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Coaxial Contrarotating Lift

  1. Oct 23, 2014 #1
    Quick question - is the lift generated by coaxial contrarotating blades equal to the lift generated by each individual lifting surface (simplified to lift x 2), or is there a more comprehensive formula available?

    Currently, I am using the following formula to calculate the lift of a single disc.
    Lift = (8.6859/((Horsepower / Area) ^ 0.3107)) * Horsepower

    The remainder of my values are Diameter of 8 1/2 feet and 100 HP. Using simplified value to ensure the calculation is correct.
     
  2. jcsd
  3. Oct 23, 2014 #2

    boneh3ad

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    Well, the total lift is going to be exactly equal to the sum the lift generated by each individual lifting surface or rotor, however there is no guarantee that the rotors are generating the same amount of lift as each other. Additionally, it is not likely that each rotor will be generating individually the same lift as a rotor of identical design in a single-rotor design. In other words, if your equation up there (which is clearly some kind of correlation and not a general equation) applies to some rotor design in a single-rotor vehicle, it will not apply if you use two counter-rotating rotors of the same design. After all, the flow field through which each rotor moves is not the same as in the single-rotor case.
     
  4. Oct 24, 2014 #3

    Right, that makes sense. I was wondering about air flow distortion between two sets of blades and what effect it would have. Can you suggest a lift calculation that I could look at to more accurately judge my design?

    In order to give some context, what I am designing/researching is an updated, and hopefully more successful, version of the Hiller flying platform. The two coaxial blades are contained within a single air chamber. The blades are capable of altering their angle-of-attack, but are supported at both the outer and inner edge, thus preventing coning. The overall diameter of the chamber/blades is 8.5 feet, with the main core axle and rotor hub consuming 6 inches of that diameter, resulting in a net rotor surface area of 8 feet (per), each rotor has 4 blades. I am shooting for approximately 100 hp and if all go well around 1000 lbs of lift. At the moment the project is theoretical and power requirements are not a major factor, that's the next step.
     
  5. Oct 24, 2014 #4

    NTW

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    In an ideal momentum approach, for 1000 lbs = 4445 N you would need, for hovering out of ground effect, a power of:

    P = 4445 * sqr [4445/(2 * 1,2250 * 4,6670)] where 1,2250 is the air density in kg/m3, and 4,6670 the disk area of the rotor system in sq. m.

    So, P would be 87,6 kW, about 117 hp...

    In the real world, you're probably going to need more...
     
  6. Oct 24, 2014 #5
    Ok, I understand the conversion to N and the static air density. However the disk area of 4,6670 sq m confuses me. An 8.5 foot diameter rotor (per) converted to meters 10.507 m^2.

    upload_2014-10-24_9-46-33.png
     
  7. Oct 24, 2014 #6
    Here are a couple sketches of my design direction, please forgive the look....
    The first pic is a top-down ortho of the platform's concept. The central air chamber and top blades are displayed, as well as structural support arms.
    upload_2014-10-24_9-49-54.png

    The second sketch is a (not to scale) concept of the centre support axle
    upload_2014-10-24_9-51-21.png
     
  8. Oct 24, 2014 #7

    NTW

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    A diameter of 8,5 feet = 2,59 m, then radius = 1,2954 m. and area = pi *r2 = 5,27178 m2

    You substract the central 8 inches, dia. = 0,2032 m, then r = 0,1060 => area = 0,03243

    So, it's really a net area of 5,23935 m2

    It's only the swept surface of the system that counts here, because the approximate calculation of the momentum assumes a mass of air flowing through a given area. How you achieve that flow is not pertinent...

    Repeating the calculation with 5,239 m2

    =>P = 4445 * sqr [4445/(2 * 1,2250 * 5,239)] = 82,7 kW = 111 hp...

    It's somewhat less than in the earlier calculation because using larger rotors is 'cheaper' in terms of power: for the same change of momentum you need less power.
     
    Last edited: Oct 24, 2014
  9. Oct 24, 2014 #8
    I get it. I was seeing the comma, not a period.
     
  10. Oct 24, 2014 #9

    NTW

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    You're probably thinking in a ducted-prop hovering platform, like this:

    14995437154_7186ea86f2_n.jpg

    Already invented. In the 1950s or so... This is a Hiller prototype.
     
  11. Oct 24, 2014 #10
    Yeah, that was the inspiration for what I am trying to design. Unfortunately the Hiller was plagued with a variety of issues. I wanted to see if they could be resolved with modern technology. Additionally, that is one BIG system, for any practical use it would need to be scaled down/made much smaller. After I have established the thrust and blade diameter requirements, I will be designing a new power system.

    Specifically, I am looking to move away from a gas driven system to a brushless DC motor design. I realize that energy density of current battery technology is lacking, and have a few ideas about electrical generation to compensate. One thing at a time.

    I have a few sketches indicating the direction I am going with that. I'll post them later (today I hope).
     
  12. Oct 24, 2014 #11

    NTW

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    Electric aircraft are difficult... However, one of the first electric helicopters, if not the first one (tethered, electric-powered, but supplied from a ground generator) was built and flown almost a century ago, during WW1...
     
  13. Oct 24, 2014 #12

    FactChecker

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    An effort is made to avoid transonic air flow at the tips of helicopter blades. I think that would greatly complicate the interaction of coaxial contrarotating blades and reduce the power a lot.
     
  14. Oct 24, 2014 #13

    NTW

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    In the summer, in Spain, I often see Kamov coaxial-rotor helicopters fighting forest fires. Those rotors seem to work quite well...
     
  15. Oct 24, 2014 #14

    FactChecker

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    Interesting. My comment was more to address whether the individual lifts from the rotors could be added together for preliminary sizing. I think it would be more complicated than that because of the interaction of the rotors.
     
  16. Oct 24, 2014 #15

    NTW

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    Sure there is an important loss in that interaction... But the power for a torque-compensating rotor is also lost power. Coaxial rotors seem, however, to be even more complex that conventional rotors, that are far from simple...
     
  17. Oct 28, 2014 #16
    In terms of the complexity of a coaxial rotor, much of the mechanical complexity has been removed (based on my concepts). The rotor itself lies within the diameter of the stator/rotor core and is connected to both the inner axle and the outer rotor core. The design precludes dissymmetry of lift and coning. Additionally, since the rotor is based on a electric power source, there is no mechanics drive mechanisms directly linking the upper and lower blades. Angle-of-attack control is achieved with a pair compact piezoelectric hexapod, one hexapod controlling the upper rotor and the second the lower. Each rotor, although typically they would have equal torque, could theoretically have separate torque applied.
     
  18. Oct 28, 2014 #17

    NTW

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    Rotor aerodynamics is a highly complex matter. There are different solutions to different problems and situations, and the accumulated knowledge since the 1930s is very considerable.

    There are excellent books on the subject (for example, Ray Prouty's 'Helicopter Aerodynamics'). There is also a wealth of information in the internet, for example in the 'Rotary Wing Forum'.
     
  19. Apr 2, 2015 #18
    This is for James9653. Could you please direct me to the origin of the formula for coaxial rotor lift you stated on oct 23 2014. From simple practical experiments it seems that for a given power, total lift for the two rotors is at least 1.4 times that of just one the rotors. Unfortunately, most theoretical approaches meander off into the woods, studiously avoiding a simple first order rule of thumb.
    Your simple formula jibes very well with my findings for co-axes over a wide range total of lift and available power.

    Thanks.
     
  20. Apr 2, 2015 #19
     
  21. Apr 2, 2015 #20
    https://www.physicsforums.com/threads/coaxial-contrarotating-lift.777815/threads/coaxial-contrarotating-lift.777815/#post-5061921 [Broken]​
    https://www.physicsforums.com/threads/coaxial-contrarotating-lift.777815/members/verdi1.549593/ [Broken]

    avatar_m.png https://www.physicsforums.com/threads/coaxial-contrarotating-lift.777815/search/member?user_id=549593 [Broken]​
    This is for James9653. Could you please direct me to the origin of the formula forcoaxial rotor lift you stated on oct 23 2014. From simple practical experiments it seemsthat for a given power, total lift for the two rotors is at least 1.4 times that of just one the rotors. Unfortunately, most theoretical approaches meander off into the woods, studiously avoiding a simple first order rule of thumb.
    Your simple formula jibes very well with my findings for co-axes over a wide range total of lift and available power.

    Thanks.​
     
    Last edited by a moderator: May 7, 2017
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