the learning never stops!

# Flight Engineering

Weight and balance is a very big deal on some aircraft and it is possible to load fuel, cargo, or people so the aircraft simply will not fly. See: CL-600 N370V for an example. On other aircraft, such as our oft-used example Gulfstream G450, aircraft design makes it harder to find the airplane un-flyable but it still needs to be considered. See: Weight and Balance.

In either case, you should have complete understanding of the concepts and a very good idea of where your airplane's range of acceptable centers of gravity lie. Why? An aircraft with a conventional tail-mounted horizontal stabilizer uses that stabilizer to exert a downward force. The aircraft's "center of lift" exerts an upward force, so the center of gravity (CG) must at all times be forward of the center of lift to keep the aircraft in stable flight. There exists a range of CG positions which the horizontal stabilizer is capable of countering. If the CG is too far either way, the aircraft cannot be controlled in pitch.

Figure: G450 CG implications, from Eddie's notes.

### Weight and Balance 101 — The Beam and Fulcrum

Adapted from Air Force Manual 51-9, chapter 7.

Figure: Balance beam, from Eddie's notes.

The Air Force teaches its flight engineers weight and balance starting with a beam on a fulcrum in perfect balance. If you place the exact center point of the beam over the fulcrum, the beam should be balanced.

Figure: Balance beam with weights, from Eddie's notes.

If you add equal weights to each side, the beam will remain in balance as long as the distance from the center for both weights is the same.

Figure: Lever law, from Eddie's notes.

[Air Force Manual 51-9, ¶7-3.b.] The lever law says the product on one weight (W1) multiplied by its distance (D1) from the fulcrum is equal to the product of the other weight (W2) multiplied by its distance (D2) from the fulcrum. Each side of the beam acts as a lever, producing torque about the fulcrum equal to the other.

The Weight X Distance is indeed the torque about the fulcrum, but to a pilot it more accurately describes the moment of the weight.

Figure: Balance beam with varied weights, from Eddie's notes.

The mathematical relationship true even when using different weights. For example, to balance the beam, a 6-pound weight has been place 5 inches from the fulcrum, and a 3-pound weight, 10 inches from the fulcrum. Now, if the law of the lever is applied, we have:

${W}_{1}{D}_{1}={W}_{2}{D}_{2}$

$6\mathrm{lb}x5\mathrm{in}=3\mathrm{lb}x10\mathrm{in}$

$30\mathrm{in/lb}=30\mathrm{in/lb}$

Since the same torque -- 30 inch pounds -- applies to each side of the beam, the beam balances.

### Center of Gravity (CG) Calculation

Figure: CG of beam without a fulcrum, from Eddie's notes.

Adapted from FAA-H-8083-1A, page 2-2.

What if we don't have a fulcrum and need to determine where the center of gravity is? We need to know: the distance of each object from a common point, called the datum, the weight of each object, and the weight and length of the object itself. Then it is a matter of some math. For example:

• We have a 180" beam that weighs 50 lbs.
• We use 10" forward of the beam as our arbitrary datum line. This means the center of our beam is at (180 / 2) + 10 = 100".
• We place two 100-lb weights on the beam, one is 50" from the datum line, the other is 90" from the datum line.
• We place a 200-lb weight on the beam, 150" from the datum line.

Where along the beam is the CG?

To solve this problem:

1. Determine the moment of each object (Moment = Weight X Arm)
2. Add all the weights
3. Add all the moments
4. Divide the total moment by the total weight to find the CG

We typically do this with a table constructed thusly:

 Item Weight X Arm = Moment CG 1st Weight 100 X 50 = 5,000 2nd Weight 100 X 90 = 9,000 3rd Weight 200 X 150 = 30,000 Beam 50 X 100 = 5,000

Adding all the weights and moments:

 Item Weight X Arm = Moment CG 1st Weight 100 X 50 = 5,000 2nd Weight 100 X 90 = 9,000 3rd Weight 200 X 150 = 30,000 Beam 50 X 100 = 5,000 450 49,000

Dividing the total moment by the total weight:

 Item Weight X Arm = Moment CG 1st Weight 100 X 50 = 5,000 2nd Weight 100 X 90 = 9,000 3rd Weight 200 X 150 = 30,000 Beam 50 X 100 = 5,000 450 49,000 108.89

Based on this information, we know placing our fulcrum at 108.89" will allow this beam to perfectly balance. Note that each weight is considered to impact the beam at the center of the weight. For long weights you need to consider where the center of gravity of the object itself lies to determine its individual arm.

The beam itself is analogous to an airplane, except the airplane is not a uniform structure so you cannot simply assume its arm lies at the halfway point, as we did with the beam.

Another analogous issue is the reference datum. We normally place this fictitious line ahead of anything on the airplane that is subject to changes. Doing so eliminates negative arms and simplifies the math.

### Reference Datum

Figure: G450 fuselage coordinate system, from G450 Maintenance Manual, §06-10-00, figure 1.

[FAA-H-8083-1A, page 2-1]

• The manufacturer establishes the maximum weight and range allowed for the CG, as measured in inches from the reference plane called the datum. Some manufacturers specify this range as measured in percentage of the mean aerodynamic chord (MAC), the leading edge of which is located a specified distance from the datum.
• The datum may be located anywhere the manufacturer chooses; it is often the leading edge of the wing or some specific distance from an easily identified location. One popular location for the datum is a specified distance forward of the aircraft, measured in inches from some point, such as the nose of the aircraft, or the leading edge of the wing, or the engine fire wall.

#### Gulfstream G450 Example

[G450 Maintenance Manual, §06-10-00, ¶2.A.] FS represent edges of vertical planes perpendicular to the horizontal reference plane in side view and perpendicular to the vertical reference plane in plan view. FS dimension and location points in the FCS from FS 0, begin 4 inches aft of the forward tip of the nose radome. FS show measurement of length along the longitudinal axis (X).

[G450 Weight and Balance Manual, page 16.] Horizontal datum: FS 72.

From a pilot's perspective the use of a reference datum line that is effectively +4 -72 = 68" aft of the forward tip of the nose radome is inconsequential unless there is a change to something in the radome. If that were to happen the aircraft's empty center of gravity may have to be adjusted.

### Horizontal Arms

Figure: G450 weight and balance, from Eddie's notes.

Everything on the airplane is weighed and the effective center of that weight measured against the Reference Datum Line. If, for example, the center of a passenger seat is located 400" from the reference datum line, an occupant of that seat will be computed as being entirely at 400".

[G450 Weight and Balance Manual, ¶2.1.2.5] Horizontal Arm, measured in inches, is used for the location of components and complete aircraft weight and center of gravity (C.G.) purposes.

Figure: Mean Aerodynamic Chord, from FAA-H-8083-1A, figure 3-9.

[FAA-H-8083-1A, page 3-6] The aircraft mechanic or repairman is primarily concerned with the location of the CG relative to the datum, an identifiable physical location from which measurements can be made. But because the physical chord of a wing that does not have a strictly rectangular plan form is difficult to measure, wings such as tapered wings express the allowable CG range in percentage of mean aerodynamic chord (MAC). The allowable CG range is expressed in percentages of the MAC. The MAC, as seen in Figure 3- 9, is the chord of an imaginary airfoil that has all of the aerodynamic characteristics of the actual airfoil. It can also be thought of as the chord drawn through the geographic center of the plan area of the wing.

The center of gravity, for most aircraft, lies somewhere between LEMAC and TEMAC so you can convert the position of the CG from the reference datum to a percentage of the mean aerodynamic chord using the following formula:

$\mathrm{C.G. in Percent MAC}=\frac{\mathrm{Horizontal Arm}-\mathrm{LEMAC}}{\mathrm{MAC}}$

In the case of our example Gulfstream:

[G450 Weight and Balance Manual, ¶2.1.2.7] Mean Aerodynamic Chord MAC). The location of the airplane center of gravity as a percentage of MAC is an alternative to Horizontal Arm.

$\mathrm{C.G. in Percent MAC}=\frac{\mathrm{Horizontal Arm}-387.7}{1.6622}$

### Moment

[FAA-H-8083-1A, page 2-1] A moment is a force that tries to cause rotation, and is the product of the arm, in inches, and the weight, in pounds. Moments are generally expressed in pound-inches (lb-in) and may be either positive or negative.

The larger the aircraft in terms of inches nose-to-tail or weight in terms of pounds or hundreds of thousands of pounds, the moment calculation can become cumbersome. A Boeing 747, for example, could have a center of gravity 960" behind the reference datum line and a weight of 800,000 pounds. The aircraft's total moment would in this case be 768,000,000 lb-in. Boeing recommends all moments be divided by 100,000 to get rid of all those zeroes. This reduces the chance of error and makes calculations possible on most electronic calculators. The 100,000 factor is by no means universal. You should consult your aircraft manuals. For example . . .

[G450 Weight and Balance Manual, ¶2.1.2.6] Moment (10,000 (lb*in)), to avoid large moment numbers in balance calculations, moments (Weight X Horizontal Arm) can be divided by 10,000. Moment is given in pound-inch units.

### Aircraft Basic Weight and CG

Manufacturers have some leeway in how they determine the aircraft's empty weight and center of gravity. You should know what goes into the numbers they provide you. In our example G450, Gulfstream weighed the airplane without most of the operator provided equipment. "If you can carry it off the airplane without getting your tools," is how we think about it, "it isn't in the empty weight." The operator is provided this information on what Gulfstream calls a Weight and Balance Form A:

Figure: G450 weight and balance "Form A," from Eddie's notes.

The example aircraft was weighed at each landing gear, the weights were multiplied by their respective arms to come up with moments. The moments were added (19,520,419) and divided by the total weight (42,202) to come up with a CG of 465.5 inches behind the datum line. Since Gulfstream wants the empty weight to include unusable fuel, that is added. The published empty weight becomes 42,294 lbs with a moment arm of 19,560,163 inch-pounds. After division we see that the aircraft's empty CG is 462.5 inches behind the reference datum, or 45.0% MAC.

### Aircraft Operating Empty Weight and CG

Figure: G450 weight and balance empty, basic, and operating, from Eddie's notes.

Of course we cannot fly without pilots, so they are added on the next form (using an average weight) to give us our Basic Weight of 42,740 lbs with a new total moment and resulting CG.

Next we add everything else, except fuel, to determine our Operating Weight Empty, also known as the "Basic Operating Weight" or BOW.

### Aircraft Ramp Weight, CG, Trim Setting

Figure: Example weight and balance form, from Eddie's notes.

Procedures will vary from aircraft manufacturer but in the case of our example Gulfstream G450, we can add personnel, baggage, and fuel using a spreadsheet, slide rule, or computer application. In this case we will use the calculator available here: G450 Weight and Balance.

The calculator assumes we will divide each moment by 1,000. It tells us that with the two passengers seated as shown, 400 lbs of baggage, and 18,500 lbs of fuel, the aircraft will weigh 44,699 lbs and will require a takeoff trim setting of 4 units. The G450 weight and balance limitation asks for zero fuel CG, which is 42.8% of MAC.

### CG Limits

Figure: G450 Zero fuel gross weight CG envelope, from G450 Weight and Balance Manual, ¶2.4.

[14 CFR 25, §25.23(a)] Ranges of weights and centers of gravity within which the airplane may be safely operated must be established. If a weight and center of gravity combination is allowable only within certain load distribution limits (such as spanwise) that could be inadvertently exceeded, these limits and the corresponding weight and center of gravity combinations must be established.

It is up to the manufacturer how the ranges of weights and centers of gravity are presented. In the case of the example G450, Gulfstream presents a chart with both ranges displayed without fuel. They further stipulate that if the aircraft is within limits with no fuel, it will always be within limits with fuel. (This is uncommon in the world of jets.) In our example, the aircraft's zero fuel weight is 44,699 lbs and the center of gravity is at 42.8% MAC, which is well inside the chart's borders so our aircraft is within its weight and balance envelope.

### References

14 CFR 25, Title 14: Aeronautics and Space, Airworthiness Standards: Transport Category Airplanes, Federal Aviation Administration, Department of Transportation

Air Force Manual (AFM) 51-9, Aircraft Performance, 7 September 1990

FAA-H-8083-1A, Aircraft Weight and Balance Handbook, U.S. Department of Transportation, Flight Standards Service, 2007

Gulfstream G450 Maintenance Manual, Revision 18, Dec 12, 2013

Gulfstream G450 Weight and Balance Manual, Revision 3, March 2008

Top