Frequently Asked Questions about Carter Aviation Technologies
What does "slowed rotor/compound" or "SR/C™" mean?
"Slowed Rotor/Compound," abbreviated as SR/C™, is Carter's term to describe aircraft such as the CarterCopter and our other concepts, which unload and slow the rotor to reduce drag in high speed flight, while using a wing to provide lift in that condition.
How can SR/C™ aircraft fly so fast and efficiently? Shouldn't the rotor slow them down?
SR/C™ aircraft are a hybrid between an airplane and a rotorcraft. A rotor is a very efficient device for providing lift at low speeds, but its drag increases rapidly as the aircraft goes faster if it must continue to support the aircraft. In SR/C™ aircraft, as the aircraft speeds up and the wings begin producing more of the lift, the rotor produces less lift and can slow down given the correct control input. The reduction of rotor lift and lower rpm significantly decrease the rotor drag (in fact, a three fold reduction in rpm results in approximately a 27 fold reduction in rotational drag- drag required to just spin the rotor). The rotor drag at very low rpms and low lift basically becomes a function of its area (which is relatively small compared to an airplane wing of similar gross weight) and the forward speed of the aircraft.
Wings are very efficient at high speed, but can't provide enough lift as the aircraft slows down. In most aircraft, the wings are sized significantly larger than they need to be in cruise flight so that the pilot can fly slower for landing. Most airplanes also have some type of high lift device, such as flaps, which further decrease the minimum flight speed of the aircraft, but add weight and complexity to the wing. The CarterCopter has a very simple wing, sized much smaller than a conventional aircraft of similar size, because the wing only needs to support the aircraft at high speeds.
The combination of the slowed, unloaded rotor and the simple, small wing is what allows SR/C™ aircraft to operate so efficiently. Their efficiency is on par, or better in high speed cruise than those of fixed wing aircraft (airplanes) with a stall speed less than 70 mph, and is much more efficient at high speed than any conventional rotorcraft.
If a rotor produces less drag the slower it rotates, why don't you just stop the rotor?
A stopped rotor would probably not produce any less drag than a slowly turning rotor when all factors are considered, and it has several disadvantages.
Rotor blade centrifugal force (CF) provides a moment (force times a lever arm) that counters the blade lift to keep the blades from rising up excessively. As the rotor slows down, at some point there will not be enough CF to handle the blade lift on the advancing blade. This will cause the rotor to flap excessively. Therefore, to stop the rotor, a flapping stop must be engaged and the rotor must be stiff enough to handle any lift forces generated by the blades. At high forward speeds the locked rotor must be able to handle the lift generated by the rotor (there is no way to keep the rotor at precisely zero angle of attack, so it definitely will generate lift). As a result the rotor must have bending structural requirements similar to a wing.
For a rotor to be an efficient lifting device it must have a low disk loading (ratio of force generated by rotor to the area it sweeps as it rotates). So, as the aircraft weight increases so must its rotor diameter. However, as the rotor diameter increases, so too do its bending requirements when functioning as a wing as described above. Since those requirements on the rotor become so high, its weight plus the weight of the locked rotor hub becomes much greater than the tip weight needed on a slowly turning rotor to handle the lift on the advancing blade.
By concentrating our weight in the tips, we maximize the centrifugal force of our rotor for a given weight. Centrifugal force is an efficient way to keep the rotor rigid. It is analogous to a rock on a string. Even though the string has no rigidity to hold the rock out horizontally, once you start rotating the system, the rock will hold the string nearly horizontal.
All the structure needed to stop the rotor adds weight and complexity, not to mention the procedure of stopping and starting the rotor in flight. The reduced drag of a stopped rotor is offset by its increased weight. Considering that it only took about 5 HP to keep the rotor spinning on the original CarterCopter Technology Demonstrator at its lowest rpm in cruise, stopping the rotor just isn't worth it unless there is a requirement to travel more than 500 - 550 mph (above those speeds, the tip speed on the advancing blade becomes greater than Mach 0.9).
What are the mu ratio and the mu-1 barrier?
Mu is the English spelling of the Greek letter μ. It can be pronounced "mew" or "moo," with "mew" being the more common. It is commonly used to represent a ratio in rotorcraft engineering, sometimes called the rotor tip advance ratio. To put it about as simply as it can be put into words, the μ ratio is the ratio of the forward speed of the aircraft to the rotor tip speed relative to the aircraft. To put it into a picture, which is worth a thousand words:
where Vtip is the tip speed of the rotor, VA is the speed of the aircraft, and μ is the mu ratio.
There are a few interesting things that can be determined from the μ ratio. First, and most obviously based on how μ is defined, you can easily determine the airspeed over the tip of the rotor at the 3 and 9 o'clock positions (when the rotor is perpendicular to the aircraft). When you are at a μ of 1, the tip speed is equal to the aircraft speed. So, on the advancing blade, at the 3 o'clock position, the airspeed over the tip is twice what it is over the rest of the aircraft. On the retreating blade, at the 9 o'clock position, the airspeed over the tip is zero.
Another interesting thing that can be determined from the μ ratio is how much of the rotor is seeing reverse airflow at the 9 o'clock position. Each position on the rotor blade sees a different airspeed, which is a combination of the speed from the rotor turning and the speed from the aircraft. As you move outboard on the rotor, the rotational velocity gets higher. The μ ratio will tell you what proportion of the retreating blade is seeing reverse flow at the 9 o'clock position. For example, at a μ of 0.3, the inboard three-tenths of the retreating blade will see reverse flow, the outboard seven-tenths will see normal flow, while the point exactly at three-tenths will have an airspeed of zero. At a μ of 1, the tip of the retreating blade will have an airspeed of zero, and everything inboard of that will see reverse flow. That is the μ-1 barrier. It is the point at which the rotor first sees complete reverse flow on the retreating blade. The CarterCopter Technology Demonstrator became the first rotorcraft to achieve a μ of 1 in June 2005. The previous high of 0.92 was established around 50 years ago by the McDonnell XV-1 .
From a flapping stability standpoint, a μ of 0.75 is the most unstable. This is better explained in answer to the question, Don't SR/C™ aircraft have a problem of retreating blade stall in high mu flight?
What is an autogyro?
An autogyro is a rotorcraft that does not power the rotor directly by the engine. The rotor is made to spin by air passing up through it in much the same way as a wind turbine is powered by the wind. This is in contrast to a helicopter, where the rotor is usually driven by a drive shaft and as a result needs some method to counter this torque, such as a tail rotor. Since the rotor is unpowered in an autogyro, it requires a propeller or some other means of propulsion to push it through the air. To read more about autogyros and autogyro history, visit Autogyro History & Theory .
Can SR/C™ aircraft hover like a helicopter?
SR/C™ technology could be applied to helicopters (click here), but none of our prototypes so far incorporate the ability to hover. However, with a prerotator, high inertia rotor and collective control for the rotor, these aircraft can still takeoff and land vertically. Because of the extreme inertia in the rotor, SR/C™ aircraft can hover for ten (10) seconds. This adds a great deal of utlity to the aircraft by allowing off airport operations. Many rotorcraft operations do not require hover, only that they can safely take off and land with zero roll. For more information on jump takeoffs, see How does an autogyro perform a jump takeoff?
There is no reason that the slowed rotor technology could not be applied to an aircraft that was part helicopter, as well. The aircraft would still operate the rotor in autorotation for cruise flight, but it would also have the ability to operate as a helicopter for hovering. In fact, we already have concept aircraft that incorporate this, including the Heliplane Transport and the CH-45 Heliplane. However, the heavier gearbox and added anti-torque devices needed to hover will increase the empty weight and thus decrease the useful load of those aircraft over the autogyro versions.
How does an autogyro perform a jump takeoff?
The process of performing a jump take-off is pretty straight-forward. The prerotator consists basically of a drive shaft and a gearbox, connecting the rotor to the engine through a clutch. Because the system is used only for short periods on the ground, and not continuously in flight, it can be much lighter than the corresponding system on a helicopter. On the ground, where the landing gear can counter the torque driving the rotor, the prerotator is engaged to spin up the rotor. It is spun to an rpm much higher than the rpm the rotor will be operating at in flight (for example, the rotor on the CGD/T will need to spin at about 250 rpm in flight, but will be prerotated to 500 rpm for a jump takeoff). This extra rpm stores quite a bit of energy, and with the tip weights of the Carter rotor, makes for a very good flywheel. During prerotation, the rotor is set to a flat pitch, both to keep it from producing a force to lift the aircraft, and to reduce the induced horsepower, to allow it to be spun to the highest rpm possible. To be able to put the most power into the rotor as possible, it also helps to have a variable pitch propeller, so that during prerotation, the prop can be set to a flat pitch where it doesn't absorb much of the engine power. When the rotor is at the desired rpm, the pilot will increase the pitch of the propeller, disengage the prerotator, and then increase the pitch of the rotor. The lift created from increasing the pitch on the rotor causes the aircraft to jump into the air, while the thrust from the propeller accelerates the aircraft to its normal flight speed.
With the original CarterCopter Technology Demonstrator, since we were mostly concerned with demonstrating the high speed flight technologies, we never really concentrated on trying to develop the procedures for jump takeoffs or zero roll landings. However, they are not really all that novel of procedures for autogyros to perform. Cierva began experimenting with jump takeoffs on the C.30 autogyro in 1933, and in a little over a year had refined the procedure so that jump takeoffs were routine. The video on the left is of the Pitcairn PA-36 "Whirlwing" Autogiro of the late 1930's, showing the dramatic jump takeoffs that that aircraft could perform.
Because of the weight concentrated at the blade tips, Carter rotors will have much higher inertia than conventional rotors. Combined with the fact that modern materials allow the rotor to be spun at proportionately higher rpms than the older rotors like the one shown in the video above, Carter rotors will be able to store 2-4 times the excess energy per pound of aircraft weight as the PA-36. While the jump takeoff in the video above is certainly impressive, it was performed with a single pilot, no baggage, and only a small amount of fuel on a cool day. Autogyros and gyroplanes utilizing a Carter rotor would be able to perform a takeoff as impressive as the one shown above fully loaded on a normal day. The video on the right shows the CarterGyro Demonstrator/Trainer performing a jump takeoff straight up, and then descending vertically for a zero roll landing.
What is flapping?
Because the advancing blade of a rotor has a higher average velocity than the retreating blade, it has the potential to produce more lift. Having more lift on one side than the other would create a lift moment (lift times a lever arm) that would try to roll the aircraft. However if the rotor has a hinge so that it can teeter (like a playground teeter totter), then the rotor will teeter / "flap" such that the advancing blade can go up and the retreating blade can go down. The upward movement of the advancing blade decreases its relative airflow angle of attack, and the downward movement of the retreating blade increases its relative airflow angle of attack. This change in angle of attack decreases the lift on the advancing blade and increases the lift on the retreating blade. The rotor will automatically flap until the lift moment generated by each blade is equal. If the unbalance in blade lift moments is too high, then the flapping becomes greater than the flapping travel and bad things will happen.
Do SR/C™ aircraft have a problem of retreating blade stall in high mu flight?
No. Retreating blade stall occurs in a rotor when it is trying to support more weight than the retreating blade can support. As explained in answer to the question, What is flapping?, as the aircraft speed increases, the average air velocity increases over the advancing rotor blade but decreases over the retreating blade. In order to keep the blade from creating unequal lift and causing a rolling moment, a teetering hinge allows the advancing blade to flap up, decreasing its angle of attack while the retreating blade flaps down and increases its angle of attack. At a certain point, if flapping gets too high, the angle of attack on the retreating blade can exceed the stall angle, and the blade will experience "retreating blade stall." In the SR/C™ aircraft, the rotor is unloaded with a wing and propeller, so that the rotor does not need to produce much lift. The pitch on the retreating blade does not need to be very high to produce the required lift, so the blade does not stall.
The worst case for this condition occurs at a mu of 0.75. Recalling the discussion from What are the mu ratio and the mu-1 barrier?, as the aircraft speed increases, the air velocity initially decreases over the outboard section of the retreating rotor blade and at the same time increases the reverse flow over the inboard blade section. From mu of 0 to 0.75, this results in the average magnitude of the air velocity decreasing. However, above mu 0.75, the increasing reverse velocity on the inboard section overtakes the decreasing velocity on the outboard section, and the average magnitude of the air velocity begins increasing, albeit in the opposite direction from what it was before. Even though most of the air is now flowing over the blade in the reverse direction, and this isn't the most efficient way to produce lift, it does provide a greater margin for flapping without causing retreating blade stall.
How do you handle the problem of blade flutter/divergence on the retreating blade?
There is an instability on the retreating rotor blade caused by reverse flow shifting its aerodynamic center from the ¼ chord to the ¾ chord. The design of our rotor, which is very torsionally stiff, has the ability to very rigidly carry torsional loads across the blade hub from one blade to the other, which means that the stability of each blade is coupled to the other blade. We developed a simple equation to predict the stability of our rotor, and have verified the equation through flight testing. We designed our rotor blade to be inherently stable up to high mu ratios (mu ~1.4) by the planform and weight distribution of the rotor blade. Near the tip of each blade, there is a leading edge extension that has a lead weight, which shifts the blade CG forward. Also at the tip, there is a trailing edge extension, which shifts the AC aft. While this combination of CG & AC shift makes the advancing blade very stable, there is a decrease in stability in the retreating blade. However because the velocity is so much lower at the tip of the retreating blade, its instability is less that the advancing blade's increased stability, and since the blades are torsionally tied together, the net result is a stable rotor. At a mu of 1, the velocity at the tip of the retreating blade is zero, which results in a very small unstable pitching moment. But at some increased mu ratio the reverse flow velocity over the retreating blade will cause that blade to become more unstable than the advancing blade becomes stable and the rotor will become unstable. At this point the cyclic control will need to be very stiff and will require boosted controls. Any cyclic movement due to the rotor trying to diverge will cause one blade to increase pitch while decreasing the pitch of the other blade and this will cause the blades to go "out of track." At this point the lift on the rotor is small and if the amount the blades are out of track is small due to a sufficiently rigid cyclic control, the 1 per rev vertical lift oscillation may not be noticeable.
During flight testing, we have experienced this rotor blade instability. It manifests itself in the rotor going "out of track." One blade will be at a higher pitch than the other. As the rotor turns, the blade with the higher pitch will run higher than the other blade. The pilots will see the one blade running higher than the other, and can slowly increase speed and note how the instability increases, or slow down and note the blades going back in track. It is not a sudden, disastrous divergence.
I have heard that two-bladed, teetering rotor systems have a reputation for something called "mast bumping" in low-g flight. What is mast bumping, and why are SR/C™ aircraft not susceptible to this problem?
Mast bumping is caused in helicopters by a lack of control of the aircraft due to the rotor not producing much lift or negative lift. During a zero-g maneuver, when the rotor is unloaded, it does not create any lift force, so it does not have any force to apply to the helicopter to control it. In this situation, if the pilot moves the cyclic to try and control the aircraft, the rotor will still tilt, but since it is producing no net lift, it will not have a reaction on the rest of the aircraft. If the pilot inputs more control because he does not feel the aircraft responding, he makes the rotor tilt even more, until it hits the mast. The tail rotor can aggravate this problem. Because it is still producing a force, it pushes the aircraft to the side, and if it's not directly on the centerline, causes the aircraft to roll as well. The pilot feeling this movement will try to counter it with the stick, causing the problem outlined above, with the tail rotor pushing the fuselage in the opposite direction of the rotor.
In SR/C™ aircraft, because of the dual control system (airplane as well as autogyro), the pilot always has control of the aircraft, as long as he has sufficient airspeed. During a zero-g maneuver, with the rotor unloaded and producing no lift, the elevators and ailerons are still effective. When the pilot moves the stick, the ailerons or elevator will control the fuselage, keeping it in approximately the same relative position compared to the rotor, so mast bumping is not a problem.
What are the other technical issues associated with high mu flight?
There is a detailed discussion of the 9 technical issues that Carter had to solve on the page,
Significance of μ-1 and the Technical Issues Involved