2-5 Distance weight B is shifted =  Total weight × Δ CG Weight shifted =  500 × –22 200 = –55 inches Weight to be shifted  = Δ CG Total weight  Distance weight is shifted 200  = 22 500 55  .4 = .4 Weight shifted = Total weight × Δ CG Distance weight is shifted =  500 × 22 55 = 200 inches Δ CG =  Weight shifted × Distance it is shifted Total weight =  200 × 55 500 = 22 inches CG limits are specified forward and aft and/or lateral (left and  right) limits within which the aircraft’s CG must be located  during flight. The area between the limits is called the CG  range of the aircraft.  A third arrangement of this basic equation is used to  determine the amount the CG is shifted when a given amount  of weight is moved for a specified distance (as it was done  in  Figure 2-11 ). The following formula is used to determine  the amount the CG is shifted when 200-pound weight B is  moved from +80 to +25. If the 200-pound weight B is shifted from station +80 to  station +25, the CG moves from station +72 to station +50.  Figure 2-14.  Center of gravity expressed as percent mean  aerodynamic chord.$^{MAC}$ On some aircraft, the CG is expressed as a percentage of the  length of the mean aerodynamic chord (MAC) or “percent  MAC.”  [Figure 2-14]  In order to make such a calculation,  the position of the leading edge of the MAC must be known  ahead of time.  The CG point affects the stability of the aircraft. To ensure  the aircraft is safe to fly, the CG must fall within specified limits established by the manufacturer. When the distance the weight is to be shifted is the amount of weight to be shifted to move the CG to any  location can be determined by another arrangement of the  basic equation. Use the following arrangement of the formula  to determine the amount of weight that has to be shifted from  station 8 to station +25, to move the CG from station +72  to station +50.  The CG of the lever in  Figure 2-11  was 72 inches from the  datum. This CG can be shifted to the center of the lever as in  Figure 2-13  by moving weight B. If the 200-pound weight B  is moved 55 inches to the left, the CG shifts from +72 inches  to +50 inches, a distance of 22 inches. The equation is balanced. The problem in  Figure 2-11  can be solved by using variations  of this basic equation. First, rearrange the formula to  determine the distance weight B must be shifted: To complete the calculations, return to the original formula  and enter the appropriate numbers. Moving weight B from +80 to +25 moves the CG 22 inches  from its original location at +72 to its new location at +50  as seen in  Figure 2-13 . Solution by Formula Mean Aerodynamic Chord MAC 30% MAC 15% MAC Datum Quarter chord Half semi-span