Coning angle is the result of the integration of some of the local forces on each element of a blade.
At each element the local forces that are relevant to coning are the moments about the flapping hinge. (Ie those forces in the plane normal to the flapping hinge at their respective distance from the flapping hinge.
They arise from two forces
They are
1 the apparent outward force (generated by the reaction of a mass to Centripetal Accn), call it Centripetal Force (Cf) if you like.
AND
2 the aerodynamic forces in that plane, traditionally thought of as ‘lift’ (although it’s not strictly true, nevertheless the aerodynamic forces are still a product of local v^2 notionally)
The first can be expressed thus
Moment = ∫(from 0 to R) mxω^2 xsin(κ) dx
m is the mass of a small length of blade (dx)
x is the position along the blade up to R the radius
ω is the angular rate in radians per second, ω≈RRPM/10
κ is the coning angle in radians
MCf is the Moment about the flapping hinge causing the blade to (not) cone up
That integral gives
MCf = -1/3 mR^3 ω^2 sin(κ)≈ 1/3mR^3 ω^2 κ (for small κ)
The second moment can be expressed like this
Moment = ∫(from 0 to R) 1/2ρAİx^2 ω^2 x dx
Where
ρ is the density of air
A is the Area of a dx sized piece of blade (ie Chord)
İ is the ‘Coefficient’ of Aerodynamic force in the plane normal to the hinge (similar to CL ) but it also assumes an ideal twist (T, or ‘washout’) that reduces CL by an amount that ensures the Lift and Cf remain in proportion to minimize unnecessary longitudinal blade bending. So it embodies an x^-1 term, (so something like CL (R/(R-x))
ML this is the Moment caused by aerodynamics in the plane of the flapping hinge
x is the distance along the blade and ω is again the angular rate in Radians/second
so xω is the local v (speed of airflow, at that x)
That integral gives
ML = 1/6ρAİR^3 ω^2
Since ML + MCf = 0
Then ML= - MCf
1/3mR^3 ω^2 κ = - - 1/6ρAİR^3 ω^2
solve for κ (THE CONING ANGLE)
κ = ρAİ/2m
SO regardless of how exact that rough working is you can see clearly that the coning angle is not dependant on RRPM (equivalent to ω)
What you can see is that coning angle depends on the ρ (density) , A (the chord), the shape and Angle of Attack (CL) and the mass of the blade.
Since there is a İSTALL then there is also a κSTALL and it does not depend on RRPM
It is irritating to have to pointlessly integrate your pointless formula, especially since you obviously barely understand it yourself
Since you have chosen to align yourself with the playground oaf I seem to have no choice but to perform the pointless maths you ask for.
I say pointless because the variable cancels out prior to having to perform any integration anyway.
So your (pointless) set of ridiculous assumptions (eg twist for constant Induced Flow, and I used a similar (but better) assumption) are unimportant since.
The whole thing can be stated far more simply (which is generally a good thing) like this
Κ = arctan(L/CF)
Which for small L and Large CF is (to 4 decimal places) the same as
Κ = L/CF
Well it should be obvious that RRPM cancels out of that but here it is
L = a times RRPM^2
CF = b times RRPM^2
So
Κ = L/ CF = a times RRPM^2 / b times RRPM^2
CANCELLING OUT you get Κ = a/b
(where a is the collection of other (unimportant here) factors (density of air that sort of thing. b is another collection of other (unimportant here) factors (mass of blade etc))
Don’t blame me for giving that answer, I was goaded into it by an over self satisfied aeronautical engineering degree holder and a playground oaf.
I’m sure there’ll be all sorts of nit picking over the detail of my maths, but I don’t have time to improve it so please don’t bother. I think it is enough to prove my point though