Flight path angle concept question

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SUMMARY

The flight path angle can indeed be expressed as 356 degrees, which is equivalent to 4 degrees when considering its definition as the angle between the flight path vector and the horizon. This conversion is achieved by subtracting 360 from 356. The discussion emphasizes the importance of understanding the specific definition of flight path angle, which is typically constrained between -180 and +180 degrees. The mathematical representation of the flight path angle is given by the formula involving true anomaly and eccentricity.

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  • Knowledge of mathematical functions, specifically trigonometry
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Can a flight path angle be 356 degrees?
 
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dwsmith said:
Can a flight path angle be 356 degrees?

Hi dwsmith, :)

Depends on the way it's defined I presume. If it's defined as, angle between the flight path vector and the horizon (refer >>this<<) then 356 degrees will be the same as 360-356=4 degrees.

Kind Regards,
Sudharaka.
 
Sudharaka said:
Hi dwsmith, :)

Depends on the way it's defined I presume. If it's defined as, angle between the flight path vector and the horizon (refer >>this<<) then 356 degrees will be the same as 360-356=4 degrees.

Kind Regards,
Sudharaka.

Originally it was -4 so I added 360 to get the positive. So I can then subtract and say it is positive?
 
dwsmith said:
Originally it was -4 so I added 360 to get the positive. So I can then subtract and say it is positive?

Where did you encounter this question and can you please tell us how the flight path angle is defined in the book that you use? I presume that it is defined as an angle in-between -180 and +180.
 
Sudharaka said:
Where did you encounter this question and can you please tell us how the flight path angle is defined in the book that you use? I presume that it is defined as an angle in-between -180 and +180.

$$
\gamma = \tan^{-1}\left(\frac{e\sin\nu}{1 + e\cos\nu}\right)
$$
where $\nu$ is the true anomaly and e is the eccentricity of the orbit.
 

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