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Weight and Balance Handbook

FAA-H-8083-1B Version 2016
All Library Documents FAA Source

Chapter 2

Weight and Balance Theory

Figure 2-11. Locating balance point with three weights.
Figure 2-11. Locating balance point with three weights.

To shift weight B so the lever balances about its center, 50 inches from the CG of weight A, first determine the arm of weight B that produces a moment that causes the total moment of all three weights around this desired balance point to be zero. The combined moment of weights A and C around this new balance point is 5,000 lb-in, so the moment of weight B must be –5,000 lb-in for the lever to balance. [Figure 2-12]

Figure 2-12. Proving the new balance point is correct.
Figure 2-12. Proving the new balance point is correct.

Determine the arm of weight B by dividing its moment, –5,000 lb-in, by its weight of 200 pounds. The arm is –25 inches. To balance the lever at its center, weight B must be placed so its CG is 25 inches to the left of the center of the lever. [Figure 2-13]

Figure 2-13. Weight distribution to balance lever.
Figure 2-13. Weight distribution to balance lever.

Figure 2-14 indicates that the shift in weight depicted in Figure 2-13 allows the lever to balance as the sum of the moments is zero.

Figure 2-14. Weight shift provides correct CG.
Figure 2-14. Weight shift provides correct CG.

Basic Weight and Balance Equation

The following formulas can be used to determine the distance weight must be shifted to obtain a desired change in the CG location. The equation can also be rearranged to find the amount of weight required to be shifted to move the CG to a desired location, to find the distance the CG is moved when a specified amount of weight is shifted, or to find the total weight that would allow shifting a specified amount of weight to move the CG a given distance.

Weight to be shifted / Total weight = ΔCG / Distance weight is shifted

Total weight = (Weight shifted × Distance weight is shifted) / ΔCG

Weight shifted = (Total weight × ΔCG) / Distance weight is shifted

ΔCG = (Weight shifted × Distance weight is shifted) / Total weight

Distance weight is shifted = (Total weight × ΔCG) / Weight shifted

Solution by Formula

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:

Distance weight B is shifted = (Total weight × ΔCG) / Weight shifted

= (500 × –22) / 200

= –55 inches

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.