Matt,
I am glad that you decided to continue to share your expertise on this subject. I have in fact reread you excellent explanation of how a helicopter works in our earlier discussions on this subject and i must say that regardless of what i said earlier, i find nothing of what you have said that i disagree with regarding how the heliopter functions (not that i am any kind of an expert on that).
While i know that i am somewhat dogmatic on this subject, i am not so dogmatic that i refuse to understand obviously expert testimony. Indeed, many contributors on this board have offered many thought provoking ideas on this matter, and i respect them all.
However, here is my dilemma...
I do believe that it should be possible to place the relevant coefficent of drag for any speed of propeller rotation (forward or reverse and even locked) on a linear graph where one end is minimum drag and the other end is maximum drag. and somewhere on this graph you must put the locked prop. Does the locked prop condition go in the forward spinning area, or does it go in the reverse spinning area, or does it go between them?
We all seem to agree that a fast spinning prop spinning in the forward direction represents the minimum or least drag, and that a fast spinning prop spinning in the reverse direction represents the maximum or most drag.
We all seem to agree that as the prop loses rpms in the forward direction the drag increases, and when the prop loses rpms in the reverse direction the drag decreases. This alone seems to dictate that the locked prop position must be placed in the middle of the graph between forward and reverse.
If that is not enough, it seems apparent that while all of the forces at play on the propeller that have been identified (turbulence, vortexes, lift, water flow and so on) do in fact have some effect on the end result, it remains that the propeller is firmly attached to the shaft on all of our boats. and so, if we can learn something from the rotation of the shaft, we are at the same time learning something about the rotation of the propeller.
This is what i think we can learn.
At any rpm, in a forward spinning freewheeling condition while under sail, one must apply pressure to the shaft to slow it down, all the way down to the stopped (or locked) position. Under no circumstances, can one slow the shafts rpms without applying additional pressure. The fact that one has to apply additional pressure to slow the shaft (propeller) even with all those forces in play undoutebly means that there is more force (drag) attempting to speed up the rotation in evey case, and the most pressure needs to be applied to stop
the propeller.
It seems to me that a freewheeling propeller is releasing energy and reducing drag as it takes the path of least resistance through the water.
Exactly why the helicopters workings dont correlate with a freewheeling propeller i cant say, but i continue to believe that it does not for the reasons stated above.
If you or anyone can refute what i have stated above, i really would like to hear it.
Merry Christmas and Good Wishes To All
Darrell