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## Dynamics cheat sheet

October 14, 2017 compiled on — Saturday October 14, 2017 at 07:40 AM

## 1Vibration

### 1.1Modal analysis for two degrees of freedom system

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).

#### 1.1.1Step one. Finding the equations of motion in normal coordinates space

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 semi-definite 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 semi-definite. It depends on the application being modeled.

#### 1.1.2Step 2. Solving the eigenvalue problem, finding the natural frequencies

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 non-normalized shape vectors also called the eigenvectors associated with and

#### 1.1.3Step 3. Finding the non-normalized eigenvectors

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 non-normalized shape vectors are found, the next step is to perform mass normalization

#### 1.1.4Step 4. Mass normalization of the shape vectors (or the eigenvectors)

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

#### 1.1.5Step 5, obtain the modal transformation matrix

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

#### 1.1.6Step 6. Applying modal transformation to decouple the original equations of motion

The EOM in normal coordinates is

Applying the above modal transformation on the above results in

pre-multiplying 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

#### 1.1.7Step 7. Converting modal solution to normal coordinates solution

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

#### 1.1.8Numerical solution using 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 non-normalized 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

phi =
-0.3803   -0.9249
-0.5340    0.2196

0.4380
1.8641

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.

### 1.2Fourier series representation of a periodic function

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.

### 1.3Generating Transfer functions for different vibration systems

#### 1.3.1Force transmissibility

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.

#### 1.3.2vibration isolation

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

#### 1.3.3Accelerometer

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.

#### 1.3.4Seismometer

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.

#### 1.3.5Summary of vibration transfer functions

For good isolation of mass from ground motion, rule of thumb: Make damping low, and stiffness low (soft spring).

Isolate base from force. transmitted by machine

Equation used

Transfer function

Isolate machine from motion of base

Equation used. Use absolute mass position

Transfer function

Accelerometer: Measure base acc. using relative displacement

Equation used. Use relative mass position

Transfer function

Seismometer: Measure base motion using relative displacement

Equation used. Use relative mass position

Transfer function

#### 1.4.1common definitions

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

ArcTan[1 - r^2, 2 Zeta r]]

and in Matlab use

atan2(2 Zeta r,1 - r^2)

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

and the problem with the denominator going to zero does not show up here. The amplitude when steady state response is maximum can be found as follows. The amplitude of steady state motion is . This is maximum when the magnification factor is maximum or when or is minimum. Taking derivative w.r.t. and equating the result to zero and solving for gives

We are looking for positive , hence when the under-damped 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

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.

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.

##### Summary table

 roots

 roots

 roots

 roots

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

 roots

 roots

 roots

 roots

The solution is

##### under-damped free vibration

The solution is

Applying initial conditions gives and . Therefore the solution becomes

##### critically damped free vibration

The solution is

where are found from initial conditions ,, hence

##### over-damped free vibration

The solution is

where are found from initial conditions.

where and are the roots of the characteristic equation

 roots

 roots

 roots

 roots

##### Roots of characteristic equation

The roots of the characteristic equation for are given in this table

 roots time constant (which to use? the bigger?)

##### impulse input

Undamped system with impulse

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

under-damped with impulse

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

over-damped 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

Summary table

 roots

 roots

 roots

 roots

The impulse response can be implemented in Mathematica as

##### Impulse sin function

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

under-damped with sin impulse

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)

over-damped 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 over-damped given by

where

Input is given by where

Summary table

 roots

 roots time constant

 roots

 roots

#### 1.4.6Tree view look at the different cases

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.

#### 1.4.7Cycles for the peak to decay by half its original value

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.

#### 1.4.8references

1. Vibration analysis by Robert K. Vierck
2. Structural dynamics theory and computation, 5th edition by Mario Paz, William Leigh
3. Dynamic of structures, Ray W. Clough and Joseph Penzien
4. Theory of vibration,volume 1, by A.A.Shabana
5. Notes on Diffy Qs, Differential equations for engineers, by Jiri Lebl, online PDF book, chapter 2.6, oct 1,2012 http://www.jirka.org/diffyqs/

## 2Dynamics equations, kinematics, velocity and acceleration diagrams

### 2.1Derivation of rotation formula

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.

### 2.2Miscellaneous hints

1. When finding the generalized force for the user with the Lagrangian method (the hardest step), using the virtual work method, if the force (or virtual work by the force) ADDS energy to the system, then make the sign of the force positive otherwise the sign is negative.
2. For damping force, the sign is always negative.
3. External forces such as linear forces applied, torque applied, in general, are positive.
4. Friction force is negative (in general) as friction takes energy from the system like damping.

### 2.5Velocity and acceleration of rigid body 2D

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.

### 2.6Velocity and acceleration of rigid body 3D

#### 2.6.23D Not Using vehicle dynamics notations

##### Derivation for in 3D

Let then using the rule

Then  can be found for the general case

##### Derivation for in 3D using principle axes

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!

### 2.8References

1. Structural Dynamics 5th edition. Mario Paz, William Leigh

### 2.9Misc. items

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.

## 3Astrodynamics

### 3.2Table of common equations

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)

### 3.3Flight path angle for ellipse

To find , if is given then use

If is given, then use

### 3.4Parabolic trajectory

This diagram shows the parabolic trajectory

### 3.5Hyperbolic trajectory

This diagram shows the hyperbolic trajectory

This diagram below from Orbital mechanics for Engineering students, second edition, by Howard D. Curtis, page 109

### 3.6showing that energy is constant

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.

### 3.7Earth satellite Transfer orbits

##### Hohmann transfer

This diagram shows the Hohmann transfer

### 3.8Rocket engines, Hohmann transfer, plane change at equator

two cases: Hohmann transfer, 2 burns, or semi-tangential. All burns at equator.

### 3.10interplanetary transfer orbits

##### interplanetary hohmann transfer orbit, case one

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 fly-by 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.

1. planet one standard gravitational parameter
2. planet two standard gravitational parameter
3. standard gravitational parameter for the sun km
6. original satellite altitude above planet one. For example, for LEO use 300 km
7. satellite altitude above second planet. (since goal is to send satellite for circular orbit around second planet)
8. mean distance of center of first planet from the sun. For earth use km
9. mean distance of center of second planet from the sun. For Mars use
10. sphere of influence for first planet. For earth use km
11. sphere of influence for second planet.

Given the above input, there are the steps to achieve the above maneuver

1. Find the burn out distance of the satellite
2. Find satellite speed around planet earth (relative to planet)
3. Find Hohmann ellipse
4. Find speed of satellite at perigee relative to sun
5. Find speed of earth (first planet) relative to sun
6. Find escape velocity from first planet
7. Find burn out speed at first planet by solving the energy equation for
8. Find needed at planet one
9. Find the eccentricity of the escape hyperbola
10. Find the angle with the path of planet one velocity vector
11. Find the dusk-line angle

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.

1. Find the apogee speed of the satellite
2. Find speed of second planet
3. Find entering the second planet sphere of influence
4. Find burn in radius where the satellite will be closest to the second planet.
5. Find burn out speed at second planet by solving the energy equation for
6. Find impact parameter on entry to second planet SOI
7. Find the required satellite speed around the second planet
8. Find needed at planet two
9. Find the eccentricity of the approaching hyperbola on second planet
10. Find the angle with the path of planet two velocity vector
11. Find the dusk-line angle for second planet

#### 3.10.1rendezvous orbits

##### Two satellite, walking rendezvous using Hohmann transfer

 Algorithm 1: Hohmann Walking Rendezvous Orbit, case 1
1:  function hohmann_walking_rendezvous()
2:   convert from degrees to radian
3:
4:   period of circular orbit
5:
6:   done:=false
7:   while not(done) do
8:   TOF :=
9:
10:
11:   if  then
12:
13:   else
14:   done:=true
15:   end if
16:   end while
17:
18:
19:
20:   return
21:  end function

An example implementation is below

And calling the above

gives

{7123.89, -0.0467913}
##### Two satellite, separate orbits, rendezvous using Hohmann transfer, coplaner

 Algorithm 2: Hohmann rendezvous algorithm, case 1
1:  function hohmann_rendezvous_1()
2:   convert from degrees to radian
3:   Hohmann orbit semi-major axes
4:   time of flight on Hohmann orbit
5:   required phase angle before starting Hohmann transfer
6:   angular speed of lower rad/sec
7:   angular speed of higher satellite rad/sec
8:   if  then adjust initial angle if needed
9:
10:   end if
11:   how long to wait before starting Hohmann transfer
13:   return wait_time
14:  end function

An example implementation is below (in Maple)

And calling the above for two different cases gives (times in hrs)

And

##### Two satellite, separate orbits, rendezvous using bi-elliptic transfer, coplaner

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 semi-major axes distance required. One can start with and adjust as needed.

 Algorithm 3: Hohmann rendezvous algorithm, case 2
1:  function hohmann_rendezvous_2()
2:   convert from degrees to radian
3:   Find Hohmann ideal phase angle before transfer
4:   if  then adjust for special case
5:
6:
7:   else
8:   angular speed of slower satellite in rad/sec
9:   find time of light of the slower satellite
10:
11:
12:   time of flight for the fast satellite
13:   Solve numerically for
14:   end if
16:  end function

An example implementation is below in Maple

And calling the above for two different cases gives

#### 3.10.5References

1. Orbital mechanics for Engineering students, second edition, by Howard D. Curtis
2. Orbital Mechanics, Vladimir Chobotov, second edition, AIAA
3. Fundamentals of Astrodynamics, Bates, Muller and White. Dover1971

## 4Dynamic of flights

### 4.1Wing geometry

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 stick-fixed static margin Must be positive for static stability

### 4.2Summary of main equations

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

##### Writing the equations in linear form

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 wing-body 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

#### 4.2.1definitions

1. Remember that for symmetric airfoil, when the chord is parallel to velocity vector, then the angle of attack is zero, and also the left coefficient is zero. But this is only for symmetric airfoil. For the common campbell airfoil shape, when the chord is parallel to the velocity vector, which means the angle of attack is zero, there will still be lift (small lift, but it is there). What this means, is that the chord line has to tilt down more to get zero lift. This extra tilting down makes the angle of attack negative. If we now draw a line from the right edge of the airfoil parallel to the velocity vector, this line is called the zero lift line (ZLL) see diagram below.
Just remember, that angle of attack (which is always the angle between the chord and the velocity vector, the book below calls it the geometrical angle of attack) is negative for zero lift. This is when the airfoil is not symmetric. For symmetric airfoil, ZLL and the chord line are the same. This angle is small, or so. Depending on shape. See Foundations of Aerodynamics, 5th ed, by Chow and Kuethe, here is the diagram.

2. stall from http://en.wikipedia.org/wiki/Stall_(flight)
In fluid dynamics, a stall is a reduction in the lift coefficient
generated by a foil as angle of attack increases.[1] This occurs when
the critical angle of attack of the foil is exceeded. The critical
angle of attack is typically about 15 degrees, but it
may vary significantly depending on the fluid, foil, and Reynolds number.
3. Aerodynamics in road vehicle wiki page
4. some demos relating to airplane control http://demonstrations.wolfram.com/ControllingAirplaneFlight/
5. http://www.americanflyers.net/aviationlibrary/pilots_handbook/chapter_3.htm
6. Lectures Helicopter Aerodynamics and Dynamics by Prof. C. Venkatesan, Department of Aerospace Engineering, IIT Kanpur http://www.youtube.com/watch?v=DKWj2WzYXtQ&list=PLAE677E56C97A7C7D
7. http://avstop.com/ac/apgeneral/terminology.html has easy to understand definitions airplane geometry. ”The MAC is the mean average chord of the wing”
8. http://www.tdmsoftware.com/afd/afd.html airfoil design software

### 4.3images and plots collected

These are diagrams and images collected from different places. References is given next to each image.

Images from http://adamone.rchomepage.com/cg_calc.htm and Flight dynamics principles by Cook, 1997.

From http://www.solar-city.net/2010/06/airplane-control-surfaces.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 FAA pilot handbook and http://www.youtube.com/watch?v=8uT55aei1NI

### 4.4Some strange shaped airplanes

Image from http://www.aerospaceweb.org/question/aerodynamics/q0130.shtml ”Boeing Pelican ground effect vehicle”