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"The helicopter rotor is modeled as an infinitely thin disc with an infinite number of blades that induce a constant pressure jump over the disk area and along the axis of rotation."

This is spectacularly not true of common propeller disks. Most of the area in propeller/rotor disk is empty space, through which air flows freely from top to bottom.

So why do we take this mostly empty disk and use it's area in aerodynamic calculations, when it would seem that we should rather be using the area of the blades themselves?

It seems to permit "fooling" disk loading equations by, for example, decreasing the blade chord without increasing the span, but still getting the same results from equations. So, what a I missing?

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    $\begingroup$ Which aerodynamics text book are you using as your reference point? Or even "which helicopter aerodynamics textbook" are you using as a reference. $\endgroup$ – KorvinStarmast May 7 at 15:49
  • $\begingroup$ This feels like a question best answered by: read ahead in your textbook until you come across the blade-element theory, then compare its pros and cons with the disk approach. $\endgroup$ – AEhere supports Monica May 7 at 16:27
  • $\begingroup$ I'll be honest. I'm just an enthusiast. I usually use Wikipedia. $\endgroup$ – Abdullah May 7 at 17:07
  • $\begingroup$ What are you missing? It works! Why should we make things overly complicated when simple is enough? $\endgroup$ – Peter Kämpf May 7 at 20:07
  • $\begingroup$ <snark> whatever you do, don't study electronics and electricity ... is it hole flow or electron flow? </snark> $\endgroup$ – CGCampbell May 7 at 21:45
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The disc model is used wherever generalities of behaviour are studied without the need to know individual blade characteristics. It captures the mathematics of those generalities in a simple way that leaves out extraneous detail which would make the maths impossibly laborious.

Typically the aircraft designer will use the disc model in the design of their airframe, leaving the blade details for the propeller or rotor designer to figure out.

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  • $\begingroup$ then the statement A low disc loading is a direct indicator of high lift thrust efficiency. doesn't appear to make sense, because the lower disk loading=greater efficiency argument appears to draw on the fact that higher bypass ratio=higher efficiency, but as opposed to bypass fans, which have full disks, propellers and rotors generally have very empty disks, giving less blade area compared to the disk area, meaning that most of the disk is not at all affected by the blades. $\endgroup$ – Abdullah May 8 at 15:39
  • $\begingroup$ Hmm... then again, more blades, more interference between blades. Disk loading takes out the blade count and individual characteristics from the equation, thus enabling simpler calculations, as stated above. Disk loading model does not work for everything, but isa nice useful simplification. For example, e does not actually equal mc^2, but that's close enough for most instances... btw, I don't quite understand how disk loading and bypass ratio are related... $\endgroup$ – Jpe61 May 9 at 7:33
  • $\begingroup$ Disc loading and solidity (your "empty/full") are very different things and they are not directly related; the designer has to decide how they will relate, taking several other factors into account - not least existing designs as a baseline. Also propellers, rotors and ducted fans are very different conditions with very different design solutions. There is no space in a comment to elaborate usefully. $\endgroup$ – Guy Inchbald May 9 at 7:38
  • $\begingroup$ "I don't quite understand how disk loading and bypass ratio are related" - there seems a lot that you do not understand. Maybe some reading up on the subject would help, StackExchange does not do taught courses. $\endgroup$ – Guy Inchbald May 9 at 7:40
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The disc approach is just a calculational tool of convenience that contains simplifying assumptions which make the math easier. It works well enough in certain circumstances to justify its use.

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  • $\begingroup$ What are these "certain circumstances"? It seems to me that the disk idea only makes sense for a fan. $\endgroup$ – Abdullah May 8 at 10:22
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    $\begingroup$ @ABJX What's the difference between a fan and a propeller or rotor? $\endgroup$ – vidarlo May 8 at 14:44
  • $\begingroup$ well, "fan" in the "turbofan" context has propeller blades taking up most or all of the disk area. propellers and rotors have a lot of free disk area. $\endgroup$ – Abdullah May 8 at 15:41

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