Chapter 7 | Large Aircraft Weight and Balance
The following consists of general guidelines for the weighing procedures of airplanes weighing over 12,500 pounds. Also included are several examples of center of gravity determination for various operational aspects of these aircraft. Persons seeking approval for a weight and balance control program for aircraft operated under Title 14 of the Code of Federal Regulations (14 CFR) part 91, subpart K, or parts 121, 125, and 135 should consult with the Flight Standards District Office (FSDO) or Certificate Management Office (CMO) having jurisdiction in their area.
Weighing Procedures
When weighing large aircraft, compliance with the relevant manuals, operations specifications, or management specification is required to ensure that weight and balance requirements specified in the aircraft flight manual (AFM) are met in accordance with approved limits. This will provide information to the flightcrew that allows the maximum payload to
The aircraft shall be weighed in an enclosed building after the aircraft has been cleaned. Check that the aircraft is in a configuration for weighing with regard to flight controls, unusable fuel, ballast, oil, and other operating fluids, and equipment as required by the controlling weight and balance procedure.
Large aircraft are not usually raised off the floor on jacks for weighing, they are weighed on ramp-type scales. The scales must be properly calibrated, zeroed, and used in accordance with the manufacturer’s instructions. Each scale should be periodically checked for accuracy as recommended in the manufacturer’s calibration schedule either by the manufacturer, or by a recognized facility such as a civil department of weights and measures. If no manufacturer’s schedule is available, the period between calibrations should not exceed 1 year.
Determining the Empty Weight and EWCG
When the aircraft is properly prepared for weighing, roll it onto the scales, and level it. The weights are measured at three weighing points: the two main wheel points and the nose wheel point.
The empty weight and EWCG are determined by using the following steps, and the results are recorded in the weight and balance record for use in all future weight and balance computations.
- Determine the moment index of each of the main-wheel points by multiplying the net weight (scale reading less tare weight), in pounds, at these points by the distance from the datum, in inches. Divide these numbers by the appropriate reduction factor.
- Determine the moment index of the nose wheel weighing point by multiplying its net weight, in pounds, by its distance from the datum, in inches. Divide this by the reduction factor.
- Determine the total weight by adding the net weight of the three weighing points and the total moment index by adding the moment indexes of each point.
- Divide the total moment index by the total weight, and multiply this by the reduction factor. This gives the CG in inches, from the datum.
- Determine the distance of the CG behind the leading edge of the mean aerodynamic chord (LEMAC) by subtracting the distance between the datum and LEMAC from the distance between the datum and the CG.
Distance CG to LEMAC = Datum to CG - datum to LEMAC
- Determine the EWCG in % MAC by using this formula:
EWCG in % MAC = CG in inches from LEMAC x 100
MAC
Determining the Loaded CG of the Airplane in Percent MAC
The basic operating weight (BOW) and the operating index are entered into a loading schedule like the one in figure 7-1 and the variables for the specific flight are entered as are appropriate to determine the loaded weight and CG.
Use the data in this example:
Basic operating Weight.......................... 105,500 lbs.
Basic operating index (total moment/1,000)...98,837.0
MAC .............................................................180.9 in
LEMAC ............................................................ 860.5
figure 7-1. Loading tables.
Use figure 7-2 to determine the moment indexes for the passengers (PAX), cargo, and fuel.
The airplane is loaded in this way:
Passengers (nominal weight 170 pounds each)
Forward compartment.......................... 18
Aft compartment..................................95
Cargo
Forward hold............................. 1,500 lbs
Aft hold..................................... 2,500 lbs
Fuel
Tank 1 & 3 ..................... 10,500 lbs each
Tank 2 ..................................... 28,000 lbs
Determine the location of the CG in inches aft of the datum by using this formula:
[INSERT FORMULA/COMPUTATION HERE]
Determine the distance from the CG to the LEMAC by subtracting the distance between the datum and LEMAC from the distance between the datum and the CG:
The location of the CG in percent of MAC must be known in order to set the stabilizer trim takeoff. Use this formula:
On Board Aircraft Weighing System
Some large transport airplanes have an on board aircraft weighing system (OBAWS) that, when the aircraft is on the ground, gives the flightcrew a continuous indication of the aircraft total weight and the location of the CG in % MAC.
The system consists of strain-sensing transducers in each main wheel and nose wheel axle, a weight and balance computer, and indicators that show the gross weight, the CG location in percent of MAC, and an indicator of the ground attitude of the aircraft.
The strain sensors measure the amount each axle deflects and sends this data into the computer, where signals from all of the transducers and the ground attitude sensor are integrated. The results are displayed on the indicators for the flightcrew.
figure 7-2. Loading schedule for determining weight and CG.
Determining the Correct Stabilizer Trim Setting
It is important before takeoff to set the stabilizer trim for the existing CG location. There are two ways the stabilizer trim setting systems may be calibrated: in % MAC, and in Units ANU (Airplane Nose Up).
Stabilizer Trim Setting in Percent of MAC
If the stabilizer trim is calibrated in units of % MAC, determine the CG location in % MAC as has just been described, then set the stabilizer trim on the percentage figure thus determined.
Stabilizer Trim Setting in Percent of ANU (Airplane Nose Up)
Some aircraft give the stabilizer trim setting in Units ANU that correspond with the location of the CG in % MAC. When preparing for takeoff in an aircraft equipped with this system, first determine the CG in % MAC in the way described above, then refer to the Stabilizer Trim Setting Chart on the Takeoff Performance page of the AFM. figure 7-3 is an excerpt from such a page from the AFM of a Boeing 737.
Consider an airplane with these specifications:
CG location......... station 635.7
LEMAC ................. station 625
MAC ...........................134.0 in
First determine the distance from the CG to the LEMAC by using this formula:
= 635.7-625.0
= 10.7 inches
Then determine the location of the CG in percent of MAC by using this formula:
Refer to figure 7-3. For all flap settings and a CG located at 8% MAC, the stabilizer setting is 73/4 Units ANU.
Determining CG Changes Caused by Modifying the Cargo
Large aircraft carry so much cargo that adding, subtracting, or moving any of it from one hold to another can cause large shifts in the CG.
Effects of Loading or Offloading Cargo
Both the weight and CG of an aircraft are changed when cargo is offloaded or onloaded. This example shows the way to determine the new weight and CG after 2,500 pounds of cargo is offloaded from the forward cargo hold.
Consider these specifications:
Loaded weight .............................. 90,000 lbs
Loaded CG................................ 22.5% MAC
Weight change...............................- 2,500 lbs
Fwd. cargo hold centroid .......... station 352.1
MAC .................................................141.1 in
LEMAC .................................. station 549.13
- Determine the CG location in inches from the datum before the cargo is removed. Do this by first determining the distance of the CG aft of the LEMAC:
- Determine the distance between the CG and the datum by adding the CG in inches aft of LEMAC to the distance from the datum to LEMAC:
CG (in. from datum) = CG in. aft of LEMAC + datum to LEMAC
= 31.84 + 549.13
= 580.97 inches
- Determine the moment/1,000 for the original weight:
1,000
= 90,000 x 580.97
1,000
= 52,287.30
- Determine the new weight and new CG by first determining the moment/1,000 of the removed weight.
Multiply the amount of weight removed (-2,500 pounds) by the centroid of the forward cargo hold (352.1 inches), and then divide this by 1,000.
1,000
= -2,500 x 352.1
1,000
= -880.25
- Subtract the removed weight and its moment/1,000 from the original weight and moment/1,000.
- Determine the location of the new CG by dividing the total moment/1,000 by the total weight and multiplying this by the reduction factor 1,000.
- Convert the new CG location to % MAC. First, determine the distance between the CG location and LEMAC:
CG (in. aft of LEMAC) = CG (in. from datum) - LEMAC
= 587.5-549.13
= 38.37 inches
- Then, determine new CG in % MAC:
Offloading 2,500 pounds of cargo from the forward cargo hold moves the CG from 22.5% MAC to 27.1% MAC.
Effects of Onloading Cargo
The previous example showed the way the weight and CG changed when cargo was offloaded. This example shows the way both parameters change when cargo is onloaded.
The same basic airplane is used in the following example, but 3,000 pounds of cargo is onloaded in the forward cargo hold.
Weight before cargo is loaded....... 87,500 lbs
CG before cargo is loaded ........ 27.1% MAC
Weight change..............................+ 3,000 lbs
Fwd. cargo hold centroid .......... station 352.1
MAC .................................................141.5 in
LEMAC .................................. station 549.13
- Determine the CG location in inches from the datum before the cargo is onloaded. Do this by first determining the distance of the CG aft of the LEMAC:
[INSERT COMPU/FORMULA]
- Determine the distance between the CG and the datum by adding the CG in inches aft of LEMAC to the distance from the datum to LEMAC:
CG (in. from datum) = CG in. aft of LEMAC + datum to LEMAC
= 38.35 + 549.13
= 587.48 inches - Determine the moment/1,000 for the original weight:
Moment/1,000 = Weight x Arm
1,000
= 87,500 x 587.48
1,000
= 51,404.5
- Determine the new weight and new CG by first determining the moment/1,000 of the added weight. Multiply the amount of weight added (3,000 pounds) by the centroid of the forward cargo hold (352.1 inches), and then divide this by 1,000.
Moment/1,000 = Weight x Arm
1,000
= 3,000 x 352.1
1,000
- Add the onloaded cargo weight and its moment/1,000 to the original weight and moment/1,000.
- Determine the location of the new CG by dividing the total moment/1,000 by the total weight and multiplying this by the reduction factor of 1,000.
[INSERT COMPU/FORMULA HERE]
- Convert the new CG location to % MAC. First, determine the distance between the CG location and LEMAC:
CG (in. aft of LEMAC) = CG (in. from datum) - LEMAC
= 579.68 - 549.13
= 30.55 inches - Then, determine new CG in % MAC:
[INSERT COMPU/FORMULA HERE]
Onloading 3,000 pounds of cargo into the forward cargo hold moves the CG forward 5.51 inches, from 27.1% MAC to 21.59% MAC.
Effects of Shifting Cargo from One Hold to Another
When cargo is shifted from one cargo hold to another, the CG changes, but the total weight of the aircraft remains the same.
Loaded weight........................... 90,000 lbs
Loaded CG.......................... station 580.97
(which is 22.5% MAC)
Fwd. cargo hold centroid ......... station 352
Aft cargo hold centroid ......... station 724.9
MAC..............................................141.5 in
LEMAC.................................... station 549
To determine the change in CG, or __CG, caused by shifting 2,500 pounds of cargo from the forward cargo hold to the aft cargo hold, use this formula:
Since the weight was shifted aft, the CG moved aft, and the CG change is positive. If the shift were forward, the CG change would be negative.
Before the cargo was shifted, the CG was located at station 580.97, which is 22.5% MAC. The CG moved aft 10.36 inches, so the new CG is:
Convert the location of the CG in inches aft of the datum to % MAC by using this formula:
The new CG in % MAC caused by shifting the cargo is the sum of the old CG plus the change in CG:
Some aircraft AFMs locate the CG relative to an index point rather than the datum or the MAC. An index point is a location specified by the aircraft manufacturer from which arms used in weight and balance computations are measured. Arms measured from the index point are called index arms, and objects ahead of the index point have negative index arms, while those behind the index point have positive index arms.
Use the same data as in the previous example, except for these changes:
Loaded CG........index arm of 0.97, which is 22.5% MAC
Index point......................................fuselage station 580.0
Fwd. cargo hold centroid ........................-227.9 index arm
Aft cargo hold centroid ..........................+144.9 index arm MAC......................................................................141.5 in LEMAC...................................................-30.87 index arm
The weight was shifted 372.8 inches (-227.9 to +144.9 = 372.8).
The change in CG can be calculated by using this formula:
Since the weight was shifted aft, the CG moved aft, and the CG change is positive. If the shift were forward, the CG change would be negative.
Before the cargo was shifted, the CG was located at 0.97 index arm, which is 22.5% MAC. The CG moved aft 10.36 inches, and the new CG is:
The change in the CG in % MAC is determined by using this formula:
The new CG in % MAC is the sum of the old CG plus the change in CG:
Notice that the new CG is in the same location whether the distances are measured from the datum or from the index point.
Determining Cargo Pallet Loads with Regard to Floor Loading Limits
Each cargo hold has a structural floor loading limit based on the weight of the load and the area over which this weight is distributed. To determine the maximum weight of a loaded cargo pallet that can be carried in a cargo hold, divide its total weight, which includes the weight of the empty pallet and its tiedown devices, by its area in square feet. This load per square foot must be equal to or less than the floor load limit.
In this example, determine the maximum load that can be placed on this pallet without exceeding the floor load limit.
Pallet dimensions ......................................36 by 48 in
Empty pallet weight .......................................... 47 lbs
Tiedown devices................................................ 33 lbs
Floor load limit ............... 169 pounds per square foot
The pallet has an area of 36 inches (3 feet) by 48 inches (4 feet). This is equal to 12 square feet. The floor has a load limit of 169 pounds per square foot; therefore, the total weight of the loaded pallet can be 169 x 12 = 2,028 pounds.
Subtracting the weight of the pallet and the tiedown devices gives an allowable load of 1,948 pounds (2,028 - [47 + 33]).
Determine the floor load limit that is needed to carry a loaded cargo pallet having these dimensions and weights:
Pallet dimensions ........... 48.5 by 33.5 in
Pallet weight ................................. 44 lbs
Tiedown devices ........................... 27 lbs
Cargo weight ........................... 786.5 lbs
First determine the number of square feet of pallet area:
Then determine the total weight of the loaded pallet:
Pallet
|
44.0 lbs
|
Tiedown devices
|
27.0 lbs
|
Cargo
|
786.5 lbs
|
|
857.5 lbs
|
Determine the load imposed on the floor by the loaded pallet:
The floor must have a minimum load limit of 76 pounds per square foot.
Determining the Maximum Amount of Payload That Can Be Carried
The primary function of a transport or cargo aircraft is the carry payload. This is the portion of the useful load, passengers or cargo, that produces revenue. To determine the maximum amount of payload that can be carried, follow a series of steps, considering both the maximum limits for the aircraft and the trip limits imposed by the particular trip. In each step, the trip limit must be less than the maximum limit. If it is not, the maximum limit must be used.
These are the specifications for the aircraft in this example:
Basic operating weight (BOW)...............100,500 lbs
Maximum zero fuel weight.....................138,000 lbs
Maximum landing weight.......................142,000 lbs
Maximum takeoff weight........................184,200 lbs
Fuel tank load ...........................................54,000 lbs
Est. fuel burn en route...............................40,000 lbs
-
Compute the maximum takeoff weight for this trip. This is the maximum landing weight plus the trip fuel.
Max. LimitTrip Limit142,000Landing weight142,000+ trip fuel+ 40,000184,200Takeoff weight182,000 -
The trip limit is the lower, so it is used to determine the zero fuel weight.
Max. Limit | Trip Limit | |
184,200 | Takeoff weight | 182,000 |
- fuel load | -54,000 | |
138,000 | Zero fuel weight | 128,000 |
- The trip limit is again lower, so use it to compute the maximum payload for this trip.
Max. Limit Trip Limit
138,000 Zero fuel weight 128,000
- BOW - 100,500
Payload (pounds) 27,500
Under these conditions 27,500 pounds of payload may be carried.
Determining the Landing Weight
It is important to know the landing weight of the airplane in order to set up the landing parameters, and to be certain the airplane will be able to land at the intended destination.
In this example of a four-engine turboprop airplane, determine the airplane weight at the end of 4.0 hours of cruise under these conditions:
Takeoff weight ........................................140,000 lbs
Pressure altitude during cruise.................16,000 feet
Ambient temperature during cruise .................-32°C
Fuel burned during descent and landing.....1,350 lbs
Determine the weight at the end of cruise by using the Gross Weight Table of figure 7-4 and following these steps:
- Use the U.S. Standard Atmosphere Table in figure 7-5 to determine the standard temperature for 16,000 feet. This is -16.7°C.
- The ambient temperature is -32°C, which is a deviation from standard of 15.3°C. (-32° – -16.7° = 15.3°). It is below standard.
- In figure 7-4, follow the vertical line representing 140,000 pounds gross weight upward until it intersects the diagonal line for 16,000 feet pressure altitude.
- From this intersection, draw a horizontal line to the left to the temperature deviation index (0°C deviation).
- Draw a diagonal line parallel to the dashed lines for “Below Standard” from the intersection of the horizontal line and the Temperature Deviation Index.
- Draw a vertical line upward from the 15.3°C Temperature Deviation From Standard.
figure 7-4. Gross Weight Table
- Draw a horizontal line to the left from the intersection of the “Below Standard” diagonal and the 15.3°C temperature deviation vertical line. This line crosses the “Fuel Flow-100 Pounds per Hour per Engine” index at 11.35. This indicates that each of the four engines burns 1,135 (100 x 11.35) pounds of fuel per hour. The total fuel burn for the 4-hour cruise is:
= 1,135 x 4 x 4
= 18,160 pounds
- The airplane gross weight was 140,000 pounds at takeoff, and since 18,160 pounds of fuel was burned during cruise and 1,350 pounds was burned during the approach and landing phase, the landing weight is:
140,000 - (18,160 + 1,350) = 120,490 pounds
figure 7-5. Standard atmosphere table.
Determining the Minutes of Fuel Dump Time
Most large aircraft are approved for a greater weight for takeoff than for landing, and to make it possible for them to return to landing soon after takeoff, a fuel jettison system is sometimes installed.
It is important in an emergency situation that the flightcrew be able to dump enough fuel to lower the weight to its allowed landing weight. This is done by timing the dumping process.
In this example, the aircraft has three engines operating and these specifications apply:
Cruise weight ..........................................171,000 lbs
Maximum landing weight .......................142,500 lbs
Time from start of dump to landing .........19 minutes
Average fuel flow during
dumping and descent ........................ 3,170 lb/hr/eng
Fuel dump rate ...................2,300 pounds per minute
Follow these steps to determine the number of minutes of fuel dump time:
-
Determine the amount the weight of the aircraft must be reduced to reach the maximum allowable landing
171,000 lbs cruise weight
- 142,500 lbs maximum landing weight
28,500 lbs required reduction -
Determine the amount of fuel burned from the beginning of the dump to touchdown:
Fuel flow = 3,170 lb/hr/engine
60
= 52.83 lb/min engine
For all three engines, this is 52.83° 3 = 158.5 lbs/min.
The three engines will burn 158.5° 19 = 3,011.5 pounds of fuel between the beginning of dumping and touchdown.
- Determine the amount of fuel needed to dump by subtracting the amount of fuel burned during the dumping from the required weight reduction:
-3,011.5 lbs fuel burned after start of dumping
25,488.5 lbs fuel to be dumped
- Determine the time needed to dump this amount of fuel by dividing the number of pounds of fuel to dump by the dump rate:
2,300 lb/min
Weight and Balance of Commuter Category Airplanes
The Beech 1900 is a typical commuter category airplane that can be configured to carry passengers or cargo. figure 7-6 shows the loading data of this type of airplane in the passenger configuration, and figure 7-14 on Page 7-16 shows the cargo configuration. Jet fuel weight is affected by temperature, the coder the fuel, the more dense and therefore the more pounds of fuel per gallon.
figure 7-6. Loading data for passenger configuration.
figure 7-7. Determining the loaded weight and CG of a Beech 1900 in the passenger configuration.
figure 7-8. Weights and moments - occupants.
figure 7-9. Weights and moments - baggage.
figure 7-10. Density variation of aviation fuel.
figure 7-11. Weights and moments -usable fuel.
figure 7-12. Weight and balance diagram.
figure 7-13. Change in CG caused by shifting passenger seats.
Determining the Loaded Weight and CG
As this airplane is prepared for flight, a manifest like the one in figure 7-7 is prepared.
-
The crew weight and the weight of each passenger is entered into the manifest, and the moment/100 for each occupant is determined by multiplying the weight by the arm and dividing by 100. This data is available in the AFM and is shown in the Weight and Moments- Occupants table in figure 7-8 on Page 7-12.
-
The weight of the baggage in each compartment that is used is entered with its moment/100. This is determined in the Weights and Moments- Baggage table in figure 7-9 on Page 7-12.
-
Determine the weight of the fuel. Jet A fuel has a nominal specific gravity at +15°C of 0.812 and weighs 6.8 pounds per gallon, but at +25°C, according to the chart in figure 7-10 on Page 7-13, it weighs 6.75 lbs/ gal.
Using figure 7-11 on Page 7-14, determine the weights and moment/100 for 390 gallons of Jet A fuel by interpolating between those for 6.7 lbs/gal and 6.8 lbs/ gal. The 390 gallons of fuel at this temperature weighs 2,633 pounds, and its moment index is 7,866 lb-in/100.
- Add all of the weights and all of the moment indexes. Divide the total moment index by the total weight, and multiply this by the reduction factor of 100. The total weight is 14,729 pounds, the total moment index is 43,139 lb-in/100. The CG is located at fuselage station 292.9.
- Check to determine that the CG is within limits for this weight. Refer to the Weight and Balance Diagram in figure 7-12 on Page 7-14. Draw a horizontal line across the envelope at 14,729 pounds of weight and a vertical line from the CG of 292.9 inches aft of datum. These lines cross inside the envelope verifying the CG is within limits for this weight.
Determining the Changes in CG When Passengers are Shifted
Consider the airplane above for which the loaded weight and CG have just been determined, and determine the change in CG when the passengers in rows 1 and 2 are moved to rows 8 and 9. figure 7-13 shows the changes from the conditions shown in figure 7-7. There is no weight change, but the moment index has been increased by 1,155 pound-inches/100 to 44,294. The new CG is at fuselage station 300.7.
This type of problem is usually solved by using the following two formulas. The total amount of weight shifted is 550 pounds(300 + 250) and both rows of passengers have moved aft by 210 inches (410 - 200 and 440 - 230).
The CG has been shifted aft 7.8 inches and the new CG is at station 300.7.
Determining Changes in Weight and CG When the Airplane is Operated in its Cargo Configuration
Consider the airplane configuration shown in figure 7-14.
The airplane is loaded as recorded in the table in figure 7-15. The basic operating weight (BOW) includes the pilots and their baggage so there is no separate item for them.
The arm of each cargo section is the centroid of that section, as shown in figure 7-14.
The fuel, at the standard temperature of 15ºC weighs 6.8 pounds per gallon. Refer to the Weights and Moments - Usable Fuel in figure 7-11 on Page 7-14 to determine the weight and moment index of 370 gallons of Jet A fuel.
The CG under these loading conditions is located at station 296.2.
Determining the CG Shift When Cargo is Moved From One Section to Another
When cargo is shifted from one section to another, use this formula:
[INSERT FORMULA here]
If the cargo is moved forward, the CG is subtracted from the original CG. If it is shifted aft, add the CG to the original.
figure 7-14. Loading data for cargo configuration.
figure 7-15. Flight manifest of a Beech 1900 in the cargo configuration.
Determining the CG Shift When Cargo is Added or Removed
When cargo is added or removed, add or subtract the weight and moment index of the affected cargo to the original loading chart. Determine the new CG by dividing the new moment index by the new total weight, and multiply this by the reduction factor.
Determining Which Limits Are Exceeded
When preparing an aircraft for flight, you must consider all parameters and check to determine that no limit has been exceeded.
Consider the parameters below, and determine which limit, if any, has been exceeded.
-
The airplane in this example has a basic empty weight of 9,005 pounds and a moment index of 25,934 pound-inches/100.
- The crew weight is 340 pounds, and its moment/100 is 439.
- The passengers and baggage have a weight of 3,950 pounds and a moment/100 of 13,221.
- The fuel is computed at 6.8 lbs/gal:
The ramp load is 340 gallons, or 2,312 pounds.
Fuel used for start and taxi is 20 gallons, or 136 pounds.
Fuel remaining at landing is 100 gallons, or 680 pounds.
- Maximum takeoff weight is 16,600 pounds.
- Maximum zero fuel weight is 14,000 pounds.
- Maximum landing weight is 16,000 pounds.
Take these steps to determine which limit, if any, is exceeded:
- Determine the zero fuel weight, which is the weight of the aircraft with all of the useful load except the fuel onboard.
The zero fuel weight of 13,295 pounds is less than the maximum of 14,000 pounds, so this parameter is acceptable.
- Determine the takeoff weight and CG. The takeoff weight is the zero fuel weight plus the ramp load of fuel, less the fuel used for start and taxi. The takeoff CG is the (moment/100 ÷ weight) x 100.
The takeoff weight of 15,471 pounds is below the maximum takeoff weight of 16,600 pounds, and a check of figure 7-12 on Page 7-14 shows that the CG at station 298.0 is also within limits.
- Determine the landing weight and CG. This is the zero fuel weight plus the weight of fuel at landing.
The landing weight of 13,975 pounds is less than the maximum landing weight, of 14,000 to 16,000 pounds. According to figure 7-12, the landing CG at station 297.5 is also within limits.