I saw this question asked the other day, and thought I could get a good discussion going on this topic.

 

Two helicopters, exactly the same in weight, type rotor speed, etc. have entered an autorotation. One has no wind and one has a 20 knot headwind. Which one would reach the ground first?

I'll guess...

 

Same time.  Difference will be how far is "the glide".

I think the one with no wind as the one with headwind will give it lift hence keep it airborne a bit longer??? IDK its just a guess based on no knowledge.

I think the one with no wind as the one with headwind will give it lift hence keep it airborne a bit longer??? IDK its just a guess based on no knowledge.

Remember, an airfoil (rotor, in this case) operates on lift, more wind=more lift at same speeds. Oops, you didn't mention speeds!  Given the same speed, I suppose in theory they would reach the ground at the same time, but the headwind will be a bit more forgiving on the bottom....in other words, if there is no wind, you may have to slide it on, whereas with a wind, you might be able to sit is down with zero airspeed. Of course, experience makes a difference too! The headwind provides more options, including being able to stretch the glide a bit further if you go to the bottom of the green....

 The headwind provides more options, including being able to stretch the glide a bit further if you go to the bottom of the green....

That's what I was thinking, surely if the distance travelled is further(with helo in headwind) - it has to take longer?

No idea what the answer is/ don't care but remember that an increase in wind speed causes a decrease in air pressure.

Given that less headwind - less lift, the helo with no headwind will reach the ground first...

But not by much.. :-)

 

It is my understanding that for an auto rotative landing you pretty much convert your vertical velocity in to air force to spin your rotor... then when closer to the ground you raise collective to convert that rotational force to lift to slow your decent and forward velocity.

 

My son who is a black hawk pilot set me up for an autoratative landing in the UH-60 simulator at Ft. Rucker, which i performed with out crashing...

 

Just my 2 cents... but it is all about velocity and converting you forward and vertical velocity to wind force to spin the rotors...

Helicopter autorotative aerodynamics are a complex subject!

 

The autorotative force is derived from the relative airflow gained from the vector between the rotational plane and the inflow airflow from below the disc. This resultant vector is further affected by horizontal inflow with forward speed.

 

The S61 used to have an 'ideal' autorotation speed of 70kts IAS which produced a rate of descent of approximately 2500fpm. If the aircraft were to fly the autorotation faster or slower than this then the position on the drag curve would produce a higher rate of descent. Therefore, the closer to the 'ideal' you can get then the lower the rate of descent. If both aircraft are descending 'vertically' then the still air aircraft 'should' land first (for land read crash!) however if both are flying a correct profile then they would both have the same airspeed just the ground speed would be different thus the angle of descent would be different and the rate of descent the same.

The one w/o a headwind would land first(if done perfectly).  during the decel the rotor is tilted aft and with the headwind it would speed up the rotor beyond the limitation, forcing a collective input to slow the rotor to within limitations.  This would generate a bit of lift, making you float a second (figuratively) longer. 

 

If done properly (atleast in a blackhawk) if you are within trim and maintain 80 kts. at 75-50 feet you decel (aft cyclic input) to slow down.  level the nose, then increase collective in order to use all available energy left in the rotor system.

Into a headwind = steeper angle  of decent and a lower groundspeed to dissipate in the flare. Preferable at the bottom.