Comparing Righting Moments

I was just looking at two righting arm curve graphs.
One is in Inches/Pounds, the Other is in Metres/Kilos

Is the conversion as simple as described in the attached image ?
eg 27.5 inch-pounds = .31 Kilo-metres of righting moment ?

 

Please ignore this post, see post 4

If 1 kgf = 2.2046226218 lbs and 1 metre = 39.37008 inches then 1 kgf-metre will be 2.2046226218 x 39.3700787 = 86.79616............

Seems to be right although I was always uncomfortable with kilogram force and kilogram weight and introducing g for gravity.

 

In the curves it appears, in abscissas the angles in degrees, and in ordinate the righting arm, in feet or in meters. It is about the GZ values and, therefore, no value appears in pounds-feet or kg-meters. So you only need to convert feet into meters (1 foot = 0.3048 meters)

 

What TANSL says makes sense, it just shows the dangers of posting before breakfast

 

If the term "Righting Arm" in the left diagram is correctly used, it is only a distance between two points (G and Z), measured in inches as it seems (not feet).
Righting arm times deplacement (in pound) is mostly called righting moment. (Because a moment of this kind is force times length, it should be poundforce instead of pound. Also the righting moment in SI units is correctly given as Newton * Meter = displacement mass in kg times accerelation of gravity in m/s² times righting arm in m. In spite of this is mostly used kilogramm * meter or tonne * meter.)

So I would understand the 27.5" (maximum left diagram) is 27.5 inch* 0,0254 m/inch = 0.6985 m

 

In addition to all that has already been mentioned, the curve on the right (in post #1) is quite interesting: from 176º, more or less, the GZ returns to have positive values. For -25º, however, the values are negative. Nor do the maximum values, in such a large angular range, seem normal.
Maybe there is an explanation but at this moment I am not able to find it. What kind of ship does this curve refer to, @rwatson?

 

I would read GZ = 0 at 180 deg. This would be the common instable equilibrium at the position "keel upwards".
The negative GZ values corresponding to the negative angle 0 deg. to -25 deg. are absolutely normal for boats or ships, while the "plateau" from 25 deg. to 75 deg., as you said, is not.

 

You're right, I measured 176º when in fact it was 180º. In any case, should not the values be equal to -25º than to 175º?

 

No, a normal (symmetric) boat or ship hull gives a rotational symmetric GZ curve. So the value of -25 deg. is the negative of the 25 deg. value, same to 175 deg.

 

In English

 

You are very right. A new stupid mistake on my part.
Let's see if someone has an explanation for the upper part of the curve.

 

Sorry. I missed the conversation here.
The high flat top in the Kg/Metre is caused by the high cabin top.

I still haven't figured out if my conversion technique was correct or not.

What's the consensus ?

 

That's correct
1 inch = 0.0254 m
1 pound = 0.4535924 kg
27.5 inch-pounds = 27.5*0.0254*0.4535924 m-kg = 0,3168342914 m-kg

I'm not sure that's the explanation. Could we see a drawing of that ship?

 

The math is correct: 27.5 inch * pound = 0.3168 kg * m
But the value of 27.5 corresponds to the unit inch only not the unit inch * pound. Because the quantity righting arm (or righting lever) is a length only, not a moment.

Therefore the righting arm in Meters corresponding to the left diagram is 27.5 inch * 0.0254 inch/m = 0.6985 m .

If you multiply the 27.5 inch with the displacement of that boat in pounds, you will get the righting moment in inch * pound.

 

That's very helpful Heimfried, thank you.

Therefore, that little conversion chart isn't doing anything useful in converting inch-pounds to Kilos-Metre ?

It sounds like you are saying that both graphs are only showing moment lengths. I was assuming that the righting moment was being expressed in units of "Force" (pounds-inches and Kilos/Metre) , but you are saying we have to multiply it by the weight of the boat.

Is it then correct to say that two boats could have similar righting moments, but because they have different ballast weights that the actual strength of wind to tip them over them would vary ?

Many Thanks