Wikiposts
Search

Notices
Flying Instructors & Examiners A place for instructors to communicate with one another because some of them get a bit tired of the attitude that instructing is the lowest form of aviation, as seems to prevail on some of the other forums!

Engine out-glide ratio

Thread Tools
 
Search this Thread
 
Old 20th February 2001 | 13:07
  #1 (permalink)  
DeltaTango
Guest
 
Posts: n/a
Post Engine out-glide ratio

Interresting thread on the "wannabees":
total engine failure

How would you explain to a student that the weight of the aircraft does not change the glide ratio, and what happenes to L/D ?


DT
 
Old 20th February 2001 | 14:22
  #2 (permalink)  
grade_3
Guest
 
Posts: n/a
Unhappy

G'day,

There's a couple of good ways I would explain it. One way is to fly a heavy acft then a light acft and note the difference, but that's not really practical . Fortunaely, the theory is pretty straight forward too......

(1) Explain how angle, rate of descent and TAS/GS effect each other. (e.g. rate increases with an increase in TAS for a given angle etc. etc.)

(2) Draw a diagram of the forces acting on the aircraft during the glide; include resolving the Weight Vector into a vertical and horizontal component.

(3) Prove / Explain to the student that the angle between the Weight Vector and the Vertical Component of Weight = the Angle of Descent. (if you get my meaning, it's kind of hard to show it without a diagram).

(4) It should now be straight forward enough to show to the student that as the total weight increases, the gliding angle remains unchanged. You could do it vectorally (i.e. with pictures on the board ) or by simple explaining it to her/him.(The patter would go something like this: "Well, if we increase the weight of the aircraft, we also need increased lift to support the aircraft. This increased lift is created by flying slightly faster, but still at the same AoA [and thus the same L/D ratio]. The extra "Apparent Thrust" needed to fly faster is provided by the extra weight we flying at, so note that our gliding angle has remained unchanged provided we glide a little faster to acheive our best L/D AoA" )

If you have a Flight Manual for a higher performance aircraft available, show the student how the Best Glide Speed reduces as the aircraft's weight reduces. This kinda backs up the theory I just outlined above.


As for explaining that L/D ratio does not change with an increase in weight, the best way I can think of to do this would be to develop a graph of L/D at various AoA's and show to the student what actually happens!

HTH and Hope it is what you were after,

Cheers,


Grade 3
 
Old 21st February 2001 | 03:44
  #3 (permalink)  
Luftwaffle
Guest
 
Posts: n/a
Post

For someone more daunted by physics diagrams than grade 3's hypothetical student, how about:

The air provides two things: lift to hold the airplane up, and drag to slow the airplane down. Coax the student to deduce that we want to maximize the first and minimize the second. Remind the student that lift increases with speed and angle of attack, and that drag increases with speed and with the area presented to the air.

Take a model airplane in your hand. In an extreme nose up attitude (exaggerate the attitude for slow flight) lift is high, but drag is very high. Remind them they had to increase the power to maintain slow flight. In an extreme nose down attitude (point the nose straight down -- even drop it on the floor if you have a durable model), you're not getting much in the way of range. get the student to agree that the best attitude is somewhere in between those extremes.

Have the student consider just the outside shape of the airplane, moving through the air. Does the angle that provides the best compromise between lift and drag depend on how much the airplane weighs, or just the outside shape? Once you get them to agree that there is one best angle, then just draw the airplane, the angle of descent and the eventual landing spot.

If the aircraft is heavy, it will need to go faster, and if it is lighter it will go slower, but either way it will follow the same glide path to the same spot.
 
Old 22nd February 2001 | 15:40
  #4 (permalink)  
grade_3
Guest
 
Posts: n/a
Thumbs up

Interesting reply Luftwaffle, always great to see a different point of view from one's own.

I agree that the way you've explained the question is a far simpler and more commonly understandable way of explaining it, my only 2 comments are:

(1)
<font face="Verdana, Arial, Helvetica" size="2">Remind the student that lift increases with speed and angle of attack, and that drag increases with speed and with the area presented to the air.</font>
My comment here is that drag is also dependant on AoA (Induced drag), so I would think that this needs to be emphasised too, not just the aircraft's attitude (area).


(2)
<font face="Verdana, Arial, Helvetica" size="2">. Does the angle that provides the best compromise between lift and drag depend on how much the airplane weighs, or just the outside shape? Once you get them to agree that there is one best angle, then just draw the airplane, the angle of descent and the eventual landing spot.......

If the aircraft is heavy, it will need to go faster, and if it is lighter it will go slower, but either way it will follow the same glide path to the same spot. </font>
My own question following this response would be "if we are going faster, then we are producing more parasite drag, so why does our angle remain constant?".

We both know the answer, of course, but I would be interested to see how you could phrase the answer, as I am unsure of how to express it without resorting to vector theory again


Interesting discussion as always,

Cheers,


Grade 3

 
Old 23rd February 2001 | 03:47
  #5 (permalink)  
chicken6
Guest
 
Posts: n/a
Post

"Ah, but if we go faster by lowering the nose a bit we reduce the induced drag don't we, this compensates for the increased parasite drag. Well spotted though."

Something like that.

One other point for grade_3, surely the weight vector is all vertical? If so, do you mean, instead of

"...include resolving the Weight Vector into a vertical and horizontal component."

...resolve the Weight vector into a forward component and a right-angled component. Label the FCW. You can see this by putting your whiteboard pen/chalk on the ruler or (plank of wood) you are using to draw your lines and tilting it. "The FCW is what makes the pen move."

 
Old 23rd February 2001 | 14:45
  #6 (permalink)  
grade_3
Guest
 
Posts: n/a
Thumbs up

Nice answer Chicken6, but I still have a question about it

<font face="Verdana, Arial, Helvetica" size="2">"Ah, but if we go faster by lowering the nose a bit we reduce the induced drag don't we, this compensates for the increased parasite drag. Well spotted though."</font>
During a max range glide, the AoA remains constant so as to maintain the best L/D ratio, correct? If so, the induced drag cannot decrease, or at least that is my understanding of it. (Incidentally, the nose attitude is identical for a best range glide at a heavy acft weight as at a light acft weight, only the EAS changes.)

Also, we don't actually go faster because we lower the nose (this would imply we are no longer maintaining the best L/D ratio), we go faster because a greater Thrust Component of Weight (TCW) is acting along our flightpath.

So, my question still stands (unless I am missing some logic here, which is possible ):

<font face="Verdana, Arial, Helvetica" size="2">
If the aircraft is heavy, it will need to go faster, and if it is lighter it will go slower, but either way it will follow the same glide path to the same spot.
</font>
"if we are going faster, then we are producing more parasite drag, so why does our angle remain constant?".



And your point about the Weight vector is a valid, thankyou for pointing it out to me !


Cheers,


Grade 3


(edited to fix a formatting error)

[This message has been edited by grade_3 (edited 24 February 2001).]
 
Old 23rd February 2001 | 17:50
  #7 (permalink)  
Noggin
Guest
 
Posts: n/a
Post

Good discussion. Why does a glider carry ballast?
 
Old 24th February 2001 | 04:57
  #8 (permalink)  
Luftwaffle
Guest
 
Posts: n/a
Post

A racing glider carries ballast to go faster. If pleasure gliders carry ballast (I don't know if they do), it might be for stability (increased Va).

Yes, there is a greater drag force acting on the faster glider as it descends, but it is acting against a heavier mass, so it has the same effect.

I can prove this with trigonometry if provoked.
 
Old 24th February 2001 | 05:38
  #9 (permalink)  
Luftwaffle
Guest
 
Posts: n/a
Post

I don't like the statement that "the FCW is what the makes the pen move." It might make the student think that the forces on the aircraft are unbalanced in a constant speed glide.

The aircraft is moving because it's moving, and Newton's first law says it's going to keep on moving unless a force acts to change its motion. Yes, there is a component of weight in the direction of the aircraft's motion, but that force is exactly balanced by drag and the component of lift acting in the opposite direction.
 
Old 26th February 2001 | 02:10
  #10 (permalink)  
DeltaTango
Guest
 
Posts: n/a
Post

Thanks alot for the reply's.....I never had any problem understanding the physics of it just the logic-if that sentence makes any sense
 
Old 26th February 2001 | 02:42
  #11 (permalink)  
bookworm
Guest
 
Posts: n/a
Unhappy

If you really want to confuse the student you could mention that the heavier aircraft actually does a little better than the lighter one.

All the (good) analysis so far assumes that the skin friction drag coefficient is independent of speed. It's not quite so, and it actually decreases slightly with speed (Reynolds number).

An aircraft that is 20% heavier, and therefore gliding 10% faster, has a skin friction drag coefficient between 2 and 5% lower, depending on the type of boundary layer. So it should get a better glide ratio, by a whisker...

That's one of the reasons why the big jets have such good glide ratios. They're big and fast, both of which raise Re.
 
Old 26th February 2001 | 04:50
  #12 (permalink)  
grade_3
Guest
 
Posts: n/a
Exclamation

That's a fascinating little bit of information Bookworm! That will make an interesting question for my colleauges at work to answer


Cheers,

Grade 3
 
Old 27th February 2001 | 00:49
  #13 (permalink)  
DeltaTango
Guest
 
Posts: n/a
Post

I'm looking forward to their reply, grade_3
 
Old 27th February 2001 | 05:45
  #14 (permalink)  
askcv
Guest
 
Posts: n/a
Post

The glide when heavy or light is complicated when a fixed airspeed is flown (eg the C152 60 knots) which means that if the best speed changes with the weight but you fly a fixed IAS then you will not necessarily be flying the right speed. If the variation is great enough then you will see that the performance will not always support the statement that weight does not affect range.
An example is the 737-200 which flies normal descent at 250IAS. At a light weight the best speed might be 200 thus the gradient will be steeper than optimum. If the airplane is heavy though, the speed will be closer to best glide so the gradient will be less steep.
Nowadays we fly ECON speeds which change with the weight, so the effect is not so pronounced.
 
Old 2nd March 2001 | 14:06
  #15 (permalink)  
chicken6
Guest
 
Posts: n/a
Post

Luftwaffle

I didn't quite finish where I should have!

"The FCW is what makes the pen move, and it will keep speeding up until the drag increases enough to balance it".

Grade 3

Good point about the nose attitude! Same A/A at same glide ratio = same attitude. I still maintain ID is reduced though - the faster airflow is straighter over the wings =&gt; less spanwise flow etc.etc.

Far out, I thought I knew this stuff inside out.

I thought gliders carried ballast so they could actually get back down without exceeding Va?

 
Old 5th March 2001 | 15:57
  #16 (permalink)  
Oktas8
Guest
 
Posts: n/a
Question

I too have been taught that higher TAS - faster chordwise airflow - less time subject to spanwise forces - lower ID.

However, consider this: higher TAS - more lift (at same AoA) - greater pressure differential - greater spanwise forces - higher ID.

So it would be a balance methinks. And since time subject to spanwise forces decreases in direct proportion to increasing TAS, but spanwise forces themselves are driven by pressure differential, and thus change by the square of TAS, surely higher TAS means more ID?

I await enlightenment...

And to the other issue of glider ballast, no, heavier aircraft means faster glide without range penalty. So if you want to win the glider race, you'd better carry some sand. But it also gives you less time in any patch of airspace, so you'd better be a pretty good at finding large areas of rising air that don't require tight turning.

[This message has been edited by Oktas8 (edited 05 March 2001).]
 
Old 12th March 2001 | 12:52
  #17 (permalink)  
chicken6
Guest
 
Posts: n/a
Post

please please please disregard! not thinking when posting, and I can't delete that last one.

hmmmmmmmm. that's an interesting point, will have a read of Kermode soon
 
Old 12th March 2001 | 13:11
  #18 (permalink)  
flyingwigwam
Guest
 
Posts: n/a
Post

"And to the other issue of glider ballast, no, heavier aircraft means faster glide without range penalty. So if you want to win the glider race, you'd better carry some sand."

Hmmm, all the gliders I flew carried water ballast. The theory behind carrying water ballast is that the increased weight shifts the drag curve to the right and thus minimum drag speed and thus best L/D occurs at a higher airspeed. Ballast is used early in the day when lift is strong and fast glide speeds are more important than being able to climb fast. (remember increased weight increased the descent rate). Later on in the day if lift strength decreases, the water ballasts is dumped to decrease the descent rate at the expense of speed at best L/D.

I maintain that the statement that weight does not affect the glide angle is an oversimplification. This is true if one considers only induced drag and the vector diagrams. However, this is only part of the story and does not take the increase in profile drag (form, skin friction, interference drag) into account at all.

For relatively small increases in weight it it generally true that the glide angle is unaffected. However, if you were to increase the weight by say, a factor of 4, I am sure that you would see a significant decrease in the glide performance.
 
Old 13th March 2001 | 18:11
  #19 (permalink)  
bookworm
Guest
 
Posts: n/a
Unhappy

<font face="Verdana, Arial, Helvetica" size="2">I maintain that the statement that weight does not affect the glide angle is an oversimplification. This is true if one considers only induced drag and the vector diagrams. However, this is only part of the story and does not take the increase in profile drag (form, skin friction, interference drag) into account at all.</font>
It is, as I pointed out previously, a slight oversimplification, but it certainly does take into account profile drag (the term I'm more familiar with for form, skin friction, interference is 'parasite drag' but that may be a US/UK difference).

The important characteristic of drag is that, for a particular angle of attack, the drag is proportional to the speed squared: i.e.

D = 1/2 rho v^2 Cd(a)

where Cd is the drag coefficient which may be a function of angle of attack, a.

This is true of the induced drag, where Cd is approximately proportional to a squared, but equally true of parasite drag, where Cd is approximately independent of a.

We've also got lift:

L = 1/2 rho v^2 Cl(a)

where Cl is approximately proportional to a.

So L/D = Cl(a)/Cd(a) which depends entirely on the angle of attack a.

Whatever angle of attack maximises Cl(a)/Cd(a) is the best glide speed. The speed that corresponds to this depends on the weight: the higher the weight the more lift is required and therefore the higher the v required for a given a.

Only if the D = 1/2 rho v^2 Cd(a) relationship breaks down does the angle of attack for best glide vary with speed. In the case I alluded to in a previous reply, Cd itself turns out to be weakly dependent on v, so the v terms don't cancel when you divide L/D.

But that may not cut much ice with the students unless they have degrees in physics.


 

Posting Rules
You may not post new threads
You may not post replies
You may not post attachments
You may not edit your posts

BB code is On
Smilies are On
[IMG] code is On
HTML code is Off
Trackbacks are Off
Pingbacks are Off
Refbacks are Off



Contact Us - Archive - Advertising - Cookie Policy - Privacy Statement - Terms of Service

Copyright © 2026 MH Sub I, LLC dba Internet Brands. All rights reserved. Use of this site indicates your consent to the Terms of Use.