Procedures for Weighing an Aircraft

in Aircraft Weight and Balance

The most important reason for weighing an aircraft is to find out its empty weight (basic empty weight), and to find out where it balances in the empty weight condition. When an aircraft is to be flown, the pilot in command must know what the loaded weight of the aircraft is, and where its loaded center of gravity is. In order for the loaded weight and center of gravity to be calculated, the pilot or dispatcher handling the flight must first know the empty weight and empty weight center of gravity.


Earlier in this chapter it was identified that the center of gravity for an object is the point about which the nose heavy and tail heavy moments are equal. One method that could be used to find this point would involve lifting an object off the ground twice, first suspending it from a point near the front, and on the second lift suspending it from a point near the back. With each lift, a perpendicular line (90 degrees) would be drawn from the suspension point to the ground. The two perpendicular lines would intersect somewhere in the object, and the point of intersection would be the center of gravity. This concept is shown in Figure 4-4, where an irregular shaped object is suspended from two different points. The perpendicular line from the first suspension point is shown in red, and the new suspension point line is shown in blue. Where the red and blue lines intersect is the center of gravity.

Figure 4-4. Center of gravity determined by two suspension points.

Figure 4-4. Center of gravity determined by two suspension points.

If an airplane were suspended from two points, one at the nose and one at the tail, the perpendicular drop lines would intersect at the center of gravity the same way they do for the object in Figure 4-4. Suspending an airplane from the ceiling by two hooks, however, is clearly not realistic. Even if it could be done, determining where in the airplane the lines intersect would not be possible.

A more realistic way to find the center of gravity for an object, especially an airplane, is to place it on a minimum of two scales and to calculate the moment value for each scale reading. In Figure 4-5, there is a plank that is 200″ long, with the left end being the datum (zero arm), and 6 weights placed at various locations along the length of the plank. The purpose of Figure 4-5 is to show how the center of gravity can be calculated when the arms and weights for an object are known.

Figure 4-5. Center of gravity for weights on a plank, datum at one end.

Figure 4-5. Center of gravity for weights on a plank, datum at one end.

To calculate the center of gravity for the object in Figure 4-5, the moments for all the weights need to be calculated and then summed, and the weights need to be summed. In the four column table in Figure 4-6, the item, weight, and arm are listed in the first three columns, with the information coming from Figure 4-5. The moment value in the fourth column is the product of the weight and arm. The weight and moment columns are summed, with the center of gravity being equal to the total moment divided by the total weight. The arm column is not summed. The number appearing at the bottom of that column is the center of gravity. The calculation would be as shown in Figure 4-6.

Figure 4-6. Center of gravity calculation for weights on a plank with datum at one end.

Figure 4-6. Center of gravity calculation for weights on a plank with datum at one end.

For the calculation shown in Figure 4-6, the total moment is 52,900 in-lb, and the total weight is 495 lb. The center of gravity is calculated as follows:

cg1An interesting characteristic exists for the problem presented in Figure 4-5, and the table showing the center of gravity calculation. If the datum (zero arm) for the object was in the middle of the 200″ long plank, with 100″ of negative arm to the left and 100″ of positive arm to the right, the solution would show the center of gravity to be in the same location. The arm for the center of gravity would not be the same number, but its physical location would be the same. Figure 4-7 and Figure 4-8 show the new calculation.

cg2

Figure 4-7. Center of gravity for weights on a plank, datum in the middle.

Figure 4-7. Center of gravity for weights on a plank, datum in the middle.

Figure 4-8. Center of gravity calculation for weights on a plank with datum in the middle.

Figure 4-8. Center of gravity calculation for weights on a plank with datum in the middle.

In Figure 4-7, the center of gravity is 6.9″ to the right of the plank’s center. Even though the arm is not the same number, in Figure 4-5 the center of gravity is also 6.9″ to the right of center (CG location of 106.9 with the center being 100). Because both problems are the same in these two figures, except for the datum location, the center of gravity must be the same.

The definition for center of gravity states that it is the point about which all the moments are equal. We can prove that the center of gravity for the object in Figure 4-7 is correct by showing that the total moments on either side of this point are equal. Using 6.87 as the CG location for slightly greater accuracy, instead of the rounded off 6.9 number, the moments to the left of the CG would be as shown in Figure 4-9.

Figure 4-9. Moments to the left of the center of gravity.

Figure 4-9. Moments to the left of the center of gravity.

The moments to the right of the CG, as shown in Figure 4-7, would be as shown in Figure 4-10.

Figure 4-10. Moments to the right of the center of gravity.

Figure 4-10. Moments to the right of the center of gravity.

Disregarding the slightly different decimal value, the moment in both of the previous calculations is 10,651 in-lb. Showing that the moments are equal is a good way of proving that the center of gravity has been properly calculated.