Engineering Design

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From our President/CEO/Captain/Lead Engineer

This journey began when I first purchased and began flying a Blade CP. I've had a lot of experience designing/building/flying RC airplanes, but helicopters presented a new challenge for me. I was quickly disappointed with the fragile nature of the balsa wood blades and desired blades made of a better material. This became even more important to me when I stopped have mishaps while flying and found that the fragile material covering would tear and rip without impact, under normal flying conditions, completely unacceptable. As an engineer, I considered many materials and decided that some kind of plastic would be best. Carbon fiber blades exist as an option for the Blade CP, but they are expensive and while very durable, transfer a lot of force into the head assembly when mishaps occur. This is evident in some RC planes when comparing wood props to some of the APC brand props, I've seen crankshafts bent under severe impacts. I was hoping for a compromise between price/durability/force, which I believe I have found in the plastic used.

The next step was the decision to make a blade exactly like the wood ones or to redesign it from scratch. It would be much easier to replicate the wood ones, but as an engineer the challenge of coming up a better design is far too tempting. My initial intent was an optimized design for both the flat bottom and symmetrical style blades. At this point I had only flown with the beginner flat bottom blades, though I upgraded to a 3 cell li-po and purchased the aerobatic kit for the lower tooth motor and heat sinks. I saved the symmetrical blades as spares, as I figured I had to wait for improvement before trying these blades out. I read in many forums about the symmetrical blades performing better, even for a beginner trying to learn, but still keep flying the flat bottom blades. I also noticed that the Honey Bee 2, another beginner helicopter, came with the symmetrical blades standard. As a result, I decided to begin redesigning symmetrical blades first. I will note that the symmetrical blades on horrible using the stop setup and battery, the flat bottom blades do get more lift. The largest advantage of the symmetrical blades is the wind seems to affect them less.

I started by researching our little Blade CP’s bigger brothers, various models of full size helicopters. I immediately found that many full scale helicopters have symmetrical blades, including Bell 209 AH-1 Cobra, Bell 212, Bell 48, Bell 61, Bell 47G-2, Bell 47J, Vertol, and many others. A helicopter blade is basically a spinning wing; however, a wing doesn’t experience as many different relative airspeeds as a helicopter blade. A hovering helicopter blade faces airspeeds that increase from the rotor hub to the blade tip. For each section of the blade, the airspeed is constant. Now, when a helicopter begins forward flight, the airspeeds are no longer constant as the relative airspeeds begin to change. If the blades turn clockwise as seen from above, such as the Blade CP, in forward flight the left side blade will experience an increase in airspeed, while the right side blade experiences a decrease in airspeed. The same principle applies when flying left, right, or backwards. Climbing and descending while moving also effects the relative airspeed as well as the pitch and roll of the helicopter. Because of all these possibilities, a symmetrical blade typically results in the optimal performance, as it is affected less directionally by all the relative airspeeds a helicopter will encounter. This is one reason they are used on so many full size helicopters. Additionally as we all know, they are also optimal for inverted flying. As a result, we first developed the symmetrical blades and then developed the flat bottom blades for the stock setup.

We followed similar steps for the development of both the symmetrical and flat bottom blades. I will now discuss the other parameters and how they affect the flight characteristics.

Chord Length: This is the length from the leading edge of the blade to the trailer edge.

Angle of Attack: This is the angle between the relative airspeed and the chord of the wing. In general, a greater angle of attack is associated with an increase in lift.

Cl: Cl is coefficient of lift, which is usually determined experimentally, but can be calculated as well. A lot of data is available for different airfoils that exist today. Cl is affected by the shape of the airfoil, the angle of attack, and the Reynolds Number.

Reynolds Number Reynolds number is a dimensionless number that is the ratio of inertial force divided by viscous force. It is typically used to account for dynamic similarity. One of main differences between full scale aircraft/helicopters and model aircraft/helicopters is this difference in Reynolds number. When testing an engineering scale model, it is usually important the Reynolds number and the Mach number are the same for the model and the full size design. Reynolds number=D*V*r/mu where D is a charactersitic length (chord), V is velocity, r is density, and mu is viscosity.

Mach Number Mach number is a dimensionless number that is the ratio of an objects speed divided by the speed of sound in that medium. It is typically used to account for dynamic similarity. When testing an engineering scale model, it is sometimes important the Reynolds number and the Mach number are the same for the model and the full size design. Mach number=V/Vsound where V is velocity and Vsound is the speed of sound in that medium. The closer the model/full size design is to 1 or higher, the more important it is to match the Mach number. This is because as the speed of an object apporachs the speed of sound, the density of the fluid increases greater where it contacts the object.

Cd: Cd is coefficient of drag, which is usually determined experimentally, but can be calculated as well. For an airfoil, Cd=Cd0+Cdi where Cd0 is drag coefficient at zero lift and Cdi is the induced drag coefficient. Cdi is the unfortunate byproduct of lift described above, the equation for Cdi is Cdi=Cl^2/(pi*AR*e), where Cl is the coefficient of lift, AR is aspect ratio, and e is an efficiency factor.

Lift: Lift is just what it sounds like; lift is created by a pressure differential between the top and bottom of the blade. For typical airfoils, lift is directly related to the air speed and angle of attack. The equation to calculate lift is L=.5*Cl*r*v^2*A, where L is lift, Cl is coefficient of lift, r is density, v is relative air velocity, and A is wing area.

Drag: Drag is also like it sounds; drag is the force opposing motion of you helicopters blades and is also an unfortunate byproduct of lift. The drag from the blades spinning is why a tail rotor is required. A lot of people mistakenly believe that the tail rotor is there just to oppose the main motor’s torque; however, in a vacuum not only would a helicopter not fly, but it always would not have a constant rotation. It would rotate slightly as the blades accelerated/decelerated, but after they reached a constant speed, the helicopter would not try to turn. In air, the constant drag of the air against the blades drive the requirement of a tail rotor. You may have noticed that if you damage your main blades your helicopter now turns more to the left (counter clockwise), this is because there is an increase in air drag force. Therefore, any reduction in the drag will result in less force needed to keep the helicopter straight. The equation to calculate drag for airfoil is D=.5*Cd*r*v^2*AR, where L is lift, Cd is coefficient of drag, r is density, v is relative air velocity, and AR is aspect ratio.

Lift to Drag Ratio: This is a very important number, it measures the ratio of lift (yeah, cheers!!) to drag (boo, hiss!!). There is a constant compromise in aerodynamics between lift and drag, it is difficult or impossible to increase lift without some kind of increase in drag. The question becomes how much more lift can you get versus how much more drag.

Aspect Ratio: This is a ratio of the length of the blade divided by its chord length. High aspect ratio wings have long thin wings, while low aspect ratio wings have shorter thick wings. An example of an aircraft with a higher aspect ratio is a glider, while most fighter jets have low aspect ratios. A higher aspect ratio has a greater lift to drag ratio, which makes it well suited for gliders. Since fighter jets travel at high speeds, reducing drag is an important part of their design.

Camber: Camber is another measurement that defines the shape of the airfoil. If you created a line with all the points halfway between the top and bottom of the airfoil, you could create the mean camber line. Camber is the difference between this line and the chord line. In a symmetrical airfoil, the camber is 0 and the mean camber line is the same as the chord line. An increase of camber will typically create more lift at a lower angle of attack/lower speed, along with more drag.

Airfoil: A section of the blade, this section clearly shows the chord length, the thickness of the section, as well as the camber. The shape of the airfoil has a large effect on the Cl and Cd.

Now that we have defined all the aerodynamic parameters, I will continue the discussion with frequent references to them. You can open a separate window that displays these parameters by clicking here.

As stated above, our first major design decision was to optimize a symmetrical airfoil design. The next steps were to examine some of our constraints. The first obvious one was being able to fit into the stock blade grips as well as the correct diameter hole. Another obvious requirement is the cutout at the rear of the blade so that when the blade folds back it doesn't contact parts of the rotor head, we actually opted for slightly more clearance for this feature. With the constraints out of the way, the team could now explore the many remaining parameters.

Before simply guessing at what profiles would be required, the team needed to develop a computer simulation that could determine the lift and drag at various rotor speeds and pitch angles. The lift equation, L=.5*Cl*r*v^2*A, presented two issues. The first required the ability to calculate Cl, while the second issue was the fact the the velocity changed from the blade grips to the tip. Through research of wind tunnel testing and many other sources, the team was able to create a large database of Cl for many different airfoils, including the symmetrical wood blades. Since Cl is depended on air foil shape and angle of attack, a 3-D map was created to lookup the Cl for a given air foil shape and angle of attack.

On to the issue of varying airspeed across the blade. The first step is correlating the airspeed to distance from center and rotor RPM. We'll declare x as the distance from center to a point on the blade and RPM as the rotor's RPM. Converting RPM to radians/sec (w), w=2*pi*rpm/60. It follows that airspeed=x*w...not so hard right? Those that fear calculus should leave now ;).

If you picture one section of the blade, the lift at that section will be L=.5*Cl*r*v^2*A=.5*Cl*r*(x*w)^2*A since we determined that airspeed (v)=x*w. Since we are looking at a single section, the area is a constant(chord length, lchord) times the thickness of the section, which in this case is infinitely small, dx (see picture below). So, A=dx*lchord. So, the lift over a section of the blade is equal to:

To calculate the total lift you would divide the blade into sections of constant profile and use the Cl for each profile; or if you could find an equation for Cl(x) you could substitue that as well. For a constant section, Cl would be constant. In most cases you would calculate the lift for one blade and double it for the other blade.

Next we need to calculate the drag forces on the blades. The drag equation is very similar to the lift equation, D=.5*Cd*r*v^2*A, except that Cd=Cd0+Cdi and Cdi=Cl^2/(pi*AR*e). Therefore, D=(Cd0+Cl^2/(pi*AR*e))*.5*r*v^2*A. As before with lift, v=x*w and A=dx*lchord. So the drag over a length of blade is:

Finally we need to determine the torque required by the main motor to determine the force required by the tail rotor. Since in the previouse drag equation we determined the drag force on an infinitely small section using A=dx*lchord, if we multiple this force by the distance from the center of rotation, x, we will have the drag torque. So the drag torque is:

Now that we can represent the blade mathmatically, the next step is to create a model in matlab, excel, etc.. that will calculate the change in lift, drag, and drag torque based on our design. Once we created this model, we created the chart below for the stock wood blades to verify that our model was valid. The chart represents the stock symmetrical wood blades at 1800 RPM, similar to idle up position. It shows the drag, lift/drag ratio, torque, tail force, and weight. The helicopter will begin to lift off (become light on the skids) with 3 degrees of pitch (where lift and weight cross), which corresponds very well to our real world measured data. The drag torque and tail force also correlate as well.

Using our model the Colorado Rocks team was able to analysis many possible designs and created the below 3-D chart of lift/drag ratios as a function of pitch angle and airfoil shape. This data along with other simulations and tools was used to come up with the 7 Prototypes that were real world tested to determine the final design.

Finally, we need to ensure that the material properties are adequate and safe for flight. The blades experience many forces, they face the upward force of lift which holds the helicopter in the air (bending), a torque force from drag (bending), and the centripetal force induced by the blades rotation (tension/compression). The centripetal force is the largest force and in most cases would be the fail mode. The equation for centripetal force is, F=m*v^2/r, where m is mass, v is velocity, and r is radius. As we discussed earlier, the v is simply r*w and w=2*pi*rpm/60. In our above equations x was the distance from center, so r=x. From this we can see that F=m*v^2/r=m*(x*w)^2/x=m*x*w^2. As an example, we will calculate the centripetal force for the symmetrical wood blades. Since our blades weigh .009 kg, the radius of the overall rotor is .265 m, w is 209.44 rad/sec; F=.009 kg * .265 m * 209.44^2 rad^2/sec^2 = 104.03 N or 23.39 lbs. Does that sound right to you? Each blade is pulling outward with that much force? That would mean that the shaft connecting the two blades has 2*23.39 lbs =4 6.78 lbs of tension... The answer here is no!! The compressive stress near the screw would be near 5.95 Mpa and the tensile stress would be around 1.17 Mpa at sections near the screw. Since low density balsa wood has a compressive strength of 4.70 Mpa and a tensile strength of 7.60 Mpa, you can see that the design would be inadequate. The screw does have the plastic reinforcement pieces in place to help distribute the forces. Can you find the flaw in the above example?

The key is the varying velocity over the length of the blade, the above example assumed all the mass at the tip. In the real world, only the tiny slice of mass at the tip experiences that velocity, the tiny slices of mass near the blade grips are at a much lower velocity and therefore contribute less force. So, how do we determine the real centripetal force? You guessed it, calculus to the rescue with another integral... In this case we have the above equation, F=m*x*w^2, which already has the velocity as a function of x built in, v(x)=x*w. We can change mass to be the density of the material times the volume, F=m*x*w^2=r*V*x*w^2. Now, we change the volume to be the 2-D area of a section times an infinitely small slice, dx, so F=r*Asection*dx*x*w^2. All of these steps with a picture are summed up below:

Recalculating using integration yields a force of 75.19 N or 16.90 lbs. Although this stress is less, it is considerable when you think about it. The corresponding compressive stress near the screw is around 4.30 Mpa and the tensile stress drops to .85 Mpa at sections near the screw. From these values you can see that the safety factor for low density balsa is only 1.09 (4.70 Mpa / 4.30 Mpa). Don't worry too much though, the plastic pieces add a lot of strength and the balsa is probably a higher density balsa wood with a higher stength (up to 19.5 Mpa for high density balsa); however I would expect that without the plastic pieces, the hole for the screw would elongate over use (perhaps I will test this sometime). Other things to keep in mind is that a reduction in RPM will reduce the forces, changing to 1800 RPM from 2000 RPM changes the force to 60.91 N or 13.69 lbs, the compressive stress to 3.48 Mpa, and the tensile stress drops to .68 Mpa.

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