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Posted

From the manual P.40

"In the center of the display is the course deviation indicator. The course deviation dots show the deviation of the current aircraft position from the required course line. Each dot represents a 5-degree deviation from the set course. During an Instrumented Landing System (ILS) landing, the bars show the aircraft deviation from the landing course." P.40

 

I understand when on final approach down the centre line of the ILS the second dot to the left and right = 10 degrees which represents the outside edge of the ILS cone.

What exactly do the 5 or 10 degree dots represent when flying waypoint to waypoint? If they represent 5 or 10 degrees to the course line how do you calculate it without a reference vertex like what you have when in ILS, the vertex being the threshold of the runway?

 

Thank you.

Posted

The reference is in this case the waypoint you are going to.

So you draw a line between the waypoint you are going to and the one you are coming from. Then you draw from the waypoint you are going to and your current position. If those two lines are not perfectly alligned then the angle between those lines is your deviation.

So if you have a deviation of, say 5 degrees and you are heading straight towards the next waypoint your deviation will stay 5 dgrs.

You will overfly the waypoint but 5 dgrs off, of the intended heading.

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Win10 64, Asus Maximus VIII Formula, i5 6600K, Geforce 980 GTX Ti, 32 GB Ram, Samsung EVO SSD.

Posted (edited)
The reference is in this case the waypoint you are going to.

So you draw a line between the waypoint you are going to and the one you are coming from. Then you draw from the waypoint you are going to and your current position. If those two lines are not perfectly alligned then the angle between those lines is your deviation.

So if you have a deviation of, say 5 degrees and you are heading straight towards the next waypoint your deviation will stay 5 dgrs.

You will overfly the waypoint but 5 dgrs off, of the intended heading.

 

Thanks for your response.

When you say:

"You will overfly the waypoint but 5 dgrs off, of the intended heading"

To clarify, I actually won't overfly the waypoint if my waypoint bearing pointer is pointing to the actual WP I'm heading to...I'll just be flying 5 degrees off from the course line the entire time it takes to get to the WP...correct? Only if I'm flying parallel to the course line all the time will I end up being off course by 5 degrees...correct?

Not sure but it looks like you just made an edit.

Edited by fitness88
Posted

As long as you are headed towards your waypoint the deviation will not change. If it was 10° 20nm ago it will still be 10° if you are still lined up to intercept your waypoint.

If you go parallel to your route your deviation will increase the closer you get to you get to your waypoint, but this way you will never intercept it but pass it on your left or right.

 

Hope I got your question right..

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Win10 64, Asus Maximus VIII Formula, i5 6600K, Geforce 980 GTX Ti, 32 GB Ram, Samsung EVO SSD.

Posted
As long as you are headed towards your waypoint the deviation will not change. If it was 10° 20nm ago it will still be 10° if you are still lined up to intercept your waypoint.

If you go parallel to your route your deviation will increase the closer you get to you get to your waypoint, but this way you will never intercept it but pass it on your left or right.

 

Hope I got your question right..

 

Yes I got it 100%...with the parallel course the deviation will open up as I approach the designated WP.

Thanks.

Posted
:thumbup:

 

 

BTW, it's nice to know how many degrees you are off course line but how do I convert the degrees off course into distance off course?

Posted

pic158.jpg

 

Use this: a=SinA*c

 

a is the distance you want to find.

A is your deviation.

c is distance from you to WP

[sIGPIC][/sIGPIC]

Win10 64, Asus Maximus VIII Formula, i5 6600K, Geforce 980 GTX Ti, 32 GB Ram, Samsung EVO SSD.

Posted (edited)
pic158.jpg

 

Use this: a=SinA*c

 

a is the distance you want to find.

A is your deviation.

c is distance from you to WP

 

If I'm 10 miles from WP and 5 degrees deviation from course line then the distance to course line is 5*10=50 miles to course line? What am I missing here, it can't be that easy? What does the Sin represent before 'A'?

Edited by fitness88
Posted

No, your deviation is 5000' in this situation. Just under 1nm. Or if you prefer, 6deg at 10nm is 1nm.

 

If I'm 10 miles from WP and 5 degrees deviation from course line then the distance to course line is 5*10=50 miles to course line? What am I missing here, it can't be that easy? What does the Sin represent before 'A'?

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I used to play flight sims like you, but then I took a slammer to the knee - Yoda

Posted

Typically you don't much care about how far off course you are and more on how to get back onto course.

 

For that you use the direction indicator (the one that points directly to the waypoint). If it's a EGI/GPS/INS/whatever waypoint it will point straight to those coordinates as calculated by the nav system. If it's a TACAN it uses the radio signal to produce a bearing to the station.

 

Typical intercept deviation is 30*, so turn 30* towards the deviation line/needle. If the needle is pegged on the extreme right or left you will need to use the directional needle. Turn until it is on the opposite side to the deviation. So if the deviation is MAX right, turn right until the direction to station/WP is to the left of your heading by 10-30*.

 

If you are really concerned you are way, way off course and have no idea by how much you can always go for a full 90* intercept. Just turn towards the deviation and put the deviation bar horizontal across the HSI. This is the fastest way back on course, however, when the deviation needle unpegs you will not have the time required to turn the 90* onto course before you shoot through the course. Not without a 6g turn anyway.

 

Not sure how much of this is simulated in the 15 though regarding bearing to station/tacan.

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