Hello,
I got during my IR rating exam an interesting PSR\PNR question where I am going right now crazy to get to a solution. I asked now already everyone at flight schools, looked in different forums, but could not find an answer. I even asked already the authority, but they were not authorized to help me...
Hopefully you guys can help me here with a solution.
The question simple:
 distance A to B 2000NM
 TAS, wind, endurance constant
 5000kg fuel (+500kg extra)
 PET from A 1200NM
what is the PNR\PSR?
answeres: I cant remember the values any more but for the approach process it should be not an issue. There were 5 answeres and I chose c.) which was 1200NM (same as PET and which is wrong).
My approach was the following, due to the fact that there are to less variables for calculations, no given fuel flow,..., it's more a theoretical question. TAS, wind, endurance are constant, so I thought it might be one of the rare cases where PET=PSR, which leads me to another question:
 Can PSR be equal PET and\or is PSR always more\less then PET?
I assume that when there is no wind and PET is exactly half of the distance, is there also PSR half of the distance or more\less?
In the example above I know I have headwind doe to the fact that PET is closer to B, so if PSR is maybe always more then PET then there might was only one answer which was more then 1200nm...
Or maybe I am completely wrong...,
anyways, appreciate your help in this case!
BR, Rusty
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Instrument rating exam  PSR/PNR question
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Instrument rating exam  PSR/PNR question
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Originally posted by JJ Jones View Post
My best guess is 300 kts.
Firstly the term GS Home divided by GS Out + GS Home is the same in both equations.
So for the PET 720 = 1200 x 0.6 (the 60 percent in the question).
So GS Home divided by GS Out + GS Home is equal to 0.6
For the PSR the time to the PSR is the endurance x GS Home divided by GS Out + GS Home
PSR = 8.4 hrs x 0.6 = 5.04 hrs
84 percent of AB is 1008 miles
1008 miles over 5.04 hours requires a speed of 200 knots
So the GS out is 200 Knots and for the GS Home divided by the GS Out + GS Home to be 0.6, the GS home needs to be 300 knots.
Best I can do with what you've given me.
BR
JJ
At least you showed me a way to come to results.
I can only remember my answer which was 200kts...what I got somehow with calculation of unknown variables.
But your way...hey, really I think for a poor rookie just trying to get the exam done it's pretty heavy...
Thanks, Rusty

This type of question hasn't appeared in an EASA ATPL exam recently. Which exam was it in and where did you take it?
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My best guess is 300 kts.
Firstly the term GS Home divided by GS Out + GS Home is the same in both equations.
So for the PET 720 = 1200 x 0.6 (the 60 percent in the question).
So GS Home divided by GS Out + GS Home is equal to 0.6
For the PSR the time to the PSR is the endurance x GS Home divided by GS Out + GS Home
PSR = 8.4 hrs x 0.6 = 5.04 hrs
84 percent of AB is 1008 miles
1008 miles over 5.04 hours requires a speed of 200 knots
So the GS out is 200 Knots and for the GS Home divided by the GS Out + GS Home to be 0.6, the GS home needs to be 300 knots.
Best I can do with what you've given me.
BR
JJ
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Hi
Did you have a list of possible answers ? I can't see that we have enough data to answer this from what you have quoted.
BR
JJ
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Got again at the last exam a question, maybe someone can help me with this one. Even this one I was not able to figure out or find any solution hints in the internet.
Dist. A  B = 1200Nm
PSR is 84% of AB
PET is 60% of AB
Endurance = 8h24mins
What is the Groundspeed from PSR to A?
I am going already crazy with these questions...everyone I ask, no one has a clue
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