Pilot's Handbook of Aeronautical Knowledge
Chapter 10: Weight and Balance
Operating an aircraft within the weight and balance limits is critical to flight safety. Pilots must ensure that the center of gravity (CG) is and remains within approved limits throughout all phases of a flight.
If an aircraft's CG is outside the approved limits, positive control will be difficult or impossible, resulting in a hazardous flight condition. In addition, operating above the aircraft's maximum weight limitation compromises its structural integrity and adversely affects performance.
Weight Control
Weight attracts all bodies to the earth. It is a product of the mass of a body. Weight also is the result of acceleration acting on the body.
Loading an aircraft beyond the manufacturer's recommended weight must be avoided. The force of lift, created by the wings, counteracts weight and sustains an aircraft in flight. However, lift is limited by several factors, including airfoil design, angle of attack (AOA), airspeed, and air density. Pilots must understand and respect these limits.
Manufacturers attempt to make aircraft as light as possible without sacrificing strength or safety. Therefore, any item aboard an aircraft that increases the total weight is undesirable for performance.
If an aircraft is not properly loaded, the initial indication of poor performance usually takes place during takeoff. An overloaded aircraft may not be able to leave the ground. If it does become airborne, it may exhibit unexpected and hazardous flight characteristics.
Preflight planning should include a check of performance charts to determine if the aircraft's weight may contribute to hazardous flight operations.
Pilots who manage to depart in an aircraft loaded beyond its gross weight limit can expect:
- Higher takeoff speed
- Longer takeoff run
- Reduced rate and angle of climb
- Lower maximum altitude
- Shorter range
- Reduced cruising speed
- Reduced maneuverability
- Higher stalling speed
- Higher approach and landing speed
- Longer landing roll
- Excessive weight on the nose wheel or tail wheel.
The operating weight of an aircraft can be changed by simply altering the fuel load. Gasoline weighs six (6) pounds per gallon, which is considerable. The weight of one passenger can range from 20 to 40 gallons of fuel. When taking passengers and cargo, reducing total fuel on board can keep the aircraft within gross weight limits. Flight planning should factor the total range or flight time available based on the reduced amount of total fuel.
During routine flight, fuel burn is the only weight change that takes place. As fuel is used, an aircraft becomes lighter and performance is improved.
Balance, Stability, and Center of Gravity
Balance refers to the location of the center of gravity (CG) of an aircraft. It is critical to stability and safety in flight.
The CG is a point at which the aircraft would balance if it were suspended at that point.
The CG can be moved along the aircraft's longitudinal axis, based on the distribution of weight in the aircraft. The fore and aft location of the CG is the primary concern when determining an aircraft's balance.
The distance between CG's forward and back limits — called the CG range — is certified for an aircraft by the manufacturer.
If the CG is displaced too far forward on the longitudinal axis, a nose-heavy condition will result. If the CG is displaced too far aft on the longitudinal axis, a tail-heavy condition results. A center of gravity outside of the permissible CG range can result in an unstable or hazardous flight condition.
The position of the lateral CG is not computed in all aircraft. However, pilots must be aware that adverse effects arise as a result of a laterally unbalanced condition. A lateral unbalance occurs if the fuel load is mismanaged by supplying the engine(s) unevenly from tanks on one side of the airplane. This can be offset with rudder trim or constant control pressure. The result will be an out-of-streamline aircraft with more drag and decreased efficiency.
Adverse Balance
Just as excess weight affects flight characteristics, adverse balance conditions operate in much the same manner. Stability and positive control are impacted by improper balance.
Loading in a nose-heavy condition causes problems in controlling and raising the nose, especially during takeoff and landing.
Loading in a tail-heavy condition impacts longitudinal stability and reduces the pilot's ability to recover from stalls and spins. Tail-heavy loading also produces very light control forces.
The CG's fore and aft limits are published for each aircraft in the Type Certificate Data Sheet (TCDS), or aircraft specification and the AFM or POH. The CG should not exist beyond these limits before flight. If the CG is beyond the CG range, reloading the aircraft or removing items may be necessary before flight.
Manufacturers purposely place the forward CG limit as far rearward as possible to aid pilots in avoiding damage when landing. This also permits sufficient elevator/control deflection at minimum airspeed.
As the CG moves aft, a less stable condition occurs, which decreases the ability of the aircraft to right itself after maneuvering or turbulence.
For some aircraft, both fore and aft CG limits may be specified to vary as gross weight changes.
Pilots can undertake several actions to relocate the CG before flight, including relocating baggage and cargo items, and placing heavier passengers in forward seats.
While fuel burn can affect the CG, this is based on the location of the fuel tanks. Most small aircraft carry fuel in the wings very near the CG. Therefore, while fuel burn will reduce the aircraft's gross weight, it will have little effect on the loaded CG.
Before any flight, the pilot should determine the weight and balance condition of the aircraft. Charts and graphs are provided in the approved AFM/POH to enable pilots to make weight and balance computations.
A typical general-aviation aircraft (two or four seats) cannot remain within the approved weight and balance limits with full passengers, fuel, and baggage compartments. If the maximum passenger load is carried, the pilot typically will need to reduce the fuel load or baggage.
Adverse Balance: The Aerobat vs. the Racer
Knowledge test-banks include several questions on adverse balance. These can be difficult to answer without an effective foundational paradigm. Memorizing the correct answers isn't a good strategy.
In general, these questions include the terms "CG forward/CG backward" and "forward limit/aft limit."
The correct answers identify various flight characteristics: controllability, stability, stall speed, cruise speed, and wing load.
Start with an aerodynamic fundamental: Aerobatic aircraft have inherently nose-heavy CGs. This is to ensure that the aircraft will pitch down in a stall and avoid entering an unrecoverable spin. This is a characteristic of longitudinal stability. As airspeed decreases, the nose will tend to drop. If stalled, the nose will fall toward the earth, and the aircraft will rotate about its longitudinal axis. This is an ideal characteristic for an airplane that is meant to be stalled frequently.
However, if the CG is excessively forward, the downforce of the horizontal stabilizer must increase in order to balance the aircraft along the longitudinal axis. Think of this as if two adults replaced two toddlers on a see-saw. The total load on the fulcrum would be substantially larger, even though the overall performance characteristics don't appear different. However, with an airplane, the increased load caused by forward CG is being imposed upon the wings. And just as load factor in a turn increases stalling speed, overloading the wings with a forward CG also increases the stalling speed.
It then stands to reason that overloading the aircraft itself — exceeding the permitted gross weight — also increases the stalling speed.
From there, it's not hard to conclude that an aircraft that is flying with a high gross weight, or a high load imposed upon the wings, will not fly as fast as a properly loaded and balanced aircraft.
Aerodynamically, this is due to the wing flying with a higher angle of attack when CG is excessively foward. This creates more drag — induced drag. And the indicated airspeed will be lower, since the pitot tube will be measuring the ram air pressure at an upward angle.
If we return to the analogy of two adults on a over-burdened see-saw, it's easy to understand that an aircraft with excessive forward CG is incurring multiple performance penalties.
- It creates excessive induced drag.
- It flies slower.
- The indicated airspeed is slower still.
- The stall speed is higher.
- It's harder to land, since the landing flare requires that the nose can be easily raised.
The only advantage of forward CG is to the aerobat — which requires increased longitudinal stability for safety. And aerobatic pilots aren't worried about the attendant performance issues. They aren't flying cross-country airplanes.
Conversely, if the CG is aft of the aft limit, the aircraft's performance isn't being penalized, because it has low stalling speeds and high cruising speeds. However, it has critical stability penalties.
While the aerobat wants to nose over and recover from a stall, the aft-loaded aircraft wants to tumble out of the sky if airspeed is not maintained. With excessive aft CG, the longitudinal stability is greatly diminished. The airplane is no longer a nose-heavy, hard-to-land aerobat, like a Pitts Special, but instead a nimble racer with reduced wing-loading, like a Hughes H-1 — quick, but potentially dangerous. *
Therefore, when the CG is at the other end — aft of the aft limit — there are stability costs:
- The aircraft has reduced longitudinal stability.
- The aircraft is less stable and less controllable.
- Recovery from a stall becomes progressively more difficult as the CG moves rearward.
- If stalled, a flat spin may develop.
- Stall and spin recovery may be impossible.
And that's a pretty terrible airplane.
If you can remember these contrasts, you won't have any problems with adverse-balance questions on a knowledge test.
* Spoiler alert for those of you who haven't seen the original Top Gun — Maverick and Goose's devastating crash in their very fast F-14 happens because it enters an unrecoverable flat spin. It won't nose over in the stall, which is to say that it has deficient longitudinal stability.
Weight and Balance Computations
Determining the total weight of a loaded aircraft simply requires totalling the aircraft's empty weight, the fuel and oil, and everything loaded on the aircraft.
Determining the distribution of mass around the CG requires additional calculations.
The point at which an aircraft balances can be determined by locating the CG — the imaginary point at which all the weight is concentrated. To provide the necessary balance between longitudinal stability and elevator control, the CG is usually located slightly forward of the center of lift. This loading condition causes a nose-down tendency in flight, which is desirable during flight at a high AOA and slow speeds.
The CG range limits are usually specified in inches, along the longitudinal axis of the airplane, measured from a reference point called a datum. The datum reference is an arbitrary point, established by aircraft designers that may vary in location between different aircraft.
The fulcrum is where the airplane would balance were it placed on a single, raised point. For sample illustrations, the datum can be located at the fulcrum. However, in many aircraft it's at a different location along the longitudinal axis, and it's only used for reference.
The arm is the distance from the datum to any component part or any object loaded on the aircraft. When the object or component is located aft of the datum, it is measured in positive inches; if located forward of the datum, it is measured as negative inches or minus inches
The station is the location of passengers, fuel, oil, baggage, and additional items.
The moment is the weight of a station multiplied by the arm. Each occupied station has a calculated moment.
The moment is is expressed in inch-pounds (in-lb). It is the measurement of the gravitational force that causes a tendency of the weight to rotate about a point or axis.
A classic example of moment weight (or moment effectiveness) is a 50-lb weight and a 100-lb weight on a playground see-saw — a long board balanced on a fulcrum. The large weight is placed closer to the fulcrum, while the smaller weight is more distant from the fulcrum. The lesser weight has a greater arm. It exhibits as much potential energy as the greater weight because of this greater arm.
In classical physics, this mechanical advantage is called leverage, proven as "the law of the lever" by Archimedes. The further a weight is from the fulcrum, the farther it travels when it is moved around the axis. Objects closer to the fulcrum do not travel as far around the axis. Thus, an object farther from the fulcrum has a greater mechanical advantage — less force is required to change its position.
If small weight's station is 100 inches from the fulcrum, the moment (downward force of the station) can be determined by multiplying 50 pounds (weight) by 100 inches (station). The specific location on the board produces a moment of 5,000 inch-pounds.
In order to balance the board, the large weight's moment also should be 5,000 inch-pounds. The weight of the object cannot be adjusted, but the station is variable. Thus, in order for the large weight to produce 5,000 in-lb of downward force, the station will need to be 50 inches from the fulcrum.
| Weight | Arm | Moment | |
|---|---|---|---|
| Small item | 50 pounds | 100 inches | 5,000 inch-pounds |
| Large item | 100 pounds | 50 inches | 5,000 inch-pounds |
These objects in these locations create equal downward forces.
If additional objects are added to the board, the stations of each object can be adjusted so that the total moments on either side of the fulcrum are equal.
There are several test questions that require identifying a missing weight or a missing distance, with all other values presented as known. This can be solved with a simple algebraic method, where the known values and X are presented in a table format, with "equals" (=) used to identify the fulcrum. Continuous reduction of complexity will provide X.
Let's remove a value from Figure 10-4. We'll replace the amount of weight at the far right with X and reduce complexity (and because the weight is unknown, the moment becomes another variable):
| 50 * 50 | + | 100 * 25 | = | 100 * X |
| 2,500 | + | 2,500 | = | 100 * X |
| 5,000 | = | 100 * X | ||
| 5,000 ÷ 100 = 50 |
100 multiplied by X must equal 5,000. Therefore, if we divide 5,000 by 100, we will identify the variable, i.e. "solve for X." You can eliminate any weight or arm in Figure 10-4 and practice the formula. FAA test questions will use different values and graphics, but the method is identical.
Determining Loaded Weight and CG
There are various methods for determining the loaded weight and CG of an aircraft. There is the computational method as well as methods that utilize graphs and tables provided by the aircraft manufacturer.
Computational Method
A simple computational method, using a table format (also called a "schedule"), can be used to determine total weight and center of gravity.
- Start by listing the weight of the aircraft, occupants, fuel, and baggage. Fuel weighs six (6) pounds per gallon.
- Consult the loading diagram in the AFM/POH to determine the station of each item, as referenced from the datum established in the AFM/POH.
- Enter the moment for each item listed. This will be the weight of each item multiplied by its arm (weight x arm = moment).
- Find the sum of all weights.
- Find the sum of all moments.
- Divide the total moment by the total weight.
| Item | Weight | Arm | Moment | ||
|---|---|---|---|---|---|
| Aircraft Empty Weight | 2,100 | x | 78.3 | = | 164,430 |
| Front Seat Occupants | 340 | x | 85.0 | = | 28,900 |
| Rear Seat Occupants | 350 | x | 121.0 | = | 42,350 |
| Fuel | 450 | x | 75.0 | = | 33,750 |
| Baggage Area 1 | 80 | x | 150.0 | = | 12,000 |
| Total | 3,320 | 281,430 | |||
| Calculate CG: | 281,430 ÷ 3,320 = 84.8 | ||||
The result is the total arm. This is the location of the CG, as referenced from the datum.
Consult the AFM/POH to determine if the aircraft is at or under the gross weight. In the above example, total aircraft weight is 3,320 pounds.
Also consult the AFM/POH to determine if the CG is within the allowable CG range. In the above example, the CG is 84.8 inches aft of the datum.
If either figure does not permit safe flight operations, the aircraft may need to be loaded differently, and/or some items may need to be removed from the aircraft.
Graph Method
For some aircraft, loaded weight and CG can be determined using graphs provided by the manufacturers. These will be in the AFM/POH. Because moment values can be large numbers, the moment may sometimes be divided by 100, 1,000, or 10,000 to simplify calculations.
To determine the moment using the loading graph, find the weight (along the left-side metric) and draw a line straight across until it intercepts the item for which the moment is to be calculated. These diagonal lines represent stations. Then draw a line straight down to determine the moment.
In the example below, the red line depicts a pilot and passenger weighing a combined 340 pounds, which has a total moment of 127,000 — or 12.7 for purposes of calculation.
Once this has been done for each item, total all weights and all moments. Draw a line for both weight and moment on the CG envelope graph. If the lines intersect within the envelope, the aircraft is loaded within limits.
In the example below, the red line depicts an airplane with a loaded aircraft weight of 2,367 pounds and a total moment of 105.2. The aircraft's CG is within the permissible CG range.
Table Method
The table method uses information and limitations are contained in tables provided by the manufacturer. It applies the same principles as the computational and graph methods. For each station, a range of moments are provided for a range of weights. Minimum and maximum moments are then provided for the aircraft's gross weight.
Weight-Shift Calculations
The Pilot's Handbook of Aeronautical Knowledge covers "Shifting, Adding, and Removing Weight," while the FAA's Weight and Balance Handbook discusses CG shifts and weight changes in more detail. Questions regarding this subject matter are found on the Commercial Pilot, Ground Instructor, and Flight Instructor written tests.
The proof for all weight-shift, weight-add, and new-CG problems is presented as three equations on a grid. The left column is the weight ratio, while the right column is the shift ratio. The quotients of both ratios are equal.
| Cargo (weight to be shifted) | Δ CG (shift) | |
| ÷ | = | ÷ |
|---|---|---|
| Aircraft (total weight) | Δ Distance (that weight is shifted) |
One simple mnemonic for this grid is "CACD," for "Cargo, Airplane, Center of Gravity (change), Distance (that weight is shifted)." Note that Δ stands for "change" and does not represent the aircraft's initial CG, nor its resultant CG. Both values in the shift ratio are range lengths, not a specific station or CG. You might also think of these as transit lengths, so as not to confuse either variable with the CG or a station. The shift ratio does not capture the aircraft's CG nor any stations. The values in the shift ratio are relative.
Another recall method is to remember that weights are to the left, while ranges/distances are to the right. The aircraft weight is the largest number and is on the lower row. The Δ Distance between and old and new station will be larger than the Δ CG, and it also is on the lower row.
This table permits identifying any missing value, simply by multiplying two corresponding values, and then dividing the result by the third known value. However, the weight of the aircraft is always known in test questions, so total weight will never be presented as X. Pilots are concerned with:
- How much weight should be moved over a given distance to bring the CG within limits?
- How much distance should a known weight be moved to bring the CG within limits?
- By what distance will the CG change when a known weight is moved over a given distance?
- How much weight can be added at a given station while keeping the CG within limits?
- If a known weight is added at a given station, will the CG remain within limits?
Each question above can be solved with a two-step equation (with the exception of removing weight).
To illustrate, let's start with an aircraft with a loaded gross weight of 4,800 lbs. The aft limit of the CG is 97 inches. However, the loaded aircraft has a CG of 99 inches aft of the datum, or two (2) inches beyond the aft CG limit. The aircraft has luggage at the rear station (130 inches), which weighs 150 pounds. All of this luggage can be relocated to a forward station (45 inches).
If the distance between the current station of the luggage and the proposed new station is known, how much weight should be transferred to the new station?
The aircraft weight (4,800 lbs.), distance between the arms (85 inches), and desired CG change (2 inches) are known. Start with the aircraft weight and multiply by the inches the CG must transit. Then, divide the result by the third known factor, which is the distance between the stations.
4800 x 2 = 9600.
9600 ÷ 85 = 112.94.
About 113 lbs. of luggage must move to the forward station to move the CG forward two (2) inches. Moving all 150 lbs to the forward station will move the CG even closer to the datum.
| Cargo (weight to be shifted)
112.94 lbs. |
Δ CG
2 inches |
|
| x ↗ | ÷ ↓ | |
|---|---|---|
| Aircraft (total weight)
4,800 lbs. |
Δ Distance (that weight is shifted)
85 inches |
Confirm that both quotients are equal to prove the result:
4800 ÷ 112.94 = 42.5
90 ÷ 2.25 = 42.5
If the weight of the luggage that is to be moved is known, by what distance should be relocated?
In this case, let's again use 150 pounds of luggage. The aircraft still weighs 4,800 lbs, and the CG is two (2) inches beyond the aft limit. Again, we will multiply the total weight by the Δ CG and divide the result by the third known factor — this time, the weight of the luggage.
4800 x 2 = 9600.
9600 ÷ 150 = 64
The luggage must move 64 inches forward to move the CG forward two (2) inches.
| Cargo (weight to be shifted)
150 lbs. |
← ÷ | Δ CG
|
|---|---|---|
| x ↗ | ||
| Aircraft (total weight)
4,800 lbs. |
Δ Distance (that weight is shifted)
64 inches |
Confirm that both quotients are equal to prove the result:
4800 ÷ 150 = 32
64 ÷ 2 = 32
If the weight and transit of the cargo is known, how much will the new loading shift the CG?
Let's move 120 pounds of luggage to a station 90 inches closer to the datum. The aircraft still weighs 4,800 lbs. For this problem, we will multiply the weight shifted by the inches of transit, and then divide the result by the aircraft's weight:
120 x 90 = 10800
10800 ÷ 4800 = 2.25
The aircraft's CG will move 2.25 inches.
| Cargo (weight to be shifted)
120 lbs. |
Δ CG
2.25 |
|
| x ↘ | ||
|---|---|---|
| Aircraft (total weight)
4,800 lbs. |
← ÷ | Δ Distance (that weight is shifted)
90 inches |
Confirm that both quotients are equal to prove the result:
4800 ÷ 120 = 40
90 ÷ 2.25 = 40
While this may not seem dead-simple at first, note that each formula begins with either the numerator or denominator of the weight ratio multiplied by the the cross-factor of the distance ratio (if using the denominator, multiply by the numerator). The result is then divided by the final known factor. Which is to say:
- Make a grid
- Start on the left
- Multiply diagonally
- Divide
"Divide Twice": An alternate pattern is to divide the known ratio to get a positive integer, and then divide that result by the remaining known factor. It's just as reliable, and for some pilots it might be easier to remember. However, the FAA doesn't present this pattern in their texts.
- Make a grid
- Divide the known ratio
- Use this result to divide again
Weight-Add & Weight-Change Calculations
Note: Consider using the computational method for both weight-add calculations. See next section.
To calculate how additional weight affects the CG, we have to adjust the formula slightly. In this case, the "distance" factor is the distance between the the aft limit of the CG range and the station where weight is added. Instead of using this value to represent a transit, the value now represents a range.
The weight of the aircraft is the loaded weight before the additional cargo is added. (A test question won't provide the aircraft weight as an unknown variable.)
| Cargo (weight to be added) | Δ CG (to reach limit) | |
| ÷ | = | ÷ |
|---|---|---|
| Aircraft (total weight) | Δ Distance (between station and CG limit) |
If luggage were added to a compartment with a station in front of the CG, then we'd compare the new station to the CG's fore limit. Baggage compartments in the nose section are typically found on large turboprop aircraft.
What is the maximum weight that can be added at a given station while keeping the CG within limits?
In a 2,800-pound loaded aircraft, let's say a passenger wants to add luggage to a station 112 inches from the datum. The current CG is 80.5, and the aft CG limit is 82, which permits 1.5 inches of aft CG transit. The station is 30 inches from the CG limit. How much weight can be added at the station?
2800 x 1.5 = 4200
4200 ÷ 30 = 140
| Cargo (weight to be added)
140 lbs. |
Δ CG (maximum)
1.5 |
|
| x ↗ | ↓ ÷ | |
|---|---|---|
| Aircraft (total weight)
2,800 lbs. |
Δ Distance (between station and CG limit)
30 inches |
Any weight beyond 140 pounds at the 112-inch station will place the CG beyond the aft limit.
Confirm that both quotients are equal to prove the result:
2800 ÷ 140 = 20
30 ÷ 1.5 = 20
How will a change in known weight at a given station affect the CG? Will the CG remain in limits?
In our 2,800-pound loaded aircraft, let's say a passenger wants to add 290 pounds of luggage to a station 108 inches from the datum. The current CG is 80.5, and the aft CG limit is 82, which permits 1.5 inches of aft CG transit. The station is 26 inches from the CG limit. How far will the CG shift when the new cargo is added to the station?
290 x 26 = 7540
7540 ÷ 2800 = 2.69
| Cargo (weight to be added)
290 lbs. |
Δ CG
2.69 |
|
| x ↘ | ||
|---|---|---|
| Aircraft (total weight)
2,800 lbs. |
← ÷ | Δ Distance (between station and CG limit)
26 inches |
290 pounds of additional weight at the 108-inch station will shift the CG 2.69 inches, to a new location of 83.19 inches. This is greater than the aft limit of 82 inches (only 1.5 inches exist between the current CG and the aft limit). Thus, the new CG is invalid.
Confirm that both quotients are equal to prove the result:
2800 ÷ 290 = 9.655
26 ÷ 2.69 = 9.66
Weight-Add & Weight-Remove Calculations: The Computational Method
Problems that address adding or removing weight can be solved using the computational method, which is the schedule-based formula used for weight-and-balance calculations. While this works for both adding and removing weight, this is the preferred method for determining a new CG when weight is removed.
Because the algebraic method works for adding and and changing weight, it might be tempting to use it for weight-removal questions. However, when using negative values in the equation, the result will be incorrect — and the incorrect answer can be presented on the test as a distractor.
The computational method method is simple, because the test question wil be providing a new weight and station (arm), which becomes a new line-item. The only other line item is the aircraft's current weight and CG (total arm).
Let's revisit the previous question, but now with the computational method:
How will a change in known weight at a given station affect the CG? Will the CG remain in limits?
In our 2,800-pound loaded aircraft, let's say a passenger wants to add 290 pounds of luggage to a station 108 inches from the datum. The current CG is 80.5, and the aft CG limit is 82, which permits 1.5 inches of aft CG transit. The station is 26 inches from the CG limit. How far will the CG shift when the new cargo is added to the station?
| Item | Weight | Arm | Moment | ||
|---|---|---|---|---|---|
| Aircraft Loaded Weight | 2,800 | x | 80.5 | = | 225,400 |
| Δ Weight | 290 | x | 108 | = | 31320 |
| Total | 3,090 | 256720 | |||
| Calculate CG: | 256200 ÷ 3,080 = 83.08 | ||||
The new CG is 83.08, which exceeds the aft limit of 82. Thus, the new CG is invalid.
Note for math nerds: 83.08 - 80.5 = 2.58 inches of CG transit. This represents a 0.11-inch differential from the algebraic method. When the aircraft weight in that forumula is increased to 3,090 pounds, the result is 2.44, which is a 0.25-inch differential (with a new CG of 82.94). Therefore, the results do not converge when the aircraft weight is increased in the algebraic method. Also, FAA test questions suggest using the "old" aircraft weight for the equations and not increasing the weight.
How will removing a known weight at a given station affect the CG? And will the CG remain in limits?
What is the location of the CG if 146 pounds are removed from Station 150?
Total weight: 7,152 lb
CG location: Station 82
When using the computational method for removing weight, use a negative value at for the Δ Weight. The corresponding moment also will be a negative value, and it will be deducted from the sum of all moments.
Answer: 80.58
| Item | Weight | Arm | Moment | ||
|---|---|---|---|---|---|
| Aircraft Loaded Weight | 7,152 | x | 82 | = | 586,464 |
| Δ Weight | −146 | x | 150 | = | −21,900 |
| Total | 7,006 | 564,564 | |||
| Calculate CG: | 564,564 ÷ 7,006 = 80.58 | ||||
Terms and Definitions
- Arm (moment arm): The horizontal distance in inches from the reference datum line to the CG of an item. The algebraic sign is plus (+) if measured aft of the datum and minus (–) if measured forward of the datum.
- Basic empty weight (GAMA): The standard empty weight plus the weight of optional and special equipment that have been installed.
- Center of gravity (CG): The point about which an aircraft would balance if it were possible to suspend it at that point. It is the mass center of the aircraft or the theoretical point at which the entire weight of the aircraft is assumed to be concentrated. It may be expressed in inches from the reference datum or in percent of MAC. The CG is a three-dimensional point with longitudinal, lateral, and vertical positioning in the aircraft.
- CG limits: The specified forward and aft points within which the CG must be located during flight. These limits are indicated on pertinent aircraft specifications.
- CG range: The distance between the forward and aft CG limits indicated on pertinent aircraft specifications.
- Datum (reference datum): An imaginary vertical plane or line from which all measurements of arm are taken. The datum is established by the manufacturer. Once the datum has been selected, all moment arms
and the location of CG range are measured from this point.
- Delta: A Greek letter expressed by the symbol Δ to indicate a change of values. As an example, ""ΔCG" indicates a change (or movement) of the CG.
- Floor load limit: The maximum weight the floor can sustain per square inch/foot as provided by the manufacturer.
- Fuel load: The expendable part of the load of the aircraft. It includes only usable fuel, not fuel required to fill the lines or that which remains trapped in the tank sumps.
- Licensed empty weight: The empty weight that consists of the airframe, engine(s), unusable fuel, and undrainable oil plus standard and optional equipment as specified in the equipment list. Some manufacturers
used this term prior to GAMA standardization.
- Licensed empty weight: the empty weight that consists of the airframe, engine(s), unusable fuel, and undrainable oil plus standard and optional equipment as specified in the equipment list. Some manufacturers
used this term prior to GAMA standardization.
- Maximum ramp weight: The total weight of a loaded aircraft including all fuel. It is greater than the takeoff weight due to the fuel that will be burned during the taxi and run-up operations. Ramp weight may also be referred to as taxi weight.
- Maximum takeoff weight: The maximum allowable weight for takeoff.
- Maximum weight: The maximum authorized weight of the aircraft and all of its equipment as specified in the TCDS for the aircraft.
- Maximum zero fuel weight (GAMA): The maximum weight, exclusive of usable fuel.
- Mean aerodynamic chord (MAC): The average distance from the leading edge to the trailing edge of the wing.
- Moment: The product of the weight of an item multiplied by its arm. Moments are expressed in pound-inches (in-lb). Total moment is the weight of the airplane multiplied by the distance between the datum and the CG.
- Moment index (or index): A moment divided by a constant such as 100, 1,000, or 10,000. The purpose of using a moment index is to simplify weight and balance computations of aircraft where heavy items and long arms result in large, unmanageable numbers.
- Payload (GAMA): The weight of occupants, cargo, and baggage.
- Standard empty weight (GAMA): Aircraft weight that consists of the airframe, engines, and all items of operating equipment that have fixed locations and are permanently installed in the aircraft, including fixed ballast, hydraulic fluid, unusable fuel, and full engine oil.
- Standard weights: Established weights for numerous items involved in weight and balance computations. These weights should not be used if actual weights are available. Some of the standard weights are:
- Gasoline: 6 lb/US gal
- Jet A, Jet A-1: 6.8 lb/US gal
- Jet B: 6.5 lb/US gal
- Oil: 7.5 lb/US gal
- Water: 8.35 lb/US gal.
- Station: a location in the aircraft that is identified by a number designating its distance in inches from the datum. The datum is, therefore, identified as station zero. An item located at station +50 would have an arm of 50 inches.
- Useful load: the weight of the pilot, copilot, passengers, baggage, usable fuel, and drainable oil. It is the basic empty weight subtracted from the maximum allowable gross weight. This term applies to general aviation (GA) aircraft only.
For additional information on weight, balance, CG, and aircraft stability refer to the FAA handbook appropriate to the specific aircraft category.
Commercial Pilot & Flight Instructor Test Questions
The center of gravity of an aircraft is computed along the longitudinal axis.
The center of gravity of an aircraft can be determined by dividing the total moment by the total weight.
With respect to using the weight information given in a typical aircraft owner's manual for computing gross weight, it is important to know that if items have been installed ni the aircraft in addition to the original equipment, the allowable useful load is decreased.
If the nosewheel of an airplane moves aft during gear retraction, how would this aft movement affect the CG location of that airplane? It would cause the CG location to move aft.
If the landing gear on an airplane moves forward during retraction, the total moment will decrease.
— The forward movement along the longitudinal axis decreases the landing gear's arm, and thus its moment, and thus the aircraft's total moment.
How does increased weight affect the takeoff distance of the airplane? The airplane will accelerate more slowly with the same power output, and a higher airspeed is required to generate necessary lift for takeoff.
To maintain level flight in an airplane which is loaded with the CG at the forward limit, an additional download must be imposed on the horizontal stabilizer. This in turn produces an additional load which the wing must support.
An aircraft is loaded with the CG aft of the aft limit. What effect will this have on controllability? Stall and spin recovery may be impossible.
An aircraft is loaded with the CG at the aft limit. What are the performance characteristics compared with the CG at the forward limit? The aft CG provides the lowest stalling speed, the highest cruising speed, and least stability.
— The CG at the aft limit relieves the tail surfaces from imposing the normal down-loading force. The wing is then relieved of load. The wing flies at a lower angle of attack, reducing stall speed and drag while increasing cruising speed. However, longitudinal stability is decreased.
As the CG moves aft, the aircraft becomes less stable and less controllable.
As the CG location is changed, recovery from a stall becomes progressively more difficult as the CG moves rearward.
What is the effect of center of gravity on the spin characteristics of a fixed-wing aircraft? If the CG is too far aft, a flat spin may develop.
The stalling speed of an aircraft will be highest when the aircraft is loaded with a high gross weight and forward CG.
— The horizontal tail surfaces must provide increased downward force to counter a forward CG, which increases the load imposed on the wing. Stall speed is increased.
If the CG of an aircraft is moved from the aft limit to beyond the forward limit, how will it affect the cruising and stalling speed? Decrease in the cruising speed and increase in the stalling speed.
— Forward CG increases load factor, due to the increased down-force required from the horizontal tail surfaces. Crusising flight will require a higher angle of attack than normal, which will reduce the cruising speed.
When an aircraft's forward CG limit is exceeded, it will affect the flight characteristics of the aircraft by producing higher stalling speeds and more longitudinal stability.
— An advantage of forward CG is greater longitudinal stability. As airspeed decreases, the nose will tend to drop.
What is a characteristic of the indicated airspeed if the CG is at the most forward allowable position and constant power and altitude are maintained? The indicated airspeed will be less than it would be with the CG in the most rearward allowable position.
— With the CG at the most-forward position, the wing will fly at a higher angle of attack and generate more induced drag. Indicated airspeed will be lower, compared to the same airplane with the CG at the maximum aft position.
Under which condition is a forward CG most critical? On landing.
— The landing flare requires enough elevator authority to raise the nosewheel.
Calcuations
Balance Computations
"If 50 lbs. of weight is located at Station X and 100 pounds at Station Z, how much weight must be at Station Y to balance the board?"
After quick calculations, the question provides two variables. The Weight X moment is 2,500 (50 x 50), and the Weight Z moment is 10,000 (100 x 100). Therefore, the left side is missing 7,500 inch-pounds of required moment. Weight Y is 25 inches left of the datum, and it must provide 7,500 inch-pounds of moment to balance the board.
Thus, 7,500 inch-pounds ÷ 25 inches = 300 lbs.
"If Weight A is 1,000 lbs. and Weight B is 500 lbs. how should Weight A be relocated to balance the board?"
Weight A's moment is 50,000 inch-pounds, while Weight B's moment is 12,500 inch-pounds. Thus, the left side of the board has an excess moment of 37,500 inch-pounds.
Only Weight A can be moved. Its weight cannot be changed, and its moment must equal 12,500 inch-pounds. Therefore, it must be 12.5 inches from the fulcrum (12,500 ÷ 1,000).
Because Weight A currently is 50 inches from the fulcrum, it must shift 37.5 inches to the right (50 – 12.5).
"If the three weights are 10 pounds each, how must weight C be shifted to balance the plank on the center of gravity?"
This question is potentially confusing because the graphic isn't presented as weights on either side of a fulcrum, but instead weights with arms measured from a datum.
- The fulcrum (CG) is 72 inches from the datum.
- Weight A overlaps the datum, but by how much isn't specified.
- Weight B is 80 from the datum.
- Weight C is 100 from the datum.
While Weight A isn't centered on the datum, the test question expects the math to determine that the moment for Weight A is 720 inch-pounds (72 x 10).
Weight B overlaps the CG, but the math expects that we will calculate its position as 8 inches to the right of the CG, with a moment of 80 inch-pounds.
Weight C is the least ambiguous of the three weights. It's 28 to the right of the CG, with a moment of 280 inch-pounds.
Therefore, the left-side moment is 720 inch-pounds and the sum of the right-side moments is 360 inch-pounds. The right side is missing 360 inch-pounds of required moment for balance, and if Weight C weighs 10 lbs., then it must move another 36 inches to the right.
While the math for this test question is simple, the graphic suggests additional complexity that test-takers must disregard.
"Based on this information, the CG would be located how far aft of datum?
- Weight X: 120 lb at 16" aft of datum
- Weight Y: 112 lb at 109" aft of datum
- Weight Z: 65 lb at 215" aft of datum"
Weight X has a moment of 1,920 inch-pounds. Weight Y has a moment of 12,208 inch-pounds. Weight Z has a moment of 13,975 inch-pounds. The sum of all moments is 28,103 inch-pounds. The sum of all weights is 297 lbs.
Thus, 28,103 ÷ 297 = 94.6 inches aft of the datum.
Shifting weight
Could 100 pounds of weight be shifted from Station 30.0 to Station 120.0 without exceeding the aft CG limit?
Total weight: 4,750 lb
CG location: Station 115.8
Aft CG limit: Station 118.0
Answer: Yes; the new CG would be located at Station 117.69.
Use the formula to find the new CG and the compare it to the published limit. The permitted Δ CG is 2.2 (118.0 - 115.8). The calculated differential is 1.89, which is within the permitted CG range. The answer is "Yes." (100 x 900 = 9,000; 9,000 ÷ 4,750 = 1.89. Or, 4,750 ÷ 100 = 47.5; 90 ÷ 47.5 = 1.89.)
Proof:
4,750 ÷ 100 = 47.5
90 ÷ 1.89 = 47.6
| 100 | X |
| 4,750 | 90 |
Adding weight
How much weight could be added at Station 120 without exceeding the aft CG limit?
TotalÊweight: 9,500 lb
CG location: Station 90.0
Aft CG limit: Station 90.5
Answer: 161.0 lb.
Use the formula to find the maximum weight permitted by the maximum CG shift. The CG cannot move more than 0.5 inches aft (90.5 - 90.0), so this is the Δ CG. The distance range is 29.5, since add/remove questions require factoring the distance between the CG limit and the variable station (120 - 90.5). The permitted weight is 161. (9,500 x 0.5 = 4,750; 4,750 ÷ 29.5 = 161. Or, 29.5 ÷ 0.5 = 59; 9,500 ÷ 59 = 161.)
Proof:
9,500 ÷ 161 = 59
29.5 ÷ 0.5 = 59
| X | 0.5 |
| 9,500 | 29.5 |
Removing weight
What is the location of the CG if 60 pounds are removed from Station 70?
Total weight: 8,420 lb
CG location: Station 85
Answer: 85.1
The solution for removing weight can be found via the schedule-based computational method. Multiply the aircraft's weight and CG (8,420 x 50 = 715,700). Since weight is being removed, multiply the negative value by the station (−60 x 70 = −4,200). Factor the new weight (8,420 − 60 = 8,360) and the new moment (715,700 − 4,200 = 711,500). Finally, divide the results to determine the resultant CG (711,500 ÷ 8,360 = 85.1).
| Item | Weight | Arm | Moment | ||
|---|---|---|---|---|---|
| Aircraft Loaded Weight | 8,420 | x | 50 | = | 715,700 |
| Δ Weight | −60 | x | 70 | = | −4,200 |
| Total | 8,360 | 711,500 | |||
| Calculate CG: | 711,500 ÷ 8,360 = 85.1 | ||||