1
$\begingroup$

I'm practicing for my EASA PPL(A) exam and one of the questions I came across sound like this:

Calculated take-off mass = 1082 kg, calculated CG = 0.254 m, fuel burn = 55 l on station 0.40 m.

A. 24.6
B. 25.2
C. 25.4
D. 24.8 <- this is the correct answer

My question is how do I actually get to this answer?

This is what I tried:

  1. 14,6 cm would be the difference between the fuel burn station and calculated CoG (in cm)
  2. 1027 kg would be the difference between calculated T/O mass and fuel burned
  3. 803 cm/kg would be the removed mass
  4. 0,78 cm would be the CoG shift calculated
  5. New CoG should be 25,4 (old CoG) - 0,78 which would mean 24,6

But yeah... that is not correct. So what am I missing here?

Improve this question
$\endgroup$
6
  • 3
    $\begingroup$ As @Bianfable has already explained to you, 55 liters of fuel do not correspond to 55 kilogrammes so your final weight is not 1082-55=1027kg... $\endgroup$
    – sophit
    Jan 19 at 20:05
  • $\begingroup$ but that's the thing, in another question on the same test, when i calculated 55 x 0,72 (density of avgas), the numbers i got were correct... so that's why here I assumed that I should not convert it. $\endgroup$
    – Tudor Ravoiu
    Jan 19 at 20:10
  • $\begingroup$ @sophit, so I should calculate 0,72 for avgas density? $\endgroup$
    – Tudor Ravoiu
    Jan 19 at 20:12
  • $\begingroup$ @TudorRavoiu: you cannot mix liters with kilogrammes. The former measures a volume, the latter a mass, they are two very different things. 1kg of cast iron is not heavier than 1kg of feathers. The two volumes anyway are going to be very different. Viceversa 1l of air is going to be much lighter than 1l of lead. $\endgroup$
    – sophit
    Jan 19 at 20:13
  • 2
    $\begingroup$ @TudorRavoiu When the metric system was created, the original definition of one kilogram is "the mass of one liter of water". Now, this is a perfectly fine and workable definition, but it seems to me (granted, this is based on only two data points) that this has created a kind of false equivalence in your head. You're so used to thinking that "kilogram = liter" that you think there's never any need to convert from one to the other. But that equation is ONLY true for pure water (and even then, it's only an approximation, since we've moved on to more precise definitions of units). $\endgroup$
    – HiddenWindshield
    Jan 20 at 22:41

1 Answer 1

Reset to default
6
+50
$\begingroup$

You are doing 1082 - 55 = 1027. The point that people are trying to make is that you have skipped the step you did correctly in the other problems you described.

You want 1082 - (55 * 0.72) = 1,042.4kg. You are subtracting the 55 as if it is the weight of the fuel in kg, but it is not. 55 liters of AVGAS weighs 39.6kg. Being a PPL test I will take the liberty of assuming the density required is for AVGAS, and using that density the problem works:

Weight (kg) Arm (m) Moment Kg-M
TO Wt/Bal 1082 (given) 0.254 (given) 274.828 (step 1)
Fuel Burn -39.6 (step 2) 0.4 (given) -15.84 (step 3)
Ldg Wt/Bal 1042.4 (step 4) 0.248 (step 6) 258.988 (step 5)

Steps:

  1. Moment(TO) = Weight(TO) * Arm(given) = 1082 * 0.254 = 274.828 kg-m
  2. Weight(fuel burn) = Volume(fuel burn) * Density(fuel) = -55l * 0.72 kg/l = -39.6 kg
  3. Moment(fuel burn) = Weight(fuel burn) * arm(fuel) = -39.6 kg * 0.4m = 15.48 kg-m
  4. Weight(ldg) = Weight(TO) - Weight(fuel burn) = 1082 - 39.6 = 1042.4 kg
  5. Moment(ldg) = Moment(TO) - Moment(fuel burn) = 274.828 - 15.48 = 258.988
  6. Resulting CG Arm = Moment(ldg) / Weight(ldg) = 258.988/1042.4 = 0.248m = 24.8cm

24.8 = You pass the test = you become a pilot = jackpot.

Improve this answer
$\endgroup$
4
  • 1
    $\begingroup$ Ehm, actually you passed the test 😉 $\endgroup$
    – sophit
    Jan 20 at 21:36
  • $\begingroup$ Thank you for taking the time to write this nice answer and actually helping out. $\endgroup$
    – DeltaLima ♦
    Jan 20 at 22:47
  • $\begingroup$ Thank you, I honestly thought about converting to Avgas after an answer from a previous question I had on aviation.stackexchange. But seeing that there is no specification to what kind of plane it's being used, I though... I shouldn't.... Maybe I should not think so much :). Thank you @DeltaLima $\endgroup$
    – Tudor Ravoiu
    Jan 21 at 6:47
  • $\begingroup$ "Maybe I should not think so much": don't think, just do (Maverick) $\endgroup$
    – sophit
    Jan 26 at 6:11

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .