Point of EqualTime (PET)

The Effect of Wind on the Position of the PET:

Let A to B total distance D = 500 NM and TAS = 300 kt.

STILL AIR	X	=	500		×
					+
		=	500		×	300
			300		+	300
		= HALFWAY
60 kt HEADWIND	X	=	500		×
					+
		=	500		×	360
			240		+	360

=250 NM

=300 NM

= Greater than HALFWAY

60 kt TAILWIND	X	=	500		×
					+
		=	500		×	240		= 200 NM
			360		+	240

= Less than HALFWAY

•In still air the PET is HALFWAY

•If there is a wind then the PET moves INTO WIND

•The stronger the wind the greater the movement INTO WIND

•A gross error check. If you have a headwind component outbound the PET has to be more than halfway between departure and destination

13

Point of Equal Time (PET) 13

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13 Point of EqualTime (PET)

Single Sector All-engine PET

Fill in the ground speed rectangles at Figure 13.3 and calculate the distance and time to the all-engine PET.

Wind Component A to the PET is +45 kt

Wind Component PET to B is -10 kt

The all-engine TAS	475 kt
Engine failure TAS	380 kt
The route distance	2050 NM

(PET) Time Equal of Point 13

		Figure 13.3 Example all-engine single leg PET
GS H	475	- 45 = 430
GS On	475	- 10 = 465
GS out to PET	475	+ 45 = 520

	X =	2050 × 430	= 985 NM
		465 + 430

985 NM @ GS out 520 = 113.5 min

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Point of EqualTime (PET) 13

Engine Failure PET

Figure 13.4 Engine failure Point of Equal Time

The loss of a power unit will necessitate invariably a “drift down” to a stabilizing pressure altitude where the aircraft will either continue ON, or return HOME, at the reduced engine failure TAS/GS, depending on whether the failure occurred before or after the ETA (computed at the all-engine TAS/GS) for the engine failure PET. If the engine failure happened at the PET then, in theory, the pilot could choose to fly to either airfield as the flight times are equal.

Point of Equal Time (PET) 13

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13 Point of EqualTime (PET)

With reference to Figure 13.5 fill in the ground speed rectangles and calculate the distance and time to the engine failure PET.

(PET) Time Equal of Point 13

Figure 13.5 Example engine failure PET single leg

GS H	380	- 45 = 335
GS On	380	- 10 = 370
GS out to PET	475 + 45 = 520

	X =	2050 × 335	= 974 NM
		370 + 335

974 NM @ GS out 520 = 112.5 min

The difference in distance between an all-engine and engine failure PET can be seen to be very small, even though in these two examples there was a difference in all-engine and engine failure TAS of 95 kt. Thus an engine failure PET is normally constructed, which may then be used for serious occurrences other than power unit failure.

To calculate the distance X to an engine failure PET use the engine failure TAS to calculate O and H in the formula.

To calculate the distance X to an all-engine PET use the all-engine TAS to calculate O and H in the formula.

To calculate the time to fly to an all-engine or an engine failure PET use the all-engine TAS to calculate the ground speed from the departure point to the PET.

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Questions - 1

1.Given:

Distance from A to B 1200 NM

GS On	230 kt
GS Home	170 kt

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

a.	600 NM	2 h 37 min
b.	510 NM	2 h 13 min
c.	690 NM	3 h
d.	510 NM	3 h

2.Given:

Distance from A to B 3200 NM

GS On	480 kt
GS Home	520 kt

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

a.	1664 NM	3 h 12 min
b.	1600 NM	3 h 20 min
c.	1664 NM	3 h 28 min
d.	1536 NM	3 h 12 min

3.Given:

TAS

400 kt

Distance from A to B 2000 NM

A 40 kt headwind is forecast from A to B

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

	a.	1100 NM	3 h 03 min
	b.	1100 NM	2 h 30 min
	c.	900 NM	2 h 30 min
	d.	1000 NM	2 h 47 min
4.	Given:
	TAS	165 kt
	W/V	090°/35
	A to B	1620 NM
	Course	035°

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

a.	903 NM	6 h 04 min
b.	810 NM	5 h 42 min
c.	708 NM	5 h
d.	912 NM	6 h 26 min

Questions 13

Questions 13

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13 Questions

Questions 13

5.	Given:
	TAS	500 kt
	W/V	330°/50
	A to B	2600 NM
	Course	090°
	What is the distance and time to the PET from “A”?
	a.	1365 NM	2 h 36 min
	b.	1235 NM	2 h 22 min
	c.	1235 NM	2 h 36 min
	d.	2012 NM	3 h 53 min

Questions 6, 7 and 8 are engine failure cases

6.	Given:
	GS On		300 kt
	GS Out		350 kt
	GS Home		250 kt
	Distance from A to B 1200 NM
	What is the distance and time to the PET from “A”?
	a.	545 NM	1 h 34 min
	b.	654 NM	1 h 52 min
	c.	500 NM	1 h 40 min
	d.	545 NM	1 h 49 min
7.	Given:
	2 Engine TAS		450 kt
	1 Engine TAS		350 kt
	Distance from A to B		3000 NM with a 50 kt tailwind component.

What is the distance and time to the engine failure PET?

a.	1285 NM	3 h 12 min
b.	1333 NM	2 h 40 min
c.	1714 NM	3 h 43 min
d.	1285 NM	2 h 34 min

8.Given:

2 Engine TAS 480 kt

1 Engine TAS 400 kt

W/V	330°/80
A to B	3500 NM
Course	200°

What is the distance and time to the engine failure PET from “A”?

a.	1515 NM	3 h 23 min
b.	1558 NM	2 h 56 min
c.	1515 NM	2 h 51 min
d.	1985 NM	3 h 44 min

250

Questions 13

Given the following data answer questions 9 &10

CAS	190 kt cruising
Pressure altitude	9000 ft
Temperature	ISA -10°C
W/V	320/40 kt
A to B is a distance	350 NM
Course	350°
Endurance	3 hours

9.What is the distance to the PET?

a.220 NM

b.311 NM

c.146 NM

d.204 NM

10.Given an actual time of departure (ATD) of 11:05, what is the ETA for the PET?

a.12:49

b.12:13

c.11:55

d.12:26

Questions 13

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13

Answers 13

Answers

Answers - 1

1. 510 NM		2 h 13 min
		X	=	1200 × 170	= 510 NM @ 230 kt = 2 h 13 min
				230 + 170
2.	1664 NM	3 h 28 min
		X	=	3200 × 520	= 1664 NM @ 480 kt = 3 h 28 min
				480 + 520
3.	1100 NM	3 h 03 min
		X	=	2000 × 440	= 1100 NM @ 360 kt = 3 h 03 min
				360 + 440
4.	912 NM	6 h 26 min

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

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

There are no short cuts!

	X =	1620 × 183	=	912 NM @ 142 kt = 6 h 26 min
		142 + 183

5. 1235 NM 2 h 22 min

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

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

	X	=	2600 × 472	= 1235 NM @ 522 kt = 2 h 22 min
			522 + 472
Engine Failure Case
6. 545 NM	1 h 34 min
	X	=	1200 × 250	= 545 NM @ 350 kt = 1 h 34 min
			300 + 250
7. 1285 NM	2 h 34 min
	X	=	3000 × 300	= 1285 NM @ 500 kt = 2 h 34 min
			400 + 300

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Answers 13

8. What is the distance and time to the engine failure PET from “A”

1515 NM 2 h 51 min

Use your Navigation Computer to get ground speed on, out and home. Remember to balance the drift for both outbound and the reciprocal home legs.

	There are no short cuts!
	X =	3500 × 342	= 1515 NM @ 530 kt = 2 h 51 min
		448 + 342
	Information for Qs 9 & 10
	ISA at 9000 ft		-3
	ISA deviation		-10
	OAT		-13°C

Navigation Computer for TAS 214 kt

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

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

Ground speed On & Out

Ground speed Home

9. 204 NM

	X =	350 × 249
		178 + 249

=178 kt

=249 kt

=204 NM

10.12:13

204 NM @ 178 kt = 1 h 08 min + 11:05 = 12:13

Answers 13

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13 Questions

Questions - 2

Questions 13

1.Given: Track 355°T

W/V 340°/30 kt TAS 140 kt

Total distance A to B 350 NM

What are the time and distance to the point of equal time between A and B?

a. 75 min	211 NM
b. 75 min	140 NM
c. 50 min	140 NM
d. 114 min	211 NM

2.Given:

Course A to B 088°(T) Distance 1250 NM Mean TAS 330 kt W/V A to B 340°/60 kt

The time from A to the Point of Equal Time between A and B is:

a.1 h 54 min

b.1 h 44 min

c.1 h 39 min

d2 h 02 min

3.Distance between airports = 340 NM True track = 320°

W/V = 160°/40 TAS = 110 kt

Distance to PET is:

a.121 NM

b.219 NM

c.112 NM

d.228 NM

4.Flying from A to B, 270 NM True track 030°

W/V 120°/35

TAS 125 kt

What are the distance and time to the Point of Equal Time?

a. 141 NM	65 min
b. 141 NM	68 min
c. 135 NM	68 min
d. 150 NM	65 min

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Questions 13

Questions 13

255

13 Answers

Answers - 2

1	2	3	4
d	b	c	c

Answers 13

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