


Joined Nov 2004
3,843 Posts

Quote:
Quote:
Quote:






[QUOTE=mnowell129]Geometrically they look negative, but aerodynamically they are around 0.
I define pitch as in relation to the plane of rotation, I'm pretty sure I'm on the same page there. In maintenance they refer to the pitch rigging which to me is just setting up the min  max travels. I complety get the difference between AOA and pitch. What do you mean by geometically and aerodynamically? Do you have washout in the blades? How do they look negative? 


Joined Nov 2004
3,843 Posts

Quote:





My blades look negative on the model but are close to 0 aerodynamically.[/QUOTE]
Mmmmm. I've never noticed the zero lift angle being used like that, just a zero lift line or zero lift axis. So for a cambered airfoil would be a negative angle from the chord line to the axis of rotation. Ok, will mill that over  feels like I'm getting somewhere. But the angle of attack will change quite a bit during flight. And I don't have a problem with AOA being positive  I'm looking specifically at the chord line in relation to the axis of rotation. If it looks negative on your model then what would a measurment of the chord line be in relation to the plane of rotation? A guess is ok. If I were to say a rotor or propeller is at a particilar rpm, and pitch, how would you know which direction the propwash or rotor wash was going? The pitch? It still seems like the pitch determines the direction of flow through the rotor disk to me. Unless in some extreme sitiuation of decelleration. So it still seems like to me you decide a few factors such as pitch, rpm and blade dimentions and the arrows get figured out from that. It seems like the popular explanation is getting the answer it want's by arbitrarily drawing the RW vector from below even though the pitch is positive. I just seems to go agains the grain  or airflow in this case. But I'll keep working on it........... 


Joined Nov 2004
3,843 Posts

Quote:
Quote:
If you pick a 1500 fpm descent, then recompute based on a upflow of 1500 fpm, figure out the rpm, drag, etc and compute the power required is negative, meaning you can take power from the rotor. Or at a different sink rate you get power required = 0, i.e. autorotation, etc. You can't just look at the pitch and RPM in a vacuum, you have to establish other conditions that you are evaluating for, then you get to draw vectors for that situation. You're trying to make it simpler than it can be. 



Joined Nov 2004
3,843 Posts

[QUOTE=yak55x]
Quote:





Following rate formula
Mike
writing down the explanations for the metric version of the following rate formula as a normal post was too complicated for me. I put it all in a Word document (Word 97 to 2003) and attached it to this post. Jochen 



A Copy of Jochens word document (just copy & paste) first, to have it somewhere else then in an attachment:
<< START QUOTE JOCHEN_K >> According to John Drake the formula for the following rate N of the main rotor of a model helicopter is: N = ω • a • c • ρ • R4 / (16 • I) Simple parts of the formula: ω is the angular velocity of the rotor in s1. You can calculate the rotor’s angular velocity using the following formula: ω = n / 60 • 2 • π (corrected according to JOCHEN_K) With n being the rotor’s number of rotations per minute (which, in the formula, is divided by 60 to get the number of rotations per second). c is the length of the blade chord in m. ρ is the density of the air in kg/m³ R is the tip radius of the rotor in m. The difficult parts of the formula: a = δ CL / δ α The δ in this formula denotes a difference. You have a graph, in this case the polar graph of your profile, and you want to know how steep this graph is at a certain point. On the graph you mark two points, one a bit to the left and one a bit to the right of your point if interest. From those two points draw vertical lines down to the xaxis (that’s where α is) and horizontal lines to the yaxis (that’s where you find CL). Now take a reading of the difference between the two CLvalues (δ CL) and the two αvalues (δ α). Before you divide δ CL by δ α, you have to convert the δ α value which is normally given in degrees, into radians. This is about the same as converting rpms to ω. Use the formula δ α [radians] = δ α [degrees] • 2 • π / 360 to do this. Now calculate a. I’ve had a look at the polars of the SG 6042 profile we’re often using. For a Reynolds number of 40000 (that’s the region the Micromum’s rotor is working in) I get something like a = 5.7, for the rotor of the RPG with a Reynolds number of about 120000, I get a = 6.9. I is the blade’s flapping moment of inertia in kgm². The basic formula for the moment of inertia is: I = ∫ r2 • dm With dm being the mass (in kg) of a very, very, very thin slice of the profile at a distance of r (in m) from the rotor axis. The ∫ is the sign for an integral, which is a mathematical expression for summing up something, in this case the masses of the thin slices multiplied by the squares of their distances to the rotor axis. We can express dm in another way: dm = m / R • dr The thin slice of the rotor blade has a thickness of dr (in m). If we divide the total mass m (in kg) of the blade by the blade length R (in m), we get something like the specific weight of the blade’s cross section, and when we multiply this specific weight with the thickness of the cross section we again get the mass of the slice. The formula for I now looks like I = m / R • ∫ r2 • dr We now integrate this expression from the blade root (r = 0) to the outer blade radius r = R (we are summing up all the tiny (r2 • dr) values) and get: I = m / R • R³ / 3 = m / 3 • R² That’s it. << END QUOTE JOCHEN_K >> Hello Jochen………. Now I understand this thing with the following rate. You made a very good job with explaining this. The integration of the inertia of mass assumes a constant density of the blade. If we use some tip weights, the resulting moment of inertia will be higher, because the radius is going quadratic to the equation! Or in other words: For showing the influence of the tip weights on the following rate, we will need a formula, which will calculate the moment of inertia with respect to the mass of the tipweight and its position. There will be some integration needed, too. Or just making a CADmodel of it, the CAD calculates it automatically. Concerning the followingrate, it will be better and more effective to make a tipweight really concentrated in the tip than just making the blades heavier with some heavy covering on full length that’s not needed! One question: Where did you get the dragpolars for the SG 6042 at low Reynolds numbers from? I searched for dragpolars sometime ago, only thing I found with some importance to modelism were profilecuts and dragpolars for Fi 156 STORCH and the suctionplanes (ABSAUGFLUGZEUG) AF 1 and AF 2. The value for a, which describes the steepness of the dragpolar at a certain point, will be different for every angle of attack you extract it. It’s a curve, and not a line, this dragpolar. I assume you calculated a for some point on the curve you think it would be in the right range, but the angle of attack changes along the blade, depending on the rotorRPM, the airspeed of the gyro, the angle of the spindle relatively to the airstream and last but not least the radius of the point on the blade you are actually looking at. I don’t even want to think about the twist of the blade. And one thing I really believe in is the following: Construction of good gyroblades doesn’t only include shaping the right profile, it also includes having the construction in a way that makes them twist in the right way when the blades are working…….. Bye Mike 



Mike,
here's an addition to my last post. It integrates tip weights into the formula for the moment of inertia. Polars: I got the values for Clift from a program called Profili v2.2. It’s shareware and costs something like 10 Euros. Jochen 



Hello Jochen.........
Good job. If I calculate the following rate of a given rotor, one time with adding a brass wire on the whole lenght of the leading edge and the other time with the same mass concentrated on the tip, I get some significant differences....... So i suppose that it would be best to use tungstenwire for the tip weights. Tungsten has almost the double density of lead and more than double density than the solderingwire commonly used. So it will help to concentrate the mass just a little bit more on the tip. But where to get some tungstenwire? Go for welding electrodes from TIGwelding. Its tungstenwire, diameter from 0.7mm up to 4.0 mm available, length about 200 mm, price: about two euros for diameter 1.5 mm. On your first calculation or the angular velocity you wrote ω = n / 60 • 2 • π / 360. Where does the 360 comes from? To my opinion it should be ω = n / 60 • 2 • π........... Bye Mike P.S.: Have you already tried the formulaeditor included in word? Its the perfect tool for writing formulas....... 



Hello Jochen..........
I changed the copy of your document. When I use the formula for a rotorblade something like the RPG with blade weight 60 grams, tip weight 10 grams, 800 revolutions per minute, Radius 0,65 meters and a equals 6.9 like you calculated, I have the following results: 1) blade without tipweight, mass 60 grams: J=0.00845 kg/m², N=46.7 (following rate) 2) blade with brass wire on full lenght, total mass 70 grams: J=0.00986 kg/m², N=40.0 3) blade with 60 grams, tip weight concentrated on tip with 10 grams: J=0,01268 kg/m², N=31.1 Okay, its clear: tip weights are better then a brass wire on full lenght for lowering the following rate. But: The values I get for the following rate are much to high to control this rotor. But the RPG flies with this rotor. Whats wrong? I checked my calculations twice, no error found. Some mistake with the moment of inertia? I checked it with the CAD, same result. The RPG is in the air, no doubt about this fact......... Possible reasons: Mistake in calculation, mistake in formula, or formula is okay for helicopters, but a well designed autogyro like the RPG doesn´t care about the control rate too much. I don´t know...... Bye Mike 



Mike,
I'm not sure what the factor a really means. John Drake had an original factor a of 5.73 which he then reduced to 3.75 and I don't know how he got this factor. That's one thing, the other one's that the formula is for helicopter blades. From what I've seen, these blades can't move 'vertically' as our gyro blades do around the flapping hinges, and that may influence the absolute values calculated. For relative calculations the formula seems good enough. Jochen 