2
$\begingroup$

I have a puzzle which asks me to consider a (engineless) glider descending at a constant rate and flying at a constant forward velocity (in still air). The question asks how to derive sink rate, $v_s = \frac {Av^{-2}} {mg} $(vertical velocity), assuming that the drag acts horizontally (because the descent angle is small).

Question 1:

My best stab at a force diagram (below) indicates that a constant forward speed is impossible! By the way I'm only considering the drag in order to produce lift (induced drag). Force and acceleration diagram

How could forward velocity be constant? Is it a reasonable approximation? Are there any other forces on the glider?

Question 2 - Trying to derive sink rate,$v_s$

Leaving that aside and assuming speed v is constant, I can see drag, $D = Av^{-2}$ is the rate of energy loss (with a small $v_s$ so that $v \approx \frac {dx} {dt}$ where x is position) because $$ Power, P = \frac{d} {dt} \int{D dx} = \frac{d} {dt} \int{Av^{-2} dx} = Av^{-2} \frac {dx} {dt} \approx Av^{-1}$$

The power developed by the weight of the object is the rate of change of G.P.E. $ = mgv_s$ $$\Rightarrow mgv_s \approx Av^{-1} \Rightarrow v_s \approx \frac {Av^{-1}} {mg}$$

Where have I gone wrong?

$\endgroup$
9
  • 2
    $\begingroup$ Your free body diagram is for a plane that is flying level. If the plane's flight surfaces are pointed slightly downwards (as it is descending), the lift will have a horizontal forward component, which can balance the drag force. $\endgroup$
    – Brionius
    Oct 25 '15 at 22:57
  • $\begingroup$ In addition to what Brionius said, forget about 'power' for a minute. Just set up equations of motion in $x$ (forward) and $y$ (vertical) directions. $\endgroup$
    – Gert
    Oct 25 '15 at 23:18
  • $\begingroup$ By the way, where do you get $D$ as proportional to $v^{-2}$? That would say that if you double the speed, the drag would decrease by a factor of 4. I'm sure that's not what you meant. $\endgroup$ Oct 26 '15 at 19:08
  • 1
    $\begingroup$ @Arty: Same question. Why would doubling $v$ divide $D$ by four? I think that negative sign in $D = Av^{-2}$ is wrong, and that ripples through all your math. $\endgroup$ Oct 28 '15 at 12:26
  • 2
    $\begingroup$ @enbinzheng I think what you meant to ask was "why do objects move both downwards AND horizontally on a ramp if gravity has no horizontal component". That's a good question, but should really be a separate question (and probably already is if you look around). In short, gravity has no horizontal component, but the normal force from the ramp surface does. In combination, they make objects move both downwards and horizontally. Any basic mechanics textbook will have a treatment of "inclined planes" (fancy word for ramp). $\endgroup$
    – Brionius
    Mar 14 '20 at 23:15
3
$\begingroup$

As the comments say, you need some horizontal component of lift to overcome drag. And your drag is more than only the induced component - the friction losses are of the same magnitude.

Forces acting on glider

Forces acting on glider (own work)

Note that I shifted the vectors such that it can be shown that the force triangle is closed, and the forces are in equilibrium. $v_{\infty}$ is the speed of air relative to the glider caused by the glider's motion. The inclination is needed to tilt the lift vector (which per definition is orthogonal to the airspeed vector) forward, resulting in a horizontal component which equals drag.

For a very crude first estimate of minimum sink speed of a glider, just take 100 and subtract the best glide ratio. The result is very close to its minimum sink expressed as centimeters per second. Works only for gliders!

On a more serious note, in order to get to a useful solution for the sink speed $v_z$, you start with potential energy loss over time: $$\frac{dE_{pot}}{dt} = m\cdot g \cdot\frac{dh}{dt} \approx W\cdot v$$

With the equilibrium of forces ($L$ = lift, $D$ = drag) $$L = -m\cdot g\cdot cos\gamma \;\text{and}\; D = -L\cdot tan\gamma $$

you can write $$m\cdot g \cdot\frac{dh}{dt} = m\cdot g \cdot cos\gamma\cdot v\cdot tan\gamma \;\text{and}\; tan\gamma = -\frac{D}{L} = -\frac{c_D}{c_L}$$ $$v_z = -v\cdot sin\gamma = v\cdot\frac{c_D}{\sqrt{c_L^2+c_D^2}} \approx v\cdot\frac{c_D}{c_L}$$

Next you need some approximation for $c_D$: $$c_D = c_{D0}+\frac{c_L^2}{\pi\cdot AR\cdot\epsilon}$$

so you can write $$v_z \approx v\cdot\left(\frac{c_{D0}}{c_L}+\frac{c_L}{\pi\cdot AR \cdot \epsilon} \right)$$

Nomenclature:
$g\;\;\;\;\;\;$Gravitational acceleration
$v\;\;\;\;\;\;$Flight speed
$c_{D}\;\;\;\;$Drag coefficient
$c_{D0}\;\;\;$Zero lift drag coefficient
$c_{L}\;\;\;\;$Lift coefficient
$\gamma\;\;\;\;\;\;$Flight path angle, positive when pointing up from the horizontal
$\pi\;\;\;\;\;\;$3.14159…
$AR\;\;\;$Wing aspect ratio (span squared over area)
$\epsilon\;\;\;\;\;\;\,$Oswald factor, normally between 0.7 and 1. For gliders 0.98 is typical

Yes, sink speed is positive downwards.

$\endgroup$
2
  • $\begingroup$ The speed of the glider is given by gravity? $\endgroup$
    – enbin
    Mar 11 '20 at 6:06
  • $\begingroup$ @enbinzheng: The potential energy of the glider grows linearly with gravitational acceleration. Also, more lift must be produced in proportion to gravity. While a range of speeds can be selected with elevator deflection, the minimum and optimum glide speeds increase with gravity. $\endgroup$ Mar 11 '20 at 8:10
2
$\begingroup$

@PeterKämpf is right. Let me just try what might be a simpler way to think about it.

The plane has a certain speed $v$ and a certain drag $D$, so the energy loss due to drag is $vD$.

The plane has a certain weight $mg$, and is descending at sink rate $v_s$, so it is changing potential energy into kinetic energy at a rate $mgv_s$.

These two energies have to be equal. It's as simple as that.

By the way, at a given speed, induced drag $D$ is just a certain fraction of lift force, so it is a certain fraction of $mg$. Most good gliders have a glide ratio of about 30:1, so the drag should be about $mg/30$, and that includes parasitic as well as induced.

$\endgroup$
0
-2
$\begingroup$

enter image description here

As shown, $AB$ is the wing of the glider. The yellow arrow is the speed direction $V$ of the glider. $G$ is the weight of the glider. $G$ has two components, one is a component $G_t$ parallel to the velocity direction, and one is a component $G_n$ perpendicular to the velocity direction. There are also two components of the air force applied to the wing, one is the component $D$ parallel to the speed direction, and the other is the component $L$ perpendicular to the speed direction.

If the speed $V$ of the glider is low, then $G_t> D$ and $G_n> L$. Thus, under the thrust of $G_t$, the speed $V$ of the glider increases. As the speed $V$ increases, $D$ will increase, and $L$ will increase. Finally, $G_t = D$, $L = G_n$, and the glider will glide along the yellow line at a uniform speed.

From this analysis of the glider, we can see that the flight of the glider is a thrust flight. Without gravity as the thrust, the glider cannot fly and cannot generate lift.

enter image description here

Why does an object move down an inclined plane? Because the component of gravity is parallel to the inclined plane. It pushes the object down an inclined plane. Can't "lift" push an object down an inclined plane here? It can't, because lift is perpendicular to the inclined plane, and it has no component in the parallel direction of the inclined plane. Here, gravity is the force that pushes an object down an inclined plane. Because gravity has components in the direction parallel to the inclined plane. The same is true of gliders. The glider is also driven downward by gravity.

$\endgroup$
2
  • 1
    $\begingroup$ Comments are not for extended discussion; this conversation has been moved to chat. $\endgroup$
    – ACuriousMind
    Mar 14 '20 at 11:28
  • 2
    $\begingroup$ This answer uses incorrect terminology. Gravity is not considered to be “thrust” by the scientific community. I recommend that it be revised to remove the non-standard personal misuse of standard terms. I have discussed this with the author and he/she has shown no willingness to fix it. $\endgroup$
    – Dale
    Mar 14 '20 at 12:04

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