ppragman
FLIPY FLAPS!
Maybe because of long thin and weird wings combined with things like water ballast?
%MAC and LEMAC: why?
- Thread starter killbilly
- Start date
inigo88
Composite-lover
Swept wing is the key. The reason it’s “Mean aerodynamic chord” and not “aerodynamic chord” is because for a swept wing your leading and trailing edge fuselage station (“arm” distance measured aft from the datum) vary with wing span location (aka “butt line”) so you have to take the average of each. Hence “mean”…
This explanation was pretty short and sweet:
aviation.stackexchange.com
Here’s an example of where the “average” LEMAC and TEMAC are on a swept wing like a 747:
Why this matters is all about pitching moment (Cm) and stability like Pat said earlier.
The definition of Moment (aka torque) is an applied force (in this case weight as a downward vector) times a distance (“arm”).
Your wing’s chord wise lift distribution is something complicated like the picture on the right:
But that parabolic-ish looking thing can be approximated into a single equivalent upward lift force vector (by taking the area under the curve through integration, boom calculus!) and that single upward vector (known as the “center of pressure” or CoP) usually lands around the 25% MAC station, aka the “quarter chord”.
To understand why we have to care about where the CG is relative to the quarter chord lift vector we simply have to draw a side view of the forces acting on the airplane. Unlike the standard Thrust = Drag, Lift = Weight picture where all the forces are equal and opposite about a single point, in real life the forces of lift and weight are offset from each other. Lift acts at the CoP and weight acts at the CG, like this:
In level flight lift may equal weight, but the resulting distance between the equivalent lift vector at the CoP (“quarter chord” for the most part) and the weight vector at the CG results in a pitching moment. In the picture above this is a nose up (clockwise) pitching moment, and to fly in nose level flight the horizontal stabilizer must create a lift vector of its own which when multiplied by the moment arm of the empennage creates an equal and opposite moment as the one created by the relative vectors of wing lift at the CoP and weight at the CG. This is what elevator deflection does, and what the pitch trim wheel is actually doing every time you adjust it for hands off level flight (you can’t change the arm, but you can change the lift force applied at the tail by changing the camber of the airfoil via elevator deflection). Note that horizontal stabilizer lift required can be positive or negative depending on the position of the CG relative to the CoP at the wing quarter-chord.
Finally this last part is an educated guess, but I think given the fact that the CoP hovers around 25% aerodynamic chord, it’s all well and good to use absolute fuselage station numbers on a straight rectangular wing airplane, but once you get into swept wing territory where the leading vs trailing edge position changes with span, representing it as a percentage of the average becomes more meaningful. Ultimately what you’re trying to communicate is whether the CG is within the range allotted relative to the lift vector at the average CoP, and when you sweep the wings a %MAC becomes more meaningful than a station number in inches.
Finally finally, when sizing an empennage you want your resulting Cm pitching moment curve to be negative. This means that as you increase angle of attack you get linearly less positive pitching moment and eventually it fights you with a negative force feedback on the stick (the “stick force per G” measurement we recently brought up in the Lancair thread):
In this picture either pushing or pulling on the stick should result in a reactive moment that returns the aircraft to the “trimmed” angle of attack. If you do it wrong and the Cm vs alpha curve is linear positive, that means the pitching moment will INCREASE as you pitch away from the in trim AOA and is statically unstable and a horizontal flat line is neutrally stable.
For a cool nerdy paper on how you “size” empennages on new airplanes given these stability constraints see this paper:
This explanation was pretty short and sweet:
What does %MAC mean?
A document said that the optimum CG position is from 13% MAC to 33% MAC. I'd like to know what that means, and what would this look like on an actual plane.
aviation.stackexchange.com
Here’s an example of where the “average” LEMAC and TEMAC are on a swept wing like a 747:
Why this matters is all about pitching moment (Cm) and stability like Pat said earlier.
The definition of Moment (aka torque) is an applied force (in this case weight as a downward vector) times a distance (“arm”).
Your wing’s chord wise lift distribution is something complicated like the picture on the right:
But that parabolic-ish looking thing can be approximated into a single equivalent upward lift force vector (by taking the area under the curve through integration, boom calculus!) and that single upward vector (known as the “center of pressure” or CoP) usually lands around the 25% MAC station, aka the “quarter chord”.
To understand why we have to care about where the CG is relative to the quarter chord lift vector we simply have to draw a side view of the forces acting on the airplane. Unlike the standard Thrust = Drag, Lift = Weight picture where all the forces are equal and opposite about a single point, in real life the forces of lift and weight are offset from each other. Lift acts at the CoP and weight acts at the CG, like this:
In level flight lift may equal weight, but the resulting distance between the equivalent lift vector at the CoP (“quarter chord” for the most part) and the weight vector at the CG results in a pitching moment. In the picture above this is a nose up (clockwise) pitching moment, and to fly in nose level flight the horizontal stabilizer must create a lift vector of its own which when multiplied by the moment arm of the empennage creates an equal and opposite moment as the one created by the relative vectors of wing lift at the CoP and weight at the CG. This is what elevator deflection does, and what the pitch trim wheel is actually doing every time you adjust it for hands off level flight (you can’t change the arm, but you can change the lift force applied at the tail by changing the camber of the airfoil via elevator deflection). Note that horizontal stabilizer lift required can be positive or negative depending on the position of the CG relative to the CoP at the wing quarter-chord.
Finally this last part is an educated guess, but I think given the fact that the CoP hovers around 25% aerodynamic chord, it’s all well and good to use absolute fuselage station numbers on a straight rectangular wing airplane, but once you get into swept wing territory where the leading vs trailing edge position changes with span, representing it as a percentage of the average becomes more meaningful. Ultimately what you’re trying to communicate is whether the CG is within the range allotted relative to the lift vector at the average CoP, and when you sweep the wings a %MAC becomes more meaningful than a station number in inches.
Finally finally, when sizing an empennage you want your resulting Cm pitching moment curve to be negative. This means that as you increase angle of attack you get linearly less positive pitching moment and eventually it fights you with a negative force feedback on the stick (the “stick force per G” measurement we recently brought up in the Lancair thread):
In this picture either pushing or pulling on the stick should result in a reactive moment that returns the aircraft to the “trimmed” angle of attack. If you do it wrong and the Cm vs alpha curve is linear positive, that means the pitching moment will INCREASE as you pitch away from the in trim AOA and is statically unstable and a horizontal flat line is neutrally stable.
For a cool nerdy paper on how you “size” empennages on new airplanes given these stability constraints see this paper:
guywhoflies
Y'NO WUT IM SAYIN
Swept wing is the key. The reason it’s “Mean aerodynamic chord” and not “aerodynamic chord” is because for a swept wing your leading and trailing edge fuselage station (“arm” distance measured aft from the datum) vary with wing span location (aka “butt line”) so you have to take the average of each. Hence “mean”…
This explanation was pretty short and sweet:
What does %MAC mean?
A document said that the optimum CG position is from 13% MAC to 33% MAC. I'd like to know what that means, and what would this look like on an actual plane.
Here’s an example of where the “average” LEMAC and TEMAC are on a swept wing like a 747:
View attachment 67249
Why this matters is all about pitching moment (Cm) and stability like Pat said earlier.
The definition of Moment (aka torque) is an applied force (in this case weight as a downward vector) times a distance (“arm”).
Your wing’s chord wise lift distribution is something complicated like the picture on the right:
View attachment 67250
But that parabolic-ish looking thing can be approximated into a single equivalent upward lift force vector (by taking the area under the curve through integration, boom calculus!) and that single upward vector (known as the “center of pressure” or CoP) usually lands around the 25% MAC station, aka the “quarter chord”.
To understand why we have to care about where the CG is relative to the quarter chord lift vector we simply have to draw a side view of the forces acting on the airplane. Unlike the standard Thrust = Drag, Lift = Weight picture where all the forces are equal and opposite about a single point, in real life the forces of lift and weight are offset from each other. Lift acts at the CoP and weight acts at the CG, like this:
View attachment 67251
In level flight lift may equal weight, but the resulting distance between the equivalent lift vector at the CoP (“quarter chord” for the most part) and the weight vector at the CG results in a pitching moment. In the picture above this is a nose up (clockwise) pitching moment, and to fly in nose level flight the horizontal stabilizer must create a lift vector of its own which when multiplied by the moment arm of the empennage creates an equal and opposite moment as the one created by the relative vectors of wing lift at the CoP and weight at the CG. This is what elevator deflection does, and what the pitch trim wheel is actually doing every time you adjust it for hands off level flight (you can’t change the arm, but you can change the lift force applied at the tail by changing the camber of the airfoil via elevator deflection). Note that horizontal stabilizer lift required can be positive or negative depending on the position of the CG relative to the CoP at the wing quarter-chord.
Finally this last part is an educated guess, but I think given the fact that the CoP hovers around 25% aerodynamic chord, it’s all well and good to use absolute fuselage station numbers on a straight rectangular wing airplane, but once you get into swept wing territory where the leading vs trailing edge position changes with span, representing it as a percentage of the average becomes more meaningful. Ultimately what you’re trying to communicate is whether the CG is within the range allotted relative to the lift vector at the average CoP, and when you sweep the wings a %MAC becomes more meaningful than a station number in inches.
Finally finally, when sizing an empennage you want your resulting Cm pitching moment curve to be negative. This means that as you increase angle of attack you get linearly less positive pitching moment and eventually it fights you with a negative force feedback on the stick (the “stick force per G” measurement we recently brought up in the Lancair thread):
View attachment 67252
In this picture either pushing or pulling on the stick should result in a reactive moment that returns the aircraft to the “trimmed” angle of attack. If you do it wrong and the Cm vs alpha curve is linear positive, that means the pitching moment will INCREASE as you pitch away from the in trim AOA and is statically unstable and a horizontal flat line is neutrally stable.
For a cool nerdy paper on how you “size” empennages on new airplanes given these stability constraints see this paper:
inigo88
Composite-lover
guywhoflies
Y'NO WUT IM SAYIN
In all seriousness, that was a great explanation. Thank you.
BravoHotel
Well-Known Member
Any current or former load masters in the house?
ahw01
Well-Known Member
Anyone wanna Airdrop something
RAF’s A400M Atlas airdrops record-breaking heavy cargo load
An A400M Atlas aircraft deployed with the RAF delivered a cargo load during a test conducted to evaluate the aircraft’s next-generation capabilities.
www.airforce-technology.com
CFI A&P
Exploring the world one toilet at a time.
The other aspect of having %MAC is for trim setting prior to takeoff (expecting the OEI pitching forces) is a value relative to %MAC. This is to make it simpler for us meat servos when configuring for takeoff. Whichever method you use to calculate W&B you’ll get a number for pitch trim setting. In light GA airplanes there’s just a small window or indent for take-off trim, whereas part 25 jets have such a wide W&B and using %MAC is a way to identify the trim setting needed.
Here are photos of the two most common NB airliners’ and their trim wheels, note the CG/ trim indices:
