User:Egm6341.s11.team2/hwk6

=Problem 6.3: ADD TITLE HERE!!!!!!!!!!!!=

Solution
=Problem 6.4: ADD TITLE HERE!!!!!!!!!!!!=

Solution
=Problem 6.5: ADD TITLE HERE!!!!!!!!!!!!=

Solution
=Problem 6.6 - Understanding Kessler's Code=

Given
Kessler's code, given in his paper :

Find
In order to better understand Kessler's code and the HOTRE, improve upon the work of s10 and explain the code line by line. Then, run Kessler's code for $$i=1,2,3$$ to reproduce his results. The collective work of s10 is referenced here under HW5, Problem 10.

Solution
The following box presents Kessler's code commented line-by-line to describe the algorithm for using the HOTRE.

Running Kessler's code with n = 3 results in the following output:

These with the additional last line written into the code, the output matches the tables presented in Kessler's paper:

=Problem 6.7: ADD TITLE HERE!!!!!!!!!!!!=

Solution


=Problem 6.8: Derive Arc Length Equation=

''' This problem was solved without referring to S10. '''

From the lecture slide Mtg 33-3

Given
The formula for arc length from lecture 32-4:


 * {| style="width:100%" border="0"

$$\displaystyle
 * style="width:95%" |
 * style="width:95%" |

\widehat{PQ} = \int_{\theta_{P}}^{\theta_{Q}}d\theta\left[r^2+\left(\frac{dr}{d\theta}\right)^{2}\right]^{\frac{1}{2}}. $$     (8.1)
 * 
 * }

The law of cosines:

Triangle OAB:


 * HWK6 P8.jpg

where

Find
Derive the formula for arc length of ellipse using triangle OAB and the law of cosines.

Solution
Applying the law of cosines formula to the OAB triangle case gives:

Substituting in Eqs 8.3-5 into Eq 8.6 gives

Taylor series expansion of $$\displaystyle \cos (d\theta) $$ is

Because theta is assumed to be small, we only need to take the first two terms of the Taylor series expansion of $$\displaystyle \cos (d\theta) $$ and sub into Eq 8.8.

The last term in Eq 8.11 $$\displaystyle rdrd\theta^2$$ is zero because the three derivatives shrink faster than the two derivative elements $$\displaystyle dr^2$$ and $$\displaystyle d\theta^2$$.

The remainder is

Taking the square root of both sides of Eq 8.12 then integrating from $$\displaystyle\theta_P$$ to $$\displaystyle\theta_Q$$ gives,

= Problem 6.9: Identification of Basis Functions $$\displaystyle \overline{N_i}$$ =

''' This problem was solved without referring to S10. '''

From the lecture slide Mtg 37-3

Given
The relationship between $$\displaystyle Z(s)$$ and $$\displaystyle d_i$$(degree of freedom):


 * {| style="width:100%" border="0"

$$\displaystyle
 * style="width:95%" |
 * style="width:95%" |

\begin{align} Z(s)&=\sum_{i=0}^{3}c_i s^i \\ &= \sum_{i=1}^{4} \overline{N_i}(s) \, d_i\\ \end{align}

$$     (9.1-2)
 * 
 * }

where,


 * {| style="width:100%" border="0"

$$\displaystyle
 * style="width:95%" |
 * style="width:95%" |

\begin{align} \overline{N_i}(s) &= basis\,\, function\\ d_i(s) &= degrees\,\, of\,\, freedom\\ d_1&=Z_i \\ d_2&=\dot{Z_i} \\ d_3&=Z_{i+1} \\ d_4&=\dot{Z}_{i+1}\\ \end{align}

$$
 * 
 * }

and
 * {| style="width:100%" border="0"

$$\displaystyle
 * style="width:95%" |
 * style="width:95%" |

\dot{Z}=\frac{dZ}{dt} = \frac{dZ}{ds} \frac{ds}{dt}

$$    (9.3)
 * 
 * }

Find
Identify and plot the the basis function $$\displaystyle \left\{\overline{N_i}(s) \right\}\,\, ,i=1,2,3,4 $$.

Solution
From the Eq 1 of 35-4(lecture slide)


 * {| style="width:100%" border="0"

$$\displaystyle
 * style="width:95%" |
 * style="width:95%" |

\begin{align} \begin{Bmatrix} c_0 \\ c _1 \\ c _2 \\ c_3 \end{Bmatrix} &= \begin{bmatrix} 1 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 \\ -3 & -2 & 3 & -1 \\ 2 & 1 & -2 & 1 \end{bmatrix} \begin{Bmatrix} Z_i \\ Z_i^' \\ Z_{i+1} \\ Z_{i+1}^' \end{Bmatrix} \\ &=\begin{bmatrix} 1 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 \\ -3 & -2 & 3 & -1 \\ 2 & 1 & -2 & 1 \end{bmatrix} \begin{Bmatrix} d_1 \\ hd_2 \\ d_3 \\ hd_4 \\ \end{Bmatrix} \\ &=\begin{bmatrix} 1 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 \\ -3 & -2 & 3 & -1 \\ 2 & 1 & -2 & 1 \end{bmatrix} \begin{Bmatrix} \overline{d}_1 \\ \overline{d}_2 \\ \overline{d}_3 \\ \overline{d}_4 \\ \end{Bmatrix} \\

\end{align}

$$     (9.3-5)
 * 
 * }

where,


 * {| style="width:100%" border="0"

$$\displaystyle
 * style="width:95%" |
 * style="width:95%" |

\begin{align} \overline{d}_1&=d_1 \\ \overline{d}_2&= h \, d_2 \\ \overline{d}_3&=d_3 \\ \overline{d}_4&= h \, d_4 \\ \end{align}

$$
 * 
 * }

Expand the matrix equation.


 * {| style="width:100%" border="0"

$$\displaystyle
 * style="width:95%" |
 * style="width:95%" |

\begin{align} &c_0= \overline{d}_1 \\ &c_1= \overline{d}_2 \\ &c_2= -3\overline{d}_1 -2\overline{d}_2+3\overline{d}_3-\overline{d}_4 \\ &c_3= 2\overline{d}_1 +1\overline{d}_2-2\overline{d}_3+\overline{d}_4 \\

\end{align}

$$     (9.6-9)
 * 
 * }

Substitute $$\displaystyle c_i$$ to Eq 9.1


 * {| style="width:100%" border="0"

$$\displaystyle \begin{align} Z(s)&=\sum_{i=0}^{3}c_i s^i \\ &= c_0 + c_1 s^1 + c_2 s^2 + c_3 s^3 \\ &= \overline{d}_1 + \overline{d}_2 s^1 + (-3\overline{d}_1 -2\overline{d}_2+3\overline{d}_3-\overline{d}_4) s^2 + (2\overline{d}_1 +1\overline{d}_2-2\overline{d}_3+\overline{d}_4) s^3 \end{align}
 * style="width:95%" |
 * style="width:95%" |

$$     (9.10-12)
 * 
 * }

Rearrange in terms of $$\displaystyle \overline{d}_i $$


 * {| style="width:100%" border="0"

$$\displaystyle \begin{align} Z(s)& = (2s^3-3s^2+1)\overline{d}_1+(s^3-2s^2+s)\overline{d}_2+(-2s^3+3s^2)\overline{d}_3+(s^3-s^2)\overline{d}_4 \\ &=\overline{N}_1(s) \, \overline{d}_1+\overline{N}_2(s) \, \overline{d}_2+\overline{N}_3(s) \, \overline{d}_3+\overline{N}_4(s) \, \overline{d}_4 \end{align} $$     (9.13-14)
 * style="width:95%" |
 * style="width:95%" |
 * 
 * }

Comparing the LHS and RHS of the equation gives
 * {| style="width:100%" border="0" cellpadding="5" cellspacing="0"

(9.15-18)
 * style="width:10%" |
 * style="width:80%" |
 * style="width:80%" | 
 * style="width:80%" | 
 * }

Plot of $$\displaystyle \overline{N_i} $$.

=Problem 6.10: ADD TITLE HERE!!!!!!!!!!!!=

Solution
=Problem 6.11: ADD TITLE HERE!!!!!!!!!!!!=

Solution
=References=

=Contributing Members & Referenced Lecture=