Detailed steps to perform modal analysis are given below for a standard undamped two degrees of freedom system. The main advantage of solving a multidegree system using modal analysis is that it decouples the equations of motion (assuming they are coupled) making solving them much simpler.
In addition it shows the fundamental shapes that the system can vibrate in, which gives more insight into the system. Starting with standard 2 degrees of freedom system
In the above the generalized coordinates are and . Hence the system requires two equations of motion (EOM’s).
The two EOM’s are found using any method such as Newton’s method or Lagrangian method. Using Newton’s method, free body diagram is made of each mass and then is written for each mass resulting in the equations of motion. In the following it is assumed that both masses are moving in the positive direction and that is larger than when these equations of equilibrium are written
Hence, from the above the equations of motion are
or
In Matrix form
The above two EOM are coupled in stiffness, but not mass coupled. Using short notations, the above is written as
Modal analysis now starts with the goal to decouple the EOM and obtain the fundamental shape functions that the system can vibrate in. To make these derivations more general, the mass matrix and the stiffness matrix are written in general notations as follows
The mass matrix and the stiffness matrix must always come out to be symmetric. If they are not symmetric, then a mistake was made in obtaining them. As a general rule, the mass matrix is PSD (positive definite matrix) and the matrix is positive semidefinite matrix. The reason the is PSD is that represents the kinetic energy of the system, which is typically positive and not zero. But reading some other references ^{1} it is possible that can be positive semidefinite. It depends on the application being modeled.
The first step in modal analysis is to solve the eigenvalue problem in order to determine the natural frequencies of the system. This equations leads to a polynomial in and the roots of this polynomial are the natural frequencies of the system. Since there are two degrees of freedom, there will be two natural frequencies for the system.
The above is a polynomial in . Let it becomes
This quadratic polynomial in which is now solved using the quadratic formula. Then the positive square root of each root to obtain and which are the roots of the original eigenvalue problem. Assuming from now that these roots are and the next step is to obtain the nonnormalized shape vectors also called the eigenvectors associated with and
For each natural frequency and the corresponding shape function is found by solving the following two sets of equations for the vectors
and
For , let and solve for
Which gives one equation now to solve for (the first row equation is only used)
Hence
Therefore the first shape vector is
Similarly the second shape function is obtained. For , let and solve for
Which gives one equation now to solve for (the first row equation is only used)
Hence
Therefore the second shape vector is
Now that the two nonnormalized shape vectors are found, the next step is to perform mass normalization
Let
This results in a scalar value , which is later used to normalize . Similarly
For example, to find
Similarly, is found
Now that are obtained, the mass normalized shape vectors are found. They are called
Similarly
The modal transformation matrix is the matrix made of of in each of its columns
Now the is found, the transformation from the normal coordinates to modal coordinates, which is called is found
The transformation from modal coordinates back to normal coordinates is
However, therefore
The next step is to apply this transformation to the original equations of motion in order to decouple them
The EOM in normal coordinates is
Applying the above modal transformation on the above results in
premultiplying by results in
The result of will always be . This is because mass normalized shape vectors are used. If the shape functions were not mass normalized, then the diagonal values will not be as shown.
The result of will be .
Let the result of be Therefore, in modal coordinates the original EOM becomes
The EOM are now decouples and each can be solved as follows
To solve these EOM’s, the initial conditions in normal coordinates must be transformed to modal coordinates using the above transformation rules
Or in full form
and
Each of these EOM are solved using any of the standard methods. This will result is solutions and
The solutions found above are in modal coordinates . The solution needed is . Therefore, the transformation is now applied to convert the solution to normal coordinates
Hence
and
Notice that the solution in normal coordinates is a linear combination of the modal solutions. The terms are just scaling factors that represent the contribution of each modal solution to the final solution. This completes modal analysis
This is a numerical example that implements the above steps using a numerical values for and . Let and let and . Let initial conditions be , hence
and
In normal coordinates, the EOM are
In this example and and and
step 2 is now applied which solves the eigenvalue problem in order to find the two natural frequencies
Let hence
The solution is and , therefore
And
step 3 is now applied which finds the nonnormalized eigenvectors. For each natural frequency and the corresponding shape function is found by solving the following two sets of equations for the eigen vectors
For
This gives one equation to solve for (the first row equation is only used)
Hence
The first eigen vector is
Similarly for
This gives one equation to solve for (the first row equation is only used)
Hence
The second eigen vector is
Now step 4 is applied, which is mass normalization of the shape vectors (or the eigenvectors)
Hence
Similarly, is found
Hence
Now that are found, the mass normalized eigen vectors are found. They are called
Similarly
Therefore, the modal transformation matrix is
This result can be verified using Matlab’s eig function as follows
Matlab result agrees with the result obtained above. The sign difference is not important.
Now step 5 is applied. Matlab generates mass normalized eigenvectors by default.
Now that is found, the transformation from the normal coordinates to modal coordinates, called is obtained
The transformation from modal coordinates back to normal coordinates is
However, therefore
The next step is to apply this transformation to the original equations of motion in order to decouple them.
Applying step 6 results in
The EOM are now decoupled and each EOM can be solved easily as follows
To solve these EOM’s, the initial conditions in normal coordinates must be transformed to modal coordinates using the above transformation rules
and
Each of these EOM are solved using any of the standard methods. This results in solutions and Hence the following EOM’s are solved
and also
The solutions , are found using basic methods shown in other parts of these notes. The last step is to transform back to normal coordinates by applying step 7
Hence
and
The above shows that the solution and has contributions from both nodal solutions.
Given a periodic function with period then its Fourier series approximation using terms is
Where
Another way to write the above is to use the classical representation using and . The same coefficients (i.e. the same series) will result.
Just watch out in the above, that we divide by the full period when finding and divide by half the period for all the other coefficients. In the end, when we find we can convert that to complex form. The complex form seems easier to use.
Let steady state
Then
Hence
So TR or force transmissibility is
If then we want small to reduce force transmitted to base. For , it is the other way round.
We need transfer function between and Equation of motion
Let and let , hence the above becomes
Hence and where .
Hence for good vibration isolation we need to be small. i.e. to be small. This is the same TR as for force isolation above.
For small , we need small and small (the small is to make ) see plot
In Matlab, the above can be plotted using
We need transfer function between and where now is the amplitude of the ground acceleration. This device is used to measure base acceleration by relating it linearly to relative displacement of to base.
Equation of motion. We use relative distance now.
Let Notice we here jumped right away to the itself and wrote it as and we did not go through the steps as above starting from base motion. This is because we want the transfer function between relative motion and acceleration of base.
Now, , hence the above becomes
Hence and
When system is very stiff, which means very large compared to , then , hence by measuring we estimate the amplitude of the ground acceleration since is known. For accuracy, need at least.
Now we need to measure the base motion (not base acceleration like above). But we still use the relative displacement. Now the transfer function is between and where now is the base motion amplitude.
Equation of motion. We use relative distance now.
Let ,and let , hence the above becomes
Now, , hence the above becomes
Hence and
Now if is very large, which happens when , then since is the dominant factor. Therefore now becomes therefore measuring the relative displacement gives linear estimate of the ground motion. However, this device requires that be much smaller than , which means that has to be massive. So this device is heavy compared to accelerometer.
For good isolation of mass from ground motion, rule of thumb: Make damping low, and stiffness low (soft spring).
Equation used
Transfer function
Equation used. Use absolute mass position
Transfer function
Equation used. Use relative mass position
Transfer function
Equation used. Use relative mass position
Transfer function
These definitions are used throughout the derivations below.
Since there is no damping in the system, then there is no steady state solution. In other words, the particular solution is not the same as the steady state solution in this case. We need to find the particular solution using method on undetermined coefficients.
Let By guessing that then we find the solution to be
Applying initial conditions is always done on the full solution. Applying initial conditions gives
Where
The complete solution is
 (1) 
Example: Given force then rad/sec, and . Let , then rad/sec. Hence , Let initial conditions be zero, then
Resonance forced vibration When we obtain resonance since in the solution given in Eq (1) above and as written the solution can not be used for analysis. To obtain a solution for resonance some calculus is needed. Eq (1) is written as
 (1A) 
When but less than , letting
 (2) 
where is very small positive quantity. And since let
 (3) 
Multiplying Eq (2) and (3) gives
 (4) 
Eq (1A) can now be written in terms of Eqs (2,3) as
Since the above becomes
Using the above becomes
From Eqs (2,3) the above can be written as
Since the above becomes
This is the solution to use for resonance.
The solution is
where
and
where
Very important note here in the calculations of above, one should be careful on the sign of the denominator. When the forcing frequency the denominator will become negative (the case of is resonance and is handled separately). Therefore, one should use that takes care of which quadrant the angle is. For example, in Mathematica use
and in Matlab use
Otherwise, wrong solution will result when The full solution is
 (1) 
Applying initial conditions gives
Another form of these equations is given as follows
Hence the full solution is
 (1.1) 
Applying initial conditions now gives
The above 2 sets of equations are equivalent. One uses the phase angle explicitly and the second ones do not. Also, the above assume the force is and not . If the force is then in Eq 1.1 above, the term reverse places as in
Applying initial conditions now gives
When a system is damped, the problem with the divide by zero when does not occur here as was the case with undamped system, since when when or , the solution in Eq (1) becomes
We are looking for positive , hence when the underdamped response is maximum.
The solution is
Where and where (making sure to use correct definition). Hence
where are found from initial conditions
The solution is
where
and
hence
where and
Hence the solution is
Let load be harmonic and represented in general as where is the complex amplitude of the force.
Hence system is represented by
Let Hence , therefore the differential equation becomes
Dividing numerator and denominator gives
Where hence the response is
Therefore, the phase of the response is
Hence at the phase of the response will be
So when is real, the phase of the response is simply
Undamped case
When the above becomes
For real force this becomes
The magnitude and phase zero.
damped cases
Hence for real force and at the phase of displacement is
lag behind the load.
When then goes from to Therefore phase of displacement is to behind force. The minus sign at the front was added since the complex number is in the denominator. Hence the response will always be lagging in phase relative for load.
For
Now is negative, hence the phase will be from to
When
Now phase is
Examples. System has and subjected for force find the steady state solution.
Answer , rad/sec, hence under the response is
The equation of motion can also be written as .
The following table gives the solutions for initial conditions are and under all damping conditions. The roots shown are the roots of the quadratic characteristic equation . Special handling is needed to obtain the solution of the differential equation for the case of and as described in the detailed section below.








Where and , the solution is
Applying initial conditions gives
And complete solution is
The general solution is
From initial conditions
Hence the solution is
The general solution is
Where from initial conditions
The solution is
Where now
Hence the solution is
Where








The solution is
The solution is
Applying initial conditions gives and . Therefore the solution becomes
The solution is
where are found from initial conditions ,, hence
The solution is
where are found from initial conditions.
where and are the roots of the characteristic equation








The roots of the characteristic equation for are given in this table
roots  time constant  
(which to use? the bigger?)  
with initial conditions and Assuming the impulse acts for a very short time period from to seconds, where is small amount. Integrating the above differential equation gives
Since is very small, it can be assumed that changes is negligible, hence the above reduces to
since we assumed and since then the above reduces to
Therefore, the effect of the impulse is the same as if the system was a free system but with initial velocity given by and zero initial position. Hence the system is now solved as follows
With and . The solution is
If the initial conditions were not zero, then the solution for these are added to the above. From earlier, it was found that the solution is , therefore, the full solution is
with initial conditions and Integrating gives
Since is very small, it can be assumed that changes is negligible as well as the change in velocity, hence the above reduces to the same result as in the case of undamped. Therefore, the system is solved as free system, but with initial velocity and zero initial position.
Initial conditions are and then the solution is
applying initial conditions gives and , hence
If the initial conditions were not zero, then the solution for these are added to the above. From earlier, it was found that the solution is , therefore, the full solution is
critically damped with impulse input with initial conditions and then the solution is
where are found from initial conditions and , hence the solution is
If the initial conditions were not zero, then the solution for these are added to the above. From earlier, it was found that the solution is , therefore, the full solution is
overdamped with impulse input With initial conditions are and the solution is
where are found from initial conditions and
Hence the solution is
where
Hence
If the initial conditions were not zero, then the solution for these are added to the above. From earlier, it was found that the solution is , therefore, the full solution is








The impulse response can be implemented in Mathematica as
Now assume the input is as follows
given by where
undamped system with sin impulse
with and For the solution is
where where is the natural period of the system. , hence the above becomes
 (1) 
When and then
The above Eq (1) gives solution during the time
Now after the force will disappear, the differential equation becomes
but with the initial conditions evaluate at From (1)
since . taking derivative of Eq (1)
and at the above becomes
since Now (2) and (3) are used as initial conditions to solve . The solution for is
1.4.5.0.0.0 Resonance with undamped sin impulse When and we obtain resonance since in the solution shown up and as written the solution can’t be used for analysis in this case. To obtain a solution for resonance some calculus is needed. Eq (1) is written as
Now looking at case when but less than , hence let
 (2) 
where is very small positive quantity. and we also have
 (3) 
Multiplying Eq (2) and (3) with each others gives
 (4) 
Going back to Eq (1A) and rewriting it as
Since the above becomes
now using the above becomes
From Eq(2) and hence the above becomes
or since
Now hence the above becomes
This can also be written as
since in this case. This is the solution to use for resonance and for
Hence for , the above equations is used to determine initial conditions at
but and and , hence the above becomes
Taking derivative of Eq (1) gives
and at
Now the solution for is
or
For Initial conditions are and and then the solution from above is
 (1) 
Applying initial conditions gives
For . From (1)
 (2) 
Taking derivative of (1) gives
at
(3) 
Now for the equation becomes
which has the solution
where and
critically damped with sin impulse For Initial conditions are and then the solution is from above
 (1) 
Where are found from initial conditions
For the solution is
 (2) 
To find from Eq(1)
taking derivative of (1) gives
 (3) 
at
 (4) 
Hence Eq (2) can now be evaluated using Eq(3,4)
overdamped with sin impulse For Initial conditions are and then the solution is
where (make sure you use correct quadrant, see not above on ) and
and
leading to the solution where and
is
For . From Eq(1) and at
 (2) 
Taking derivative of Eq (1)
At
 (3) 
Equation of motion now is
which has solution for overdamped given by
where
Input is given by where
This tree illustrates the different cases that needs to be considered for the solution of single degree of freedom system with harmonic loading.
There are cases to consider. Resonance needs to be handled as special case when damping is absent due to the singularity in the standard solution when the forcing frequency is the same as the natural frequency. When damping is present, there is no resonance, however, there is what is called practical response which occur when the forcing frequency is almost the same as the natural frequency.
The following is another diagram made sometime ago which contains more useful information and is kept here for reference.
This table shows many cycles it takes for the peak to decay by half its original value as a function of the damping . For example, we see that when then it takes cycles for the peak (i.e. displacement) to reduce to half its value.
This formula is very important. Will show its derivation now in details. It is how to express vectors in rotating frames.
Consider this diagram
In the above, the small axis is a frame attached to some body which rotate around this axis with angular velocity (measured by the inertial frame of course). All laws derived below are based on the following one rule
 (1) 
Lets us see how to apply this rule. Let us express the position vector of the particle . We can see by normal vector additions that the position vector of particle is
 (2) 
Notice that nothing special is needed here, since we have not yet looked at rate of change with time. The complexity (i.e. using rule (1)) appears only when we want to look at velocities and accelerations. This is when we need to use the above rule (1). Let us now find the velocity of the particle. From above
Every time we take derivatives, we stop and look. For any vector that originates from the moving frame, we must apply rule (1) to it. That is all. In the above, only needs rule (1) applied to it, since that is the only vector measure from the moving frame. Replacing by and by , meaning the velocity of and , Hence the above becomes
and now we apply rule (1) to expand
 (3) 
where is just
The above is the final expression for the velocity of the particle using its velocity as measured by the moving frame in order to complete the expression.
So the above says that the absolute velocity of the particle is equal to the absolute velocity of the base of the moving frame + something else and this something else was
Now we will find the absolute acceleration of . Taking time derivatives of (3) gives
 (4) 
As we said above, each time we take time derivatives, we stop and look for vectors which are based on the moving frame, and apply rule (1) to them. In the above, and qualify. Apply rule (1) to gives
 (5) 
where just means the acceleration relative to moving frame. And applying rule (1) to gives
 (6) 
Replacing (5) and (6) into (4) gives
Eq (7) says that the absolute acceleration of is the sum of the acceleration of the base of the moving frame plus the relative acceleration of the particle to the moving frame plus
Hence, using Eq(3) and Eq(7) gives us the expressions we wanted for velocity and acceleration.
Finding linear acceleration of center of mass of a rigid body under pure rotation using fixed body coordinates.
In the above is the speed of the center of mass in the direction of the axis, where this axis is fixed on the body itself. Similarly, is the speed of the center of mass in the direction of the axis, where the axis is attached to the body itself.
Just remember that all these speeds (i.e. ,) and accelerations (, ) are still being measured by an observer in the inertial frame. It is only that the directions of the velocity components of the center of mass is along an axis fixed on the body. Only the direction. But actual speed measurements are still done by a stationary observer. Since clearly if the observer was sitting on the body itself, then they will measure the speeds to be zero in that case.
Let then using the rule
Then can be found for the general case
The above derivation simplifies now since we will be using principle axes. In this case, all cross products of moments of inertia vanish.
Hence
So, we can see how much simpler it became when using principle axes. Compare the above to
So, always use principle axes for the body fixed coordinates system!
The Jacobian matrix for a system of differential equations, such as
is given by
For example, for the given the following 3 set of coupled differential equations in
then the Jacobian matrix is
Now to find stability of this system, we evaluate this matrix at where is a point in this space (may be stable point or initial conditions, etc...) and then become all numerical now. Then we can evaluate the eigenvalues of the resulting matrix and look to see if all eigenvalues are negative. If so, this tells us that the point is a stable point. I.e. the system is stable.
If X is distributed then is distributed.
The following table contains the common relations to use for elliptic motion. Equation of ellipse is
term to find 
relation 
conversion between and 

position of satellite at time Solve for , then find . here is time at and is mean satellite speed. 

eccentricity 

Maor axes 

Minor axes 







specific angular momentum 

Total Energy 

velocity 

(closest) 

(furthest) 

magnitude of 

period 

mean satellite speed 

eccentric anomaly 

area sweep rate 

equation of motion 

spherical coordinates relation 
where is the inclination and is the azimuth and is latitude ^{1} 





Notice in the above, that the period of satellite depends only on (for same )
In the above, where is the mass of the body at the focus of the ellipse and is the gravitational constant. is the specific mass angular momentum (moment of linear momentum) of the satellite. Hence the units of is length.
To draw the locus of the satellite (the small body moving around the ellipse, all what we need is the eccentricity and , the major axes length. Then by changing the angle the path of the satellite is drawn. I have a demo on this here
See http://nssdc.gsfc.nasa.gov/planetary/factsheet/earthfact.html for earth facts
This table below is from my class EMA 550 handouts (astrodynamics, spring 2014)
To find , if is given then use
If is given, then use
This diagram shows the parabolic trajectory
This diagram shows the hyperbolic trajectory
This diagram below from Orbital mechanics for Engineering students, second edition, by Howard D. Curtis, page 109
Showing that energy is constant.
Most of such relations starts from the same place. The equation of motion of satellite under the assumption that its mass is much smaller than the mass of the large body (say earth) it is rotating around. Hence we can use and the equation of motion reduces to
In the above equation, the vector is the relative vector from the center of the earth to the center of the satellite. The reason the center of earth is used as the origin of the inertial frame of reference is due to the assumption that where is the mass of earth (or the body at the focal of the ellipse) and is the mass of the satellite. Hence the median center of mass between the earth and the satellite is taken to be the center of earth. This is an approximation, but a very good approximation.
The first step is to dot product the above equation with giving
And there is the main trick. We look ahead and see that but and we also see that but Hence equation 1 above can be written as
Hence
Where is a constant, which is the total energy of the satellite.
This diagram shows the Hohmann transfer
two cases: Hohmann transfer, 2 burns, or semitangential. All burns at equator.
The following are the steps to accomplish the above. The first stage is getting into the Hohmann orbit from planet 1, then reaching the sphere of influence of the second planet. Then we either do a flyby or do a parking orbit around the second planet. These steps below show how to reach the second planet and do a parking orbit around it.
The input is the following.
Given the above input, there are the steps to achieve the above maneuver
The above completes the first stage, now the satellite is in the Hohmann transfer orbit. Assuming it reached the orbit of the second planet ahead of it as shown in the diagram above. Now we start the second stage to land the satellite on a parking orbit around the second planet at altitude above the surface of the second planet. These are the steps needed.
An example implementation is below
And calling the above
gives
An example implementation is below (in Maple)
And calling the above for two different cases gives (times in hrs)
And
In this transfer, the lower (fast satellite) does not have to wait for phase lock as in the case with Hohmann transfer. The transfer can starts immediately. There is a free parameter that one select depending on fuel cost requiments or any limitiation on the first transfer orbit semimajor axes distance required. One can start with and adjust as needed.
An example implementation is below in Maple
And calling the above for two different cases gives
below is the core chord of the wing.
This is a diagram to use to generate equations of longitudinal equilibrium.
This distance is called the stickfixed static margin Must be positive for static stability
This table contain some definitions and equations that can be useful.
# 
equation 
meaning/use 
1 

is lift coefficient. is angle of attack. is slope which is the same as 
2 

wing lift coefficient 
3 

drag coefficient 
4 

pitching moment coefficient due to wing only about the C.G. of the airplane assuming small This is simplified more by assuming and 
5 

simplified wing Pitching moment 
6 

simplified pitching moment coefficient due to wing and body about the C.G. of the airplane. is the angle of attack 
7 

is the lift coefficient generated by tail. is the tail area. is airplane air speed 
8 

total lift of airplane. is lift due to body and wing and is lift due to tail 
9 

coefficient of total lift of airplane. is coefficient of lift due to wing and body. is lift coefficient due to tail. is the total wing area. is tail area 
10 

pitching moment due to tail about C.G. of airplane 
11 

pitching moment coefficient due to tail. is called tail volume 
12 

introducing bar tail volume which is but uses instead of . Important note. depends on location of C.G., but does not. 
13 

pitching moment coefficient due to tail expressed using . This is the one to use. 
14 

pitching moment coefficient due to propulsion about airplane C.G. 
15 

total airplane pitching moment coefficient about airplane C.G. 
16 

simplified total Pitching moment coefficient about airplane C.G. 
17 

derivative of total pitching moment coefficient w.r.t airplane angle of attack 
18 

location of airplane neutral point of airplane found by setting in the above equation 
19 

rewrite of in terms of . Derived using the above two equations. 
20 

static margin. Must be Positive for static stability 





The following equations are derived from the above set of equation using what is called the linear form. The main point is to bring into the equations the expression for written in term of This is done by expressing the tail angle of attack in terms of via the downwash angle and the angle. in the above equations are replaced by and is replaced by . This replacement says that it is a linear relation between and the corresponding angle of attack. The main of this rewrite is to obtain an expression for in terms of where is expressed in terms of , hence do not show explicitly. The linear form of the equations is what from now on.
# 
equation 
meaning/use 
1 

is constant, represents and is propulsion pitching moment coeff. at zero angle of attack 
2 

main relation that associates with . is the wingbody angle of attack, is downwash angle at tail, and is tail angle with horizontal reference (see diagram) 
3 

Lift due to tail expressed using and (notice that do not show explicitly) 
4 

defined for use with overall lift coefficient 
5 

overall airplane lift using linear relations 
6 

overall angle of attack as function of the wing and body angle of attack and tail angles 
7 

overall airplane pitch moment. Two versions one uses and one uses 
8 

Two versions of one for and one one uses 
9 

is total pitching moment coef. at zero lift (does not depend on C.G. location) but is total pitching moment coef. at (not at zero lift). This depends on location of C.G. 
10 


11 

Used to determine 




These are diagrams and images collected from different places. References is given next to each image.
This below from http://www.grc.nasa.gov/WWW/k12/UEET/StudentSite/dynamicsofflight.html
http://www.grc.nasa.gov/WWW/k12/airplane/alr.html
From http://en.wikipedia.org/wiki/Lift_coefficient and http://en.wikipedia.org/wiki/File:Aeroforces.svg
from http://adg.stanford.edu/aa241/drag/sweepncdc.html
Images from http://adamone.rchomepage.com/cg_calc.htm and Flight dynamics principles by Cook, 1997.
From http://chrusion.com/BJ7/SuperCalc7.html
From http://www.willingtons.com/aircraft_center_of_gravity_calcu.html
From http://www.solarcity.net/2010/06/airplanecontrolsurfaces.html nice diagram that shows clearly how the elevator causes the pitching motion (nose up/down). From same page, it says ”The purpose of the flaps is to generate more lift at slower airspeed, which enables the airplane to fly at a greatly reduced speed with a lower risk of stalling.”
Images from flight dynamics principles, by Cook, 1997.
Images from Performance, stability, dynamics and control of Airplanes. By Pamadi, AIAA press. Page 169. and http://www.americanflyers.net/aviationlibrary/pilots_handbook/chapter_3.htm
Image from http://www.americanflyers.net/aviationlibrary/pilots_handbook/chapter_3.htm
Image from http://www.americanflyers.net/aviationlibrary/pilots_handbook/chapter_3.htm
Image from FAA pilot handbook and http://www.youtube.com/watch?v=8uT55aei1NI
Image http://www.youtube.com/watch?v=8uT55aei1NI and http://www.youtube.com/user/DAMSQAZ?feature=watch
Image http://edition.cnn.com/2014/01/16/travel/insideairbusbeluga/index.html?hpt=ibu_c2
Image from http://edition.cnn.com/2014/01/16/travel/insideairbusbeluga/index.html?hpt=ibu_c2
Image from http://www.nasa.gov/centers/dryden/Features/super_guppy.html
Image from http://www.aerospaceweb.org/question/aerodynamics/q0130.shtml ”Boeing Pelican ground effect vehicle”