031 Mass and Balance topic guide
Centre of Gravity Calculations
Every centre of gravity calculation in Mass and Balance comes down to two steps. First, convert each item of load into a moment about a chosen datum: mass multiplied by its arm, the distance from the datum to where that mass acts. Second, divide the total moment by the total mass. Once every row on the loading list has been converted this way, the method keeps the same shape no matter how many items the table holds.
The same logic covers what happens after loading. Add mass, remove mass, or move it between two positions, and the CG shifts by an amount worked the same way: for a move, the mass moved multiplied by the distance moved, divided by the total aircraft mass. Examiners favour clean numbers for a reason: a clean answer confirms the method, while a messy one usually means the wrong mass, arm, or direction has slipped into the working.
Moments about a datum
A datum is simply a reference point chosen for the calculation; in most large aeroplane loading manuals it sits at or ahead of the nose, so every arm in the loading table comes out positive and no item needs a minus sign. Build the table one row at a time: basic mass or dry operating mass first, then every item of traffic load, each with its own mass and its arm from the datum.
Multiply mass by arm on each row to get that row's moment, then add every mass down one column and every moment down the other. Divide the moment total by the mass total and the result is the CG position, expressed as an arm from the same datum. The method works with any number of rows: three items with masses of 6000 kg, 3000 kg and 1000 kg at arms of 4 m, 10 m and 6 m give moments of 24000, 30000 and 6000 kg.m, a total mass of 10000 kg, a total moment of 60000 kg.m, and a CG of exactly 6.0 m from the datum.
Shifting the CG when load changes
Adding or removing mass is an extension of the same table: fold the new item's mass and moment into the running totals, or subtract them if the mass leaves the aircraft, then divide again. Moving mass between two positions without changing the total mass is different enough to earn its own formula: the CG shift equals the mass moved multiplied by the distance it moves, divided by the aircraft's total mass, and the CG moves in the same direction as the mass.
For example, moving 200 kg of baggage 5 m aft on an aircraft with a total mass of 20000 kg shifts the CG by (200 x 5) / 20000, which is 1000 / 20000, or 0.05 m aft. The formula only needs three numbers, but every one of them has to be the right one: the mass that actually moved, the distance it moved rather than the arm of either position on its own, and the mass of the whole aircraft, not just the load.
Worked example
Worked example: CG from a three-item moment table
Not quite. The correct answer is B.
Correct answer: B
- A. This adds only two of the three moments (12000 + 24000) while still dividing by the full mass of 10000 kg, silently dropping the 2000 kg item's moment.
- B. Correct: the moments are 4000 x 3 = 12000, 2000 x 9 = 18000 and 4000 x 6 = 24000, a total moment of 54000 kg.m over a total mass of 10000 kg, giving 5.4 m.
- C. This is the plain arithmetic mean of the three arms, (3 + 9 + 6) / 3, which ignores the masses entirely and would only be correct if all three items weighed the same.
- D. This keeps all three moments (54000 kg.m) but drops the 4000 kg item from the mass total, dividing by 6000 kg instead of 10000 kg.
Step by step
- Calculate each moment: 4000 kg x 3 m = 12000 kg.m, 2000 kg x 9 m = 18000 kg.m, 4000 kg x 6 m = 24000 kg.m.
- Add the masses: 4000 + 2000 + 4000 = 10000 kg.
- Add the moments: 12000 + 18000 + 24000 = 54000 kg.m.
- Divide total moment by total mass: 54000 / 10000 = 5.4 m from the datum.
- Sanity check: 5.4 m sits between the arm of the lightest item (9 m) and the arms of the two heavier items (3 m and 6 m), leaning towards the heavier side, which is what a mass-weighted answer should do.
Common mistakes
Averaging the arms instead of weighting them by mass
An unweighted average treats every item as equally heavy, so a light item at a distant arm drags the answer just as hard as a heavy item close to the datum, and the exam builds this exact wrong number into the options.
Dropping an item's mass from the total while its moment stays in the sum
This shrinks the denominator without touching the numerator, which pushes the calculated CG well beyond where it can plausibly sit, and it happens easily when a loading table has more rows than the working space allows for.
Assuming a load shift always moves the CG towards the nose
The CG moves in whatever direction the mass moves, aft or forward. Assuming one default direction turns a correctly sized shift calculation into a wrong final answer with the sign flipped.
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Last reviewed July 2026