are typical computations to use in finding the c.g. 4. c.g. in c MAC= 11.2% location in relation to MAC: NoTE.-Steps 2 and 3 can be eliminated by the use of an aeronautical computer to solve the propor- Example 3.
Given: tion in step 1 (fig. 16). With the use of the proportion formula presented MAC-Sta. 100 to Sta. 250 c.g.-Sta. 130 changed from c MAC to "inches from datum." The above, the expression of a location can be easily Find: c.g. in c MAC following example illustrates a typical problem with Solution: a computer solution.
1. MAC=250 - 100 = 150 in.
2. Distance of c.g. from LEMAC = 130 - 100 = Example 6.
30 in. Given:
3. c.g. in % MAC = Distance of c.g. from LEMAC MAC-175 in. c.g.-20% MAC MAC LEMAC-Sta. 380 30 Find: c.g. in inches from datum = 150 =20% MAC Solution:
Use the following method to convert locations ex- 1. Set up the proportion on the computer (fig.
pressed in % MAC to locations expressed in inches 17).
from datum: 2. Add to LEMAC:
Example 4. 380 + 35 = 415.0 in.
Given: NOTE.-It is easy to check the computer solution MAC-170 in. (fig. 17) by arithmetical means. The arithmetical LEMAC-Sta. 500 solution is:
c.g.-27.5% MAC 1. MACX% MAC:
Find: c.g. in inches from datum 175X.20 =35.0 in. (c.g. aft of LEMAC) Solution: 2. Add to LEMAC:
1. MACX% MAC= inches aft of LEMAC 380 + 35.0 = 415.0 in.
170 X.275 = 46.75 in. FAA written tests often make use of a graphic
2. LEMAC + 46.75= c.g. aft of datum presentation of the information needed to solve cen- 500 + 46.75 = 510.75 in. ter of gravity problems (fig. 18). The following is a Proporfion formulas can be readily adapted to the typical example which combines some of the princonversion of % MAC to inches from datum. ciples explained in this chapter:
A typical problem solved by the use of a propor- Example 7.
tion formula follows: Given: The aircraft in figure 18 was weighed in Example 5. the empty weight condition and was found to have the following readings at the three scales:
c.g.-Sta. 410.2 Pounds MAC-180 in. Nose wheel weight. ..20,500 Leading adge of MAC (LEMAC) -Sta. 390 Right wheel weight , 70,000 Find: c.g. in % MAC Left wheel weight . 70,500 Find: The c.g. location expressed in % MAC.
1. Use the proper proportion formula: Solution:
c.g. in inches from LEMAC MAC 1. Find c.g. in inches from datum using pro-
410.2-390 180 Nose wheel weight
c.g. in c MAC 100% Total weight C.g. in C MẶC 100 20.2 Distance from main wheel to c.g.
Distance between main and nose wheels
2. Cross multiply: 20.500 Distance from main wheel to c.g.
180 c.g. in % M.1C=2020 161,000 - 480 in.
3. Divide both sides of the equation by 180 = 61.1 in.
SETUP THESE NUMBERS READ ANSWER ON INNER SCALE 180 100 C.G. %MAC = 11.2% 20.2 NOTE:
OUTER SCALE = INCHES FROM LEMAC INNER SCALE = %MAC AIR NAVIGATION. DEAD RECKONING 1200 150 AIR TEMPERATURE FAR A18340d n
MERS PART MO FELSEMTHAL TYPE MB.4 50 1C4.41.155 LATITUDE
- 6-54198 O0°S FIGURE 16. Computer solution-c.g. in percent MAC.
SETUP THESE NUMBERS READ ANSWER ON OUTER SCALE 175 C.G. AFT LEMAC = 35" 100 20 DEAD RECKONING VES 1:50 COMPUTER. AIR NAVIGATION. TYPE MB.4 LATITUDE - AT TEMPERATURE °C SAMOS OFT 27I519 PARTE OM LOYd SHIM SEALAL HO sona SPECIFICATION MIL-C-5 FS# 6605 - 526-5342 ORDER NO P.11625-61-1 FIGURE 17. Computer solution: erting percent MAC to inches.
MAC DATUM STA 500 STA 690 / COLLE 120" 600" FIGURE 18. Large airplane weight and balance calculation diagram.
2. 600.0-61.1 = 538.9 in. aft of datum.
3. 538.9-500=38.9 in. aft of LEMAC.
4. Convert to % MAC by using the proportion formula (see example 5) :
38.9 in.
% MAC- 100%
c.g. in % MAC=20.5% By using information on the diagram (fig. 18), we can determine c.g. limits in inches from datum when they are expressed in % MAC.
Example 8.
Given: The aircraft illustrated in figure 18 has its forward c.g. limit located at 12% MAC and its rearward c.g. limit located at 32% MAC.
Find: What are the c.g. limits of this aircraft in inches from datum?
Solution:
1. Multiply MAC times the given percentages (in decimal form) :
190 X.12=22.8 in.
190 X.32 = 60.8 in.
2. Add to LEMAC:
Forward limit (500 + 22.8) = 522.8 in. aft of datum.
Aft limit (500 + 60.8) =560.8 in. aft of datum.
AIRCRAFT MODIFICATIONS After alteration of an aircraft or after the removal or installation of equipment, it is necessary to establish that the authorized weight and c.g. limits as shown on the FAA aircraft type certificate data sheet or specification are not exceeded when the aircraft is properly loaded. The owner should assure that this determination has been made and that the repair agency has entered appropriate changes in the weight and balance records of the aircraft. If equipment alterations are made without preparation of weight and balance records, all subsequent calculations by operating personnel would be in error. The effect of weight and balance calculation errors upon the safety of flight is po tentially tragic; therefore, strict adherence to regulations and ethical practices by the owner and repair agency is essential.
The original basis for weight and balance calculations pertaining to alterations of the aircraft are the FAA aircraft type certificate data sheets or specifications. They provide the essentials for calculation of c.g. changes due to aircraft modifications: weights, arms, and limitations. These essen tials are illustrated in the excerpts from a typical FAA aircraft type certificate data sheet shown in figure 19. It should be noted that all details listed in the type certificate data sheet may not be appropriate for an aircraft which has been modified The manufacturer is required to provide documents which show the certificated empty weight and
c.g. for each new aircraft. This weight aud balance data may also include a schematic diagram which illustrates the fixed dimensions for all aircraft of the particular model (see fig. 20). The continued validity of weight and balance records during the life of the aircraft depends upon the maintenance of a series of similar documents which show the calculations-for each successive weight change. This series of documents starts with the manufacturer's data and continues in chronological order to the latest weight and balance report. When a new weight and balance report is prepared for an aircraft, the previous report should be marked superseded and reference the date of the new document. This would preclude the necessity to search for the current report.
Data prepared by the repair agency for each modification should indicate that the maximum weight of the aircraft will be within the maximum allowable weight with anticipated loads. The new empty weight is derived from the empty weight recorded on the most recent weight and balance report plus the weight of the items added, minus the weight of items removed. When load items are added to this new empty weight, the total weight can be compared to the limit listed in the aircraft specifications.
IV - Model PA-24-260, 4 PCIM (Normal Category) Approved June 19, 1964 Engine Lycoming 0-540-E4A5 (See Item 109(f) for optional engines) Fuel 91/96 min. grade aviation gasoline Engine limits A11 operations 2700 I.P.m. (260 hp.)
Airspeed 11mits (CAS) Vne Never Exceed 227 m.p.h. (197 knots) THo Max. structural cruising 180 m.p.h. (156 knots) Landing gear extended 150 m.p.h. (130 knots) Maneuver ing 144 =.p.h. (125 knots) Vie Flaps extended 125 m.p.h. (108 knots) C.G. range (+86.0) to (+93.0) at 2900 1b.
(gear extended) (+82.5) to (+93.0) at 2600 1b.
(+80.5) to (+93.0) at 2000 1b. or less Straight line variation between points given.
Moment due to retracting of landing gear (1266 in.-1b.)
3100 2900 Aircraft Weight (1b.)
2600 2000 Inches Aft of Datum Empty Weight None C.G. range Maximum weight 2900 1b.
No. of seats 4 (2 at +85, 2 at +120.5) Maximum baggage 200 1b. (Rear Compartment) (+142) Fuel capacity 56 gal. (Two 28 gal. wing tanks) (+90) (See Note 1 for unusable fuel) (See Item 112 for auxiliary fuel tanks) Oil capacity 3 g81. (+28) NOTE 1. Current weight and balance report including list of equipment included in certificated empty weight certification. and loading instructions when necessary, must be provided for each ai:craft at the time of original The certificated empty weight and the corresponding center of gravity location must include unusable fuel (not included in fuel capacity) as follows:
24 Ib. (+90) for Model PA-24-250 Serial Nos. 24-2563, 24-2844 and up and PA-24-260, and 36 1b. (+90) for Model PA-24-400, Serial Nos. 26-2 and up.
FIcURE 19. FAA aircraft type certificate data sheet excerpts.
WEIGHT AND BALANCE DATA
DATUM (FIREWALL, FRONT FACE) STA 0.0 → 20.2"- MAC 59.9"
LEVEL ON WINDOW SILL C.G L&R
SCALE POSITION SCALE READING TARE SYMBOL NET WEIGHT
LEFT WHEEL L
RIGHT WHEEL R
NOSE WHEEL N
AIRCRAFT EMPTY WEIGHT (AS WEIGHED) W X = ARM (C.G.) = (A) - (N) x (B) X= ( ) - = ( ) IN.
FIcURE 20. Typical weight and balance data.
WEIGHT AND BALANCE REPORT Detailed instructions on repair and alteration procedures are contained in Advisory Circulars
43.13-1A and 43.13-2. Generally, the repair agency should prepare weight and balance data to show that the aircraft does not exceed maximum weight limits, in various load combinations, after the alteration has been made. The owner should assure that the data has been provided by the repair agency.
The repair agency also includes in the weight and balance data, information showing that the c.g. of the aircraft (usually in the fully loaded condition) falls between the specified c.g. limits when loaded in one of the extreme conditions. The weight and balance extreme conditions represent the maximum forward and rearward c.g. position for the aircraft.
The computations are known as the forward and rearward extreme conditions check.
When a forward extreme condition check is made, the objective is to establish that neither the maximum weight limit nor the forward c.g. limit listed in the aircraft specifications is exceeded. Normally, in the case of a four-place airplane, this check must be made assuming both front seats are occupied and the rear seats empty. If the baggage compartment is in the rear, it is also assumed to be empty. If the fuel tanks are located forward of the forward limit, they are assumed to be full. If they are located aft of the forward limit, they are assumed to be empty.
However, a minimum fuel load is always included in the calculation. This minimum fuel load for a small aircraft with reciprocating engines is calculated by:
Minimum fuel (lb.)
METO (maximum except takeoff) horsepower = 2 For jet engine aircraft, the minimum fuel for extreme conditions check is specified by the manufacturer.
When a rearward weight and balance check is made, the objective is to establish that neither the maximum weight limit nor the rearward c.g. limit listed in the aircraft specifications is exceeded. The 21
ITEM CHK DESCRIPTION WT ARM 001 Engine, Continental, 0-200-A 200.0 - 18.5 002 Propeller, McCauley IA 100 20.0 - 32.0 003 Spinner, Propeller 1.0 - 34.5 00ЗА Spinner, Propeller, Large 2.0 - 34.5 004 Generator, 35 Amp, 14 Volt 12.5 - 6.0 005 Regulator, Voltage, 35 Amp, 14 Volt 1.0 - 1.0 006 Battery, 12 Volt, 24 AH 24.5 - 4.5 007 Filter, Carburetor Air 0.5 - 23.5 008 Heating System, Carburetor and Cabin 10.5 -20.0 009 Wheel, Brake & Tire Assy, (two), 6.00×6, 4 Ply Rating, Main 35.5 48.5 010 Wheel & Tire Assy, 5.00×5, 4 Ply Rating, Nose 9.0 - 10.5 FICURE 21. Typical equipment list.
loading conditions are obviously opposite to those used for the forward check. For a typical four seat airplane, the rearward check is made with one pilot, maximum rear passengers, maximum rear baggage, and full fuel loaded in tanks behind the rear c.g.
limit. After making these checks, the repair agency completes records in the form of a weight and balance report, loading schedule, or placard to inform the owner and operator about the permissible load combinations.
A list of the equipment (fig. 21) included in the aircraft during calculation of the certificated empty weight may be found in either the approved airplane flight manual or the weight and balance report. The repair agency should enter in the weight and balance report all required, optional, and special equipment installed in the aircraft at the time of weighing and when equipment changes are made. The owner should assure that the person making an equipment change completes an entry on the equipment list to indicate items added, removed, or relocated. The entry should also include the date accomplished, identity of the person making the change, and certificate number of that person.
Suggested methods of tabulating the various data and computations for determining the c.g., in the empty weight condition and the forward and aft extreme loaded conditions, are given in figure 22.
Ballast is sometimes permanently installed for
c.g. balance purposes as a result of installation or removal of equipment items and is not used to correct a nose-up or nose-down tendency of an aircraft.
It is usually located as far aft or as far forward as possible in order to bring the c.g. position within acceptable limits with a minimum of weight increase.
WEIGHT AND BALANCE REPORT EMPTY WEIGHT C.G.
Scale алеї Net Left Jack Foint 514 2 512 Right Jack Point 515 2 513 Nose Wheel 70 1095 C.G. +37.5 - (70x +58.5) <60T C.G. +3705 3•7 C.G. = +33.8
WEIGHT AND BALANCE EXTREME CONDITIONS Forward Check Rearward Check Wt. Arm Mom • Wt. Arm Mom • Airplane Empty 1095 +33.8 37011 1095 +33.8 37011 Pilot 170 +39.0 6630 170 +39.0 6630 Passenger - 170 +39.0 6630 Fuel 50 +42.0 2100 135 +42.0 5670 Oil 11 -12.0 - 132 11 -12.0 - 132 Baggage 120 +64.0 7680 1326 45609 1701 63489 45609 1326 = 34.4" 63489 1701 = 37•3" Most Forward C.G. Location Most Rearward C.G. Location Limits are 34.1" and 38.0" Max. Weight - 1,750 Lbs.
FIGURE 22. Weight and balance report.
Chapter 4
INDEX AND GRAPHIC LIMITS MAX TO. LIMIT
FWD LIMIT INCHES FROM DATUM Aviation has been one of the most dynamic of industries since its beginning. New aircraft are continually being developed and always represent an improvement over older models. Improvements in design have, in many cases, tended to increase aircraft complexity. However, considerable effort has been spent to keep the new designs simple so that operations and maintenance procedures can be accomplished by the average airman. In accordance with this design philosophy, weight and balance engineers have developed some simplified methods, which they have applied to the weight and balance problems of most modern aircraft. Index numbers and graphic presentation of limits are features of simplified methods in common use.
INDEX NUMBERS The use of index numbers and a reduction factor greatly simplifies weight and balance calculations, especially for large aircraft. The index is a moment divided by a reduction factor and may be found by this formula:
Index = Weight × Arm (moment) Reduction factor 24 The moments with which we are concerned on transport aircraft and the larger general aviation aircraft will be large numbers because of the large weights and arms which are involved. In a transport aircraft, a fuel tank located at station 500 which is loaded with 5,000 pounds of fuel would represent a moment of 2,500,000. This moment when divided (or reduced) by a reduction factor of 10,000 becomes a more manageable index of 250.0. The same problem exists to a smaller degree for general aviation aircraft. In this case, a reduction factor of 100 or 1,000 produces manageable numbers.
A simple way to change the moment into the index is to count the number of zeros in the reduction factor and move the decimal point of the moment the same number of places to the left.
Reduction factor-10,000 (4 zeros) Moment -2,500,000 move the decimal point 4 places to the left
250.0000 = Index of 250.0 If the moment is a number with sufficient zeros, for instance a moment of 5,600,000, simply cross out
the same number of zeros as you find in the reduc- In typical problems involving index numbers, the tion factor. For example: weights and index numbers of particular aircraft Redı load items are given and the fully loaded c.g. must Moment be found.
Example 10.
Follow Given:
from ind Aircraft empty weight-65,000 lb.
Redı Empty weight c.g. -Sta. 400.0 Index Reduction factor -10,000 Item Weight (Ib.) Index Crew 770 15.4 Full oil 900 27.0 You Passengers ..8,000 360.0 would no Baggage . 1,000 45.2 Fuel ..6,000 240.0 were not a reduct Find: What is the c.g. location when the aircraft is loaded with the items given above?
Solution:
1. Find empty weight index by use of formula:
Empty weight index = Reduction factor Weight X Arm Empty weight index = 65,000 × 400.0 10,000 Empty weight index =2600.0
2. Add loaded weights and index units.
Item Weight (lo.) Index Aircraft empty ..65,000 2600.0 7,200,0DP = Index of 360.0 zeros and divide by 2 Full oil Crew .. 900 770 15.4 27.0 2 Baggage Passengers 8,000 1,000 360.0 45.2 Reduction factor-30,000 (4 zeros) Fuel 6,000 240.0 Index - 492.1-move decimal 4 Total .81,670 3287.6 places to the right and 3. Divide the index by the weight:
multiply by 3 4,921,000.0×3= Moment 3287.6 = .04025 81,670 tem can be applied to total aircraft moments as It should be apparent that the index number sys- of 14,763,000 4. To find c.g., move the decimal point to the right the same number of places as there are zeros in the reduction factor:
readily as to individual load items. This principle is c.g. =0402.5 = 402.5 in. aft of datum illustrated in the following example: (.04025 × 10,000 = 402.5) Example 9. Given: the problem above are easily obtained from tables The index units given for the particular items in Aircraft total weight-105,000 lb. in the weight and balance reports for the aircraft.
c.g. (average arm) -Sta. 500 Tables are provided for all types of items which Reduction factor -10,000 may be loaded on the aircraft. A typical table is Find: Total weight index shown in figure 23.
Solution: Index units may be used during flight to calculate Weight × Arm the effect of fuel consumption upon the c.g. of the Use index formula = Reduction factor aircraft. The index units for the fuel used are sub- 105,000 × 500 tracted from the aircraft total index which is calcu- 10,000 - = 5250.0 lated for takeoff conditions. The new c.g. is then Total weight index = 5250.0 obtained by dividing the new total index by the new 25
OIL TABLE Example 11.
CAPACITY = 9.4 GALLONS Given:
Aircraft total weight - 125,000 lb.
ARM = 77" Total weight index -6250.0 U.S. Weight Moment Reduction factor -10,000 Gals. (Pounds) 100 Average fuel tank location-Sta. 550
7.5 6 Find: The c.g. of the aircraft after consuming 6,000 lb. of fuel.
2 15.0 12 Solution:
22.5 17
1. Determine the index of the fuel used:
4 5 30.0 37.5 23 29 Index = Reduction factor Weight Arm 6,000 × 550
45.0 35 Fuel index = 10,000
52.5 40 Fuel index = 330.0 8 60.0 46 2. Subtract fuel weight and index from the 9 67.5 52 original aircraft weight and index:
9.4 70.5 54 Weight (lb.) Index (units) 125,000. 6250.0 FIGURE 23. Oil index table. - 6,000 - 330.0 119,000. 5920.0 total aircraft weight after fuel burnoff. Fuel consumption index units may either be obtained from a table, such as that shown in figure 24, or by calculations using weight, arm, and reduction factors.
3. Multiply index by reduction factor and divide by weight:
5,920.0×10,000 = 497.5 in. aft of datum 119,000
TOTAL FUEL IN GALLONS AND INDEX Fuel = 6 Ibs/gal.
Gals. Index Gals. Index 1200 325 3400 920 1400 379 3600 974 433 3800 487 4000 541 4200 595 4400 649 4600 704 4800 758 5000 812 5200 866 5400 1467 FIGURE 24. Fuel index table.
Example 12. 6. Multiply index by reduction factor and Given: divide by weight:
Aircraft total weight--101,000 lb. 3250.9 × 10,000
c.g. -20.0% MAC 77,000 - = 422.2 in. c.g. aft of datum MAC —Sta. 395 to Sta. 565 7. Determine c.g. in % MAC:
Reduction factor -10.000 c.g. - LEMAC = % MAC Find: What is the location of the c.g. in /c MAC MAC after consuming 4,000 gallons of fuel? Lse the 422.2 - 395.0 fuel index table (fig. 24) : 170 - = 16% MAC Solution: CENTER OF GRAVITY
1. Determine original c.g. in inches from VARIABLE LIMIT GRAPH datum: Many aircraft are designed with c.g. limits which
c.g. = LEMAC + (MACX%) vary with changes of weight and certain other
c.g. =395.0 + (170.0 in. X.20) operational factors. The limits are presented in the
c.g. = 395.0 + 34.0 = 429.0 in. aircraft type certificate data sheets (fig. 19) or
2. Determine total weight index: specifications and other publications in a graphic form and are usually expressed in inches from Total weight index = Reduction factor Weight X Arm datum or % MAC. A typical graphic presentation Total weight index = 101.000 × 429.0 of aircraft c.g. limits is shown in figure 25. It is apparent from inspection of the graph that 10,000 limits for various weights would include:
Total weight index = 4332.9 Weight Forward Aft
3. Find fuel burned index on table (fig. 24) : Fuel burned index for 4,000 gals. = 1082.0 100,000. 70,000. .20.0% .20.0% 27.2% 26.6%
4. Convert fuel gallons to pounds: 130,000.. .23.0% 26.0% 4,000 gal. X6.0 lb./gal. = 24,000 lb. Limits for intermediate weights may be deter-
5. Subtract fuel burned weight and index from mined by visual or mathematical interpolation ot the original total weight and index: graph. The graph should be read to the nearest one- Weight (lb.) Index (units) tenth of a percent of MAC. As an example, the rear 101,000. +332.9 limit for a weight of 90,000 lb. is interpolated to be -24,000 - 1082.0 26.8% MAC.
77,000. 3250.9 The limits may be converted to "inches from
MAX TO LIMIT
- 130,000
- 120,000 LIMIT AFT LIMIT - 110,000 = - 100.0000 FWD 90,000
INCHES FROM DATUM
- 80,000 70,000 20 21 22 23 24 25 26 27
PERCENT OF MAC MAC = STA 400 to STA 560 FIGURE 25. C.G. variable limit graph.
datum" by computation methods previously explained. A typical problem using the limits in the graph (fig. 25) follows:
Example 13.
Given:
Graphic c.g. limits (fg. 25) MAC Sta. 400 to Sta. 560 Find: Forward c.g. limit for 115,000 lb. in inches from datum.
Solution:
1. Determine forward c.g. limit in % MAC by use of graph for 115,000 lb.:
Fwd. c.g. limit =21.5% (.215)
2. Find length of MAC:
560-400 = 160 in.
3. Find limit in "inches from datum":
160X.215 = 34.4 in. aft of LEMAC 400 + 34.4 = 434.4 in. aft of datum
INDEX ENVELOPE C.G. limits may be expressed graphically in the aircraft weight and balance reports by means of an index envelope. The envelope defines the forward 2300 and aft limits and also the maximum weight limit in terms of index units.
The envelope permits the rapid determination of the weight and balance condition of an aircraft when the weight and total index units are known.
Thus, the procedure of computing the c.g. location from datum or in relation to MAC is simplified. The envelope informs the pilot that the c.g. is within acceptable limits without actually locating it on the longitudinal axis. In most cases, this is all the pilot needs to know. The pilot needs only to be assured that the c.g. is within approved limits.
A typical index envelope is shown in figure 26.
The similarities and differences between this graphic form and the variable limit graph (fig. 25) should be noted.
It is apparent from the envelope that a loading of 2,000 lb. with an index of 110 is within limits (point A) while a loading of 2,000 lb. with an index of 130 is not within limits (point C). Index information which the operator must gather in order to utilize the envelope is obtained from index charts or tables in the weight and balance report. Various load items and associated index numbers are added to obtain the totals.
The following is a simplified loading check making use of the index envelope:
MAX WEIGHT LIMIT 2200 2100 • B TOTAL WEIGHT 1900 2000 - FORWARD LOADING LIMIT • C 1800 AFT LOADING LIMIT 1600 1500 90 95 100 105 110 115 125 TOTAL INDEX OR MOMENTS/1000 FICURE 26. Index envelope.