Single main rotor helicopter

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The discussion revolves around the modeling of lift in a helicopter simulation based on a NASA report. The user is confused about why lift is calculated as zero when the angle of attack is 90 or -90 degrees, despite the expectation that lift should be at its maximum at these angles. Responses clarify that at these angles, the airflow hits the blades broadside, resulting in drag rather than lift, as lift is defined perpendicular to the inflow. The user is currently modeling the fuselage alone and is trying to understand the reference axis for the angle of attack, which is assumed to be related to the fuselage motion. Ultimately, the conversation highlights the complexities of accurately modeling aerodynamic forces in helicopter dynamics.
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Hello everyone!
I am currently developing a thesi to become an aeronautical engineer.
I am trying to reproduce the NASA report "a mathematical model of a single main rotor helicopter for piloted simulation".
I already built a simulink model but i am having a bit of trouble looking at the equations.
I uploaded a picture of the equation's set. I can't explain the L equation highlighted.
How can the lift be 0 when the angle of attack is 90 or -90 deg? cos(90) is zero, but at that alpha the lift should be maximum.

Thank to who will answer!
 

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Lift is always defined perpendicular to the inflow. So if at cos(90), then I guess the blades are not rotating and your chopper is falling straight down. But the lift is now defined perpendicular to the inflow, which is directed in the horizontal plane, so you will not ever get any lift from this 'lift'... :) Only some from the drag component, which is directed vertical in this case.
 
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What angle does alpha represent in that equation?
 
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Marco9518 said:
I am trying to reproduce the NASA report "a mathematical model of a single main rotor helicopter for piloted simulation".
Here is a link to that report for background info for everybody:

https://ntrs.nasa.gov/citations/19830001781
 
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Marco9518 said:
How can the lift be 0 when the angle of attack is 90 or -90 deg? cos(90) is zero, but at that alpha the lift should be maximum.
Why do you say this? At that angle of attack, there will only be drag. That would be the air hitting the blade broadside. (Is this angle of attack with respect to the blade motion or to the vehicle motion?)
 
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Arjan82 said:
Lift is always defined perpendicular to the inflow. So if at cos(90), then I guess the blades are not rotating and your chopper is falling straight down. But the lift is now defined perpendicular to the inflow, which is directed in the horizontal plane, so you will not ever get any lift from this 'lift'... :) Only some from the drag component, which is directed vertical in this case.
Thank you a lot for your answer. So at the moment i am modelling the fuselage alone and i am not taking into account any other component. As soon as i figure this out i will add the rotor inflow. For what i understand it is than possible for the lift to be zero at that angle of attack.
 
FactChecker said:
Why do you say this? At that angle of attack, there will only be drag. That would be the air hitting the blade broadside. (Is this angle of attack with respect to the blade motion or to the vehicle motion?)
Thank you a lot for your answer. So at the moment i am modelling the fuselage alone and i am not taking into account any other component. Talking about the angle of attack, they did not make very clear the reference axis. I supposed it to be with respect to the fuselage motion.
 
Lnewqban said:
What angle does alpha represent in that equation?
The way i formulated it, it represents the angle between the direction of the flow and an ipotetical "chord" that goes from the nose to the tail of the fuselage.
 
Marco9518 said:
The way i formulated it, it represents the angle between the direction of the flow and an ipotetical "chord" that goes from the nose to the tail of the fuselage.
It seems to be the lift contribuited by the fuselage while the helicopter is vertically descending for landing, which should equal its drag.
In that case, I can’t understand that equation; sorry.
 
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