Relining my Matco Aircraft Brakes.
24:43
Rotax 912 Carburetor Synchronization
11:22
Rotax 912 Oil System Purge
11:45
3 жыл бұрын
Easy Weight and Balance for Pilots
7:53
Пікірлер
@Dinovass
@Dinovass 8 күн бұрын
Use 2 flat dies and roll over the area your going to bead roll. That will prestretch the metal without bending or forming the work surface
@АндрейБаштовой-щ7в
@АндрейБаштовой-щ7в 15 күн бұрын
Great review, thank you. In the old Lycoming carbureted engines, the pilot can change the mixture quality from the cockpit, which can save a lot of fuel at cruise. Is this capability provided in the Rotax / Bing ?
@LightAndSportyGuy
@LightAndSportyGuy 15 күн бұрын
No, there is no mixture control out of the factory on the carbureted Rotax. Some people have developed aftermarket devices, but I have no experience with them.
@michaelc5378
@michaelc5378 22 күн бұрын
Thanks for the detailed video. I recently took mine apart but after reassembling the engine bogs down trying to go higher rpm. What could it be?
@LightAndSportyGuy
@LightAndSportyGuy 22 күн бұрын
Assuming that it was fine before disassembly, it doesn't seem likely that varnish / debris is the problem. If the slide isn't lifting properly, that would make it run lean at high airflows - could the diaphragm have been damaged or not aligned correctly? Did you remove the needle from the slide? Correct C clip position - screwed down fully? Does it just bog for a moment then run OK? That would suggest accelerator pump issues.
@stpOwner
@stpOwner 23 күн бұрын
Clear as mud
@panagiotisathanasiou7918
@panagiotisathanasiou7918 Ай бұрын
Correct!!!
@littlebittybert
@littlebittybert Ай бұрын
Old backyard mechanic here. I stumbled on this video while looking for some way to familiarize myself with the Bing carb on my 500cc Rotax powered MuZ motorcycle. I've cleaned out and refreshed dozens of carburetors over the decades. I've always had a fair grasp of how the things work, and I've had excellent luck making them run well in spite of my ignorance of the intricate details, but if someone asked me some specific questions about their function, I really didn't have precise answers. That's a little better now, thanks to your well done video. Thank you.
@madmarkstoys
@madmarkstoys Ай бұрын
had the same problem always had a rough running 912 at idle I had a fractured wire right at the CDI box lucky I had enough wire left coming out of the CD box to solder and Shrink tape and isolate the wires from vibration I sync my carbs with the broken wire so make sure you resync your carbs after you fix a fractured wire
@trutrek913
@trutrek913 Ай бұрын
This is a great explanation. I hear curves is what makes lift, then they can't explain why a plane with flat wings can fly.
@metodyx7507
@metodyx7507 Ай бұрын
Very good video, explained in detail! I have a question, would it be possible to drive the Himalayas, the highest mountain pass at 6200 meters, without additional adjustments on a BMW motorcycle (bmw r 100 rt) with such carburetors?
@LightAndSportyGuy
@LightAndSportyGuy Ай бұрын
I really don't know. I've been up to about 3000 meters. Clearly the power would be greatly reduced - but beyond that, I would look for others who have driven vehicles over the pass - has anyone managed it without re-jetting the carburetor? Even vehicles other than one with the Bings?
@metodyx7507
@metodyx7507 Ай бұрын
@@LightAndSportyGuy I hope that someone else will read our correspondence and write their own experiences...
@kmwongeworldusa
@kmwongeworldusa Ай бұрын
hi,is this considered as digital pid or analogue pid? something in between?
@LightAndSportyGuy
@LightAndSportyGuy Ай бұрын
I suspect that the controller is analog, but I don't know for sure.
@kmwongeworldusa
@kmwongeworldusa Ай бұрын
more on analog...​@@LightAndSportyGuy
@danvanhuben2316
@danvanhuben2316 Ай бұрын
If I purchase a previously flying experimental with a airworthness certificate issued, but the plane is currently disassembled down to the fuselage and will require airframe repairs, wings re installed, engine to be installed ect. if I can complete 51% can i still qualify for a repairman certificate even if its already been a completed airplane are one point?
@LightAndSportyGuy
@LightAndSportyGuy Ай бұрын
Seems unlikely. There is only one repairman certificate issued per airworthiness certificate. So you would have to satisfy the FAA / DAR that you built a "new" aircraft. Any parts that are re-used (even if they require significant rework) would be "repaired". You could, for example, likely re-use the wings if you fabricated a new fuselage from new material, etc. No different than using parts from a production aircraft. Somewhere in the advisory circulars referenced, it is suggested that you contact the FAA before re-using large assemblies in your E-AB (assuming you are going for a new certificate)
@CraigKeenan-lp8tk
@CraigKeenan-lp8tk 2 ай бұрын
I agree with the comment below, what's the small hole that's opposite the idle outlet hole on the vacuum side? (forward of the transfer ports and visible at about minute 29.33 on this video)
@LightAndSportyGuy
@LightAndSportyGuy 2 ай бұрын
Not sure. Could be a second idle fuel port - there could be a cross drilling from the port I point to in the video. But, without a carb in my hand and the ability to blow air (or carb cleaner) into the idle mixture adjust I couldn't say for sure. It is odd that the hole for the port appears to be plugged with a screw instead of the more typical lead ball...
@LTDan-hu5fq
@LTDan-hu5fq 3 ай бұрын
I bet 50 bucks the reason your pump was leaking is that the stainless bolt with copper washer was not in the hole with the journal running through it. Put it back in the right hole and change the copper washer was probably all you needed!
@aaronpetry715
@aaronpetry715 3 ай бұрын
What i never understood is $3-5k more just get a PPL. Your a just a couple more inches away and 20 hours away from getting it all.
@LightAndSportyGuy
@LightAndSportyGuy 3 ай бұрын
Assuming that a 3rd class medical is not an issue... But, for many people getting that third class can be difficult, expensive, and time consuming.
@tapioca7115
@tapioca7115 3 ай бұрын
"if I've seen further, it's from standing on the shoulders of giants". Thank you for sharing this !
@whoanelly737-8
@whoanelly737-8 3 ай бұрын
Very helpful!
@brgranger
@brgranger 4 ай бұрын
Jim, i just bought a Merlin GT and am looking for other owners to chat with, are you aware of any user groups?
@LightAndSportyGuy
@LightAndSportyGuy 4 ай бұрын
The only group I know of is on facebook. facebook.com/groups/2356797377881273
@brgranger
@brgranger 3 ай бұрын
@@LightAndSportyGuy thanks!
@aerospacedoctor
@aerospacedoctor 4 ай бұрын
Hi Jim, circulation is not a strange mathematical condition, although most engineers tend to express it that way. The circulation is the manifestation of the viscosity. The use of the term “inviscid” lift is why engineers think that; and another justification for the existence of circulation is needed, KJ theory, the Kutta condition. But if people accept that D’Alembert’s paradox, as confirmed by Euler, which predicated no drag and no lift, is the result of no viscosity, then it become much clearer. Fundamental Eulerian flow is inviscid and hence lacks any rotation. Now, in CFD you can have what is called mathematical dissipation which emulates viscosity, which is a dissipative force, it is diffusion, and is non conservative, but that is a mathematical construct that comes from errors in numerically solving Euler’s equations. But, the potential flow solutions you have shown, without circulation, are analytic solutions, and no numerical uncertain is involved, hence there is no rotation (no turbulence). Potential flows, analytic Eulerian flows, are irrotational. And with all due respect to Prof. Fidkowski, while an infinite number of solutions are possible, the uniform flow direction and relative position of the aerofoil dictate the single solution that is allowed. This is the symmetric solution. That symmetric solution, BTW, gives symmetric pressure and velocity, and hence equal transit... Also, with all due respects to my colleagues who are physicists and mathematicians, while an equation can have a singularity in it, reality does not. So, even a wing on your aircraft, when it starts moving, has a symmetric potential flow solution, and the acceleration is not infinite at the TE, but it is significant. It is this with viscosity that starts the formation of a vortex at the trailing edge; this grows, and that centre of rotation is a low pressure. This then sucks the stagnation point down from that symmetric point to the trailing edge. That is, the Kutta condition is caused by viscosity. This is accompanied by a vortex being shed from the trailing edge in a counter clockwise direction. Noting you have an appreciation of physics, it will be easy to accept that conservation of angular momentum is also required, hence some other part of the fluid must rotate in the clockwise direction. This is the circulation, or the bound vortex around the aerofoil. Effectively, the viscosity of the fluid has made it such that fluid more easily flows over the upper surface (it moved the stagnation point, so this should be intutive), so it has a higher velocity, and the converse is true for the bottom surface. That is, the viscosity term in the Navier Stokes equation, the correct form of Newton’s laws for fluids, alters the balance between the pressure and acceleration on the upper surface relative to the bottom surface, and this persists as an asymmetry in the boundary layer. So, viscosity causes the circulation, which results in a difference in flow speed, which through Bernoulli we can show as a pressure difference, which is the actual force pushing up and down on the aerofoil. Yes, if you take away or try to simplify one of those steps, “the wheels come off the car”. So, don’t, and if people try, tell them not to. There are really only 3 terms in Navier Stokes, acceleration, pressure, and viscosity, so no more or no less is needed to explain lift.
@wandateems2096
@wandateems2096 4 ай бұрын
Skymaster 1964 fixed gear
@Aaditya_Kumar_2004
@Aaditya_Kumar_2004 4 ай бұрын
great video thanks a lot
@MotorboatToo
@MotorboatToo 5 ай бұрын
I realize this video is 3 years old but, I had noticed a couple of thing that hopefully had been corrected or might currently be an answer to a random gremlin electrical problem.. In the close up of the connector to the voltage regulator, it looks like a few wires are swollen with corrosion. May be worth a look. Secondly, the fuel supply hose at the carb was not seated all the way on. The worm clamp did not appear to be clamping on the hose itself. That may end up leaking or potentially slipping off.
@LightAndSportyGuy
@LightAndSportyGuy 5 ай бұрын
Thank you for bringing that to my attention.
@bluehornet6752
@bluehornet6752 5 ай бұрын
Just happened upon this video a day or so ago. I'm an A&P, as well as a CFII, and must say that I definitely learned a couple of things listening to your presentation. More correctly I suppose, some of the things you stated didn't sound familiar to me so I was compelled to go research it. Excellent! I'm planning a tube & fabric EAB build myself, starting later this year. So this video was very interesting to me, and I'm sure I will revisit it again. Then I looked at your other videos and saw the stuff on statics--instant SUBSCRIBE for me. I don't have an engineering degree, but have a few other degrees...but I have also had calc, diff EQ, statics, dynamics, some fluids, etc, etc. But that stuff is 15-20 years ago for me, so I was very happy to see it in your channel, especially with aircraft used in your examples. Thanks for making the video!
@LightAndSportyGuy
@LightAndSportyGuy 5 ай бұрын
Thank you, and have fun with your project!!!
@ozelot250
@ozelot250 6 ай бұрын
I have a camloc that I’m trying to replace. It comes with the quarter turn stud and one washer. I couldn’t see from video how to install?
@LightAndSportyGuy
@LightAndSportyGuy 6 ай бұрын
The washer pushes onto the end of the quarter turn stud once the stud is in place through the hole in whatever you are installing it in - that's what you use the two sockets for - one to push on the outside end of the stud, and one that slips over the stud and pushes the washer into place.
@LightAndSportyGuy
@LightAndSportyGuy 6 ай бұрын
You may find this helpful: kzbin.info/www/bejne/d4ucp398pZl7mbMsi=19F0GuhXPEDBeqVC
@m.crenshaw2095
@m.crenshaw2095 6 ай бұрын
The floats have the B on the side. Is there an up or down orientation for the letter?
@LightAndSportyGuy
@LightAndSportyGuy 6 ай бұрын
It doesn't matter. FWIW, I've switched to the Marvel Schebler replacement floats. MS80-430
@RationalDiscourse
@RationalDiscourse 6 ай бұрын
And still no response the request for you to present the physics and maths of your explanation. You dismissed my explanation because i wasn't using "the larger flow field". Come on now, it's time to show your explanation.
@johnb.6245
@johnb.6245 7 ай бұрын
Nice job.
@seanyounk1
@seanyounk1 7 ай бұрын
Lol. Yes, I am still awake. Going to bed now.
@badenhall6291
@badenhall6291 7 ай бұрын
Great Video, even after a few years, The float arm height spec you were looking for without the gauge is 10.5mm from the the bowl lip edge.
@LightAndSportyGuy
@LightAndSportyGuy 7 ай бұрын
Thank you.
@RationalDiscourse
@RationalDiscourse 7 ай бұрын
So many misconceptions, so much confusion. 1. YOU CAN'T GET LIFT WITHOUT AN EXTERNAL SOURCE OF ENERGY. 0:37 "No energy added/removed" does not apply to the generation of lift! An aeroplane needs to burn fuel to generate the velocity through the air. A glider, in still air, has to convert potential energy of vertical height into kinetic energy of horizontal motion to generate the velocity. When it's hill soaring or working a thermal it's converting the pressure gradient from the vertical component of the airflow. Bernoulli's principle does not apply to the generation of lift from an aerofoil. There are many other reasons, but one good reason should suffice to disprove the theory. P.S.The force enabling a kite to fly is provided in the tension in the string. Break the string and it tumbles to the ground.
@LightAndSportyGuy
@LightAndSportyGuy 7 ай бұрын
The energy goes into the circulation vortex. But I suspect you already knew that. Or, is your definition of Bernoulli's principle different than the fluid mechanics definition?
@RationalDiscourse
@RationalDiscourse 7 ай бұрын
Bernoulli doesn't apply with an external force. In your "Fundamentals IV" video, kzbin.info/www/bejne/lYO0XoGll6xmiMU, you acknowledge the need to invoke Bernoulli to get lift from Circulation. BERNOULLI DOES NOT APPLY IN THE PRESENCE OF AN EXTERNAL FORCE.
@RationalDiscourse
@RationalDiscourse 7 ай бұрын
I have demonstrated the momentum theory comes nowhere near generating enough force to keep a light plane in the sky. You were dismissive of my mathematical treatment so I have invited you to offer your alternative. Can repeat my request for you to present the physics and maths of your explanation please?
@justplanefred
@justplanefred 7 ай бұрын
When is Mosaic supposed to be final?
@LightAndSportyGuy
@LightAndSportyGuy 7 ай бұрын
Per the EAA, "The FAA will now review the thousands of comments it has received during the just-concluded public comment period. The agency is expected to issue a final rule sometime in 2024 or early 2025." www.eaa.org/eaa/news-and-publications/eaa-news-and-aviation-news/news/2024-01-22-mosaic-comments
@RationalDiscourse
@RationalDiscourse 7 ай бұрын
A simple back of the envelope calculation clearly shows that there is not nearly enough rate of change in the momentum of the air above and below the wing a of an aircraft to create sufficient force to resist the weight of the aircraft. Try it: a C172: weight 1,100 Kg, wing span 11 m, chord 1.5 m, AoA around 5°, airspeed say 100 kts, (50 m/sec) air density 1.023 kg/m^3. Force required: 11,000 Newtons. Force generated from moving the air vertically maybe 200 Newtons. Not even in the ballpark!
@LightAndSportyGuy
@LightAndSportyGuy 7 ай бұрын
Not sure how you came up with 200N. Not even close. And, if Newton's laws don't apply, what does? Note: An easy experiment is to put a multicopter in a box sitting on a scale. As the copter lifts off, the reading of the scale does not change - the airfoils accelerate the air downward which then hits the bottom of the box generating a force equal to the weight of the copter - all as you would expect from Newton.
@RationalDiscourse
@RationalDiscourse 7 ай бұрын
@@LightAndSportyGuy I'm not sure what youTube is doing, but my response 10 hours ago has disappeared.
@RationalDiscourse
@RationalDiscourse 7 ай бұрын
@@LightAndSportyGuy I will try again and show you my logic. This is "back of the envelope" stuff so I'm sticking to 2-digit precision. Firstly the geometry: Visualise the wing of the C172 in flight as a thin flat surface, a chord length (1.5 m) wide and a wingspan (11 m) long and an angle of attack (AoA) of, say 5° to the horizontal. Now, visualise the triangular prism of air under the wing: 11 m long, 1.5 m on the diagonal at 5° to the horizontal base. The height of its vertical face, its "pitch" is (chord * sin (5°)) = 0.13 metres. The horizontal width of the prism is (1.5 * cos(5°)) is also 1.5 m since the cosine of 5° is so close to 1.0. So, the volume of the prism of air is 11 * 0.13 /2 = 0.72 cubic metres (m^3). The density of air at NTP is 1.2 Kg/m^3, so the mass of air in this volume is 0.86 kg. Now, the mechanics: The plane, moving at 100 kts or 50 m/sec, is moving at 50/1.5 = 33 chords per second, so every 1/33 seconds, or 0.03 seconds, it moves forward by one chord length. Momentum: Once the wing has passed, this volume has moved downwards by the pitch of 0.13 m. It has displaced the 0.86 kg of air by 0.13 metres. It moved the volume down in 0.03 seconds, so its vertical speed was 0.13 metres in 0.03 seconds, or 4.3 metres/sec. Its momentum therefore was mass * velocity = 0.86 * 4.3 = 3.7 kg m/sec. Initially, it was stationary, so its momentum was zero, so the momentum change was 3.7 kg m/sec. This momentum change occurred over 0.03 seconds, so the rate of change of momentum was 3.7/0.03 = 123 kg m/sec. This is 123 Newtons. Since the same mass of air was similarly displaced from above the wing, the total force is 2* 123 = 250 Newtons. 250 Newtons is way short of the 11,000 Newtons required to support the aircraft against gravity.
@LightAndSportyGuy
@LightAndSportyGuy 7 ай бұрын
@@RationalDiscourse The effects of the airfoil extend far above and below the wing (and in front and behind) - it is not just a small triangle of air that is deflected. An example is figures 36 and 37 in this paper by Prandtl - ntrs.nasa.gov/api/citations/19930091180/downloads/19930091180.pdf - that show the pressure effect on the ground.
@alans172
@alans172 7 ай бұрын
But that is a totally different theory. It is not your explanation. Come on , you can’t move the goalposts!!!!!
@grimspyder0001
@grimspyder0001 7 ай бұрын
I have always wanted to get my pilot license, but it seems so expensive. That I'm not sure if it will make sense, i feel like i will never be able to afford to fly a plane, even with my license.
@alans172
@alans172 7 ай бұрын
3:26 "particle of fluid" What's that? A molecule? No? So what is it? Its existence is fundamental to your explanation. Please explain.
@LightAndSportyGuy
@LightAndSportyGuy 7 ай бұрын
Imagine a small virtual ball of fluid of an arbitrary size. It's easier to understand something like water than air because you can pretty much ignore the buoyant forces which make it seem like lifting a bubble of air would not be adding any potential energy.
@alans172
@alans172 7 ай бұрын
@@LightAndSportyGuyIs it like we do in mechanics, where we take an object of a known mass and imagine all of its mass concentrated at an infinitely small point at its centre of mass?
@LightAndSportyGuy
@LightAndSportyGuy 7 ай бұрын
@@alans172 Sort of - the "problem" with fluids is that you don't have the known mass because you don't have a defined size (typically) but you look at things like pressure, density, and velocity as properties at a point.
@alans172
@alans172 7 ай бұрын
​@@LightAndSportyGuy You say "sort of", and I understand it's not precisely like that because when the mass of an object is concentrated at a point, it doesn't really have a density. Its density would be infinite, its pressure also would be infinite. So, in what way is the fluid particle like the infinitely small point of mass concentration in mechanics? Or is it something else completely?
@LightAndSportyGuy
@LightAndSportyGuy 7 ай бұрын
@@alans172 When you assume a "particle" of a fluid you assume the density and pressure as a property at that point - those are really distributed through your area of interest. That's why you use parameters like pressures and density in place of forces and mass.
@emjizone
@emjizone 7 ай бұрын
see also: en.wikipedia.org/wiki/Coand%C4%83_effect
@LightAndSportyGuy
@LightAndSportyGuy 7 ай бұрын
Yes. Much better diagrams than what I drew. Thank you.
@Nehmo
@Nehmo 7 ай бұрын
The is too long, and I couldn't spend a half an hour on it. I skimmed through and couldn't find an answer.
@LightAndSportyGuy
@LightAndSportyGuy 7 ай бұрын
The answer is in the title. Bernoulli does not explain the paper rising when you blow over it because Bernoulli's principle does not tell us that the pressure in the jet of air is lower than the ambient pressure. The paper rises due to the flow patterns set up around the jet due to viscosity. Think Coanda effect.
@davidgreentree133
@davidgreentree133 7 ай бұрын
Well Done!! ... as a former engineer, it has irritated me to see this phony "experiment" on TV documentaries over the years misrepresenting Bernoulli. I am so pleased that you have done this video
@michaelandreas2177
@michaelandreas2177 7 ай бұрын
My takeaway is that the lifting of the foil is not DIRECTLY caused by the lowered pressure in the jet described by the Bernoulli principle, but rather the entrainment of the air between the jet and the foil. Agreed. However, entrainment of the air is a result of lowered pressure in the jet. So the initial lifting of the foil is INDIRECTLY caused by the lowered pressure in the jet. And directly held in place by the lowered pressure in the jet once it rises too it. 28 minutes to make a pedantic point. Congratulations.
@LightAndSportyGuy
@LightAndSportyGuy 7 ай бұрын
How does Bernoulli's principle explain a pressure in the jet that is lower than the ambient pressure?
@petrtomsej6064
@petrtomsej6064 7 ай бұрын
I think part of lifting force on piece of pape can be actually caused by Bernoulli, but most of the force is caused by Coandă effect you briefly mentioned at the beginning. I didn't watch this through tho. You may say it there, i just didn't see it, as i don't have that much time😃 Anyway, have a nice day.
@satanaz
@satanaz 7 ай бұрын
Great video. So, the low pressure, that causes the paper the paper to raise, is due to the kinetic energy of the fast moving molecules "pulling" on the stationary molecules, because of the fluid's viscosity. BUT, even if the viscosity was ZERO, taking the example on 25:00, the net kinetic energy of the molecules exiting the nozzle would still cause most of the other molecules to move along with the flow to the right, dragging the molecules on top of the sheet of aluminum along with the flow, thus causing a low pressure area to arise on top of the sheet. Even when considering ZERO viscosity (zero friction) the molecules bouncing on each other would still net a velocity to the right, over the whole flow surrounding the nozzle, no? Of course the effect will be much less noticeable, but the paper would still experience a force pulling it "upward" if viscosity = 0, due to the pressure difference.
@LightAndSportyGuy
@LightAndSportyGuy 7 ай бұрын
In an ideal fluid, with no viscosity, when the jet starts, it would have to move some of the ambient fluid out of the way. But Bernoulli's equation only applies to steady state flow - ignoring any transient effects. Then, in an ideal fluid (which does not exist in real life), you would have a virtual "tube" of fluid moving through the stationary ambient fluid and nothing really would happen. If the paper moves, then energy was transferred and not conserved - contrary to the assumptions used to derive Bernoulii's equation. Bernoulli's principle dates back to the late 1700's when this was very new theoretical stuff and Navier's work to incorporate the effects of viscosity into fluid mechanics didn't happen until the early 1800's.
@Ryomichi
@Ryomichi 7 ай бұрын
You are demonstrating Bernoulli's principle
@LightAndSportyGuy
@LightAndSportyGuy 7 ай бұрын
When? Could you be more specific?
@Ryomichi
@Ryomichi 7 ай бұрын
@@LightAndSportyGuy when? The whole video.
@JasonAStillman
@JasonAStillman 7 ай бұрын
Are you also saying that Bernoulli doesn't explain the lift on a wing?
@LightAndSportyGuy
@LightAndSportyGuy 7 ай бұрын
Depends on your starting point. One of the old (1904ish) mathematical models of lift is circulation. Circulation says that a vortex around the wing (back over the top, forward under the bottom, and around) is superimposed on the flow past the wing giving the higher velocity over the top and relatively lower speed underneath. If you start with that theory then you can apply Bernoulli's principle to explain the pressures. However, if you don't start with circulation, then you can't really explain the differences in speed in a manner that is consistent with reality, so that makes it hard to apply Bernoulli's principle... Things like "equal transit", or "an airfoil is like a cross section of a venturi" are pretty much nonsense. And, I suspect they came into being as a way to attempt to explain circulation without explaining circulation - but I have never been able to track down the origins of those myths.
@JasonAStillman
@JasonAStillman 7 ай бұрын
Thanks for the detailed reply. Why we were taught circulation around the wing now makes sense. I've been using Bernoulli to explain all kinds of phenomenon over the years that now seem questionable. Here's one. A friend put hot soup in a blender and there was an "explosive" reaction. I assumed it was Bernoulli lowering the pressure around the blades, and the lower pressure lowing the boiling point, and boom a sudden transition from liquid to gas. Seemed an easy way to conceptualize what happened. Is that at all useful? Because that's not conservation of energy along a streamline...? @@LightAndSportyGuy
@LightAndSportyGuy
@LightAndSportyGuy 7 ай бұрын
@@JasonAStillman Well, the soup sounds interesting :-) One could invoke the concept of lift around the blades, perhaps. But I suspect that with the soup having stuff mixed in I probably would have suggested cavitation behind the chunks of chicken.
@JasonAStillman
@JasonAStillman 7 ай бұрын
Let's keep the chicken out of this :)@@LightAndSportyGuy
@JasonAStillman
@JasonAStillman 7 ай бұрын
so was the phase transition "seeded" by the cavitation? @@LightAndSportyGuy
@PBMS123
@PBMS123 7 ай бұрын
You haven't actually proved that the pressure isn't actually lower. Regardless of whether the total pressures are different, if the static pressure is lower over the top of the paper, than the ambient static pressure, there will be a resultant force. High pressure on the bottom, lower pressure on top.... Of course there is a confluence of factors causing it, including viscosity boundary layer effects etc. but where do you actually show that the pressure must not be lower? Even if what you say is true, it is also true that the Jet streamline will have a lower static pressure, causing a force... If what you're saying is correct, that would turn all of aerospace on its head. The lift equation specifically includes Bernoulli L = 1/2 C_L*A*(rho)V^2
@LightAndSportyGuy
@LightAndSportyGuy 7 ай бұрын
The pressure over the paper is lower. That's why the paper moves up. And, the pressure in the jet is going to be pretty close to the pressure around it. But, Bernoulli's principle does not tell us that the pressure in the jet is lower than ambient pressure and this is not a demonstration of Bernoulli's principle.
@PBMS123
@PBMS123 7 ай бұрын
@@LightAndSportyGuy but isnt it true that once the jet leaves the mouth, and enters the ambient, is equivalent to the ambient pressure. Just as the pressure in a water stream from a hose, is under pressure inside the hose, but isn't once it leaves the nozzle? The jet is not separated from the ambient pressure. The air is moving faster than the outside air, I don't how this isn't just like a venturi?
@LightAndSportyGuy
@LightAndSportyGuy 7 ай бұрын
@@PBMS123 Yes, once a jet leaves your mouth, it is pretty close to ambient pressure. so the pressure in the jet is lower than in your mouth. That is Bernoulli (if you ignore losses). But the point of the video is that the claim that the jet is at a lower than ambient pressure just because it is moving faster is not supported by Bernoulli's principle since Bernoulli applies only along a particular streamline. If you want to use Bernoulli to compare between streamlines, you need to know the total pressures of each streamline or be in a situation where all of the streamlines originated at an approximately equal velocity / pressure (like the flow through a venturi). But that is not the case when you are comparing air propelled by a high pressure source (your mouth) and the ambient air. The lower pressure in a venturi can be modeled by using two principles - first is conservation of mass - the mass of air that flows through the constriction is the same as the mass of air that flows through the larger diameter. If we assume that the density of the fluid remains constant (a generally reasonable assumption in many cases) then the volumetric flow rate through the constriction is the same as through the larger diameter. To get the same volume of fluid through a constriction, the velocity has to be faster. Given the higher velocity, conservation of energy (Bernoulli) within the venturi gives us the lower pressure. When you are blowing out of your mouth, the flow is not constricted once it leaves your mouth so while mass is conserved, we can't make the claim that the flow is faster given the volumetric flow rate since the area is no longer constricted. The venturi ends at your lips and the jet expands and slows down to approximately match the ambient pressure. Blow through your lips and feel around with your finger to guestamate the diameter of the jet and compare to the size of the gap between your lips. You should be able to see that the "diameter" of the jet is larger.
@PBMS123
@PBMS123 7 ай бұрын
@@LightAndSportyGuy if the atmosphere is still any flow is faster though right? There is some constriction of the flow due to the atmosphere and streamlines, exactly the way a plane flies. I just can't see how Bernoulli is not related to this phenomena at all. Obviously its a combination of things like viscosity and boundary layer but is ultimately due to a lower pressure region, from the flowing air, as the flow viscosity is just causing further acceleration of the flow, if the page moves up, the air must move down.
@LightAndSportyGuy
@LightAndSportyGuy 7 ай бұрын
@@PBMS123 The explanation typically given is that the paper moves up because the jet of air is at a lower pressure than ambient due to it's motion as predicted by Bernoulli. That simply is not true. Bernoulli only predicts that the pressure in the fast moving jet is lower than the slower moving air in your mouth. As far as how an airplane flies - the streamline squeezing due to the thickness of an airfoil happens both above and below the airfoil and claims that airfoils look like, or work like, a cross sections of a venturi are pretty much nonsense. I touch on it briefly here: kzbin.info/www/bejne/jH7KfYaFi8-WoZo
@Bunny99s
@Bunny99s 7 ай бұрын
The explanation seems a bit weak and unspecific. You do know about the Venturi effect? It's also based on the bernoulli's principle. Same goes for Bernoulli-grippers. Like shown here: ( phSlc8B4kFo ) Those grippers work by blowing air out onto the surface you want to lift and due to the bernoulli principle the static pressure actually creates a suction effect, even though the air actually blows onto the surface. so it's not a vacuum. These grippers are great because they don't cause a high pressue onto delicate surfaces.
@LightAndSportyGuy
@LightAndSportyGuy 7 ай бұрын
Yes, I am familiar with how a venturi works. An actual venturi is a good way to demonstrate Bernoulli since you are working with streamlines that all originate at the same total pressure. But, the idea that Bernoulli explains a lower pressure in a streamline that came from a higher pressure source than the ambient air (with a lower total pressure) is not supported by Bernoulli's principle. Bernoulli only tells us that faster = lower pressure in the context of a single streamline (or streamlines with the same total pressure). So, yes, while things like the grippers you mention, floating balls in the outlet of a vacuum cleaner, etc. are often attributed to Bernoulli, the phenomena observed is not really caused by the conservation of energy along a streamline (Bernoulli's Principle).
@jaredmercer392
@jaredmercer392 7 ай бұрын
What if you buy an almost complete aircraft with no build logs. Are you still able to complete the build and get it permited?
@LightAndSportyGuy
@LightAndSportyGuy 7 ай бұрын
I would say it is likely, but really that would be a question for a DAR or your FSDO. It may cost you a bit to get a DAR to review the particular circumstances, but I would suggest that it would be money well spent.
@h2oski1200
@h2oski1200 7 ай бұрын
¡¡¡¡¡¡¡¡¡¡¡BERNOULLI!!!!!!!!!!!
@LightAndSportyGuy
@LightAndSportyGuy 7 ай бұрын
The man, the myth, the legend
@SoloRenegade
@SoloRenegade 8 ай бұрын
2:38 doesn't matter, irrelevant.
@PeterClymans-n1l
@PeterClymans-n1l 8 ай бұрын
Very educative and clear clarification on the whole functionality, this helps to understand troubleshooting !
@LightAndSportyGuy
@LightAndSportyGuy 8 ай бұрын
Thank you!
@darrylwbraun
@darrylwbraun 8 ай бұрын
That was, hands down, the best explanation of a Bing carb I have ever watched! Thank you.
@LightAndSportyGuy
@LightAndSportyGuy 8 ай бұрын
Thank you for the kind comment. I'm glad you found it interesting.
@darrylwbraun
@darrylwbraun 9 ай бұрын
Ok... I don't know if this is my problem but I'm sure glad I found this video. I have the same problem but it's not one mag over the other, it's both. I was leaning toward the Rotec TBI I've been trying to get running but maybe this is the problem. I did find a charging wire with bad insulation, that I fixed, but maybe it's broken somewhere else too. Thanks, I've got some more work to do now
@LightAndSportyGuy
@LightAndSportyGuy 9 ай бұрын
Check the ground wires as well. The service manuals give the expected resistance readings for the coils in the ignition system.
@darrylwbraun
@darrylwbraun 9 ай бұрын
@@LightAndSportyGuy I have the heavy maintenance manual and it doesn't match my wiring. The colors are all wrong and I don't really know which pins to check across. I did clean all the ground wires and had a realization yesterday. The engine was still running very rough, till I pulled the mixture about 3/4 out. And, closer to idle I have to pull it all the way out. My problem seems to be a massively rich mixture, like running on full choke. This Rotec TBI, specifically the fuel regulator, is just not working right. I'm now rebuilding my carbs just for comparison. If it runs, I'll dump the TBI. If I can get the TBI working I'll see the carbs.