Item Weight Index Lockheed L-188-Index Aircraft Empty • 1,300 67.8 = 100 - Weight × (598.2- Sta.)
Pilot 170 10.2 30,000 Oil Passengers . Fuel . 310 210 60 22.5 13.6 2.4 Convair 880M-Index = 100 + Weight X (Sta. -849.0) Baggage 50 5.0 100,000 Total . ... 2,100 121.5 Douglas DC-6B-Index The intersection of the above total weight and total index values falls well within the index envelope (point B), therefore, the airplane in the example is considered to be within its operating limitations as far as weight and balance is concerned.
The index number system can be modified by applying selected constant factors to the moments of load items. The selected constants are chosen to make the index system less complex, and so that the system can be used in conjunction with special loading charts. In these cases, special formulas are used to obtain index units. Typical formulas are:
= 100- + Weight X (Sta. - 430) 20,000 It should be noted that these special formulas are refinements of the standard formula:
Index = Reduction factor Weight X Arm The constants that are used in the special formulas do not affect the accuracy in determining c.g. locations, as long as the same formula is applied to all weights and arms. These modifications to the index formula permit the index numbers to be related easily to other important numbers, such as % MAC and stabilizer setting.
Chapter 5
CHANGE OF WEIGHT A pilot must be able to solve accurately and rapidly problems which involve the shift, addition, or removal of weight. For example, the pilot may load the aircraft within the allowable takeoff weight limit, then find a c.g. limit has been exceeded. The most satisfactory solution to this problem is to shift baggage, or passengers, or both. The pilot should be able to determine the minimum load shift needed to make the aircraft safe for flight. Pilots should also be able to determine if the shifting of a load to a new location will correct an out-of-limit condition.
There are some standardized and simple calculations which can help make these determinations.
WEIGHT SHIFTING When weight is shifted from one location to another, the total weight of the aircraft is unchanged The total moments, however, do change in relation and proportion to the direction and distance the weight is moved. When weight is moved forward, the total moments decrease; when weight is moved aft, total moments increase. The moment change is proportional to the amount of weight moved. Since many aircraft have forward and aft baggage compartments, weight may be shifted from one to the other to change the c.g. If we start with a known aircraft weight, c.g., and total moments, we can 30 calculate the new c.g. (after the weight shift) by dividing the new total moments by the total aircraft weight.
Example 14.
To determine the new total moments, find out how many moments are gained or lost when the weight is shifted.
The weight shift conditions indicated for the aircraft illustrated in figure 27 show that 100 lb. has been shifted from Sta. 30 to Sta. 150. This movement increases the total moments of the aircraft by 12,000 lb.-in.
Baggage moment when at Sta. 150 = 100 1b. ×150 in. = 15,000 lb.-in.
Baggage moment when at Sta. 30 = 100 Ib. X 30 in. = 3,000 1b.-in.
Moment change = 12,000 lb.-in.
By adding the moment change to the original moment (or subtracting if the weight had heen moved forward instead of aft), we obtain the new total moments. We can then determine the new c.g. by dividing the new moments by the total weight:
Total moments = 616,000 + 12,000 = 628.000
c.g. = 628,000 8.000 = 78.5 in.
100 LBS SHIFTED ORIGINAL C.G. = STA 77.0 STA 30 STA 150 20 %..-:40 60 80 100 120 1401...-:160 180 200 DATUM C.G. LIMITS
75.2" TO 80.5"
14.8% TO 30.0% MAC MAC 70.0" TO 105.0" TOTAL WEIGHT = 8000 LBS × OLD C.G. (STA 77.0) = 616,000 LB IN FIGURE 27. Weight shifting diagram.
The shift of the baggage has caused the c.g. to shift to Sta. 78.5.
A simpler solution may be obtained by using the aercnautical computer and a proportion formula
2. The change of c.g. is added to (or subtracted from) the original c.g. to determine the new c.g.:
77 + 1.5 = 78.5 in. aft of datum (fig. 28). This can be done because the c.g. will shift A possible point of difficulty arises in the computera distance which is proportional to the distance the type solution when an attempt is made to locate the weight has shifted. decimal point in the answer. How do we make sure Example 15. the dc.g. in the above problem is not .15 in. or
1. 15.0 in.? The answer you get can always be checked Weight shifted = mistance weight is shifted Ac.g. (change of c.g.) by cross multiplying. Substitute the answer in step Total weight 1 (example 15) :
8,000 100 = Ac.g. 120 1.5×8.000 = 12,000 Ac.g. = 1.5 in. 100 × 120 = 12,000
SETUP THESE NUMBERS READ ANSWER ON OUTER SCALE 100 4 C.G. = 1.5 8000 120 RECRONING 1:40 mans 1:50 15 ICAL 1643117 192:9H1
SET CON COMPUTATIONS 20 COMPUTER. AIR NAVIG - C-54148 Se FSM 6805-528-5342 ORDER NO. P-1145-61-1 00°9 00L FIGURE 28. Computer solution-shifting weight.
If the cross multiplication answers are not the same, you have selected the wrong decimal location for the Ac.g. and the decimal should be relocated ac.
cordingly.
Finding the decimal location is primarily a matter of observation; remember that the proportions on either side of the equal sign (example 15, step 1) are similar.
The shifting weight proportion formula can also be used to determine how much weight must be shifted to achieve a particular shift of the c.g. The following problem illustrates a solution of this type.
Example 16.
Given:
Aircraft total weight-7,800 lb.
c.g. -Sta. 81.5 in.
Aft c.g. limit -80.5 in.
Find: How much cargo must be shifted from the aft cargo compartment at Sta. 150 to the forward cargo compartment at Sta. 30 to move the c.g. to exactly the aft limit?
Solution:
1. Use the shifting weight proportion:
Weight (to be) shifted - 120 in. Weight shifted Total weight -= Dist. wt. shifted 1.0 in. Ac.g.
7,800 Weight to be shifted = 65 lb.
SETUP THESE NUMBERS
2. Cross multiply to check for accuracy of decimal point location in the answer.
7,800 × 1.0 = 7,800 65×120 = 7,800 A combination problem may involve the shifting of weight when the c.g. and the c.g. limits are expressed in % MAC.
Example I7.
Given :
Aircraft total weight-7,200 lb.
c.g. -12% MAC Fwd. c.g. limit -14.8% MAC MAC -Sta. 70 to Sta. 105 = 35 in.
Find: How much cargo must be shifted from the front baggage compartment at Sta. 30 to the aft baggage compartment at Sta. 150 to move the c.g. to exactly the forward limit?
Solution:
1. Convert the % MAC locations to inches from datum by using the aeronautical computer (fig. 29).
c.g. =LEMAC (70 in.) + % MAC in inches (4.2 in.) = 74.2 in.
Fwd. limit=LEMAC (70 in.) + % MAC in inches (5.2 in.) = 75.2 in.
READ ANSWER ON OUTER SCALE MAC • OR 35 INCHES = 4.2"' INCHES = 5.2" 100% 100 12% 14.8% CORAL 18-52011-d OM 83080 24C5-925-509909 - 55 MANT. 70 COMPUTER.
→ FOR AIRSPEED AND for density alliede 7o 60 30 0oi₴ FIGURE 29. Computer solution- limits in percent MAC.
2. Determine Ac.g. (distance c.g. must be used, the distance must be calculated from the moved) : "old" c.g. Just the opposite is true if the "old"
75.2 - 74.2 = 1.0 total weight is used.
Ac.g. = 1.0 in. aft A typical problem may involve the calculation of a new c.g. for an aircraft which, when loaded and
3. Use the shifting weight proportion: Weight shifted = Dist. wt. shifted Дc.g ready for flight, receives some additional cargo or passengers just before departure time.
Weight (to be) shifted - 120 Total weight Example 18.
7,200 Given:
Weight to be shifted = 60 lb. Aircraft total weight-6,860 lb.
c.g. -Sta. 80.0 WEIGHT ADDITION OR REMOVAL Find: What is the location of the c.g. if 140 lb.
In many instances the weight and balance of the of baggage is added to station 150?
aircraft will be changed by the addition or removal Solution:
of weight. When this happens a new c.g. must be 1. Use the added weight formula:
calculated and checked against the limitations to Added weight see if the location is acceptable. This type of weight New total weight Dist. between wt. and old c.g.
and balance problem is commonly encountered when 140 1C.Z.
the aircraft burns fuel in flight, thereby reducing 6,860 + 140 - 150-80 the weight located at the fuel tanks. Most aircraft 7,000 = 4c.g. 140 are designed with the fuel tanks positioned close to 70 the c.g., therefore, the consumption of fuel does not affect the c.g. to any great extent. However, large Ac.g. = 1.4 in. aft.
jet aircraft with fuel tanks located in the swept- 2. Add Ac.g. to the old c.g.:
back wings require careful planning on each flight New c.g. =80.0 in. + 1.4 in.= 81.4 in.
to prevent the c.g. shifting out of limits due to the Example 19.
consumption of fuel. Given:
The addition or removal of cargo presents a c.g.
change problem which may have to be calculated rapidly before flight. The problem may always be Aircraft total weight-6,100 lb. c.g. -Sta. 78 solved by calculations involving total moments. Find: What is the location of the c.g. if 100 Ib. is However, a shortcut formula which can be adapted removed from station 150?
to the aeronautical computer may be used to simplify Solution:
computations: 1. Use the removed weight formula:
Weight added (or removed) New total weight Ac.g.
= Distance between wt. and old c.g.
In this formula the terms "new" and "old" refer to conditions before and after the weight change.
It is often more convenient to use another form of this formula when required to find the weight change needed to accomplish a particular c.g. change (Ac.g.). In this case we use:
Weight (to be) added (or removed) Old total weight дC.g.
= Distance between wt. and new c.g.
Notice that the terms "new" and "old" are not found on both sides of the equation in either of the above proportions. If the "new" total weight is Weight removed New lotal weight = Dist. between wt. and old c.g 100 Jc.g.
6,100-100 - 150-78 5,000 = 4c.g. 100 Ac.g. = 1.2 in. forward
2. Subtract Ac.g. from old c.g.:
New c.g. =78 in. - 1.2 in. = 76.8 in.
NOTE.--In the above two examples, the Ac.g. is either added to or subtracted from the old c.g Deciding which to accomplish is best handled by mentally calculating which way the c.g. will shift for the particular weight change. If the c.g. is shifting aft, the Ac.g. is added to the old c.g.; if it is shifting forward, the Ac.g. is subtracted from the old c.g.
To summarize c.g. movement:
Weight added fwd. of old c.g. Je.g. moves fwd. new c.g. will be exactly on the aft limit; if the Weight removed aft of old c.g. weight is located any further to the rear, the aft Weight added aft of old c.g. Weight removed fwd. of old c.g. / c.g. moves aft c.g. limit will be exceeded.
Example 20. Example 21. Given:
Given: Aircraft total weight-6,400 lb.
Aircraft total weight-7,000 lb. c.g. -Sta. 80.0 Cog. -Sta. 79.0 Aft c.g. limit -Sta. 80.5 Find: How far aft can additional baggage weigh- ing 200 Ib. be placed without exceeding the Rear c.g. limit -Sta. 80.5 Find: How much baggage can be located in the aft baggage compartment at station 150 with- out exceeding the aft c.g. limit?
rear c.g. limit?
Solution: Solution:
1. Use the added weight formula: Use the added weight formula.
Added weight Ac.g. NoTE.—In this problem, the new total weight New total weight - Dist. between wt. and old c.g. is not given, therefore, it is more convenient
1.5 to use the version of. the formula which makes 7,200 Dist. between wt. and old c.g. use of the old total weight.
2. Add to old c.g.: Distance between wt. and old c.g. =54 in. 79 in. +54 in.= 133 in. aft of datum Old total weight Added weight Added weight - 150-80.5- = Dist. between wt. and new c.g. Ac.g.
When the 200 Ib. is located at Sta. 133, the Added weight = 46 lb. 6,400 34
Chapter 6
CONTROL OF LOADING-
GENERAL AVIATION Before any flight, the pilot should determine the weight and balance condition of the aircraft. In the early days of flying, aircraft were loaded by guess and intuition. On occasion, the results were grim. Through trial and error the early pilots learned about weight and balance. Today there is no excuse for following this method. Simple and orderly procedures based on sound principles have been devised by aircraft manufacturers for the determination of loading conditions. The pilot, however, must use these procedures and exercise good judgment. In many modern aircraft, it is not possible to fill all seats, baggage compartments, and fuel tanks and still remain within the approved weight and balance limits. If the maximum passenger load is carried, the pilot must often reduce the fuel load or reduce the baggage.
USEFUL LOAD CHECK A simple and fundamental weight check should always be made by general aviation pilots before flight. This check should determine if the useful load is exceeded. The check may be a mental calculation if the pilot is familiar with the aircraft's limits and knows that unusually heavy loads are not aboard.
But when all seats are being occupied, fuel tanks are full, and some baggage is aboard, the pilot should do some careful calculations.
The pilot needs to know the useful load limit of the particular aircraft. This information may be
EMPTY AIRCRAFT PILOT AGGAGE
PSGR PSGR FIGURE 30. Empty weight + useful load = takeoff weight.
WEIGHT & BALANCE DATA AIRCRAFT SERIAL NO. 15556480 FAA REGISTRATION NO. N3248X
ITEM WEIGHT ARM MOMENT STANDARD AIRPLANE 975.0 32.0 31200. 0 OPTIONAL EQUIPMENT 89.0 26.1 2322.9 PAINT 15.5 85.3 1322.2 UNUSABLE FUEL 20.0 43.0 860.0 LICENSED EMPTY WEIGHT 1099.5 32.5 35705.1 (GROSS WT) - (LICENSED EMPTY WT) = USEFUL LOAD (1800 LB) -( 1099.5 LB) = 700.5 LBS
IT IS THE RESPONSIBILITY OF THE OWNER AND PILOT TO ENSURE THAT THE AIRPLANE IS PROPERLY LOADED. THE DATA ABOVE INDICATES THE EMPTY WEIGHT, C.G., AND USE-
FUL LOAD WHEN THE AIRPLANE WAS RELEASED FROM THE FACTORY. REFER TO THE LATEST WEIGHT AND BALANCE RECORD WHEN ALTERATIONS HAVE BEEN MADE.
SAMPLE LOADING PROBLEM
ITEM WEIGHT ARM MOMENT (LBS) (IN) (LB-IN/1000) LICENSED EMPTY WEIGHT 1099.5 35.7 OIL 12 - 15.0 -0.2 PILOT & PASSENGER 340 40.0 13. 6 FUEL 188.5 43.0 8.1 BAGGAGE 160 65.0 10. 4 TOTAL LOADED AIRPLANE 1800 67. 6 FIGURE 31. Weight and balance data.
found in the latest weight and balance report, in a log book, or on a major repair and alteration form, located in the aircraft. If useful load is not stated directly, simply subtract empty weight from maximum takeoff weight. Be especially weight conscious of aircraft which have a limited useful load because 36 they are the ones which cause weight and balance troubles.
• The check is simple enough-just be sure to include all the load items included in the useful load--then check the total against the limit. The calculations might look like this:
Example 22.
Pounds Mr. Jones (instructor) • 175 Pilot .. 180 Fuel-30 gallons 180 Oil-8 quarts 15 Baggage 5 Total . 555 Useful Joad limit is 575 Ibs.
The calculations indicate that the useful load is not exceeded and the flight can take place.
Now suppose that Mr. Jones, in the example, is replaced by a new instructor who weighs 210 lb. A useful load check will show that the aircraft is too heavy. The pilot in our example must reduce the load to the specified useful load limit. There is no alternative in this small aircraft but to reduce the fuel load, even if all the baggage has been removed.
Pilots should be aware of, and on the alert for, unusual loadings. They should remember that the manufacturer's initial weight and balance calculations and some examples in the owner's manual make the assumption that the pilot and passengers weigh a standard 170 Ib. each. Heavyweight passengers can overload a small aircraft seriously. A student and instructor may easily weigh 220 Ib. each in winter clothing; this represents a potential overload of 100 lb. The baggage compartment is another place where pilot vigilance should be directed-the maximum compartment load placard must be obeyed. Frequently, a restriction is placed on rear seat occupancy with the maximum baggage aboard.
WEIGHT AND BALANCE RESTRICTIONS Be sure to follow your aircraft's weight and balance restrictions. The loading conditions and empty weight of your particular aircraft (fig. 31) may differ from those in the owner's manual due to modifications or equipment changes. Sample loading problems in the owner's manual are intended for guidance only; each aircraft must be treated separately for weight and balance. The pilot should understand that although the aircraft is certified for a specified maximum gross weight, it will not safely take off with this load under all conditions. Conditions which affect takeoff and climb performance such as high elevations, high temperatures, and high humidity (high-density altitudes), may require operation at reduced weight. Other factors to consider are runway length, runway surface, runway slope, surface wind, and the presence of obstacles.
Pilot experience and proficiency should always be considered-if in doubt, reduce the load.
Some small aircraft are designed so that it is not possible to load them in a condition which will place the c.g. outside the fore or aft limits if standard load schedules are observed. These aircraft have the seats, fuel, and baggage accommodations located very near the c.g. limits. They also have special empty weight c.g. limits listed in their specifications. Loads can be added to or removed from any location within the c.g. range with complete freedom from concern about c.g. movement.
Such action cannot cause the c.g. to move beyond the c.g. limits of these aircraft (see fig. 32), but maximum weight limits can still be exceeded.
ADDING OR REMOVING THIS WEIGHT CANNOT MOVE THE C.G. BEYOND LIMITS
FORWARD LIMIT AFT LIMIT FIGURE 32. Changing weight between c.g. limits.
Most aircraft, however, can be loaded in a manner which will place the c.g. beyond limits. Even though the useful load is not exceeded, an out-of-balance condition is serious from a stability and control standpoint. The pilot can quickly determine if the load is within limits, if the aircraft is simple enough to make use of a loading schedule. This schedule may be found in the weight and balance report, the aircraft log book, the owner's manual, or may be posted in the form of a placard. A typical placard may appear similar to the one shown in figure 33.
LOADING SCHEDULE
FUEL PASSENGERS BAGGAGE FULL 2 REAR 100 LBS 39 GAL 1 FRONT AND NONE 2 REAR FULL 1 FRONT AND FULL 1 REAR
INCLUDES PILOT AND FULL OIL FIGURE 33. Loading schedule placard.
The loading schedule should be treated as a suggested loading plan only. The pilot should make a check by means of weight and balance calculations to see if limitations are not being exceeded. The assumption in the use of the loading schedule is that each passenger weighs approximately the standard weight of 170 lb. It is obvious that passenger weights could vary widely from the assumed standard.
AIRPLANE FLIGHT MANUAL Each airplane of over 6,000 lbs. maximum weight is furnished with an airplane flight manual. An airplane of less than 6,000 Ibs, may have information furnished in the form of placards, markings, or manuals. When an airplane flight manual is furnished, the following is included:
a. Limitations and data:
(1) The maximum weight.
(2) The empty weight and c.g. location.
(3) The useful load.
(4) The composition of the useful load, including the total weight of fuel and oil with full tanks.
b. Load distribution:
The established c.g. limits are furnished in the airplane flight manual. If the available loading space is adequately placarded or arranged so that no reasonable distribution of the useful load will result in a c.g. outside of the stated limits, the airplane flight manual may not include any information other than the statement of c.g. limits. In other cases, the manual includes enough information to indicate loading combinations that will keep the c.g. within established limits.
LIGHT SINGLE-ENGINE AIRCRAFT
LOADING PROBLEMS Aircraft manufacturers use one of several available systems to provide the aircraft loading information. The following weight and balance problems will show how the pilot can determine if the maximum weight limit is exceeded or the c.g. is located beyond limits.
Assume you are a pilot planning a flight in a light single-engine, four-place aircraft. Your load consists of yourself, one front seat passenger and two rear seat passengers, full fuel and oil, and 60 lb. of baggage (fig. 34). Here is how the critical weight and balance problems are solved for this case by two different methods (examples 23 and 24).
Example 23.
Solution by index table:
1. From the manual or weight and balance report, determine the empty weight and empty weight c.g. (arm) of the aircraft.
2. Determine the arms for all useful load items.
3. Determine the maximum weight and c.g.
range. (For this case-Max. TOGW= 2,400 lb., c.g. range - Sta. 35.6 to 45.8.)
4. Calculate the actual weights for the useful load items.
5. Construct a table as follows (pg. 39), and enter the appropriate values. Multiply each individual weight and arm to obtain moments.
OIL PILOT & FRONT PASS TWO REAR PASS BAGGAGE -20" + 35" +72" + 92" C.G. LIMITS NEGATIVE ARMS POSITIVE ARMS FUEL +35.6" to +45.8" +48" DATUM EMPTY WEIGHT 1340 LBS MAX T.O. WEIGHT 2400 LBS EMPTY WEIGHT C.G. +38.5" FIcURE 34. General aviation aircraft-weight and balance diagram.
Weight X Arm = Moment (lb.-in.) Weight and balance computations are greatly Aircraft simplified by two graphic aids the loading graph Oil ......... (empty) .. 1,340 15 - 20.0 38.5 51,590 - 300 and the center of gravity moment envelope. The Pilot and front loading graph (fig. 35) is typical of those found in passenger .. 320 35.0 11,200 general aviation aircraft owner's manuals. This Fuel .... 241 48.0 11,568 graph, in effect, multiplies weight by arm giving Rear passengers. 300 72.0 21,600 moment, then divides the moment by a reduction Baggage •..• 60 92.0 5,520 factor, giving an index number. Weight values ap- Total 2,276... ............... pear along the left side of the graph. The moment/ NoTE.-Observe that the oil tank for this aircraft is located forward of the datum. Care must be taken to subtract the negative oil moment when totaling the moment column.
6. Adding the weights produces a total of 2,276 lb., and adding the moments produces a total of 101,178 lb.-in. The c.g.
is calculated by dividing the total moment by the total weight:
101,178 = 44.5 in. aft of datum 2,276
7. The total weight of 2,276 lb. does not ex ceed the maximum weight of 2,400 lb., and the computed c.g. of 44.5 falls within the allowable c.g. range of 35.6 to 45.8 in. aft of datum.
1,000 or index numbers are along the bottom. In this example, each line representing a load item is labeled. To determine the moment of any load item, find the weight along the left margin, then project a line right to a point of intersection with the appropriate load item line. For example, the index number of a pilot weighing 170 lb. is 6.1. The c.g.
moment envelope (fig. 36) allows the pilot to bypass the computation of a c.g. number. It gives an acceptable range of index numbers for any aircraft weight from minimum to maximum. If the lines from total weight and total moment intersect within the envelope, the aircraft is within weight and balance limits. In solving the sample problem, follow this procedure:
360 320 PASSENGER 280 240 LOAD WEIGHT (LBS) PILOT & FRONT 200 FUEL PASSENGERS
REAR A BAGGAGE MOMENT/1000 (LB INS) 10 12 14 16 18 20 22 24 26 28 FIGURE 35. Loading graph.
2400 2300 2200 LOADED AIRCRAFT WEIGHT (LBS) -NORMAL CATEGORY
21.00 2000 1900 CATECORY LOADED AIRCRAFT MOMENT/1000 (LB INS) 1500 50 55 60 65 70 75 80 85 90 95 100 105 110 115 FIGURE 36. C.G. moment envelope.
Example 24. Sample Loading Problem
1. Determine the aircraft empty weight and the ltem Weight Moment/1,000 empty weight index from the weight and 1. Empty aircraft weight .1,340 51.6 balance report. 2. Oil 3. Pilot and front passenger 320 15 - 0.3 112
2. Construct a table such as the one that follows. 4. Fuel 241 11.6 In the left column, enter the actual weights of 5. Rear seat passengers 300 21.6 the empty aircraft, oil, pilot and front seat 6. Baggage 60 5.5 passenger, fuel, rear seat passenger, and bag- 7. Total aircraft weight •2,276 101.2 gage. In the right column, enter the aircraft empty weight index (moment/1,000).
3. From the loading graph, fig. 35, determine the index number (moment/1,000) of each useful load weight item and enter it in the table.
4. Add the weight and moment columns and write in the totals.
5. Refer to the c.g. moment envelope, fig. 36, and find the point of intersection of a line projected right from total weight (2,276 Ibs.) and of a line projected up from total moment/1,000 (101.2).
6. The point of intersection falls within the envelope, therefore, the weight and c.g. are within limits.
LIGHT TWIN-ENGINE AIRCRAFT Modern light twin-engine aircraft are larger than most single-engine aircraft; accordingly, their useful load is almost always greater than that found in the smaller aircraft. In these aircraft, it is possible to have many different loading combinations. Their large baggage compartments may be full or empty, and there may be wide variations in the number of seats being occupied. These variations are to be expected and are normal for the types of operations for which the aircraft are used. However, the c.g.
is bound to range backward and forward as the loads are varied; therefore, weight and balance control is essential.
If a variety of loads can be placed aboard an aircraft in a number of locations, the pilot must be 40