Why Maneuvering Speed Changes With Weight
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- Опубликовано: 6 авг 2024
- In the previous video titled, "Understanding Maneuvering Speed," I explained how maneuvering speed helps prevent structural damage to the airplane. In this video, I explain why maneuvering speed changes with a change in the airplane's weight. You must watch the first video for this video to make sense.
This video vividly demonstrates why Rod Machado is one of the world's great teachers.
Thank you sir!
You are most welcome, Gregg. And thank you for that very very nice comment. That's very nice of you to say that.
Best,
Rod
Rod Machado: Master Airman and of Martial Arts!
YES!! Thanks Rod! ALWAYS LEARNING and refreshing! jb.
6:27 Finally! Someone addresses the elephant in the room of equal maximum lift, regardless of weight and/or g.
In many cases, the entire fuselage might be considered a fixed-weight component (e.g. aircraft with wing-mounted fuel tanks). For a given lift force, a heavy fuselage with light wings would stress the wing roots more than a heavy fuselage with heavy wings.
If we assume that the wing-root joint is the primary structural limitation of an aircraft, the ideal configuration for maximum maneuvering speed would be a light fuselage with heavy wings, so long as such weight/fuel is stored somewhat inboard (i.e. not tip tanks).
An F-14 can safely carry 2 engines, 4 Phoenix missiles, 2 external fuel tanks, and some internal fuel on its fuselage, because the fuselage is wide enough to contribute a significant portion of lift, while also reducing the span of each (cantilever-beam) wing. Structurally and aerodynamically, such aircraft with lifting-body fuselages can be thought of as a hybrid between a conventional aircraft design and a flying wing.
This is really great Rod. I am continually amazed at your ability to simplify tough concepts.
So nice and I do appreciate the comment. I somehow managed never to see this one until this evening. Thank you again.
Rod
This is probably the best and simplest explanation I've heard. Obviously,this person has great knowledge and knows how to pass it on. Great job and keep up the good work.
I consulted many scientific literature, EASA handbooks, but this man is a legend!
Great video, as an engineer this was all intuitively obvious to me - but my CFI looked at me like I had grown horns when I said it to him back when I was earning my pilots license. So the wings are no more likely to break off at a given speed when the plane is lighter, but it makes for a very bad day if the engine breaks off and the center of gravity is now behind the wing...
Greetings Clark:
That's correct. As long as the airplane is at or below its present operational maneuvering speed, the wings aren't in danger of damage. It's the fixed-weight components that present a structural damage risk at lighter weights. I always love hearing about it when a student teaches his CFI something.
Best,
Rod
Yes!!! I have been waiting on Part 2 . Great job Rod, love your work. You have helped me so much in my flight training.
Great material (easy math) and presentation! I will be using this explanation on my CFI checkride
Success on your ride, Josh!
Best,
Rod
Bummed out that this video is only on one audio channel :'(
connecting stall speed with Va is what finally made it click for me, the lightbulb moment! Thanks Rod, awesome as always and great explanations + graphics.
Thank you, Veni, Vidi, Amavi (Julius would be proud of you). And yes, you got it. It's all about stall speed and when the airiplane will stall.
Best,
Rod
Totally sharing this with my students. Thank you!
This might be the first video on RUclips that I have ever seen with no dislikes.
TechyMark You had to say something...
@@freakfly23 LOL My thoughts exactly!
Still loving Mr Rod Machado. He tought me all the basics when I started my interest in flying.❤
Wow, what a clear and comprehensive explanation. Thank you.
This is the best and clearest explanation I've heard. I've been very aware of Va and Vo of late for some reason. Your video makes the point very well! Thanks!
Thank you, John.
Rod
This is fantastic. Thank you !!!
You’re a hero Rod. That explanation was excellent.
Love all your videos, thanks
Thank you for the Great Explanation
You are most welcome, Khatib.
Best,
Rod
Very useful and informative video! and humorous gentleman! Thanks a lot. :)
The BEST video about the maneuvering speed! Thank you sir!
Mr. Machado, you have earned another Like and Subscriber. Thank you for this incredible video!
Nicely explained , thank you .
Very nice explanation. I couldnt have understood why the speed of airplane decreases with the weight decreases before watch your video.
Thanks alot sir !
Thank you very much. I have my flight instructor test in a weeks time. This helps immensely.
You're the man!!!!! Finally I can say I truly comprehend this. Thanks so much.
Thank you, Katie (a late reply, sorry).
Best,
Rod
This was a *fantastic* explanation. Thank you so much!
Thank you, Gabe!
Best,
Rod
Amazing explanation.
I just watched a channel before explaining this and got lost big time but Mr Rod Machado made it clear as day😊
Came here from blancolirio to better understand maneuvering speed. As a non-plane person but an aviation Fan-Boy, I got it. THX
This is great! the other one too! thanks!
A very nice explanation. You covered the essentials in a timely manner. .
Very nice explanation! "Fixed Weight Components," made the light bulb go on. Thank you for finally providing an answer to my ten yr long question!
Ten years? I do understand this. I've spent more than a year researching some topics only to find an answer in some obscure corner of a library somewhere. Glad you found this useful.
@@Flight-Instructor Fixed Weight Components, that was exactly what I'd been missing. I'm not one of the 5 in 4 people who can't do math :-) and I always understood the relationships between AoA, lift, and load factor, but couldn't work out why the structural limits were defined by load factor and not purely by lift (force) - because I'd only ever been thinking about the wings. Bingo. So simple in the end. Many thanks!
Thanks really easy and clearly to understand
Thank you!
There's one thing in this video that's missing that helped it make sense for me. The lift equation is L = C_L 1/2 rho v^2 S. Meaning that at a given AoA, an increase in the velocity of relative wind over the wing will increase the lift produced by that wing. This helps explain why you need more AoA at a lower airspeed to maintain an equivalent amount of lift as a low AoA with high relative wind velocity, regardless of weight. It's not mentioned here, but for me it was critical to all of this. It's relevant to the section around 4:55.
Very clear explanation
Thanks rod.. I'm currently in training for atc, but i'm really struggeling with this subject. Could you possibly do a video about flight envelope?
Thank you for enligtenment!
Rod does such a human job of explaining!
Very well done
Well explained as part 1.
Great explanation
Thank you, Girum.
Best,
Rod
Great video. One of the things that tends to confuse students (is the limit fixed weight components or wing loading, well, depends on what you’re flying). Very well explained sir.
Thank you, Samu!
Awesome video very good👍👍👍
Amazing explanation! wish he was teaching my CPL
Thank you, Evan.
Best,
Rod Machado
I had to watch this twice. He carefully avoided the fact that *lift* is proportional to airspeed *squared* ;-)
Thanks!!1
Love your vids! I’m still training for CFI. Where can I donate to you?
Thank you, Stephen. That's very kind of you. Just share these videos when you can. I would appreciate it very much.
Best,
Rod
If I'm flying an air tanker at the edge of its maneuvering speed, about to drop a full load of water on a forest fire. If my throttle suddenly becomes stuck (the obvious way to reduce airspeed), would it be better to pull up or down, to minimize aircraft damage?
My speed would unavoidably increase, so which maneuver would accelerate my speed the least? And, would a pull-up in order to stall, then recover be a solution? Otherwise, my speed would continue to increase until structural damage occurs, as the aircraft settles to the new level flight speed.
I know it's a contrived scenario, but I'm curious about the dynamic physics of the sudden weight change.
The one thing that isn't making sense to me is: the AOA is directly proportional with the lift produced? E.g not sure if you have 3º in AOA, if you get a gust and it takes it to 6º does it really double the Lift?
So basically, as weight decreases we need less lift/lower aoa to counteract that weight? So to “fix/put” our aoa back to a place where our c-aoa would be reached before exceeding the limit load factor….we slow the plane down/have a higher aoa for 1g flight?
At 7:10 he says the wings have been designed to handle the extra G force. How much G force are wings designed to handle? If you were able to fly the plane remotely (in this example) at its empty weight, would the G force go higher than 6 with this strong gust of wind?
Hey Rod...age old question.....if an airplane is full of parakeets and is too heavy to take off......if you fired a cap gun and got all of them to fly at the same time, would it then be light enough to get off the ground? hehe Lay it on me baby. Met you at a CFI renewal a long time ago...hope all is well!
Hello Rob 👋🏻 Why does a gust increase AOA? A horizontal gust increases or decreases my IAS speed (depending of the direction) and a vertical gust lifts or pushes the entire airplane down. Why does a gust increase my AOA as long as i hold the same pitch attitude?
I understand, that the speed increases of a horizontal guest will increase IAS and therefore lift and g-load and a vertical gust increases g-load by pushing the plane upwards, but how and why the change in AOA?
It's because the direction of travel of the aircraft does not change instantaneously, due to inertia. The wing does not know anything about orientation (horizontal/vertical) but only about airflow. Change the relative direction of airflow and you change the AoA immediately.
What if the plane is loaded with a most aft CG or most forward CG? That would change your angle of attack during cruise flight thus reducing or increasing the total degrees from your critical angle. In theory can’t this change your Va somewhat?
Greetings Trek:
Yes, it can, but not much. There is, however, another fact that can make a larger difference, in my opinion. And that is the use of power. Thrust reduces stall speed and Va is calculated at zero thrust. This is why I always recommend that pilots fly 10 to 15 knots below Va or Vo (depending on weight) when encountering turbulence.
Best,
Rod Machado
@@Flight-Instructor thanks for the explanation! love your videos.
I am confused at how the angle of attack would be increased to 18 degrees with a gust coming from the same direction. From 4.5 to 18 means a 13.5 degree gust. If a gust hit the plane at the 3 degree AOA from the same direction, it’s new AOA would be 16.5 degrees wouldn’t it? I guess the point is still demonstrated but when I drew it out that’s where I stumbled a bit.
Greetings Jonathan:
Ahh, the issue here is with your assumption about the "...gust coming from the same direction." The gust doesn't come from the same direction as the wind striking the wing. It comes from below the wing. Watch the video again with this in mind.
Best,
Rod
I just heard Ron describe a linear relationship between change in AoA and change in load factor (G force). He said, and I paraphrase... if AoA was quadrupled from 4.5 degrees to 18 degrees, then the amount of G force is quadrupled from 1 to 4. ( by way of Lift quadrupled from 2500 to 10000 which yields 4 G's by way of Lift /Weight). I had not head of the relationship of AoA to G force before. I have not been too successful in finding any other supporting information. Is that what all of you heard as well? Ron, are you out there... could you comment? Thank you.
Greetings Tom:
Here's what I'd like you to do. Click the link below and examine the "coefficient of lift" chart. Now tell me if the relationship between angle of attack and the CL "coefficient of lift" is basically a line or a curve up to the point where a stall occurs. Let me know by replying here when you do this.
www.grc.nasa.gov/www/k-12/WindTunnel/Activities/lift_formula.html
Rod, Thank you very much for your reply. I did follow your link and while the Velocity Relationship Curve graph does show a linear relationship between CL and AoA (of course until one approaches the Critical AoA where everything falls apart), It does not look like a 1-to-1 ratio relationship. if it were 1-to-1, and the appropriate scale is used on the X and Y axis, the line of the graph would be at 45 degrees. And one could take any given AoA(A1) and plot the corresponding CL(C1), then go to the doubled (2 * A1) AoA value(A2) and plot the corresponding CL(C2) and hope to find that C2 = 2 * C1. But I'm not seeing that on the chart. Now, CL is not lift... There are other factors to consider to derive lift from CL (L = (1/2) d v(squared) s CL). So I'm still not seeing the relationship where double (or triple or quadruple) AoA also doubles (or triples or quadruples) load factor (G's). Am I missing something here (I'm sure I am)? I'm thrilled that some with your experience and standing in the aviation community has taken the time to reply. I would really appreciate any further guidance you could share (I have been wrestling with a good explanation of this topic for quite a while). If you prefer, we could continue offline via email, or continue this thread here. Again, thank you so much.
Greetings Tom:
Almost there. Not only is the CL-AoA relationship one-to-one, it’s also linear. Linear doesn’t imply, suggest or even require equality, either. It only means that a change in one variable results in a “proportional” change in another variable. If I start at an AoA of 4 degrees (the typical AoA value for straight and level flight at Va), and instantaneously double the AoA to 8 degrees, the CL increases proportionally by (.3). Given that CL is a direct multiplier for lift, how many times can I increase the CL by .3 before reaching stall?
CL = Coefficient of LIft
Rod, thanks again for the continued conversation. If we're almost there, you must have a better view than I :) I understand and agree with what you say about one-to-one and linear relationships. I was only focusing on the equality of the magnitude of the change because I heard you to say in your video that the quadrupling of AoA causes quadrupling of lift, which in turn causes a quadrupling of load factor (LF). I'm with you on the quadrupling of LF caused by the quadrupling of lift via LF = L/W. But I'm still grasping for the relationship of AoA to Lift. I understand the relationship of AoA to CL by examining the graph, and I can see that CL is a factor in deriving lift by way of the formula L = (1/2) d v(squared) s CL. I'm not sure what you mean when you called CL a direct multiplier. To answer your question from above, it looks like CL can increase by .3 about 4.5 times before reaching stall. Besides CL there are other multipliers... I supposed that with everything else remaining constant, CL becomes a direct multiplier and in our scenario is the only thing changing Lift. But I'm not seeing how the .3 incremental changes in CL would have the effect of doubling, tripling and quadrupling Lift (as in the video). I really like the journey we're on and I can't wait to see the finish line. I only hope I recognize it when we get there :) ...but wait, as I read my on reply back to myself before sending, I see that quadrupling AoA is almost quadrupling CL (CL increase a bit more)... that definitely seems like part of the puzzle, but I'm not seeing the whole picture just yet (though it's probably right in front of me... kind of like that airport you look for at night). Another hint please?
Greetings Tom:
Busy, but I'm back. So in response to your original question, there is indeed a linear relationship between AoA and G force. The fact that the CL-AoA graph is a line should make this quite clear (see previous link). And the line on that graph does indeed identify a one-to-one relationship. Keep in mind that a one-to-one relationship doesn't require equality between variables (look up the definition). And finally, the easiest way I can complete this explanation is to offer the following idea.
Assume an airplane produces 1,000 lbs of lift in straight and level unaccelerated cruise flight at 100 knots at a 4 degree AoA. The CL at 4 degrees AoA is appx. (.4). If we double the angle of attack to 8 degrees, the approximate CL increases to (.8). Since lift has a direct linear relationship with AoA (all other lift variables held constant here), then this equation show what lift we can expect for a CL of (.8).
1000 lbs/.4 = X lbs/.8
Answer = 2000 lbs. Therefore, doubling the angle of attack over its starting value doubles the lift and, therefore, doubles the load factor (based on the parameters of the video above).
Good fun,
Best,
Rod
I know this is the standard explanation and the result even shows on the FAA exam and in operating handbooks so their is a good chance I have missed something, but it seems wrong. Does anyone have a link to a more strict physics treatment of this issue, possibly with an empirical example [like models breaking or not breaking in a wind tunnel]?
Specifically it seems like load factor is being confused with acceleration or force. If an airframe has a load factor of 4 and a max gross weight of 1000 this is a load on the wings of 4000(neglecting the distribution of the mass fraction in the wings themselves) and is achieved at 4g acceleration(pulling up or in a turn). So far so simple. However the strength of the various airframe components is not measured in units of acceleration, the strength is measured in units of absolute force and the force generated by an airfoil is dependent only on speed and angle of attack, the payload mass is irrelevant. If a wing is at 18 degrees and 100 knots in a fixed density atmosphere it produces exactly X lift force before stalling, the lift force needed for level flight has no bearing on this force and no bearing on the stall AoA. So a gross weight of 500 would require 8g acceleration to create the same wing stress of 4000 (from the above setup) and the same speed and 18deg of AoA would be need to create that 4000 units of lift force needed for either either acceleration and gross weight combination.
At this point I must conclude that this formula was derived to comply with bureaucratic operational classification rules rather than actual engineering and physics. I will abide by it as it is the more conservative theory but academically I am not convinced.
MyTech F=ma. If the wing loading (total aerodynamic force) would be the limit for structural damage (wings ripping off), then lighter plane could handle more ”g”. This is the case for example for Extra300 aerobatic plane (max allowable load factor changes acc to TOW. Lighter plane, more ”g” for the same FORCE).
However if the structural damage occurs to fixed weight components e.g. engine mounts (engine mass is always the same, tow makes no difference), then the load factor is the limit (fixed weight components produce certain force at cerrain ”g”).
Take-off distance, landing distance, stall characteristics, climb rate, etc all are affected by aircraft weight. Physics rules.
Aren't you neglecting spar stress loads of the wing against the fuselage in addition to the fixed mounts you mentioned?
Nope. It's just as I say in the video. Don't read anything else into this.
Best,
Rod
Could you elaborate on fixed-wings breaking free from fuselages in reference to maximum allowable maneuvering speeds? @@Flight-Instructor
Greetings Musashi:
The question you're asking indicates that you might not understand what the definition of maneuvering speed actually is. Have you had a chance to watch the first video in this series titled, "How Maneuvering Speed Is Determined?" Take a look at it here (ruclips.net/video/o_KdHEzIJkk/видео.html).@@musashi4856
Best,
Rod
Fudge 40 years in aviation and I was ignorant of this!
i am confused. g=lift/weight. but then in step turn we increase load which is almost same as weight and we still have high G. if the weight increase in steep bank. then g=lift/weight. that mean g force should decrease with steep bank. can someone please explain?
Greetings Abdulaziz:
Actually, load isn't the same as weight as discussed here. Weight doesn't change (unless you're throwing things out of the airplane:). It's the lift that is required to increase in a turn if you want to maintain level flight (assumed here). When lift increases and weight remains the same, the g-force has to increase. It's implicit in the equation.
I hope this helps.
Best,
Rod
do you have any videos on why flaps decrease stall speed. by difinition stall (exceed the critical angle of attack). and flaps cause increase angle of attack. so it should stall easier than with no flaps.
Greetings Abdulazzi:
No, I do not. That's more of a "book learning" topic. But let me ask you a question. You said, "...so it should stall easier than with no flaps.
" What do you mean by "easier"? Do you mean slower? Let me know.
Rod
i am sorry. I meant (stall easier with flaps) because we increased the angle of attack when we use the flaps.
Greetings Abdulaziz:
Well, I can't say that it will stall "easier" because I'm not quite sure what you mean here I can say that the airplane will stall at a slower speed. And that's the real value of adding flaps during landing.
Best,
Rod Machado
So in summary when your lighter fly slower or your plane will break up before stalling.
wouldn't exceeding the 4G limit load factor break off the planes wings, rendering you with no time to slow down once said turbulence hits?
Greetings Alex:
If the airplane is flown at or below its present maneuvering speed (Va, at max gross weight) or its operating maneuvering speed (Vo, at a lesser weight than max gross), neither the wings or the airplane's fixed-weight components could be damaged.
What happened to his voice?
So basically, you want the airplane to stall before it breaks. Well, best to avoid both by slowing down!
That is correct.
Best,
Rod
Hahahahaha 3:25