Now try these — Page 50, Lesson 68

Right, let's get into the calculations. These are the classic CG problems you'll be expected to solve quickly and accurately. Question 1: We have a MAC of 82 inches. The leading edge of the MAC is 103 inches aft of the datum. The CG is at 14.7% MAC. We need the CG distance from the datum. First, convert the percentage to a decimal: 14.7% is 0.147. Multiply that by the MAC length: 0.147 × 82 inches = 12.054 inches. That's the CG's position aft of the leading edge of the MAC. Since the leading edge is already 103 inches aft of the datum, we add: 103 + 12.054 = 115.054 inches. So the CG is 115.05 inches aft of the datum. Question 2: Now we work backwards. CG is 21% MAC, MAC is 73 inches, and the CG datum is 26 inches aft of the leading edge of the MAC. Find the CG position relative to the datum. Again, convert: 21% is 0.21. Multiply by the MAC: 0.21 × 73 = 15.33 inches. That's the CG aft of the leading edge. Add the 26 inches: 15.33 + 26 = 41.33 inches. The CG is 41.33 inches aft of the datum. Question 3: This one asks for limits as a percentage. The CG limits are from 5 inches forward to 7 inches aft of the datum. The MAC is 41 inches, and its leading edge is 15 inches forward of the datum. So the leading edge is at −15 inches relative to the datum. The forward limit is at −5 inches. The distance from the leading edge to the forward limit is: −5 − (−15) = 10 inches. As a percentage of MAC: 10 ÷ 41 = 0.2439, or 24.4% MAC. The aft limit is at +7 inches. Distance from leading edge: 7 − (−15) = 22 inches. As a percentage: 22 ÷ 41 = 0.5366, or 53.7% MAC. So the limits are from 24.4% to 53.7% MAC. Question 4: MAC is 58 inches. Limits are 26% to 43% MAC. CG is found at 45.5% MAC. How many inches out of limits? The CG is above the aft limit of 43%. The excess is 45.5 − 43 = 2.5% MAC. Convert to inches: 0.025 × 58 = 1.45 inches. The CG is 1.45 inches aft of the aft limit. Question 5: An aircraft of mass 62,500 kg has the leading and trailing edges of the MAC at body stations +16 and +19.5 metres. What is the arm of the CG if the CG is at 30% MAC? The MAC length is 19.5 − 16 = 3.5 metres. The leading edge is at station +16. The CG is at 30% MAC: 0.30 × 3.5 = 1.05 metres aft of the leading edge. So the arm is 16 + 1.05 = 17.05 metres. The mass of 62,500 kg is given but not needed for this calculation — it's there to remind you that the CG arm is independent of mass. Now, the next section: Repositioning of the Centre of Gravity. The rule is absolute: if the CG is out of limits for any part of the flight, the aircraft must not take off until the load is adjusted to bring it back into limits. There are two ways to do this. First, by repositioning mass already on board — usually baggage or passengers. Second, by adding or removing mass. Mass put on purely to correct the CG position is called ballast. The minimum mass to move, load, or off-load is the amount that just brings the CG onto the nearest limit. You may, of course, bring it further inside the limits for safety margin. Once you've calculated the adjustment, you must verify it makes the aircraft safe for both take-off and landing. Let's look at the figures for these repositioning problems. shows the basic setup. illustrates repositioning mass. And shows a specific problem — how much load should be transferred from No. 2 hold. The key principle for repositioning: when you move mass, the CG shifts in the direction of the movement. The formula you'll use is based on the moment change — mass moved times the distance moved equals the total mass times the CG shift. That's the core relationship you'll apply in these problems.