| my account | login-logout | resources | support | catalog | home | get webcard |

Online Classroom


Post New Topic  Post A Reply
search | help desk | commons
  next oldest topic   next newest topic
» Online Classroom   » Inland and Coastal Navigation   » Public Discussion of Inland and Coastal Navigation   » Question 6-4

   
Author Topic: Question 6-4
Sandy Englehart


 - posted February 17, 2005 09:07 AM      Profile for Sandy Englehart           Edit/Delete Post 
I used the Luminous Range chart, entered the values, and came up with something that looks like 13. Am I using the chart wrong? Thanks.
David Burch


 - posted February 22, 2005 01:41 PM      Profile for David Burch           Edit/Delete Post 
No you are not using it wrong. The problem is we have two different diagrams in the materials. There is a short note on this in the Tech Support section.

The Luminous range diagram in the printed Chart Problems book is from the US Light List. We have since added a much better one (I think from a Canadian pub) and you can find it at this Luminous range discussion.

A section is below with this problem marked, and i would have to say that something like 11 miles would be the best luninous range for this example. The original diagram had the curve at 5.5 nm visibility, so that is why the problem is stated that way, and the new one shows the curve at 5 nm. In practice we would never know the distinction, because we cannot know the visibility to that precision.

Note that our trick formula for luminous range gives: 5.5/10 x 18 + 1 = 10.9, so 11 is pretty good guess for this.

 -

From: Starpath, Seattle, WA


All times are Pacific  
Post New Topic  Post A Reply Close Topic    Move Topic    Delete Topic next oldest topic   next newest topic
Hop To:

Starpath School of Navigation

Copyright, 2003-2021, Starpath Corporation

Powered by Infopop Corporation
UBB.classicTM 6.3.1.1