansl. Your diagram was a great help.

The boat that was calculated for the "right" curve was this one. Very boxy in cross section, so its near maximum GZ persists for many of the heeling angles.

8. Joined: Sep 2011
Posts: 6,947
Likes: 542, Points: 123, Legacy Rep: 300
Location: Spain

### TANSLSenior Member

I believe, modestly and I can be wrong, that it is not possible. Would you mind showing me the body lines plan, or a 3d model, so that I can study them with my programs?. Excuse my insistence but I find that curve "intriguing". Thank you.

9. Joined: Oct 2008
Posts: 7,276
Likes: 1,165, Points: 113, Legacy Rep: 2488
Location: Japan

KN is related to the shape of the hull and its waterplane area, i.e its metacentre. It is not dependent upon the KG.

The KG dicates the final value of the GM/GZ, since GZ=KN - KG.sin(Theta)

10. Joined: Sep 2011
Posts: 6,947
Likes: 542, Points: 123, Legacy Rep: 300
Location: Spain

### TANSLSenior Member

@Ad Hoc, I can not believe what I'm reading. What I want to say, and that is what I said, is that the value of the KN varies with the displacement. Therefore, if a weight is added or removed from the ship, not only does the value of the KG change, but when changing the displacement, the value of the KN is no longer the same. Do you know what cross curves of stability are? Well, if you do not know, I'll explain it to you. They are the curves that indicate, for each angle of heeling, how the KN value varies when the displacement varies. For more information, consult with any text of the ship's theory.
I hope I've helped.

11. Joined: Oct 2008
Posts: 7,276
Likes: 1,165, Points: 113, Legacy Rep: 2488
Location: Japan

Then perhaps rather than usually jumping to the wrong conclusion you should, as always, first seek clarification. I shall quote for clarity:
So does this make your statement invalid:

No it doesn't. Your statement is correct.

Why (both can be correct) - Reason being is that adding or removing a weight may or may not actually influence the value of KN, since adding 1kg on a vessel that is over 1000 tonnes has no practical effect. Other than the number of decimal places one wishes to go - which is not the MO of the cross curves.

However,

Again, you have to be careful in your explanations as it is not wholly true for all vessel types and shapes. A one liner doesn't satisfy all types and can lead to confusion...

Below is shown the cross curves of a monohull we designed many years ago. It ranges from well below lightship to way above full load, merely to satisfy Class at that time. The normal operating range of the vessel is highlighted by the red box:

As you can see in the normal operating range the value of KN change is essentially negligible - especially at low angles of inclination. (Only at the extreme angles does it make a real difference). It is only measurable by several decimal places on a computer print out. Yet, for ease of simplicity the Master may assume for a quick calc to be the same between 50-60 tonne, if time does not permit a very detailed calculation - if the stability is marginal.

If we now look at a multihull, which coincidentally also has a set of cross curves similar to that of the monohull:

As above, within the vessel's normal operating range defined by the red box, the change in KN is negligible. Because as above the hull shape and also its WPA changes very little.

Thus whilst the displacement is varying, for these vessels, within its normal operating range, the KN values are pretty much identical, save for a several decimal places., i.e. mm.

Outside these ranges, where the hull shape varies greatly the change in KN warrants a recalculation for each and every angle. Which is the reason for this statement:

Since the change in displacement can be seen in the normal operating to have almost no influence.

However if we now examine the monohull's KN curves and apply some actual loading scenarios to this, what occurs?. It is clear that the final KG is more of an influence within the 50-60tonne displacement range, rather than 'just' adding or removing a weight which affects its displacement within the 50-60tonne range. Because depending upon the location of the added weight it can significantly effect the final KG and thus the new GZ. In other words, the location of the weight being added is critical, not the final displacement per se. Since if the vessel is floating on a draft that has a displacement of 50 tonne and a 10 tonne weight is added, one could assume the value of KN to be the same for the 50 tonne and then its new 60 tonne displacement - see above. Since upon first glance, it pretty much is the same, bar several decimal places if one is being pedantic. However, if the 10 tonne is placed above the KG of the vessel it shall greatly influence the final stability checks, and very negatively so. Yet if it placed below the KG it wont, it shall be a positive addition.

Thus the KN and the final GZ are linked but not directly proportional to each other in the sense you are trying to suggest for all vessel types. It is where the weight is placed ...the CoG of the weight added relative to the CoG of the vessel in that condition ...is the major influencing factor, not necessarily the new/final displacement and its influence on the KN value.

However, it must be noted that on different vessel types, other than those shown above, the value of KN may change significantly over a small change in displacement. In which case, for these hull form types, your statement would be valid. But only a detailed analysis would confirm this for each hull form.

Trust the more detailed explanation clarifies for you, to save you referring back to textbooks.

12. Joined: Sep 2011
Posts: 6,947
Likes: 542, Points: 123, Legacy Rep: 300
Location: Spain

### TANSLSenior Member

There is no doubt that now you know what cross curves of stability are. I'm glad. But, if you have any doubts about it, do not hesitate to consult me.
I did not say that you were not right because you were not saying anything. After all that huge argument (honestly, I have not had the courage to read it. English is not my mother tongue and so long paragraphs overwhelm me) that, frankly, I do not know who asked for it or for what purpose you do it, it turns out that my statement "When you add or remove a weight, not only does the position of the center of gravity change but the value of the KN also changes .... " is still as valid as before and I still do not know why you have quoted it. But you do not need to give me more masterful lessons. Thanks however for I do not know what.
Cheers.

Forum posts represent the experience, opinion, and view of individual users. Boat Design Net does not necessarily endorse nor share the view of each individual post.
When making potentially dangerous or financial decisions, always employ and consult appropriate professionals. Your circumstances or experience may be different.