NavList:
A Community Devoted to the Preservation and Practice of Celestial Navigation and Other Methods of Traditional Wayfinding
Re: Rolling Cruise Ship
From: Frank Reed CT
Date: 2006 Jul 26, 23:44 -0500
Red wrote:
"So I have a base side of 2' and hypotenuse at 15'...
Sin* = opposite/hypotenuse = 2/15= 0.1333
And the angle becomes about 8 degrees if I've done that right.
Admittedly VERY crude and done with a fast eye instead of proper measurements
from the photos, but at eight degrees that's still only half of the
fifteen(?)
degrees that the cruise line states now."
No, I don't think that's the right geometry. The actual angle, given your
inputs of 2 ft and 15 ft works out to 15 degrees almost exactly. If the actual
depth below the rim is 2.5 ft, then the angle is 18 degrees.
You work this out by preserving water volume in the pool. Draw yourself a
picture with the pool tilted by some angle X such that the water is just
touching the upper rim on one side. Now draw the pool level and call the distance
from the rim to the level water D. Calling the length L and preserving volume
from tilted back to level, you should be able to derive this equation:
tan(X)=2*D/L. As you can see, if D is 2 ft and L is 15 ft, then X is 15 degrees.
One of the more extremist reports claimed that the vessel tilted 38 degrees.
This would require the pool's water level to be 6 feet below the rim after it
righted itself, which was clearly not the case.
Incidentally, in case anyone is suspicious of the images in the web site
which Patty pointed out to the list (please note, this is NOT her web site), I
watched the live helicopter video feed from a local tv station as the ship was
returning to Port Canaveral, and the water level in the pools was similar to
what you can see in those web site photos --a couple of feet of water spilled
out.
-FER
42.0N 87.7W, or 41.4N 72.1W.
www.HistoricalAtlas.com/lunars
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From: Frank Reed CT
Date: 2006 Jul 26, 23:44 -0500
Red wrote:
"So I have a base side of 2' and hypotenuse at 15'...
Sin* = opposite/hypotenuse = 2/15= 0.1333
And the angle becomes about 8 degrees if I've done that right.
Admittedly VERY crude and done with a fast eye instead of proper measurements
from the photos, but at eight degrees that's still only half of the
fifteen(?)
degrees that the cruise line states now."
No, I don't think that's the right geometry. The actual angle, given your
inputs of 2 ft and 15 ft works out to 15 degrees almost exactly. If the actual
depth below the rim is 2.5 ft, then the angle is 18 degrees.
You work this out by preserving water volume in the pool. Draw yourself a
picture with the pool tilted by some angle X such that the water is just
touching the upper rim on one side. Now draw the pool level and call the distance
from the rim to the level water D. Calling the length L and preserving volume
from tilted back to level, you should be able to derive this equation:
tan(X)=2*D/L. As you can see, if D is 2 ft and L is 15 ft, then X is 15 degrees.
One of the more extremist reports claimed that the vessel tilted 38 degrees.
This would require the pool's water level to be 6 feet below the rim after it
righted itself, which was clearly not the case.
Incidentally, in case anyone is suspicious of the images in the web site
which Patty pointed out to the list (please note, this is NOT her web site), I
watched the live helicopter video feed from a local tv station as the ship was
returning to Port Canaveral, and the water level in the pools was similar to
what you can see in those web site photos --a couple of feet of water spilled
out.
-FER
42.0N 87.7W, or 41.4N 72.1W.
www.HistoricalAtlas.com/lunars
--~--~---------~--~----~------------~-------~--~----~
To post to this group, send email to NavList@fer3.com
To , send email to NavList-@fer3.com
-~----------~----~----~----~------~----~------~--~---
