2
$\begingroup$

Please note that the question is "Why", not "How".

For sure I'm getting this wrong, but I'm stuck and I don't know why! If a reverse thrust is applied during an aircraft landing wouldn't the total amount of work necessary to pull the air inside the turbine (hence pushing the aircraft forward) be canceled by the amount of work used to push the air forward (i.e. pushing the aircraft backwards)? I mean, why does it work? Why does the net work push the aircraft backwards and help to stop the plane?

$\endgroup$
  • $\begingroup$ What distinction do you see between forward and reverse thrust? Why "should" forward thrust work, while reverse not? $\endgroup$ – Wasabi Jun 30 '16 at 23:39
  • $\begingroup$ Forward thrust works by pushing air mass backwards using a fan resulting in a force that pushes the aircraft forward. Whereas reverse thrust also pushes the air mass backwards using the same fan, which creates the same force as the forward thrust. The difference here is that afterwards it reverses the air mass velocity direction to create a force that brakes the aircraft. But wouldn't this force be canceled by the forward force previously created by the fan? $\endgroup$ – PDuarte Jun 30 '16 at 23:57
  • $\begingroup$ At a simple level, if pushing a mass of air backwards using the fan generates a force of F, then turning that air round so it moves forwards generates a force bigger than F in the opposite direction. The forces would only "cancel out" if the air flow from the fan was stopped, but not reversed. $\endgroup$ – alephzero Jul 1 '16 at 0:42
  • $\begingroup$ @alephzero interesting point $\endgroup$ – PDuarte Jul 1 '16 at 1:01
  • 2
    $\begingroup$ @CarlWitthoft: It's a valid comment, but the edit was inappropriate. We all understand the distinction that the OP is making. $\endgroup$ – Dave Tweed Jul 1 '16 at 14:21
1
$\begingroup$

Imagine that the engine is a person throwing a 1 kilogram ball backward at a speed of 1 meter per second. The reverse thrust system can be modeled as a wall attached to the plane that bounces the ball so it goes forward.

The forward impulse caused by each ball is, obviously. 1 kg m/s. Then when it bounces, its velocity changes from 1 m/s backward to 1 m/s forward, for a net change of 2 m/s, applying a reverse impulse to the plane of 2 kg m/s. So the net total impulse is 1 kg m/s backwards.

| improve this answer | |
$\endgroup$
  • $\begingroup$ So, are you saying that the Coyote's fan-thrusted sail actually works? rebrn.com/re/… $\endgroup$ – PDuarte Nov 24 '16 at 10:55
  • $\begingroup$ @PDuarte Yes - the Mythbusters even did it and showed that it worked. See physics.stackexchange.com/questions/135548/… - it's less efficient, because unlike in the model you don't get all the impulse with no losses, but the principle works. $\endgroup$ – Random832 Nov 24 '16 at 20:24
  • $\begingroup$ Great! Thanks for the simple and clear explanation! $\endgroup$ – PDuarte Nov 24 '16 at 20:27
5
$\begingroup$

It works because burning the fuel inside the engine adds a lot of energy to the exhaust air that more than compensates for the work done sucking in and compressing the intake air.

If you shut off the fuel supply, it won't work — just as you suspect.

| improve this answer | |
$\endgroup$
  • $\begingroup$ The exhaust air is not used in reverse thrust, it is pushed to the back of the aircraft. The amount of air that brakes the aircraft is the one that usually passes through the turbine bypass section. $\endgroup$ – PDuarte Jul 1 '16 at 11:24
  • 1
    $\begingroup$ @PDuarte: That's just an implementation detail. It doesn't negate the fact that all air coming out of the engine has more energy (momentum) than it had going in. $\endgroup$ – Dave Tweed Jul 1 '16 at 11:32
  • $\begingroup$ @PDuarte and more thrust comes from the bypass air than from the exhaust. so diverting all bypass will result in a netto negative forwards thrust. $\endgroup$ – ratchet freak Jul 1 '16 at 12:26
  • 1
    $\begingroup$ @CarlWitthoft: That isn't at all relevant here. Auto-rotation in a helicopter relies on the the fact that a significant amount of kinectic energy is "stored" in the rotating mass of the blades themselves, and there's a significant amount of potential energy in the mass of the helicopter due to its altitude. You use the potential energy to keep the blades turning as you descend at a controlled rate, and then use the kinectic energy to produce extra lift at the end and achieve a soft landing. In terms of the aerodynamics, it's no different from landing a fixed-wing glider. $\endgroup$ – Dave Tweed Jul 1 '16 at 13:34
  • 2
    $\begingroup$ @CarlWitthoft: No, it's completely different. Thrust reversal does not rely on stored energy (gravity or kinectic) at all. $\endgroup$ – Dave Tweed Jul 1 '16 at 13:41

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.