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.
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: