Thank you Jim for posting this video…I think I owe you a nice lunch! I was doing my usual run-up when all of a sudden the (L on switch) A module completely cut out then was ok a few switch turns later then cut out again. I did indeed find the break in the wire as you described and that saved me a ton of time tracing all the circuitry or assuming a bad module. I did check the ignition switch first but it checked out solidly. Thanks again for taking the time to post this video. 👍😁
@LightAndSportyGuy24 күн бұрын
Glad it was helpful! Also check the ground wires where the wires connect to the engine.
@antonhenriksson462027 күн бұрын
Clear! Thanks
@charosenzАй бұрын
Your video was great. I especially appreciate you showing some of the parts of the process that can be a bit challenging. Too often video makers only show the easy parts that go well. Thank you. I really like your shirt slogan too!!!
@cmalbertosub2 ай бұрын
I saw below where you mentioned part number 877 295 for the impeller puller - what about the flywheel puller and possible the Sprag Clutch housing part numbers? TIA....
@whoanelly737-83 ай бұрын
I do not understand the language that is used for the 912 carb sync adjustments. I see it this way. There are two adjustments: idle and off idle. These "mechanical" and "pneumatic" terms make no sense (at least to me, they are both mechanical adjustments). Adjust the "idle" balance by turning idle stop screws. Adjust the "off idle" balance (say at 2500 to 3000 RPM) by adjusting the length of the throttle cable with the stop nuts. High vacuum means the airflow is restricted (low RPM, throttle plate closed), low vacuum means the airflow is not restricted (high RPM, throttle plate open). So, pick a side, to lower vacuum, raise RPM, to increase vacuum, decrease RPM. Think opposite. It's a bit easier with a dual needle manifold pressure gauge rather than a vacuum gauge. To increase manifold pressure, increase RPM, to decrease manifold pressure, lower RPM.
@LightAndSportyGuy3 ай бұрын
Yes, both adjustments are "mechanical" - but when adjusting the cables you are measuring vacuum/pressure so that is more pneumaticish... Manifold pressure would be easier as you suggest - one less "inversion" in the thinking process.
@HorsepowerHouse3 ай бұрын
8:51 if you have a heavy-duty impact driver, you can often remove the flywheel along with that puller tool. Similar to many powersport vehicles, atv, motorcycle, etc.
@HorsepowerHouse3 ай бұрын
8:29 what size puller is that? 50mm?
@LightAndSportyGuy3 ай бұрын
The impeller wrench? I have no idea - one of those Rotax special tools 877 295
@HorsepowerHouse3 ай бұрын
@@LightAndSportyGuy I believe it is 46mm x 1.5
@scottmarquardt87704 ай бұрын
It strikes me that the reason the paper lifts is because the flow becomes attached to it progressively more and more as the angle of attack decreases and further attachment is possible as the paper lifts and continues to decrease the angle of attack (as well as proximity to the faster flow) and so forth.
@LightAndSportyGuy4 ай бұрын
Yes, the flow attaches to the paper - and as the paper moves up, it remains attached further along.
@madmarkstoys4 ай бұрын
great videos
@DesiChhore4 ай бұрын
Well explained 🎉🎉
@Dinovass5 ай 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-щ7в5 ай 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 ?
@LightAndSportyGuy5 ай 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.
@michaelc53785 ай 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?
@LightAndSportyGuy5 ай 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.
@stpOwner5 ай бұрын
Clear as mud
@panagiotisathanasiou79185 ай бұрын
Correct!!!
@littlebittybert6 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.
@madmarkstoys6 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
@trutrek9136 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.
@metodyx75076 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?
@LightAndSportyGuy6 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?
@metodyx75076 ай бұрын
@@LightAndSportyGuy I hope that someone else will read our correspondence and write their own experiences...
@kmwongeworldusa6 ай бұрын
hi,is this considered as digital pid or analogue pid? something in between?
@LightAndSportyGuy6 ай бұрын
I suspect that the controller is analog, but I don't know for sure.
@kmwongeworldusa6 ай бұрын
more on analog...@@LightAndSportyGuy
@danvanhuben23166 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?
@LightAndSportyGuy6 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-lp8tk7 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)
@LightAndSportyGuy7 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-hu5fq7 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!
@aaronpetry7158 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.
@LightAndSportyGuy8 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.
@tapioca71158 ай бұрын
"if I've seen further, it's from standing on the shoulders of giants". Thank you for sharing this !
@whoanelly737-88 ай бұрын
Very helpful!
@brgranger8 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?
@LightAndSportyGuy8 ай бұрын
The only group I know of is on facebook. facebook.com/groups/2356797377881273
@brgranger8 ай бұрын
@@LightAndSportyGuy thanks!
@aerospacedoctor9 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.
@02vLxcZF9 күн бұрын
How about those who invoque the Coanda effect as the source of lift: can that do away with the necessity of a circulation flow? kzbin.info/www/bejne/Y6C4p5xmmbeffLc
@wandateems20969 ай бұрын
Skymaster 1964 fixed gear
@Aaditya_Kumar_20049 ай бұрын
great video thanks a lot
@MotorboatToo10 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.
@LightAndSportyGuy10 ай бұрын
Thank you for bringing that to my attention.
@bluehornet675210 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!
@LightAndSportyGuy10 ай бұрын
Thank you, and have fun with your project!!!
@ozelot25010 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?
@LightAndSportyGuy10 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.
@LightAndSportyGuy10 ай бұрын
You may find this helpful: kzbin.info/www/bejne/d4ucp398pZl7mbMsi=19F0GuhXPEDBeqVC
@m.crenshaw209510 ай бұрын
The floats have the B on the side. Is there an up or down orientation for the letter?
@LightAndSportyGuy10 ай бұрын
It doesn't matter. FWIW, I've switched to the Marvel Schebler replacement floats. MS80-430
@RationalDiscourse11 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 Жыл бұрын
Nice job.
@seanyounk1 Жыл бұрын
Lol. Yes, I am still awake. Going to bed now.
@badenhall6291 Жыл 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 Жыл бұрын
Thank you.
@RationalDiscourse Жыл 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 Жыл 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 Жыл 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 Жыл 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 Жыл бұрын
When is Mosaic supposed to be final?
@LightAndSportyGuy Жыл 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 Жыл 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 Жыл 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 Жыл бұрын
@@LightAndSportyGuy I'm not sure what youTube is doing, but my response 10 hours ago has disappeared.
@RationalDiscourse Жыл 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 Жыл 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 Жыл бұрын
But that is a totally different theory. It is not your explanation. Come on , you can’t move the goalposts!!!!!
@grimspyder0001 Жыл 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 Жыл 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 Жыл 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 Жыл 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 Жыл 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 Жыл 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 Жыл 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 Жыл бұрын
see also: en.wikipedia.org/wiki/Coand%C4%83_effect
@LightAndSportyGuy Жыл бұрын
Yes. Much better diagrams than what I drew. Thank you.
@Nehmo Жыл 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 Жыл 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 Жыл 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 Жыл 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 Жыл бұрын
How does Bernoulli's principle explain a pressure in the jet that is lower than the ambient pressure?
@petrtomsej6064 Жыл 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 Жыл 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 Жыл 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.