CHAPTER 5 - CHANGE OF WEIGHT

A pilot must be able to solve accurately and rapidly problems which involve the shift, addition, or removal of weight. For example, the pilot may load the aircraft within the allowable takeoff weight limit, then find a c.g. limit has been exceeded. The most satisfactory solution to this problem is to shift baggage, or passengers, or both. The pilot should be able to determine the minimum load shift needed to make the aircraft safe for flight. Pilots should also be able to determine if the shifting of a load to a new location will correct an out-of-limit condition. There are some standardized and simple calculations which can help make these determinations.

WEIGHT SHIFTING

When weight is shifted from one location to another, the total weight of the aircraft is unchanged. The total moments, however, do change in relation and proportion to the direction and distance the weight is moved. When weight is moved forward, the total moments decrease; when weight is moved aft, total moments increase. The moment change is proportional to the amount of weight moved. Since many aircraft have forward and aft baggage compartments, weight may be shifted from one to the other to change the c.g. If we start with a known aircraft weight, c.g., and total moments, we can calculate the new c.g. (after the weight shift) by dividing the new total moments by the total aircraft weight.

EXAMPLE 14.

To determine the new total moments, find out how many moments are gained or lost when the weight is shifted. The weight shift conditions indicated for the aircraft illustrated in figure 27 show that 100 lb. has been shifted from Sta. 30 to Sta. 150. This movement increases the total moments of the aircraft by 12,000 lb-in

Baggage moment when at Sta. 150

= 100 lb. x 150 in. = 15,000 lb-in

Baggage moment when at Sta. 30

= 100 lb. x 30 in.= 3,000 lb-in

Moment change = 12,000 lb-in

By adding the moment change to the original moment (or subtracting if the weight had been moved forward instead of aft), we obtain the new total moments. We can then determine the new c.g. by dividing the new moments by the total weight:

Total moments = 616,000 + 12,000 = 628,000

628,000

c.g. = ------- = 78.5 in.

8,000

The shift of the baggage has caused the c.g. to shift to Sta. 78.5. A simpler solution may be obtained by using the aeronautical computer and a proportion formula (fig. 28). This can be done because the c.g. will shift a distance which is proportional to the distance the weight has shifted.

EXAMPLE 15.

1. Weight shifted
c.g. (change of c.g.)

-------------- = ----------------------------

Total weight
Distance weight is shifted

100
c.g.

----- = ----------

8,000
120

c.g. = 1.5 in.

2. The change of c.g. is added to (or subtracted from) the original c.g. to determine the new c.g.:

77 + 1.5 = 78.5 in. aft of datum

A possible point of difficulty arises in the computer-type solution when an attempt is made to locate the decimal point in the answer. How do we make sure the c.g. in the above problem is not 0.15 in. or 15.0 in.? The answer you get can always be checked by cross multiplying. Substitute the answer in step 1 (example 15):

1.5 x 8,000 = 12,000

100 x 120 = 12,000

If the cross multiplication answers are not the same, you have selected the wrong decimal location for the c.g. and the decimal should be relocated accordingly.

Finding the decimal location is primarily a matter of observation; remember that the proportions on either side of the equal sign (example 15, step 1) are similar.

The shifting weight proportion formula can also be used to determine how much weight must be shifted to achieve a particular shift of the c.g. The following problem illustrates a solution of this type.

EXAMPLE 16.

Given:

Aircraft total weight - 7,800 lb.

c.g.
- Sta. 81.5 in.

Aft c.g. limit
- 80.5 in.

Find: How much cargo must be shifted from the aft cargo compartment at Sta. 150 to the forward cargo compartment at Sta. 30 to move the c.g. to exactly the aft limit?

Solution:

1. Use the shifting weight proportion:

Weight shifted
c.g.

-------------- = -----------------

Total weight
Dist. wt. shifted

Weight (to be) shifted 1.0 in.

---------------------- = -------

7,800
120 in.

Weight to be shifted = 65 lb.

2. Cross multiply to check for accuracy of decimal point location in the answer.

7,800 x 1.0 = 7,800

65 x 120 = 7,800

A combination problem may involve the shifting of weight when the c.g. and the c.g. limits are expressed in % MAC.

EXAMPLE 17.

Given:

Aircraft total weight - 7,200 lb.

c.g.
- 12% MAC

Fwd. c.g. limit - 14.8% MAC

MAC
- Sta. 70 to Sta. 105 = 35 in.

Find: How much cargo must be shifted from the front baggage compartment at Sta. 30 to the aft baggage compartment at Sta. 150 to move the c.g. to exactly the forward limit?

Solution:

1. Convert the % MAC locations to inches from datum by using the aeronautical computer (fig. 29).

c.g. = LEMAC (70 in.) + % MAC in inches (4.2 in.) = 74.2 in.

Fwd. limit = LEMAC (70 in.) + % MAC in inches (5.2 in.) = 75.2 in.

2. Determine c.g. (distance c.g. must be moved):

75.2 - 74.2 = 1.0

c.g.= 1.0 in. aft

3. Use the shifting weight proportion:

Weight shifted
c.g.

-------------- = -----------------

Total weight Dist.
wt. shifted

Weight (to be) shifted 1.0

---------------------- = ---

7,200
120

Weight to be shifted = 60 lb.

WEIGHT ADDITION OR REMOVAL

In many instances the weight and balance of the aircraft will be changed by the addition or removal of weight. When this happens a new c.g. must be calculated and checked against the limitations to see if the location is acceptable. This type of weight and balance problem is commonly encountered when the aircraft burns fuel in flight, thereby reducing the weight located at the fuel tanks. Most aircraft are designed with the fuel tanks positioned close to the c.g., therefore, the consumption of fuel does not affect the c.g. to any great extent. However, large jet aircraft with fuel tanks located in the sweptback wings require careful planning on each flight to prevent the c.g. shifting out of limits due to the consumption of fuel.

The addition or removal of cargo presents a c.g. change problem which may have to be calculated rapidly before flight. The problem may always be solved by calculations involving total moments. However, a shortcut formula which can be adapted to the aeronautical computer may be used to simplify computations:

Weight added (or removed)
c.g.

------------------------- = --------------------------------

New total weight
Distance between wt. and old c.g.

In this formula the terms "new" and "old" refer to conditions before and after the weight change. It is often more convenient to use another form of this formula when required to find the weight change needed to accomplish a particular c.g. change ( c.g.). In this case we use:

Weight (to be) added (or removed)
c.g.

--------------------------------- =
--------------------

Old total weight
Distance between wt.

and new c.g.

Notice that the terms "new" and "old" are not found on both sides of the equation in either of the above proportions. If the "new" total weight is used, the distance must be calculated from the "old" c.g. Just the opposite is true if the "old" total weight is used.

A typical problem may involve the calculation of a new c.g. for an aircraft which, when loaded and ready for flight, receives some additional cargo or passengers just before departure time.

EXAMPLE 18.

Given:

Aircraft total weight - 6,860 lb.

c.g. - Sta. 80.0

Find: What is the location of the c.g. if 140 lb. of baggage is added to station 150?

Solution:

1. Use the added weight formula:

Added weight
c.g.

---------------- = --------------------------------

New total weight Distance between
wt. and old c.g.

140
c.g.

----------- = ----------

6,860 + 140 150 - 80

140
c.g.

----- = ----------

7,000 70

c.g. = 1.4 in. aft.

2. Add c.g. to the old c.g.:

New c.g.= 80.0 in. + 1.4 in. = 81.4 in.

EXAMPLE 19.

Given:

Aircraft total weight - 6,100 lb.

c.g. - Sta. 78

Find: What is the location of the c.g. if 100 lb. is removed from station 150?

Solution:

1. Use the removed weight formula:

Weight removed
c.g.

---------------- = ------------------------------

New total weight Dist.
between wt. and old c.g.

100
c.g.

----------- = ----------

6,100 - 100 150 - 78

100
c.g.

----- = ----------

6,000 72

c.g. = 1.2 in. forward

2. Subtract c.g. from old c.g.:

New c.g. = 78 in. - 1.2 in. = 76.8 in.

NOTE - In the above two examples, the c.g. is either added to or subtracted from the old c.g. Deciding which to accomplish is best handled by mentally calculating which way the c.g. will shift for the particular weight change. If the c.g. is shifting aft, the c.g. is added to the old c.g.; if it is shifting forward, the c.g. is subtracted from the old c.g. To summarize c.g. movement:

Weight added fwd. of old c.g

c.g. moves fwd.

Weight removed aft of old c.g

Weight added aft of old c.g

c.g. moves aft

Weight removed fwd. of old c.g

EXAMPLE 20.

Given:

Aircraft total weight - 7,000 lb.

c.g. - Sta. 79.0

Rear c.g. limit - Sta. 80.5

Find: How far aft can additional baggage weighing 200 lb. be placed without exceeding the rear c.g. limit?

Solution:

1. Use the added weight formula:

Added weight
c.g.

---------------- = --------------------------------

New total weight Distance
between wt. and old c.g.

200
1.5

----- = -----------------------------

7,200 Dist. between wt. and old c.g.

Distance between wt. and old c.g. = 54 in.

2. Add to old c.g.:

79 in. + 54 in. = 133 in. aft of datum

When the 200 lb. is located at Sta. 133, the new c.g. will be exactly on the aft limit; if the weight is located any further to the rear, the aft c.g. limit will be exceeded.

EXAMPLE 21.

Given:

Aircraft total weight - 6,400 lb.

c.g. - Sta. 80.0

Aft c.g. limit - Sta. 80.5

Find: How much baggage can be located in the aft baggage compartment at station 150 without exceeding the aft c.g. limit?

Solution:

Use the added weight formula.

NOTE - In this problem, the new total weight is not given, therefore, it is more convenient to use the version of the formula which makes use of the old total weight.

Added weight
c.g.

---------------- = --------------------------------

Old total weight
Distance between wt. and new c.g.

Added weight 0.5

------------ = ---------- =

6,400
150 - 80.5

Added weight = 46 lb