Need assistance
 

Hi everyone;

I need an urgent assistance / question to be answered with its source.I thought that this is one of the best forum that can help me.
I would be appreciative af any information provided by your skills n experiences regarding with the below question.I am just a new member,so i didnt have too much time to read all posted.
My question is as follows:

An aircraft 60 miles from a vor station has a CDI indication of one-fifth deflection this represents a course centerline deviation of approximitaley .......?

I need the answer and the source if available to be utılised at my class homework.

Thank u all for ur kind intention of helping.

With my best regards,

VOR CDI Deflection
 

This web site should help.

http://www.flightsimaviation.com/avi...on_part_1.html

Well to be plain with u ,i am a rookie at this subject.And unfortuantely i couldnt undertand what is mentioned there.Is there any1 here who can give the answer.This answer is my fortune.

1-60 Rule
 

Hi,I`ll try and have a go:}Thinks it 1nm made easy being 60nm
60nm brings 1-60 to mind.You know your 2 degrees off from the CDI.
For future questions did it like this.
Lots of people use the whiz-wheel but for the written exams i wrote out a
Formula: dis gone/dis off x 60 + dis to go/dis off x 60=c.a.
Kind of makes sense "gone" then what you have "to go"
You only need to use the first part,ie using track error.Which is 2 degrees!
Now solve by moving the formula around and you see it equals 2nm off i believe.By canceling the 60`s out.
Is this the correct answer???

i really dont no theanswer..it is supposed to be 2 or 3..but 10x for ur effort and help.
appreciate u.

I thought a 1/5th deflection on the cdi equals 2 degrees? At 60nm, 2degrees = 2nm course centreline deviation?

10x a lot..i suppose it is 2...u r so kind.10x for ur effort and help.
regards