PET/PSR Questions
 

The equation needed to work out the pet and psr confuses me. ExOxH over O+H. How do we work out the endurance part (i.e. 5 hours for example)?

Its probs a very simple solution. Thanks in advance

Heres an example
For a distance of 1860 NM between Q and R, a ground speed "out" of 385 kt, a ground speed "back" of 465 kt and an endurance of 8 HR (excluding reserves) the distance from Q to the point of safe return (PSR) is

You do
E*H/H+O
So, 8*465/850 (Which is 465+ 385)
which gives you an answer of 4.37 this is your endurance if you wanted the distance all you do is multiply 4.37 by your outbound speed.
Hope this helps,
P.s the correct answer to the above question was 1685Nm
Here is a PET question,
Given: Distance 'A' to 'B' 2484 NM Groundspeed 'out' 420 kt Groundspeed 'back' 500 kt The time from 'A' to the Point of Equal Time (PET) between 'A' and 'B' is:
so you do
2484*500/500+420 =1350nm so i divide this by my outbound speed 420 which gives me 193mins (3 hours and 12 mins) which is the right answer

I have looked at a few questions in the QB and most questions already have the endurance stated so I dont think it will cause as many problems as i thought.


cheers anyway

I only had one of each on both FPL and Gen Nav in March - be warned , both were worth a 2 and 3 mark respectively ! make sure you are able to recognise and calculate each one as in my exam they put the psr answer on a pet question incase you worked it out wrong , being the little charmers that they are :}