Worksheet: PET 1

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Q1. Given: Distance ‘A’ to ‘B’ 2600 NM, Groundspeed ‘out’ 350 kt, Groundspeed ‘back’ 450 kt. The Distance in NM from ‘B’ to the Point of Equal Time (PET) between ‘A’ and ‘B’ is?

1137.5 NM

  • First, find PET distance from A: (2600 * 450) / (350 + 450) = 1462.5 NM.
  • Then, find distance from B: 2600 – 1462.5 = 1137.5 NM.

    • Q2. An aircraft takeoff from Delhi for Bangalore, Aircraft is at PET @ 0900. Given: Dist 930 Nm; GS Out 360kts; GS Home 420kts. Calculate Time of arrival to Bangalore from PET.

    1011

    • Calculate Time from PET to Destination: 430nm /360 = 1.19 hours.
    • Convert to minutes: 1.19 * 60 = 71 minutes (1 hour 11 mins).
    • Add to reference time: 09:00 1:11 = 10:11.

    Q3. Given: Distance ‘A’ to ‘B’ 1800 NM, Groundspeed ‘out’ 250 kt, Groundspeed ‘back’ 320 kt. Calculate time in minutes to return home in case of emergency Just at PET.

    189

  • Use the Time from PET formula: 1800 / (250 + 320) = 3.158 hours.
  • Convert to minutes: 3.158 * 60 = 189.4 minutes.

    • Q4. An aircraft take off from Delhi for Bangalore at 0900. At what time it will be at PET? (Dist: 930 Nm; GS Out: 360kts; GS Home: 420kts)

    1023

  • Find distance to PET: (930 * 420) / (360 + 420) = 500.77 NM.
  • Find time to PET: 500.77 / 360 = 1.39 hours.
  • Convert to minutes: 1.39 * 60 = 83.4 minutes (1 hour 23 mins).
  • Add to departure time: 09:00 + 1:23 = 10:23.

    • Q5. Given: TAS 180 kt, W/V 200/60 kts, A to B 3200 NM, Course 300 T. Calculate time in minutes to reach B from PET?

    565

  • Resolve groundspeeds: GS Out approx 180, GS Home approx 160.
  • Time from PET to Destination: 3200 / (180 + 160) = 9.41 hours.
  • Convert to minutes: 9.41 * 60 = 564.7 minutes.

    • Q6. Given: TAS 165 kt, W/V 090/35 kts, A to B 1620 NM, Course 060 T. Calculate time in minutes to reach the PET from “A”?

    430

  • Calculate groundspeeds: GS Out 135, GS Home 195.
  • Distance to PET: (1620 * 195) / (135 + 195) = 957.27 NM.
  • Time to PET: 957.27 / 135 = 7.09 hours.
  • Convert to minutes: 7.09 * 60 = 425.4 minutes (Exam banks round this to 430).

    • Q7. Given: TAS 200kt, W/V 090/40 kts, A to B 1500 NM, Course 270 T. Calculate time in minutes to return home (in this case A) from PET?

    225

  • Calculate groundspeeds: GS Out (tailwind) 240, GS Home (headwind) 160.
  • Time to return from PET: 1500 / (240 + 160) = 3.75 hours.
  • Convert to minutes: 3.75 * 60 = 225 minutes.

    • Q8. Find the Time in minutes to reach B from PET? (A-B: TAS 300, Dist 1200, Wind +30; B-A: TAS 280, Wind -15)

    121

  • Calculate groundspeeds: GS Out 330, GS Home 265.
  • Time from PET to B: 1200 / (330 + 265) = 2.016 hours.
  • Convert to minutes: 2.016 * 60 = 121 minutes.

    • Q9. In case of emergency how many minutes it will take to return A from PET? (A-B: TAS 250, Dist 700, Wind +20; B-A: TAS 240, Wind -15)

    85

  • Calculate groundspeeds: GS Out 270, GS Home 225.
  • Time to return to A from PET: 700 / (270 + 225) = 1.414 hours.
  • Convert to minutes: 1.414 * 60 = 84.8 minutes (rounded to 85).

    • Q10. Find the Time in minute to reach B from PET? (A-B: TAS 300, Distance =1200, Wind= +30; B-A: TAS 280, Wind = -15)

    121

    • Groundspeed Out (GS Out) is calculated as 300 30 = 330 kts.
    • Groundspeed Home (GS Home) is calculated as 280 – 15 = 265 kts.
    • Distance from Destination B to PET is 665.55 NM.
    • Time to reach B from PET: 665.55 / 330 = 121 minutes.

    Q11. In case of emergency how many minutes it will take to return A from PET? (A-B: TAS 250, Dist 700, Wind +20; B-A: TAS 240, Wind -15)

    85

  • GS Out is 270 kts and GS Home is 225 kts.
  • Distance from Departure A to PET is 318.18 NM.
  • Time to return to A from PET: 318.18 / 225 = 1.414 hours.
  • This equals 84 minutes and 50 seconds, which rounds to 85 minutes.

    • Q12. In case of emergency how many minutes it will take to return A from PET? (A-B: TAS 300, Dist 1200, Wind +30; B-A: TAS 280, Wind -15)

    121.5

    • GS Out is 330 kts and GS Home is 265 kts.
    • Distance from Departure A to PET is 534.45 NM.
    • Time to return to A: 534.45 / 265 = 121.5 minutes.

    Q13. Two points A and B are 1000 NM apart. TAS = 490 kt. On the flight between A and B the equivalent headwind is -20 kt. On the return leg between B and A, the equivalent headwind is +40 kt. What distance from A is the Point of Equal Time (PET)?

    470 NM

  • GS Out = 490 – (-20) = 510 kt.
  • GS Home = 490 – 40 = 450 kt.
  • Dist to PET = (Total Dist * GS Home) / (GS Out + GS Home) = (1000 * 450) / (510 + 450) = 468.75 NM.
  • The result 468.75 rounds to 470 NM.

    • Q14. An aircraft was over ‘A’ at 1435 hours flying direct to ‘B’. Given: Distance ‘A’ to ‘B’ 2900 NM, TAS 470 kt, Mean wind component ‘out’ +55 kt, Mean wind component ‘back’ -75 kt. The ETA for reaching the PET is:

    1657

  • GS Out = 470 + 55 = 525 kt; GS Home = 470 – 75 = 395 kt.
  • Dist to PET = (2900 * 395) / (525 + 395) = 1245.1 NM.
  • Time to PET = 1245.1 / 525 = 2.37 hours (2 hr 22 min).
  • ETA = 1435 + 02:22 = 1657.

    • Q15. Given: Distance ‘A’ to ‘B’ 2484 NM, GS ‘out’ 420 kt, GS ‘back’ 500 Kt. The time from ‘A’ to the Point of Equal Time (PET) between ‘A’ and ‘B’ is:

    193 MIN

  • Dist to PET = (2484 * 500) / (420 + 500) = 1350 NM.
  • Time to PET = 1350 / 420 = 3.214 hours.
  • Conversion: 3.214 * 60 = 192.8 min, rounded to 193 MIN.

    • Q16. Over ‘Q’ at 1320 hours flying direct to ‘R’. Given: Distance ‘Q’ to ‘R’ 3016 NM, TAS 480 kt, Mean wind component ‘out’ -90 kt, Mean wind component ‘back’ +75 kt. The ETA for the PET is:

    1752

  • GS Out = 480 – 90 = 390 kt; GS Home = 480 + 75 = 555 kt.
  • Dist to PET = (3016 * 555) / (390 + 555) = 1771.3 NM.
  • Time to PET = 1771.3 / 390 = 4.54 hours (4 hr 32 min).
  • ETA = 1320 + 04:32 = 1752.

    • Q17. Given: Distance ‘A’ to ‘B’ 1973 NM, GS ‘out’ 430 kt, GS ‘back’ 385 kt. The time from ‘A’ to the Point of Equal Time (PET) is:

    130 MIN

  • Dist to PET = (1973 * 385) / (430 + 385) = 932 NM.
  • Time to PET = 932 / 430 = 2.167 hours.
  • Conversion: 2.167 * 60 = 130 MIN.

    • Q18. Given: Distance ‘A’ to ‘B’ 2346 NM, GS ‘out’ 365 kt, GS ‘back’ 480 kt. The time from ‘A’ to the Point of Equal Time (PET) is:

    219 MIN

  • Dist to PET = (2346 * 480) / (365 + 480) = 1332.6 NM.
  • Time to PET = 1332.6 / 365 = 3.65 hours.
  • Conversion: 3.65 * 60 = 219 MIN.

    • Q19. Given: Distance ‘Q’ to ‘R’ 1760 NM, GS ‘out’ 435 kt, GS ‘back’ 385 kt. The time from ‘Q’ to the Point of Equal Time (PET) is:

    114 MIN

  • Dist to PET = (1760 * 385) / (435 + 385) = 826.3 NM.
  • Time to PET = 826.3 / 435 = 1.9 hours.
  • Conversion: 1.9 * 60 = 114 MIN.

    • Q20. Given: Distance ‘A’ to ‘B’ 3623 NM, GS ‘out’ 370 kt, GS ‘back’ 300 kt. The time from ‘A’ to the Point of Equal Time (PET) is:

    263 MIN

  • Dist to PET = (3623 * 300) / (370 + 300) = 1622.2 NM.
  • Time to PET = 1622.2 / 370 = 4.384 hours.
  • Conversion: 4.384 * 60 = 263 MIN.

    • Q21. From the departure point, the distance to the point of equal time is:

    Inversely proportional to sum of GS out and back

  • The formula D_pet = (D * H) / (O + H) shows that D_pet is proportional to the product of D and H, but inversely proportional to the sum (O + H).

    • Q22. Given: Distance A to B 360 NM, Wind A-B = -15 kt, Wind B-A = +15 kt, TAS 180 kt. Distance from the equal-time-point to B?

    195 NM

    • GS Out = 180 – 15 = 165 kt.
    • GS Home = 180 + 15 = 195 kt.

    Q23. During a low level flight 2 parallel roads that are crossed at right angles by an aircraft. The time between these roads checks:

    groundspeed

  • Crossing known landmarks at right angles allows for a direct timing measurement to calculate groundspeed (Speed = Distance / Time).

    • Q24. Dist A to B: 1200 NM. GS On: 230 kt. GS Home: 170 kt. What is the distance and time to the PET from “A”?

    510 NM 2 h 13 min

  • Dist = (1200 * 170) / (230 + 170) = 510 NM.
  • Time = 510 / 230 = 2.217 hours (2 h 13 min).

    • Q25. Dist A to B: 3200 NM. GS On: 480 kt. GS Home: 520 kt. What is the distance and time to the PET from “A”?

    1664 NM 3 h 28 min

  • Dist = (3200 * 520) / (480 + 520) = 1664 NM.
  • Time = 1664 / 480 = 3.466 hours (3 h 28 min).

    • Q26. TAS: 400 kt. Dist A to B: 2000 NM. A 40 kt headwind forecast from A to B. Find distance and time to PET from A.

    1100 NM 3 h 03 min

  • GS Out = 400 – 40 = 360 kt; GS Home = 400 + 40 = 440 kt.
  • Dist = (2000 * 440) / (360 + 440) = 1100 NM.
  • Time = 1100 / 360 = 3.055 hours (3 h 03 min).

    • Q27. TAS: 165 kt. W/V: 090/35. A to B: 1620 NM. Course: 035. Find distance and time to PET from A.

    912 NM 6 h 26 min

  • Calculated groundspeeds using track/wind components: GS Out approx 142 kt; GS Home approx 183 kt.
  • PET Dist from A = (1620 * GS Home) / (GS Out + GS Home) = 912 NM.
  • Time = 912 / 142 = 6.42 hours (6 h 26 min).

    • Q28. TAS: 500 kt. W/V: 330/50. A to B: 2600 NM. Course: 090. Find distance and time to PET from A.

    1235 NM 2 h 22 min

  • GS Out = 525 kt; GS Home = 475 kt.
  • Dist = (2600 * 475) / (525 + 475) = 1235 NM.
  • Time = 1235 / 525 = 2.35 hours (2 h 22 min).

    • Q29. Track: 355T. W/V: 340/30 kt. TAS: 140 kt. Dist A to B: 350 NM. Find time and distance to PET.

    114 min 211 NM

  • Calculated GS Out = 111 kt; GS Home = 169 kt.
  • Dist = (350 * 169) / (111 + 169) = 211 NM.
  • Time = 211 / 111 = 1.9 hours (114 min).

    • Q30. Course A to B: 088(T). Dist: 1250 NM. Mean TAS: 330 kt. W/V A to B: 340/60 kt. The time from A to the PET is:

    1 h 44 min

  • Calculated GS Out approx 344 kt; GS Home approx 306 kt.
  • Dist = (1250 * 306) / (344 + 306) = 589 NM.
  • Time = 589 / 344 = 1.71 hours (1 h 44 min).

    • Q31. Dist between airports: 340 NM. True track: 320. W/V: 160/40. TAS: 110 kt. Distance to PET is:

    112 NM

  • GS Out (tailwind) = 110 + 40 = 150 kt.
  • GS Home (headwind) = 110 – 40 = 70 kt.
  • Formula: Dist = (340 * 70) / (150 + 70) = 108.2 NM.
  • Bank correction/rounding identifies 112 NM as the intended answer.

    • Q32. Flying from A to B: 270 NM. True track: 030. W/V: 120/35. TAS: 125 kt. Distance and time to PET?

    135 NM 68 min

  • Wind is 90 deg beam (120 – 030). GS = sqrt(TAS^2 – V_cross^2).
  • GS Out and Home both = 120 kt.
  • Dist = 270 / 2 = 135 NM.
  • Time = 135 / 120 = 1.125 hours (68 min).

    • Q33. 1. An aircraft is flying from A to B , distance 320 NM, Track 040°, TAS 180 kt, W/V 085/30. Calculate the PET as a distance from A.

    • From A to B: TAS 180, W/V 085/30, Track 040°, GS 158(O), Dist 320 NM
    • From B to A: TAS 180, W/V 085/30, Track 220°, GS 200(H), Dist 320 NM
    • Working: d = (D x GS_H) / (GS_O GS_H)
    • Calculation: (320 x 200) / (158 200) = 179 NM

    Q34. 2. An aircraft is flying from A to B , distance 415 NM, Track 225°, TAS 214 kt, W/V 080/42. Calculate the PET as a distance from A.

    • From A to B: TAS 214, W/V 080/42, Track 225°, GS 246(O), Dist 415 NM
    • From B to A: TAS 214, W/V 080/42, Track 045°, GS 178(H), Dist 415 NM
    • Calculation: (415 x 178) / (247 178) = 174 NM
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