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Welcome to the CPL Navigation question and answer forum. Please feel free to post your questions but more importantly also suggest answers for your forum colleagues. Bob himself or one of the other tutors will get to your question as soon as we can.

- boeing777mark
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Hi fellow Aviators

I have my CNAV exam this Friday and people have been telling me that a few questions pop up in regards to working out distances between 2 airports using latitude and longitude. Just wondering if anyone would be able to give me an example on how to work these questions out using 2 airports. My confidence on these types of question is not great and may get me stuck this Friday

Cheers

I have my CNAV exam this Friday and people have been telling me that a few questions pop up in regards to working out distances between 2 airports using latitude and longitude. Just wondering if anyone would be able to give me an example on how to work these questions out using 2 airports. My confidence on these types of question is not great and may get me stuck this Friday

Cheers

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- bobtait
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The only such questions I can imagine would be for locations on the same meridian. Then every degree of latitude difference would represent sixty nautical miles. There is no way CASA would ask you to calculate the distance between locations with different longitude and different latitude coordinates. That would require advanced mathematics including spherical geometry. Your GPS can do that but CASA would not expect you to do it in an exam.

The only other possible question type could be something like this.

Two towns are located on 30° south latitude. Town A is on 151°E and town B is on 152°E. The distance between the two towns is

(a) 60nm

(b) less than 60nm

(c) greater than 60nm

In that case the answer would be 'less than 60nm' because the meridians converge towards the pole, one degree of longitude is not a constant distance. The distance between meridians gets less and less as you move towards the pole. However, the distance between parallels of latitude remains the same where ever it is measured.

The only other possible question type could be something like this.

Two towns are located on 30° south latitude. Town A is on 151°E and town B is on 152°E. The distance between the two towns is

(a) 60nm

(b) less than 60nm

(c) greater than 60nm

In that case the answer would be 'less than 60nm' because the meridians converge towards the pole, one degree of longitude is not a constant distance. The distance between meridians gets less and less as you move towards the pole. However, the distance between parallels of latitude remains the same where ever it is measured.

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- boeing777mark
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Thank you so much Bob for the quick response. That has made it a little easier to understand. Hopefully I can get pass this Friday and then move onto AGK. Appreciate all the effort you guys put into these forums and also the online practice exams.

Cheers

Cheers

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- Carello
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Two towns are located on 30° south latitude. Town A is on 151°E and town B is on 152°E. The distance between the two towns is

(a) 60nm

(b) less than 60nm

(c) greater than 60nm

In that case the answer would be 'less than 60nm' because the meridians converge towards the pole, one degree of longitude is not a constant distance. The distance between meridians gets less and less as you move towards the pole.

Way back in the 70's when I did my SCPL NAV (now ATPL) the syllabus covered a topic on Departure - the distance between to points on the same parallel.

In short, the formula for departure (in nm) was relatively simple:

**Departure (nm) = ChLon x Cos (Lat)** *where ChLon was the change in longitude in minutes.*

In the example above, the distance (departure) between Town A and Town B can be calculated as follows:

Dep = ChLon x Cos(Lat)

Dep = 60 Cos 30 = 51.9 nm

If the towns were on the equator the distance between them would be: 60 Cos 0 = 60 nm

If the towns were at 90S the distance between them would be: 60 Cos 90 = 0 nm (meridians converge at the pole)

Apart from the SCPL NAV exam I don't recall ever using departure to calculate the distance between two points!

(a) 60nm

(b) less than 60nm

(c) greater than 60nm

In that case the answer would be 'less than 60nm' because the meridians converge towards the pole, one degree of longitude is not a constant distance. The distance between meridians gets less and less as you move towards the pole.

Way back in the 70's when I did my SCPL NAV (now ATPL) the syllabus covered a topic on Departure - the distance between to points on the same parallel.

In short, the formula for departure (in nm) was relatively simple:

In the example above, the distance (departure) between Town A and Town B can be calculated as follows:

Dep = ChLon x Cos(Lat)

Dep = 60 Cos 30 = 51.9 nm

If the towns were on the equator the distance between them would be: 60 Cos 0 = 60 nm

If the towns were at 90S the distance between them would be: 60 Cos 90 = 0 nm (meridians converge at the pole)

Apart from the SCPL NAV exam I don't recall ever using departure to calculate the distance between two points!

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