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-   -   Ace the Technical interview by G.Bristow (https://www.pprune.org/tech-log/208011-ace-technical-interview-g-bristow.html)

1pudding1 23rd May 2006 10:23

Lift does equal Cl*0.5*rho*V^2*S, where S is the wing area. :hmm:

Wizofoz 23rd May 2006 10:29

It isn't necessary to include wing area or any other factor into the formula as CL is just that-a co-efficient.

If one wing produces twice the lift of another wing at the same speed and air pressure, it has twice the CL. Could be because of wing area, aerofoil section, planform, angle of attack etc. etc.

In any case you multiply, not add, and "Wing span area" is still a nonsense term.

Mercenary Pilot 23rd May 2006 10:30

Agreed it needs a a revision but i think its a good book and gives simple straight to the point answers for most stuff. Good for refreshing the memory without having to dust off the old textbooks IMHO:ok:

Jetavia 23rd May 2006 10:38

The book is published through Mcgraw-Hill any errors should be brought to the attention of their editorial contacts, which for aviation related materials is Steve Chapman, e-mail: [email protected]

john_tullamarine 23rd May 2006 12:18

If one wing produces twice the lift of another wing at the same speed and air pressure, it has twice the CL

.. now, that statement should generate some discussion ....

Wizofoz 23rd May 2006 12:21

How about from you John? Am I right or have I missed a fundemental?

Rivet gun 23rd May 2006 12:34


Originally Posted by Wizofoz
It isn't necessary to include wing area or any other factor into the formula as CL is just that-a co-efficient.
If one wing produces twice the lift of another wing at the same speed and air pressure, it has twice the CL. Could be because of wing area, aerofoil section, planform, angle of attack etc. etc.
.

CL is a dimensionless coefficient. If you analyse the dimesions in the equation you will find that you DO need area in the right hand side in order that CL be dimensionless.

Do the analysis to prove it to yourself.

InSoMnIaC 23rd May 2006 13:20

Wizofoz

S is required as CL is only applicable to 1 section of airofoil. ie the wing is made up of many sections of airofoil so therefore S is required to find how much lift a wing with 'X' meters wingspan is producing.

I put it to u that CL is the same for a peice of airofoil with the same characteristics that is multiplied by S to determine how many peices of that particular airofoil is present.

barit1 23rd May 2006 13:21

Hmmm. A biplane has twice the Cl as a monoplane. What a concept! :ooh:

Wizofoz 23rd May 2006 13:43

Well, now you mention it....
 
Yep. Quite right. To get a force (KG*M/S^2) you need an extra M^2 on the right of the equation. I stand corrected.

barit- Absolutely not, that's why I resisted the idea that area should be part of the equation as twice the area does not mean twice the lift.

That being said, I think we hold the record for getting off topic the fastest as

1) He still should have known the difference between addition and multiplication.
2) "Wing span area" still doesn't mean anything, and
3) That was only one of the errors I highlighted, and my list was not exhaustive.

Rainboe 23rd May 2006 14:08

Wizofoz- you are right. Some of the information is outrageous. I found myself getting increasingly annoyed with it and dumped it 1/3 of the way through. There are some very dodgy areas and i concluded better to not trust anything in the book than try and sort the wheat from the chaff. There has been a previous thread about it.

False Capture 23rd May 2006 21:47

I almost bought the book but after speaking to a few friends who did I decided against it. I'd buy the revision if they act on the feedback they're receiving.

At least I didn't have an interview otherwise it would have been a case of "Ar$e the Pilot Interview".:}

FougaMagister 23rd May 2006 22:47

I reckon there were some answers in that book that I just didn't understand :confused:

If you have an interview, I would recommend reading "Aerodynamics, Engines and Systems for the Professional Pilot" at Airlife. Very thorough - sometimes more in-depth than needed for JAR (it's written by a group of Oz pilots). Well worth the 30 quid or so.

Cheers :cool:

john_tullamarine 23rd May 2006 22:55

A few comments on CL (now, if I could write in sub- and super-scripts, wouldn't that be wonderful ..) ...

CL, and a bunch of other terms (eg Re, M, etc.) are derived from a math technique called "dimensional analysis" - a typical story is given here but, unless you fancy brain strain, I wouldn't worry too much about the detail.

This technique arose as a result of the empiricism of the Industrial Age when some bright folk figured that there had to be an easier, more compact way of presenting data for engineering use.

Looking at forces generated in fluid flow, one considers all the parameters which might reasonably be involved ..

(a) mass (density is more appropriate for fluids)
(b) some sort of size measure (a big sheet of ply is harder to hold in a breeze than a little sheet)
(c) airflow speed
(d) viscosity (sticky-ness of the fluid eg compare honey with water)
(e) compressibility

.. viscosity and compressibility reflect the fact that we want to deal with real fluids in the real world.

The technique uses an exponential form of expression and, for force, gives a result that force is proportional to

dynamic pressure * Re * M * a representative area

If we hide Re (relevant to low speed) and M (relevant to high speed) within a constant of proportionality (this doesn't mean they go away .. have a look at any of the text books which will show the effect on CL slope at higher M values .. just that it would be a routine and not very useful complication to have them in the expression for normal use) we might write

force = constant * dynamic pressure * area

or, in the more usually seen form

L = CL * (1/2*rho*V2) * S

and, similarly, we can run up an expression for drag forces.

The area, S, need not be wing area, but that is probably the most important shape (area) consideration so it is conventional so to choose.

The advantage of DA to we boring old engineering folk is that we can put a huge amount of data into a very simple form rather than having reams of empirical data looking at the relationship between one variable and another ..

It is held, generally, that all the meaningful dimensionless values have been discovered .. find yourself a new one and you, too, could get your name in text book lights ...

As an aside, be aware that this use of the term "dimensional analysis" tends to be the engineers' view of life .. other meanings may be seen, including the simple dimensional checks of equations to make sure that the dimensions all balance out ...

I've never seen the book referred to in this thread so I can have no view on its usefulness etc.

Wizofoz 24th May 2006 04:09


The area, S, need not be wing area,
True.

When you consider that in flight Lift=Weight, then lift is the total reaction of all surfaces (E.G Tail and Fuselage) producing a vertical force in either direction.

In fact the wind area in isolation is probably going to have to produce MORE lift than the aircrafts weight to counter the down force of the tail.

john_tullamarine 24th May 2006 04:42

The point is that the S thing is just a reference area ... doesn't matter what area ... and depending on the actual size chosen, the CL values will fall out accordingly. Not really related to tails and wings as such ...

Mad (Flt) Scientist 24th May 2006 04:52

In fact, it's not the real area of the wing anyway - most analysis uses some form of idealised wing area.

And often, when producing a derivative aircraft, it's simpler to keep the same reference area, even if the wing is physically increased in size.

Wizofoz 24th May 2006 12:38

In light of the above two responses, is there any reason why S could not be assumed to be 1m^2 for all aircraft, thus making my original contention that the different lifting ability of different wings could be compared exclusivley by different values of CL?

Mad (Flt) Scientist 24th May 2006 13:06

You can define it to be anything you want, so it's theoretically possible, but pretty pointless; the usefulness of the coefficients is in comparing the efficiency of different wing DESIGN FEATURES - by a more representative non-dimensionalisation, you can eliminate the influence of purely area changes and compare, say, the effect of sweep, or t/c, on the lift.

Knowing that a wing that is the same design but twice the size generates twice the lift isn't really insightful. Knowing that a change in t/c by 1% changed the CL by 3% for two DIFFERENT SIZED wings allows you to understand the influence of t/c on CL.

SR71 24th May 2006 13:16


A few comments on CL (now, if I could write in sub- and super-scripts, wouldn't that be wonderful ..) ...
John,

Was it you I emailed about a year ago (or whenever) about getting TeX support for the Forum to make this alot easier?

I used to write my more mathematically inclined posts in TeX script, and various posters were/are conversant with it...

It should be possible for the administrators of the BB to get vBulletin to recognise [TEX] tags and voila.

My $0.02.

Wizo


barit- Absolutely not, that's why I resisted the idea that area should be part of the equation as twice the area does not mean twice the lift.
Twice the area does mean twice the lift.

It does not mean twice the lift coefficient.

The point of C_L is exactly as you suggest - to compare different objects ability to produce lift.

Wizofoz 24th May 2006 14:31

SR71,

The suggestion was that a bi-plane would automatically have twice the lift of a mono-plane with half the wing area- Clearly not true as the airflow interference of the bi-plane configuration reduces it's efficiency.

In any case, doesn't Reynolds affect amongst other things mean that simply doubling the area of a wing DOESN'T automatically double the lift (A of A and Rho being equal)?

John, interesting CL discussion (I'm certainly learning a lot), but obviously completely off topic. Could we carve it off to a seperate thread?

JonaLX 24th May 2006 15:20

add to your list of grievance the explanation of Dutch Roll. I imagine guys explaining it that way at an interview :}

SR71 24th May 2006 20:29

Wizo,
Typically the kinematic viscosity of air is ~10^-5, which means that a commercial aircraft tends to fly in a regime where the Reynolds number is ~10^7.

You can see as a result, that you'd need to change the velocity of a flow by a hell of a lot to get a significant change in Re.

As John said, Re is a dimensionless number used to denote similarity of flow regimes. If you are testing a model of aforementioned commercial aircraft in a wind tunnel, you need to match the Re during testing.

Depending on what you're trying to measure you may need to match the Mach Number as well. Mach is just another dimensionless number like Re.

Re denotes the ratio of inertial to viscous forces in a flow.

In the regime we are typically interested in, inertial forces are large compared to viscous ones.

You won't see much dependence of lift coefficient with Re but, conversely, it will be important to match Re if you want a good drag estimate.

I remember when doing my PhD some colleagues trying to obtain laminar flow over wings at high Re with the intention of applying this technology to commercial aircraft.

They laser drilled thousands of minute holes in the upper surface of their wing and attempted to suck away the boundary layer, thereby delaying transition.

Of course, in any technical undertaking like this one has to deal with bugs.

In this case, the bugs kept blocking the laser drilled holes and the exercise was less than satisfactory.

NASA claim to have made progress on this front though:

http://www.nasa.gov/centers/langley/...HSR-Wings.html

with their F-16XL:

http://www.nasa.gov/centers/langley/...Wings-fig1.gif

Wizofoz 24th May 2006 21:24

SR71

Thanks- but we are now officially in WAY over my head!!

The reasearch on laminar flow is interesting though. At least one glider has reached production with a similar system, a venturi being used to provide suction to a series of small holes on the surface.

john_tullamarine 25th May 2006 01:13

SR71,

.. as I recall ... indeed we discussed TeX (now that takes me back to the late 60s/early 70s if recollection serves me correctly) .. I'll raise the matter with those further up the totem pole ..

John

barit1 28th May 2006 11:44

Wizofoz isn't the first to critique this book. Take a look. :rolleyes:

On speed on profile 28th May 2006 16:45

Just read a question on the use of the ILS and its limits. Page 297. The question: When, on the ILS can you descend on the GP?

1st part: When cleared for the ILS procedure. ..... Correct.
2nd part: When you have captured the localiser within +/- 5 degrees C...... absolutely incorrect.

Now the second part is obviously a huge mistake but, if the author meant +/- 5 degrees of the centreline (omitting the reference to temperature), which I think he did, isnt he still incorrect.

Please correct me if I am wrong here but I was taught that the approach should be thrown away when you are descending on GP and you go outside either half scale fly up or down or 1/2 scale fly left or right. And ......

On page 145, he says one dot on the ILS indicator = 1/2 degree (which I think is correct) so therefore on a normal HSI when flying the ILS, his answer should be +/- 1.25 degrees (or within half scale). This is a big contradiction to his answer (forgetting the reference to temperature).

I know what I used to know, but the numbers now have me confused big time!

To clarify this..........

Should the answers for the limitis of lateral navigation, when asked when one can descend with the procedure be:

+/- 2.5 degrees for a VOR approach. (full scale deflection = 5 degrees)
+/- 1.25 degrees for an ILS approach (full scale deflection = 2.5 degrees)

This book just gets more and more confusing!

barit1: It has been discussed in depth on Prune as well and the problems still crop up!!!

santiago15 31st May 2006 12:49


Should the answers for the limitis of lateral navigation, when asked when one can descend with the procedure be:

+/- 2.5 degrees for a VOR approach. (full scale deflection = 5 degrees)
+/- 1.25 degrees for an ILS approach (full scale deflection = 2.5 degrees)
For a VOR app, is full scale deflection not 10 degrees? Therefore, should read +/- 5 degrees for a VOR approach.

On speed on profile 31st May 2006 12:58

Yes, I was nutting it out with a colleague last night. On reflection I think it should be +/- 5 degrees for the VOR and +/- 2.5 degrees for the ILS or halfscale deflection for both.

Rainboe 31st May 2006 18:01

I think that book is up to one major mistake per page. It was so outrageous I couldn't read it all. One volubly expletes trying to read it- the dog was hiding in the corner. I don't know how he got away with it. It's the Benny Hill Show of the aviation world. We should sue- a class action for fraud!

On speed on profile 31st May 2006 19:20


Originally Posted by Rainboe
We should sue- a class action for fraud!

Thats a bit harsh. We are all obviously smart enough to spot the mistakes. I still think its helpful for the questions (and for the most part the answers.)

I wonder if the man reads this site and the numerous threads on his book and is feeling a teeny bit embarassed. Or maybe he is laughing all the way to the bank! :hmm:

Wizofoz 2nd June 2006 09:21


We are all obviously smart enough to spot the mistakes.
Well, WE might be. But as this is a reference book intended for people to find things out, it should be incumbent on the Author to ensure at least a reasnoble degree of accuracy.


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