Area under GZ curve

I strongly recommend that you do one calculation using paper and pencil.
For example, use a small number of nodes and try the trapezoidal rule.

Once you understand what you are doing, try implementing it in Excel, or use some other computer program. Otherwise, you are just another disaster waiting to happen.

Good luck!
Leo.

 

"add the Y values and multiply by the increment" only works in general if one Y value per interval is added. If the Y values at each "node" between intervals, ie N+1 values for N intervals with both ends added then (in general) the result will be too large in magnitude.

 

Also recommended is using the calculation method to determine the area of several simple cuves for which the answer is known. For example:
- straight, horizontal line. Area is height of line times length of line.
- a semi-circle: Are is 0.25 * Pi * Height * Length (or 0.5 * Pi * Radius ^ 2)

Another way to estimate the area is to overlay the curve on graph paper and count the number of squares under the curve. Estimate the fractions of the squares bisected by the curve. Then multiply the number of squares by the area of a square.

 

Except that GZ curves start at the origin so the first value is always zero. Then whether you sum the y value times the x increment or add all the y values and multiply by the x increment you'll get the same result. But it always pays to apply some common sense and actually think about what is really a very simple operation.

Don't lose sight of what this is actually about. It's not about numerical methods but about GZ curves, they are not absolutes at the best of times. They are calculated for still water with all items aboard fixed.
I presumed the initial poster had a string of values in excel from a computer program that spat out a calculated GZ for each 5 degrees of heel or so. That's the usual scenario and then they want to find the area from that data up to say 90 degrees as a rule of thumb for the energy required to roll the vessel to that point. High levels of accuracy in applications like this are really useless.

 

That is usually true, however, for some functions which are exponentially damped at the endpoints the trapezoidal rule is more accurate than Simpson.
Also, for some (all? can't remember) cyclic functions, the trapezoidal rule has the same accuracy as Simpson's rule.

 

ah simpson yes, wonder same rule applies in sulfer hexafluoride and if you could swim in it..

science is beautifull aint it

dumb question i actually forgot and was looking for is density
water beeing 400 or 700 times denser than air at sea level?

 

Don't recall excatly but 1m3 of air was smth like 1.4kg. You could calculate it using the mol weight of gases /22,4 litres. Both oxygen and nitrogen are O2 and N2 so the they weight accordingly 32 and 28g/22,4l (if nitrogen was number 14?). O2 being 21vol% and N2 79vol% (Ar 0,9vol%, 0,1 for the rest)

 

Teddy you must have a few mendlejeff charts still on the wall
it were the mol's special at temperatures and more that made me a drop-out
on the net i read various sources from 400 to 784 and 829 times denser

 

Excactly "The molar volume of an ideal gas at 1 atmosphere of pressure is 22.414 L/mol at 0 °C"
And checked my diving source and it says 1,2kg/m3 at 15C. So for fresh water it's 833 times or sea water 854 times heavier than air.

 

When designing to operate in BOTH mediums ,one should have a very good idea of the mass involved.

Here is one that was designed for Specific mass. The running surface and the Tunnel.

http://www.youtube.com/watch?v=h1csGLgX4tk

 

last night thought back on the weight of your cubed air, oxi nitro, than realised a liter is a kilo and how simple it was but thanks for the brush up