Why cannot the airplane accelerate for ever in the sky?

[albertlee]

These 2 possibilities is what bi can come up:

1) According to Newton's first law, if the force of the airplance is equaled to the force of the air, it goes in a straight line.... which means the air force (I mean not the millitary stuff :-) ) also increases as the the force of the airplance increases.... but how can the airplane accelerate in the bigginning...?

2) According to Newton's third law, when the force exirting on the air, the air exerts back, but therefore the plane cannot even move, since the plane also exerting backwards and the air exerting back.......

Anyway, ithink the reason for this is all about pressure of both (air + plane) and its transformation of force resulting the rate of air resistance........

Any way, back to my topic, please explain why..And tell me some specific ways of calculation to it

Thx

Albert

[blike]

I'm not sure what you're asking?

[Pinch Paxton]

Push your hand down on bath water slowly, and your hand easily moves through it. The faster you slam your hand through the water the more resistance you feel. I think it is like a concertina effect, the water needs time to push the next layer out of the way. Same with air I guess.

Pincho.

[Sayonara³]

I think something along these lines is what you need to be thinking about.

The weight is the resultant of gravity on the plane's mass. Or something.

[ski_power]

Firstly, at low speeds the plane is able to accelerate because there is very less air resistance(How are you able to sprint?, Same type). But as Pincho said, the faster you go, more the resistance(Drag). Hence, you are able to accelerate to normal speeds of around 0.7-1.0 Mach. Also, the aerodynamics play an important factor over here. More streamlined is the design, lesser is the air-resistance, reater is the possibility to accelerate to higher speeds.
Also, the Newton laws thing, it's crazy. The air doesn't offer any reaction(atleast that's what I think, correct me if wrong), because you aren't applying any action on the air. You're pushing the plane forwards, not the air backwards.

[albertlee]

Also, the Newton laws thing, it's crazy. The air doesn't offer any reaction(atleast that's what I think, correct me if wrong), because you aren't applying any action on the air. You're pushing the plane forwards, not the air backwards.
From skipower

When u push the plane forwards, is'nt it also pushing air forwards? Then if so, does Newton's law still apply,here? if not, how and why?

On the other hand, back to my first message, Can anybody give some mathematical calculations towards this topic? because all i got is just concept....but to me, i think we always measure in Physics........

Will Apreciate for the responds of all my questions

Thx

[Skye]

Two seperate forces, each with their own 'equal and opposite reaction'.

1) Thrust. Force of the propeller on the air, pushing BACKWARDS, with an equal and opposite force FORWARDS.

2) Drag. Force of the whole plane travelling through the air, pushing FORWARDS, with an equal and opposite force BACKWARDS.

Thrust and drag aren't equal and opposite forces of each other, they are only related in that thrust accelerates the plane which increases it's velocity, which then means that it shoves more air out of the way leading to more drag.

[albertlee]

Thx for the reply froM skye, It is very useful.

On the other hand, can any one explain to Me More clearly about the difference of Thrust and Drag about their forces and relationship?

Say it More clearly, how are thrust and drag not equaled?

very apreciate for the responds

[dave]

Like skye said, thrust is the force pushing the plane through the air, which comes from the propeller or jet engine on the plane. Drag is effectively the air resistance from the plane travelling through the air.

(this is quite a basic description)

[ski_power]

Yeah, but still we don't know how thrust and drag are equalled(if they actually are, or not)?

[albertlee]

I really wonder why drag force is less than the thrust? just as simple as that......
but it really takes a big time on thinking.........

Any help?

[Skye]

The thrust depends on the power of the engine, more or less under control of the pilot by controlling the throttle and thus the revs per minute. The drag depends on the speed of the plane. So if you have a plane that is stationary, then switch the engine on to a certain rpm, it will have a certain amount of thrust, and no drag. The plane will then accelerate, and it's drag will increase gradually, untill it reaches a speed at which drag equals thrust. It will then continue at a constant speed. If you increase the rpm some more, the thrust will again be higher than the drag, and the plane will accelerate till drag equals thrust again. If you reduce the rpm to it's original value, drag will now be greater than thrust and the plane will deaccelerate back to it's original speed.

[albertlee]

Sorry, still confusing.......
It is almost telling the laws in more useful in Physics.....
Can anybody tell me why not the thrust equalled to the drag? since according to the Newton's third law..

Any help?

[swansont]

Sorry, still confusing.......
It is almost telling the laws in more useful in Physics.....
Can anybody tell me why not the thrust equalled to the drag? since according to the Newton's third law..

Any help?

Drag isn't the reaction force to the thrust. Thrust action/reaction is the force the propeller exerts on the air passing through it, and the force the air exerts on the propeller. Drag action/reaction is the force the plane exerts on the air by moving through it, and the force the air exerts on the whole plane.

You can have thrust with no drag (e.g. plane bolted down with the prop running) and drag with no thrust (e.g. a glider).

[kpcoder]

swansont is correct. Drag is not equal to thrust because these 2 forces are independant. You were mistaking Drag and Thrust as action/reaction which they are not. The correction action reaction pairs are as follows.

1) action: Engine Thrusts gas in a south direction
equal and opposite reaction: Jet is accelerated North.

2) Action: air pushed out of the way by the jet (drag)
equal and opposite reaction: Jet is accelerated South (meaning it slows down)

This is why Drag does not equal thrust, they are not an action/reaction pair, they are independant.

[albertlee]

Ok....So, can anybody tell me the equation of drag? with also some explanations?

Apreciate

[kpcoder]

Drag = .5 * Coefficientdrag * densityair * Velocity^2 * Area

Coefficientdrag = coefficient of drag. This value accounts for the shape of the body, its orientation/inclination, air viscosity, and compressibility.
Click here for examples of drag coefficients for various shapes (http://wright.nasa.gov/airplane/shaped.html)

[albertlee]

As a conclusion..........Why drag and thrust are independable? since the stronger the thrust, the stronger the drag........therefore i think they both apply Newton's third law in some ways..........

Any helps? or suggestions?

Apreciate

[YT2095]

when you think about it logicaly,it has to be more or less entirely about Air friction/resistance.

imagine that plane could fly in space, it`s a special type of plane with it`s own oxygen supply onboard (in otherwords a rocket).
this rocket can burn for months! (this is only hypothetical for the point), it has a thrust of say 10 newtons (only a small single thruster), the rocket will continue to accelerate until the fuel burns out, still using that 10 newton thrust, it doesn`t matter how much the mass of the craft it, it will still keep acceletating, faster and faster :)

a "Normal" planes engine in Air, set to a constant thrust, will reach a maximum speed then stay there?

so break the problem down. what is the difference between the 2 scenarios (factoring out the engine types as that`s only to present thrust) ?

maybe that will help elucidate the problem a little more clearly? :)

[Kedas]

I didn't read all of above but:
air resistance is determent by speed and shape of the airplain and surrounding air.
So completely independent of the trust.
and action = reaction meaning the trust will be equal to the sum of F1=m.a and the resistence of the air F2
meaning if speed=constant then trust=resistance/drag
(m=mass of the airplain, a=acceleration of the )

[albertlee]

Ok.....Let me sort it out..........So....., is it possible to have drag without thrust? and secondly, if so, where does the force come from to have the drag? since drag is a mechanical force......

Any help?

[YT2095]

no, it`s not possible to have ANY resistance, be it from air or in an electric circuit without force or energy being applied to "recognise" it :)

[albertlee]

Yes, that's it, YT, if the drag needs thrust, which means this apply to Newton's third law....... Simply, what i wonder is why not Thrust is the action, and Drag is the reaction in one way or another? Sorry for being confused so long, but what I really confuse is always about the Newton's third law as role in "force and flight"......

Any help?

[Kedas]

Ok.....Let me sort it out..........So....., is it possible to have drag without thrust? and secondly, if so, where does the force come from to have the drag? since drag is a mechanical force......
Any help?
Yes, that is completely possible.
The force comes from the air that doesn't want to move easily.
If trust = 0 then resistance = m.a
a bullet that is horizontally flying through the air has no trust and still a lot of drag until it stops. (gravity force ignored)
(keep in mind a falling object has a vertical 'trust' of F=m.g)

maybe your problem is that you don't understand F=m.a very well?

no, it`s not possible to have ANY resistance, be it from air or in an electric circuit without force or energy being applied to "recognise" it :)

Correct but you don't need trust to be able to talk about energy being transfered.
so it's "yes,....."

[swansont]

Yes, that's it, YT, if the drag needs thrust, which means this apply to Newton's third law....... Simply, what i wonder is why not Thrust is the action, and Drag is the reaction in one way or another? Sorry for being confused so long, but what I really confuse is always about the Newton's third law as role in "force and flight"......

Any help?

Newton's third law states that every force has a reaction. That is, FA-B = FB-A. A exerts the same force on B that B exerts on A.

But thrust and drag are both exerted on the plane - the same object. The reaction force to either has to be a force that the plane exerts on the air.

[albertlee]

ok......Now I get 50% of my question.......The Drag = .5 * Coefficientdrag * densityair * Velocity^2 * Area. This also means the force exert by the plane to the air in front, right?

I have thought about this equation for some time, and I finally realize that the "velocity^2" is the key of relationship between Thrust and Drag, right?
But I wonder why it is written as square of velocity? since it means not the acceleration............According to my conjecture, does the drag equation is after simplified? For example, K.E. = 1/2mv^2, but before simplified, it is = (m*v/t)(1/2v*t).....

Secondly, what does the term "area" in the drag equation mean?

Many thx for the responds of all my problems

[YT2095]

maybe a search for the deffinition of Reynolds (sp) numbers will help, maybe a look at laminar flow also.
my airoplane knowledge is limited to say the least, but I know these 2 are taken into account regarding drag and air molecules (laminar flow also applies to water).
these might help?

[albertlee]

Thx YT for relevant information..........But simply i just want to know what is the "meaning" of squared velocity....or in another saying, why is it squared velovity? and secondly, The term "area" in the equation is for what? eg, area of the airplane, air, or etc...............

Any help?

[YT2095]

that`s where it leave me cold, sorry, that kinda stuff is not my area (excuse the pun). some of the others will know though :)

[wolfson]

v^2 = u^2 + 2as

Sq root of U^2 + 2as = v

Where u =inital velocity, a = acceleration and s = displacement.

[albertlee]

Ok......, thx anyway to wolfson, but although i see the expansion of v^2, i still cannot realize the "concept" behind it................since the term 'v" means the final velovity right? And the final velocity = acceleration*time + initial velovity, where initial velovity is almost always 0, because in 0 second, there is no velocity......which means:
v^2 = (at)^2.............

What i really need is the "concept" of the squared final velocity..........
Secondly, I was also asking what does the term "area" refer to?

Please help.....thx

[Skye]

The area is the cross sectional area of the object, in the direction it is travelling in. For a plane this would be the area you see when you look at it front on.

[albertlee]

Ok...... I might think it is very hard to explain the drag equation very completely in a concept..........It must be at a very high level in physics.......Since i am just too curious to know, so i ask..........
I think no one here would know what is the concept behind the velocity^2.......Anyway, thx to every one who answers my questions..........

[dave]

When you're dealing with constant acceleration, there's one very important equation, and that is obtained from a velocity-time diagram. As you may know, the displacement travelled can be worked out from one of these diagrams by calculating the area under the graph, so for the simple case of constant acceleration, we get a trapezoid shape (trapesium). Assuming we start off at an initial speed u and end up at a final velocity v in a time t, then the area which is the same as the displacement, s = t*(v+u)/2. You also know that a = (v-u)/t.

Now by using these two equations, you can derive the rest of the uvast equations - by re-arranging the second one, we have v = u + at. Then by substituting for v into the first equation, we have s = ut + 0.5*a*t^2. To get the one we want, we need to eliminate the variable t (time), so by rearranging the second equation, we get that t = (v-u)/a, so by substituting into the first equation, we have s = (v-u)(v+u)/(2s), which implies that v^2 - u^2 = 2as, and hence v^2 = u^2 + 2as as required.

Basically, all I'm trying to say that the only one that has a direct link to some physical interpretation is our original first and second equations. The other equations are literally the same equations, but with different variables missing each time. They only aid us in obtaining numerical solutions to problems with constant accelleration. Hope this helps.

[albertlee]

Thx Dave.....Now I know the calculation of v^2, but I am still confused why the drag equation use this term?, since it is just a "part" of equation of displacement, which is quite meaningless.......

Any help?

[wolfson]

look here:

http://www.lerc.nasa.gov/WWW/K-12/airplane/dragco.html

[albertlee]

Thx wolfson, but......sorry, it does only tell about what is the square of velocity as a concept in drag equation .......But I want to know about the significance of v^2, since you cannot feel it such as force, acceleration, etc.

Or in another saying, what is the expansion of drag equation? since it must be after cancelation which finally shows an odd term like v^2........

Any ideas?

[dave]

I'd help, but I don't really know what you're trying to ask.

[albertlee]

Ok Let me say it more clearly: Why the drag equation use the term "v^2" from the simultaneous equations of constant acceleration?Since, The term has no physical meaning......

Therefore, the drag equation I knew must be after the cancelation....Hence we get the odd term "v^2" which has no physical meaning.............

So What is the drag equation before cancelation?

many thanks for furthur responds.

[dave]

Okay, I don't exactly know the answer to your question. I don't really know the significance of using v^2 instead of v in the equation, and perhaps you were using a different form of the drag equation.

[swansont]

The drag equation uses v2 because the force depends on the square of the speed. If you go twice as fast, the drag is 4 times bigger. That's what has been observed to hold. You don't use v because it doesn't depend linearly on v. There is probably a theory that shows why the term is v2, but if anyone here knew it they'd probably have posted it by now.

If I had to guess, I'd say that it's a combination of imparting momentum to air molecules by collision, which depends on the relative speed, and trying to compress/do work on the air (or whatever fluid) which would also depend on the relative speed. So you end up with a combination of the two effects, which is nonlinear and thus depends on v2. Again, this is a guess.

[dave]

I can't think of a direct reason why it would be v2 to be totally honest (apart from the point you made about it being an observed quantity). After all, Kepler did a similar thing when it came to planetary motion.

[albertlee]

Is it really sure the term "v^2" does come from v^2 = u^2 + 2as in the drag equation? since v^2 means initial velocity squared and 2 times acceleration times distance.......which just meaning nothing..............

Any ideas?

[albertlee]

I think the "v^2" must come up with a different way of calculation......

[dave]

I don't think it comes from the uvast equations.

[albertlee]

So......Any suggestions?

[dave]

The only reason I can think of is what swansont said above; the v^2 was obtained from observational methods.

[albertlee]

This is how i think...........
The drag always equal to the the thrust according to Newton's first law, 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?

[Skye]

My guess is that the equation is basically modelling the transfer of kinetic energy.

[albertlee]

Skye, transfer of kinetic energy to what?......

[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, 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?

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.

[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

[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

[SniperMax]

Ok I hope you are still interested.

The reason for the v^2 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:

p=mv

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:

m_{Air}=A*\rho*v*t

When we put this into the momentum equation, we get

p=m*A*\rho*t*v^2

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 F_{Drag}=\frac{1}{2}*A*C_D*\rho*v^2

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:

a=\frac{F}{m}

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:

a_{Net}=\frac{F_{Thrust}-F_{Drag}}{m}

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:

F_{Thrust}=F_{Drag}

F_{Thrust}=\frac{1}{2}*A*C_D*\rho*v^2

and solve it for v.

v=\sqrt{\frac{2*F_{Thrust}}{A*C_D*\rho}}

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

I hope this helps. Or else just ask

Edit: typos

[Kedas]

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

m_{Air}=A*\rho*v*t

When we put this into the momentum equation, we get

p=m*A*\rho*t*v^2


should be p=A*\rho*t*v^2 I assume.
but otherwise interesting to see it written out.

[SniperMax]

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