Lift - what a drag....
Thread Starter
Joined: Sep 2006
Posts: 7
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From: Here, there & everywhere
Lift - what a drag....
Can anyone help me clarify this...
Having been asked the question "How is my lift drag ratio affected when you have 4 pax on board as apposed to just the pilot?" I set about blowing the cobwebs off my books and finding the best answer.
We have had a conflicting point of view over the crew room table.
The common factors are
Best lift drag ratio is approx 4 degrees
With 4 pax you have a higher requirement for power
Therefore you have a higher blade angle
Which equals greater induced flow...
This is where the differences begin, one manual says;
Each example has a similar angle of attack
So - they have a similar lift to drag ratio
BUT with just the pilot on board you have a better ROTOR DRAG to ROTOR THRUST ratio and more rotor efficiency.
There is a disputing argument that states;
Due to the increase in weight and the the requirement for a greater blade angle you have an increase in drag, which angles your total reaction towards a decreased lift drag ratio.
Anyone else got a spin to put on this?
Having been asked the question "How is my lift drag ratio affected when you have 4 pax on board as apposed to just the pilot?" I set about blowing the cobwebs off my books and finding the best answer.
We have had a conflicting point of view over the crew room table.
The common factors are
Best lift drag ratio is approx 4 degrees
With 4 pax you have a higher requirement for power
Therefore you have a higher blade angle
Which equals greater induced flow...
This is where the differences begin, one manual says;
Each example has a similar angle of attack
So - they have a similar lift to drag ratio
BUT with just the pilot on board you have a better ROTOR DRAG to ROTOR THRUST ratio and more rotor efficiency.
There is a disputing argument that states;
Due to the increase in weight and the the requirement for a greater blade angle you have an increase in drag, which angles your total reaction towards a decreased lift drag ratio.
Anyone else got a spin to put on this?
Avoid imitations
Joined: Nov 2000
Aviation Qualifications: ATPL
Posts: 15,110
Likes: 1,083
From: Wandering the FIR and cyberspace often at highly unsociable times
As we all know, helicopters are powered by money. The equation is
Money = lift - drag.
If no money remains after a flight, there was insufficient lift. With only the pilot on board, no money remains so there was a poor lift/drag/money ratio.
With four pax on board, a little money can sometimes remain. So the lift/drag/money ratio must have improved with 4 pax.
Money = lift - drag.
If no money remains after a flight, there was insufficient lift. With only the pilot on board, no money remains so there was a poor lift/drag/money ratio.
With four pax on board, a little money can sometimes remain. So the lift/drag/money ratio must have improved with 4 pax.
Joined: Dec 2001
Posts: 1,835
Likes: 3
From: Philadelphia PA
In all the years I've been flight testing helicopters and teaching flight testing and performance, the subject of lift to drag ratio has never come up.
It's a fixed wing thing, and may be of relevance for helicopters with wings.
It's a fixed wing thing, and may be of relevance for helicopters with wings.
Joined: Dec 2001
Posts: 1,835
Likes: 3
From: Philadelphia PA
ATN:
The thread was about L/D.
VX is a completely different matter - standby for a column in Vertical Magazine about VX and the elusive and mythical "Maximum Performance Takeoffs"
There is no VX published for helicopters because it's a variable that ranges from zero to VY, and I don't know of any way to make the calculation, let alone make it easy and repeatable. Until there is 40 Kts on the airspeed indicator, the indications are unreliable (at best).
The thread was about L/D.
VX is a completely different matter - standby for a column in Vertical Magazine about VX and the elusive and mythical "Maximum Performance Takeoffs"
There is no VX published for helicopters because it's a variable that ranges from zero to VY, and I don't know of any way to make the calculation, let alone make it easy and repeatable. Until there is 40 Kts on the airspeed indicator, the indications are unreliable (at best).
Joined: Feb 2005
Posts: 340
Likes: 1
From: KPHL
The derivation of Vx considers the component of thrust against gravity. If you can vector thrust, then Vx would require the most downwards deflection of your thrust vector. In a helicopter the most downwards deflection is fully vertical, and the obvious consequence of Vx=0 is revealed.
Of course, this assumes you have excess power at V=0.
Of course, this assumes you have excess power at V=0.
Fleet Manager
Joined: Aug 2006
Aviation Qualifications: CPL
Posts: 7,083
Likes: 2,939
From: Ontario, Canada
This is probably more a question, than an answer, so comments are welcome...
You refer to an angle of attack of 4 degrees being optimum, and the obvious requirement for more power with more load. As the power is increased, the rotor RPM is not. So, it must be that the increased power is being consumed by an increased blade angle (hence the need for more pedal).
So we agree that an increase in blade angle is required, to produce greater induced airflow, to carry a greater load, but what is the reference for that blade angle, the plane of rotation, or the angle of attack relative to the oncoming air?
Perhaps the blade angle is measured relative to the oncoming air, which is moving faster becasue of the increased power, to create thrust to oppose weight. If the oncoming air is moving faster, the blade angle could be the same relative to the oncoming air. Climbing or descending would also be a factor in this.
I'm sure that someone wiser than I can support this theory, or help me understand better too...
In the mean time, I better go back to my books...
Pilot DAR
You refer to an angle of attack of 4 degrees being optimum, and the obvious requirement for more power with more load. As the power is increased, the rotor RPM is not. So, it must be that the increased power is being consumed by an increased blade angle (hence the need for more pedal).
So we agree that an increase in blade angle is required, to produce greater induced airflow, to carry a greater load, but what is the reference for that blade angle, the plane of rotation, or the angle of attack relative to the oncoming air?
Perhaps the blade angle is measured relative to the oncoming air, which is moving faster becasue of the increased power, to create thrust to oppose weight. If the oncoming air is moving faster, the blade angle could be the same relative to the oncoming air. Climbing or descending would also be a factor in this.
I'm sure that someone wiser than I can support this theory, or help me understand better too...
In the mean time, I better go back to my books...
Pilot DAR
Joined: May 2001
Posts: 563
Likes: 1
From: queensland australia
what a totally irrelevant question and why, why, why would anyone care one iota. you must have the most boring crew room in the history of helicopters.
get a life you blokes, drink rum or something
get a life you blokes, drink rum or something
