Hey guys, I've been out of school for quite some time but I've run into a situation that involves a bit of calculus and I kinda want to wrap my head around it and figure it out mathematically.. I need some help coming up with an equation for the following scenario.
My boat has a 7' draft, the harbour entrance is 5.5' and the only way to get the boat out of the harbor is to pitch the boat to the right, at a certain degree that will bring the keel from 7' to 5.5' in depth. I need to find out what Degree that is. The keel is 7' deep when the boat is at 90 degrees, so what degree would the keel be at 5.5' deep?
I hope this is the right place to post this and any help would be appreciated. Also if there are any other variables that I need to include but haven't please let me know.
Apr 19th 2010, 09:52 PM
Let me make sure I understand what's happening. If the boat is upright, the lowest part (the keel, of course) is 7' under water. You want to rotate the boat around what I'll call the centerline - going from bow to stern. So the top of the mast goes one way and the bottom of the keel goes the other way. (I assume left would work just as well as right.) If you go just a few degrees, the depth of the keel won't rise much, but as you rotate more it starts to rise faster, and if you could go all the way to 90 degrees the depth of the keel would go to zero.
If I have it right, the answer is given by 5.5 = 7 cos x, where x is the angle of rotation (0 = upright, 90 degrees = flat on it's side). So x = 38.2 degrees.
There is one thing to consider - does the water level with respect to the boat change as you rotate it? That is, if you draw a line from the bottom of the keel toward the mast, does always meet the level of the water at 7 feet?
Also, wouldn't the depth of the water at the harbor entrance change with the tide?
Hope this helps. If you're still having trouble, go ahead and post again in this thread.