A&P Weight and Balance Practice: 3 Worked CG Problems
Last updated October 5, 2026
To calculate a center of gravity (CG) for the A&P General test, multiply each weight by its arm to get a moment, add up the weights and the moments, then divide the total moment by the total weight. That one rule solves empty weight CG, equipment changes and adverse-loading checks (Aircraft Weight and Balance Handbook, FAA-H-8083-1B, ch. 2, pp. 2-2 to 2-3). Below are three original practice problems, worked in tables, plus common mistakes to avoid.
How much weight and balance is on the test
Weight and Balance is subject C of the General test. The FAA blueprint gives it 5 to 10 percent of the 60 questions, which is 3 to 6 questions (Companion Guide to the mechanic ACS, FAA-G-ACS-1, Change 2, ch. 3, p. 13). The Aviation Mechanic ACS, FAA-S-ACS-1 lists what you must know, including (p. 4):
- AM.I.C.K3: weighing procedures and preparations.
- AM.I.C.K4: calculating arm, positive or negative moment, CG, or moment index.
- AM.I.C.K7: adverse loading, and how to calculate whether it causes an out-of-limit condition.
- AM.I.C.K8: the proper empty weight configuration.
Weight and balance is chapter 6 of the Aviation Maintenance Technician Handbook - General, FAA-H-8083-30B. For definitions and more worked examples, use the Aircraft Weight and Balance Handbook, FAA-H-8083-1B. As of October 2026, the FAA treats its October 20, 2025 MOSAIC addendum as part of the current edition (addendum, p. 1). It rewrites chapter 4 and a few sentences in chapters 2 and 7, but no formula used below.
The terms, as the handbook defines them
All definitions are from the Weight and Balance Handbook glossary (FAA-H-8083-1B, pp. G-1 to G-5) unless another page is given.
| Term | Definition |
|---|---|
| Datum | An imaginary vertical plane or line from which all measurements of arms are taken. The manufacturer sets it (p. G-2). |
| Arm | The horizontal distance from the datum to the CG of an item. Plus (+) if aft of the datum, minus (-) if forward of it (p. G-1). |
| Moment | The weight of an item multiplied by its arm (p. G-4), usually in pound-inches, lb-in (ch. 2, p. 2-2). |
| Center of gravity (CG) | The point at which the airplane would balance if suspended. Its distance from the datum is the total moment divided by the total weight (p. G-2). |
| Empty weight | The airframe, engines, all permanently installed equipment and unusable fuel. Undrainable oil or full oil is included, depending on the rules the aircraft was certificated under (p. G-2). |
| Useful load | The difference between takeoff weight (or ramp weight, if applicable) and basic empty weight (p. G-5). |
| Percent MAC | The distance in inches of the CG from LEMAC (the leading edge of the mean aerodynamic chord) divided by the MAC (p. G-4). Multiply by 100 to get a percentage (ch. 3, p. 3-9). |
| Tare weight | The weight of chocks or other devices that hold the aircraft on the scales. Subtract it from the scale reading to get net weight (p. G-5). |
The aircraft in all three problems below is made up. Its datum is the wing leading edge, so some arms are negative.
Problem 1: Empty weight CG from three scale readings
A tricycle-gear airplane is weighed in its empty weight condition, level, with chocks on every scale. The nose wheel weighing point is 41.5 inches forward of the datum and both main wheel weighing points are 22.0 inches aft of it. Scale readings: nose 368 lb, left main 651 lb, right main 657 lb. Tare: 12 lb of chocks on the nose scale and 6 lb on each main scale. Find the empty weight and the empty weight CG.
Subtract the tare first, then multiply each net weight by its arm (ch. 3, p. 3-6).
| Weighing point | Scale reading (lb) | Tare (lb) | Net weight (lb) | Arm (in) | Moment (lb-in) |
|---|---|---|---|---|---|
| Nose | 368 | 12 | 356 | -41.5 | 356 x -41.5 = -14,774.0 |
| Left main | 651 | 6 | 645 | +22.0 | 645 x 22.0 = +14,190.0 |
| Right main | 657 | 6 | 651 | +22.0 | 651 x 22.0 = +14,322.0 |
| Total | 1,652 | +13,738.0 |
CG = total moment / total weight = 13,738.0 / 1,652 = +8.32 inches aft of the datum (8.316 before rounding).
Check it a second way. The handbook’s EWCG formulas start from the main wheels: they find the moment of the nose (or tail) wheel, divide it by the total weight, and add or subtract the result from D, the distance between the datum and the main wheels (ch. 3, p. 3-6). For a nosewheel airplane: CG = D - (F x L) / W, where F is the net nose weight, L is the distance between the main and nose weighing points, and W is the total net weight (Figure 3-9). Here L = 41.5 + 22.0 = 63.5 inches, so F x L = 356 x 63.5 = 22,606 and 22,606 / 1,652 = 13.68. Then 22.0 - 13.68 = +8.32 inches. Same answer, because the CG does not depend on where the datum is, as long as every measurement comes from the same one (ch. 3, p. 3-7).
As percent MAC. Suppose this airplane has a rectangular wing with a 60-inch chord, and its leading edge is the datum, so LEMAC is station 0. For a rectangular wing of constant airfoil section, MAC is just the chord (ch. 2, p. 2-6). Percent MAC = (CG - LEMAC) x 100 / MAC (ch. 3, p. 3-9) = 8.316 x 100 / 60 = 13.9 percent MAC.
Problem 2: New CG after removing and adding equipment
Start from the airplane in Problem 1 (1,652 lb, moment +13,738.0 lb-in). A mechanic removes a 7.6 lb radio at -12.0 inches and a 4.4 lb ELT at +128.0 inches. They install a 4.8 lb radio at -12.0 inches, a 2.9 lb ELT at +128.0 inches and a 3.5 lb engine monitor at -15.0 inches. Find the new empty weight and empty weight CG.
Removed items get a negative weight, installed items a positive one (ch. 7, p. 7-5, Figure 7-4). Let the signs do the work.
| Item | Weight (lb) | Arm (in) | Moment (lb-in) |
|---|---|---|---|
| Airplane before the change | 1,652.0 | +8.32 | +13,738.0 |
| Radio removed | -7.6 | -12.0 | -7.6 x -12.0 = +91.2 |
| ELT removed | -4.4 | +128.0 | -4.4 x 128.0 = -563.2 |
| Radio installed | +4.8 | -12.0 | 4.8 x -12.0 = -57.6 |
| ELT installed | +2.9 | +128.0 | 2.9 x 128.0 = +371.2 |
| Engine monitor installed | +3.5 | -15.0 | 3.5 x -15.0 = -52.5 |
| New totals | 1,651.2 | +13,527.1 |
New empty weight CG = 13,527.1 / 1,651.2 = +8.19 inches aft of the datum.
Check it. The changes alone add up to -0.8 lb and -210.9 lb-in. Then 1,652.0 - 0.8 = 1,651.2 and 13,738.0 - 210.9 = 13,527.1. Same totals.
Removing weight forward of the datum gives a positive moment, because a negative weight times a negative arm is positive (AMT General handbook, ch. 6, p. 6-3). The CG moved forward (from +8.32 to +8.19) because 1.5 lb net came off the tail while 0.7 lb net was added forward of the datum.
Adding or removing an item on the equipment list is a minor alteration. It needs an entry in the maintenance records and a change to the weight and balance record (FAA-H-8083-1B, ch. 7, p. 7-2). The revision should show the new empty weight, empty weight arm or moment index, and the new useful load (ch. 7, p. 7-5). With a maximum weight of 2,550 lb, the useful load here is 2,550 - 1,651.2 = 898.8 lb, using the AMT General handbook’s method of subtracting empty weight from maximum weight (ch. 6, p. 6-4).
Problem 3: Adverse-loading check against both limits
An adverse-loaded CG check finds out whether any legal loading can move the CG outside its limits (FAA-H-8083-1B, p. G-1). You must do it after a change when the Type Certificate Data Sheet lists the empty weight CG range as “None” (ch. 7, p. 7-6).
The airplane from Problem 2 (1,651.2 lb, moment +13,527.1) has a CG range of +6.0 to +17.0 inches at all weights up to its 2,550 lb maximum. Front seats are at +12.0, rear seats at +48.0, fuel at +20.0 (40 gallons usable), the rear baggage area at +72.0 (100 lb maximum) and a nose baggage area at -30.0 (40 lb maximum). The engine has 180 METO (maximum except takeoff) horsepower. Is the airplane within limits at both extremes?
The handbook’s rules (ch. 7, p. 7-7):
- Forward check: load everything in front of the forward limit. Behind it, carry only what flight needs: the pilot and minimum fuel.
- Aft check: load everything behind the aft limit to the maximum. In front of it, carry only the pilot.
- Standard figures: 170 lb per occupant (nominal), 6 lb per gallon of aviation gasoline, and minimum fuel in pounds = METO horsepower / 2. Here that is 180 / 2 = 90 lb.
Forward check. The pilot (+12.0) and fuel (+20.0) are behind +6.0, so use the pilot and minimum fuel. The nose baggage (-30.0) is in front of the limit, so load it full.
| Item | Weight (lb) | Arm (in) | Moment (lb-in) |
|---|---|---|---|
| Airplane (empty) | 1,651.2 | +8.19 | +13,527.1 |
| Pilot | 170.0 | +12.0 | +2,040.0 |
| Minimum fuel | 90.0 | +20.0 | +1,800.0 |
| Nose baggage (full) | 40.0 | -30.0 | -1,200.0 |
| Total | 1,951.2 | +16,167.1 |
CG = 16,167.1 / 1,951.2 = +8.29 inches. That is behind the +6.0 forward limit, so the forward check passes.
Aft check. The fuel (+20.0), rear seats (+48.0) and rear baggage (+72.0) are behind +17.0, so load them all full: 40 gal x 6 = 240 lb of fuel and 2 x 170 = 340 lb in the rear seats. The pilot is in front of the limit but must be included.
| Item | Weight (lb) | Arm (in) | Moment (lb-in) |
|---|---|---|---|
| Airplane (empty) | 1,651.2 | +8.19 | +13,527.1 |
| Pilot | 170.0 | +12.0 | +2,040.0 |
| Full fuel | 240.0 | +20.0 | +4,800.0 |
| Rear seats (2) | 340.0 | +48.0 | +16,320.0 |
| Rear baggage (full) | 100.0 | +72.0 | +7,200.0 |
| Total | 2,501.2 | +43,887.1 |
CG = 43,887.1 / 2,501.2 = +17.55 inches. That is 0.55 inches behind the +17.0 aft limit, so the aft check fails, even though the weight (2,501.2 lb) is under the maximum.
How much rear baggage would fit with both rear seats full? Without baggage the totals are 2,401.2 lb and +36,687.1 lb-in (CG +15.28). Solve (36,687.1 + 72B) / (2,401.2 + B) = 17.0 and you get B = 75.15 lb. Check: with 75 lb, (36,687.1 + 5,400.0) / 2,476.2 = +16.997 inches, just inside the limit.
When a legal loading can put the CG out of limits, the handbook’s answers are placards and loading instructions that tell the pilot the restriction, or ballast (ch. 7, pp. 7-7 to 7-8).
Common mistakes
- Losing the minus sign forward of the datum. In Problem 1, treating the nose arm as +41.5 instead of -41.5 gives a CG of +26.20 inches instead of +8.32. Arms forward of the datum are negative (FAA-H-8083-1B, p. G-1).
- Forgetting the tare. Using raw scale readings in Problem 1 gives an empty weight of 1,676 lb, 24 lb too heavy, and a CG of +8.06. That error then flows into every later loading calculation. The tare must be subtracted (p. G-5).
- Mixing units. Convert fuel from gallons to pounds before multiplying. Do not mix a moment index (moment divided by a reduction factor, usually 100 or 1,000) with full moments in one table (p. G-4).
- Mixing inches and percent MAC. A CG of 8.32 inches and 13.9 percent MAC are the same point. Convert with the formula; never compare one with the other.
- Wrong items in an adverse check. The forward check uses minimum fuel when the tanks are behind the forward limit. The aft check uses full fuel when the tanks are behind the aft limit (ch. 7, p. 7-7).
- Weighing out of level or with temporary ballast aboard. The aircraft must be in level flight attitude (ch. 3, p. 3-5) and temporary ballast must be removed (ch. 3, p. 3-4).
A short study plan
- Read FAA-H-8083-1B chapter 2 (theory) and chapter 3 (weighing, tare, EWCG formulas, percent MAC).
- Read chapter 7 (equipment changes, adverse-loaded checks and ballast), then AMT General handbook chapter 6.
- Rework the three problems without looking, then change one number and solve again.
- Practice on the Weight and Balance topic page and brush up signed numbers on the Mathematics topic page.
For the test format and the other 11 subjects, see the A&P General written test guide.
Common questions
What formula do I need for A&P weight and balance?
CG = total moment / total weight, where each moment is weight x arm (FAA-H-8083-1B, ch. 2, pp. 2-2 to 2-3). Every problem type above is that formula with a different list of items.
Can I use a calculator on the test?
Aviation-oriented calculators and small electronic calculators that only do arithmetic are allowed, but test centers “may provide calculators and/or deny the use of personal calculators” (Companion Guide, p. 16).
How often must an aircraft be weighed?
Regulations do not require periodic weighing of privately owned and operated aircraft. The handbook says an aircraft should be reweighed after extensive repair or a major alteration, when the pilot reports nose or tail heaviness, and when the records are suspected to be wrong (FAA-H-8083-1B, ch. 3, p. 3-2).
Does the location of the datum change the answer?
No. The datum can be anywhere, as long as all measurements are made from it (ch. 3, p. 3-7). Problem 1 shows this.
What is minimum fuel?
For weight and balance, the fuel for half an hour at rated maximum continuous power: 1/12 gallon per METO horsepower, which works out to METO horsepower / 2 in pounds (FAA-H-8083-1B, ch. 7, p. 7-7).