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Refer to figure or CAP758 figure 2.16.
Given the following information, what is the Take Off Mass (TOM) and lateral Centre of Gravity (CG) for the Single Engine Piston Helicopter (SEPH)?

BEM: 1 165 lb with lateral arm +0.43
Pilot: 85 kg
RH pax: 75 kg
Center pax: 80 kg
Fuel 20 USG
(Assume 6 lb/USG)

  • A
    1 813 lb; -1.34 in
  • B
    1 813 lb; +1.34 in
  • C
    1 525 lb; +0.34 in
  • D
    1 525 lb; -0.34 in

Refer to figures.
To calculate the position of the Centre of Gravity (CG), the mass and the arm of the stations in the helicopter must be known. We can find out the total Centre of Gravity position by adding the total moments and the total masses on board, thereafter, using the formula:

  • Moment = Mass x Arm

The question is asking for the lateral CG at TOM and therefore, we need to extract the individual moments from the attached figure by:

  1. Convert the weights given in KG to lbs by factor of 2.205.
  2. Draw a line from the line showing mass up the individual masses line.
  3. Thereafter, draw a line to the left to reach the moment of that mass.

Repeat the process for each individual weight and then make a table with the weights and moments as shown in attached figure.

The TOM is:

  • TOM = BEM + Pilot + Pax + Fuel = 1 165 lbs + 187 lbs + 176 lbs + 165 lbs + 120 lbs
  • TOM = 1 813 lbs

Find out the TOM moment by adding them and thereafter, calculate the lateral CG by:

  • CG = TOM Moment / TOM mass
  • CG = +2 430.95 lbs in / 1 813 lbs = 1.34 in

NOTES:

  1. The negative value of the arm shows that the CG is to the laterally to left of the datum and longitudinally aft of the datum. This is a standard way of calculating the CG in aviation.
  2. It’s important to draw the lines as accurate as possible to get the most accurate value.

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