Ground effect/Urban myth
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Well, I never. I joined this string when it had grown to page 4. I even started reading from page 1. But then I skipped to page 4 and to discover we are still on the same topic (although there were a few attempts to divert the discussion
Many of the points made about the pressure under the disc reducing the induced airflow, as seen on a vector diagram of the airflows and forces acting on a rotor blade were agreeable to me.
However, when I did my Instructor rating a few years ago, a wise old instructor once taught me:
I.F.O.I.T.A.
Does this ring any bells with anyone?
Just
DOINMABEST
Many of the points made about the pressure under the disc reducing the induced airflow, as seen on a vector diagram of the airflows and forces acting on a rotor blade were agreeable to me.
However, when I did my Instructor rating a few years ago, a wise old instructor once taught me:
I.F.O.I.T.A.
Does this ring any bells with anyone?
Just
DOINMABEST
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Going right back to basics - Since the helicopter 'weighs' the same, it takes the same amount of lift force to hold the helicopter at 1 foot above the ground as it does at 100 feet above the ground at the same atmospheric conditions. It also takes the same amount of lift to hold the helicopter at 10 feet above the ground whether you're at sea level or 10,000' above sea level.
The power required to overcome the drag of the blades and produce the lift is the thing that changes.
The power required to overcome the drag of the blades and produce the lift is the thing that changes.
TeT - if you fly a cushion creep style transition in ground effect, you will notice that the maximum power required is just before the onset of translational lift - this is because the rotors are battling their way through the tip vortices and are suffering loss of effectiveness at the tips.
The same thing happens when you transition from an OGE hover - you don't fall of a bubble of air you just need more power to accelerate the aircraft through the translational stage.
Grass and confined areas create recirculation which increases tip losses and requires more power to compensate for.
The same thing happens when you transition from an OGE hover - you don't fall of a bubble of air you just need more power to accelerate the aircraft through the translational stage.
Grass and confined areas create recirculation which increases tip losses and requires more power to compensate for.
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Shawn, building on your well laid foundation:
Since the lift is the same, at 1' and 100' , the Relative Angle of Attack that each blade section requires is the same. Since the induced velocity to produce this lift is lower near the ground ("mirror image" helicopter is generating a gentle descent flowfield) this means that the pitch goes down to maintain the same Rel AOA. So the same lift force (and hence total pressure) is generated from a lower collective position. Thus a lower induced torque, hence power, is required.
Since the lift is the same, at 1' and 100' , the Relative Angle of Attack that each blade section requires is the same. Since the induced velocity to produce this lift is lower near the ground ("mirror image" helicopter is generating a gentle descent flowfield) this means that the pitch goes down to maintain the same Rel AOA. So the same lift force (and hence total pressure) is generated from a lower collective position. Thus a lower induced torque, hence power, is required.
Last edited by Graviman; 27th May 2009 at 11:47.
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Helisphere. i don't get the time to check rotorheads so often these days..
The total pressure under the disk is what is holding you up - this is referred to in textbooks as disk actuator theory. Total pressure is the sum of static and dynamic pressure, so the rotor generates either an increase in downwash velocity (hence wake contraction) or an increase in static pressure (since the actual streamlines are revectored as the aerofoil passes). Since the weight of the heli stays the same so does the total pressure. The rotor is accelerating air down to generate the lift, which just either side of the rotor produces a step in static pressure.
In ground effect you are changing the 3D picture around heli. So for the same static pressure there can be a greater static pressure step and a smaller dynamic pressure step. In other words the air needs be accelerated less. This comes about because the ground "blockage" has the effect of superimposing a "mirror image" of heli below the ground. So heli is operating in climbing air.
The root of my arguement comes from vortex panel theory, which is still used as a pleliminary prediction for aircraft performance. The subject can, and does, still boggle my mind.
The total pressure under the disk is what is holding you up - this is referred to in textbooks as disk actuator theory. Total pressure is the sum of static and dynamic pressure, so the rotor generates either an increase in downwash velocity (hence wake contraction) or an increase in static pressure (since the actual streamlines are revectored as the aerofoil passes). Since the weight of the heli stays the same so does the total pressure. The rotor is accelerating air down to generate the lift, which just either side of the rotor produces a step in static pressure.
In ground effect you are changing the 3D picture around heli. So for the same static pressure there can be a greater static pressure step and a smaller dynamic pressure step. In other words the air needs be accelerated less. This comes about because the ground "blockage" has the effect of superimposing a "mirror image" of heli below the ground. So heli is operating in climbing air.
The root of my arguement comes from vortex panel theory, which is still used as a pleliminary prediction for aircraft performance. The subject can, and does, still boggle my mind.
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Ok, I think the confusing part is static pressure and total pressure. I've read some different things and it doesn't seem real black and white.
If we look at Bernoulli's p + q = p0
And p0 is total pressure and doesn't change, p is static pressure and q is dynamic pressure.
So IGE: q is smaller and p is bigger
OGE: q gets bigger and p gets smaller but p0 remains the same for both.
So then what is the static source measuring? I assume it's measuring p, but if this is the case then OGE/IGE will show up in the altimeter. But if the static source is measuring p0 this makes no sense because it couldn't read as high as a pitot tube vertically placed in the rotorwash.
Do you see my confusion?
If we look at Bernoulli's p + q = p0
And p0 is total pressure and doesn't change, p is static pressure and q is dynamic pressure.
So IGE: q is smaller and p is bigger
OGE: q gets bigger and p gets smaller but p0 remains the same for both.
So then what is the static source measuring? I assume it's measuring p, but if this is the case then OGE/IGE will show up in the altimeter. But if the static source is measuring p0 this makes no sense because it couldn't read as high as a pitot tube vertically placed in the rotorwash.
Do you see my confusion?
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What gets considered as static pressure or total pressure depends on the context.
For an aircraft flying through the atmosphere, then static pressure is the same as the ambient pressure, and total pressure is the sum of this static pressure plus the dynamic pressure resulting from the aircraft accelerating the surrounding air to the same velocity as the aircraft.
For an aircraft in a wind tunnel (open return) with the air traveling past it, then total pressure is the pressure of the ambient air in the room containing the wind tunnel and the static pressure is a pressure below this ambient pressure value based on how fast the wind tunnel is moving the air.
The situation gets complicated beneath a rotor because the rotor adds energy to the flow, hence increasing the total pressure.
For an aircraft flying through the atmosphere, then static pressure is the same as the ambient pressure, and total pressure is the sum of this static pressure plus the dynamic pressure resulting from the aircraft accelerating the surrounding air to the same velocity as the aircraft.
For an aircraft in a wind tunnel (open return) with the air traveling past it, then total pressure is the pressure of the ambient air in the room containing the wind tunnel and the static pressure is a pressure below this ambient pressure value based on how fast the wind tunnel is moving the air.
The situation gets complicated beneath a rotor because the rotor adds energy to the flow, hence increasing the total pressure.
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Helisphere, what you have to remember is that total pressure (static + dynamic) only stays constant along a streamline. So in an ideal world the pitot-static ports both operate on the same streamline.
In a rotor the inflow "streamlines" gets muddled up as each blade passes, so if you are only looking at the larger picture then it appears that there is a step increase in total pressure across the rotor "disk". Since engineers (like myself) are trying to produce a useable model then this is all nicely wrapped up in the name "disk actuator theory". Both above and below the rotor, along each streamline, Bernoulli applies as normal...
In a rotor the inflow "streamlines" gets muddled up as each blade passes, so if you are only looking at the larger picture then it appears that there is a step increase in total pressure across the rotor "disk". Since engineers (like myself) are trying to produce a useable model then this is all nicely wrapped up in the name "disk actuator theory". Both above and below the rotor, along each streamline, Bernoulli applies as normal...
Last edited by Graviman; 24th Feb 2011 at 11:48.
