Point of Equal Time (PET) — Page 250, Lesson 256

I want to walk you through the Point of Equal Time, or PET — sometimes also called the Critical Point. This is a key concept in flight planning, and I'll start by showing you how the formula is derived and how wind affects its position. Let's begin with the effect of wind on the position of the PET. Imagine a route from point A to point B, with a total distance D of 500 nautical miles, and a true airspeed, or TAS, of 300 knots. In still air, the PET is exactly halfway. Let me show you the calculation: X equals D multiplied by the ground speed home, divided by the sum of the ground speed out plus the ground speed home. In still air, ground speed out and ground speed home are both equal to TAS, so X equals 500 times 300, divided by 300 plus 300, which gives 250 nautical miles — exactly halfway. Now, if we introduce a 60-knot headwind on the outbound leg, the ground speed out becomes TAS minus the headwind, so 300 minus 60 equals 240 knots. The ground speed home, with the wind now behind you, becomes TAS plus the wind, so 300 plus 60 equals 360 knots. Plugging into the formula: X equals 500 times 360, divided by 240 plus 360, which gives 300 nautical miles — greater than halfway. If instead we have a 60-knot tailwind on the outbound leg, the ground speed out becomes 300 plus 60 equals 360 knots, and the ground speed home becomes 300 minus 60 equals 240 knots. Then X equals 500 times 240, divided by 360 plus 240, which gives 200 nautical miles — less than halfway. So the key rules are: in still air, the PET is halfway. If there is a wind, the PET moves into wind. The stronger the wind, the greater the movement into wind. And here's a useful gross error check: if you have a headwind component outbound, the PET has to be more than halfway between departure and destination. Now let's move to a single-sector all-engine PET example. I want you to look at Figure 13.3, which shows an example all-engine single leg PET. In this example, the wind component from A to the PET is plus 45 knots — that means a tailwind outbound. The wind component from the PET to B is minus 10 knots — a headwind on that leg. The all-engine TAS is 475 knots. The engine failure TAS is 380 knots. The route distance is 2050 nautical miles. We fill in the ground speed rectangles. Ground speed home, which I'll call GS H, is the all-engine TAS minus the wind component from PET to B: 475 minus 10 equals 465 knots. Ground speed on, which is the ground speed from the PET onward to B, is the same — 465 knots. Ground speed out to the PET is the all-engine TAS plus the wind component from A to PET: 475 plus 45 equals 520 knots. Now we calculate the distance X to the all-engine PET using the formula: X equals D times GS H, divided by GS out plus GS H. So X equals 2050 times 465, divided by 430 plus 465. Wait — let me check that carefully. In the excerpt, GS H is given as 430? No — I see the excerpt shows GS H as 475 minus 45 equals 430. Let me re-read. The excerpt says: "GS H: 475 - 45 = 430" and "GS On: 475 - 10 = 465" and "GS out to PET: 475 + 45 = 520". So GS H is 430 knots, and GS On is 465 knots. Then X equals 2050 times 430, divided by 465 plus 430, which gives 985 nautical miles. The time to fly to the PET is 985 nautical miles at a ground speed out of 520 knots, which equals 113.5 minutes. Now let's talk about the engine failure PET. When an engine fails, the aircraft will invariably drift down to a stabilizing pressure altitude. At that altitude, the aircraft will either continue on to the destination or return home, using the reduced engine failure TAS and ground speed. The decision depends on whether the failure occurred before or after the estimated time of arrival, or ETA, computed at the all-engine TAS and ground speed, for the engine failure PET. If the engine failure happens exactly at the PET, then in theory the pilot could choose to fly to either airfield because the flight times are equal. Let's work through the example in Figure 13.5. The wind components are the same: plus 45 knots from A to the PET, and minus 10 knots from PET to B. The all-engine TAS is still 475 knots, but now for the engine failure PET we use the engine failure TAS of 380 knots to calculate the ground speeds. Ground speed home, GS H, is engine failure TAS minus the wind component from PET to B: 380 minus 10 equals 370 knots. Ground speed on, GS On, is the same — 370 knots. Ground speed out to the PET is still the all-engine TAS plus the wind component from A to PET: 475 plus 45 equals 520 knots. Why do we use all-engine TAS for the outbound ground speed? Because the engine failure hasn't happened yet — you're still flying out with all engines operating until you reach the PET. Now calculate the distance X to the engine failure PET: X equals 2050 times 335, divided by 370 plus 335. Wait — the excerpt shows GS H as 380 minus 45 equals 335? Let me check. Yes: "GS H: 380 - 45 = 335" and "GS On: 380 - 10 = 370". So GS H is 335 knots. Then X equals 2050 times 335, divided by 370 plus 335, which gives 974 nautical miles. The time to fly to the PET is 974 nautical miles at a ground speed out of 520 knots, which equals 112.5 minutes. Notice that the difference in distance between the all-engine PET and the engine failure PET is very small — only 11 nautical miles in this example — even though the all-engine TAS and engine failure TAS differed by 95 knots. This is why an engine failure PET is normally constructed, and it can then be used for serious occurrences other than just power unit failure. Here's the key rule to remember: to calculate the distance X to an engine failure PET, use the engine failure TAS to calculate the ground speeds 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. But to calculate the time to fly to either an all-engine or an engine failure PET, always use the all-engine TAS to calculate the ground speed from the departure point to the PET — because you're flying that leg with all engines operating.