• Posts: 27

### jukzizy created the topic: Cpl Navigation

Please i need to know how to calculate the greatest: I got a question about it on exam last friday and i'm planing to sit again this friday.

What is the distance between Alpha (S15 E120) andBravo (S15 E150)?

Please anyone with an explanation help me..
a)Greater than 900nm
b)900nm
c)less than 900nm

I chosed chalie n got it wrong

• Posts: 84

### Carello replied the topic: Cpl Navigation

What is the distance between Alpha (S15 E120) andBravo (S15 E150)?

Please anyone with an explanation help me..
a)Greater than 900nm
b)900nm
c)less than 900nm

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 Alpha and Bravo can be calculated as follows:
Dep = ChLon x Cos(Lat)
Dep = 30x60 Cos 15 = 1738.7 nm

So the answer you want is > 900nm i.e. the answer is a

If the A&B were on the equator the distance between them would be: 1800 Cos 0 = 1800 nm - recall that one minute of longitude at the equator equals 1nm.
If A&B were at 90S the distance between them would be: 60 Cos 90 = 0 nm - meridians converge at the pole

With the above in mind, if you don't know the formula or can't use a calculator in the exam, you can reason that the distance between A&B will be a little less than 1800nm as A&B are on a parallel a little south of the equator.