I'm not a pilot, just asking out of curiosity. What is the typical descent rate for a helicopter when landing? I found the answer of 300 feet per minute on the web, but I want to confirm this. If this is true, than a helicopter descending from 2400' to land would take 8 minutes. Is this correct?
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2 Answers
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That is of course dependent on airspeed at approach, and the approach angle. I have a flight manual of an older type helicopter, I cannot reproduce the contents, but it shows True Airspeed on the horizontal axis, Rate of Descent on the vertical, and a set of inclined lines for the angle of Descent.
In the Emergency Procedures - Normal Approach, an Angle of Descent of 8 - 10º is specified, and a typical TAS of 65 knots. This amounts to a Rate of Descent between 800 and 1100 ft/min. Maximum allowable ROD at that speed is 1600 ft/min which is marked as Approximate Autorotation.
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$\begingroup$ Thanks. Doesn't "Emergency Procedures" imply higher than typical rate of descent? This would suggest that normal rate of descent is lower. $\endgroup$ Commented Dec 11, 2017 at 3:26
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$\begingroup$ Yes it would, although use of Normal Approach would suggest, well, normalcy. The normal angle of descent seems to be 8 - 10 deg for this helicopter, a Huey. Could not find any other references to a normal descent rate in it. $\endgroup$– KoyovisCommented Dec 11, 2017 at 3:29
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$\begingroup$ It also depends upon the exact aircraft make/ model/ equipment as a "normal descent" can put you in the bad area of the height-velocity diagram for potentially a long period of time in certain circumstances. Managing the risk of that versus a descent that is not particularly stabilized is the art of flying. In the particular helicopter I had trained in, with a stabilized approach from about 50' AGL to 6' AGL you were in the bad area of the H-V diagram and if the engine fails it's really going to hurt. $\endgroup$– RudyBCommented Dec 23, 2017 at 23:53
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The 300 feet per minute rate is the maximum descent rate for a hover. In helicopters, it is the rotor that is providing airflow downwards to keep the helicopter in the air. The air that has traveled through the rotor, downwards, keeping the helicopter aloft has become very turbulent air with a downward velocity. This means that it until it settles, it is not going to provide lift. If I'm in a hover, and I descend at a rate higher than 300fpm, then I can potentially descend into my own down-wash, and then the rotor stops providing lift. This is known as vortex ring state, or sometimes settling with power (Vortex Ring State is more precise, settling with power can occur for various reasons and in different conditions). If I'm traveling in some direction at a reasonable speed, I'll be moving away from the turbulent air that just went through the rotor, and the low descent speed no longer applies.
The calculations from the earlier post are reasonably accurate for an emergency descent. In an emergency landing (an autorotation, or power-off landing), the R22 emergency procedures state to establish your glide at 65 knots, with your vertical descent being controlled by what is required for me to keep the rotor spinning.
In a normal descent, my forward airspeed would be closer to 90 knots, and 8-10 degrees would be on the higher end for descent rates.
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$\begingroup$ Thanks. So what does a typical helicopter (e.g. for sightseeing) descending from 2400' do usually? Hover down for 8 minutes or adopt some kind of a spiral path to speed up the descent? $\endgroup$ Commented Dec 11, 2017 at 6:42
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2$\begingroup$ Helicopters use far more power in a hover than they do in forward flight. In forward flight, anything above about 30 knots, the helicopter starts developing translational lift, where the entire spinning rotor (the rotor disc) acts like a wing and generates extra lift. Due to both the translational lift, the restrictions on the descent speed in a hover, and the ability to merge with regular airport traffic, in most situations, helicopters will land in the exact same pattern as an airplane, except that we'll generally circle around in the opposite direction (making right turns instead of left). $\endgroup$– Tom KCommented Dec 11, 2017 at 22:42