Longitudinal Aircraft Dynamics #8 – worksheet implementation of the dynamics equations (c)

This section continues the worksheet implementation of the dynamics formulas for aerodynamic forces and momenta.


Longitudinal Aircraft Dynamics #8- worksheet implementation of the real dynamics

by George Lungu

– This section continues with the dynamics formulas governing our 2D plane.
Worksheet implementation of the force calculation formulas:

– We will calculate these forces in a new area of the worksheet. The formulas are shown to the right of
this page. Next page contains the detailed spreadsheet formulas used. A snapshot of the force formula
area is shown below (column N contains labels):

Lift wing x  c l _ wing  q d  Chord wing Span wing  sin(alpha speed)
Lift wing y  c l _ wing  d q Chord wing Span wing  cos(alpha speed)
Lift stabilizer x  c l _ stabilizer  d q Chord stabilizer Span stabilizer  sin(alpha speed)
Lift stabilizer y  c l _ stabilizer  d q Chord stabilizer Span stabilizer  cos(alpha speed)
Drag wing x  c d _ wing  d q Chord wing Span wing  cos(alpha speed)
Drag wing y c d _ wing  d q Chord wing Span wing  sin(alpha speed)
Drag stabilizer x  c d _ stabilizer  d q Chord stabilizer Span stabilizer  cos(alpha speed)
Drag stabilizer y c d _ stabilizer  d q Chord stabilizer Span stabilizer  sin(alpha speed)
Drag fuselage x  c d _ fuselage  d q Frontal area fuselage  cos(alpha speed)
Drag fuselage y c d _ fuselage  d q Frontal area fuselage  sin(alpha speed)
Gravity x 0
Gravity y  g  m

Where q is the dynamic pressure expressed as:

q   Air_density (v  v )
d plane_ x plane_ y


These are the spreadsheet formulas used:
– Let’s start with the dynamic pressure: O66, “=B4*(D52^2+E52^2)/2”
– Lift_wing_x, O67: “=T65*O66*Chord_wing*Span_wing*SIN(RADIANS(O64))”
– Lift_wing_y, O68: “=T65*O66*Chord_wing*Span_wing*COS(RADIANS(O64))”
– Lift_stabilizer_x, O69: “=W65*O66*Chord_stabilizer*Span_stabilizer*SIN(RADIANS(O64))”
– Lift_stabilizer_y, O70: “=W65*O66*Chord_stabilizer*Span_stabilizer*COS(RADIANS(O64))”
– Drag_wing_x, O71: “=U65*O66*Chord_wing*Span_wing*COS(RADIANS(O64))”
– Drag_wing_y, O72: “=-U65*O66*Chord_wing*Span_wing*SIN(RADIANS(O64))”
– Drag_stabilizer_x, O73: “=X65*O66*Chord_stabilizer*Span_stabilizer*COS(RADIANS(O64))”
– Drag_stabilizer_y, O74: “=-X65*O66*Chord_stabilizer*Span_stabilizer*SIN(RADIANS(O64))”
– Drag_fuselage_x, O75: “=B40*O66*B39*COS(RADIANS(O64))”
– Drag_fuselage_y, O76: “=-B40*O66*B39*SIN(RADIANS(O64))”
– Gravity_y, O77: “=-Mass*9.81”

The general moment-of-force calculation formula in a 2D Cartesian reference:

– In a Cartesian system of coordinate we need the

coordinates of the moment reference (O), the force origin (P)

F and the x-y components of the force vector involved.

arm for Fx
– If we have all of them, the following formula applies (the
yP  signs in the formula are so that a momentum which produces

x a clockwise rotation has a positive sign just like in Xflr5):
Moment arm for Fy


An inventory of the momenta acting on the airplane:

– There are a total of four moment-producing forces involved (they are the lift of the wing, the drag
of the wing, the lift of the horizontal stabilizer and the drag of the horizontal stabilizer). They have a
total of two components each (x and y) and are applied about the quarter chord point of the wing
(half of them) and the quarter chord point of the stabilizer (the other half).

– There are also two additional pure momenta to add to the resultant airplane moment. They are
produced by the wing and the horizontal stabilizer and their normalized interpolated formulas have
been extracted from Xflr5 curves. Being pure momenta (torques) there are no moment arms needed in
any computation and they just require a scale-back to be used in the final moment formula.

– The resultant moment will be calculated about the center of mass or gravity (CG) of the plane. The
center of mass is considered placed on the median fuselage line. The drag of the fuselage and the
gravity produce no moment since they are aligned to the center of mass (gravity) at all times,
regardless of the airplane conditions (for the drag this is a simplifying assumption).

– We already have the formulas for the eight x-y force components (mentioned on the top of this page
and implemented in the first two pages of this section) and we will use their outputs to further
calculate the moment components they generate.

– All we need to be able to do that are the two force arms (one for the wing and one for the
stabilizer) let’s call them levers. The first lever will connect the CG with the quarter point of the wing
and the second lever will connect the CG with the quarter chord point of the horizontal stabilizer.

-In order to be able to use the Cartesian moment calculation formulas from the previous page, we
need the x and y components for the two levers, therefore we need to calculate a total of four levers
(two levers will be parallel to the x axis and to will be parallel with the y axis of the global reference).

Calculating the arms of the momenta (levers):

– Let’s estimate the components of the levers in the original “static” system of reference, then using a
rotation by the angle of the airplane find the new lever components in the final “global” system of

– We can remember that the static system of reference was centered in the center of gravity (CG) of the
plane and in this system of reference the plane was positioned horizontally.

-Let’s estimate x-y coordinates of the levers in the static system of reference (since the static angles of
attack are small, for calculating the y-components of the levers, we will still keep a very good precision if
we assume that the wing and stabilizer are parallel with the fuselage):
y (static)
Center_Gravity* Length_Fuselage/100 Static_Alpha_Stabilizer
Height_wing x (static)
0 (center of mass/gravity)
Leading_Edge_Wing* Length_Fuselage/100
Chord_wing/4 (3/4)*Chord_stabilizer

The lever components in the static x-y reference:

– The origins of the lever vectors are in the origin (coordinates [0,0]) of the system of coordinates, since
the static system is centered in the center of gravity.
– The tip of the levers are at the quarter-chord point of the wing and horizontal stabilizer respectively.
Because of this fact, the coordinates of the lever tips (which we will soon calculate) are equal to the
lengths of the levers x-y components.
– Analyzing the diagram in the previous page we can see the lever vectors as dotted brown arrows. The
quarter-chord points (where the tips of the lever vectors are) are represented as green dots on the
wing and stabilizer respectively.
– We can write the following relationships valid in the static x-y system of reference:
(Leading_edge_wing Center_Gravity) Chord_wing
Lever_wing_static_x  Length_fuselage  
100 4
Lever_wing_static_y  Height_wing
Lever_stabilizer_static_x  Length_fuselage   Chord_stabilizer
Lever_stabilizer_static_y  Height_stabilizer
– Since the levers originate in the center of gravity, in order to find the lever lengths in the global
system of reference we will only need to rotate them by “-Alpha_Plane” (the minus sign comes from the
fact that the plane faces left and a positive pitch corresponds to a negative trigonometric angle).

Calculating the lever components in the global x-y reference:

– The following general rotation formulas were derived on this blog before. If we rotate a point
(position vector tip) around origin in a x-y system of coordinates we can write the following formulas:
– Using the rotation formulas to the right, we can further
x  x 
write the final formulas for the x-y lever components in the rotated  cos() sin() initial
   
  
 
global reference system (the formulas take in consideration    
sin() cos()
y   y
 rotated   initial 
the fact that the angle is measured clockwise in our model):
 (Leading_edge_wing Center_Gravity) Chord_wing 
Lever_wing_x  Length_fuselage    cos( _ plane)  Height_win g  sin( _ plane)
 
 100 4 
 (Leading_edge_wing Center_Gravity) Chord_wing 
Lever_wing_y  Length_fuselage    sin( _ plane)  Height_win g  cos( _ plane)
 
 100 4 
 3 
Lever_stabilizer_x  Length_fuselage   Chord_stabilizer  cos( _ plane)  Height_stabilizer  sin( _ plane)
 
 4 
 3 
Lever_stabilizer_y  Length_fuselage   Chord_stabilizer  sin( _ plane)  Height_stabilizer  cos( _ plane)
 
 4 
Pay attention to the signs and the previous page
formulas. Remember that “cosine” is an even function
and “sine” is an odd function! In trigonometry the angles
are measured counterclockwise (we do the opposite).
To be continued…

Leave a Reply

Your email address will not be published.