Screening of Gs

I am surprised at the slowness of change of RPM announced by my simulation of rotor: More than 10 seconds. Is this possible?
( Rotor 10° in cruise to 400 r.p.m , +5° added give 1,19g instantaneously, and 2g, 10 second later to 570 r.p.m)

This seems excessively long. Did you get the moment of inertia right?

-- Chris.

Sounds like thems VERY heavy blades.
How long dose it take them to settle back to 1G rpm?

Chris, Birdy : 400 r.p.m à G=1 (466pounds 50 mph) diameter= 21 feet One blade = 16 pounds

What's the cruising speed, chord of the blade and the built in balde pitch angle and what induced velocities does your simulation predict for the two rotor states?

kolibri, Cruise speed: 80 km/h Chord:180 mm Blade pitch : 2,7° (aerodynamic) Induced velocities: 0,52 m/s through the disk for 10° (rpm stabilised in cruise) and 1,15m/s for 15° (after new rpm stabilised)

Induced velocity

Induced velocity

Jean Claude,

it has been found that the induced velocity in high speed
flight may be approximated by

vi= T/(2*ro*A*Vg)
(see file RaM558----.pdf on page 563, see link below)

where T is the rotor thrust (assumed to be the mass of the
gyro times gravity), ro is the density of air, A is the rotor
disk area and Vg is the forward speed of the gyro.

using your values gives vi=1.17 m/s for the first case
(rrpm=400 1/min) and 2.35 m/s for rrpm=570 1/min
(it seems our results differ by a factor 2, that seems
to indicate our formulae are different by that factor)

Calculating and comparing the angle of attack of a blade
section at the three quarter (0.75) station of the rotor
gives a difference of 20% for case 1 and 25% for case 2
I feel that this may be the cause for the strange results
you get.

Please check my calculations in the spread sheet.

Another idea is to check the torque applied to the rotor
since this will decide the rotational acceleration. I
would like to suggest to split up the torque into accele-
ratin and decelerating torque as in naca report 716.
That way you would be able to check if your acceleration
time fits your blade inertia and torques.

Since the reports are to large to upload to the forum
I have packed a nice parcel and placed it here:


http://www.divshare.com/download/10187946-836

Cheers,

Juergen

Juergen, In my simulation, the induced velocity is not a prior calculation. Here, the lift and rpm are détermnés only by calculating the resultant of lift and drag for 10 sections of blades (between the center and tip) for 24 positions of rotation (every 15 °). Then the total torque C is calculated for different rpm and the values of rpm which gives C = 0 is the autorotation. The addition of 20 blades sections gives the full force of the rotor.
I had already observed that the induced velocity differs by a factor of 2, and obviously it hurts me because I can not understand why.
But since it is the couple who give the autorotation and who I obtains the rpm correct, I assume that the acceleration is correct.
I would however confirmation by experiment, because it seems too slow. Early my goal is to vériier improvement (or deterioration) with the rotor blade pitch, and twisting. I could see the danger of too improve the rotor: the risk of stalled in flight if CL blade is too large (pitch angle exceeds 3.7 ° for example). Furthermore, I can observe the vibrations of aerodynamic origin (long fuzzy to me)
I will read the report NACA 716
Thanks
Jean Claude

Jean Claude wrote:

In my simulation, the induced velocity is not a prior calculation. Here, the lift and rpm are détermnés only by calculating the resultant of lift

Click to expand...

If you calculate your lift from balde element theory (as seems to be the case since you mention balde stations) you need the local angle of attack. As I mentioned there is
quite some difference presently so you may want to check, whether the lift you calculate is correct.

Cheers,

Juergen


PS: what programming language do you use?

Juergen, I use a simple Excel spreadsheet: 10 columns (blade sections) and 24 lines (angular position)
Table 1 for the velocity component tangential to the blade
Table 1 for the axial velocity component of disk
Table 1 for the attack angles of sections
Table 1 for the Cl
Table 1 for the lift
Table 1 for Cd
Table 1 for the drag
Table 1 for the resulting drag + lift
Table 1 for the angle of the resultant in relation to the shaft bearings
Table 1 for torque rotary
Table 1 for torque beat
I give: Diameter, Chord, Cd min foil blade, rpm, flight speed, disk angle, flapping angle, pitch blade section
I Obtain: The torque of rotating, flapping torque, lift, drag the rotor.
Giving rpm and flapping angle which gives 0 for torques, I obtains flight conditions
Curiously, taking as axial velocity through the disk, just flight speed x Sin (disk angle) results are already consistent with those of Beaty!
Subsequently, I completed with the cone angle and assumed an distribution of the induced velocity (from front to rear) to check the side effects and vibrations.
Little differences
Jean Claude

If I get you right you omit the induced velocity altogether, in which case I would be at quite a loss if I had to explain how the lift could be correct,
while your local angle of attack would be off by 30%.

It surely is a good idea to check your calculations against experiment and best of all there is no need to conduct the experiment yourselfe.
You will find a very comprehensive test on a gyro rotor in this report:

http://hdl.handle.net/2060/19930091587

Please drop a line to the list telling us how your calculations compare to the values measured there.

Cheers,

Juergen

Juergen,
This is what gives my calculation taking no account of the induced velocity through the disc. Just flight speed x Sin (disk angle):
With diameter: 21 '(only drag: 4') chord: 7 ", min CL: 0.015, stall: 12 °, flight speed: 50 mph, lift: 470 lbs, blades: flat 2.7 ° (aerodynamic)
I Obtain: 398 rpm and disk angle: 10.3 ° blade flapping: 1.83 ° (bearings: 8.47 °)
For the same data, with mean CL: 0.50, Chuk Beaty gives: 400t/mn, 10.0 °, 1.89 °
Just to compare my results, can anyone predict the difference in torque rotary self 400t/mn between 10 and 15 ° for the same flight speed (50 mph)
Thank you
Jean Claude

Jean Claude,

what I am trying to point out is, that it is fairly easy to tune your program to give
correct results at one point of the flight envelope, e.g. by tweaking the lift curve
slope, yet with that model you will get wired results at any point somewhat further
away from your "design" point.

I hope you don't find me overly picky, actually I just try to help answer yesterdays question.

Veuillez vous, chère Jean Claude, accepter l'expression de mes sentiments distingues.

Juergen

Juergen. I undestand. I tried to take into account the velocity induced by the overall disk, but as you probably know, Excel does not loop calculation. However, the lift changes the flux through the disk and modifies the flow angles ... and lift!
Jean Claude

Jean Claude,

it is exactly as you say, induced velocity and rotor thrust are interdependent and therefore an iterative solution would be required.
If you want to head down that road I propose to use octave or scilab (both matlab clones) since they offer routines for iterative solutions.
Alternatively you could use the iterative scheme proposed by Padfield (page 123 in his book "Helicopter Flight Dynamics: The Theory and
Application of Flying Qualities and Simulation Modeling") You could program the formulae given there in Basic (Excel comes with a built in Basic
interpreter, just press Alt+F11 when you have opened an Excel file to get access to the Basic developement environment.
Lastly you could use the formula I mentioned above without iteration. Taking induced velocity into account using just the "static" value calculated
via that formula should give *far* better results than chucking it altogether.

Happy simming!

Juergen

PS: Just in case you look for some bed time reading on vortex ring state:
http://terpconnect.umd.edu/~leishman/Aero/vring.html

Thank you very much, Juergen

In the first simulation, the air flow through the rotor was not supposed deviated by the unmatérial disk. But, as a fixed wing, I have attributed an deflection of each blade on herself (induced drag). the results were correct for normal speeds.
After discussion with Juergen, I deleted the deflection of each blade on herself, but I have supposed traveling in a general flow deflected. The results are also correct for normal speeds. The torque produced by transient variations of the disk angle is now greater.The time setting of rpm is now 6 seconds from the cruise flight (realistic?)
The results remain unsatisfactory to the vertical descent (Cx disk>2 !)
Jean Claude

6 seconds is long for the rotor I fly but might be OK for heavier rotors. My blades (8,4m Aircopter blades) adjust to normal rpms after a 2G (60° bank) turn in about 3-4 seconds.

-- Chris.

Jean Claude,

in vertical descent the rotor may be in different flow regimes. These are the wingmill brake or the vortex ring state. In both states Galuerts formula (the one in answer #7) for induced velocity (vi) does no longer apply. There are approximations to vi but investigation in these rotorstates is rather a case for a CFD analysis. If you try to get more than an approximate solution to these states without a CFD program you are, as far as I know, wasting your time. One accepted approximation is that in the windmill brake state the rotor has the characteristics of a parachute, i.e. you can assume the thrust of the rotor to be 0.5*rhoAir*cD*vDescent^2*A, where A is the area of the rotor disk (pi*R^2= pi*R*R). A ball park figure for cD is 1.33.

Cheers,

Juergen
PS: By the way, the dynamics of a rotor are, to the best of my knowledge, a pretty advanced topic. Why is it so important for your project?
PPS: Excel is, IMHO, the wrong tool for a transient analysis. You need a full fleged programming language (C++/Java) for that job to
implement a time marching algorithm.