# What is the downwash speed on a typical helicopter?

On a typical 2 or 4 seat helicopter like a Robinson R22/R44, at full throttle at 20ft ASL, if you put a Wind Speed Meter/Anemometer about 10ft below the blades and halfway along a rotor blade's span, about what speed should it read?

What you are searching for is called the induced velocity $$v_i$$. It can be calculated in a good approximation via the following formula:

$$v_i = \sqrt{\frac{F_n}{2 \rho S}}$$

With $$F_n$$ being the weight of the aircraft, $$\rho$$ the air density, and $$S$$ the disk area. This formula assumes an "ideal" rotor, therefore air is uniformly accelerated over the rotor plane without losses.

Here are some example values for light, medium and heavy weight helicopters as well as a tilt rotor at sea level for the MTOW (Maximum Take-off Weight):

• Robinson R-22 (light weight helicopter): $$7.34 \frac{m}{s} \approx 24.1 \frac{ft}{s}$$
• Airbus H135 (medium weight helicopter): $$10.01 \frac{m}{s} \approx 32.8 \frac{ft}{s}$$
• Leonard AW139 (heavy weight helicopter): $$13.55 \frac{m}{s} \approx 44.5 \frac{ft}{s}$$
• Boeing V-22 Osprey (tilt-rotor helicopter): $$20.24 \frac{m}{s} \approx 66,4 \frac{ft}{s}$$

In reality however, the rotor plane is not a continuous plane which accelerates air, as assumed by the momentum theory. In reality the airspeed over the rotor plane rises linearly outwards, as the velocity rises linearly over their span (a rotor segment spins faster the more distant it is from the rotor hub). The situation is roughly as follows:

However as you ask about about a point halfway along a rotor blade, the induced velocity is pretty much the mean induced velocity as calculated above. Additionally, the stream tube does not contract very fast, therefore the induce velocity might be a bit higher, but not by much. Sophit states in his answer that the stream tube contracts by a factor of 2 over the rotor diameter, which would be 25 feet for the R22, therefore at 10 feet it is perhaps 10ft/25ft = 0.4 about 40% faster. Given that, an the fact that I estimate the full throttle thrust to be about 20% of hover thrust at MTOW, the induced velocity of an R22 would then be about $$11.36 \frac{m}{s} \approx 37,3 \frac{feet}{s}$$.

• An R22 7.34m/s? Unless I'm misunderstanding something those numbers just don't make any sense to me(that is too slow). Also where do you take into account the 10ft distance bellow the blades on that formula?
– Gabe
Commented Jan 20 at 3:29
• @Gabe The 10 ft can only be taken into consideration with a full computational fluid dynamics solution. U_flow's answer is for a spherical cow in a vacuum (see the words 'good approximation' in the opening), which gets you most of the way there, and certainly allows comparison between disk loadings. Commented Jan 20 at 6:14
• @Gabe no the number is definitely correct. The induced velocity as calculated above stems from momentum theory, therefore it is a mean value across the rotor disk. In reality, the induced velocity rises from the hub to the tip linearly until shortly before the tip, at which point it drops of. However at the location which you pointed at, it hits exactly the mean value. Therefore, the answer is correct. I was also a bit taken back, because by intuition this number should be higher, but helicopters are built to reduce induced velocity to increase efficency. Commented Jan 20 at 9:30
• I think this only applies for horizontal flight at reasonable speed, when the rotor is constantly moving into fresh still air. In hover the air velocity will be higher as the air starts to spin around the helicopter. Here is a CFD simulation for UH-60; the formula in this answer gives 44 ft/s, which seems to agree quite well.
– jpa
Commented Jan 20 at 9:39
• @Gabe I tried to update my answer to include more real-life effects. However 7.34 m/s is accurate for the mean velocity. I get that this seems low, but this is just how it is! Helicopters are built to accelerate large cross sections of air to relatively low speed, because that is efficent hence their big rotor! The inverse would be things like the harrier jet, which accelerates a small cross-section of air to a very high velocity (due to their jet-engine) but is extremly inefficient, and therefore can only hover for a very small amount of time. Commented Jan 20 at 15:57

Airflow around rotors can be extremely complicated (NASA source):

Estimating the airspeed in exactly one point is therefore very difficult and is normally done only if the point belongs to the helicopter. Anyway a good approximation of the average flow field under a rotor can be obtained via the simple momentum theory by which the rotor is assumed as a thin disk imparting the air a jump in its pressure (source):

As visible in this picture, airspeed increase from 0 far above the rotor to $$v$$ by the rotor and to $$w$$ beneath it. This particular airspeed shape is termed "vena contracta" (dashed line). For a rotor of disk area $$A$$ generating a thrust $$T$$, the momentum theory returns a speed $$v$$ by the rotor of:

$$v=\sqrt{\frac{T}{2 \rho A}}$$

As said, under the rotor the airflow keeps on contracting and the speed increasing, reaching a value $$w=2v$$ at some one-diameter distance under it. Further below airspeed will eventually decrease due to viscosity.

Being in hover thrust equal weight, from that equation it is easy to calculate the speed at the point you indicated in your question. Considering the point halfway between $$v$$ and $$w$$ and given the main rotor area of 46m² and the MTOW of 620kg for a R22, we get at sea level:

$$w=1.5\sqrt{\frac{620 \times 9.81}{2 \times 1.225 \times 46}}=12m/s$$

• Good answer, but the air density at sea level is 1.225 kg/m^3. However I did not know that the stream tube contracts so fast! However, please take into account that the point OP is asking about is not 1 Rotor diameter below the rotor plane, but only 10 feet, therefore only 40%. Commented Jan 20 at 15:54
• @U_flow: right thanks, I'll correct the value for the density. One rotor diameter is obviously just a rule of thumb, let's say that a value halfway is maybe more correct. Commented Jan 20 at 16:07
• The instantaneous turbulent flow field is indeed extremely complicated but in these kinds of question and problems we usually care about the values averaged in the Reynolds averaging sense and then the average velocity field is much smoother and the W value at a given point shall be well defined. Commented Jan 20 at 19:54