The critical point (point of equal time)

Navigation · Chapter 18 · DGCA CPL and ATPL

The point on a route from which going on and going back take the same time, and how wind and an engine failure move it.

Remember

CP: d = D × H ÷ (O + H) from A; t = d ÷ O.
No wind or an exact beam wind: half-way. The CP moves into the wind.
Find O and H on the CX-3 (Wind Correction); home track = on track ± 180°.
Engine failure: reduced TAS for O and H in the distance; time to the CP at the normal GS.
The CP is about time, not fuel: fuel never moves it.

Where on a route is going on as quick as going back?

Trouble half-way: go on or go back?

An ocean crossing • You fly from A to B: 600 NM of open sea • There is no airfield in between
Trouble early on • 150 NM out, a passenger is taken ill (or an engine fails) • Back to A takes 0 h 45; on to B takes 2 h 15 • Turn back: it is quicker
Trouble later on • Now the trouble comes 450 NM out • Back takes 2 h 15; on takes 0 h 45 • Go on to B
Somewhere the two are equal • Watch the aircraft fly: the two times change all the way • Where they are equal is the critical point (CP), also called the point of equal time (PET or ETP) • Before it, turn back (teal); after it, go on (magenta)
No wind: the middle • Same speed both ways, so equal time means equal distance • The CP is half-way
Now add a wind • A headwind going to B: O = 160 kt going on, H = 240 kt going home • Watch again: the times are equal further out
Why it moved towards B • Going on is slow, going back is fast • So the CP sits where the slow part is shorter: 360 NM from A, 240 NM to B • Both take 1 h 30
Where the formula comes from • Let d be A to the CP. Time back = {{d // H}}; time on = {{D − d // O}} • They are equal, so d × O = (D − d) × H • d (O + H) = D × H, so d = {{D × H // O + H}}
Same time either way • Two aircraft leave the CP together: one on, one back • The slow one has less far to go: they arrive together
Not the PNR • The CP is about time: which way is quicker • The PNR (next chapter) is about fuel: how far you can go and still turn back • Same route, two different points

Solve it: distance and time to the critical point

Solve it: critical point

Solve it: distance and time to the critical point