Point of Safe Return (PSR) 14

Transposing the Formula to the Navigation Computer

Figure 14.3 Transposing formula to the navigation computer

The Effect of Wind on the Location of the PSR

	Let E = 10 h; TAS = 300 kt.
	STILL AIR	T	=	×	=	min
				+
		D	=		min @ kt
			=		NM	(Answer 1500 NM)
	50 kt HEADWIND OUT	T	=	×	=	min
				+
		D	=		min @ kt
			=		NM	(Answer 1458 NM)
	50 kt TAILWIND OUT	T	=	×	=	min
				+
		D	=		min @ kt
			=		NM	(Answer 1458 NM)

Point of Safe Return (PSR) 14

•In still air the distance to the PSR is the greatest.

•Any wind component reduces the distance to the PSR.

•This distance is the same for a HEAD or TAIL wind of the same value.

•The greater the wind component the greater the reduction in the distance to the PSR.

261

14 Point of Safe Return (PSR)

Single Leg PSR

(PSR) Return Safe of Point 14

Figure 14.4 Example PSR

Study the Figure 14.4 and, using the formula, calculate the time and distance to the PSR. The aircraft is flying towards its destination at a TAS of 220 kt with a wind component of +45 kt. Its Total Endurance is 7 h 40 min and the Safe Endurance is 6 h; use TAS 220 kt throughout.

Workings

	6 × 175	= 2 h 23 mins @ 265 kt = 632 NM
	265 + 175
		(Answer: 2 h 23 min, 632 NM)

262

Questions - Single Leg PSR

1.	Given:
	Total endurance		7 h 40 min
	Safe endurance		6 h
	GS Out		230 kt
	GS Home		170 kt
	What is the time and distance to the PSR from “A”?
	a.	2 h 33 min	587 NM
	b.	3 h 15 min	750 NM
	c.	3 h 27 min	794 NM
	d.	2 h 33 min	434 NM
2.	Given:
	Total endurance		5 h
	Reserves required		1 h
	GS On		250 kt
	GS Out		280 kt
	GS Home		320 kt

What is the time and distance to the PSR from “A”?

	a.	2 h 40 min	747 NM
	b.	2 h 15 min	629 NM
	c.	2 h 08 min	597 NM
	d.	1 h 52 min	523 NM
3.	Given:
	Total endurance		300 min
	Required reserves		45 min
	TAS		140 kt
	Course		050°
	W/V		270°/30

What is the time and distance to the PSR from “A”?

a.	148 min	401 NM
b.	125 min	338 NM
c.	90 min	242 NM
d.	106 min	287 NM

Questions 14

Questions 14

263

14 Questions

Questions 14

4.	Given:
	TAS		160 kt
	W/V		100°/30
	A to B		1620 NM
	Course		030°
	Depart A at		09:30 UTC
	Total endurance		4 h
	Safe endurance		3 h 20 min
	What are the distance, time and ETA to the PSR from “A”?
	a.	94 min	231 NM	11:04
	b.	106 min	261 NM	11:16
	c.	128 min	315 NM	11:38
	d.	106 min	296 NM	11:16
5.	Given:
	TAS		500 kt
	W/V		330°/50
	A to B		4600 NM
	Course		090°
	Total endurance		12 h
	Safe endurance		10 h

What is the time and distance to the PSR from “A”?

a.	4 h 45 min	2480 NM
b.	2 h 22 min	1235 NM
c.	5 h 42 min	2974 NM
d.	4 h 45 min	2242 NM

264

Answers 14

Answers - Single Leg PSR

1.	2 h 33 min	587 NM
	T =		6 × 170		= 2 h 33 min @ 230 kt = 587 NM
		230 + 170
2	2 h 08 min	597 NM
	T =		4 × 320		= 2 h 08 min @ 280 kt = 597 NM
		280 + 320
3.	106 min		287 NM

Use your Navigation Computer to get ground speed out and home.

Remember to balance the drift for both outbound and the reciprocal home legs.

There are no short cuts!

	T =	255 × 116		= 106 mins @ 162 kt = 287 NM
		162 + 116
4.	106 min	261 NM		11:16

Use your Navigation Computer to get ground speed out and home.

Remember to balance the drift for both outbound and the reciprocal home legs.

Answers 14

	T =		200 × 168	= 106 min @ 148 kt = 261 NM	ETA 11:16
		148 + 168
5.	4 h 45 min	2480 NM

Use your Navigation Computer to get ground speed out and home.

Remember to balance the drift for both outbound and the reciprocal home legs.

	T =	10 × 472	= 4 hr 45 min @ 522 kt = 2480 NM
		522 + 472

265