-1
$\begingroup$

I am wondering about the g-load and its relation to the so called gliding ratio.

I read this really good contribution here on this forum:

Does lift equal weight in a climb?

So the g load factor is defined as: g=Lift / Weight and the gliding ratio= L/D

  1. In a descend flight: what does a g-load factor of 0.5 have for a meaning, when there is no thrust force.

L=Lift ; D=Drag ; W=Weight ; gamma=gliding angle

When gliding, the vertical force components (_v) have to balance the Weight, so: L_v + D_v = W with L_v=Lcos(gamma) and D_v=Dsin(gamma)

So:

Lcos(gamma)+Dsin(gamma) = G (1)

AND

g=L / G = 0.5 (2)

Insert (2) in(1) it comes to:

L/D=sin(gamma)/(2-cos(gamma))

a) Is this correct ?

b) When I want to have a gamma=10° this would mean that L/D=0.17 and for gamma=40° L/D=0.52087.

But on the other hand gamma has a relation ship L/D=1/tan(gamma) . So for gamma=40° it comes to L/D=1.19

I am really confused about that. Can someone help me?

Thank you all experts!

Lucas

$\endgroup$
4
  • 2
    $\begingroup$ If you are at .5 G you are not gliding, you are accelerating downwards quite rapidly! $\endgroup$ – Michael Hall Feb 16 at 20:55
  • $\begingroup$ Lucas, you mean a steady-state glide, right? Not just that the prop stopped turning? $\endgroup$ – Camille Goudeseune Feb 16 at 21:35
  • 1
    $\begingroup$ I think W=G? Seems you switched symbols halfway $\endgroup$ – Sanchises Feb 16 at 21:53
  • $\begingroup$ I’m not totally clear on what you are trying to accomplish here, but to expand on my earlier comment: If you are expecting some sort of steady state L/D glide output using .5 G you won’t get it. .5 G is not steady state, nor is it gliding. You are accelerating rapidly downwards. Falling in other words, at an ever increasing speed at just half the rate of free fall. Could this be the problem with your math? $\endgroup$ – Michael Hall Feb 17 at 17:09
1
$\begingroup$

I am wondering about the g-load and its relation to the so called gliding ratio. ...

So the g load factor is defined as: g=Lift / Weight

We'll adopt that definition for the purposes of this answer. Note that that is not the entire apparent g-load felt by the aircraft and pilot. It is just the component that is acting in what we might loosely call the "upward" direction in the aircraft's reference frame. More precisely, it is the component acting perpendicular to the flight path. Which is approximately the same as the component that would be registered by a conventional (mechanical) panel-mounted g-meter. No need to get too caught up in semantics here, as long as we clearly define what we are talking about.

(Another minor point would be to suggest that you settle on "g-load" or "load factor", rather than combining the terms into "g-load factor". Even though the two terms arguably have the same meaning.)

and the gliding ratio= L/D

Yes, that's certainly true

  1. In a descend flight: what does a g-load factor of 0.5 have for a meaning

If L/W is 0.5, the glide angle is (arccos) 0.5, or 60 degrees below horizontal. For more, see the related ASE answer What produces thrust along the line of flight in a glider?

L=Lift ; D=Drag ; W=Weight ; gamma=gliding angle

When gliding, the vertical force components (_v) have to balance the Weight, so: L_v + D_v = W with L_v=Lcos(gamma) and D_v=Dsin(gamma)

So:

Lcos(gamma)+Dsin(gamma) = G (1)

All true except the last-- it should be W, not G.

AND

g=L / G = 0.5 (2)

Same-- change "G" to "W" and it's correct

Insert (2) in(1) it comes to:

L/D=sin(gamma)/(2-cos(gamma))

a) Is this correct ?

I don't understand how you did this substitution. Can you edit the question to explain better? Anyway, it is not correct. In our example above where gamma is 60 degrees, when I plug that into your equation, I get an L/D of 0.57778, when it should be 0.5.

b) When I want to have a gamma=10° this would mean that L/D=0.17

From What produces thrust along the line of flight in a glider? we can see that D/L = tan (gamma), where gamma is the glide angle, so L/D = 1 / tan (gamma). If gamma is 10 degrees, L/D is 1 / tan (gamma) = 1 /.1763 = 5.672. The inverse of what you said.

and for gamma=40° L/D=0.52087.

No, the tangent of 40 degrees is 0.8391, and 1/.8391 = 1.1918. So that's the L/D associated with a 40-degree glide angle.

But on the other hand gamma has a relation ship L/D=1/tan(gamma) . So for gamma=40° it comes to L/D=1.19

Yes

I am really confused about that. Can someone help me?

I'm not sure I follow exactly where the confusion arises, but I hope that helps.

$\endgroup$
1
  • $\begingroup$ Some recent corrections made to math. $\endgroup$ – quiet flyer Feb 16 at 23:37

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.