Time to Turn+α
Heading?
Speed
Dr. Mr. Munski
CG
100
Performing a maintenance flight, a pilot has to fly for 4 hours and 20 minutes in cruise flight. If the winds aloft are 240° at 19 kts, the true course out is 021°, and the TAS in cruise is 145 kts, what is the time to turn back and what is that distance out?
Answer:116 minutes and 309 NM
(130+159=289):4h20min=130:x
time to turn x= 116 minutes
159 GS for 116 minutes=309NM
(National 2011 Q20)
100
You are currently 118 NM into your 300 SM trip. You decide to do a position check and determine that you are currently 3.6 SM off course. How many degrees do you need to turn to proceed direct to your destination?
Answer: 2.9°
118 NM=135.79 SM
3.6/135.79=X/🔼
3.6/(300-135.79)=Y/🔼
X 1.59+Y 1.31=2.9°
(National 2021 Q3)
100
Departing Runway 34 at KAVL, field elevation 2165’ MSL, you follow the Asheville One Departure procedure that requires a 300’ per nm climb gradient to 4800’ MSL. If you hold the climb rate of 300’ per nm and it takes you 6 minutes 12 seconds to reach 4800’ MSL, what is your goundspeed?
Answer: 97.8 mph (85KTS)
4800’-2165’=2635’
2635’/300’ per nm=8.78nm
8.78nm takes 6.2 min. →85 KTS GS!
(Regional 2009 Q32)
100
Flying direct from Tokyo to San Fransisco at Flight Level 410, the DME shows you are at 88.5 nm from San Fransisco. Directly below is a ship that is 88.0 nm from San Fransisco, how many nautical miles above the earth’s surface are you?
Answer:6.74 nm
41,000 feet / 6076feet =6.74 nm
(Regional 2004 Q35)
100
You’re 42 lbs over gross weight with a CG of 103.4”. To bring the aircraft within limits, we decide to remove the weight from station 155 which moves the CG forward .6”. What is the new aircraft weight?
Answer: 3612 lbs
42 lbs:X=.6”:(155”-103.4”)
X=3612
(National 2010 Q37)
200
You and your buddy Roy are each taking an airplane for a trip to get some lunch at an airport 230 NM away. What a perfect opportunity to test the range of your new pocket-size emergency transceiver. You have a TAS of 108 knots and Roy’s TAS is 161 knots. The useable range for your transceiver is listed at 35 NM. Assuming you both leave the airport at the same time, how far will you fly before you should exceed the transceiver’s range? (Assume calm wind)
Answer:71 NM
53 knots difference, 35 NM at 53 knots will take 39.6 minutes.
108 knots for 39.6minutes = 71NM
(National 2007 Q19)
200
After flying for 37 minutes on a compass heading of 115° you find yourself 13 km off course to the right. If your groundspeed is 138 knots, what CH must you fly to parallel your intended course?
Answer:110°
Convert 13 km to 7 NM.
7:85=x:60
x=4.9
114-4.9=110°
(National 2007 Q32)
200
Passing by the Manchester, NH airport (KMHT), a pilot wishes to check his/her groundspeed using the VOR. If the pilot is 54 NM from the VOR and it takes 4 minutes 24 seconds to travel between the 180° and 168° radials, what is the groundspeed?
Answer: 147 kts
4.4 Min:12°=54NM:X
x=147 GS
(National 2011 Q23)
200
A 757 (two-engine airliner) has its 11,480 gallon tanks completely filled early in the cool morning when the Jet A density is 6.74 lbs./gal. The aircraft is delayed and the rising temperature causes the Jet A to expand and subsequently leak out the fuel vents. At its peak temperature, the Jet A had a density of 6.62 lbs./gal. How much fuel was lost through the vents?
Did you say it was too easy? -500 points 😡
Answer: 1378 lbs.
11,480×6.74 lbs./gal=77375.2
11,480×6.62 lbs./gal=75997.6
77375.2-75997.6=1377.6 lbs.
(National 2015 Q1)
200
If the CG range of an aircraft is 81.2 to 89.2, with a weight of 2700 lbs, and the current CG is at 91.7, what is the minimum weight that needs to be moved from the aft cargo area (Station 120) to the forward cargo area (Station 25) to bring the aircraft within limits?
Answer: 70.8 lbs
X:2700=(91.7-89.2=2.5):(120-25=95)
X=71 lbs
(National 2011 Q25)
300
You have 20 Imp. Gallons onboard and burn fuel at 5.8 US Gal./hr. The winds aloft are 220° at 22 kts. You sustain a 144 KTAS. You are on a true course of 100° out and 280° back. If you need 45 minutes reserve fuel upon returning to your base, how far out can you travel before having to return to base?
Answer:95 minutes (1.58 hours)
20 Imp=24 US Gal, 24 US Gal÷5.8 US Gal./hr=4.14 hours
4.14 h - 0.75 h=3.39 h
(131.7GS+153.7GS=285.4):3.39h=131.7:X
X=1.56h=93.6 minutes
(Regional 2007 Q20)
300
On a flight between two airports 200 NM apart and on a heading of 120°, there is a ridgeline exactly halfway between the two airports, which is perpendicular to the course. The pilot wishes to deviate to the left at a 45° angle, to intercept the ridgeline. The pilot wishes to deviate 13.3 NM before the ridgeline. Assuming no wind. If the pilot then wishes to re‐intercept the course 50 NM immediately after the ridgeline, what is the new heading to intercept the original course?
Answer: 132°
X:13.3=45:△60
Distance off X=10 NM
10:50=Y:△60
Degrees to intercept Y=12°
120°+12°=132°
(National 2011 Q38)
300
Two pilots are discussing the virtues of high-speed cruise flight. One pilot operates an airplane that cruises at Mach .74. The other operates an airplane that cruises at Mach .90. To better understand what speed advantage the faster airplane has in real-world terms, the pilots turn to their flight computers. Cruising at a hypothetical 36000 feet for 2500 nautical miles with an outside temperature of -67 degrees F and no wind, how much shorter will the faster airplane’s cruise flight be?
Answer: 62 minutes
Mach .74 aircraft →352 minutes
Mach .90 aircraft →289 minutes
352-289=62 minutes
(National 2015 Q8)
300
A pilot is departing Ketchikan, AK (30 feet above sea level) on a true course of 196°, where the altimeter setting is 29.92”. There is a hill 6.2NM away that is 2607 feet above field elevation. If the winds aloft are 230° at 41 kts, and OAT of +30°C, and the indicated airspeed is 120 kts in the climb, what rate of climb is required to clear the obstacle by at least 100 feet?
Answer: 650 FPM (632 FPM)
123 TAS, 87 GS
2707feet/6.2nm=436feet/nm
87GS:60=x:(436feet/nm)
x=632 FPM
(National 2001 Q36)
300
After completing your weight and balance, you find that your gross weight is 910 kg, and you are 1.5” aft of your center of gravity. How many 20 lb. boxes of cargo will you have to move to the next cargo hold 20” forward to come within limits?
Answer: 8
X:(910kg=2006lb.)=1.5”:-20”
X=150.4lb.
150 lb.÷20 lb.=7.5→8 boxes
(National 2014 Q17)
400
With a fuel flow of 48 kilos per hour you have an endurance of 6:09 hours. The winds are 098° at 29 knots and your outbound course is 211° with an indicated airspeed of 176 knots. How far can you fly outbound before having to turn around so you don’t run out of gas? Your pressure altitude of choice is 8,500’ and the temperature is 11° C below standard.
Answer: 592 nm
-13℃@8500’→ 176IAS=195 TAS
(205 GS+182 GS):369min=205:X
X=195 minutes
182GS for 195 minutes=592 nm
(National 2009 Q37)
400
Rec the aircraft!!
You can earn 1,000 points if you answer the engine type!
Piper Archer III PA-28-181 (P28A)
Lycoming IO-360-B4A
Four Cylinder, Direct Drive,
Horizontally Opposed with Fuel Injection
180 BHP
400
After returning home to the USA, you’re out flying with your alma-mater’s NIFA team helping them prepare for nationals. While riding along on a practice navigation route at 7500 feet MSL with an altimeter setting of 30.12, you decide to see if you still have your old skills. The pilot is flying a heading of 069 magnetic while maintaining a course of 057 magnetic, the variation is 12E, the winds aloft are 253@24 true and the temperature is 12℃. What indicated airspeed does the pilot need to fly to maintain a planned groundspeed of 128 knots?
Answer: 92 knots (94 knots if U corrected Temp)
57MC+12=69TC
253@24, 128GS → 104TAS
12℃, 7300 feet PA
indicated airspeed=92 knots
(National 2008 Q11)
400
Your C172 has a steady electrical load of 23.2 amps. Halfway through your flight, you lose your alternator and are running off your 35 amp-hour battery. Assuming you are able to shed 8.7 amps, how many minutes can you fly before depleting your battery?
⭐️Answer: 145 minutes
Reduced electrical load:23.2 amps-8.7 amps=14.5 amps
35 amp-hour battery÷14.5 amps=2.41 hours=144.8 minutes
(National 2015 Q37)
400
You are taking your friends for a $100 hamburger. Your aircraft, fully loaded, weighs 3844 lbs. with a moment of 323,000. On the way home, your 163 lb. friend decides he wants to let your other 221 lb. friend sit up front this time. If the front and rear seats are 39 inches apart, what is your new CG?
Answer: 83.5”
221 lb-163 lb=58 lb (WT shifted)
58 lb:3844 lb=X:-39”
X=-0.58 (CG change)
weighs 3844 lbs. with a moment of 323,000→84.0CG
84.0-0.58=83.42”
(National 2015 Q39)
500
The operator of a 5200 mAh (milliamp-hour) battery-powered UAV programs it for an aerial photography flight to a location 4 SM away. Once airborne the UAV travels at an average speed of 10.7 meters per second with a steady electrical draw of 4.1 amps. Assuming the battery drain is linear, how long can the UAV remain on-station if its auto-return setting states that at 20% battery life it will fly back to the launch point?
☺️Answer: 50.9 MINS
5.2 amp/hour÷4.1 amps×80%=1.014h=60.84min
4SM=6400 meters
6400 meters÷10.7 meters per sec=598sec≒10min
60.84min(The flight time before auto-return)-10min(airborne time)=50.84 minutes(remain on-station)
(National 2016 Q35)
500
You’re flying FedEx 1348, a Boeing 727 from Billings to Denver at a TAS of 483 knots with winds aloft of 310@78 true. While over Casper, Wyoming, the company calls and says they need to reroute you to Salt Lake City to pick up some extra freight. ATC clears you direct to Salt Lake City. Assuming a variation of 12E, what magnetic heading do you need to fly to maintain a direct course of 244 magnetic?
Answer: 251°
244MC+12E=256°TC
483TAS, 256°TC, 310@78
244MC+7WCA=251°MH
(National 2008 Q1)
500
Stanley departs KABC at 8:45 AM and instantly accelerates to his cruise speed of 226 KTS. He flies directly towards KXYZ. Gordon departs KXYZ at 9:05 AM and immediately accelerates to his cruise speed of 170 MPH. He flies directly toward KABC. If the two airports are 420 NM apart, at what time will Stanley and Gordon pass each other? Both airports are located within the same time zone. ??:??AM
☺️Answer: 10:00 AM
KABC 8:45AM→ 420nm ←9:05AM KXYZ
KABC 9:05AM→ 344.7nm ←9:05AM KXYZ
use equal time point
(226KTS+147.7KTS=373.7):344.7nm=226KTS:X
X=208.46nm from KABC
226KTS for 208.46nm minutes=55.34 minutes
9:05AM+55.34 minutes=10:00AM!
(National 2016 Q8)
500
An air traffic controller sees two aircraft on his radar scope flying the same jet route at the same altitude 5 NM apart. He has 0:16:40 to get 30 NM separation between the two aircraft. The first jet is told to maintain max forward speed, which the controller sees from his scope is 13 NM every 2 minutes. Winds at altitude are a 60 knot headwind and temperature is -42°C. What maximum speed (MACH) can the controller tell the second, trailing plane to fly in order to get 30 NM separation? Assume the time and speed for the second plane to slow down is negligible.
Answer: Mach 0.61
☺️You should be a team captain! +777points!
The first jet:13 NM every 2 min→390GS
0:16:40=16.66min
16.66 minutes after,
the first jet goes 108.29 NM
the trailing plane goes 108.29 NM-30NM+5NM=83.29NM
83.29NM takes 16.66 minutes→300GS→360TAS
360TAS,-42°C→Mach 0.61
(National 2014 Q21)
500
The empty weight of a 55-liter fuel drum is 32 pounds. Your aircraft is capable of transporting 3250 pounds. Your customer prefers to ship full drums of heating oil that weighs 0.905 kilograms per liter, but will ship a single partial drum to comply with your limits. How many full drums can you carry to still be in the aircraft’s capability?
Answer: 22
1L:0.905KG=55L:X KG
X=49.775KG=109.74 pounds
109.74 pounds+32 pounds(empty weight of fuel drum)=141.74 pounds
3250 pounds/141.74 pounds=22.92→22drums
(National 2015 Q23)