The lift formula explained

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  • เผยแพร่เมื่อ 27 ส.ค. 2020
  • Airbus 380 or Cessna 172, big or small, they all adhere to the same physical laws. This video shows how the formula for lift can be used to demonstrate the relationship between aircraft weight, airspeed, and angle of attack.
    Weight is commonly, but not 100 % correctly, used to describe mass in kilograms (kg). “If my mass is 60 kg, my weight is 60 kg on planet Earth, but only 10 kg on the Moon.” But to be 100 % scientific, we have to use newton to measure weight, because weight is the gravitational force acting on a mass. On planet Earth, 1 kg mass equals 9.8 newtons. Enough said. After this video, I will, like everybody else, refer to weight as kg.
    This video is a remake of a video I made some months ago. I have simplified the calculations and corrected some errors. When making the first video, I stumbled into a problem: How could I find the density of the atmosphere when the air is dry? There are dozens of different density calculators online, some are plain wrong, and others cannot calculate the density in kg per cubic meters when the altitude is given in feet. For this video, I selected to use the Atmospheric properties calculator (see the link below).
    Attributions:
    Airbus S.A.S: Airbus 380 departure.
    Gevers Aircraft: Genesis aircraft with telescopic wings.
    Jeffrey A. Scott: Drag Coefficient & Lifting Line Theory www.aerospaceweb.org/question/...
    John McIver: Cessna Skyhawk performance assessment www.temporal.com.au/c172.pdf
    Recommended links:
    “Bernoulli or Newton?” • Video
    Atmospheric properties calculator: www.aerospaceweb.org/design/sc...
    Recommended reading:
    John T. Lowry: Performance of light aircraft. ISBN 1563473305, 9781563473302.
    Follow Fly with Magnar: / 1140864699685560
  • วิทยาศาสตร์และเทคโนโลยี

ความคิดเห็น • 67

  • @cruxader27
    @cruxader27 3 ปีที่แล้ว +15

    The most common question asked in about every pilot job interview. This video is very helpful to those applying for the job. Thank you!

  • @gaven6531
    @gaven6531 3 ปีที่แล้ว +7

    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?

  • @stvrstc
    @stvrstc 3 ปีที่แล้ว +6

    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.

  • @veloyozh
    @veloyozh 8 หลายเดือนก่อน

    Thanks a lot! I tried to find lift formula but everywere I saw it it was without phisical units. Your video greatly helped!

  • @thomasl7932
    @thomasl7932 3 ปีที่แล้ว +2

    Thanks for the great video, this is very useful for my learning for my ppl exams

  • @autorotate1803
    @autorotate1803 2 ปีที่แล้ว

    Best explanation on the internet, thank you!

  • @loayoraby674
    @loayoraby674 3 ปีที่แล้ว

    i had fun watching this video and i learn new things that wasn't clear to me,thanks.

  • @josephcameron530
    @josephcameron530 ปีที่แล้ว

    Great video!

  • @smwagh123
    @smwagh123 ปีที่แล้ว

    very good explanation

  • @chrissteevens442
    @chrissteevens442 3 ปีที่แล้ว

    Thanks good explanation.....

  • @dawninarizonanow9252
    @dawninarizonanow9252 ปีที่แล้ว

    This was helpful and I really didn’t understand lift coefficient.

  • @Barabyk
    @Barabyk 3 ปีที่แล้ว

    Brilliant!

  • @flywithjericho1902
    @flywithjericho1902 3 ปีที่แล้ว +1

    thank you sir

  • @modelcraft299
    @modelcraft299 ปีที่แล้ว

    Very good 😊

  • @andrewhale12345
    @andrewhale12345 3 ปีที่แล้ว +1

    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.

    • @FlywithMagnar
      @FlywithMagnar  3 ปีที่แล้ว +5

      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.

  • @dieumercikitambala3875
    @dieumercikitambala3875 2 ปีที่แล้ว

    Clear teaching

  • @alsasalsabila5781
    @alsasalsabila5781 3 ปีที่แล้ว +2

    How to calculate the engine power after solving that equation?

  • @phyarth8082
    @phyarth8082 2 ปีที่แล้ว

    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.

  • @tzachus
    @tzachus 2 ปีที่แล้ว +1

    wow wow wow ....you are stunning

  • @halfameerkat1296
    @halfameerkat1296 ปีที่แล้ว

    how do you calculate the lift coefficient from the geometry of the wing

  • @joacorodriguez3420
    @joacorodriguez3420 11 หลายเดือนก่อน

    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)

    • @FlywithMagnar
      @FlywithMagnar  11 หลายเดือนก่อน

      The relation between AoA and lift coefficient is shown at 6:22.

    • @joacorodriguez3420
      @joacorodriguez3420 11 หลายเดือนก่อน

      @@FlywithMagnar Thanks! That table is a fixed correlation valid for a cessna only I assume. I've learned a lot today :)

    • @FlywithMagnar
      @FlywithMagnar  11 หลายเดือนก่อน

      @@joacorodriguez3420 , yes, every aircraft model has a different table.

  • @slevinshafel9395
    @slevinshafel9395 ปีที่แล้ว

    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.

    • @FlywithMagnar
      @FlywithMagnar  ปีที่แล้ว

      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.

  • @blackberrykangaroo538
    @blackberrykangaroo538 3 ปีที่แล้ว +1

    How are you getting the AoA value at 7:39?

    • @FlywithMagnar
      @FlywithMagnar  3 ปีที่แล้ว +1

      The AoA value depends on the lift coefficient and is found in the graph at 6:30.

  • @rushwolf9360
    @rushwolf9360 3 ปีที่แล้ว

    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].

    • @FlywithMagnar
      @FlywithMagnar  3 ปีที่แล้ว +2

      The answer is the weight of the airplane.

  • @dragonhunter5231
    @dragonhunter5231 2 ปีที่แล้ว

    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?

    • @FlywithMagnar
      @FlywithMagnar  2 ปีที่แล้ว +2

      You reduce the angle of attack by trimming the nose down as the speed increases.

    • @dragonhunter5231
      @dragonhunter5231 2 ปีที่แล้ว

      @@FlywithMagnar Thank you

  • @cassiecalore1019
    @cassiecalore1019 ปีที่แล้ว

    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.

  • @alessio272
    @alessio272 2 ปีที่แล้ว

    This guy speaks Norsk.

  • @xxx_xxx_xxx_bobthebuilder_1597
    @xxx_xxx_xxx_bobthebuilder_1597 2 ปีที่แล้ว

    how did you get both CL and angle of attack form the same equation.

    • @FlywithMagnar
      @FlywithMagnar  2 ปีที่แล้ว +1

      When no other parameters are changed (for example flaps setting), the CL is a function of the angle of attack.

  • @Ed-ip2sg
    @Ed-ip2sg ปีที่แล้ว

    Can this formula be applied to the lift of a sail boat?

  • @flemmingstelling8079
    @flemmingstelling8079 2 ปีที่แล้ว

    Please explain, Why is it exactly 1/2 infront of air denstity ? Why not 3/4 or 1 ?

    • @FlywithMagnar
      @FlywithMagnar  2 ปีที่แล้ว +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

  • @ocny1
    @ocny1 11 หลายเดือนก่อน

    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?

    • @FlywithMagnar
      @FlywithMagnar  11 หลายเดือนก่อน

      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.

  • @husainhamid9872
    @husainhamid9872 3 ปีที่แล้ว

    👍👍👍

  • @slevinshafel9395
    @slevinshafel9395 ปีที่แล้ว

    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)

  • @taharbenseddik5607
    @taharbenseddik5607 2 ปีที่แล้ว

    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.

  • @agnitio_potentia_anima
    @agnitio_potentia_anima 9 หลายเดือนก่อน

    What is conversion from CL to aoa?

    • @FlywithMagnar
      @FlywithMagnar  9 หลายเดือนก่อน

      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.

  • @Allahhh
    @Allahhh 3 ปีที่แล้ว

    Which area is to be considered for drag force..wing area or whole aeroplane area

    • @FlywithMagnar
      @FlywithMagnar  3 ปีที่แล้ว +1

      The whole aeroplane. Total drag consists of induced drag, which increases with the angle of attack, and parasite drag, which increases with the speed.

    • @Allahhh
      @Allahhh 3 ปีที่แล้ว

      I am doing a R &D ...Can I get your mail id...I have lot of questions to ask.

    • @FlywithMagnar
      @FlywithMagnar  3 ปีที่แล้ว

      magnar.nordal@gmail.com

  • @icgulf4480
    @icgulf4480 3 ปีที่แล้ว

    Kindly also explain the Drone

    • @FlywithMagnar
      @FlywithMagnar  3 ปีที่แล้ว

      Regarding drones, I'm not an expert. However, rotors are rotating wings. Therefore, the aerodynamic forces are similar.

    • @icgulf4480
      @icgulf4480 3 ปีที่แล้ว

      @@FlywithMagnar thanks sir,

  • @EyeTouchKids
    @EyeTouchKids ปีที่แล้ว

    At around the 7 and a half minute mark, why is L the mass of the airplane multiplied by gravity?

    • @FlywithMagnar
      @FlywithMagnar  ปีที่แล้ว +1

      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.

    • @EyeTouchKids
      @EyeTouchKids ปีที่แล้ว +1

      @@FlywithMagnar Yep, I realized after commenting what it meant by going back through the video again. Thanks for the quick reply. Cheers!

  • @vlatkopopovski2685
    @vlatkopopovski2685 ปีที่แล้ว

    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).

    • @FlywithMagnar
      @FlywithMagnar  ปีที่แล้ว

      This video explains the formula for lift. Here is a video about how lift is created: th-cam.com/video/ph1HqrioLPs/w-d-xo.html

    • @vlatkopopovski2685
      @vlatkopopovski2685 ปีที่แล้ว

      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.

    • @vlatkopopovski2685
      @vlatkopopovski2685 ปีที่แล้ว

      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.

    • @vlatkopopovski2685
      @vlatkopopovski2685 ปีที่แล้ว

      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.

  • @antoniofreitas5379
    @antoniofreitas5379 3 ปีที่แล้ว

    How many rpm to lift the helicopter?

    • @FlywithMagnar
      @FlywithMagnar  3 ปีที่แล้ว +2

      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.

  • @roxout5743
    @roxout5743 3 ปีที่แล้ว

    A recent video