Definitions and Calculations — Page 58, Lesson 76

We're starting a fresh topic now — the adding or removing mass problems in the mass and balance chapter. These are the classic calculation drills you'll use constantly in the real world, so let's build them up properly. First, the core idea. Whenever you add or remove mass from an aircraft, you change two things at once: the total mass, and the position of the centre of gravity. The CG is the point where the whole aircraft balances, measured in inches or feet aft of the datum — the datum being the reference point everything is measured from. When you shift mass, the CG moves, and you need to know exactly how far it moves so you stay inside the CG limits. Let me give you the fundamental relationship we use for all of these. The change in CG position equals the mass moved, multiplied by the distance between the old CG and the new mass location, all divided by the final total mass. In symbols, that's delta CG equals m times d divided by M. Here, m is the mass you're adding or removing, d is the distance from the old CG to the hold or station where the mass sits, and M is the final total mass after the change. That single formula drives every one of these problems. Now let's walk through the actual questions, because each one tests a slightly different twist. Problem one. Three holds at 10, 100 and 250 inches aft of the datum — those are holds A, B and C. Total aircraft mass is 3500 kilograms, CG is at 70 inches aft of the datum. The CG limits run from 40 to 70 inches aft. So the CG is currently on the aft limit, and we need to move it forward to the forward limit at 40 inches. The question: how much load must be removed from hold C to get the CG onto the forward limit? So we're removing mass from hold C at 250 inches aft, and we want the CG to shift forward from 70 to 40 inches — a shift of 30 inches forward. Problem two. Mass is 5000 pounds, CG at 80 inches aft. The aft CG limit is 80.5 inches aft. So the CG is just 0.5 inches forward of the limit. We want to load mass into a hold at 150 inches aft, and we need to find the maximum mass we can put there without exceeding the limit. Here we're adding mass, and the CG will move aft, so we calculate how much mass we can add before the CG reaches 80.5. Problem three. Loaded mass is 108,560 pounds, CG at 86.3 feet aft of the datum. The aft CG limit is 85.6 feet. Notice the CG is actually aft of the limit — it's 0.7 feet too far back. So we need to add ballast in a hold at 42 feet aft of the datum to bring the CG forward onto the aft limit. The question asks how much ballast. This is the adding-mass-to-correct-a-CG case. Problem four. Aft CG limit is 80 inches aft. Loaded CG is at 80.5 inches aft — again, 0.5 inches beyond the limit. Mass is 6400 pounds. We need to remove mass from a hold at 150 inches aft to bring the CG back onto the aft limit. So here we're removing mass from behind the CG, which moves the CG forward. Problem five. Mass 7900 kilograms, CG at 81.2 inches aft. We load a 250-kilogram package into a hold at 32 inches aft of the datum. The question asks for the new CG position. Here we're adding mass forward of the CG, so the CG will move forward, and we calculate the new position directly. Problem six. CG limits from 72 to 77 inches aft. Mass 3700 kilograms, CG at 76.5 inches aft. We remove 60 kilograms from the forward hold, located at 147 inches forward of the datum. Note the direction — that hold is forward of the datum, so it's at minus 147 inches relative to the datum. Removing mass from forward of the CG moves the CG aft. The question asks for the change to the CG position. Problem seven is cut off mid-sentence — it starts with an aeroplane having a zero fuel mass of 47,800 kilograms and a performance limited take-off... and that's where the excerpt ends. So we'll stop there and pick that one up when we see the rest. The key thing to remember across all of these: always identify whether you're adding or removing mass, whether the hold is forward or aft of the current CG, and whether the CG needs to move forward or aft. The sign of the distance d in the formula depends on that. When you remove mass, the final mass M is the original minus the removed mass. When you add, it's the original plus the added mass. Get those signs right and the formula does the rest. Let's work through problem one together as a model, because it shows the full method. We want the CG to move from 70 to 40 inches — that's a forward shift of 30 inches. We're removing mass from hold C at 250 inches aft. The distance d from the old CG at 70 to the hold at 250 is 180 inches. The final mass M will be 3500 minus the unknown removed mass. So we set up delta CG equals m times d over M, with delta CG equal to 30 inches, d equal to 180 inches, and M equal to 3500 minus m. Solve for m, and that gives you the load to remove from hold C. That's the pattern for all of them. Each problem just changes which variable you're solving for — sometimes the mass, sometimes the new CG position, sometimes the change in CG. But the underlying relationship is always the same: delta CG equals m times d over M.