9/8/98
AC 43.13-1B
EXAMPLE OF THE DETERMINATION OF THE POUNDS OF FUEL AND BAGGAGE PERMISSIBLE
WITH MAXIMUM PASSENGERS
Aircraft empty
Oil
Pilot
Passenger (1) front
Passenger (2) rear
Fuel (39 gals.)
Baggage
Total
Weight (#)
+ 1169
+ 17
+ 170
+ 170
+ 340
+ 234
---
+ 2100
x Arm (“) = Moment (“#)
+ 10.6
+ 12391
- 49
- 833
+ 16
+ 2720
+ 16
+ 2720
+ 48
+16320
+ 22
+ 5148
---
---
+ 38466
Divide the TM (total moment) by the TW (total weight) to obtain the loaded center
of gravity.
TM = 38466
TW
2100
= + 18.6”
The above computations show that with the maximum number of passengers, 39
gallons of fuel and zero pounds of baggage may be carried in this aircraft without
exceeding either the maximum weight or the approved C. G. range.
This condition may be entered in the loading schedule as follows:
GALLONS OF FUEL
*Full
39
NUMBER OF PASSENGERS
*2 Rear
1(F) 2(R)
POUNDS OF BAGGAGE
* 100
None
* Conditions as entered from Figure 10-12
(F) Front seat
(R) Rear seat
FIGURE 10-13. Loading conditions: determination of the fuel and baggage permissible with maximum passengers.
Par 10-20
Page 10-17