Why cannot the airplane accelerate for ever in the sky?

[Skye]

To whatever fluid the object is travelling through. In the case of a plane, to the air.

[albertlee]

So, Skye, so what is the name of the energy of the air?

Skye, are u think about this equation: k.e. = 1/2mv^2?

[albertlee]

Any idea?

[Skye]

Yeah, that's just a guess.

 

[Edit]

Here's a history of aerodynamics that puts it into perspectivehttp://www.centennialofflight.gov/essay/Theories_of_Flight/early_aero/TH3.htm

[swansont]

This is how i think...........

The drag always equal to the the thrust according to Newton's first law' date=' therefore it is a constant speed......

The reason why the plane increases its speed is because that the plane exerts more force on the thrust, hence the drag increases, therefore the constant speed increases..........

 

Again, this is just how i think..................but i dont think it is just this simple........

 

Any suggestions?[/quote']

 

We've covered this before. The thrust is not always equal to the drag. If it were, a plane couldn't accelerate and take off, as there would never be a net force on it in the horizontal direction. You contradict yourself in the two statements about the forces always being equal and then having "more force on the thrust" and the "constant speed increases." If it's constant, it can't increase, by definition.

 

Drag is a property of a fluid interacting with a solid. Drop something (in air, in water, whatever fluid you want) there is a constant force on it due to gravity, but the drag increases until it reaches terminal speed, which is when the two forces end up cancelling. Never any thrust at all.

 

You aren't going to have success understanding this if you don't understand Newton's laws and basic definitions.

[albertlee]

So, Swansot, Do you also mean that the drag increases the force more faster than the thrust?

[swansont]

So, Swansot, Do you also mean that the drag increases the force more faster than the thrust?

 

The drag depends only on the speed of the object. If you increase your thrust, your speed will increase but then so will the drag, until you reach the speed where they are equal. Then the net force will be zero, and speed will be constant.

[albertlee]

So is the rate of drag increases faster than thrust? just this simple.....

[albertlee]

Any body?

[Kedas]

albertlee,

Why the desperate need to link drag with thrust.

I think all the information to understand it is in this thread and more.

If you don't understand it's because you don't want to let go of some wrong old assumptions. and don't try to oversimplify it.

[Dave]

I suggest we just leave this thread be now; it's more or less completely exhausted all avenues of discussion on this particular subject.

[Guest PYRONOVA]

Theoretically a plane, if made with a 0 fiction factor, would never stop accelerating. But do to gravity even if there was no friction, eternal acceleration is impossible. The plane would eventually hit the gravity barrier causing all all atom to spontaneusly combust. My type of fun but not to be tried by ppl 8 years or younger. The nucleus of the atom would lose its hold of its elctrons and as i see it 2 things can happen. 1) u go boom and have a nice day or 2) every thing loses its magnetic properties and the the effect of 0* Kalvin sets in. This would stop all progression and ud be in for one hell of power brake

[Guest beholdasun]

When there is no net force, there is no ACCELERATION. This means that when thrust equals drag, the plane stops ACCELERATING OR DECELERATING. Thus, if when the thrust became equal to the drag the plane was moving at 600mph, it would continue happily along at 600mph until the thrust or drag were again altered. If either were altered so that thrust < drag (deceleration) or drag < thrust (acceleration), the plane's velocity would again change (which changes the drag since drag is directly proportional to the square of velocity), until equilibrium was reached. To see why thrust and drag are independent of each other fundamentally, imagine an airplane is cruising at 600 mph and the thrust (engines) is suddenly cut to 0 Force. The drag at that instant would still be that drag of moving 600mph. The plane would then obviously decelerate to an eventual 0 mph if thrust were not return, since the only way to have 0 drag, and thus, equilibrium is to have 0 velocity. So, thrust will tend to increase the velocity of the plane, and it is the VELOCITY that is the all important factor in determining drag. Thrust just refers at how hard the engine is working at a given time.

 

-Mudder

[Guest SniperMax]

Ok I hope you are still interested.

 

The reason for the [math]v^2[/math] in the drag equation is as follows. The object travelling through the air has to transfer a part of its momentum to the air continuously. This is why it slows down, it loses momentum. The momentum is:

 

[math]p=mv[/math]

 

But the faster the object moves, the more air it faces in the same time, thus the mass of the amount of air the object encounters during a certain time is bigger.

 

The mass of the air it encounters during the time t is:

 

[math]m_{Air}=A*\rho*v*t[/math]

 

When we put this into the momentum equation, we get

 

[math]p=m*A*\rho*t*v^2[/math]

 

The force though, is the momentum divided by the time, thus the t in the above equation is cancelled out.

 

This is the reason why the drag force equation is [math]F_{Drag}=\frac{1}{2}*A*C_D*\rho*v^2[/math]

 

As for the terminal velocity problem:

 

Thrust has nothing to do with the velocity of the plane. Thrust exerts a constant force on the plane in the direction of flight. This thrust never changes (unless the throttle position changes). If there was no air resistance, the acceleration would be:

 

[math]a=\frac{F}{m}[/math]

 

for ever.

 

As there is air resistance, the acceleration is not constant, but decreasing with the velocity. The air resistance increases with speed squared and exerts a varying force in the opposite flight direction. The net acceleration decreases, first slowly, but more quickly when terminal velocity is reached. Net acceleration is given by:

 

[math]a_{Net}=\frac{F_{Thrust}-F_{Drag}}{m}[/math]

 

It is obvious that when both forces are equal, the net acceleration is 0.

 

To find the terminal velocity of an airplane, you have to set up this equation:

 

[math]F_{Thrust}=F_{Drag}[/math]

 

[math]F_{Thrust}=\frac{1}{2}*A*C_D*\rho*v^2[/math]

 

and solve it for v.

 

[math]v=\sqrt{\frac{2*F_{Thrust}}{A*C_D*\rho}}[/math]

 

Here you are. No other equations required. Try it with different values for [math]F_{Thrust}[/math],[math]A[/math],[math]C_D[/math] and [math]\rho[/math].

 

I hope this helps. Or else just ask

 

Edit: typos

[Kedas]

The mass of the air it encounters during the time t is:

 

[math]m_{Air}=A*\rho*v*t[/math]

 

When we put this into the momentum equation' date=' we get

 

[math']p=m*A*\rho*t*v^2[/math]

 

should be [math]p=A*\rho*t*v^2[/math] I assume.

but otherwise interesting to see it written out.

[Guest SniperMax]

Yes of course, it's [math]p=A*\rho*t*v^2[/math]. My bad.