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Old Oct 02, 2014, 08:36 PM
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Attempts at sub-120g autogyros.

Hello.

In this thread I'm going to post my experiences with making light ( sub 120 gram ) autogyros out of materials and equipment I have stored over the last few years. A lot of what I will post will be in retrospect as I've already done it, but the last updates of this thread will be more recent. I will slowly post all the steps I've taken up to the point I get something to fly.

This is the second time I've ever tried making a model autogyro, my first attempt was in 2008 with a micro ( sub 30g ) RET single rotor gyro. It never flew. Since then I've played a lot with helicopters, and this has certainly helped me with autogyros.

One of the things I did (and still do ) is play a video game called Garry's mod. Although it isn't purposely designed for this, one can build and play with mechanical systems that are mostly accurate with real life. I spent a lot of time attempting, and eventually getting an autogyro fly to in the game. One of the things I learned was the critical importance of rotor blade lift to drag ratio. This was the key to successful flight ingame.

Based on that experience, it seemed reasonable to me that it should be the first thing I should focus on. After reading comments online, many of those in Mnowell's aerodynamics thread (Thank you for posting this), I made up my mind that designing the blades should be the first step.

The first thing I thought was: At what reynolds number range will the blades be operating in? I didn't have any park flyer sized receivers, and really no batteries or speed controllers for that size either. All I had to build autogyros with was very small, micro sized equipment. Most of the receivers I had were from a walkera 4#6.


The 4#6 has a weight of around 75 - 80g ready to fly. Seeing as I would probably use it's electronics to make an autogyro, it seemed reasonable to make one that weighed about the same. This was important as I could find out the reynolds numbers a 4#6 blade operated in. When It was still operational, I recorded the sound it's main rotor made to find out it's RPM. I learned you could do this by watching this video:

R/C helicopter - DIY hatcam tachometer
R/C helicopter - DIY hatcam tachometer (6 min 49 sec)

The recording ( which is attached to the post) showed a frequency of spike of 91 hz, that's to say, the rotor blades passed in front of the microphone 91 times a second ( fast! ). To double check this, I examined the audio graph for a very small time interval, from 0 to 0.1 seconds. By multiplying by 10 (as there ten 0.1 seconds in 1 second) and counting the number of crests in the interval, I could get an approximate value for the sound's frequency:

F ~ C/T

where:

H = Crests in the sound interval
T = As measured from zero, the time until crest counting stops. In other words, the upper limit of the measured time interval, assuming it is smaller than 1 and the interval starts at zero.
There where around 9 humps so:

F ~ 10*9 = 90 hz

where:
F = Rotational frequency.

90 hz is very close to the audacity's measure of 91hz. Since the program is more accurate than the simple crest count, I used 91 for further calculations.

Note
: Further in my posts I'll measure RPM in the same way as I did here.

Knowing the recoding frequency, RPM could be calculated. The rotor had two blades and for one revolution, the blades would sweep the microphone two times. This made the calculation:

RPM = R*60/N = 91*60/2 = 2730 rpm.


where:
R = recording frequency
N = number of blades

Given that velocity for a rotating object is calculated with rotational frequency, the actual value of importance is:

F = R/N = 91/2 = 45.5 hz


As I knew the rotor frequency, and could measure the blade length and chord, I could calculate the reynolds number for any spanwise position of a blade. Reynolds number is defined as:

Re = V*L/U

where:
V = velocity
L = Characteristic length. In this case the blade chord.
U = Kinematic viscosity of medium

To find U I looked at this chart from the engineering toolbox: Dry Air Properties. I did the recording at room temperature ( about 20 C/70 F ), meaning a kinematic viscosity between 1.34*10^-5 and 1.66*10^-5 . I guessed at a value of around 1.50*10^-5. With this value the Re expression can be rewritten:

Re = V*L/(1.5*10^-5) = (10^5)*V*L/1.5

For the velocity term I substituted the tangential velocity of a blade:

V = F*r*2*pi

where:
r = spanwise position (radius ) on blade

Making Re:
Re = (10^-5)*F*r*L/1.5

With this I calculated the reynolds number at the tip of the rotor blade:

r = distance from bolt hole to tip + distance from center of shaft to bolt = (12.3 + 1.5)cm = 0.138 m
L = 1.5 cm = 0.015 m
V = 0.138*2*3.14*45.5 = 39.43 m/s ~ 142 km/h
<- pretty fast! I thought it would be slower.
Re = 0.015*(39.43)*(10^5)/1.5 = 39430

So a Re of around 40k at the tips.

My next thought was, If I was not be able get the autogyro as light as the 4#6, the blades may have to operate at a higher Reynolds number as a result. If so, how quickly does reynolds number rise with size and weight? I did exact calculation for a Trex 450 with a rotor diameter of around 678 cm and a mass of 780g. Although RPM can vary greatly with this model, an efficient setting for endurance flight is 1000 2000 rpm for 335 cm blades.

F = 2000/60 = 33.33 hz
r = 335 + 4 = 339 cm = 0.339 m
V = 6.28*0.339*33.33 = 71 m/s = 255.6 km/h
L = 3.3 cm = 0.033 m
Re = 0.033*71*(10^5)/1.5 = 156200


Around 80k 160k Re. That's about four times the reynolds number for a helicopter almost 10 times heavier ( 780g/75 g = 10.4 ) with a rotor more than twice as big ( 339cm/138cm ~ 2.4 ). Since the Re increases so slowly, I thought it safe to assume the autogyro blades would operate at reynolds numbers bellow around or slightly higher to 50k.
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Old Oct 03, 2014, 05:30 PM
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Choosing an airfoil.

After knowing the reynolds number range, it was time to select an airfoil. I didn't find many papers reguarding 50k Re. Most of what I found related to airfoils at Re < 10k or Re > 200k, both far away from my point of interest. I did find one paper, by Thomas J. Mueller, which was very good and targeted the 50k range in detail. He experiments with the effect of camber as well as trailing and leading edge shape. By his tests a circular arc, 4% cambered plate with 1.93% thickness performed well at 60k. This is very noteworthy as it wouldn't be the first time someone made the same comment.The rest of my research came from reading rc groups. Here are the threads that stood out:

I need a glider airfoil for very low Reynolds number
Very low Re airfoil wind tunnel data?
Micro airfoils
The perfect airfoil for micro stuff..
Airfoil Performance at Very Low Reynolds Numbers
Why is flat airfoil good for indoor 3D????
Sources for Peanut and Bostonian Airfoils <- This thread contains some very good data on thin airfoils. Highly recommended.
Airfoil for 24" wingspan biplane <- This thread contains Jef Raskins description of the 4% cambered airfoil. Very good read.

The paper by Mueller can be found here:
Aerodynamic measurements at low reynolds numbers for fixed wing micro-air vehicles

Very slight camber seemed to provide the best results. To prove this for myself, I set out to do an experiment.
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Old Oct 06, 2014, 02:01 PM
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The effects of airfoil camber on rotor performance.

The force an autogyro rotor produces can be assumed to be perpendicular to the plane of the disk. This force increases with the angle of the disk relative to the velocity of the air past it. I'll call this angle of attack.


This force can be divided into two components, one perpendicular to the velocity (lift) and one parallel to it (drag ). As a result, at very low angles, most of the force it will produce lift. At larger angles, more of the force is redirected backwards, increasing the drag.


The lift to drag ratio compares how much lift is produced for a given drag, as such, it is a direct indicator of rotor efficiency. Consider this:

LD= L/D

where:

L = Lift force
D = Drag force
LD = Lift to drag ratio

For a gyroplane to fly, the lift must equal the weight of the machine:

L = w = m*g


where:

m = Mass of gyroplane
g = Acceleration due to gravity

The drag must be counteracted by an equal force acting in the opposite direction. Otherwise the machine would slow to a halt and fall. As it needs to be moving in order to fly, it will require power:

Power = Force * Velocity

P = F*V

F = D -> P = D*V

where:

P = Power required to fly
F = Force being applied

Looking at the definition for LD, it can be seen drag can be written in terms of lift:

LD = L/D -> L/LD = D


D = L/LD

As power depends on drag:

P = V*D = V*L/LD

As can be seen, LD is diving the whole expression. Thus if one increases LD, less power is required to maintain flight.

I wanted to see how blade camber affected the lift to drag ratio. As a gyroplane rotor is pneumatically driven, it extracts power from an airstream. That power extraction is seen as a reduction in the speed of the air behind the rotor disk. Going back to power and rewriting in terms of Force:

P = F*V

F = P/V


It can be seen that if the rotor can extract more power from the air, it will produce more force as a result. If not all this force is used to counteract weight, then the angle of attack can be reduced, improving the lift to drag ratio, and reducing the power required to fly. It is in everyone's best interest to make the rotor as efficient as possible at extracting power from the air.

I compared 3 sets of blades that had identical planform, area and twist ( of which there was none ) . The only aspect I changed was the camber, which was roughly constant across the whole blade. The maximum camber point somewhere around the 40 - 50 % chord. These were the cambers I tried:

- A flat plate ( 0% camber )
- A slightly cambered plate ( roughly 4% camber )
- A highly cambered plate ( roughly 10% camber )

The rotor blades that were most efficient would be those with the slowest air behind them.

To measure the speed of the air, I took a thin paper strip and taped it to a straw. The idea was, the long, flexible strip would be easily bent by the drag it produced if placed in an airstream. The speed of the airstream could be measured by watching the deflection of the strip. If it bent a lot, the airspeed was high, if it barely moved, the airspeed was low. The most efficient profile would be the one that bent the paper less.

I made a rotor hub to mount the blades on, and some fixtures to hold the rotor in front of a fan. The fan would be used to create an airflow and spin the rotor. Before I measured the airspeed, I waited for both the fan and the rotor to reach a steady speed.

These were the results
Effect of airfoil camber on small autogyro rotors (3 min 49 sec)

- The flat blades reached an RPM of 2550 rpm. Although the rotor speed was high, the airflow behind it was noticeably fast. The paper was substantially bent. By placing my hand close to the tips of the blades, I could feel a smooth transition to the free stream created by the fan. By getting closer to the center of the rotor, the airspeed would slow down, up to a point, but overall, it was smooth. This indicated to me a poor lift to drag ratio. The drag of the blades was low allowing them to spin quickly, but the poor lift they produced barely disturbed the air. The reynolds number at the tips was around 37000.

- The slightly cambered blades reached an RPM of 2820 rpm. The airflow behind it was very slow. It was so slow in fact that if I moved my hand close to the tips, the transition between the free airstream and the rotor disk was almost abrupt. there was a very sharp increase in airspeed once you left the wake of the rotor. This indicated to me a very good lift to drag ratio as the air swept by the blades was highly slowed down. The reynolds number at the tips was around 40700.

- The Highly cambered blades reached an rpm of around 2220 rpm. It had the lowest rpm of all the profiles. The air behind it was much slower than that of the flat blades, about as slow as those of the slightly cambered blades. The transition close to the tips was also abrupt. To me this indicated an improvement in lift to drag ratio over the flat blades, but the increase in drag due to the high camber hindered the lift gains. The overall lift to drag ratio was lower then that of the slightly cambered blades as they did not spin as fast, yet the increased lift allowed the air to be slowed as much as the slightly cambered blades did. The reynolds number at the tips was around 32000.

In conclusion, all the comments people have made are right. A roughly circular 4% cambered airfoil works very well at Reynolds bellow 50k. The blades that used this camber performed the best. Camber can also drastically increase the lift to drag ratio of a rotor. As the power required to fly decreases with this ratio, cambered blades are necessary to achieve efficient flight. This is true only up to a point. Beyond it, the rotor may still produce the same lift, but at a lower rotor RPM. I imagine at very high cambers, rotor rpm would decrease drastically due to the increase in airfoil drag, decreasing overall rotor efficiency as a result.

The last tests I did with these blades, especially the 4% camber ones, was placing them outside the window of a moving car. I moved the rotor though the air at model flying speeds ( bellow 30 - 40 km/h ). I noticed that I needed a large angle of attack, around 30 - 40 degrees to produce substantial lift. I found this unacceptable. I needed larger blades as the extra lift they produced could be converted to a lower angle of attack. Thus my next step was making a large set of rotor blades using the 4% camber airfoil.

Note: JochenK's rotor head design.
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Old Oct 09, 2014, 02:01 AM
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Structural challenges of thin blades and pipe construction.

A low reynolds number airfoil should be very thin. While this is wise from an aerodynamic point of view, it's a poor choice from a structural perspective. A rotor blade can be thought of as a long, thin strip. This shape has the characteristic of being very torsionally flexible about it's length. Making this strip even thinner, as the aerodynamics recommend, makes the strip even more flexible. This poor torsional rigidity makes the choice of airfoil a critical consideration. As was seen before, a cambered airfoil is desirable for good rotor performance. This type of airfoil produces a torque that will bend a rotor blade to a lower angle of attack, the effect increasing with airspeed.


This reduction in angle can be very detrimental to the power a rotor can extract. As this power extraction directly affects how much lift a rotor can produce, the blades should be as rigid as possible to avoid this effect. There are two solutions for this: The blades are constructed out of a very rigid material, or they are somehow made thicker. Two questions arise from this though:

1) What material is rigid enough, yet light enough to be used sizeably, to make a thin cambered strip stiff in torsion?

This directly points to a composite material, carbon fiber seeming like the first choice. As I didn't have carbon cloth and resin, the only sources of sizeable CF I had were a 0.5mm thick sheet, and a set of broken Trex 450 blades. I decided to use both sources to make a set of blades, then compare which produced the best results.

2) How can the recommended ( 4% camber, circular arc ) airfoil be modified to allow greater thickness, without substantially affecting performance?

The mean camber line is a critical factor for airfoil design. It is the mean distance between the top and bottom surface of a section. I had to make sure my modifications did not change this parameter, as to produce the least alterations in performance. I was likely to build this airfoil, thus I wanted to make the modifications very simple. One change is to add flat bottom to the section. This makes it substantially thicker, while making construction no more complicated. It affects the mean camber line halving the maximum camber from 4% to 2%. To regain the 4% camber line, I needed to change the thickness of the airfoil. By increasing the maximum thickness from 4% to 8%, the camber line could be restored.


This increase in thickness further helps torsional rigidity, yet the airfoil is substantially different from the recommended one. I didn't know how it would perform, so I decided to make a set of blades to find out. Carbon fiber wasn't required for this, I could use a softer material instead. Balsa wood seemed like a good choice. It's light and easily shaped, and relatively rigid too. An alternative way to make blades I considered came from DIY wind turbines. It's common practice in this area to make blades out of PVC pipe. It gives a circular cross section, and depending on how it's cut, it can provide twist. These characteristics along with the ease of building such a blade ( it involves drawing a template on a pipe, then cutting it) made it very appealing.

It seemed very straight forward, but what wasn't so obvious was the size of the pipe. What diameter should the pipe have for a given blade chord and camber? To solve this problem I looked at saggita of a circular arc. The wikipedia article on this shows the equation for the radius of the circle, given arc depth and width.

r = (s^2 + l^2)/(2*s)

where:
s = saggita of arc ( depth )
l = half of the chord ( width )

For the purpose of a blade, these parameters can be substituted. As we don't know the half chord of a blade, but it's total chord, l can be rewritten:

l = c/2

where:
c = blade chord

As the recommended airfoil has 4% camber, and ideally would have no thickness, we can use this percentage to find find the depth of the arc, in terms of blade chord:

s = c*t

where:
t = maximum camber in terms of chord.

Substituting in the equation for r:
r = (s^2 + l^2)/(2*s) = ( (c*t)^2 + (c/2)^2 )/(2*c*t)
= (c^2)*( t^2 + 1/4 )/(2*c*t)
r = c*( t^2 + 1/4 )/(2*t)


Having found the radius, the diameter of the circle can be found:
d = 2*r
d = 2*c*( t^2 + 1/4 )*(2*t)
d = c*( t^2 + 1/4 )/t

where:
d = circle diameter

If one wants to find the diameter based on camber percentage, the expression becomes:
d = c*( (y/100)^2 + 1/4 )*100/y

where:
y = camber percentage

A curious question that comes from knowing the diameter is, how many blades can one make with a single pipe? I thought this was a great idea as untested autogyros are bound to crash, and having many spare blades would be very convenient; If the blades are straight ( constant chord ) and are cut vertically on the pipe, the number of blades can be expressed as the ratio of pipe circumference divided by the arc length of a given chord:

D = pi*d
n = D/A

where:
n = number of blades
D = pipe circumference
A = chord arc length

Circular arc length is defined as the angle swept by the arc times the radius of the circle that forms it:
A = j*r

where:
j = angle swept by arc

In this case, r is the one previously obtained:
A = j*c*( t^2 + 1/4 )/(2*t)

By looking at the diagram for a saggita, one sees that the swept angle is half the angle between l and r.

i = j/2
j = 2*i

i = angle between l and r.

This allows the relationship:
sin(i)*r = l

by substituting r and l for previous definitions one obtains:
sin(i)*c*( t^2 + 1/4 )/(2*t) = c/2
sin(i)*( t^2 + 1/4)/t = 1


Solving for i:
i = asin( t/( t^2 + 1/4 ) )

Having found i, one can substitute in A:
A = j*c*( t^2 + 1/4 )/(2*t)
= 2*i*c*( t^2 + 1/4 )/(2*t)
A = asin( t/( t^2 + 1/4 ) )*c*( t^2 + 1/4)/t


Substituting once more in the definition for N we obtain:
n = D/A
= pi*d/A
= ( pi*c*( t^2 + 1/4 )/t )/( asin( t/( t^2 + 1/4 ) )*c*( t^2 + 1/4)/t )
= pi*c*( t^2 + 1/4 )*t/( (t^2 + 1/4)*c*asin( t/( t^2 + 1/4 ) ) )
n = pi/asin( t/( t^2 + 1/4 ) )


With a pipe 19 cm in diameter, one can make about 20 straight blades. That's a lot of blades for 1 pipe!

I experimented some with the idea but found plastic, be it PVC or otherwise, too heavy be used in a gyroplane. Plastic wasn't the way to go. Using a large composite tube would have been much better, but seeing as I didn't have the means to do this, I abandoned the concept altogether. It was easier to just make blades out the materials I already had.
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Old Oct 15, 2014, 05:58 PM
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Building larger blades.

Carbon fiber blades

The first set of large blades I constructed came from a set of broken carbon fiber 335mm blades. It was a fairly straightforward process. These were the steps I took:

1. Using a dremel cutoff wheel, I cut the blades where any major break happened.

2. With a thin, sharp knife ( a boxcutter in my case), I slowly cut down the length of the broken blades, doing so on the leading edge and trailing edge. I was careful to keep the knife as centered as possible, especially on the trailing edge as the carbon laminates get very thin once separated.

3. Once the blades were cut in two, I took each half and removed any wood or foam glued to the newly exposed underside.

4. I then heavily sanded the tips and underside of each half, removing any large bumps, imperfections and loose carbon strands.

Note: For anyone who attempts this, please cover your hands and arms before sanding. I didn't and the carbon dust that got stuck on my skin caused terrible itching. It will go away in a day but it's really uncomfortable.

5. I took a 6x1mm carbon strip and drilled a 1/16 inch ( approximately 1.5mm ) hole at one end. To reinforce this area I glued a sheet of ripstop (any other fabric can be used, carbon cloth or fiberglass will work better. I just used ripstop since I had some from a kite) over the hole and repeated the drilling.

6. The sanded blade halves were now very thin and relatively flexible. To reinforce them, I took the strip and glued it across the entire leading edge using CA. To enhance the bond, I compressed the pieces together with some pliers, holding them down with rubber bands.

Note: I made sure some of the strip extended some distance away from a blade tip.

7. I heavily sanded the underside once more, smoothing the transition from carbon strip to blade half. At this point I started balancing the blades. Using a gram scale, I measured the weight of each blade, sanding whichever was heaviest until both blades weighed the same.

8. Next I sanded the underside yet again using fine sanding paper, doing so until the bottom of both blades was very smooth. I also sanded the ripstop reinforcement at the end of the carbon strip.

9. To remove the carbon dust made by sanding, I washed the blades using water and a small amount of soap. I was careful not compress any soap into the carbon, only lightly smearing it on.

10. The last step was removing the large positive pitch angle caused by gluing the strip to the leading edge. This step had to be done very carefully as it could easily damage the blades. I heated center portion of the exposed carbon strip using a candle. While this happened, with my hands I twisted the strip as to decrease blade pitch and make it slightly negative. As soon as I felt the carbon soften, I immediately moved it away from the flame and let it cool, all while holding the twist. Once strip had fully cooled, I let go and watched as it kept the bend.

Note: This heat bending process decreases the strength of the carbon fiber. Do it with care.

Pictures of this build are attached to the bottom of the post; For those interested, Hobbyking offers a very similar set to the ones I used: http://www.hobbyking.com/hobbyking/s...in_Blades.html
Balsa wood blades

The next set of blades I constructed were made of balsa wood. Instead of using the 8% thick airfoil I described before, I decided to approximate the 4% camber airfoil using line segments. This would make construction very simple as little sanding would be needed. By cutting a sheet of balsa to the right dimensions, and gluing the resulting pieces at the right angle, I could easily change the maximum camber as well as it's position.


I experimented some with this concept, making 3 blade sections out 1.5mm thick balsa. They all had approximately 4% maximum camber. The camber positions I tried were: 25%, 40% and 50% chord as measured from the leading edge. Constructing these test sections was very easy and with some sanding could be made into a nice curves.

Next I concerned myself with waterproofing the wood. I considered many things but the simplest solution I found was covering the blades with clear contact paper. It's fairly light, easy to mold around curves and doesn't require a hot iron to adhere. I could simply lay it on the wood and carefully press it down. This made blade construction very quick which was important for making spares. To further waterproof the blades, I would add a thin layer of 5 minute epoxy to the uncovered tips. This would completely seal the wood.


To make the bolt hole of each blade, I drilled some distance away from the root. The only reinforcement I made was soaking the balsa wood in CA. This plus the covering would add enough strength for the blade to resist centrifugal load, so long as the blades weren't loaded heavily (which I hoped would the case as I wanted to make a light autogyro).

One last concern was impacts. Model rotor blades are destined to hit something. Bare balsa wood can't stand localized impacts, and covering it won't do much. To reinforce the leading edge I glued with CA a 1.5mm carbon rod across entire length of the blade.

Pictures of these blades are also attached to the post.
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Old Oct 25, 2014, 08:43 AM
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Testing the first blades and flexplate design.

To test the blades I had constructed I needed a rotor head of some sort. The carbon fiber blades, due to the strip I used in reinforcing them, would easily slide into the blade holder of a 4#6 rotor head. I had the option between a bell-hiller rotor and a simpler hiller rotor system. I knew from my experience with the 4#6 that the bell-hiller rotor was much better, with control inputs being more sudden and consistent; The Hiller rotor would at times, feel slow to react and have a noticeable lag; This was of course, as used on a helicopter, which needs quick and responsive controls when hovering. An autogyro on the other hand, with it's inability to hover can do without such aggressive control.

I did some investigating and found multiple examples of hiller controlled autogyros. One that I consistently ran across was the Begi by Mnowell. This model seemed to fly well with authorative control.
Begi: Tailess autogyro with a flybar
Begi RC Autogyro (1 min 21 sec)

With this in mind I set out adjust the 4#6 hiller rotor. Normally it uses a relatively short, steel flybar to control blade cyclic. I decided instead to use a much longer rod of 1mm carbon fiber. As the carbon blades I made were much longer than the stock 4#6 blades, it seemed resonable the flybar should be longer too.

Using the main frame of my 4#6, I mounted the blades onto the hiller rotor and took it outside. The rotor would spin up well, but once at speed the blades would refuse to track! Worse still, they would rotate about the blade holder if their rpm was high enough, causing extreme vibration due to the offset mass.


Puzzled as to why this happened, I placed the rotor aside and concentrated on the wooden blades. Here I took a more conventional approach and decided to make a flexplate hub. Most RC autogyros seem to use the system, proving it works. A flexplate hub is just a thin plate with a central bearing support and some holes to mount the blades on. Extremely simple to make, but it has a severe flaw:

As I see it, the thin flexible plate has two purposes: It's easy to make, and the flexibility reduces the torque required to rotate the rotor ( The rotor is a very large gyroscope. If the blades are placed on a flexible element, the flexplate, they can orbit around it, and the torque required is that to bend the plate, and not to directly move the blades ). This causes a severely detrimental tradeoff.

A flat plate can be made flexible about one axis by making it very thin in all other directions. This is beneficial for control load reduction, but the plate becomes extremely flexible about the blade pitch axis. Couple this with cambered blades and one gets huge changes in pitch. To reduce these variations, the flexplate must be made thicker or wider, both of which increase the rigidity of the plate about the blade flapping axis, increasing control load.


Knowing my wood blades were cambered, I was worried if a flexplate would even work. As long as the blades didn't spin too fast, the pitching moment would be low enough to not cause a large pitch rotation. I was hoping this would be the case.

The flexplate I made was a long thin strip. The central bearing support used the same design I used for the cambered blade test ( to JochenK's credit ). Once I was done building it, I tested it in some breeze and the results were not good. The rotor seemed to work well at low rpms, but once the incoming wind was fast enough, the pitching moment would torque the blades to a very large, negative pitch angle, slowing down the rotor tremendously. Even worse, the blades would flutter once this happened!


This was completely unacceptable. It would be very hard to make something fly with such a rotor. To solve these issues, one partial solution is to make the flexplate more rigid (something I did not want to do). I made a new flexplate, much wider than before to see if any improvement happened. By testing the rotor in some breeze again, it was obvious the blade moment drastically affected pitch less. Although this was a plus, the blades would still flutter; I needed a way to avoid this.


Thinking my blades were too flexible, I made a third rotor. This time the common 3 blade system. The blades were all balsa with no carbon fiber, just a layer of CA around the leading edge and tips. I also added a bolt hole reinforcement. This consisted of two plastic tabs glued to the root. They would provide an increase in strength which would be needed for the higher rpm's flutter occurred at. It's noteworthy that this is not the best solution for increasing blade root strength.Ideally one would want to pull on all wood fibers at the root, distributing the centrifugal load in tension.


This is very difficult as there is really no simple way to grab onto all the fibers. Instead one is forced to carry the load via shear. This is a very inefficient way of carrying loads as only a tiny fraction of the fibers actually support any of it. On top of that, the load must be passed through the adhesion present between surface fibers and whatever is attached to them. This can be much weaker than the fibers themselves, so their potential strength might not be used. To increase the load transferred via shear, one has to increase the area in which fibers are being pulled in. Ideally covering as much as possible for maximum strength.


When the time came to mount the blades on the flexplate rotor. I noticed an interesting detail. Depending on how the blades are placed on the flexplate, be it above the plate, or bellow them, load on the blade bolts can be drastically reduced:

- If the blades are placed above the flexplate, the bolts must counteract two loads: The vertical load caused by the lift of the blades, and the load caused the by root of the blades pressing against the flexplate.

- By placing the blades bellow the flexplate, the load on the bolts can be greatly decreased. In this configuration, the lift of the blades is counteracted by pressing against the flexplate. As they will only contact the flexplate in the area adjacent to the bolt hole, no other contact force is present. The only load carried the bolts are the centrifugal loads.


Knowing this, there is really no reason to mount the blades on the top surface of the flexplate. The bottom surface is a much better option. There is a problem that comes from this though:

The way I constructed the balsa blades, the top surface is curved throughout their whole length. The plastic tab glued to the root will not be level as compared to one on the bottom surface. This will offset the pitch depending on where the upper tab is placed. By mounting the blades on the top surface of the flexplate, the flat, even bottom of the blades ensures no major pitch offset.

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Old Jan 19, 2015, 12:13 AM
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Solving flutter and the importance of chordwise mass balance.

Woah, quite some time without posting! Sorry for the large bump everyone, I've been working on other projects and I kind of forgot about this thread Anyhow back on track:

The 3 blade rotor. Blades with 40% chordwise balance

As with the previous two rotors. I tested the 3 blade system in some strong breeze. It was different in that it presented very little vibration. The rotor could be tilted in any direction and the resistive torque it produced was very smooth. This was not the case with the two bladed systems. A definite thumping could be felt as those rotors were forced to change orientation.

The 3 blade rotor did nonetheless flutter. It did so at a substantially higher rpm than the line segment rotor, so there was an improvement. This can be explained by the higher torsional rigidity of the blades and the flexplate. Before flutter presented itself, the rotor would make a loud and sharp thumping sound. Once the blades did flutter, the sound would suddenly become much quieter, with the pitch of the thumping decreasing accompanied by a slight burbling noise. When viewed from the side, blade tracking would seem to change, with random twitching motions seen in the path the blade tips took:


Faced with the problem of flutter I did some research on the issue. These were some of the websites I visited:

http://www.unicopter.com/Blade_Balance.html
http://www.rwas.com.au/blade-balancing.html
http://www.cavalrypilot.com/fm1-514/Ch3.htm
http://w3mh.co.uk/articles/html/csm9-11.htm
http://www.rotaryforum.com/forum/showthread.php?t=24773
http://www.arc.vt.edu/ansys_help/cfx...tbr_case3.html
http://rotorcraft.arc.nasa.gov/Resea...or_design.html. <- The videos here show just how much of a difference mass balance can make. I highly recommend watching them.

After browsing those sites, the solution appeared to be very simple. For constant chord blades, all that's required is that the center of mass be placed at the quarter chord point, or rather, the quarter chord line:


Note:
Once I was done testing, I took the blades off the 3 blade flexplate and measured where they balanced. They seemed to do so at the 40% mark as measured from the leading edge.

The 3 blade rotor worked well so long as it spun bellow a certain RPM. Above this critical RPM, flutter would suddenly present itself. This was an important observation. A rotor blade will not flutter bellow a certain airspeed, and if this airspeed is never encountered in-flight, it makes no difference to use a balanced or unbalanced blade balanced or unbalanced blades will create useable force bellow this speed. This is only possible if the blades are rigid enough in torsion. If they're too floppy, they'll flutter at an airspeed seen in-flight. This is no different than a piece of paper held in some fast moving air, it requires little motion to make it flutter. Likewise, a sheet of wood with identical shape will also eventually flutter, but at a much higher airspeed.

There are many examples of model autogyros with unstable blades, and none of them seem to have reports of in-flight flutter. In all these cases, the blades are operating bellow their flutter speed. The best example I think are foam autogyros. Foam is not very stiff and there are a plethora of these models, all seemingly operating fine:


Still, if I could completely eliminate the possibility of flutter just adding some weights, It seemed worth it If nothing else for the peace of mind.

Two blade rotor with rigid hub. Blades with 33% chordwise balance

To test this out in practice I made a new rotor. Using the same construction process of before, I made a new set of blades with the 8% thick airfoil. These unlike previous sets, these had a steel rod imbedded to into the leading edge near the tip. The rods were not placed exactly at the leading edge as they were relatively thick. Rather they were placed somewhat inboard as to not change the airfoil shape:


Additionally they were placed some distance away from the tips. By cutting a grove in the wood for the rods to sit in, one returns to the same situation of reinforcing a bolt hole. Centrifugal loads cannot be supported by directly pulling the fibers next the rods, but must be carried via shear and some adhesive. Structurally this is a very poor situation as steel is fairly dense and the centrifugal load created by the rods is concentrated in a very small area, precisely what isn't needed to support high loads.

The only simple way to increase load carrying area is to move the rods inwards from the tips. If the rods contact the block of wood precisely in front of them when viewed radially outwards, the entire length of wood from the contact area to the tip of the blade can carry some of the centrifugal load. The layer of wood precisely between the block of wood and the rest of the blade will present an area with which shear loads can be supported. By moving the rods inwards, the area will increase. Not only this, but the net centrifugal force felt by the rods will decrease as the radius from the center of rotation is smaller:

I guesstimated the distance from which to cut the groove for the weights and proceeded to cut out the hole. Once done, I snugly fit the weights inside the grooves and soaked them with CA. I added two layers to be sure it penetrated as much wood as possible. This area had to be strong.

Once I was done constructing the blades I measured their chordwise balance point. It was somewhere around the 33% chord as measured from the leading edge. I was not pleased with the results. I was looking for balance around the 25% chord. As it was too much effort to add new tip weights I simply left the blades as they were. Not only was this a motivation, but the amount of steel rod that would be required to achieve 25% balance would have been substantial. A wooden blade without an imbedded rod would weigh around 10g. By adding the steel, a rod of around 7g would have been required. That's close to doubling the weight of the blade. I felt this increase was too drastic. Maybe the 33% balance would be enough to avoid flutter. Testing would reveal if this was true.

For the hub I didn't go with a flexplate. Instead I glued two carbon sheets to a wooden block. With this design, the blades would be sandwiched in between the sheets making a very stiff structure.

This hub would transfer huge loads to any servos trying to tilt it, thus it's not a practical design with which to control an autogyro. It's only real use is as a rigid fixture to experiment with different blades.

As was the trend, I took this rotor outside and exposed it to some strong breeze. This rotor reached much higher rpms than any of the previous ones. The frequency of the thumping made by the rotor was noticeably higher as well as the intensity of the sound. At a glance it seemed it was flutter resistant, yet with a very strong breeze, slightly variations rpm could be heard. The variations were much more subtle than before, with no major visual change apparent. With each rpm variation one could hear a slight pitch decrease in the sound produced by the rotor.

To completely verify if the rotor would flutter. I placed it outside the window of a moving car. By slowly increasing the speed of the car, the rotor would spin up until eventually, at round 35 km/h, the sound produced by the blades would suddenly decrease in pitch. This was the same change heard on the 3 blade system, albeit at a higher wind speed. With this test it seemed the 33% chord balance was not enough to achieve complete flutter resistance. It only increased the airspeed required for the instability to present itself.

External tip weights. Blades with 25% chordwise balance

The next set of wooden blades I made were equipped with external tip weights. Instead imbedding the steel rod, why not place it at the tip of the blade, with some of it poking forward? This way much more mass is simply placed in front of the blade. More importantly, one can use less rod for the same balance!. In fact, one doesn't even have to use a heavy material. Anything rigid enough to not deform under the centrifugal load, can be used, so long as it's long enough. For the first set of blades like this. I used two thin, 1.5mm carbon rods. To not make the rods too long, I glued some rounded steel cylinders to their exposed ends. This was much lighter than a heavy steel rod stuck inside the blade. 5grams vs 1.2g! That's a huge difference for 10g blades!


To know how well they worked, I mounted them on the two layer hub I made earlier. As always, I waited for some strong breeze and nothing seemed to happen. I ran with the rotor into the strong breeze! Still, it would do nothing but rev up to tremendous rpms (It was dangerous as hell, but what a rush! The power that was there was amazing! ). I repeat this test, but this time with a car. Holding the rotor into the breeze, the car accelerated up to around 35 km/h. Unlike the Blades with an imbedded counterweight, These did not have any sound variation. It was very consistent. It was so consistent that I got overzealous and went faster than 35/h. Well... Shortly after the thin carbon rods couldn't take the load and one of them Ripped off the blade! Instantly this whole thing started shaking, with one blade fluttering. It didn't take long for the car to stop after that

Curiosly, this idea has been used in manned autogyros. Here's a Ken Wallis flying his autogyro with plainly visible external weights.

Very pleased with these results, I made a second pair as it was too much hassle to repair the early set. For these I used a slightly shorter steel rod. This was done for two reasons:

- Steel will bend and not snap like carbon fiber. This gives a measure of safely in the event or rotor overspeed.
- Steel will bend in the event of an impact. Unlike carbon fiber it will not shatter. As all model blades destined to hit something, better prepare for it

To complement these successful blades, I used the wide flexplate I tested on the line segment blades with. This allowed for some flapping, making it much more suited for controllable rotor.

Lastly, Here is a video showing the progression from the very flutter prone line segment blades, to the flutter proof ones with external tip weights:

Effect of chordwise balance and rotor hub rigidity on blade flutter
Effect of chordwise balance and rotor hub rigidity on blade flutter (1 min 49 sec)

The video is a bit shoddy at showing what fluttering blades look like close up. I'll work on a new video that does a better job.
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Old Jan 19, 2015, 12:18 PM
Mickey from Orlando. Really.
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Originally Posted by XXXmags View Post
A rotor blade will not flutter bellow a certain airspeed, and if this airspeed is never encountered in-flight, it makes no difference to use a balanced or unbalanced blade.
I disagree, just because it doesn't flutter doesn't mean it does not cause other control/stability issues. The unbalanced blade will be more sensitive to gust disturbance and control input. If the response rate of the blade reactions to these forcing functions is too high it can result, in the least case, a difficult to fly aircraft, and in the worst case, an unstable one that is uncontrollable by a human pilot.
The unbalanced blade can cause control system stress as well, feeding aerodynamic forces back into the control system (servos), causing it's own set of problems.
Flutter has been previously considered in model gyrocopters and the use of CG correcting tip weights and torsion-ally rigid blades is well established.

Also important is the effect that the blade center of gravity has on the lead/lag angle and the resulting control system issues associated with inappropriate amounts of lead or lag in the blade motion. John Burkam, a model helicopter pioneer, used blade lead to force mechanical feedback into a flybar from a gust to apply corrective cyclic pitch to recover from the gust in his free flight helicopters.
Misunderstood, the effects of lead and lag can have mysterious deleterious effects.
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Old Jan 20, 2015, 01:19 AM
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Certainly stability can be affected by unbalanced blades. The external tips I talked about coupled with rather flexible blades may even act like a flybar. The advantages of balance are unquestionable. But what of a model that flies well with unbalanced blades? One without major control or stability issues. Is it worth adding the mass and complexity of tip weights? Should the machine fly more or less the same.. there really is no point. Studying why such a machine flies well is a possible way to simplify model design and construction. Unbalanced blades are not necessarily a game breaker.

As for control loads, If one uses a cambered airfoil, forces will be fed into the control system, balanced blades or not. Should the rotor have any asymmetric lift velocity across it, the advancing blade will induce a greater pitching moment than the retreating blade. In fact, if the blades are unbalanced and lead by right amount, the lift produced can create a torque that counteracts the pitching moment. It might not completely cancel it, but the mean force fed to the controls can be reduced.

It's not unreasonable to think asymmetric lift is always present. The torque it creates on the fuselage can be counteracted either by propeller thrust or by a horizontal stabilizer. Allowing the blades to freely flap doesn't eliminate the torque, the rotor will just tilt back. Further I doubt any model even has the precise amount of cyclic required to eliminate it. Even then a small airspeed change will get rid of the trim.

I completely agree with effects of lead and lag though. I had a serious problem with two flybar equipped helicopters being servilely underdamped ( HK-450 and the Wk 4#6 ). They would have a terrible wobble if slightly disturbed. I was able to achieve a nearly overdamped state by giving the blades a little lead. The effects were DRASTIC, zero wobble, super crisp. How much this transfers over to rotorcraft without a flybar I don't know, but it's certainly something to keep an eye on.
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Old Jan 20, 2015, 01:35 AM
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Construction of balsa wood blades with external tip weights.

For anyone interested in how I built my blades with external tip weights, I made a video tutorial showing the complete construction process:

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Old Jan 20, 2015, 11:14 AM
Mickey from Orlando. Really.
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But what of a model that flies well with unbalanced blades? One without major control or stability issues. Is it worth adding the mass and complexity of tip weights? .
No it's not. But this does not support your statement that if the blades don't flutter there is no effect. Only when the model with unbalanced blades is flown with balanced blades can you determine if the model design is optimum. Just because something works, does not mean it is optimally designed.

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The torque it creates on the fuselage can be counteracted either by propeller thrust or by a horizontal stabilizer.
Which torque? Roll torque or pitch torque? Your lead in comments were about asymmetric lateral lift, yet the stab is mostly for longitudinal forces, unless you are referring to using the stabilizer as a taileron. You'll need to clarify.


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Allowing the blades to freely flap doesn't eliminate the torque, the rotor will just tilt back. Further I doubt any model even has the precise amount of cyclic required to eliminate it.
I disagree. The rotor tilt back is natural nose down cyclic pitch, effectively perfectly cancelling asymmetric lift.
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Old Jan 20, 2015, 07:12 PM
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I like your way of using quotes, clear and clean. I'ma copy it
Hope you don't mind btw...

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No it's not. But this does not support your statement that if the blades don't flutter there is no effect.
You're right. Lack of flutter doesn't mean other areas aren't affected. I'll rewrite my above post to fix the misinformation.

This is my edit:
"if this airspeed is never encountered in-flight, it makes no difference to use a balanced or unbalanced blade balanced or unbalanced blades will create useable force bellow this speed"

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Originally Posted by mnowell129 View Post
Only when the model with unbalanced blades is flown with balanced blades can you determine if the model design is optimum. Just because something works, does not mean it is optimally designed.
True as well!

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Originally Posted by mnowell129 View Post
Which torque? Roll torque or pitch torque? Your lead in comments were about asymmetric lateral lift, yet the stab is mostly for longitudinal forces, unless you are referring to using the stabilizer as a taileron. You'll need to clarify.
A lot of the torque asymmetric lateral lift creates is shifted to the longitudinal axis by gyroscopic precession. This can be countered by a horizontal stabilizer or by a thrust source that's not aligned with the center of mass. The roll torque can to an extent, be countered my engine reaction torque, should one use a single propeller.

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The rotor tilt back is natural nose down cyclic pitch, effectively perfectly cancelling asymmetric lift.
Indeed, but remember, the rotor will be tilted aft... This will re-direct the net force of the rotor slightly back. Should this force be above the center of mass, a torque will be produced.
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Old Jan 20, 2015, 08:01 PM
Mickey from Orlando. Really.
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A lot of the torque asymmetric lateral lift creates is shifted to the longitudinal axis by gyroscopic precession.
Nope. Not precesssion because it's not a solid body. It's a basic resonance problem with a mass spring damper system that if undamped ends up being 90 degrees phase lag. With some kind of damping the phase lag gets less than 90 degrees.

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or by a thrust source that's not aligned with the center of mass.
Thrust/COM misalignment only provides a torque in an accelerating case. Thrust/Drag misalignment applies in the steady state non accelerating case.

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The roll torque can to an extent, be countered my engine reaction torque, should one use a single propeller.
The trimmed rotor doesn't produce any roll torque. Cyclic pitch and or flapping induced cyclic pitch takes care of this.
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Indeed, but remember, the rotor will be tilted aft... This will re-direct the net force of the rotor slightly back. Should this force be above the center of mass, a torque will be produced.
Yes, I understand this, but there is no point here. This is just a basic trim design problem.

Don't take me wrong. I'm interested in your experiements, but if we're going to extrapolate theories from experiments those theories are gonna have to align with some well known rotor physics.
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Old Jan 21, 2015, 02:12 AM
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We're going of in a tangent.

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Nope. Not precesssion because it's not a solid body. It's a basic resonance problem with a mass spring damper system that if undamped ends up being 90 degrees phase lag. With some kind of damping the phase lag gets less than 90 degrees.
The rotor can be considered a mass spring system. Reasonably so, but it's not erroneous to say the source of this lag is from gyroscopic precession:

Precession is not exactly an uncommon term:
http://en.wikipedia.org/wiki/Swashpl..._blade_control
http://image.slidesharecdn.com/helic...?cb=1383961676
http://avstop.com/ac/basichelicopterhandbook/ch2.html <- Go to figure 12.
https://www.faasafety.gov/gslac/ALC/...9&preview=true
http://www.rchelicopterfun.com/gyros...recession.html
http://www.tpub.com/air/10-3.htm

I just randomly picked these up from google search.

At any rate, however one calls it or sees it, what's of importance here is that asymmetric lift can create torques about the fuselage in the pitch and roll axes. Like you said, the phase angle can be less than 90. I agree with this.

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Originally Posted by mnowell129 View Post
1) Thrust/COM misalignment only provides a torque in an accelerating case. Thrust/Drag misalignment applies in the steady state non accelerating case.
2) Yes, I understand this, but there is no point here. This is just a basic trim design problem.
This misalignment will always cause a torque whenever there is thrust! Force times distance at an angle is torque!

Whether or not this is torque is cancelled by some other torque, is completely different. But that is precisely what gives this misalignment a use. The fact that the torque won't go away, and that it must be cancelled by another force inducing torque. This can be provided by asymmetric lift from the rotor. The fact that the phase angle must be greater than zero ( or the rotor wouldn't be spinning ) allows this to happen.

But that's the thing, you can have asymmetric lift with an aircraft that maintains level flight. Returning to why we're even discussing this; the uneven velocity of a translating rotor won't allow the pitching moment of a cambered airfoil to be cancelled between the advancing and retreating blades. Hence loads are fed to the control system. This can be alleviated allowing the blades to be unbalanced, letting them lead a little and lift does it's thing.

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The trimmed rotor doesn't produce any roll torque. Cyclic pitch and or flapping induced cyclic pitch takes care of this.
True, but rotor the can be untrimmed, and something else can counter the roll torque. It's the same interaction as above.

At any rate, the flutter stuff which was the source of this discussion, to me is settled. Let's just leave it at this. I thank you though for being scrupulous about what I'm stating. A critical eye looking at all the nonsense I've written is good.
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Old Jan 21, 2015, 06:50 AM
Mickey from Orlando. Really.
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We're going of in a tangent.
Then quit making up erroneous theories. Stick to reporting results of your excellent experiments.
If you expect to keep stating incorrect theories I am going to keep calling you out on them.
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The rotor can be considered a mass spring system. Reasonably so, but it's not erroneous to say the source of this lag is from gyroscopic precession:
It is in fact erroneous. Precession is a rigid body trying to maintain angular momentum. A rotor is not a rigid body. I refer you to Don Lodge's Model Helicopter Design for a consise mathematically explanation of the cause of the phase lag in a rotor.
Also I have the text referred to in the google article with the math describing how the rotor blade is resonant system being excited at it's natural frequency, no mention of precession either.

Of course everything on the internet is true. Pilot manuals are notorious for bogus theories. They are chock full of easily refuted "theories" for fixed wing as well as rotary wing aircraft. I suggest you research this topic in engineering texts, not pilot manuals and wikipedia entries.

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This misalignment will always cause a torque whenever there is thrust! Force times distance at an angle is torque!
Yes, but for there to be torque there has to be a fulcrum. In the accelerating case the fulcrum is the mass's inertia, in the steady state case the mass's inertia is no longer the fulcrum because as F = ma, when a is 0 there is no restraining force. When you accelerate in your car the force of the tires creates a force, the mass above the wheels resists the acceleration and the car leans backwards. At speed the force from the tires is the same but as the car is no longer accelerating so the car rides level. Using your theory if you put more people in your car it will lean back further at highway speed, something that is easily shown to be be false in practice.If you mount a sheet of plywood on top of the roof the car will again lean backward but this is due to the torque between the force of the tires and the drag of the plywood and the car's mass is irrelevant.
You are mixing the static and dynamic case of the source of the torque. The dreaded "pushover" problem of gyrocopter crashes with low centers of gravity and high mounted motors is an acceleration case where the aircraft is slow and the rotor is unloaded. The pilot applies full power to correct and the aircraft accelerates and in this case the torque is thrust to mass distance.

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. The fact that the phase angle must be greater than zero ( or the rotor wouldn't be spinning ) allows this to happen.
This doesn't make any sense at all.

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But that's the thing, you can have asymmetric lift with an aircraft that maintains level flight..
But just because you can have something doesn't mean you must have something. Basic logic.

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the uneven velocity of a translating rotor won't allow the pitching moment of a cambered airfoil to be cancelled between the advancing and retreating blades.
I have found no evidence nor have you provided any basis to couple these things together.

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This can be alleviated allowing the blades to be unbalanced, letting them lead a little and lift does it's thing.
This, I agree with. How much unbalanced is a design problem. I cannot agree that you can conclude that just any old amount of unbalance is the correct amount and/or provides the right amount of correction. It also ignores the unstable nature of a forward swept wing, which is another topic.


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At any rate, the flutter stuff which was the source of this discussion, to me is settled. Let's just leave it at this. I thank you though for being scrupulous about what I'm stating. A critical eye looking at all the nonsense I've written is good.
I agree with these conclusions you made :
-- flutter is bad
-- flutter is a speed related phenomena
-- once source of flutter is an unbalanced blade
-- lack of torsionally rigidity contributes to flutter
-- The onset of flutter can be raised by making the blade more rigid
-- balancing the blade can mitigate flutter
-- cambered airfoils have a nose down moment that can feed forces into the control system
-- the correct lead / lag can help mitigate the forces fed into the control system

One final note. The blade balance method with the steel spike sticking out of the tip of the blade, while effective, is a safety hazard, violates AMA safety rules about metal edged propellers, and would be disallowed at most club fields.
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