Thanks a lot for doing this, this is an unimaginable help for us students. I am currently on my CPL/IR/ATPL theory distance learning course and these videos help realy a great deal, sometimes it is just one small detail that makes a click in my head and i got it, like with this video here. Thank you again and my deepest respect.
Video try to explain lift just by a drag force, but CL or CD coefficient must be vectorial angle attack must be sine and cosine function accordingly, and must be between 1 and 0, they use velocity v to define that CL coefficient. We can not be defined by velocity angle of attack can be changed relative to x-axis direction, that is a big misconception because is the very optimal angle of attack depending on engine of plane and area. Or more philosophical question do wings need at all? and when wings become drags force to move freely force with higher velocity forward (x-axis) that is a very nice feedback loop math calculation. Drag coefficient CL - is independent of plane velocity or wind in the case of windmills.
9:22 But how you know the angle of atack? what is the formula ? Because i want to calculate the speed and the angle of atack for specific weight at sea lvl and at 10km altitude.
I used the graph seen at 5:06. Such graphs are the result of airfoil design, computer simulations and flight tests. As you can see, the lift coefficient and the angle of attack are linear up to near maximum lift coefficient. That makes it easy to find the relation between them.
This is defenetly one of the most interesting formulae I've ever learned. And not, because I always wanted to become a F/O on the Boeing 747-400. Now I'm older I do believe for a long time that God, Allah (swt), have a good reason for not letting me become a F/O on the most beautiful commercial airplane ever.
Thanks for the great video! I have a question regarding lift. If I am climbing in an aircraft and I am able to maintain the same indicated airspeed shouldn't my rate of climb remain constant since my increased velocity (TAS) is compensating for the fall in density? From practice and the POH I know it is not the case but trying to get my head around the theory because in my mind one would assume the same pressure is hitting the pitot tube.
Hi. It is correct that TAS increases when you climb with constant IAS. It is also correct that increased TAS compensates for reduced air density, but not 100 %. According to my calculations, the factor is 92 to 94 %. However, there's one more factor: Engine power. The less air the engine gets, the less power it delivers. This is the most limiting factor.
Hallo, pls. it means that if the Cessna will weight 3 000 kg the CL will be bigger. The more bigger CL is than the plain is lifted more? If the plain is heavier also CL is bigger ...the lift must be less...sorry I don't understand. Pls. what or how I can use these results?
Hi. CL is part of the lift formula. When the weight of the aircraft is higher, the lift must be increased. This can be done by increasing the CL (use higher angle of attack), or by increasing the speed.
Can you help me with this.... A moderate speed aircraft moving at a speed of 13 ms-1 whose wing span and aspect ratio is 150 cm and 5 respectively. Assuming suitable wing loading and density of air calculate the lift generated by the wing at cruise speed. [Take g = 10 m/s2].
That's an interesting story! At the time of the Wright brothers, the formula for lift was L=k⋅S⋅V2⋅CL. The k is a constant called Smeaton's coefficient, which accounts for the density of air. However, it did not account for variations in air density. In 1759, John Smeaton published a paper where he defined k to be 0.005. The Wright brothers calculated it to be 0.0033. Today, it is considered that k=0.5⋅ρ. That's why the lift formula has the 1/2. aviation.stackexchange.com/questions/22257/does-smeatons-coefficient-have-a-modern-accepted-value-or-it-is-dependent-of-th
Can you please check the time mark? Lift (L) equals the mass (M) of the airplane multiplied by gravitational force (G-force). In unaccelerated flight, the G-force is 1, and L=M*1. In a coordinated turn with 60 degrees bank, the G-force is 2. Therefore, L=M*2. If you push the stick forward and you get weightless, the G-force is 0. Therefore zero lift.
It depends on the wing profile and the flaps setting. Each wing profile is different. The correlation between AOA and CL on a Cessna 172 (flaps up) is shown at 6:10.
Almost every helicopter fly with constant RPM. The RPM of the main rotor depends on the diameter. The larger the diameter of the rotor, the higher the tip speed for a given RPM. When the tip speed reaches the speed of sound, the efficiency of the rotor decreases rapidly. Therefore the RPM is set to avoid that. A typical main rotor speed is 450 to 500 RPM. Lift is created by varying the pitch of the rotor blades. www.redbackaviation.com/helicopter-engine-rpm-rotorblade-pitch-management/#:~:text=Most%20helicopters%20operate%20at%20around,split%20or%20overlapping%20rpm%20gauge.
Thank you for very informative video. But I have a question. Think, we are flying a Cessna( which has an asymmetric airfoil) at 100ft of altitude. we know that asymmetric airfoil generates more lift if the aircraft increases its speed. therefore, that additional lift will cause aircraft to increase its altitude. Now if I want to maintain this altitude and want to speed up to the maximum. how do I do that?
The authors have two wrong scientific approaches: researching the creation of Lift force and Low pressure at upper side of the wing, relative to the ground surface and Earth. I explain the aerodynamic cavitation and existence of Lee side aerocavern, and creation of Aerodynamic force. Low pressure creates force normal to the cord (contact surface), and it name is "aerodynamic force" because is made from the air (aero) in motion (dynamic), or wind relative to the wing (object).
There is NO Lift "Force", it is Aerodynamic force normal to the contact surface. Next, for mathematical calculations in the coordinate system, the Lift and Drag component are projected.
I say that in the Leeward side of the solid object and air flow there is aerodynamic cavitation which creates an Aerodynamic force perpendicular/orthogonal to the contact surface, which would be said at a normal/right angle to the chord of the aero profile. Because the angle of attack in flight is small, the Aerodynamic force has a much larger Lift component than the Drag, and the vectors of Aerodynamic force and Lift component has very similar magnitude and direction.
The force object receives is always normal to the contact surface and air pressure always acts normal to the surface of the body. This has long been well known, and I don't know why in flight theories and aerodynamics books this is (mostly) omitted.
The most common question asked in about every pilot job interview. This video is very helpful to those applying for the job. Thank you!
How is it that I only find the most useful and concise videos after I have completed a course, and only when I am not directly searching for it?
Thanks a lot for doing this, this is an unimaginable help for us students. I am currently on my CPL/IR/ATPL theory distance learning course and these videos help realy a great deal, sometimes it is just one small detail that makes a click in my head and i got it, like with this video here. Thank you again and my deepest respect.
Glad it was helpful!
Thanks a lot! I tried to find lift formula but everywere I saw it it was without phisical units. Your video greatly helped!
Excellent. At 8:35 the visual shows A=843 m3 but he pronounces it correctly as m sq.
Thanks for the great video, this is very useful for my learning for my ppl exams
Video try to explain lift just by a drag force, but CL or CD coefficient must be vectorial angle attack must be sine and cosine function accordingly, and must be between 1 and 0, they use velocity v to define that CL coefficient. We can not be defined by velocity angle of attack can be changed relative to x-axis direction, that is a big misconception because is the very optimal angle of attack depending on engine of plane and area. Or more philosophical question do wings need at all? and when wings become drags force to move freely force with higher velocity forward (x-axis) that is a very nice feedback loop math calculation. Drag coefficient CL - is independent of plane velocity or wind in the case of windmills.
Drag Coeff. is CD.. not CL.!?!
This was helpful and I really didn’t understand lift coefficient.
Best explanation on the internet, thank you!
How to calculate the engine power after solving that equation?
i had fun watching this video and i learn new things that wasn't clear to me,thanks.
very good explanation
1:54 I need this to explain the ratio of spaceship. 1.0 is not fly 1.1 and above will fly and more big more fast go(escape from earth)
9:22 But how you know the angle of atack? what is the formula ? Because i want to calculate the speed and the angle of atack for specific weight at sea lvl and at 10km altitude.
I used the graph seen at 5:06. Such graphs are the result of airfoil design, computer simulations and flight tests. As you can see, the lift coefficient and the angle of attack are linear up to near maximum lift coefficient. That makes it easy to find the relation between them.
thank you sir
How are you getting the AoA value at 7:39?
The AoA value depends on the lift coefficient and is found in the graph at 6:30.
how do you calculate the lift coefficient from the geometry of the wing
Great video!
This is defenetly one of the most interesting formulae I've ever learned.
And not, because I always wanted to become a F/O on the Boeing 747-400.
Now I'm older I do believe for a long time that God, Allah (swt), have a good reason for not letting me become a F/O on the most beautiful commercial airplane ever.
Clear teaching
This guy speaks Norsk.
Hi, thanks for the video! May I ask, where from does the AoA in degrees in the formula result came? I missed that part. Thanks (7:38)
The relation between AoA and lift coefficient is shown at 6:22.
@@FlywithMagnar Thanks! That table is a fixed correlation valid for a cessna only I assume. I've learned a lot today :)
@@joacorodriguez3420 , yes, every aircraft model has a different table.
Sir , how can I contact you better I am from Nigeria and I am 18 and want to go deeper in Aviation and probably build my own plane.
Can this formula be applied to the lift of a sail boat?
Thanks good explanation.....
Very good 😊
How did you find/calculate the air density?
You can use an air density calculator like this one: www.omnicalculator.com/physics/air-density
Thanks for the great video! I have a question regarding lift. If I am climbing in an aircraft and I am able to maintain the same indicated airspeed shouldn't my rate of climb remain constant since my increased velocity (TAS) is compensating for the fall in density? From practice and the POH I know it is not the case but trying to get my head around the theory because in my mind one would assume the same pressure is hitting the pitot tube.
Hi. It is correct that TAS increases when you climb with constant IAS. It is also correct that increased TAS compensates for reduced air density, but not 100 %. According to my calculations, the factor is 92 to 94 %.
However, there's one more factor: Engine power. The less air the engine gets, the less power it delivers. This is the most limiting factor.
Brilliant!
Hallo, pls. it means that if the Cessna will weight 3 000 kg the CL will be bigger. The more bigger CL is than the plain is lifted more? If the plain is heavier also CL is bigger ...the lift must be less...sorry I don't understand. Pls. what or how I can use these results?
Hi. CL is part of the lift formula. When the weight of the aircraft is higher, the lift must be increased. This can be done by increasing the CL (use higher angle of attack), or by increasing the speed.
Can you help me with this....
A moderate speed aircraft moving at a speed of 13 ms-1 whose wing span
and aspect ratio is 150 cm and 5 respectively. Assuming suitable wing
loading and density of air calculate the lift generated by the wing at cruise
speed.
[Take g = 10 m/s2].
The answer is the weight of the airplane.
wow wow wow ....you are stunning
Please explain, Why is it exactly 1/2 infront of air denstity ? Why not 3/4 or 1 ?
That's an interesting story! At the time of the Wright brothers, the formula for lift was L=k⋅S⋅V2⋅CL. The k is a constant called Smeaton's coefficient, which accounts for the density of air. However, it did not account for variations in air density. In 1759, John Smeaton published a paper where he defined k to be 0.005. The Wright brothers calculated it to be 0.0033. Today, it is considered that k=0.5⋅ρ. That's why the lift formula has the 1/2.
aviation.stackexchange.com/questions/22257/does-smeatons-coefficient-have-a-modern-accepted-value-or-it-is-dependent-of-th
At around the 7 and a half minute mark, why is L the mass of the airplane multiplied by gravity?
Can you please check the time mark? Lift (L) equals the mass (M) of the airplane multiplied by gravitational force (G-force). In unaccelerated flight, the G-force is 1, and L=M*1. In a coordinated turn with 60 degrees bank, the G-force is 2. Therefore, L=M*2. If you push the stick forward and you get weightless, the G-force is 0. Therefore zero lift.
@@FlywithMagnar Yep, I realized after commenting what it meant by going back through the video again. Thanks for the quick reply. Cheers!
What is conversion from CL to aoa?
It depends on the wing profile and the flaps setting. Each wing profile is different. The correlation between AOA and CL on a Cessna 172 (flaps up) is shown at 6:10.
Kindly also explain the Drone
Regarding drones, I'm not an expert. However, rotors are rotating wings. Therefore, the aerodynamic forces are similar.
@@FlywithMagnar thanks sir,
Which area is to be considered for drag force..wing area or whole aeroplane area
The whole aeroplane. Total drag consists of induced drag, which increases with the angle of attack, and parasite drag, which increases with the speed.
I am doing a R &D ...Can I get your mail id...I have lot of questions to ask.
magnar.nordal@gmail.com
How many rpm to lift the helicopter?
Almost every helicopter fly with constant RPM. The RPM of the main rotor depends on the diameter. The larger the diameter of the rotor, the higher the tip speed for a given RPM. When the tip speed reaches the speed of sound, the efficiency of the rotor decreases rapidly. Therefore the RPM is set to avoid that. A typical main rotor speed is 450 to 500 RPM.
Lift is created by varying the pitch of the rotor blades.
www.redbackaviation.com/helicopter-engine-rpm-rotorblade-pitch-management/#:~:text=Most%20helicopters%20operate%20at%20around,split%20or%20overlapping%20rpm%20gauge.
Thank you for very informative video. But I have a question. Think, we are flying a Cessna( which has an asymmetric airfoil) at 100ft of altitude. we know that asymmetric airfoil generates more lift if the aircraft increases its speed. therefore, that additional lift will cause aircraft to increase its altitude. Now if I want to maintain this altitude and want to speed up to the maximum. how do I do that?
You reduce the angle of attack by trimming the nose down as the speed increases.
@@FlywithMagnar Thank you
The authors have two wrong scientific approaches: researching the creation of Lift force and Low pressure at upper side of the wing, relative to the ground surface and Earth. I explain the aerodynamic cavitation and existence of Lee side aerocavern, and creation of Aerodynamic force. Low pressure creates force normal to the cord (contact surface), and it name is "aerodynamic force" because is made from the air (aero) in motion (dynamic), or wind relative to the wing (object).
This video explains the formula for lift. Here is a video about how lift is created: ua-cam.com/video/ph1HqrioLPs/v-deo.html
There is NO Lift "Force", it is Aerodynamic force normal to the contact surface. Next, for mathematical calculations in the coordinate system, the Lift and Drag component are projected.
I say that in the Leeward side of the solid object and air flow there is aerodynamic cavitation which creates an Aerodynamic force perpendicular/orthogonal to the contact surface, which would be said at a normal/right angle to the
chord of the aero profile. Because the angle of attack in flight is small, the Aerodynamic force has a much larger Lift component than the Drag, and the vectors of Aerodynamic force and Lift component has very similar magnitude and direction.
The force object receives is always normal to the contact surface and air pressure always acts normal to the surface of the body. This has long been well known, and I don't know why in flight theories and aerodynamics books this is (mostly) omitted.
There's Coefficient of DRAG, but i don't ever BELIEVE of any CL(coef.of Lift)..No.
Maybe b'cuz CD is in-align of Flow but LIFT is perpendicular to it...
👍👍👍
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