434

Adverse-Loaded CG Checks When an alteration has been made on an aircraft whose specifications do not list an empty-weight center of gravity range. it is wise to make an adverse-loaded CG check to determine whether or not it is possible to load the aircraft in such a way that its operational CG will fall outside of its allowable limits.

For an example of adverse-loaded CG checks, we wil l use this information:

Airplane EW and EWCG .... ...... .... ... ... ............... I ,340 pounds at +37.0 Engine METO horsepower .............................................................. 165 Maximum allowable gross weight ................................... 2,400 pounds CG limits ........................................................................ +35.6 to +47.5 Pilot ......... .. .................................................. .......... 170 pounds at +34.5 Front seat passenger .............................................. 170 pounds at +34.5 Rear seat passengers (2) ........................................ 340 pounds at + 7 1.5 Fuel

Full tanks (40 gallons) ....................................... 240 pounds at +48 Minimum fue l (METO HP-:- 2) .......... .. .... ... .... 82.5 pounds at +48

Baggage (maximum) .................................................. 60 pounds at +92

Forward CG Check To conduct a forward CG check, make a chart that includes the EW and EWCG of the aircraft after the alteration. the pilot, and all passengers and equipment that wou ld move the CG ahead of the forward CG limit. Do not include any passengers or equipment located behind the forward limit. Include full fue l in all tanks located ahead of the forward limit and minimum fuel in al l tanks located behind the forward limit. M inimum fuel is no more than the quantity needed for one-half hour of operation at rated maximum continuous power, and is considered for this purpose as being 'In gallon for each maximum except takeoff (METO) horsepower. Minimum fuel. in pounds. is found by dividing the METO horsepower by 2.

Under these conditions the most forward CG of +37.0 is behind the forward limit of +35.6, and is therefore an acceptable condition.

A VIATIO:'-i M AINTENAi\CE T ECHNICIAN SERIES

Fwd CG Aft CG

Datum limit limit

Figure 6-30. Load conditions for adverse loaded CG checks

Item Weight Arm Moment

Airplane 1,340.0 37.0 49,580

Pilot 170.0 34.5 5,865

Front seat passenger 170.0 34.5 5,865

Minimum fuel 82.5 48.0 3,960

1,762.5 37.0 65,270

Aft CG Check To conduct an aft, or rearward, CG check, make a chart that includes the EW and EWCG of the aircraft after the alteration, the pilot, and all passengers and equipment that would move the CG behind the aft CG limit. Do not include any passengers or equipment located ahead the aft limit. Include full fuel in all tanks located behind the aft limit, and minimum fuel in all tanks located ahead of the aft limit.

Under these conditions the most rearward CG of +45.0 is ahead the rearward limit of +47.5, and is therefore an acceptable condition.

Item Weight Arm Moment

Airplane 1,340 37.0 49,580

Pilot 170 34.5 5,865

Full fuel 240 48.0 11 ,5~0

Rear seat passengers 340 71.5 24,310

Baggage 60 92.0 5,520

2,150 45.0 96,795

W EIGHT AND B ALANCE

Figure 6-31. Chart for finding the most forward center of gravity

Figure 6-32. Chart for finding the most rearward center of gravity

Chapter 6 435 -

STUDY QUESTIONS: ADVERSE-LOADED CG CHECKS

Answers are on Page 447. Page numbers refer to chapter text.

22. No adverse-loaded CG checks are required if the empty-weight CG falls within the empty-weight ___ _ _ _ . Page 434

23. When computing the maximum forward-loaded CG of an aircraft, minimum weights, anns, and moments should be used for items of useful load that are located aft of the (forward or aft) CG limit. Page 434

24. When computing the maximum rearward-loaded CG of an aircraft, minimum weights, arms, and moments

436

should be used for items of useful load that are located forward of the (forward or aft) CG limit. Page 435

Center of Gravity Change After Repair or Alteration There is no requirement for periodic reweighing of a general aviation air-craft operated under 14 CFR Part 9 1. but it is the responsibility of any tech-nician performing a repair or alteration that changes the empty weight and/or balance conditions to update the weight and balance information to reflect this change.

We want to find the new EW and EWCG after this alteration has been made.

1. Replace a 30-pound seat at+ 72 inches with a 20-pound seat whose arm is + 73.5 inches.

2. Install a 30-pound radio at +30 inches

We start with the aircraft EW and EWCG found in the current weight and balance information; then construct a chart to compute the total moment and total weight after the alterations have been made. When the total moment and total weight have been found , divide the total moment by the total weight to find the new EWCG. See Figure 6-33.

The weight and balance information must be updated to change the EW from 1.800 pounds to 1 ,820 pounds. and the EWCG from 31.5 inches aft of the datum to 31.3 inches aft of the datum.

AH-\TIO:O.. MAI'\TE'\A:\CE TECIINICIA:O.. SERIES GE:\ERAL

Item Weight Arm Moment

Aircraft 1,800 31.5 56,700

Seat (remove) -30 72.0 -2,160

Seat (install) 20 73.5 1,470

Radio 30 30.0 900

1,820 31.3 56,910

Figure 6-33. Chart for locating the center of gravity after a repair

STUDY QUESTIONS: CG CHANGE AFTER REPAIR OR ALTERATION

Answers are on Page 447. Page numbers refer to chapter text.

25. An aircraft with an empty weight of 1,500 pounds and an empty weight CG of +28.4 was altered as

follows: a. Two seats weighing 12 pounds each, located at +65.5, were removed. b. A structural modification increasing the weight by 28 pounds was made at+ 73. c. A seat and safety belt weighing 30 pounds were installed at+ 70.5.

d. Radio equipment weighing 25 pounds was installed at +85. The new empty weight is pounds, and the new empty weight CG is located _ ___________ inches aft of the datum.

Page 436

26. An aircraft with an empty weight of 2,100 pounds and an empty weight CG of +32.5 was altered as

follows: a. A generator weighing 18 pounds, located at -8.0, was removed. b. A new propeller which weighs 17 pounds more than the old one was installed at -27.0. c. Radio equipment weighing 35 pounds was removed from +30.0.

The new empty weight is pounds, and the new empty weight CG is located ____________ inches aft of the datum.

Page 436

W EJGHT AND B ALANCE Chapter 6

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438

Determination of Needed Ballast It sometimes becomes necessary to adjust the balance of an aircraft by add-ing ballast. This can be temporary ballast needed for certain night conditions. or it can be permanent ballast needed to bring the EWCG into the allowable range after a repair or alteration has moved it outside its allowable limit.

Temporary ballast, in the form of bags of lead shot or lead bars, is often car-ried in the baggage compartments to adjust the balance for certain flight conditions. The bags are marked "BALLAST XX POUNDS- REMOY AL REQUIRES WEIGHT AND BALANCE CHECK." Temporary ballast must be secured so it cannot shift its location in flight, and the structural limits of the baggage compartment must not be exceeded.

An airplane whose empty weight is I ,205 pounds has been al tered in such a way that its EWCG is +41.0. The forward EWCG limit is +43.2. In order to bring the EWCG into its allowable range. permanent ballast can be attached to a bulkhead at tation 260.5. We can find the weight needed for permanent ballast by using this formula:

aircraft weight x distance out of limits ballast weight = ------=----------

di stance between ballast and desired CG

1,205 . (43 .2 - 41) = -------(260.5 - 43.2)

= ] ,205 . 2.2

2 17.3

= 12.2 pounds

A lead weight weighing 12.2 pounds should be prepared, painted red. and marked as 'PERMANENT BALLAST -DO NOT REMOVE." It should be attached to the structure in such a way that it does not interfere with any control action, and attached rigidly enough that it cannot be dislodged by any fl ight maneuvers or rough landing.

AV!ATIO;\" MAI;\TEi\A:--CE TECHNICIA~ SERIES GE:--ERAL

STUDY QUESTIONS: BALLAST

Answers are on Page 447. Page numbers refer to chapter text.

27. The weight of any installed permanent ballast ____ (is or is not) considered to be part of the empty weight of the aircraft. Page 438

28. An aircraft has a loaded weight of 4,954 pounds, its forward CG limit is at fuselage station 32.0, and its CG is located at +30.5. To move the CG into the allowable range, _____ pounds of ballast will have to be attached at fuselage station 162. Page 438

Large Aircraft Weight and Balance Computations Finding the Maximum Payload To determine the maximum amount of payload that can be carried, use these values:

Basic operating weight.. .............................................. 150,000 pounds Max. zero fuel weight ...... ........................ ................... 230,000 pounds Max. landing weight ................................................... 245,000 pounds Max. takeoff weight ................................ ................. ... 320,000 pounds Fuel tank load ................................................................ 94,500 pounds Estimated fuel burn en route ......................................... 71,500 pounds

Solve this problem by using these steps:

1. Compute the maximum takeoff (T/0) weight for this trip:

a. Start with the maximum landing weight of 245,000 pounds, and add the estimated fuel that will be burned en route. This will give there-quired trip takeoff weight.

b. Compare the trip takeoff weight with the maximum takeoff weight and use the smaller value for the next step.

Maximum Limit

245,000

320,000

landing weight

+trip fuel

T/0 weight

Trip Limit

245,000

+ 71,500

316,500

Continued

W EIGHT Al\'D B ALM\CE

Basic Operating Weight (BOW). The weight of the aircraft, including the crew, ready for flight, but without payload and fuel. This term applies only to transport category aircraft.

zero fuel weight. The weight of an aircraft without fuel.

Chapter 6 439

440

2. Our trip takeoff weight is smaller than the maximum allowed, so we will use it to find the zero fuel weight:

a. Subtract the fuel load from the trip takeoff weight to find the zero fuel weight.

Maximum Limit

320,000 T/0 weight -fuel load

230,000 zero fuel weight

Trip Limit

316,500 - 94,500

222,000

b. Our trip zero fuel weight is smaller than the maximum zero fuel weight, so we will use it.

3. Subtract the basic operating weight from the trip zero fuel weight to find the maximum payload it is possible to carry.

Maximum Limit

245,000 zero fuel weight - B.O.W.

payload

Trip Limit

222,000 - 150,000

72,000

We can carry 72,000 pounds of payload and a fuel load of 94,500 pounds without exceeding our maximum takeoff weight, landing weight, or zero fuel weight.

Determining Minutes of Fuel Dump Time We will use thi information to determine the number of mi nutes of dump time needed to reduce the airplane weight to its maximum allowable landing weight:

Three operating engines 171.000 pound cruise weight 142.500 pound maximum landing weight 3.170 lb/hr/engine average fuel flow during dumping and descent 19 minutes from start of dump to landing 2.300 lb/min fue l dump rate

A VIATIO'i MAI~TENANCE T ECHNICIAN SERIES G EI':ERAL

Solve this problem by using these steps:

1. Find the needed weight reduction by subtracting the maximum landing weight from the cruise weight:

171,000 - 142,500

28,500

lb cruise weight lb maximum landing weight

lb needed weight reduction

2. Find the total amount of fuel burned from beginning of dump to touch-down by multiplying the fuel flow per engine by the number of engines:

3,170 lb/hr/engine fuel flow x 3 number of engines

9,510 lb/hr (total fuel flow per hour)

Find the total fuel flow per minute by dividing the total fuel flow by 60:

9,510 lb/hr-:- 60 = 158.5lb/min

Multiply the fuel burned per minute by the number of minutes from start of dump to landing:

158.5lb/rnin. x 19 min= 3,011.5lb fuel burned after dumping started

3. Find the amount of fuel needed to dump by subtracting the amount of fuel burned after the dumping started from the needed weight reduction:

28,500.0 -3,011.5

25,488.5

lb needed weight reduction lb fuel burned after dumping started

lb = amount of fuel to dump

4. Find the time required to dump the fuel by dividing the amount of fuel to dump by the dump rate:

25,488.5 lb-:- 2,300 lb/min = 11.08 minutes

STUDY QUESTION: LARGE AIRCRAFT WEIGHT AND BALANCE

Answer is on Page 447. Page number refers to chapter text.

29. The maximum allowable weight of a loaded aircraft without the fuel is called the _______ _

weight of the aircraft. Page 439

W EIGHT AND B ALANCE Chapter 6 -

441

442

Weight and Balance Computations with an Electronic Computer Weight and balance computations for large aircraft can be quite complex. and have for years been computed with slide rule-type devices called load adjusters. Modem electronic flight computers have weight and balance programs written into them that make computing weight and balance easy.

Consider these conditions:

Basic Operating Weight (BOW) = 70.500 pounds BOW CG = 25.0% MAC Length of MAC = 164.5 inches LEMAC = 527.0-39.6 =Station 487.4 Cargo

Compartment A Station 227.5-1 ,000 pounds Compartment B Station 3 17- 2,000 pounds Compartment C Station 407- 3,500 pounds Compartment D Station 497- 3.500 pounds Compartment E Station 587 -2.500 pounds Compartment F Station 677 - 2.500 pounds Compartment G Station 766.5-1,000 pounds

Fuel - Average location 555.0-30,000 pounds

Finding the CG in inches from the datum:

1. Length of MAC = 164.5

CG = 25% MAC = 487.4 + (1 64.5 x .25) = 528.5 inches aft of the datum.

2. Construct a chart such as the one in Figure 6-34.

3. Find the CG in inches from the datum by dividing the total moment by the total weight:

61.861.250 ..;- 116,500 = 531 inches aft of the datum

4. Find the CG location in inches aft of LEMAC:

531 - 487.4 = 43.6 inches aft ofLEMAC

A \ "IATIO' M AI"\TE"\A."\CE T ECIINICIA:-.. S ERIES GE"\ERAL

Item Weight Arm Moment

Aircraft 70,500 528.5 37,259,250

Fuel 30,000 555.0 16,650,000

Cargo A 1,000 227.5 227,500

Cargo 8 2,000 317.0 634,000

Cargo C 3,500 407.0 1,424,500

Cargo D 3,500 497.0 1,739,500

Cargo E 2,500 587.0 1,467,500

Cargo F 2,500 677.0 1,692,500

Cargo G 1,000 766.5 766,500

116,500 531 .0 61,861 ,250

Figure 6-34. Charr showing rhe effect of cargo and fuel on rhe cenrer of gravity of a large aircrafr using arms and moments

Find the CG in % MAC:

CG in % MAC = CG from LEMAC J 00 MAC

= (53 1 - 487.4) 100 164.5

43.6. 100 =

164.5

= 26.5 % MAC

W EIGl IT AND B ALANCE

-Chapter 6 443

Figurer -35. CX-Ja electronicfli.~ht comput. can solve most complex ~teight and balance problems quickly and accurately.

444

The CX- J a Pathfinder electronic flight computer produced by ASA, Inc. (see Figure 6-35), handles this type or problem quickly and accurately by using the steps shown in Figure 6-36.

1 . Press the [ON) button and the [left arrow] one time to enter the WT/ Arm program.

2. Press [ENTER] and enter the aircraft weight of 70,500 pounds.

3. Press [ENTER] and enter the arm of the aircraft. This is 528.5 inches.

4. This problem uses very large numbers for the moment, and it would be convenient to use a reduction factor of 1 ,000. Press [ENTER] and enter the reduction factor of 1 ,000.

5. Press [ENTER) and enter the weight of the fuel. This is 30,000 pounds.

6. Press [ENTER) and enter the arm of the fuel. This is 555.0 inches.

7. Press [ENTER] and enter the weight of the cargo in compart-ment A. This is 1,000 pounds.

8. Press [ENTER] and enter the arm of the cargo in compartment A. This is 227.5 inches.

9. Press [ENTER] and enter the weight of the cargo in compart-ment B. This is 2,000 pounds.

10. Press [ENTER] and enter the arm of the cargo in compartment B. This is 317.0 inches.

11. Press [ENTER] and enter the weight of the cargo in compart-ment C. This is 3,500 pounds.

12. Press [ENTER) and enter the arm of the cargo in compartment C. This is 407.0 inches.

13. Press [ENTER] and enter the weight of the cargo in compart-ment D. This is 3,500 pounds.

14. Press [ENTER] and enter the arm of the cargo in compartment D. This is 497.0 inches.

15. Press [ENTER] and enter the weight of the cargo in compart-ment E. This is 2,500 pounds.

16. Press [ENTER] and enter the arm of the cargo in compartment E. This is 587.0 inches.

17. Press [ENTER] and enter the weight of the cargo in compart-ment F. This is 2,500 pounds.

18. Press [ENTER] and enter the arm of the cargo in compartment F. This is 677.0 inches.

19. Press [ENTER] and enter the weight of the cargo in compart-ment G. This is 1,000 pounds.

20. Press [ENTER] and enter the arm of the cargo in compartment G. This is 766.5 inches.

21. Press [ENTER] and the CG of 531.0 appears on the display. This is 531.0 inches aft of the datum.

22. Press [left arrow) to return to the menu and then press [right arrow] to enter the % MAC program.

23. Press [ENTER] and enter the length of MAC. This is 164.5 inches.

24. Press [ENTER] twice and enter LEMAC. This is 487.4 inches aft of the datum.

25. Press [ENTER] and 26.5% MAC appears on the display. The CG is located at 26.5% MAC.

Figure 6-36. Step-by-step .wlution.for finding the center of gravity of a large aircraft in % MA C using an electronic flight computer

AVIATION MAii\TE:-IAI':Cl: T ECH:-.ICit\N Sl:RIES GE~ERAL I

The electronic f light computer can also be used if the information is given in terms of weight and moment index.

If the weight and balance information is furnished by charts or tables that list the weight and a moment index, use the steps on the CX-la computer shown in Figure 6-38 on Page 446.

Item Weight Moment/1 ,000

Aircraft 70,500 37,259

Fuel 30,000 16,650

Cargo A 1,000 227

Cargo 8 2,000 634

Cargo C 3,500 1,424

Cargo 0 3,500 1,739

Cargo E 2,500 1,467

Cargo F 2,500 1,692

Cargo G 1,000 766

116,500 61,861

Figure 6-37. Char/ showing the effect of cargo and fuel on the center of waviry o(a larJ!.e aircraft using weights and moment indices

WEIGHT AND B ALANCE

-Chapter 6 445

446

1. Press the [ON] button and the [left arrow] one time and the [right arrow] twice to enter the WT/Mom program.

2. Press [ENTER] and enter the aircraft weight of 70,500 pounds.

3. Press [ENTER] and enter the moment index of the aircraft. This is 37,259 pound-inches/1 ,000.

4. The reduction factor used to get the moment index is 1 ,000. Press [ENTER] and enter the reduction factor of 1 ,000.

5. Press [ENTER] and enter the weight of the fuel. This is 30,000 pounds.

6. Press [ENTER] and enter the moment index of the fuel. This is 16,650.

7. Press [ENTER] and enter the weight of the cargo in compart-ment A. This is 1 ,000 pounds.

8. Press [ENTER] and enter the moment index of the cargo in compartment A. This is 227.

9. Press [ENTER] and enter the weight of the cargo in compart-ment B. This is 2,000 pounds.

10. Press [ENTER] and enter the moment index of the cargo in compartment B. This is 634.

11. Press [ENTER] and enter the weight of the cargo in compart-ment C. This is 3,500 pounds.

12. Press [ENTER] and enter the moment index of the cargo in compartment C. This is 1 ,424.

13. Press [ENTER] and enter the weight of the cargo in compart-ment D. This is 3,500 pounds.

14. Press [ENTER] and enter the moment index of the cargo in compartment D. This is 1 ,739.

15. Press [ENTER] and enter the weight of the cargo in compart-ment E. This is 2,500 pounds.

16. Press [ENTER] and enter the moment index of the cargo in compartment E. This is 1 ,467.

17. Press [ENTER] and enter the weight of the cargo in compart-ment F. This is 2,500 pounds.

18. Press [ENTER] and enter the moment index of the cargo in compartment F. This is 1 ,692.

19. Press [ENTER] and enter the weight of the cargo in compart-ment G. This is 1 ,000 pounds.

20. Press [ENTER] and enter the moment index of the cargo in compartment G. This is 766.

21. Press [ENTER] and the CG of 531.0 appears on the display. This is 531.0 inches aft of the datum.

22. Press [left arrow] to return to the menu and then press [left arrow] to enter the% MAC program.

23. Press [ENTER] and enter the length of MAC. This is 164.5 inches.

24. Press [ENTER] twice and enter LEMAC. This is 487.4 inches aft of the datum.

25. Press [ENTER] and 26.5 % MAC appears on the display. The CG is located at 26.5% MAC.

Figu re 6-38. Step-by-step solution for finding 1he celller of grmily of a large aircraft in % MAC ll'ith an eleclronic .flight compUTer using momelll indices

A VIATION M Au\JTE ANCE T ECHNIC!At--: SERIES Gt:\ERAL

Answers to Chapter 6 Study Questions

I. arm 11. equipment list 21. 21,540, +56.7

2. moment 12. Type Certificate Data Sheet 22. CG range

3. pound-inches 13. lavatory water 23. forward

4. positive 14. full 24. aft

5. negative 15. tare 25. 1,559, +30.35

6. negative 16. stress plates 26. 2,064, + 32.4

7. positive 17. range 27. is

8. useful load 18. unusable 28. 57.16

9. +35 19. is 29. zero fuel

10. Type Certificate Data Sheet 20. 6,697, 150.1

-W EIGHT AND B ALANCE Chapter 6 447