0
$\begingroup$

Good morning everyone, I found out that the glide ratio of an airplane can be calculated by this formula : 1/tan(y) considering y the angle formed by its nose and an horizontal line. Check this photo : enter image description here

Imagine if y = 0. That means that we would divide by 0 in our formula. That's what I don't understand.

Thanks for answering by advance !

Improve this question
$\endgroup$
1
  • 1
    $\begingroup$ Is "y" the red or the green line? Normally we'd consider the trajectory (green), not the body angle (red), when thinking about glide ratio, because we want to know how much ground we can cover for the altitude we have. So a body angle of zero (horizontal) wouldn't matter so much if the trajectory is X degrees down. On the other hand, if the trajectory itself is horizontal, you don't really have a glide ratio. As that trajectory approaches horizontal, the glide ratio gets bigger & bigger; if your flight path is horizontal, then "glide ratio" is meaningless & undefinable. $\endgroup$
    – Ralph J ♦
    May 25 at 16:09

2 Answers 2

Reset to default
1
$\begingroup$

@Aditya Sharma answer is correct. However, no reason is given for why $\gamma$ will never be zero or negative for a gliding aircraft.

The reason relates to the causal relationship between L/D and the flight path angle. Thinking that a zero flight path angle would lead to a divide by zero and therefore infinite L/D is not the path of causality...

The aircraft has a finite L/D that it is operating at. That leads to a certain flight path angle that can never be zero (or negative). I.e. L/D causes FPA -- not FPA causes L/D.

Experimentally, we might choose to estimate the L/D by observing FPA in a glide -- but that does not mean causality follows that path.

Improve this answer
$\endgroup$
1
$\begingroup$

$\gamma$ is the glide angle, not the aircraft pitch angle. Other than that, the formula is correct.

For the purposes of determining the gliding capabilities of an aircraft, it is assumed that there is no vertical motion in the freestream air (which would otherwise alter the glide angle). Under such conditions, $\gamma$ will always be a positive angle - it will never be zero or negative.

In practice, there will always be some vertical motion in the freestream air, which will improve or worsen the glide angle. Specifically with strong up-currents, it is possible for the glide angle to be zero (level flight) or even negative (climb). In such cases, the aircraft would be able to stay airborne indefinitely, so using the term 'glide ratio' would be meaningless.

Improve this answer
$\endgroup$

You must log in to answer this question.

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