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	<title>Comments on: Speed limits – Part 2</title>
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	<link>http://blog.aopa.org/helicopter/?p=107</link>
	<description>AOPA's Helicopter Blog</description>
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		<title>By: Adam</title>
		<link>http://blog.aopa.org/helicopter/?p=107#comment-981</link>
		<dc:creator>Adam</dc:creator>
		<pubDate>Mon, 26 Apr 2010 18:40:45 +0000</pubDate>
		<guid isPermaLink="false">http://blog.aopa.org/helicopter/?p=107#comment-981</guid>
		<description><![CDATA[In response to my previous post.  Dont think of induced flow as a downward force, think of how it affects one blade.  As induced flow increases, airflow meeting the blade is more from above.  Which makes the resultant relative wind angle less horizontal (more from above), decreasing AoA.]]></description>
		<content:encoded><![CDATA[<p>In response to my previous post.  Dont think of induced flow as a downward force, think of how it affects one blade.  As induced flow increases, airflow meeting the blade is more from above.  Which makes the resultant relative wind angle less horizontal (more from above), decreasing AoA.</p>
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		<title>By: Adam</title>
		<link>http://blog.aopa.org/helicopter/?p=107#comment-980</link>
		<dc:creator>Adam</dc:creator>
		<pubDate>Mon, 26 Apr 2010 18:11:20 +0000</pubDate>
		<guid isPermaLink="false">http://blog.aopa.org/helicopter/?p=107#comment-980</guid>
		<description><![CDATA[I&#039;ve always explained induced flow, resultant relative wind, and AoA as you did here.  But, I was just thinking about the forces and got confused.  If there are two forces at work, one from ahead (rotational relative wind) and one from above (induced flow) that combine to form resultant relative wind, then an increase in induced flow should cause the resultant relative wind to be more from below, NOT lift it up as drawing suggest.  Why is it that as induced flow increases, all drawing show rotational relative wind staying horizontal, and resultant relative wind angle moves up to decrease the AoA?  If the induced flow force is from above, how is the resultant moving up.  I know that aerodynamically this cant make sense because what I am saying would allow both induced flow and AoA to increase at the same time.  

Can you explain why the downward force of induced flow actually moves the resultant relative wind angle up?]]></description>
		<content:encoded><![CDATA[<p>I&#8217;ve always explained induced flow, resultant relative wind, and AoA as you did here.  But, I was just thinking about the forces and got confused.  If there are two forces at work, one from ahead (rotational relative wind) and one from above (induced flow) that combine to form resultant relative wind, then an increase in induced flow should cause the resultant relative wind to be more from below, NOT lift it up as drawing suggest.  Why is it that as induced flow increases, all drawing show rotational relative wind staying horizontal, and resultant relative wind angle moves up to decrease the AoA?  If the induced flow force is from above, how is the resultant moving up.  I know that aerodynamically this cant make sense because what I am saying would allow both induced flow and AoA to increase at the same time.  </p>
<p>Can you explain why the downward force of induced flow actually moves the resultant relative wind angle up?</p>
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		<title>By: demir</title>
		<link>http://blog.aopa.org/helicopter/?p=107#comment-950</link>
		<dc:creator>demir</dc:creator>
		<pubDate>Sun, 21 Mar 2010 09:58:52 +0000</pubDate>
		<guid isPermaLink="false">http://blog.aopa.org/helicopter/?p=107#comment-950</guid>
		<description><![CDATA[I have a question about the Vne in helicopters. Could you explain why it is decreasing with density altitute and the weight of the helicopter? Is it due to the increased loading on the rotor blades and the vibration due to blade stall?]]></description>
		<content:encoded><![CDATA[<p>I have a question about the Vne in helicopters. Could you explain why it is decreasing with density altitute and the weight of the helicopter? Is it due to the increased loading on the rotor blades and the vibration due to blade stall?</p>
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		<title>By: Kurt McKibben</title>
		<link>http://blog.aopa.org/helicopter/?p=107#comment-298</link>
		<dc:creator>Kurt McKibben</dc:creator>
		<pubDate>Mon, 29 Jun 2009 03:59:10 +0000</pubDate>
		<guid isPermaLink="false">http://blog.aopa.org/helicopter/?p=107#comment-298</guid>
		<description><![CDATA[Yes Ehud,

That was a bar room description of an airfoils stall characteristics. Bernoulli&#039;s principle refers to &quot;any fluid.&quot; Newtons 3rd law is the force acting on the underside of the airfoil.

Kurt]]></description>
		<content:encoded><![CDATA[<p>Yes Ehud,</p>
<p>That was a bar room description of an airfoils stall characteristics. Bernoulli&#8217;s principle refers to &#8220;any fluid.&#8221; Newtons 3rd law is the force acting on the underside of the airfoil.</p>
<p>Kurt</p>
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		<title>By: Kurt McKibben</title>
		<link>http://blog.aopa.org/helicopter/?p=107#comment-294</link>
		<dc:creator>Kurt McKibben</dc:creator>
		<pubDate>Sun, 28 Jun 2009 23:49:50 +0000</pubDate>
		<guid isPermaLink="false">http://blog.aopa.org/helicopter/?p=107#comment-294</guid>
		<description><![CDATA[Thank you for responding Ehud,

I&#039;m always trying to review learn something more.

 I am also a rated fixed wing pilot, land and sea. The actual separation of air flow over the airfoil happens at around 12 to 16 degrees between the relative wind and the airfoils cord line. I have practiced stall spin recovery&#039;s over and over. Not recommended or required until the CFI training, but fun if your ship is classified in the utility or acrobatic categories, and you clear an area with plenty of altitude. It&#039;s an E ticket ride. (5,000 AGL minimum)

I shared this power on &quot;Blade Stall&quot; at sea level question with a group of CFI&#039;s locally and their response was this. Power on blade stall can occur at any altitude, even at sea level when pitch command inputs are larger than horse power available to maintain rotor RPM. Abrupt collective pitch inputs can drag down rotor RPMs. If there is insufficient power to maintain rotor RPMs, high pitch angle and low rotor RPM  (STALL)  can happen at any altitude. This has caused enough fatalities that the FAA has written the &quot;Special SFAR 73&quot; in the FAR&#039;s to address training requirements for pilots flying in the low inertia // low horsepower Robinson&#039;s 22 and 44. Since excessive pitch and a lack of horsepower is the cause,, recovery is impossible,, and the result is a catastrophic boom chop.  Common causes in training for getting into a low rotor RPM situation are rolling the throttle the wrong way, or &quot;death&quot; gripping the throttle. 

Thanks for your help Ehud,,, I&#039;ll read your article----- http://thehood.livejournal.com/42360.html
 
Kurt McKibben]]></description>
		<content:encoded><![CDATA[<p>Thank you for responding Ehud,</p>
<p>I&#8217;m always trying to review learn something more.</p>
<p> I am also a rated fixed wing pilot, land and sea. The actual separation of air flow over the airfoil happens at around 12 to 16 degrees between the relative wind and the airfoils cord line. I have practiced stall spin recovery&#8217;s over and over. Not recommended or required until the CFI training, but fun if your ship is classified in the utility or acrobatic categories, and you clear an area with plenty of altitude. It&#8217;s an E ticket ride. (5,000 AGL minimum)</p>
<p>I shared this power on &#8220;Blade Stall&#8221; at sea level question with a group of CFI&#8217;s locally and their response was this. Power on blade stall can occur at any altitude, even at sea level when pitch command inputs are larger than horse power available to maintain rotor RPM. Abrupt collective pitch inputs can drag down rotor RPMs. If there is insufficient power to maintain rotor RPMs, high pitch angle and low rotor RPM  (STALL)  can happen at any altitude. This has caused enough fatalities that the FAA has written the &#8220;Special SFAR 73&#8243; in the FAR&#8217;s to address training requirements for pilots flying in the low inertia // low horsepower Robinson&#8217;s 22 and 44. Since excessive pitch and a lack of horsepower is the cause,, recovery is impossible,, and the result is a catastrophic boom chop.  Common causes in training for getting into a low rotor RPM situation are rolling the throttle the wrong way, or &#8220;death&#8221; gripping the throttle. </p>
<p>Thanks for your help Ehud,,, I&#8217;ll read your article&#8212;&#8211; <a href="http://thehood.livejournal.com/42360.html" rel="nofollow">http://thehood.livejournal.com/42360.html</a></p>
<p>Kurt McKibben</p>
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		<title>By: Ehud Gavron</title>
		<link>http://blog.aopa.org/helicopter/?p=107#comment-293</link>
		<dc:creator>Ehud Gavron</dc:creator>
		<pubDate>Sun, 28 Jun 2009 04:54:37 +0000</pubDate>
		<guid isPermaLink="false">http://blog.aopa.org/helicopter/?p=107#comment-293</guid>
		<description><![CDATA[I did spend a paragraph discussing rotorcraft blade stall at http://thehood.livejournal.com/42360.html

The gist of it is that if you increase the blade pitch so that the angle of attack  (angle between the chord of the rotor blade and the relative wind) increases then there exists a point where -- just like in a fixed-wing aircraft -- the air will separate from the airfoil.

At that point it has vortexes which prevent it from &quot;reattaching&quot; to the blade even if you lower collective pitch.  (Note that I said &quot;blade pitch&quot; above because you can get a good stall region going with cyclic inputs alone if the conditions are right.)

Blade stall in its simplest form is the same for fixed-wing as it is for rotary wing.  The angle of attack can get beyond the point where air will adhere to the surfaces.  At that point there is a stall condition.  In a fixed-wing a pilot can aim down and hope the resultant high velocity will get the airflow back on the airfoil.  Sadly in a rotorcraft when we stall we have no airfoil and we have no ability to generate forward speed.

I hope that&#039;s helpful... if not I hope to learn more by reading what all you people write :)  

Regards,

Ehud
Tucson, AZ USA]]></description>
		<content:encoded><![CDATA[<p>I did spend a paragraph discussing rotorcraft blade stall at <a href="http://thehood.livejournal.com/42360.html" rel="nofollow">http://thehood.livejournal.com/42360.html</a></p>
<p>The gist of it is that if you increase the blade pitch so that the angle of attack  (angle between the chord of the rotor blade and the relative wind) increases then there exists a point where &#8212; just like in a fixed-wing aircraft &#8212; the air will separate from the airfoil.</p>
<p>At that point it has vortexes which prevent it from &#8220;reattaching&#8221; to the blade even if you lower collective pitch.  (Note that I said &#8220;blade pitch&#8221; above because you can get a good stall region going with cyclic inputs alone if the conditions are right.)</p>
<p>Blade stall in its simplest form is the same for fixed-wing as it is for rotary wing.  The angle of attack can get beyond the point where air will adhere to the surfaces.  At that point there is a stall condition.  In a fixed-wing a pilot can aim down and hope the resultant high velocity will get the airflow back on the airfoil.  Sadly in a rotorcraft when we stall we have no airfoil and we have no ability to generate forward speed.</p>
<p>I hope that&#8217;s helpful&#8230; if not I hope to learn more by reading what all you people write <img src='http://blog.aopa.org/helicopter/wp-includes/images/smilies/icon_smile.gif' alt=':)' class='wp-smiley' />   </p>
<p>Regards,</p>
<p>Ehud<br />
Tucson, AZ USA</p>
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		<title>By: Kurt McKibben</title>
		<link>http://blog.aopa.org/helicopter/?p=107#comment-292</link>
		<dc:creator>Kurt McKibben</dc:creator>
		<pubDate>Sun, 28 Jun 2009 02:49:05 +0000</pubDate>
		<guid isPermaLink="false">http://blog.aopa.org/helicopter/?p=107#comment-292</guid>
		<description><![CDATA[I&#039;m comfortable explaining dissymmetry of lift and the correction needed by way of  blade flapping which then leads to the topic of retreating blade stall and VNE. What I find moore difficult to understand is &quot;Blade Stall.&quot; Not retreating blade stall, but &quot;blade stall.&quot; They are two completely different issues with &quot;blade stall&quot; being unrecoverable and catostrophic. Blade stall ocures from &quot;over pitching&quot; the blades. This makes sense to me at the service celing and gross weight, but I can&#039;t understand how this issue pertains a to a pilot at sea level. I got hammered on a check ride one day about  this. I understood power off blade stall, but couldn&#039;t put together a power on &quot;blade stall&quot; lesson plan. 

Any help out there???]]></description>
		<content:encoded><![CDATA[<p>I&#8217;m comfortable explaining dissymmetry of lift and the correction needed by way of  blade flapping which then leads to the topic of retreating blade stall and VNE. What I find moore difficult to understand is &#8220;Blade Stall.&#8221; Not retreating blade stall, but &#8220;blade stall.&#8221; They are two completely different issues with &#8220;blade stall&#8221; being unrecoverable and catostrophic. Blade stall ocures from &#8220;over pitching&#8221; the blades. This makes sense to me at the service celing and gross weight, but I can&#8217;t understand how this issue pertains a to a pilot at sea level. I got hammered on a check ride one day about  this. I understood power off blade stall, but couldn&#8217;t put together a power on &#8220;blade stall&#8221; lesson plan. </p>
<p>Any help out there???</p>
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		<title>By: Dan Olson</title>
		<link>http://blog.aopa.org/helicopter/?p=107#comment-290</link>
		<dc:creator>Dan Olson</dc:creator>
		<pubDate>Sat, 27 Jun 2009 04:35:47 +0000</pubDate>
		<guid isPermaLink="false">http://blog.aopa.org/helicopter/?p=107#comment-290</guid>
		<description><![CDATA[If the blades flap up on the advancing side and down on the retreating side during forward flight, how does the rotor stay horizontal?
Check the video on My Space &quot;NH-90 NATO Frigate Helicopter&quot;.  The rotor is horizontal, parallel with the horizon that shows directly behind it.  The helicopter is in forward flight at probably 40-50 knots as it approaches the ship&#039;s deck.
Dan Olson]]></description>
		<content:encoded><![CDATA[<p>If the blades flap up on the advancing side and down on the retreating side during forward flight, how does the rotor stay horizontal?<br />
Check the video on My Space &#8220;NH-90 NATO Frigate Helicopter&#8221;.  The rotor is horizontal, parallel with the horizon that shows directly behind it.  The helicopter is in forward flight at probably 40-50 knots as it approaches the ship&#8217;s deck.<br />
Dan Olson</p>
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		<title>By: Jacob Draisen</title>
		<link>http://blog.aopa.org/helicopter/?p=107#comment-270</link>
		<dc:creator>Jacob Draisen</dc:creator>
		<pubDate>Tue, 09 Jun 2009 02:33:25 +0000</pubDate>
		<guid isPermaLink="false">http://blog.aopa.org/helicopter/?p=107#comment-270</guid>
		<description><![CDATA[Tim,

As always, great article.I always look forward to your articles.]]></description>
		<content:encoded><![CDATA[<p>Tim,</p>
<p>As always, great article.I always look forward to your articles.</p>
]]></content:encoded>
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		<title>By: Adam Finn</title>
		<link>http://blog.aopa.org/helicopter/?p=107#comment-268</link>
		<dc:creator>Adam Finn</dc:creator>
		<pubDate>Mon, 08 Jun 2009 19:28:14 +0000</pubDate>
		<guid isPermaLink="false">http://blog.aopa.org/helicopter/?p=107#comment-268</guid>
		<description><![CDATA[Perfect article, just going on a tangent here...  Second paragraph &quot;When the rotor blades stay in the reference plane of rotation the pitch angle and angle-of-attack are the same&quot;. Pitch angle = A mechanical angle between the plane of rotation and the blade chord line. Angle of attack = A aerodynamic angle between resultant relative wind and the blade chord line. A blade rotating with flat pitch is creating a resultant relative wind striking the blade right on the leading edge,. A blade rotating with some pitch provided by the collective input is now creating a induced flow, or a verticle movement of air through the rotor disc, think of a fan pointed downward. This vertical movement of air makes the resultant relative wind strike the blade higher up making the angle of attack  different from the pitch angle anytime the blades are pitched.]]></description>
		<content:encoded><![CDATA[<p>Perfect article, just going on a tangent here&#8230;  Second paragraph &#8220;When the rotor blades stay in the reference plane of rotation the pitch angle and angle-of-attack are the same&#8221;. Pitch angle = A mechanical angle between the plane of rotation and the blade chord line. Angle of attack = A aerodynamic angle between resultant relative wind and the blade chord line. A blade rotating with flat pitch is creating a resultant relative wind striking the blade right on the leading edge,. A blade rotating with some pitch provided by the collective input is now creating a induced flow, or a verticle movement of air through the rotor disc, think of a fan pointed downward. This vertical movement of air makes the resultant relative wind strike the blade higher up making the angle of attack  different from the pitch angle anytime the blades are pitched.</p>
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