Critical Point

Question 1

Critical Points (CP)

Answer 1

● This is the point at which it takes the same amount of time to GO ON to the airport that is ahead of the aircraft, as it will to GO BACK to the airport that is behind the aircraft.
● Critical points also go by the names Equi-Time Point (ETP) and Point of Equal Time (PET)

Total distance x GSout/ (GSout+GSback)
G/S out = ground speed towards destination
G/S back = ground speed towards departure aerodrome.
CP = distance of critical point from the destination

Example
Distance from departure to destination (total) = 1 000 NM
TAS = 250 KT
Headwind = 50 KT

First calculate ground speed out and ground speed back.
GS (out) = 250 – 50 = 200 KT
GS (back) = 250 + 50 = 300 KT

1000 x 200/ (200+300) = 400 nm
CP is 400 NM short of the destination.

Double Check
400 NM / 200 KT = 2 hours to continue to destination
600 NM / 300 KT = 2 hours to turn back

Question 2

CP Situations

Answer 2

● The situation for which the CP is being calculated will determine the speeds that are used in the formula.
● You may also need to calculate time to the critical point, in which case you would use the ground speed OUT.
● Example.
➢ If the total distance is 1 000 NM and the CP is at 400 NM (from the destination), then the CP is 600 NM from the departure aerodrome (1 000 - 400 = 600).
● If the ground speed out is 200 KT, calculate the time it will take to travel from the departure aerodrome to the CP

Question 3

● Three situations:

Answer 3

➢ Normal Operations use the normal TAS.
➢ Depressurized CP use the depressurized TAS
➢ One Engine Inoperative use the TAS for one engine inoperative.
● If the aircraft depressurised prior to the depressurized CP it is faster to turn back.
➢ and vice versa

Question 4

One Engine CP Example

Answer 4

● The same formula is used to calculate the critical point distance from destination for the scenarios given on the previous slide
● The GS out and GS back would both be based on the TAS for flying at a lower altitude (depressurized) or with one engine as applicable

● Example:
TAS at planned FL250 is 250 Knots One engine TAS at 15,000’ to return or proceed is 160 Knots Headwind to destination 1000 NM away is 50 Knots

Solution: (1000 * 110) / (110 + 210) = 344 NM
● It will take the same time to go to destination as to turn around on one engine when 344 NM from destination with an engine failure

Question 5

Wind and CP

Answer 5

● When nil wind conditions exist, the CP will be exactly halfway between the two airports.
● If there is a headwind, the CP will be closer to destination (since travelling back is faster due to tw).
● If there is a tailwind, the CP will be closer to departure aerodrome.

Question 6

Remember

Answer 6

● The fuel burn back from the CP equals the fuel burn out from the CP.
● The flight out to the CP does not enter the equation at all.

Question 7

Critical Point and Point of No Return

Answer 7

● Some may ask what is the difference between the critical point and the Point of No Return (PNR)?
➢ Good question
● The CP is simply a distance calculation.
➢ It doesn’t think at all about the fuel.
● The PNR, however, is a fuel based question.
➢ It will ensure that we get home with enough fuel－and a reserve when we get there.
● Although the equations may initially appear to be similar, you will note that the PNR has endurance in place of distance