SailNet Community - View Single Post - Navigation problem
View Single Post
post #21 of Old 02-22-2007
CapnHand
humble pie rat
 
CapnHand's Avatar
 
Join Date: Nov 2006
Location: Location, Location
Posts: 867
Thanks: 0
Thanked 0 Times in 0 Posts
Rep Power: 9
 
Ok, so this is simpler than I first thought. This is way I should have done it the first time.


First of all, this is the situation. The distance the friendís boat will travel is 5t (t = hours) NM on a course of 036 Deg T. You are 84NM from him at a bearing of 096 Deg T relative to his boat. In order to meet, you will travel a distance of 7t NM at some unknown heading.



Itís obvious that the angle between the track of your friendís boat and the bearing from his boat to yours is 96 Ė 36 = 60 Deg
http://www.sailnet.com/photogallery/...ll=no&dthumbs=


Remember the law of sines? Itís handy. It says that for any triange the ratios of the length of each side to the angle opposite that side are equal.

If A is the angle between the heading you need and the bearing from your boat to your friends,

that means: sin60 / 7t = sinA / 5t

or sin Ė1 (5/7 * sin 60) = A

A = 38.2 Deg

Your heading then should be 96 + 38.2 + 180 = 314.2 Deg T

How long to get there?

We all remember that the sum of all the internal angles of a triangle = 180 so the remaining angle is 180 Ė 38.2 Ė 60 = 81.8 Deg

From the law of sines, 7t = 84 * sin 60 / sin 81.8

t =10.5 hours

There are 10 kinds of people. Those who understand binary and those who don't.

Last edited by CapnHand; 02-22-2007 at 09:22 PM.
CapnHand is offline  
Quote Share with Facebook
 
 
For the best viewing experience please update your browser to Google Chrome