Vmd implies equal form and induced drag

[Conan the Vulgarian]

In the Bristol GS manuals I have here, it says that at Vmd, the induced and zero-lift drag are equal. It says this as a generality, as though it is always true, but never derives it anywhere that I can see - can someone explain if this is true and if so, why?

[BEagle]

Think of Drag as the sum of 'profile' and 'lift dependent' drag. Or whatever the latest terms are.

'Profile' is proportional to the square of velocity (A x V squared), whereas 'lift dependent' is inversely proportional to the square of velocity (B / V squared). So total drag has these 2 components.

Differentiate the expression and equate to zero and you will find that, at V for min. drag, these 2 components are equal:

http://i14.photobucket.com/albums/a341/nw969/drag.jpg

[sycamore]

I thought BEags was going to show the total drag curves diagram,or which you may well have,otherwise `google around and you will find them . they relate to straight and level ,unaccelerated flight only,as any turning will affect both.

[BEagle]

Graphs can be made to show anything - simple proof by diferential calculus is more convincing.

Obviously the expression assumes straight and level unaccelerated flight.

My calculus is a bit rusty, but that's the proof 2 of us gave to the bumbling ignoramus who attempted to teach us aerodynamics when I joined the RAF 37 years ago - we both had a BSc in Aeronautical Engineering which helped.

[Keith.Williams.]

Conan,

If you are familiar with differential calculus, then the above explanation will have convinced you that profile drag is equal to induced drag at Vmd. But if you are not familiar with differential calculus, then you will be none the wiser. If this is the case, it might help to start with a less mathematical approach.

You are probably happy with the facts that:

1. Profile drag increases as airspeed increases.
2. Induced drag decreases as airspeed increases.
3. Total drag is equal to Profile Drag + Induced Drag.
4. The total drag curve looks like a bucket shape.

Looking at the total drag curve we can see that it has a steep downward slope at the low speed end, a steep upward slope at the high speed end, and is flat where its value is lowest at Vmd, somewhere between the two ends. The important thing to note is that total drag is minimum at the speed at which the slope of the curve is zero.

Total drag = profile drag + induced drag, so the slope of the total drag curve is the sum of the slopes of the profile drag and the induced drag curves.

The total drag curve is zero when the upward slope of the profile drag curve is exactly equal to the downward slope of the induced drag curve. This occurs at the bottom of the total drag curve, where the speed is Vmd.

In differential calculus the process of differentiation provides a means of finding the slopes of these curves.

In BEagle’s explanation:

The first line states that Total Drag = Profile Drag + Induced Drag

The second and third lines derive the conditions in which the slope of the total drag curve is zero. Remember this is at Vmd at the bottom of the total drag curve

The subsequent lines develop this to show that at Vmd Profile Drag is equal to Induced Drag.

[Conan the Vulgarian]

Thanks gents. The differential solution is a much more straightforward explanation than I had expected - I find it a slightly nonintuitive result, but there it is in black and white. Perhaps if I think on it a little more it will seem more intuitive for me.

Conan

[John Alcock]

I agree with all the Beagle has written here, but by simply putting the differential equal to zero, doesn't his proof also imply that Induced and Profile drag are equal to each other at the maximum drag speed also?

[BEagle]

It's blindingly obvious, by inspection, that the expression refers to a minimum!

However, for pedants, consider the second derivative:

http://i14.photobucket.com/albums/a341/nw969/Internet/drag2.jpg

:hmm:

[BillieBob]

simple proof by diferential calculus is more convincingNot to those who have managed to complete a perfectly satisfactory career in aviation without ever having learned differential calculus, to whom it is entirely unconvincing.

[BEagle]

Well, if you don't understand basic calculus, you'll just have to believe those who do, I guess.

Why do you consider such a proof 'entirely unconvincing'?

I think we studied calculus at the age of 14 or 15 when I was at school - that and co-ordinate geometry were the first parts of the Additional Mathematics 'O' level syllabus.

[Keith.Williams.]

BEagle,

No explanation can ever be convincing if the audience do not understand it.

The reason for my earlier post is the fact that the vast majority, and I really do mean the vast majority, of current JAR ATPL students in the UK know absolutely nothing about calculus. Many of them have great difficulty in dealing with what you and I would regard as basic arithmetic.

To understand how they would have viewed your initial explanation, we can use your statement;

I think we studied calculus at the age of 14 or 15 when I was at school - that and co-ordinate geometry were the first parts of the Additional Mathematics 'O' level syllabus.

Just cast your mind back to when you were 13 3/4 (before you first met calculus). Now imagine how convincing any explanation using differentiation would have been to you then. You may have been intrigued, you may have been impressed, but I doubt if you would have been convinced. A more probable response from many current ATPL students would be "Oh Bollox, this maths is all far too difficult, I just want to be an airline pilot".

[BEagle]

You mean the "Want it now, don't care how" generation.....

I get very annoyed at the current generation of thickos who view simple mathematic proof as some sort of inherent challenge to their very existence. The fact that they were too bone idle to learn mathematics shouldn't mean that they have a right to decry those who did actually work harder at school.....

[Conan the Vulgarian]

At the risk of going a mile off topic here, I wonder don't any of this "vast majority" actually want to understand what they are studying? I mean, we're not talking subtleties here - surely people don't just read stuff like this in these manuals and take it at face value ??? Have people no curiosity whatsoever? (or indeed mistrust of unproven assertions in manuals likely to have been written/proof-read/printed by the lowest bidder)

I suppose I really ought to be old and crusty enough by now to presume they don't teach anything at school any more.

Words fail me.

[Alex Whittingham]

Of course it is very fashionable to smite your forehead and mutter that education ain't what it used to be but, if you are providing training material, there are two problems with the concept of trying to offer a proof of every statement you make. The first is that the proofs often become so complex that they obscure the simple idea you are trying to get across, the second is that what we often regard as 'truths' on inspection turn out only to be approximations that hold good in certain sets of circumstances.

Our policy is to avoid complex explanations when they are not required by the syllabus, but always be ready to justify a statement with a full proof if a student is interested. As GB might have said, I agree with Keith.

[wheelbrace]

I get very annoyed at the current generation of thickos who view simple mathematic proof as some sort of inherent challenge to their very existence. The fact that they were too bone idle to learn mathematics shouldn't mean that they have a right to decry those who did actually work harder at school.....

This is a bit harsh, maybe?

BEagle - it would seem that you were predisposed to understanding these functions from the womb, and were 'streamed' in the mathematical sense to then go on to design the wonderful flying machines in which we now indulge our innermost freedom fantasies.

I, for one, prefer to look at the chart and trust that august societies have put together the graphs to inform, not deceive! Do you consult PANSOPS before any approach to check the veracity of the DA/H?

It's blindingly obvious, by inspection, that the expression refers to a minimum!


How so? Once again, I prefer the graphical representation. Then again, I am too bone idle and not prepared to learn mathematics.

Wheelbrace, 2(i) BSc Computing Science and Information Systems (Hons).

:8

[WELCO]

With all the respect to the instructors posting in here, It's just a very simple concept that has been explained in a bit indirect puzzling way for starters.

Zero-lift drag is virtually the same thing as parasite drag i.e the drag produced in straight and level flight. Once the the angle of attack is increased, lift is produced along side with the induced drag (drag due to lift), while "zero-lift" drag is decreasing.

Vmd, at the Drag vs TAS curve, is the speed corresponding to the point were induced drag is equal to parasite (zero-lift) drag. So basically "Zero-lift" and "Parasite" drag are only 2 different terms describing the same thing, and that's it!

I assume that your reference is mentioning the term "Zero-lift Drag" just with the intention to get you thinking more thoroughly.


Good luck