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Lift in a helicopter increases with airspeed (from zero IAS up to some Vmax). But in contrast to fixed-wing aircraft a helicopter can generate lift at zero airspeed.

Is there a formula for the airspeed at which a helicopter achieves it maximum rate of climb?

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  • $\begingroup$ "a helicopter can generate lift at zero airspeed" the speed you mention is not the airspeed, but the forward speed. To be at zero airspeed you would need to stop the forward motion, but also to stop to rotor, the rotor is a rotating wing creating its own airspeed and lift. Actually this forward motion increases lift on the advancing blade but decreases it on the retreating blade, creating an unbalanced rotor load. $\endgroup$
    – mins
    Jun 21, 2021 at 20:30

3 Answers 3

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There are a couple of things at play here:

First, the increase in overall lift of a helicopter as it correlates to an increase in forward airspeed actually has a lot to do with the establishment of Effective Translational Lift (ETL) or the airspeed at which the entire rotor system realizes the benefit of the horizontal air flow. This happens when the helicopter's rotor disc moves completely out of its own downwash and into undisturbed air.

Second, beyond the ETL improvement, forward airspeed in a helicopter only produces an increase in asymmetrical lift, rather than the the nice, tidy symmetrical lift dispersal of fixed wing aircraft. This is because of a principle known as (surprise, surprise) Dissymmetry of Lift. The forward airspeed is added to the rotational speed of the advancing rotor blade on one side of the aircraft and subtracted from the retreating rotor speed on the opposite side of the aircraft. While this is counteracted to some extent by mechanical adjustments (flapping, feathering, leading-lagging) to equalize the inherent rolling tendency, this means that the forward speed alone, as a catalyst for lift production, is not the be-all and end-all.

You are correct that a helicopter, under the right conditions and possessing a surplus of available power can generate lift at zero airspeed. This is why a helicopter, under these conditions, effectively has a Vx of zero, since its best angle-of-climb over distance would be a straight vertical climb.

For Vy, the best rate-of-climb over time performance, it is important to plan for torque settings at level flight. Climb performance charts show the change in torque, above or below torque, required for level flight under the same gross weight and atmospheric conditions to obtain a given rate of climb or descent.

Here is a sample calculation taken from the FAA Helicopter Flying Handbook:

Determine torque setting for cruise or level flight using Figure 7-7.

Pressure Altitude = 8,000 feet, Outside Air Temperature = +15 °C, Indicated Airspeed = 80 knots, Maximum Gross Weight = 5,000 lb.

enter image description here

With this chart, first confirm that it is for a pressure altitude of 8,000 feet with an OAT of 15°. Begin on the left side at 80 knots indicated airspeed (point A) and move right to maximum gross weight of 5,000 lb (point B). From that point, proceed down to the torque reading for level flight, which is 74 percent torque (point C). This torque setting is used in the next problem to add or subtract cruise/descent torque percentage from cruise flight.

Determine climb/descent torque percentage using Figure 7-8.

Rate of Climb or Descent = 500fpm, Maximum Gross Weight = 5,000lb.

enter image description here

With this chart, first locate a 500-fpm rate of climb or descent (point A), and then move to the right to a maximum gross weight of 5,000 lb (point B). From that point, proceed down to the torque percentage, which is 15 percent torque (point C). For climb or descent, 15 percent torque should be added/subtracted from the 74 percent torque needed for level flight. For example, if the numbers were to be used for a climb torque, the pilot would adjust torque settings to 89 percent for optimal climb performance.

As you can see on chart 7.7, the green line represents the Vy speed and is relatively low, because it represents the speed at which the lowest percentage of maximum torque is required to sustain level flight in the given conditions, and thus, the speed at which there is the maximum amount of surplus torque available to climb, all while remaining in ETL flight.

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  • $\begingroup$ Above certain weight and altitude, $V_x$ is not zero, because the helicopter is no longer able to sustain hover out of ground effect. In the first diagram the 5,000 lb gross weight line is way above the max torque available indicating hover is not possible at the weight and altitude. $\endgroup$
    – Jan Hudec
    Jun 19, 2021 at 16:19
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The above explanation is really good. Another way to simplify the answer is to consider the two main areas of drag associated with helicopters and the power required to overcome this drag. In the hover Induced Drag is a maximum and this reduces with increased speed whilst Parasite Drag is a minimum in the hover and increases with an increase in speed. Where these two graphs cross, would represent the maximum power margin and the speed for the maximum rate of climb (fig.7.7). For most helicopters this normally occurs between 60-70kts. To answer your question, yes there could be a formula to calculate the best rate of climb for a specific helicopter, but as there are many variables associated with this calculation such as density altitude and AUM etc a performance graph is normally used.

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I find that the easiest way to visualize the answer is with a "power available" chart, which shows both the power available and the power required.

enter image description here

Above is a chart from an article by Frank Lombardi, "Understanding Helicopter Power Requirements: The Power Struggle", Rotor & Wing International (Oct 10, 2017)

The best rate of climb is where the total power required curve is at the lowest. As the article explains:

"Total power [required] continues to decrease with increasing airspeed, until you reach the “bucket” speed. This is the point of greatest difference between power required and power available, which can be translated into maximum rate of climb." (word in brackets added)

This kind of chart takes into account things like translational lift, which reduces the power required as forward speed increases.

Coming up with a universal equation to compute the speed at which total power required is at the minimum could be a nightmare. This is because - as the chart shows - a helicopter has several components which consume power (the FAA refers to them all as drag). These are: (1) profile drag; (2) induced drag; (3) parasitic drag; and (4) miscellaneous drag. Each component has a different equation, some of which are very complex.

But even if you go to all the work to create an equation which includes all these components, you should be aware that, with helicopters, even the best equations are notoriously inaccurate. As one author noted: "Hover performance cannot, even today, be predicted to within ±5 percent" [Franklin D. Harris, "Introduction to Autogyros, Helicopters, and Other V/STOL Aircraft", Vol. II Helicopters, p. xi]

So, if you are going to fly a helicopter, you should use the published tables described elsewhere. These contain adjustments for altitude, which are critical for a helicopter.

If you just want to solve an equation for fun, you could use curve fitting to create an equation that closely matches the bucket in the power required curve and then solve for the minimum value.

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