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Find the position of the centre of gravity in % of the MAC (mean aerodynamic chord) if the distance from the datum to the LEMC (leading edge of mean chord) is 13 meters, the distance to the TEMC (terminal edge of mean chord) is 19 meters and the distance from the datum to the centre of gravity is 15 meters?

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    $\begingroup$ How far have you gotten so far? Show your work. $\endgroup$ – abelenky Dec 8 '18 at 18:32
  • $\begingroup$ I'd know exactly $\endgroup$ – Enki Sumerian Dec 8 '18 at 19:21
  • $\begingroup$ How is calculating? $\endgroup$ – Enki Sumerian Dec 9 '18 at 1:54
  • $\begingroup$ @EnkiSumerian, you use to many abbreviation. You should put clearly. Provide with drawing or picture as well so it clear. $\endgroup$ – AirCraft Lover Dec 10 '18 at 1:35
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I believe that the Length of the Aerodynamic Chord is $19\ m - 13\ m = 6\ m$, and that the CG is $2 m$ from the Leading Edge of the Mean Chord (LEMC).

Therefore, I believe that the CG in percent is: $$ \mathrm{\frac{CG}{Length\ of\ Chord}} = \frac{CG - LEMC}{TEMC - LEMC} = \frac{15-13}{19-13} = \frac{2}{6} = 33\% $$

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  • $\begingroup$ @abelensky: In a wing, where the length of chord we measure? Is that At the tip? At the middle? At the root? or where? $\endgroup$ – AirCraft Lover Dec 14 '18 at 6:57
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    $\begingroup$ The key term is Mean Chord, meaning average chord. If the wing edge is straight, you can just take the middle. If it is a curved, sweeping shape, you may need to do a calculus integral to get an exact answer. In this example, the LEMC (Leading Edge of Mean Chord) is at 13. So part of the leading edge may be less and part may be more, but the average leading edge is at 13. $\endgroup$ – abelenky Dec 14 '18 at 13:17
  • $\begingroup$ Thank you my friend for the nice explanation. $\endgroup$ – AirCraft Lover Dec 14 '18 at 21:03

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