Seaworthiness

Paulo,
You have said:

"About the POGO40 STIX, I will post the relevant part of the official document that certifies the boat. No need to make any calculations. They have already been made correctly and ISO certified."

That statement may very well not be true, if the STIX at MOC condition is lower than the one at MTL, which is the condition used for the posted calculation (By the way: Are those two pieces you posted, taken from the same document?)

"5.3T is the lightship weight.....The one (stability curve) I have posted is the curve in MinSC..."

So it seems you posted the calculations for MTL condition but then the curve for MOC. I find that confusing.

"The Rm at 90º of heel goes from 5100 to 6132 "

You seem to have both curves and both calculations. So I insist you please post the two complete things for our full understanding and better judgement. You wouldn't like we suspect you're trying to hide something, would you?

Cheers.

 

Randy,
Although not coinciding with all that's said there, I find your #425 post a very nice one.

By the way: I still think we cannot compare seaworthiness of boats using equal displacements criteria only, as their moments of inertia (for instance) can be very, very different (which is the case between Catalina and Pogo 40).

On the other hand: if we compare Catalina 30 against Pogo 40, should we compare their prices too...? Is that the same comparative criteria as for, let's say, Valiant against Pogo...?

Cheers.

 

Rayk,
Only for claryfying purposes, it seems you are comparing the loaded condition for the Catalina against the MOC condition for the Pogo 40 (Paulo's posted stability curve seems to be the one for that condition).

Cheers.

 

Since this is a thread on seaworthiness, and not a thread specifically to discus the attributes and encumbrances between two yachts, I think it best to allow a flow of thoughts on the subject at hand. A brainstorming session bogged down by the premature concern over cost is a good way to bury creative thinking in mud.

 

I agree.
Cheers.

 

Yes!
You may have a fast boat like the Pogo 40 to quickly cross oceans in a racing style mood, and then living at hotels when at destination, totally depend on local supplies, or take a flight back home in a couple of weeks because you never intended to do extensive cruising....
Cheers

P.S.: I know this statement is an exaggeration, but.....

 

Guillermo,

I dont put much emphasis on the numbers.
Much of the time they are incorrect, incomplete, or false anyway.
I have a bit of faith in Rhoughs offerings on the Catalina, but Vega seems a bit shady (that means full of ****). His cut and paste posts suggest that he does the same thing with data.

Regarding comparison by displacement, I find the contrasts well suited for observing difference in design.
If you were asked to draw a boat of displacement X, you would learn very quickly the delicacy of extra this and extra that.

To improve the inverted characteristics of the Pogo 40 would slow it down.
And that is a fact.

 

I'm almost speechless (I'm sure some people wish I were )

The measure most people use is the ratio of + area to - area as an indicator of resistance to capsize. By that measure the Catalina's 2.8 vs the Pogo's 6.0 ratio makes the Pogo much more capsize resistant.

You state that the Pogo is as hard to right after inversion as the Catalina is to knock down. Earlier in this thread, it was pointed out to me that the area under the curve is the force required ... thus (from only these data points) we have 7,405 Kgfm to invert the Catalina and 20,100 Kgfm to invert the Pogo. Almost 3 times the energy. If you total the numbers under the line you get 4023 for the Catalina and 3120 for the Pogo. It takes more energy to re-right the Catalina.

Your statement "The force required to knock down the Catalina is about the same to re-right the Pogo." is not correct. It takes more than twice the energy to knock down the C30 as to re-right the Pogo (7405 vs 3120).

As far as making the Catalina go faster ... here are some numbers:

True WS______ Vs C30____ VMG C30_____ Vs P40c____ VMG P40c
6____________ 4.075 _____ 2.746 _______ 5.46 ______ 3.98
8____________ 4.862 _____ 3.311 _______ 6.40 ______ 4.83
10___________ 5.295 _____ 3.654 _______ 6.78 ______ 5.39
12___________ 5.474 _____ 3.848 _______ 7.02 ______ 5.69
14___________ 5.575 _____ 3.953 _______ 7.32 ______ 5.89
16___________ 5.630 _____ 3.998 _______ 7.49 ______ 6.04
20___________ 5.698 _____ 3.976 _______ 7.69 ______ 6.25
25___________ ?.??? ______ ?.??? _______ 7.84 ______ 6.38
30___________ ?.??? ______ ?.??? _______ 7.96 ______ 6.44
35___________ ?.??? ______ ?.??? _______ 8.05 ______ 6.44

Note that in wind speed over 10 knots true, the Pogo's VMG is higher than the Catalina's boat speed! Even if the Catalina could sail directly into the wind, the Pogo is faster.

Since "Gentleman don't sail to weather" and no sane cruiser is going to make passages at best VMG upwind, so I'll include the VMG downwind also:

True WS______ Vs C30____ VMG C30_____ Vs P40c____ VMG P40c
6____________ 3.941______3.177 _______ 5.78 ______ 4.41
8____________ 4.697______4.010 _______ 6.70 ______ 5.50
10___________ 5.129______4.813 _______ 7.49 ______ 6.38
12___________ 5.693______5.532 _______ 8.03 ______ 7.15
14___________ 6.108______6.055 _______ 8.79 ______ 7.82
16___________ 6.474______6.443 _______ 9.32 ______ 8.48
20___________ 7.109______7.082 _______12.35______ 10.38
25___________ ?.??? ______?.??? _______15.07______13.24
30___________ ?.??? ______?.??? _______17.24______ 15.72
35___________ ?.??? ______?.??? _______19.22______ 18.33

Upwind, the Pogo's VMG at 8 knots true wind speed is better than the best VMG the Catalina can make at any wind speed. Downwind, the Pogo is faster at 12 knots true than the Catalina will ever be. Trying to make the Catalina faster is a very expensive and nearly pointless exercise (I can show you the bills). I almost forgot to add that the polars for the Catalina are in full racing trim with 1225 pounds crew. I don't know what crew weight was used to generate the Pogo polars. The polars for the Racing version are better and the Class40 racing trim is 1 or 2 crew, so I feel it is a safe assumption that the Cruising version polars are for the same 1 or 2 crew. I am just about to order a new set of polars for the Catalina to be run with a crew of 1 (200 Pounds) and an A-Sail that can be tacked off center. This is the current configuration of my boat.

What these abstract numbers mean to me is that the Pogo is much more seaworthy than the Catalina. If I was so unlucky/stupid as to find myself in a Force 10+ storm, I would much rather be on the Pogo. Combine the much better stability of the Pogo (remember that the Catalina has better than average numbers) with the speed potential and the likelihood of facing Force 10+ conditions is even more remote. Faced with a storm that puts me achored on a lee shore (Cabo San Lucas in 87'?) many "seaworthy", overloaded, safe when hove-to, boats could not escape and ended up on the beach. A Cal40 (light, fin keel, spade rudder) was able to sail off the beach to weather and survive. The Pogo makes the Cal look silly, there is no doubt in my mind that the Pogo could sail off the lee shore in the conditions that trashed the cruising fleet.

I'm sure those folk were secure in their cockpits behind their dodgers as their boats went on the beach.

 

• Looking at a graph of righting moments...
  • Area within the curves is the minimum amount of energy required to roll the boat.
    If the energy required to roll the boat is not present in the conditions, the boat will not roll.
  • The max force at the peak of the curves must be present in the conditions to roll the boat.wrong
    If the max force is not present in the conditions, then the boat will not roll.wrong
• Looking at the sea...
  • A triangle doubled in dimensions has six times the area...a rouge wave twice the height of the others is a fairly impressive thing.
  • If a rouge wave twice the height of the others has the peak force to flip the Pogo, then it will wallow around upside down in the normal waves for a long time.
  • The Catalina might be merrily flipping over and over in the same conditions, but it will be flipping upright again and again.

These are both 6T boats. Their different forms have different behaviours. Both behaviours sound familiar.

RHough, when I suggested that the Catalina could be made faster, I was suggesting it could be stretched to 40' and given an outrageous beam and deep high aspect ratio bulbed keel and a high aspect ratio rig. Then it would go fast.

But I also asked what the Pogo would have to change to reduce its inversion moment....:?:

 

Rayk, I believe the logic of what you say about the Pogo/Catalina comparison is flawed.

The force needed to knock down a boat is not the Max RM moment force, but the energy of all positive RM moments, at all points of heel till the AVS point. This value is the area behind the positive part of the RM curve (not the GZ curve).

The same happens to the energy needed to re-right a boat. This time it is the value of the area behind the negative part of the RM curve.

Modern boats don’t re-right by themselves, their AVS is smaller than 180º. This is because the same kind of sea motion that is needed to capsize them should bring them easily to the upright position, assuming that the negative area under the RM curve is much smaller than the positive area under the RM curve.

The sea condition needed to capsize a Pogo is incomparable worst than the one needed to capsize the Catalina. Thus what matters, regarding the ability to recover from a capsize (while comparing different boats), is the relation between the positive area and the negative area of the RM curve. The Pogo has an unusually high relation between the Positive and Negative areas (6/1) a relation that I am sure is bigger than the one from the Catalina.

A boat that is much harder to capsize, normally a bigger boat, will usually need also more energy to re-right itself, even if its stability curve is a very good one.

 

No, I don't have both curves; I have only the MOC condition curve.

Yes , the two pieces I have posted are from the same document.

Yes, the Stix calculations I have are for the Max. Load Condition and that is plainly said in the document I have posted (I don’t have the STIX at MinSailCond).

I have the RM at 90º in the MaxLoadcondition (Nm 61320).

Contrary to what you imply (on post 86), in this kind of boats there is not a big difference in the STIX at MinSalCondition and in MaxSailCondition. For example, for the RM1200, that difference is from 38.33 and 38.7.

Besides as I have said, I consider the STIX an important global stability Index but mainly for the typical boat buyer. For the ones that know more, the information that you can take from the stability curve is much more relevant.

I don’t care about what you suspect. If I cared about that I would have to care about you calling me ignorant and dishonest.


Cheers

 

Can you clarify this please? What do you mean with "cut and paste posts"?

Are you saying that I have misquoted someone? If so, please, tell of what are you talking about?

You have no right to slander any other member without saying of what you are talking about.

 

It is important to separate stability due to form, and stability due to ballast. Remember that capsizing and re-righting is a dynamic state. therefore, you have to consider the force over time and distance to determine how much energy is needed in both of the previously mentioned situations.

It is not instructive to look at righting arms and not consider the following:
1. How is the force being applied
2. Is the force constant, or variable
3. Over what duration is the force being applied
4. Over what distance is the force being applied
What should be considered, is not force alone, or even the moment arm alone, but both torque and moment of inertia. ‘Force’ often times being used in various forms in this discussion, is more correctly applied to linear velocity; torque is necessary to change angular velocity.

Torque is somtimes expressed as follows:
Where t is torque

t=(moment arm) x Force

This definition is sometimes used, and it is being used here in verbal expression, but it falls short because it only considers torque as a single magnitude without direction. In addition it does not consider torque as a function of time. Before we discuss torque in greater depth, the concept of angular momentum must be understood:

Angular momentum of an object rotating about a point is the measure of the degree to which the object will continue to rotate about that point unless acted upon by an external torque. (Note, that is a torque, not force) It can be expressed as follows:

L = r x p


L is the angular momentum of an object of given mass
r is the position of the mass/object expressed as a displacement vector from a point of origin
p is the linear momentum of the mass/object, which is a product of mas times velocity.

p = m x v

m is the mass
v the velocity

To express torque as the rate of change of angular momentum, you would use the following calculus formula:



Where t = time
This can also be called the time derivative of angular momentum.

Although discussing this subject verbally facilitates the expression of ideas quickly, at some point, a more thorough analytical approach must be taken to decide conclusively the virtues of a given vessel.

 

Hum, before we enter on a more detailed analysis I think it is better to define better the “basics” .

I know that you know this is correct, since for the stability curve both the stability due to form, and stability due to ballast are taken into account, but I believe that there will be some that think that you are saying something else. I would like you to clarify this point before going further.

Saying this, I am very curious about what you are saying and I confess that it seems very interesting.

 

Yes, this I guess s a good way to say it verbally, as it states energy over a distance indicated by "all points of heal" to an end maximum of AVS.(with an upright starting point inferred.)

I guess this I am unsure of, and maybe I need to study it more; but there are a couple of points that I am not seeing reflected in the stability curve.

1 Upon rolling, form stability is predominant and diminishes after some point 'x' which is determined by the shape of the hull, combined with its longitudinal volume and how far that volume extends out from the axis of rotation.
2. Ballast stability starts at zero, and increases to maximum at about 90 degrees of heal before diminishing back to zero.
3. Total energy required is not indicated it seams, as inertia is not taken into account, for the reasons just stated in statenment 1, namely, that form is not explicitly defined.
4.Since velocity is not considered, momentum can not be considered. and without calculating momentum how do we determine torque that is required to start a roll, maintain a roll, or stop a roll?

Although your first statment that I have highlighted is true, the second concluding statement does not encompass all that the first statement is suggesting.