Disk Loading comparison of MV-22 Osprey Tiltrotor vs CH-47 Chinook Helicopter

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The above is comparison table of several rotating wing aircraft's disk loading, especially I highlighted the disk loading of MV-22 Osprey tiltrotor versus CH-47 Chinook helicopter. As in the table, the disk loading of MV-22 = 129.63kg/m2 with 38ft rotor diameter each while disk loading of CH-47 = 43kg/m2 with rotor diameter is 60ft each. Both of them come with 2 rotors and 3 blades each rotor. If we see that table, the disk loading of MV-22 is three times of disk loading of Chinook's with their blades' length is more than double/half one of another. Means, the lifting thrust of the MV-22 is very high. So my question, what makes the MV-22 has lift thrust and its disk loading is so high?


2 Answers 2


Your question considers both the CH47 and MV22 rotating wing aircraft, but this is only partially true for the MV22 which also operates as a fixed wing aircraft with rotors tilted forward acting as props.

A prop requires a higher velocity through the disc than the aircraft's airspeed or it ceases to produce thrust. The MV22 flies at 240kt so the airflow needs to be higher than 240kt. The CH47's rotors even at speed see very low inflow velocities, so it uses a much lower disc loading to increase lift efficiency.

This shows a rough relationship between disc loading and air velocity through the disc. As disc loading increases hover efficiency drops but downwash speed increases, so the MV22 uses a higher disc loading to get higher airflow speed through the props.

enter image description here

  • $\begingroup$ I am asking when it V/STOL as mentioned in the table. Not a case when it cruise. I am asking when it VTOL, what makes it so powerful. The rotor it self is shorter 5ft that it should be, but due to storage necessity, it made just 38ft. Just at 480 RPM the rotor's tip speed will not exceed Mach 0.85. Clearly, it can not rotate faster. Then my question, what special design in that rotor's blade so it has so powerful lift? $\endgroup$ Dec 8, 2022 at 6:19
  • $\begingroup$ CH22? Please make up your mind! $\endgroup$ Dec 8, 2022 at 10:54
  • $\begingroup$ According to your plot, the MV-22's disc loading of 130kg/m² gives an induced velocity of some 100km/h which is definitely well below the MV-22 flying speed of 450km/h (240kts). How is this compatible with your statement that "the airflow needs to be higher than 240kts"? $\endgroup$
    – sophit
    Dec 8, 2022 at 21:42
  • $\begingroup$ @PeterKämpf fixed, thanks $\endgroup$
    – Pilothead
    Dec 8, 2022 at 22:52
  • $\begingroup$ @sophit This plot illustrates the principle for the case in the reference, which describes a different aircraft in hover. It is not the MV22. It would have to be adjusted to apply to any other aircraft and particularly in cruise, but the principle still holds. $\endgroup$
    – Pilothead
    Dec 8, 2022 at 23:19

What makes the MV-22 has its disk loading is so high?

Compromises. As usual in the aerospace world.

In particular, here we have the following two contrasting requirements the equilibrium of which drives the design of a tiltrotor resulting in its relative high disc loading in respect to a pure helicopter:

  1. in helicopter configuration the blade span should be as big a possible to have an efficient design requiring less power to work, but;
  2. in airplane configuration that same blade cannot be too big in order to limit its tip speed under the speed of sound.

Disc loading influences the power $P$ needed by a rotor to generate a certain amount of thrust $T$ according to the following simple equation:

$P=\sqrt{\text{(disc loading)}\frac{T^2}{2 \rho}}$

Being the thrust kept constant, a higher disc loading implies more power which has to be supplied to the rotor to give that thrust. So, in order to keep power required low, disc loading should be also low i.e. disc area as big as possible. For example, at the extremes of the table you posted, on one end there might be a man-powered helicopter with a very big rotor $\Rightarrow$ low disc loading $\Rightarrow$ low needed power; and on the other end an F-35 with its very high disc loading $\Rightarrow$ very high power required.

Anyway the rotor of a tiltrotor cannot be too big since it has to work also as a propeller when the aircraft is in airplane configuration. Propellers also obey to the same previous equation but with an additional limitation: the speed seen by a propeller is the sum of the speed due to its rotation (just like a rotor in hover) plus the flight speed. This sum is highest at the tip of the blade and it must be lower than the speed of sound in order to avoid transonic effects that increase drag and reduce thrust.

Obviously there are many other requirements/constraints which have to be considered in the design of a tiltrotor, like for example:

  • rotor assembly should be kept slender to reduce weight; this help limiting both CG excursion and aeroelastic effects during transition phase;
  • blade span is limited by obvious reasons of clearance with the fuselage when in airplane configuration; longer blade would imply a wider wing to keep the blade far from the fuselage giving an higher wing weight;
  • airflow velocity under/behind the disc rotor (aka induced velocity) is also proportional to disc loading, according to the following equation: $v_i=\sqrt{\frac{\text{(disc loading)}}{2 \rho}}$ and with its 80kts the induced velocity under the MV-22 can be classified as a hurricane; that means that the MV-22 cannot for sure land on some unprepared field; high induced velocities increase also the risk of brownout/whiteout; and the fuselage is better accessed from behind instead from the side where the high induced velocities can make it difficult.
  • $\begingroup$ when you introduce formulas, please introduce also the symbols therein contained. T is the thrust {\displaystyle \rho }\rho is the density of air (or other medium) A is the area of the rotor disc P is power $\endgroup$
    – EarlGrey
    Dec 6, 2022 at 13:45
  • 1
    $\begingroup$ @EarlGrey: done $\endgroup$
    – sophit
    Dec 7, 2022 at 10:04
  • $\begingroup$ Unlimited power is not always solution. If the rotor length of the MV-22 is just, i.e., 2m (diameter=4m), even with unlimited power will not be enough to lift that aircraft in maximum load. There is limitation such as sonic boom. I am asking specific MV-22, not general. The rotor of the MV-22 itself is shorter 5ft than it should be. Then my question was, what makes it lifting force so high? There must be specific design. $\endgroup$ Dec 8, 2022 at 6:00
  • $\begingroup$ @AirCraftLover I actually wrote the opposite: the rotor should be big to reduce power but on the MV-22 the rotor cannot be as big as needed since there are other limitations like supersonic speed at its tip and clearance with the fuselage. Sonic boom happen when the entire aircraft flies at supersonic speed, I really don't think that the MV-22 can fly supersonic. Please don't downvote before understanding things. $\endgroup$
    – sophit
    Dec 8, 2022 at 7:33
  • 1
    $\begingroup$ Another constraint for rotor diameter in the MV-22 is span: Larger rotors require a larger wing, which is heavier. $\endgroup$ Dec 8, 2022 at 10:57

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