What is the lift to empty weight ratio of some popular modern helicopters?

  • 4
    $\begingroup$ This question cannot be answered as it is. It's too broad for one thing, and you are asking for people's opinions rather than facts. Plus, I don't have any idea what you are actually asking. $\endgroup$ – GdD Oct 1 '17 at 19:09
  • $\begingroup$ Your question about pros and cons is mostly a legal and financial one, not an aviation one. We might be able to help with technical issues for your prototype, though, if you have any specific questions. $\endgroup$ – Pondlife Oct 1 '17 at 22:38
  • $\begingroup$ I have edited my question. Thank you. $\endgroup$ – David Walden Oct 2 '17 at 12:53
  • 1
    $\begingroup$ I have edited your question further to really focus on the aviation part of the question, making it more likely you get a good answer. You can rollback my edit if you disagree. $\endgroup$ – Sanchises Oct 2 '17 at 14:10
  • 3
    $\begingroup$ Popular according to whom? Popular for what purpose? $\endgroup$ – user Oct 7 '17 at 21:25

From Prouty: Helicopter Performance, Stability and Control: the example helicopter that is referenced throughout the book, to illustrate the equations used and computations made.

The rotor thrust T is a function of blade geometry and number of blades. The blade is just a rotating wing and maximum thrust is a function of the lift coefficient $C_l$ of the blade profile. Rotor thrust is also a function of rotor disc area and number of blades.

The example helicopter has the following characteristics:

  • Design gross weight = 20,000 lb
  • Minimum operating weight = 10,700 lb
  • Main rotor radius = 30 ft
  • Blade geometry: chord = 2 ft, airfoil = NACA0012, 4 blades

The average lift coefficient of the blade is computed on page 18 as $C_l$ = 0.52 at sea level, for the design gross weight. NACA 0012 has a maximum $C_l$ of 1.2, which makes the maximum rotor thrust at sea level:

$$ T_{max} = 20,000 * 1.2 / 0.52 = 46,153 lb $$

So maximum lift divided by minimum operating weight is 46,153/10,700 = 4.3

Since NACA0012 and 0015 can still be found in many helicopters, this would be the general order of magnitude for popular type helicopters. A max lift/empty weight of about 4. Further variations are due to disk loading: higher disk loading means a higher $C_l$ to lift off.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.