Definitions and Calculations — Page 50, Lesson 70

Let’s pick this up right where the maths gets practical. We’ve already found a centre of gravity that’s out of limits, and now I want to show you the standard fix: repositioning the centre of gravity by repositioning mass. That’s the exact title of this part, and it’s one of the most useful tools you’ll have in mass and balance work. Look at Figure 2.12. We have an aircraft whose centre of gravity has been found to be out of limits at a distance ‘a’ inches aft of the datum. The forward CG limit is ‘b’ inches aft of the datum. So the CG is sitting too far aft, beyond the forward limit — that’s the problem. To bring the CG back into limits, we’re going to move some baggage, mass ‘m’, from compartment A to compartment B. In the figure, A is the forward compartment and B is the aft one, so we’re shifting load forward to pull the CG forward. Here’s the core idea. If a mass of m pounds is moved from A to B, the change of moment is m × d. That is, the change of moment equals the mass moved, m, multiplied by the distance through which it moves, d. The distance d is the separation between the two compartments — the arm difference between A and B. Now let’s tie this to the whole aircraft. If the total mass of the aircraft is M, with the CG at ‘a’ inches aft of the datum, then the total moment around the datum is M × a. We need to move the CG to ‘b’ inches aft of the datum, so the new total moment will be M × b. The change in moment required is therefore M × b minus M × a, which is M(b − a). And here’s the beautiful part — that required change in moment must equal the change produced by moving the baggage. So we set M(b − a) = m × d. Now, b − a is exactly the change in the CG position, which we call ‘cc’. So we can write the whole thing as m × d = M × cc. In words: the mass you move, multiplied by the distance you move it, equals the total aircraft mass multiplied by the distance the CG moves through. That’s the master formula for this whole technique. Let me walk you through Example 5 so you see it in action. The CG limits of an aircraft are from −4 to +3 inches from the datum. Note the minus sign — that means the forward limit is 4 inches forward of the datum, and the aft limit is 3 inches aft of the datum. The aircraft is loaded as shown. Basic Empty Mass is 2800 lb at an arm of 2 inches, giving a moment of 5600 lb-in. Crew is 340 lb at an arm of −20 inches, moment −6800 lb-in. Fuel is 600 lb at arm 10 inches, moment 6000 lb-in. Forward Hold is 0 lb at arm −70 inches, moment 0. Aft Hold is 150 lb at arm 80 inches, moment 12,000 lb-in. Total mass is 3890 lb, total moment is 16,800 lb-in. So the CG is total moment divided by total mass: 16,800 divided by 3890, which is 4.32 inches. The aft limit is +3 inches, so the CG is 1.32 inches out of limits — too far aft. That’s our problem. We correct it by moving freight or baggage from the rear hold to the forward hold. The distance between them is 150 inches — from 80 inches aft to 70 inches forward, that’s a separation of 150 inches. Now we apply our formula: m × d = M × cc. Here m is the unknown mass to move, d is 150 inches, M is the total aircraft mass 3890 lb, and cc is the CG shift needed, 1.32 inches. So m × 150 = 3890 × 1.32. Solving for m, we get m = (3890 × 1.32) / 150, which is 34.232 lb. So we need to move about 34.2 pounds of freight or baggage from the aft hold to the forward hold. Now, the book does something very important — it checks the aircraft is safe for all fuel states after take-off. That means we calculate the CG at Zero Fuel Mass, with 35 lb of baggage moved to the forward hold. Let’s rebuild the table. Basic Mass 2800 lb at arm 2, moment 5600. Crew 340 lb at arm −20, moment −6800. Fuel is zero at zero fuel mass. Forward Hold now has 35 lb at arm −70, moment −2450. Aft Hold now has 115 lb at arm 80, moment 9200. The Zero Fuel Mass is 3290 lb, and the zero fuel moment is 5550 lb-in. So the ZFM CG is 5550 divided by 3290, which is 1.69 inches — and that’s in limits, because it’s between −4 and +3. So the whole logic is: find the out-of-limit CG, compute the required shift, use m × d = M × cc to find how much mass to move, then verify at zero fuel mass that the result is safe. That’s the complete repositioning procedure.