7
$\begingroup$

I have been working on helicopter design and autorotation for a while. An Index used for this is the so-called "Autorotation Index" Evan A. Fradenborugh derives it in this work https://doi.org/10.4050/JAHS.29.73. Basically, it describes the ratio of available energy compared to needed energy for a flare.

In Equation 10 he presents the final result, but in Equation 9, he also provides a useable index that is non-dimensional. In this Equation, an influence is ϱ * g (disregarding surface-level density as it cancels out anyways). It seems obvious why a denser atmosphere increases the "energy" available, as you can create more lift in denser parts of the atmosphere.

However, I do not understand how g is also increasing this factor. If at all, a stronger gravitational pull should lower this factor as more energy is needed in the rotor to overcome gravity and land safely.

What are your thoughts on this?

enter image description here

enter image description here

Improve this question
$\endgroup$
2
  • 2
    $\begingroup$ Please avoid pseudo-answers-in-comments. If you want to answer, do so and it can be voted on appropriately. If you want to speculate what the answer might be, feel free to but you don't need to add it as a comment. $\endgroup$
    – Jamiec ♦
    Commented Apr 19, 2023 at 11:58
  • 1
    $\begingroup$ There is an analogy with gliders: long as wing AoA is optimal, there is no weight penalty to glide ratio. The gliders potential energy, mg$\Delta$h, matches its lift requirement: mg, or pound-force. A heavier glider glides faster, enabling it to support its greater weight. $\endgroup$
    – Robert DiGiovanni
    Commented Apr 19, 2023 at 22:28

2 Answers 2

Reset to default
4
$\begingroup$

I have found the solution.

In the Nomenclature, Fradenburgh states that $W$ has the unit of lb (thus mass). Using something that is a force as a mass is not untypical in engineering and done sometimes.

However, if we look at EQ1 and do a dimensional analysis, we can see where this leads us. $ROD$ is supposed to be descent rate, so in SI-Units $\frac{m}{s}$. Thus a dimensional analysis of the formula should lead us to $\frac{m}{s}$

ROD

The unit does not match the speed. Speed requires time which is nowhere to be seen. What if we assumed $W$ was a force rather than a mass? For this we need to multiply $\frac{m}{s^2}$ by the term below the root:

ms

As expected, the units start to make sense. This propagates to EQ 9, where $W$ and $DL$ appear in the denominator. Assuming steady-state, they can be expanded to $W=mg$. Thus, the following relation to gravity can be derived from EQ 9 (disregarding all other parameters):

solution

Which is way more intuitive (to me).

The solution is that the Nomenclature is (at least) imprecise or (at worst) wrong, FPS as a Unit system was used implicitly, which is using pounds as a unit of force. Basically an oversight by myself.

Improve this answer
$\endgroup$
5
  • 2
    $\begingroup$ Thanks for coming back and including such a detailed, well laid out, answer to your question. $\endgroup$
    – Jamiec ♦
    Commented Apr 19, 2023 at 13:39
  • $\begingroup$ Of course :) no problem $\endgroup$
    – Clex
    Commented Apr 19, 2023 at 13:41
  • $\begingroup$ The author clearly employs pounds as units of force, within the coherent FPS unit system. Hence, he expresses weight and thrust in pounds, power in foot-pounds per second, and air density in slug/cubic foot... $\endgroup$
    – xxavier
    Commented Apr 19, 2023 at 14:11
  • $\begingroup$ You are right that he uses FPS. I would replace your "clearly" by "apparently", otherwise I agree. I should have seen that. $\endgroup$
    – Clex
    Commented Apr 19, 2023 at 17:47
  • 1
    $\begingroup$ Yeah, derp, nothing about using pounds as both force and mass is ever clear to the vast majority of the world. I would have made that same mistake myself. $\endgroup$
    – Kenn Sebesta
    Commented Apr 20, 2023 at 12:27
-1
$\begingroup$

Just above equation 2, the direction of $g$ is defined to be up. Therefore, this is not the normal acceleration due to gravity, but rather the opposite.

Improve this answer
$\endgroup$
3
  • 1
    $\begingroup$ Can you elaborate on that? I am clueless about how you got to that result. I can't deduct that from the text between EQ1 an 2. Furthermore, I added an answer that (to me) seems more reasonable. Do you have any thoughts on that? If I understand you correctly these ideas would compete. $\endgroup$
    – Clex
    Commented Apr 19, 2023 at 13:33
  • $\begingroup$ @Clex "Arrest the rate of descent" means the acceleration is up. $\endgroup$
    – Anonymous Physicist
    Commented Apr 19, 2023 at 16:50
  • $\begingroup$ Reading the whole sentence you are referring to I understand that acceleration provided thru thrust "arrests" (I think I use that word correctly) is the rate of descent that would otherwise be accelerated by weight. In that case, $g$ is not defined upwards. $\endgroup$
    – Clex
    Commented Apr 19, 2023 at 17:44

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .