From elswhere:
Question:
How would you calculate the rotor downwash velocity (in mph), specifically for the Hughes 500?
Answer:
Following the advice of Mr. Newton, we know that our rotor must produce an upward force, or thrust, equal to our helicopter’s weight if we are to hover. Since Force = (Mass) x (acceleration), that upward thrust must come from continually accelerating a stagnant mass of air downwards through the plane of our rotor disk to some final downwash velocity. This final velocity depends on the weight of the helicopter, the size of the rotor disk area, and the density of the air the helicopter is trying to hover in. Physics says that the energy transfer between the rotor and the air must happen at an equal rate. Equating these energy expressions and manipulating them with a little algebra results in the velocity of the downwash at the rotor disk being equal to the square root of: Weight divided by 2 x (Air density) x (Disk Area). Considering the example of a Hughes 500: Gross Weight = 3,000 lbs Disk Area = 547 square feet Air Density = .002378 slugs/feet cubed (assuming sea level) Plugging in the values into the above formula and hitting the square root key on the calculator results in the downwash velocity at the rotor being approximately 34 ft/sec, or about 23 mph. But we haven’t completely answered the question yet! Keep in mind that this speed is at the rotor disk. As the column of air is forced down below the rotor, it constricts, much like molasses being poured out of a pitcher does. In doing so, it reaches its maximum velocity at 1.5 — 2 rotor diameters below the disc. Consequently, the final fully developed downwash velocity can be shown to be 2x the above calculated amount, or 46 mph in our Hughes 500 example. Points to remember: Higher weights, smaller rotor disk areas, and higher altitudes will all produce higher downwash velocities.
Author
Frank Lombardi is a Police Helicopter Pilot,Testing & Evaluation