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3D Conversion for Amateur Game DevelopersLongitudinal Aircraft
Dynamics #8– worksheet implementation of the real dynamics by
George Lungu

- This section continues with the dynamics formulas governing our 2D plane.

- 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):

Worksheet implementation of the force calculation formulas:

)( 2

mgGravity

Gravity

speedalphafuselageareaFrontalqcDrag

speedalphafuselageareaFrontalqcDrag

speedalphastabilizerSpanstabilizerChordqcDrag

speedalphastabilizerSpanstabilizerChordqcDrag

speedalphawingSpanwingChordqcDrag

speedalphawingSpanwingChordqcDrag

speedalphastabilizerSpanstabilizerChordqcLift

speedalphastabilizerSpanstabilizerChordqcLift

speedalphawingSpanwingChordqcLift

speedalphawingSpanwingChordqcLift

y

x

dfuselagedyfuselage

dfuselagedxfuselage

dstabilizerdystabilizer

dstabilizerdxstabilizer

dwingdywing

dwingdxwing

dstabilizerlystabilizer

dstabilizerlxstabilizer

dwinglywing

dwinglxwing

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”

O

yF

xF

- In a Cartesian system of coordinate we need the coordinates of the moment reference (O), the force origin (P) and the x-y components of the force vector involved.

- If we have all of them, the following formula applies (the signs in the formula are so that a momentum which produces a clockwise rotation has a positive sign just like in Xflr5):

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

yOPxOPO FxxFyyM

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).

<www.excelunusual.com> 4

- 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

reference.

- 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):

Calculating the arms of the momenta (levers):

Static_Alpha_Wing

Length_Fuselage

g e e

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:

bilizerHeight_static_yilizer_staLever_stab

- 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).

<www.excelunusual.com> 6

- Using the rotation formulas to the right, we can further write the final formulas for the x-y lever components in the global reference system (the formulas take in consideration the fact that the angle is measured clockwise in our model):

To be continued…

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

- This section continues with the dynamics formulas governing our 2D plane.

- 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):

Worksheet implementation of the force calculation formulas:

)( 2

mgGravity

Gravity

speedalphafuselageareaFrontalqcDrag

speedalphafuselageareaFrontalqcDrag

speedalphastabilizerSpanstabilizerChordqcDrag

speedalphastabilizerSpanstabilizerChordqcDrag

speedalphawingSpanwingChordqcDrag

speedalphawingSpanwingChordqcDrag

speedalphastabilizerSpanstabilizerChordqcLift

speedalphastabilizerSpanstabilizerChordqcLift

speedalphawingSpanwingChordqcLift

speedalphawingSpanwingChordqcLift

y

x

dfuselagedyfuselage

dfuselagedxfuselage

dstabilizerdystabilizer

dstabilizerdxstabilizer

dwingdywing

dwingdxwing

dstabilizerlystabilizer

dstabilizerlxstabilizer

dwinglywing

dwinglxwing

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”

O

yF

xF

- In a Cartesian system of coordinate we need the coordinates of the moment reference (O), the force origin (P) and the x-y components of the force vector involved.

- If we have all of them, the following formula applies (the signs in the formula are so that a momentum which produces a clockwise rotation has a positive sign just like in Xflr5):

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

yOPxOPO FxxFyyM

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).

<www.excelunusual.com> 4

- 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

reference.

- 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):

Calculating the arms of the momenta (levers):

Static_Alpha_Wing

Length_Fuselage

g e e

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:

bilizerHeight_static_yilizer_staLever_stab

- 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).

<www.excelunusual.com> 6

- Using the rotation formulas to the right, we can further write the final formulas for the x-y lever components in the global reference system (the formulas take in consideration the fact that the angle is measured clockwise in our model):

To be continued…

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