Retreating blade AoA.

With blades pitch setting 3 degrees, the A.o.A disk is about 9 degrees (*), airframe horizontal, neutral stick, when the forward speed is 110 Km/h.

Now , with blades pitch setting 0 degree, the A.o.A disk should be about 14 degrees (*). Since the airframe balance requires always the GC on the perpendicular line to the disk center, then the airframe now will be +5 degrees nose up at 110 km/h, neutral stick.

To have the horizontale airframe, it requires put the GC more in front (i.e airframe to -14 degrees instead -9° during the hang test)

(*) A.o.A disk = ATAN (Drag /Lift)
 
Quote: Are you saying when you pull back the stick, you going to get a pitch down attitude? /Quote

A rotary wing aircraft is a pretty complex beast, it is often not intuitive what a change in any one parameter entails regarding all other parameters. Below is a small part of the output of my rotorcraft simulation program. I have calculated the trim values for a Magni M-16 at 60mph first using the 13.5ft rotor diameter of the actual aircraft and then increasing the diameter to 16.5ft. The aircraft actually trims slightly more nose up with the smaller rotor. (theta = 0.92°) against (theta=0.24°) for the 16.5ft rotor. The key here is the induced inflow lamda (lamI) which is less than two thirds for the larger rotor (lamI=0.00492) against (lamI =0.00674). To maintain flight equilibrium you have to move the stick slightly forward for the larger rotor (B1S = 2.57°) agianst (B1S= 1.71°) for the 13.5ft rotor. B1S is the angel between the bolt about which the blades rotate with the mast, positive forward.

Unfortunately the forum software can not handle a calc table holding all the values below, telling you that you are trying to post a message with more than 10000 characters, which, of course, is rubbish since even with the dots you have to place below to avoid that the software makes a complete mess of it, the table entries would be less than 2500 characters. Together with this explanation there are still less than 4000. This seems to be a major flaw.

"=i=i=i”....R=13.5ft....Results
xMR....0....xcg....-0.667....dxCG....0.667....iRigMR....8.7
cTs....0.0581....cT....0.00198....cLs....-4.9823....cL....-0.16963
716....aN....3.03°....a1....1.27°....b1....0.60°
mu....0.146....lamD....0.01681....lamI....0.00674....mu*a1....0.00324....lamNf....0.01357
oMR....181.76....m/s....oMR....596.31....ft/s....oM....44.17....rad/s....nRot....421.8....1/min
A1S....-0.24°....B1S....1.71°....alfaD....9.18°....alfaNf....0.14°....alSm....-0.44°
thtaN....2.50°....t75....2.50°....thtaTR....-0.03°
Trim....phi....0.03°....theta....0.92°
vBx....26.81....vBy....0....vBz....0.43
fMR(1)....-560.12....fMR(2)....-10.51....fMR(3)....-4222.29
fHS(1)....-19.55....fHS(2)....0....fHS(3)....57.8
fCG(1)....-66.77....fCG(2)....2.48....fCG(3)....4137.79
fWN(1)......0.......fWN(2).......0....fWN(3)....0
fR(1)....9.31E-09....fR(2)....-5.66E-10....fR(3)....3.41E-08
Moment....Equilibrium
mMR(1)....-11.8....mMR(2)....-228.84....mMR(3)....2.13
mHS(1)....0....mHS(2)....177.93....mHS(3)....0
mWN(1)....0....mWN(2)....0....mWN(3)....0
mR(1)....-6.27E-10....mR(2)....3.26E-10....mR(3)....1.01E-09
MR....Fx....-560.12....Fz....-4222.29....thtaClmb....0....torque....0
dxMR....0.0542....Fx/Fz....7.6°....sigma....0.034....dlt....0.79878
"=i=i=”....R=16.5ft....Results
cTs....0.05772....cT....0.00161....cLs....0.00281....cL....8E-05
716....aN....3.01°....a1....1.40°....b1....0.66°
mu....0.161....lamD....0.017....lamI....0.00496....mu*a1....0.00395....lamNf....0.01305
oMR....165.06....m/s....oMR....541.54....ft/s....oM....32.82....rad/s....nRot....313.4....1/min
A1S....-0.24°....B1S....2.57°....alfaD....7.77°....alfaNf....0.11°....alSm....-1.17°
thtaN....2.50°....t75....2.50°....thtaTR....-0.03°
Trim....phi....0.03°....theta....0.24°
vBx....26.82....vBy....0....vBz....0.11
fMR(1)....-500.34....fMR(2)....-10.51....fMR(3)....-4235.36
fHS(1)....-20.3....fHS(2)....0....fHS(3)....80.68
fCG(1)....-16.98....fCG(2)....2.49....fCG(3)....4138.29
fWN(1)....0....fWN(2)....0....fWN(3)....0
fR(1)....6.33E-09....fR(2)....-1.54E-09....fR(3)....-4.35E-08
Moment....Equilibrium
mMR(1)....-11.8....mMR(2)....-298.65....mMR(3)....2.14
mHS(1)....0....mHS(2)....249.41....mHS(3)....0
mWN(1)....0....mWN(2)....0....mWN(3)....0
mR(1)....-1.71E-09....mR(2)....-4.77E-09....mR(3)....1.63E-09
MR....Fx....-500.34....Fz....-4235.36....thtaClmb....0
dxMR....0.0705....Fx/Fz....6.7°....sigma....0.0279....dlt
 
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Quote: Now , with blades pitch setting 0 degree, the A.o.A disk should be about 14 degrees (*). /Quote
Using a full trim program gives an AoA slightly larger, about 16° but the main difference I get is, that the rotor speed and the induced inflow ratio are vastly different. Below are the values from my calculation. What are the rotor speed and inflow values you get?

thatN= 2.5°........n=421 1/min............lamI= 0.00674
thtaN= 0.0°.......n=583 1/min............lamI= 0.00525
 
Juergen,
I specify my messages #16 and #21:
Diameter x Chord: 8.26 x 0.216 m
Lift: 3640 N
Forward speed: 110 km/h
Sea Level

With aerodyn. pitch setting of 3°:
Steady Rrpm: 342
A.o.A disk : 8.85°
Longitudinal flapping angle: 2.27°

With aerodyn. pitch setting of 2.5°:
Steady Rrpm: 356
A.o.A disk : 9.27°
Longitudinal flapping angle: 1.94°

With aerodyn. pitch setting of 0°:
Steady Rrpm: 450
A.o.A disk : 13.72°
Longitudinal flapping angle: 0.65°

My finite element calculations does not use the inflow coefficient. Your value seems amazing to me.
For comparison, the rotor of Mike Goodrich's ELA 07 turns at 345 -350 rpm at this weight (300 m ASL)
 
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Thank you for providing the values, JC, I will run the sample calculations tomorrow. From my experience the disk angle depends quite strongly on the drag coefficient. Could you please post the one you are using? I take it that you use the same one for all calculations. As for the induced inflow I have so far not seen any calculation where it is omitted and all the naca/nasa papers I have seen so far, even those dating back to the 1930ies, stress how important it is to include induced inflow in the calculation. Perhaps you could explain your rational for not using it?

PS: If you are using a finite element scheme the induced inflow is included via calculating the circulation for the blade element under consideration, that is how I have implemented this in my FEM rotor program.


Merci d'avoir fourni les valeurs, JC, je vais effectuer les calculs de ces examples demain. D'après mon expérience, l'angle du disque dépend assez fortement du coefficient de traînée. Pourriez-vous afficher celui que vous utilisez ? Je suppose que vous utilisez le même pour tous les calculs. En ce qui concerne l'afflux induit, je n'ai vu jusqu'à présent aucun calcul où il est omis et tous les documents de la naca/nasa que j'ai vus jusqu'à présent, même ceux qui remontent aux années 1930, soulignent combien il est important d'inclure l'afflux (l'entrée ???) induit dans le calcul. Peut-être pourriez-vous expliquer votre raison de ne pas l'utiliser ?

PS: Si vous utilisez un schéma d'éléments finis, l'entrée (l'afflux???) induite est inclu par le calcul de la circulation pour l'élément de pale considéré, c'est ainsi que j'ai mis en œuvre cette méthode dans mon programme de rotor FEM.
 
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From my experience the disk angle depends quite strongly on the drag coefficient. Could you please post the one you are using? I take it that you use the same one for all calculations. As for the induced inflow I have so far not seen any calculation where it is omitted and all the naca/nasa papers I have seen so far, even those dating back to the 1930ies, stress how important it is to include induced inflow in the calculation. Perhaps you could explain your rational for not using it?
Juergen,
My method consists in entering the angle of attack of the disc, the rpm of the rotor, the flapping angles a1 and b1, by successive trial and error, until the required lift is obtained and cancel the moments of rotation, pitch, and roll.
When this is achieved, the values entered are the steady flight conditions.
I take into account, of course, the induced speed (here: 0.93 m/s)
This method could not be usable in the 1930's, without personal computer, nor Excel

Sans titre.png
(Continued with 5000 automatic calculation boxes in which an average of 5 operations are performed)

I take a constant Cd0 of 0.011 for the blades profile drag.
More recently, I considered that cd min of profile was variable with the angle of attack of the element, to obtain more accurate results in cases of unsteady flight (Rrpm deceleration rate under low "g"), and which gives the same steady flight results.
 
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Thank you for the additional information, JC. It took me a while until I realized that your lift value of 3640N was not an error, but that you are probably calculating the case of an aircraft with just the pilot and no passenger. The gross weight of the ELA 07 is rather something like 4400 N. After changing this the results we get are similar, but it is no surprise that we don't get the exact same values. As I said before the results depend very much on the drag coefficient. I have slightly modified that one to get the same disk angles. My rotor speeds are consistently slightly higher but I think that the values are close enough. One would now have to find results from flight tests to further tweak the models but since my ultimate goal is to calculate the flight stability I will not make that effort. The stability values are very "elastic" regarding minor changes in geometry.
Here are the results of the samples, 0.013 and 0.012 are the drag coefficients I used to get your disk angle:

theta N​
3.0​
0.013​
2.5​
0.0130​
0.0​
0.012​
JC​
JH​
JC​
JH​
JC​
JH​
V km/h​
110​
110.38​
110​
110.58​
110​
109.75​
n 1/min​
342​
351.3​
356​
368​
450​
452​
mu​
0.2​
0.194​
0.134​
lamI​
0.908​
0.91​
0.93​
alfa Nf​
6.49​
7.33​
13.61​
a1​
2.27​
2.1​
1.94​
1.82​
0.65​
0.61​
alfaD​
8.85​
8.61​
9.27​
9.15​
13.72​
14.22​
For those who would like to see how these results were calculated I have made the program available here:
https://www.magentacloud.de/lnk/85BpEZMh

password is again alfamax and you need octave to run the program (see previous posts)
 
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Quote: is there an easy calculation to work out aerodynamic pitch from geometric pitch for a said airfoil? /Quote
I am not sure I got the question right. If you want to find the angle of zero lift for a given airfoil profile you can use JAVA Foil. With that information it would be up to you to select a constant geometric pitch angle for your aircraft depending on the environment you fly in.
 
Juergen, I see that you are using the Greek letter lamI (λ ?) as Induced Velocity (Vi)
Usually the "rotorists" uses λ as the dimensionless ratio V/ Vc where V is the quotient airflow through the disc by its surface, and Vc is the tip speed.
Are you sure you haven't confused the two?
 
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A rotary wing aircraft is a pretty complex beast, it is often not intuitive what a change in any one parameter entails regarding all other parameters. Below is a small part of the output of my rotorcraft simulation program. I have calculated the trim values for a Magni M-16 at 60mph first using the 13.5ft rotor diameter of the actual aircraft and then increasing the diameter to 16.5ft.
Surely you mean radius here, not diameter...
 
Quote: Are you sure you haven't confused the two? /Quote
Quote: Surely you mean radius here, not diameter... /Quote

You are both correct, #1 I had the dimensionless ratio lamda in my table and then changed the values to induced velocity in m/s, since JC had posted that value, without correcting the designation of the row and #2 I had changed the rotor radius from 13.5 to 16.5 ft. Thank you for pointing out these glitches..;-(
 
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