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Tue 22 Aug, 2006 01:22 pm
The sailor of a ship at sea has the harbor in sight at which the ship is to dock. She spots a lighthouse that she knows is 1 mile up the coast from the mouth of the harbor and she measures the angle between the line-of-sight observations of the harbor to be 20 degrees. With the ship heading directly toward the harbor, she repeats this measurement after 5 minutes of traveling at 12 miles per hour. If the new angle is 30 degrees, how far is the ship from the harbor?
Isn't it a little early in the school year for us to be doing your homework?
Nevermind that, ships don't travel at "miles per hour." The travel in speeds measured in knots, which is a referential measurement giving a close approximation to nautical miles per hour. Six nautical miles equal seven statute miles on land. We don't have sufficient information on this problem, and no assurance of the ability of said sailor to have made reliable dead-reckoning observations.
Re: Distances at Sea
shark wrote:The sailor of a ship at sea has the harbor in sight at which the ship is to dock. She spots a lighthouse that she knows is 1 mile up the coast from the mouth of the harbor and she measures the angle between the line-of-sight observations of the harbor to be 20 degrees. With the ship heading directly toward the harbor, she repeats this measurement after 5 minutes of traveling at 12 miles per hour. If the new angle is 30 degrees, how far is the ship from the harbor?
I assume the angle you refer to is the relative bearing of the lighthouse.
Is there any set or drift due to the motion of the sea? Stated diffrerently, did the compass course of the ship change at all during the 5 minutes?
Remove the decidedly un nautical verbiage and you have merely disguised a geometrical description of a right triangle with sides 1 and 2, and an hypotenuse of the square root of three. A ninth grade problem at best.
Setanta was so kind to give you some really valuable informations. And georgeob really tried to help you, shark.
I don't like to do other people's homework - but the following pic illustrates how it could be done.
Somehow, at least:
:wink:
Re: Distances at Sea
georgeob1 wrote:
Remove the decidedly un nautical verbiage and you have merely disguised a geometrical description of a right triangle with sides 1 and 2, and an hypotenuse of the square root of three. A ninth grade problem at best.
The point here is that the bit about the first bearing and the ships motion over the 5 minutes is quite irrelevant It is sufficient to know that an object 30 degrees off the bearing to the port is one mile distant from it on a line perpindicular to the bearing -- the ship is then two miles distant from the port.
i have to take a course in navigation and book another cruise to be able to give an answer ... might be a while
hbg
What George said (his last).
(The other info would used to plot the location of the ship ... without making a bearing.)
Walter, OMt.d.R.
walter signed : "OMt.d.R."
this "obermaat" (also called 'smutje')seems to be testing a new secret weapon of the german navy
.
hbg
distance in miles =(1 mile)X tan 30. Thats all ye need know. TBut how ya gonna know when its a right triangle?
SO, if the angle is with the lighthouse to the harbor as the R trianglre, then its Distance= cot 30 x 1 mi
Farmer(reads his Chapman) Man
I just got done taking some squid parts and seaweed out of my starboard thruster so Im not one to be trifled with, even if Im wrong.
He hasn't calculated drift, and hasn't yet explained how the distance was initially calculated. This joker is so hopeless at navigation, i'm not goin' out with him in a rowboat.
Its just a simple trig function as george said. Hes in a ship making it directly to harbor. He doesnt need to worry about drift just keep his line.
Red on the right.
The only info hes failed to give us is whether the harbor is in a direct line to the lighthouse and perp to his course.
In actuality Id say look at his damn plotter and hit the "compute" button.
Red to starboard, FM . . . sheesh . . .
ITS A wee mnemonic that Chapman teaches to beginners
"red on the right when returning"
"Sinistral to starboard "would be just flat wrong
I always liked the fact that the Royal Navy continued to use starboard and larboard well into the twentieth century. What good is it to be a "sailor with salt in his eyes" if you can't completely confound the lubbers?
N squid in yer thrusters.
It's not a right triangle, and there are two possible answers. If the ship is between the lighthouse and the harbor, then the original triangle is 80, 80, 20. If the lighthouse is on the other side of the harbor, then the original triangle is 100, 60, 20.
1/sin(20) = (x+1)/sin(160-a)
1/sin(30) = x/sin(150-a)
a is angle BHL (boat, harbor, lighthouse)
The rule is "red right returning (to port)" for the location of channel bouys.
markr is correct. I assumed the ship was headed perpindicular to the shoreline ((or the baseline between harbor & lighthouse. In fact that is not stated.
George wrote-
Quote:The rule is "red right returning (to port)" for the location of channel bouys.
You can't argue with that. Not statistically at least with only one Captain who has his honour to look to.
The problem lies in the word 'directly' which has no meaning that I know of in geometry or trigonometry. If we assume that the ship is traveling on a course perpendicular to the line between the harbor and lighthouse, then the ship, at the point in question is 2 miles from the harbor. The speed and previous angle are irrelevant.
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