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Old Oct 02, 2014, 07:36 PM
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Attempts at sub-120g autogyros.


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.

Size of the 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:

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


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

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.

: 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.

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

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

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.

Thus the Re is:

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 rpm for 335 cm blades.

F = 1000/60 = 16.67 hz
r = 335 + 4 = 339 cm = 0.339 m
V = 6.28*0.339*16.67 = 35.5 m/s = 127.76 km/h

L = 3.3 cm = 0.033 m

Re = 0.033*35.5*(10^5)/1.5 = 78100

Around 80k Re. That's about double 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 50k.
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Old Oct 03, 2014, 04:30 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, 01:01 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.

Rotor disk 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.

Rotor disk force components

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:



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


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


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, 01:01 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.

Low torsional rigidity of a blade and camber pitching moment

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 may have had 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.

Effects of adding a flat bottom to mean camber line

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.

General concept of pipe blades and possible alterations

Video showing the construction of PVC pipe blades

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.

Saggita of a circle

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


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


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


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


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


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:

Number of blades present in a pipe with straight cut blades

D = pi*d

n = D/A


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


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.

Arc length and triangle

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 pipe blades 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, 04:58 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 attempt 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 are 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 have 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:

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.

Effect of line segment dimensions on airfoil camber and 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.

Clear contact paper

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 we're 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 Today, 07:43 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 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.

Offset blade imbalance

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.

Design compromise between torsional and flapping rigidity

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.

Different flexplate planforms

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.

Actual wood fiber orientation compared to ideal

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.

Effect of covered area on load carrying fibers and root 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.

Bolt load change with blade position

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.

Effect a curved root has on pitch
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