Why cannot the airplane accelerate for ever in the sky?

[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?......