text:
User:Egm6321.f11.team1/report6

= R*6.1 Special IFM to solve a nonhomogeneous L2-ODE-CC with general f(t)= == Given == A nonhomogeneous L2-ODE-CC is
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$$
 * "$\displaystyle a_2y''+a_1y'+a_0y=f(t)$"
 * $$\displaystyle (Equation\;6.1.1)
 * $$\displaystyle (Equation\;6.1.1)
 * }
 * }

Find
$$\mathbf{(1)}$$ Find the PDEs that govern the integrating factor h(x,y) for Equation(6.1.1).

$$\mathbf{(2)}$$ Trial solution for the integrating factor $$\displaystyle h(t)=e^{\alpha t}$$ where $$\displaystyle \alpha$$ is unknown to be determined.
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$$ Because of the integrating factor in exponential form, assume the LHS of Equation(6.1.2) take the form :
 * "$\displaystyle \int \, e^{\alpha t}[a_2y''+a_1y'+a_0y]dt= \int \, e^{\alpha t}f(t)dt$"
 * $$\displaystyle (Equation\;6.1.2)
 * $$\displaystyle (Equation\;6.1.2)
 * }
 * }
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$$
 * "$\displaystyle \int \, e^{\alpha t}[a_2y''+a_1y'+a_0y]dt= e^{\alpha t}[\bar a_1y'+\bar a_0y]$"
 * $$\displaystyle (Equation\;6.1.3)
 * $$\displaystyle (Equation\;6.1.3)
 * }
 * }

$$\mathbf{(2.1)}$$ Find $$\displaystyle (\bar a_1,\bar a_0)$$ in terms of $$\displaystyle (a_0,a_1,a_2)$$.

$$\mathbf{(2.2)}$$ Find the quadratic equation for $$\displaystyle \alpha$$
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$$
 * "$\displaystyle a_2\alpha ^2-a_1\alpha+a_0=0$"
 * $$\displaystyle (Equation\;6.1.4)
 * $$\displaystyle (Equation\;6.1.4)
 * }
 * }

$$\mathbf{(2.3)}$$ Reduced-order equation: Equation(6.1.3)and Equation(6.1.4) lead to
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$$
 * "$\displaystyle \bar a_1y'+\bar a_0y=e^{-\alpha t}\int d^{\alpha t}f(t)dt$"
 * $$\displaystyle (Equation\;6.1.5)
 * $$\displaystyle (Equation\;6.1.5)
 * }
 * }

$$\mathbf{(2.4)}$$ Use the IFM to solve Equation(6.1.5)
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$$ Find the solution y(t) for general excitation f(t).
 * "$\displaystyle \bar h(t)=exp[\int \, \frac{\bar a_0}{\bar a_1}dt]=e^{\beta t}$"
 * $$\displaystyle (Equation\;6.1.6)
 * $$\displaystyle (Equation\;6.1.6)
 * }
 * }

$$\mathbf{(2.5)}$$ Show that
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$$
 * "$\displaystyle \alpha\beta=\frac{a_0}{a_2}$"
 * $$\displaystyle (Equation\;6.1.7)
 * $$\displaystyle (Equation\;6.1.7)
 * }
 * }
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$$ thus $$\displaystyle (\alpha,\beta)$$ are roots of the quadratic equation:
 * "$\displaystyle \alpha+\beta=\frac{a_1}{a_2}$"
 * $$\displaystyle (Equation\;6.1.8)
 * $$\displaystyle (Equation\;6.1.8)
 * }
 * }
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$$
 * "$\displaystyle (\lambda-\alpha)(\lambda-\beta)=\lambda^2-\underbrace{(\alpha+\beta)}_{\displaystyle \frac{a_1}{a_2}}\lambda+\underbrace{\alpha\beta}_{\displaystyle \frac{a_0}{a_2}}=0$|undefined"
 *
$$\displaystyle (Equation\;6.1.9)
 *
$$\displaystyle (Equation\;6.1.9)
 * }
 * }

$$\mathbf{(2.6)}$$ Deduce the particular solution $$y_p(t)\,$$ for general excitationf(t).

$$\mathbf{(2.7)}$$ Verify result with table of particular solution for:
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$$
 * "$\displaystyle f(t)=texp(bt)$"
 *
$$\displaystyle (Equation\;6.1.10)
 *
$$\displaystyle (Equation\;6.1.10)
 * }
 * }

$$\mathbf{(2.8)}$$ Solve the nonhomogeneous L2-ODE-CC Equation(6.1.1) with the Hyperbolic function:
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$$ For the coefficients $$(a_0,a_1,a_2)$$, consider two different characteristic equations:
 * "$\displaystyle f(t)=tanht$"
 *
$$\displaystyle (Equation\;6.1.11)
 *
$$\displaystyle (Equation\;6.1.11)
 * }
 * }

$$\mathbf{(2.8.1)}$$
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$$
 * "$\displaystyle (r+1)(r-2)=0$"
 *
$$\displaystyle (Equation\;6.1.12)
 *
$$\displaystyle (Equation\;6.1.12)
 * }
 * }

$$\mathbf{(2.8.2)}$$
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$$
 * "$\displaystyle (r-4)^2=0$"
 *
$$\displaystyle (Equation\;6.1.13)
 *
$$\displaystyle (Equation\;6.1.13)
 * }
 * }

$$\mathbf{(2.9)}$$ For each case in Equation(6.1.12)and Equation(6.1.13), determine the fundamental period of undamped free vibration, and plot the solution for the excitation for about 5 periods, assuming zero initial conditions.

Solution
$$\mathbf{(1)}$$ Multiply Equation(6.1.1) by the integrating factor h(t,y): "$\underbrace{h(t,y)a_2}_{\displaystyle f}y''+\underbrace{h(t,y)a_1y'+h(t,y)a_0y-h(t,y)f(t)}_{\displaystyle g}=0 $" Recall the 2nd exactness condition for N2-ODEs: "$\displaystyle f_{xx}+2pf_{xy}+p^2f_{yy}=g_{xp}+pg_{yp}-g_y$" "$\displaystyle f_{xp}+pf_{yp}+2f_y=g_{pp} $" Calculate the following terms in the exactness condition: "$\displaystyle f_t=a_2h_t\,;\,f_{tt}=a_2h_{tt}\,;\,f_{ty}=a_2h_{ty}\,;\,f_{tp}=0$" "$\displaystyle f_y=a_2h_y\,;\,f_{yy}=a_2h_{yy}\,;\,f_{yp}=0$" "$\displaystyle g_t=a_1ph_t+a_0yh_t-f(t)\cdot h_t-h\cdot f'(t)$" "$\displaystyle g_{tp}=a_1h_t$" "$\displaystyle g_y=a_1ph_y+a_0yh_y+a_0h-f(t)h_y$" "$\displaystyle g_{yp}=a_1h_y\,;\,g_p=a_1h\,;\,g_{pp}=0$" Substituting those calculated terms into 2nd exactness condition: "$\displaystyle a_2h_{tt}+2p\cdot a_2 h_{ty}+p^2\cdot a_2h_{yy}=a_1h_t+\cancel{p\cdot a_1h_y}-\cancel{a_1ph_y}-a_0yh_y-a_0h+f(t)h_y$" "$\displaystyle 0+p\cdot 0+2\cdot a_2h_y=0$" From the second equation, we can obtain that "$\displaystyle h_y=0$" which means that integrating factor $$\displaystyle h(t,y)=h(t)$$. Substituting h(t) into the first equation of 2nd exactness condition, we can obtain an L2-ODE-CC, "$\displaystyle a_2h_{tt}-a_1h_t+a_0h=0$" Therefore, that PDE reduces into an ODE, which can be solved.

$$\mathbf{(2.1)}$$ Differentiate the RHS of Equation(6.1.3), "$\displaystyle \frac{d}{dt}\left[e^{\alpha t}(\bar a_1y'+\bar a_0y)\right]=\alpha e^{\alpha t}(\bar a_1y'+\bar a_0y)+e^{\alpha t}(\bar a_1 y''+\bar a_0 y')$" "$\displaystyle =e^{\alpha t}[\bar a_1y''+(\bar a_0+\alpha \bar a_1)y'+\alpha \bar a_0y]$" Compare the coefficient with the integrand on LHS of Equation(6.1.3), "$\displaystyle e^{\alpha t}[\bar a_1y+(\bar a_0+\alpha \bar a_1)y'+\alpha \bar a_0y]=e^{\alpha t}[a_2y+a_1y'+a_0y]$" "$\displaystyle \bar a_1=a_2$" "$\displaystyle \bar a_0+\alpha \bar a_1=a_1$" "$\displaystyle \alpha \bar a_0=a_0$"

We can find $$\displaystyle (\bar a_1,\ \bar a_0)$$ in terms of $$\displaystyle (a_0,\ a_1,\ a_2)$$ that "$\displaystyle \bar a_1=a_2$" "$\displaystyle \bar a_0=\frac{a_0}{\alpha}$"

$$\mathbf{(2.2)}$$ Substituting $$\displaystyle \bar a_0$$ and $$\displaystyle \bar a_1$$ into $$\displaystyle \bar a_0+\alpha \bar a_1=a_1$$, we can get that <\br> "$\displaystyle \frac{a_0}{\alpha}+\alpha\cdot a_2=a_1$" Rearrange it, we obtain the quadratic equation for $$\displaystyle \alpha$$: "$\displaystyle a_2 \alpha^2-a_1\alpha+a_0=0$"

$$\mathbf{(2.3)}$$ Therefore, the reduced-order equation is obtained "$\displaystyle e^{\alpha t}[\bar a_1y'+\bar a_0y]=\int e^{\alpha t}f(t)dt$" Rewrite it,
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$$ Equation(6.1.14) is a L1-ODE-CC, which can be solved by the IFM.
 * "$\displaystyle \bar a_1y'+\bar a_0y=e^{-\alpha t}\,\int e^{\alpha t}f(t)dt$"
 *
$$\displaystyle (Equation\;6.1.14)
 *
$$\displaystyle (Equation\;6.1.14)
 * }
 * }

$$\mathbf{(2.4)}$$ Recall the IFM of general non-homogeneous L1-ODE-VC, "$\displaystyle 1\cdot y'+\underbrace{\frac{Q(x)}{P(x)}}_{a_0(x)}y=\underbrace{\frac{R(x)}{P(x)}}_{b(x)}$|undefined" The integrating factor is "$\displaystyle h(x)=exp[\int^x a_0(s)ds]$" The solution is "$\displaystyle y(x)=\frac{1}{h(x)}\left[\int^x h(s)b(s)ds+k\right]$" Divided Equation(6.1.14) by $$\displaystyle \bar a_1$$, "$\displaystyle 1\cdot y'+\underbrace{\frac{\bar a_0}{\bar a_1}}_{\displaystyle \beta}y=\underbrace{\frac{e^{-\alpha t}}{\bar a_1}\, \int e^{\alpha t} f(t)dt}_{\displaystyle b(t)}$|undefined" Therefore, we can obtain the integrating factor "$\displaystyle \bar h(t)=exp[\int\,\frac{\bar a_0}{\bar a_1} dt]$" "$\displaystyle =exp[\int\,\beta dt]$" "$\displaystyle =e^{\beta t}$" Therefore, the solution of Equation(6.1.14) is "$\displaystyle y(t)=\frac{1}{\bar h(t)}\left[ \int^t h(s)b(s)ds+k\right]$" "$\displaystyle =\frac{1}{e^{\beta t}}\int^t e^{\beta s}\cdot\frac{e^{-\alpha s}}{\bar a_1}\left[\int^s e^{\alpha \tau}f(\tau)d\tau\right]ds$|undefined"

$$\mathbf{(2.5)}$$ Substitute $$\displaystyle \bar a_1=a_2\,;\,\bar a_0+\alpha \bar a_1=a_1\,;\,\alpha \bar a_0=a_0$$ into $$\displaystyle \alpha$$ and$$\displaystyle \beta$$, we can obtain that "$\displaystyle \alpha\beta=\alpha \frac{\bar a_0}{\bar a_1}=\frac{a_0}{\bar a_1}=\frac{a_0}{a_2}$" "$\displaystyle \alpha +\beta= \frac{a_1-\bar a_0}{\bar a_1}+\frac{\bar a_0}{\bar a_1}=\frac{a_1}{\bar a_1}=\frac{a_1}{a_2}$" Thus, $$\displaystyle (\alpha,\beta)$$ are the roots of the following quadratic equtation: "$\displaystyle (\lambda-\alpha)(\lambda-\beta)=\lambda^2-\underbrace{(\alpha+\beta)}_{\displaystyle \frac{a_1}{a_2}}\lambda+\underbrace{\alpha\beta}_{\displaystyle \frac{a_0}{a_2}}$|undefined" Obviously, it is the same as the quadratic equation obtained by $$\mathbf{(2.2)}$$.

$$\mathbf{(2.6)}$$ Rearrange the solution of Equation(6.1.14) "$\displaystyle \frac{1}{e^{\beta t}}\int^t e^{\beta s}\cdot\frac{e^{-\alpha s}}{\bar a_1}\left[\int^s e^{\alpha \tau}f(\tau)d\tau\right]ds$|undefined" "$\displaystyle =e^{-\beta t}[\int^t\frac{e^{(\beta-\alpha)s}}{\bar a_1}\int^s e^{\alpha \tau}f(\tau)d\tau]ds$|undefined" "$\displaystyle =\frac{e^{-\beta t}}{\bar a_1}[\int^t e^{(\beta-\alpha)s}[\int e^{\alpha s}f(s)ds+k_1]ds]$|undefined" "$\displaystyle =\frac{e^{-beta t}}{\bar a_1}[\int^te^{\beta-\alpha)s}[\int e^{\alpha s}f(s)ds]ds+\int^t e^{(\beta-\alpha)s}k_1ds]$|undefined" "$\displaystyle =\frac{e^{-\beta t}}{\bar a_1}[\int e^{(\beta-\alpha)t}[\int e^{\alpha s}f(s)ds]dt+k_2+\frac{k_1}{\beta-\alpha}e^{(\beta-\alpha)t}]$|undefined" "$\displaystyle =\underbrace{\frac{k_1}{\bar a_1(\beta-\alpha)}e^{-\alpha t}}_{\displaystyle C_1 y_H^1}+\underbrace{\frac{k_2}{\bar a_1}e^{-\beta t}}_{\displaystyle C_2 y_H^2}+\underbrace{\frac{e^{-\beta t}}{\bar a_1}\int e^{(\beta-\alpha)t}[\int e^{\alpha s}f(s)ds]dt}_{\displaystyle y_P}$|undefined" The particular solution $$\displaystyle y_P(t)$$ for general excitation $$\displaystyle f(t)$$ is "$\displaystyle y_P(t)=\frac{e^{-\beta t}}{\bar a_1}\int e^{(\beta-\alpha)t}[\int e^{\alpha s}f(s)ds]dt$|undefined" $$\mathbf{(2.7)}$$

The particular solution for an arbitrary $$\displaystyle f(x)$$ is given directly above in section 2.6. The solutions for $$\displaystyle f(x)=t^2$$, as evaluated by Wolfram-Alpha is:

$$y_P(t)=\frac{2}{\alpha^3 a_1 \beta}+\frac{2}{\alpha^2 a_1 \beta^2}+\frac{2}{\alpha a_1 \beta^3}+\frac{t^2}{\alpha a_1 \beta}-\frac{2t}{\alpha^2 a_1 \beta}-\frac{2t}{\alpha a_1 \beta^2}$$

The solution as found in Table 3.6.1 on page 1 of the following link (note use $$\displaystyle s=0$$ accordingly):

$$\displaystyle y_P(t)=A_0t^2+A_1t+A_0$$

Choosing the coefficients $$\displaystyle A$$ wisely yields a perfect match. Now check solutions for $$\displaystyle f(x)=te^{bt}$$, again use Wolfram-Alpha:

$$y_P(t)=\frac{e^{bt}(-\alpha+2\beta b t-b-2\beta+2\alpha\beta t)}{4a_1\beta^2(\alpha+\beta)^2}$$

The solution for this case also is found in the same link above:

$$\displaystyle y_P(t)=(A_1t+A_0)e^{bt}$$

Similarly, a careful choice of coefficients proves equivalency.

$$\mathbf{(2.8.1)}$$ $$\displaystyle \alpha$$ and $$\displaystyle \beta$$ is the roots of quadratic equation: "$\displaystyle (r+1)(r-2)=0.$" Therefore let $$\displaystyle \alpha=-1\,;\,\beta=2$$ As for the excitation f(t) is Hyperbolic function: "$\displaystyle f(t)=tanh(t).$" For the general solution Equation(6.1.15) "$\displaystyle y(t)=\frac{1}{6}\left[1+2e^{-2t}log(e^{2t}+1)-4e^ttan^{-1}(e^{-t})\right]$" This is calculated by www.wolframalpha.com.

$$\mathbf{(2.8.2)}$$ The quadratic equation is: "$\displaystyle (r-4)^2=0.$" Therefore let $$\displaystyle \alpha=\beta=4$$ As for the excitation f(t) is Hyperbolic function: "$\displaystyle f(t)=tanh(t).$" For the general solution Equation(6.1.15) "$\displaystyle y(t)=-\frac{Li_2(-e^{2t}\cdot e^{-4t})}{2}-\frac{e^{-2t}}{2}+\frac{1}{16}$|undefined" where $$\displaystyle Li_2(t)\,$$ is the polylogarithm function. This is calculated by www.wolframalpha.com.

$$\mathbf{(2.9)}$$ Work is in progress.

=R*6.2 Showing the form of particular solution in notes agree with the one in book by King and others =

Given
And

Find
with

Solution
- Solved on our own $$\frac{1}{h(x)}=\frac{u_1(x)u_2'(x)-u_2(x)u_1'(x)}{{u_1}^2(x)}=\frac{W(x)}{{u_1}^2(x)}$$
 (6.2.5)

Substituting the value of 'h(x)' from Equation 6.2.5 into Equation 6.2.1

$$y_P(x)=u_1(x)\int{\frac{W(x)}{{u_1}^2(x)}}\left[\int \frac{{f(x)}u_1(x)}{W(x)}\,dx \right]\,dx =u_1(x)\left[\left(\int\frac{W(x)}dx \right)\frac{u_2(x)}{u_1(x)}-\int\frac{W(x)}dx \right]$$

Applying dummy integrals:

$$y_P(x)=u_1(x)\left[\left(\int^{x}\frac{W(s)}ds \right)\frac{u_2(x)}{u_1(x)}-\int^{x}\frac{W(s)}ds \right] =u_2(x)\int^{x}\frac{W(s)}ds - u_1(x)\int^{x}\frac{W(s)}ds $$

Taking $$\displaystyle u_1(x) \,\, u_2(x)$$ inside integral.

$$y_P(x)=\int^{x}\frac{W(s)}ds - \int^{x}\frac{W(s)}ds =\int^{x}f(s)\left[\frac{u_1(s)u_2(x)-u_1(x)u_2(s)}{W(s)}\right]\,ds $$

This shows that the form of particular solution in notes agree with the one in book by King and others

=R*6.3 Invaid Solution=

Given
From [[media:Pea1.f11.mtg35.djvu|Mtg 35-4]]

The first homogeneous solution: Trial solution: Characteristic equation:

Find
Explain why $$\displaystyle r_2(x)$$ is not a valid root, i.e., $$\displaystyle u_2(x) = e^{xr_{2}(x)}$$ is not a valid solution.

Solution
- Solved on our own As $$\displaystyle u_2(x) = e^{xr_2(x)}$$

Validity of $$\displaystyle y= u_2(x) $$ can be shown by substituting it into the Eq. (6.3.1).

Substitute $$\displaystyle y, y^{'}, y^{''}$$ in the Eq. (6.3.6) into the Eq. (6.3.1), As shown in the Eq. (6.3.7), $$\displaystyle u_2(x) $$ is not a valid homogeneous solution.

= Problem R*6.4 For a L2-ODE-VC using variation of parameters find the 2nd homogeneous solution and compare =

Reference Lecture 35-4 found here:[[media:pea1.f11.mtg35.djvu|Mtg 35]]

Given
The following liner, second order, ordinary differential equation, with varying coefficients (L2-ODE-VC)

"$\displaystyle (x - 1) y'' - x y' + y = x$"
$$\displaystyle (6.4.1) $$

Find
A valid homogeneous solution and call it $$\displaystyle u_1 $$

Find the particular solution using variation of parameters method and compare it to $$\displaystyle e^{xr_2(x)} $$

Solution
Lets start by validating that the homogeneous solution selected below

$$\displaystyle u_1 = e^x $$
$$\displaystyle (6.4.2) $$

Use the trial solution:

"$ \displaystyle u_1(x) = e^{rx}$"
$$\displaystyle (6.4.3) $$

Differentiate the above to get "$ \displaystyle u'_1(x) = r e^{rx}$"
$$\displaystyle (6.4.4)$$

"$ \displaystyle u''_1(x) = r^2 e^{rx} $"
$$\displaystyle (6.4.5)$$

Plugging these derivatives and the trial solution into 6.4.1 yields "$ \displaystyle \left[(x-1) r^2 - rx + 1\right] e^{rx} = 0$"
$$\displaystyle (6.4.6)$$

Rearranging by grouping the terms with x from above yields

"$ \displaystyle \left[(r^2 - r) x + (1 - r^2)\right] e^{rx} = 0$"
$$\displaystyle (6.4.7)$$

In order for the above equation to be equal to zero, the following equations must be satisfied

"$ \displaystyle r^2 - r = r (r - 1) = 0 $"

"$ \displaystyle 1 - r^2 = -(r + 1)(r - 1) = 0 $"
$$\displaystyle (6.4.8)$$

The solution of the above equations is $$\displaystyle r=1$$. So 6.4.3 can be written as

"$ \displaystyle u_1(x) = e^x $"
$$\displaystyle (6.4.9)$$

Which is a valid homogeneous solution

So now lets find the 2nd homogeneous solution using variation of parameters

We can rewrite 6.4.1 as

$$ \displaystyle y'' + \underbrace{\frac{-x}{x-1}}_{a_1(x)} y' + \underbrace{\frac{1}{x-1}}_{a_0(x)} y  = \underbrace{\frac{x}{x-1}}_{f(x)} $$
$$\displaystyle (6.4.10) $$

Using equation (4) from page 34-5 of the class notes [[media:pea1.f11.mtg34.djvu|Mtg 34 (b)]] the following expression for the 2nd homogeneous solution is obtained:

$$\displaystyle {u}_{2}(x)= \underbrace{{u}_{1}(x)}_{{e}^{x}} \int \frac{1}{ u_1^2(x)}exp[-\int \underbrace{a_1(x)}_{\frac{-(x)}{x-1}} \ dx] \ dx $$
$$\displaystyle (6.4.11) $$

Using Wolfram Alpha to evaluate both integrals and simplify it can be shown that

$$\displaystyle {u}_{2}(x)= x $$
$$\displaystyle (6.4.12)$$

Which is the 2nd homogeneous solution

This can can then be compared to $$\displaystyle e^{xr_2(x)} $$ as requested

if $$\displaystyle r_2(x) = ln(x)/x $$ the expressions for $$\displaystyle{u}_{2}$$ are equivalent

= R*6.5: Find Two Homogeneous Solutions using a Trial Solution =

Reference lecture notes page 36-3 found here [[media:pea1.f11.mtg36.djvu|Mtg 36]]

Given
Find u1 and u2 of Equation 6.5.1 below using the trial solution given in equation 6.5.2:

"$\displaystyle (x +1) y'' - (2x+3)y' + 2y = 0 $"
(6.5.1)

"$\displaystyle y(x)=e^{rx}$"
(6.5.2)

Solution
Lets start by differentiating the trial solution as follows:

"$ \displaystyle y'(x) = r e^{rx}$"
(6.5.3)

"$ \displaystyle y''(x) = r^2 e^{rx}$"
(6.5.4)

The next step is to plug the derivatives and the original trail function into equation 6.5.1 as follows:

"$ \displaystyle e^{rx}\left[(x+1)r^2-(2x+3)r+2\right]=0 $"
(6.5.5)

Divide $$ \displaystyle e^{rx} $$ by both side and group all the $$ \displaystyle x $$ terms to obtain:

"$ \displaystyle x(r^2-2r)+(r-2)(r-1)=0$"
(6.5.6)

The following conditions must be satisfy for equation 6.5.6 to be true:

"$ \displaystyle r(r-2)=(r-2)(r-1)=0 $"
 (6.5.7)

In order for 6.5.6 to be true it can be seen that "$ \displaystyle r=2 $"
(6.5.8)

The first homogeneous solution is obtained by plugging equation 6.5.8 into the trial solution given in equation 6.5.2:

"$ \displaystyle u_1(x)=e^{2x} $"
(6.5.9)

Using equation (4) from page 34-5 of the class notes [[media:pea1.f11.mtg34.djvu|Mtg 34 (b)]] the following expression for the 2nd homogeneous solution is obtained:

"$ \displaystyle u_2(x)=u_1(x)\int{u_1^{-2}(x)e^{-\int{a_1(x)}\,dx}}\,dx$|undefined"
(6.5.10)

Equation 6.5.1 can be rearranged into the desired form shown:

"$ \displaystyle y''-\underbrace{\frac{2x+3}{x+1}}_{a_1}y'+\underbrace{\frac{2}{x+1}}_{a_0}y=0$|undefined"
 (6.5.11)

The value for $$ \displaystyle a_1 $$ from equation 6.5.11 is then substituted into equation 6.5.10, additionally the first homogeneous solution from equation 6.5.9 is substituted to solve for the second homogeneous solution as shown:

"$ \displaystyle u_2(x)=e^{2x}\int{e^{-4x}e^{\int{\frac{2x+3}{x+1}}\,dx}}\,dx$|undefined"
 (6.5.12)

Evaluating the inner most integral yields:

"$ \displaystyle u_2(x)=e^{2x}\int{(x+1)e^{-2x}}\,dx$|undefined"
(6.5.13)

Finally evaluating the last integral and simplifying yields:

"$ \displaystyle u_2(x)=\frac{-1}{4}(2x+3)$"
(6.5.14)

This is the 2nd homogeneous solution.

= R*6.6 Reverse Engineering ODE Equations =

Given
Trial solutions:

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

$$
 * $$\frac{e^{rx}}{x^2}, r=\text{constant}$$
 *
$$\displaystyle (Equation\;6.6.1)
 *
$$\displaystyle (Equation\;6.6.1)
 * }
 * }

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

$$
 * $$\frac{e^{rx}}{\sin x}, r=\text{constant}$$
 *
$$\displaystyle (Equation\;6.6.2)
 *
$$\displaystyle (Equation\;6.6.2)
 * }
 * }

Characteristic equations:
 * {| style="width:100%" border="0" align="left"

$$
 * $$\displaystyle r^2+3=0$$
 *
$$\displaystyle (Equation\;6.6.3)
 *
$$\displaystyle (Equation\;6.6.3)
 * }
 * }

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

$$
 * $$\displaystyle r_1=2$$
 *
$$\displaystyle (Equation\;6.6.4)
 *
$$\displaystyle (Equation\;6.6.4)
 * }
 * {| style="width:100%" border="0" align="left"
 * {| style="width:100%" border="0" align="left"

$$
 * $$r_2(x)=\frac{1}{x+1}$$
 *
$$\displaystyle (Equation\;6.6.5)
 *
$$\displaystyle (Equation\;6.6.5)
 * }
 * {| style="width:100%" border="0" align="left"
 * {| style="width:100%" border="0" align="left"

$$.
 * $$\displaystyle (r-r_1)[r-r_2(x)]=0$$
 *
$$\displaystyle (Equation\;6.6.6)
 *
$$\displaystyle (Equation\;6.6.6)
 * }
 * }

== Find ==

Find a homogeneous L2-ODE-VC that accepts the trial solutions Equation 6.6.1 and Equation 6.6.2, and their respective characteristic equations, Equation 6.6.3, and Equation 6.6.4 thru Equation 6.6.6.

Solution
I will mirror the approach laid out in Meeting 36.

Multiply the trial solution of Equation 6.6.1 (and its derivatives) by unknown coefficients $$\displaystyle a(x)$$.

$$\left[\frac{e^{rx}}{x^2}\right]\times a_0(x)$$

$$\left[\frac{d}{dx}\left(\frac{e^{rx}}{x^2}\right)=\left(r-\frac{2}{x}\right)\frac{e^{rx}}{x^2}\right]\times a_1(x)$$

$$\left[\frac{d^2}{dx^2}\left(\frac{e^{rx}}{x^2}\right)=\left(r^2-\frac{4r}{x}+\frac{6}{x^2}\right)\frac{e^{rx}}{x^2}\right]\times a_2(x)$$

Now equate the sum of the above three results to the product of the trial solution (Equation 6.6.1) and the characteristic equation (Equation 6.6.3):

$$\frac{e^{rx}}{x^2}\left[a_2\left(r^2-\frac{4r}{x}+\frac{6}{x^2}\right)+a_1\left(r-\frac{2}{x}\right)+a_0\right]=\frac{e^{rx}}{x^2}(r^2+3)=0$$

$$a_2r^2+\left(\frac{4a_2}{x}+a_1\right)r+\left(\frac{6a_2}{x^2}-\frac{2a_1}{x}+a_0\right)=r^2+3$$

Comparing the polynomial coefficients of $$\displaystyle r$$ yields the following:

$$a_2=1, \quad a_1=-\frac{4}{x}, \quad a_0=3-\frac{48}{x^2}$$

If we define $$f:=\frac{e^{rx}}{x^2}$$, the following L2-ODE-VC will be satisfied:

Now the same technique will be applied to Equation 6.6.2 and its characteristic equation. Even though one root is a function of x, the technique is still valid as long as one root is constant. See page (36-2) in Meeting 36.

Multiply the trial solution of Equation 6.6.2 (and its derivatives) by unknown coefficients $$\displaystyle a(x)$$.

$$\left[\frac{e^{rx}}{\sin x}\right]\times a_0(x)$$

$$\left[\frac{d}{dx}\left(\frac{e^{rx}}{\sin x}\right)=\left(r-\cot x\right)\frac{e^{rx}}{\sin x}\right]\times a_1(x)$$

$$\left[\frac{d^2}{dx^2}\left(\frac{e^{rx}}{\sin x}\right)=\left(r^2-2r\cot x+\cot^2x+\csc^2x\right)\frac{e^{rx}}{\sin x}\right]\times a_2(x)$$

Now equate the sum of the above three results to the product of the trial solution (Equation 6.6.2) and the characteristic equation (Equation 6.6.6):

$$\frac{e^{rx}}{\sin x}\left[a_2\left(r^2-2r\cot x+\cot^2x+\csc^2x\right)+a_1\left(r-\cot x\right)+a_0\right]=\frac{e^{rx}}{\sin x}(r-r_1)[r-r_2(x)]=0$$

$$a_2r^2+\left(a_1-2a_2\cot x\right)r+\left(a_0-a_1\cot x +\cot^2x+\csc^2x\right)=r^2-2r-\frac{r}{x+1}+\frac{2}{x+1}$$

Comparing the polynomial coefficients of $$\displaystyle r$$ yields the following:

$$a_2=1, \quad a_1=2\cot x-\frac{1}{x+1}-2, \quad a_0=\cot x\left(2\cot x-\frac{1}{x+1}-2\right)-\cot^2x-\csc^2x+\frac{2}{x+1}$$

If we define $$f:=\frac{e^{rx}}{\sin x}$$, the following L2-ODE-VC will be satisfied:

= R*6.7 Using variation of parameters method to get the 2nd homogeneous solution $$Q_2(x)$$=

Given
Legengre differential equation:
 * {| style="width:100%" border="0" align="left"

\displaystyle (1-x^2)y''-2xy'+n(n+1)y=0 $$ $$
 *
$$\displaystyle (Equation\; 6.7.1)
 * }
 * }

For n=2
 * {| style="width:100%" border="0" align="left"

\displaystyle (1-x^2)y''-2xy'+6y=0 $$ $$
 *
$$\displaystyle (Equation\; 6.7.2)
 * }
 * }

First homogeneous solution:
 * {| style="width:100%" border="0" align="left"

\displaystyle P_2(x)=\frac{1}{2}(3x^2-1) $$ $$
 *
$$\displaystyle (Equation\; 6.7.3)
 * }
 * }

== Find == Show the second homogeneous solution:
 * {| style="width:100%" border="0" align="left"

\displaystyle Q_2(x)=\frac{1}{4}(3x^2-1)\log \left ( \frac{1+x}{1-x} \right )-\frac{3}{2}x $$ $$
 *
$$\displaystyle (Equation\; 6.7.4)
 * }
 * }

Solution
For a Non-homogeneous L2-ODE-VC:
 * {| style="width:100%" border="0" align="left"

\displaystyle y''+a_1(x)y'+a_0(x)y=f(x) $$ $$
 *
$$\displaystyle (Equation\; 6.7.5)
 * }
 * }

Given 1st homogeneous solution u_1(x) Find 2nd homogeneous solution u_2(x). $$ \displaystyle y(x)=U(x)u_1(x) $$

$$ \displaystyle \begin{align} y&=Uu_1 \\ y'&=Uu_1'+U'u_1 \\ y&=Uu_1+2U'u_1'+U''u_1 \end{align} $$

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

\displaystyle a_0y+a_1y'+y=U[a_0u_1+a_1u_1'+u_1]+U'[a_1u_1+2u_1']+U''u_1=f(x) $$ $$
 *
$$\displaystyle (Equation\; 6.7.6)
 * }
 * }

Let $$ \displaystyle Z:=U' $$

$$ \displaystyle Z'+\frac{a_1u_1+2u_1'}{u_1}Z=\frac{f}{u_1} $$

$$ \displaystyle \begin{align} \displaystyle h(x)&=exp\left [ \int \left ( a_1(x)+\frac{2u_1'(x)}{u_1(x)} \right ) dx \right ] \\ \displaystyle &=u_1^2(x)exp\left [\int a_1(x)dx \right ] \end{align} $$

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

\displaystyle Z(x)=\frac{1}{h(x)}[k_2+\int h(x)\frac{f(x)}{u_1(x)}dx] $$ $$
 *
$$\displaystyle (Equation\; 6.7.7)
 * }
 * }
 * {| style="width:100%" border="0" align="left"

\displaystyle U(x)=k_1+\int Z(x)dx $$ $$
 *
$$\displaystyle (Equation\; 6.7.8)
 * }
 * }

$$ \displaystyle \begin{align} \displaystyle u_2(x):&=u_1(x)\int \frac{1}{h(x)}dx \\ \displaystyle &=u_1(x)\int \frac{1}{u_1^2(x)}exp \left [ -\int a_1(x)dx \right ]dx \end{align} $$

For $$ \displaystyle \begin{align} u_1(x)&=P_2(x)=\frac{1}{2}(3x^2-1) \\ \displaystyle a_1(x)&=\frac{-2x}{1-x^2} \end{align} $$

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

\displaystyle \begin{align} \displaystyle \frac{1}{h(x)}&=\frac{4}{(3x^2-1)^2}\cdot exp \left [ -\int \frac{-2x}{1-x^2}dx \right ] dx\\ \displaystyle &=\frac{4}{(3x^2-1)^2(1-x^2)} \end{align} $$ $$
 *
$$\displaystyle (Equation\; 6.7.9)
 * }
 * }

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

\displaystyle \begin{align} \displaystyle Q_2(x)=u_2(x)&=\frac{1}{2}(3x^2-1)\int \frac{4}{(3x^2-1)^2(1-x^2)}dx \\ \displaystyle &=\frac{1}{4}(3x^2-1)\left (-\frac{6x}{3x^2-1}-\log (1-x)+\log (1+x)\right ) \end{align} $$ $$
 *
$$\displaystyle (Equation\; 6.7.10)
 * }
 * }

Rearranging:

Computing with the help of WA, please refer to: WalframAlpha

= R*6.8 Find Solutions to L2-ODE-VC with Trial Solution and Variation of Parameters with Excitation=

Given
Two Non-Homogeneous L2-ODE-VC

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

$$
 * $$\displaystyle (x-1)y''-xy'+y=f(x)$$
 *
$$\displaystyle (Equation\;6.8.1)
 *
$$\displaystyle (Equation\;6.8.1)
 * }
 * }

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

$$
 * $$\displaystyle xy''+2y'+xy=f(x)$$
 *
$$\displaystyle (Equation\;6.8.2)
 *
$$\displaystyle (Equation\;6.8.2)
 * }
 * }

== Find ==

Find the 1st homogeneous solution by trial solution.

Find the complete solution by variation of parameters with the following excitation:

 * $$\displaystyle f(x)=x$$

 * $$\displaystyle f(x)=\sin x$$

 * $$\displaystyle f(x)=\exp(\alpha x)$$

Part 1
Find the 1st homogeneous solution by trial solution. Since we are looking for homogenous solution, $$\displaystyle f(x)=0$$. For Equation 6.8.1 we will assume the form $$\displaystyle y(x)=e^{rx}$$, where $$\displaystyle r$$ is an undetermined coefficient. Thus Equation 6.8.1 becomes:

$$\displaystyle (x-1)r^2e^{rx}-xre^{rx}+e^{rx}=0$$

$$\displaystyle (x-1)r^2-xr+1=0$$

$$\displaystyle [(x-1)r-1](r-1)=0$$

Thus the two roots are:

$$\displaystyle r_1=1$$

$$\displaystyle r_2=\frac{1}{x-1}$$

The second root is not valid, since it undermines the assumption of form in the trial solution due to it being a function of x. See R*6.3 for additional detail. Therefore the 1st homogeneous solution for Equation 6.8.1 is:

For Equation 6.8.2 we will assume the form $$\displaystyle y(x)=\frac{\sin rx}{x}$$, where $$\displaystyle r$$ is an undetermined coefficient. Thus Equation 6.8.2 becomes:

$$x\left(-\frac{(r^2x^2-2)\sin rx +2rx\cos rx}{x^3}\right)+2\frac{rx\cos rx -\sin rx}{x^2}+x\left(\frac{\sin rx}{x}\right)=0$$

$$-\frac{(r^2x^2-2)\sin rx +2rx\cos rx}{x^2}+2\frac{rx\cos rx -\sin rx}{x^2}+\sin rx=0$$

$$\frac{-r^2x^2\sin rx}{x^2}+\sin rx=0$$

$$\displaystyle -r^2\sin rx+\sin rx=0$$

$$\displaystyle 1-r^2=0$$

Thus the two roots are:

$$\displaystyle r_1=1$$

$$\displaystyle r_2=-1$$

Therefore, taking note that sine is an odd function, the 1st homogeneous solution for Equation 6.8.2 is either:

Section 1
Find the complete solution by variation of parameters to

$$\displaystyle (x-1)y''-xy'+y=x$$

$$\displaystyle y''-\frac{x}{x-1}y'+\frac{1}{x-1}y=\frac{x}{x-1}$$

Thus following the syntax of Meeting 34,

$$a_1(x)=-\frac{x}{x-1}$$

$$f(x)=\frac{x}{x-1}$$

We also deduced from Part 1 that

$$\displaystyle u_1(x)=e^x$$

Now via (4) on (34-5) of Meeting 34:

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

$$
 * $$u_2(x):=u_1(x)\int\frac{1}{u_1^2(x)}\exp\left[-\int a_1(x)dx\right]dx$$
 *
$$\displaystyle (Equation\;6.8.3)
 *
$$\displaystyle (Equation\;6.8.3)
 * }
 * }

$$u_2(x):=e^x\int\frac{1}{e^{2x}}\exp\left[-\int -\frac{x}{x-1}\right]dx$$

$$\displaystyle u_2(x)=-x$$

Now via (1) on (34-6) of Meeting 34:

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

$$
 * $$y_P(x)=u_1(x)\int\frac{1}{h(x)}\left[\int h(x)\frac{f(x)}{u_1(x)}dx\right]dx$$
 *
$$\displaystyle (Equation\;6.8.4)
 *
$$\displaystyle (Equation\;6.8.4)
 * }
 * }

Where

$$h(x)=u_1^2(x)\exp\left[\int a_1(x)dx\right]$$

$$\displaystyle h(x)=\frac{e^x}{x-1}$$

So

$$y_P(x)=e^x\int\frac{1}{\frac{e^x}{x-1}}\left[\int \frac{e^x}{x-1}\frac{\frac{x}{x-1}}{e^x}dx\right]dx$$

To find the complete solution we use (3) on (34-5) from Meeting 34, where constants are determined by initial conditions:

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

$$
 * $$\displaystyle y(x)=k_1u_1(x)+k_2u_2(x)+y_P(x)$$
 *
$$\displaystyle (Equation\;6.8.5)
 *
$$\displaystyle (Equation\;6.8.5)
 * }
 * }

Section 2
Find the complete solution by variation of parameters to

$$\displaystyle (x-1)y''-xy'+y=\sin x$$

$$\displaystyle y''-\frac{x}{x-1}y'+\frac{1}{x-1}y=\frac{\sin x}{x-1}$$

Thus following the syntax of Meeting 34,

$$a_1(x)=-\frac{x}{x-1}$$

$$f(x)=\frac{\sin x}{x-1}$$

$$\displaystyle u_1(x)=e^x$$

The solution is the same as that of Section 1, except now

$$y_P(x)=e^x\int\frac{1}{\frac{e^x}{x-1}}\left[\int \frac{e^x}{x-1}\frac{\frac{\sin x}{x-1}}{e^x}dx\right]dx$$

Section 3
Find the complete solution by variation of parameters to

$$\displaystyle (x-1)y''-xy'+y=\exp(\alpha x)$$

$$\displaystyle y''-\frac{x}{x-1}y'+\frac{1}{x-1}y=\frac{\exp(\alpha x)}{x-1}$$

Thus following the syntax of Meeting 34,

$$a_1(x)=-\frac{x}{x-1}$$

$$f(x)=\frac{\exp(\alpha x)}{x-1}$$

$$\displaystyle u_1(x)=e^x$$

The solution is the same as that of Section 1, except now

$$y_P(x)=e^x\int\frac{1}{\frac{e^x}{x-1}}\left[\int \frac{e^x}{x-1}\frac{\frac{\exp(\alpha x)}{x-1}}{e^x}dx\right]dx$$

Section 4
Find the complete solution by variation of parameters to

$$\displaystyle xy''-2y'+xy=x$$

$$\displaystyle y''-\frac{2}{x}y'+y=1$$

Thus following the syntax of Meeting 34,

$$a_1(x)=-\frac{2}{x}$$

$$\displaystyle f(x)=1$$

We also deduced from Part 1 that

$$\displaystyle u_1(x)=\frac{\sin x}{x}$$

Now via Equation 6.8.3

$$u_2(x):=\frac{\sin x}{x}\int\frac{1}{\frac{\sin^2 x}{x^2}}\exp\left[-\int -\frac{2}{x}\right]dx$$

Now use Equation 6.8.4, where

$$h(x)=u_1^2(x)\exp\left[\int a_1(x)dx\right]$$

$$\displaystyle h(x)=\frac{\sin^2x}{x^4}$$

So

$$y_P(x)=\frac{\sin x}{x}\int\frac{1}{\frac{\sin^2x}{x^4}}\left[\int \frac{\sin^2x}{x^4}\frac{1}{\frac{\sin x}{x}}dx\right]dx$$

To find the complete solution use Equation 6.8.5

Section 5
Find the complete solution by variation of parameters to

$$\displaystyle xy''-2y'+xy=\sin x$$

$$\displaystyle y''-\frac{2}{x}y'+y=\frac{\sin x}{x}$$

Thus following the syntax of Meeting 34,

$$a_1(x)=-\frac{2}{x}$$

$$\displaystyle f(x)=\frac{\sin x}{x}$$

$$\displaystyle u_1(x)=\frac{\sin x}{x}$$

The solution is the same as that of Section 4, except now

$$y_P(x)=\frac{\sin x}{x}\int\frac{1}{\frac{\sin^2x}{x^4}}\left[\int \frac{\sin^2x}{x^4}\frac{\frac{\sin x}{x}}{\frac{\sin x}{x}}dx\right]dx$$

Section 6
Find the complete solution by variation of parameters to

$$\displaystyle xy''-2y'+xy=\exp(\alpha x)$$

$$\displaystyle y''-\frac{2}{x}y'+y=\frac{\exp(\alpha x)}{x}$$

Thus following the syntax of Meeting 34,

$$a_1(x)=-\frac{2}{x}$$

$$\displaystyle f(x)=\frac{\exp(\alpha x)}{x}$$

$$\displaystyle u_1(x)=\frac{\sin x}{x}$$

The solution is the same as that of Section 4, except now

$$y_P(x)=\frac{\sin x}{x}\int\frac{1}{\frac{\sin^2x}{x^4}}\left[\int \frac{\sin^2x}{x^4}\frac{\frac{\exp(\alpha x)}{x}}{\frac{\sin x}{x}}dx\right]dx$$

= R 6.9 Non-homogeneous Legengre equation with $$f(x)=1$$=

Given
Non-homogeneous Legengre differential equation:
 * {| style="width:100%" border="0" align="left"

\displaystyle (1-x^2)y''-2xy'+2y=f(x) $$ $$
 *
$$\displaystyle (Equation\; 6.9.1)
 * }
 * }

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

\displaystyle f(x)=1 $$
 *

 * }
 * }

First homogeneous solution:

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

\displaystyle u_1(x)=P_1(x)=x $$ $$
 *
$$\displaystyle (Equation\; 6.9.2)
 * }
 * }

== Find == The final solution $$y(x)$$ by variation of parameters

Solution
For a Non-homogeneous L2-ODE-VC:
 * {| style="width:100%" border="0" align="left"

\displaystyle y''+a_1(x)y'+a_0(x)y=f(x) $$ $$
 *
$$\displaystyle (Equation\; 6.9.3)
 * }
 * }

Given 1st homogeneous solution u_1(x) Find 2nd homogeneous solution u_2(x). $$ \displaystyle y(x)=U(x)u_1(x) $$

$$ \displaystyle \begin{align} y&=Uu_1 \\ y'&=Uu_1'+U'u_1 \\ y&=Uu_1+2U'u_1'+U''u_1 \end{align} $$

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

\displaystyle a_0y+a_1y'+y=U[a_0u_1+a_1u_1'+u_1]+U'[a_1u_1+2u_1']+U''u_1=f(x) $$ $$
 *
$$\displaystyle (Equation\; 6.9.4)
 * }
 * }

Let $$ \displaystyle Z:=U' $$

$$ \displaystyle Z'+\frac{a_1u_1+2u_1'}{u_1}Z=\frac{f}{u_1} $$

$$ \displaystyle \begin{align} \displaystyle h(x)&=exp\left [ \int \left ( a_1(x)+\frac{2u_1'(x)}{u_1(x)} \right ) dx \right ] \\ \displaystyle &=u_1^2(x)exp\left [\int a_1(x)dx \right ] \end{align} $$

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

\displaystyle Z(x)=\frac{1}{h(x)}[k_2+\int h(x)\frac{f(x)}{u_1(x)}dx] $$ $$
 *
$$\displaystyle (Equation\; 6.9.5)
 * }
 * }
 * {| style="width:100%" border="0" align="left"

\displaystyle U(x)=k_1+\int Z(x)dx $$ $$
 *
$$\displaystyle (Equation\; 6.9.6)
 * }
 * }

For $$ \displaystyle \begin{align} \displaystyle u_1(x)&=P_1(x)=x \\ \displaystyle a_1(x)&=\frac{-2x}{1-x^2} \\ \displaystyle f(x)&=1 \end{align} $$

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

\displaystyle \begin{align} \displaystyle h(x)&=u_1^2(x)exp \left [ \int a_1(x)dx \right ] \\ \displaystyle &=x^2exp \left [ \log (1-x^2)\right ] \\ \displaystyle &=x^2 (1-x^2) \end{align} $$ $$
 *
$$\displaystyle (Equation\; 6.9.7)
 * }
 * }

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

\displaystyle \begin{align} \displaystyle Z(x)&=\frac{1}{x^2(1-x^2)}\left [ k_2+\int \frac{x^2(1-x^2)}{x}dx \right ] \\ \displaystyle &=\frac{1}{x^2(1-x^2)}\left [ k_2+\frac{1}{4}x^2(2-x^2)\right ] \end{align} $$ $$
 *
$$\displaystyle (Equation\; 6.9.8)
 * }
 * }

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

\displaystyle \begin{align} \displaystyle U(x)&=k_1+\int \frac{1}{x^2(1-x^2)}\left [ k_2+\frac{1}{4}x^2 (2-x^2) \right ]dx \\ \displaystyle &= k_1+k_2\left ( \frac{1}{2}\log \left ( \frac{1+x}{1-x}\right ) - \frac{1}{x} \right ) + \frac{1}{8} \left ( 2x+ \log \left ( \frac{1+x}{1-x} \right ) \right ) \end{align} $$ $$
 *
$$\displaystyle (Equation\; 6.9.9)
 * }
 * }

Final solution:

= References =

= Team Work Distribution =

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title: User:Egm6321.f11.team1/report6


text: File:Battery Image 2.jpg

license: Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/

title: File:Battery Image 2.jpg


text:
Orienteering/Videos/Preparation

An orienteering sign at a road junction. A ‘sporty’ kind of car approaches (eg Subaru or other 4WD), slows at the sign, and turns down the road, disappearing into the distance.

PRESENTER (V/O)

I’m on my way to my first ever orienteering competition. Orienteering is a unique sport that involves both physical and mental challenges. Kind of like doing a crossword while you’re on a treadmill.

The car pulls up alongside an official handing out event information.

OFFICIAL

Welcome to Summer Series [or whatever the event is called]. Have you been orienteering before?

PRESENTER

No, this is my first time.

OFFICIAL

Well, here’s some information about the competition today. After you’ve parked your car, go to the registration tent and someone will help you out there.

PRESENTER

Thanks.

She drives on, passing several groups of elite and/or junior orienteers, wearing full orienteering kit.

PRESENTER (V/O)

Sounds pretty simple. But I can’t help noticing that all these people seem to be wearing some pretty serious racing gear. And all I’ve brought with me is a pair of sneakers.

She parks the car and climbs out, surveying the scene. A couple of orienteers jog past, carrying dibbers and compasses.

PRESENTER (V/O) (cont’d)

And what are those funny little things they’re carrying? Think I’d better ask someone.

CUT TO Presenter approaching an elite orienteer who is chatting to a couple of others.

PRESENTER

Excuse me, you look like a pretty serious competitor. Could I ask you a few questions?

ELITE

Sure.

He moves away from the others.

PRESENTER

So what with the get up?

He laughs. Ad-lib general chit-chat, pointing to different items of clothing etc. no sound:

PRESENTER (V/O)

I’m in luck. It turns out that this is [name], an elite orienteer who has competed in many national and international competitions.

CUT TO racing footage of Elite (or other Australian elites) competing internationally or in recent national events.

PRESENTER (V/O) (cont’d)

Top orienteers use specialized orienteering clothing and shoes designed for the rough terrain and thick forest that competitors often have to run through.

Back on Elite and Presenter.

PRESENTER

But I don’t need to wear anything as fancy as this?

ELITE (ad-lib, along the lines of…)

Definitely not. Just light comfortable clothes, like you’d wear on an easy bush walk.

PRESENTER

What about specialized equipment? Do I need any of that?

ELITE (ad-lib…)

Well, most orienteers use a compass.

(CLOSE UP of his compass)

This is a thumb compass, but when they first start out, most people use a base plate compass, which is more like a traditional bushwalking compass.

PRESENTER

What if I don’t have any compass?

ELITE

You can always hire them from the registration tent at the competition. Come on, I’ll take you over there.

As they walk over towards the registration tent –

PRESENTER (V/O)

Depending on the competition, I might also need an electronic timing chip that records whether I’ve been at the check point, or ‘control’.

Footage of elite competitors punching a control with a dibber (CLOSE UP if possible)

PRESENTER (V/O)

Some competitions use these chips to record competitors’ times at every control, but others use them just for the start and finish, to record competitors’ overall time. Today I’ll also need a special card that I mark with a punch, which imprints a unique pattern on the card at each control.

Footage of someone (possibly younger juniors or a family group) punching a control with a control card.

The Presenter comes out of the registration tent, carrying a control card, map, dibber and compass.

PRESENTER (to camera)

I’ve just paid my entry fee and now I’ve got everything I need - my compass (holds up), my control card (holds up), my timing chip (holds up) and I’ve even got a map (holds up). I’ve also got a control description, which will tell me what kind of feature I’m looking for on the map. The next thing I need to decide is how long - and how difficult - I want my orienteering course to be. But first, I need to get my racing shoes on.

She heads towards the car.

title: Orienteering/Videos/Preparation


text:
Orienteering/Videos/Arriving

The Presenter opens up the boot of her car and pulls out a pair of trail running shoes. She is already wearing shorts (or track pants) and a sporty t-shirt and/or jacket. Sitting on the edge of the boot, she proceeds to take off her casual shoes and pull on the trail running shoes.

PRESENTER (to camera)

I’m here at my very first orienteering event. I’ve got my map, my compass, my control description, my timing chip and the special control card which I’ll need to take on my orienteering course today.

Close-up of map, compass and card in a pile next to the presenter.

PRESENTER (to camera)

In an orienteering race, the aim is to get through the checkpoints – called ‘controls’ – as quickly as possible. And for a lot of people, that means running. I’m not sure how much running I’ll be doing today - that will depend on just how difficult my course is.

CUT TO the presenter chatting to an official (novice instructor) at the registration table [perhaps with a ‘Need help? Ask me’ or ‘Helper’ sticker or t-shirt].

PRESENTER

So just how hard are these courses?

INSTRUCTOR (ad-lib along the lines of…)

Well, the ones with the lowest numbers are very tricky, and they’re usually the longest too. As you go down through the numbers, the courses get shorter but not necessarily easier. At every competition there are also courses with moderate navigation for people with some experience and courses with very easy navigation for beginners.

PRESENTER

So what course would you suggest that I do?

INSTRUCTOR

I’d suggest you do the easy navigation. Course 8 [or whatever].

The Instructor hands over the correct registration card.

PRESENTER

How far is that?

INSTRUCTOR

About two kilometers.

PRESENTER

And how long will that take me?

INSTRUCTOR

(laughs)

We’ll just have to wait and see.

PRESENTER

So I just fill in this card now?

INSTRUCTOR

That’s right. Put your name here, course here…[etc.]

The Presenter leans on the table and starts filling in the details. Close up of registration card, filling in the correct course no. etc. The Presenter looks up as She gets to the ‘Car Registration’ part.

PRESENTER (V/O)

Car registration? Is that in case I get a parking ticket?

INSTRUCTOR (ad lib…)

No, it’s so we can see if you’re still here at the end of the day. Sometimes people forget to hand in their control card after they’ve finished their course, and we need to know if we should send out a search party.

PRESENTER

That sounds a bit ominous. Should I take my GPS?

INSTRUCTOR

Actually, that’s not allowed.

PRESENTER

No GPS?

INSTRUCTOR

Not a GPS that gives you your position, no. Just your map and your compass.

PRESENTER

And my superior sense of direction.

INSTRUCTOR

(laughs)

Any more questions?

PRESENTER

Just one. Which way to the start?

Silently, the Instructor points to a large, un-missable START sign and arrow, pointing along a track.

PRESENTER

Ah.

She starts to jog off in that direction.

title: Orienteering/Videos/Arriving


text:
Orienteering/Videos/Start procedure

The Presenter approaches the start area, where a number of orienteers are waiting for their starts.

PRESENTER (V/O)

Orienteering is a sport that involves physical fitness and navigational challenges. I’m on my way to the start of my first race, where I’ll set off on my course at my pre-allocated start time. Like a cycling time trial, most orienteering races are run individually, against the clock. Whoever has the fastest time at the end of the day, wins.

The presenter spots the Elite (script 1) near the start area, who waves ‘Hi’. Then presenter moves over to him.

PRESENTER (V/O)

I’ve just spotted [Name], an elite Australian orienteer. Let’s see what he does to prepare for the start of a race.

PRESENTER (to Elite)

Hi [name], what course are you running today?

ELITE

I’m running Course 1

PRESENTER

And how long will that take you?

ELITE

Hopefully about 40 [or whatever] minutes if I have a good run.

PRESENTER

I’m running a course 8. How long do you think that will take me?

ELITE (laughs)

Hard to say.

PRESENTER

Why does everyone laugh when I ask that?

(more laughter)

So is there anything special you do to prepare for the start of a race?

ELITE (ad-lib, along the lines of…)

Make sure I get there in plenty of time! I also make sure I remember to clear and check.

PRESENTER

Clear and check? What’s that?

ELITE

I’ll show you.

They walk a short distance to the clear and check boxes.

ELITE

Every race, you need to clear the information from your previous race from the electronic chip in your stick. This box clears the stick [CLOSE UP of Presenter clearing] and this box checks that the chip is empty [CLOSE UP of presenter checking].

PRESENTER

Anything else I should do?

ELITE (ad-lib…)

I also make sure I’ve had a good warm up and that I’ve taped my shoelaces...

PRESENTER

You shoelaces? I never would have thought of that!

Over general chit-chat, no sound, looking at his shoes etc…

PRESENTER (V/O)

There are actually several of different types of starts in orienteering. As well as individual start, like today -

(footage of normal start in a big competition)

- there are also ‘mass starts’ that take place for relay events, where everyone starts together.

(footage of relay starts)

Back to Presenter and Elite.

PRESENTER

So what’s your favourite type of start?

ELITE (ad-lib)

[whatever it is]

STARTER

Seven o’clock starters can go, seven o-six starters to the map.

PRESENTER

That’s my start.

ELITE

Good luck!

The Presenter heads to the map.

PRESENTER (V/O)

Because this is a small local competition, today I have to draw my own course on the map. But in bigger competitions, the maps are pre-marked and competitors don’t see them until the moment they start. (if available, more footage of start procedure at championship event, focused on the competitors picking up maps)

CLOSE UP of presenter marking course on the map – she is just about done and we see her completing the task.

PRESENTER (spoken while marking on course)

I have six minutes to carefully mark my course onto my map. I want to get it exactly right, because the control flag that I will be looking for is located in the centre of these circles. The start is marked with a triangle, here –

(points to triangle)

And the finish is marked with the double circle, here –

(points to double circle).

CUT TO presenter and a few other competitors standing near the starter.

PRESENTER (to camera)

One really important thing I need to take with me on the course today is this little piece of paper (she holds it up) called a control description. Some people have special holders for these –

(CLOSE UP of a control description holder on another competitor’s arm)

But I have just taped mine to the front of my map.

(CLOSE UP)

Not only does it tell me what kind of feature I’m looking for on the map, it also tells me the unique code that I should see on the control that will tell me it’s the one I’m looking for.

STARTER

Seven o-six starters can go, seven twelve to the map.

The competitors all start, dibbing at the start control. The Presenter dibs last.

PRESENTER (V/O)

Once my chip has registered a time at this start control – I’m officially on my way.

She places the compass on the map and turns around, trying to orientate the map.

PRESENTER (V/O)

Now how did this thing work, again?

title: Orienteering/Videos/Start procedure


text:
Orienteering/Videos/Map skills

The Presenter stands at the start triangle, staring at her map, while other orienteers (a variety of ages, from juniors to super veterans) run off purposefully toward their first controls [suggest this can be in 2X or 4X speed as the Presenter studies her map with a furrowed brow].

PRESENTER (V/O)

I’ve just started my first orienteering event. Everyone around me seems to know exactly where they’re going, but I haven’t really got a clue. All I know is that I am at this little triangle on the map.

CLOSE UP of map, thumb close to the start triangle. The presenter looks up.

PRESENTER (to camera)

I think I am going to need some help.

She clicks her fingers. The Novice Instructor appears magically in the frame (JUMPT CUT - a la ‘The Goodies’).

INSTRUCTOR

Need some help?

PRESENTER

Thought you’d never ask.

CLOSE UP of the map.

PRESENTER

So what am I actually looking for?

INSTRUCTOR

You’re looking for the feature in the centre of that circle. Which in this case is a track junction [or whatever].

PRESENTER

And that’s my first control?

INSTRUCTOR

Yes.

PRESENTER

And I have to do them in order?

INSTRUCTOR (ad lib, along the lines of…)

In this kind of event, yes. There are other events (eg. Street O, rogaines) where you don’t get the controls in any set order. But today, you need to.

PRESENTER

(CLOSE UP of map)

That’s a shame, because if I just got this one before this one and this one…

INSTRUCTOR

Then you’d be disqualified.

PRESENTER

Ah.

CUT TO footage of elite orienteers running through the forest INTERCUT with SCREEN SHOT of GPS tracking on a map.

PRESENTER (V/O)

An orienteering map is a highly specialized piece of cartography. Working from detailed aerial photographs, orienteering map makers spend hours doing fieldwork that involves marking on tiny details – such individual boulders and creek beds - that orienteers use to navigate through the terrain.

Back to Presenter and Instructor.

PRESENTER

So how do I know what these symbols on the map represent in the forest?

INSTRUCTOR (ad-lib…)

Each map has a legend [CLOSE UP] that shows you what each symbol on the map means. The more you orienteer, the more you learn to recognize what each symbol represents in the forest.

PRESENTER

And what about these brown squiggly lines?

INSTRUCTOR

Those are contours.

PRESENTER

Which means I’m going up or down, right?

INSTRUCTOR

Usually up.

INSTRUCTOR

(laughs)

Right.

Over ad-lib chit chat, where Instructor points at things on the map, then at features in the terrain (no sound) –

PRESENTER (V/O)

So now I know where I have to get to and what I’m looking for when I get there. The next question is – which direction do I go in?

title: Orienteering/Videos/Map skills


text:
Orienteering/Videos/Compass skills

CONTINUOUS from script 4 Map skills - Over ad-lib chit chat (no sound), the Instructor points at things on the map, then at features in the terrain while the Presenter asks questions etc.

PRESENTER (V/O)

I’ve just started my first orienteering race. I’ve been studying the map and with the help of my expert orienteering instructor, and I’ve learned what all the navigation features in the forest look like on the map. Now I’m going to learn how to use my compass.

MID SHOT of the Presenter, addressing the instructor.

PRESENTER (to Instructor)

So my first question is, do I really need a compass?

INSTRUCTOR (ad-lib, along the lines of…)

You don’t have to use one, but they’re very, very useful. There is a former world champion from Finland who doesn’t use one, but he’s definitely the exception. Most orienteers use them.

PRESENTER

Well, luckily I have one right here.

She holds it up. The Instructor takes it and places it on the map.

INSTRUCTOR (ad-lib, along the lines of…)

So this type of compass is called a base plate compass, and the first thing you need to do is lie it flat on the map, like so (CLOSE UP). That allows you to easily orientate the map.

PRESENTER

That’s lining the compass up with the north arrows, right?

INSTRUCTOR

That’s right. Once you’ve done that, you can turn your body so you’re facing the way you want to go.

WIDE SHOT of Presenter turning, facing in the right direction.

PRESENTER

(pointing ahead)

So that’s that way, right?

INSTRUCTOR

Right. And if you look at the symbol on the map…

CLOSE UP of map.

PRESENTER

That’s a track, right?

INSTRUCTOR

And on the ground we have…

PRESENTER

A track. So we just follow the track in this direction, right?

INSTRUCTOR

Very good.

PRESENTER

Shall we go?

The camera follows as they walk down the track.

PRESENTER (V/O)

As my elite orienteering friend told me earlier, there are actually several different types of compass.

CUT TO footage of an elite using a thumb compass, close up then mid/long shot of reading the map on the run.

PRESENTER (V/O)

More experienced orienteers use a thumb compass, which attaches to the thumb and is held continually against the map so that it can be read easily on the run.

Back to Presenter and Instructor.

PRESENTER

So what exactly is “thumbing the map”?

INSTRUCTOR (ad-lib…)

Thumbing the map is when you follow your position on the map with your thumb, or with the edge of your compass. Thumb compasses are designed so that you can do this easily.

PRESENTER

Do I need to do that?

INSTRUCTOR (ad-lib…)

It’s a good habit to get into, but it’s not as vital when you are following a linear feature like a track or a fence and you don’t need to know exactly where you are all the time. Thumbing the map is important when you are running in the forest and you need to pick up individual features to let you know where you are.

As they walk along the track, an elite orienteer crosses the track in front of them, map folded and thumbing the map.

PRESENTER

Wow, that was fast.

They watch the orienteer disappear into the forest.

PRESENTER

I couldn’t help noticing he seemed to have a very small map.

INSTRUCTOR

(laughs)

It’s the same map we have, folded up into a smaller size.

PRESENTER

Is that because it’s easier to carry?

INSTRUCTOR (ad-lib…)

No, it’s to make it easier to thumb the map and to make sure the north line of your compass is lined up with the north line on the map when your change direction.

PRESENTER

Can I try that?

INSTRUCTOR

Sure.

Ad-lib Instructor showing/describing how to fold the map correctly, lining up edge of map with north lines etc.

PRESENTER

So when we go around this bend in the track, I’ll still be able to line up the north lines on my compass and my map and have the map pointing in the right direction?

INSTRUCTOR

Exactly.

They continue down the track.

PRESENTER (V/O)

I think I’m getting the hang of this. But I’ll be feeling a lot more confident after I’ve found my first control.

title: Orienteering/Videos/Compass skills


text:
Orienteering/Videos/On the ground

The Presenter and the Instructor walk down the track, with the Presenter thumbing the map as they pass a boulder [or some other feature] on the side of the track. She points it out to the Instructor, who nods.

PRESENTER (V/O)

I’ve just started my first orienteering event. I’ve learned how to align my map with my compass so that I’m heading in the right direction, and how to thumb my map. Now I’m looking for my first checkpoint, which is known as a ‘control’.

PRESENTER (to Instructor)

So what exactly am I looking for?

INSTRUCTOR (ad-lib…)

You’re looking for an orange and white flag on a stand. It should be placed clearly at the feature you’re looking for, so that when you’re at the feature, you can see the control.

PRESENTER

So the flags aren’t hidden behind bushes or anything?

INSTRUCTOR (ad-lib…)

That sometimes happens, but it’s discouraged. Especially on an easy navigation course, when you get close to the centre of your circle, you should be able to see the control.

PRESENTER

And how do I know what I’m looking for?

INSTRUCTOR

You can usually tell just from looking at the map –

(CLOSE UP of the map)

But it also will tell you on your control description.

PRESENTER

That’s this thing here?

CLOSE UP of Presenter pointing the control description [in English]

INSTRUCTOR

Right. So your first control is on a track junction.

PRESENTER

And what’s this number here?

INSTRUCTOR (ad lib…)

There’s a unique number on each control stand. If it’s the same number as the number on your control description, you’re in the right place.

PRESENTER

And if it’s not?

INSTRUCTOR

That probably won’t happen today.

They walk on…

PRESENTER (V/O)

Probably? That doesn’t exactly fill me with confidence.

...as a young girl goes running past.

PRESENTER (V/O)

At least she looks like she knows where she’s going.

They look ahead to where the kid is running. There’s the control!

PRESENTER

The control!

(to Instructor)

Mind if I run from here?

INSTRUCTOR

Sure. You don’t really need me any more.

And with a ‘click’, the Instructor disappears into a JUMP CUT.

PRESENTER (to camera)

Looks like I’m on my own.

(calling out to kid)

Wait for me!

She runs toward the control, where the kid punches quickly then runs off to her next control.

CLOSE UP of control, as Presenter runs into frame and punches the control, immediately heading off down the track a few meters…

PRESENTER (V/O)

My first control. And it’s even the right number.

…before stopping abruptly, then scratching her head.

PRESENTER (V/O)

But before I go rushing off to my next control, I probably need to work out how I’m going to get there.

She looks at the map, frowning.

PRESENTER (to self/camera)

That kid made it look so easy.

CUT TO footage of elite orienteers showing control flow.

PRESENTER (V/O)

Experienced orienteers plan which way they are going to leave the control circle, often before they’ve even found the control in the middle of it. This is called ‘control flow’ and it means exactly that: Flowing smoothly through the control and leaving the control circle in the right direction, so that you’ve set yourself up the quickest and cleanest possible route to the next control. It’s not uncommon to see elite orienteers move so quickly past a control that you’re not even sure it they’ve punched it!

Back on Presenter, studying her map.

PRESENTER (to camera)

But when you’re a beginner, it pays to take your time and really know where you’re going before you go anywhere. Now that I’ve lined up the north arrow on my compass with the lines on my map, worked out what feature I’m looking for and what I should follow to get there, and pointed myself in the right direction, I’m ready to set off to find my second control.

She sets off at right angles to the track, along a fence/other track.

PRESENTER (V/O)

Which happens to be at right angles to the direction I was just headed.

title: Orienteering/Videos/On the ground


text:
Orienteering/Videos/Control flow

The Presenter jogs across a clearing towards a control on a distinctive tree [or whatever].

PRESENTER (V/O)

This is my second orienteering competition and I’m halfway through a course with medium level navigation. I’ve learned how to plan my route, made my first navigational mistake and – luckily - learned how to relocate. I think I’m doing pretty well. But I can’t help noticing that just about everyone else is moving smoothly through their controls, without really stopping.

As she punches the control, two orienteers – a young junior and a veteran – pass through the control in quick succession.

PRESENTER (V/O)

I think I’m going to need a little help on this one.

She clicks her fingers and her Elite friend appears.

PRESENTER

Thanks for stopping by.

ELITE

No problem. How can I help?

PRESENTER

I’ve noticed that everyone seems to be able to pass through the control circle really quickly. Whereas I seem to spend a lot of time standing around.

ELITE (ad-lib, along the lines of…)

That’s called control flow, and it’s a good thing to work on even when you’re just learning. There are two parts to control flow. The first part of it is the approach. Always check your control descriptions before you’re in the circle to make sure you know what you’re looking for.

CLOSE UP of control description.

ELITE (ad-lib…) (cont’d)

You also need to check which side of the feature the control will be located on. You can check the number then too if you think you can remember it. If not, you can always quickly check at the control.

As the Elite and Presenter study her map and clue description (no sound) -

PRESENTER (V/O)

In this competition, I’m using control descriptions with international symbols.

SCREEN SHOT of a map, with a close up of a clue sheet with international descriptions.

PRESENTER (V/O) (cont’d)

Instead of describing the feature in English, international symbols denote individual features –

(circle)

The position of the feature relative to others nearby -

(circle)

And which side of the feature the control flag will be on. Other features, such as drinks and radio controls, are shown in the last column.

(circle both)

Back on Presenter and Elite.

PRESENTER (V/O)

So it’s just a matter of learning the symbols off by heart?

ELITE (ad-lib…)

Pretty much. But the symbols are often very similar to their symbol on the map, so you pick them up pretty quickly.

PRESENTER

Now you said there were two parts to control flow. What’s the second part?

ELITE (ad-lib…)

The second part is planning the direction you should leave the control – before you get there.

CUT TO footage of elites showing good control flow – approaching, slowing down and moving quickly away from the control.

ELITE (ad-lib…)

When you can see the control flag – or even the feature, if it’s obvious – you usually have at least a few seconds when you’re running towards it that you don’t need to navigate. So you can start looking at your map and planning the direction you need to head in after you’ve punched the control. You might even have time to plan your route.

Back on Presenter and Elite.

PRESENTER

Sounds pretty simple.

ELITE

(laughs)

Right.

PRESENTER

Race you to the next control?

ELITE

You’re on.

They both set off in slightly divergent directions.

title: Orienteering/Videos/Control flow


text:
Orienteering/Videos/Finish

The Presenter follows a fence at a fast walk, thumbing the map, a look of concentration on her face.

PRESENTER (V/O)

I’m on my way to the second control in my first orienteering race. Finding the first control was a breeze – although I did have a little help from my expert instructor. But now, I’m on my own.

The Presenter reaches a fence corner, stops and looks up.

PRESENTER (to camera)

I’ve just reached this fence corner –

(CLOSE UP of map)

So according to my calculations, my second control should be just down this track.

She starts to head down the track soon sees the control…just as a young boy races past her, punches the control, and takes off down an adjacent track.

CUT TO Presenter punching the control.

PRESENTER

Two down, nine to go.

After studying her map for a moment, she heads off in the same direction as the junior orienteer. The footage speeds up, so that the next eight controls are a high speed montage of an identical scenario: the presenter looks up from her map, sees the control and starts to head towards it, as some little kid (a different one every time) flies past, punches the control, and disappears into the distance.

PRESENTER (V/O)

As I complete the rest of my course, I can’t help but notice a recurring theme.

As the Presenter approaches the final control before the finish chute (normal speed), the final little kid flies past. This time the presenter narrows her eyes, and gives chase, not quite managing the catch the kid on the finish line. She dibs her e-stick (CLOSE UP) then bends over panting, while the kid heads over to his group of friends who have already finished.

PRESENTER (to camera)

Damn!

She heads toward the finish tent.

PRESENTER (V/O)

Now that I’ve finished, I need to hand my control card into the finish tent so that they can check that I’ve been to all the right controls. I also need to download the data from my electronic chip, which records my total time.

She downloads, then hands her finish card to an official.

OFFICIAL

How did you go?

PRESENTER

Pretty well, thank you, but I kept getting passed by ten year olds.

OFFICIAL

(laughs)

They’re pretty quick.

The Presenter moves away from the finish, heading towards the water containers.

PRESENTER (V/O)

So quick that I thought I’d go over and find out how they did it. But first, I’m going to take a well-earned drink.

CUT TO footage of different kinds of race finishes, at WOC if available and sprint finishes in relays. Electronic timing finishes if possible.

PRESENTER (V/O)

There are several different types of finish in an orienteering race. Where electronic timing is used, competitors often use their electronic chip on the finish line to get their final time. In many of these races, not only is the final time recorded, so is the time taken to get to each control. Although competitors are able to see who was the fastest on an individual leg, it’s still only the fasted overall time that counts.

Mid-shot of juniors discussing their course with the Presenter.

PRESENTER (to juniors) So which of you had the fastest time today?

FASTEST KID

I did!

PRESENTER

And how long did you take?

FASTEST KID

18 minutes [or whatever]. How long did you take?

PRESENTER

Ahh….I think it was just a little bit longer than that.

The kids laugh.

PRESENTER

So how did every one go to the first control?

Ad-lib extended discussion with kids, with kids pointing out route choice on maps.

PRESENTER (V/O)

So that was my first orienteering race. I learned how to orientate and thumb my map and how to recognize and translate different symbols on the map into features in the forest. I even completed my course without finding any of the wrong numbered controls. Next time, I think I’ll be ready for some more challenging navigation and maybe, with a little help from my new friends, I just might learn how to improve my control flow.

title: Orienteering/Videos/Finish


text:
Orienteering/Videos/Route choice

The Presenter’s 4WD approaches an orienteering sign on the road, slows at the sign, and turns down the road/track. This is similar to the opening of Video 1, but the venue is clearly different.

PRESENTER (V/O)

I’m on my way to my second orienteering competition. At my first competition I learned basic map and compass skills and completed a simple orienteering course entirely on my own – well, almost entirely on my own.

The car pulls up by an official who hands the Presenter a flyer, as per Video 1.

PRESENTER (to Official)

Thanks

(to camera, through car window)

Today I’m planning to do an orienteering course with more challenging navigation and hopefully learn a few more advanced orienteering skills.

CUT TO registration tent. The Presenter pays her money for the event entry and for hire of a compass and dibber. Ad-lib dialogue with Registration Official.

Presenter exits the Registration tent with compass, dibber, map and control description.

PRESENTER

I’m an old hand at this now.

CUT TO Presenter drawing up her map.

PRESENTER (to Camera)

Today I’ve chosen to do a course with moderate navigation. This means that not all the controls are on linear features, such as fences and tracks, and I’ll have to rely on my compass to navigate my way through the forest.

CLOSE UP of presenter marking her map, showing controls off tracks.

CUT TO Presenter waiting with a number of other competitors at the start.

PRESENTER (V/O)

My course is also a little bit longer today. I wonder what time it gets dark?

The starter announces the start time. The Presenter dibs last, then follows competitors down a marked route towards the start triangle.

PRESENTER (V/O)

Another difference today is that the official start point – where the triangle is marked on the map – isn’t actually where we started, but a hundred meters away down a marked route.

CUT TO Presenter arriving at the start triangle, where she stops.

PRESENTER (to camera)

During that short jog to the start triangle, everyone else worked out which way they were going to go to their first control. But I’m still not sure.

SCREEN SHOT of GPS tracking, showing three different route choices on a long leg.

PRESENTER (V/O)

Deciding which way you are going to go to a control is one of the most important components of competitive orienteering. It’s called route choice, and the fastest route choice for one competitor may not be the fastest for another.

Back on Presenter, still at the start triangle.

PRESENTER (to camera)

Since this is only my second event, I thought I’d enlist some expert help.

She clicks her fingers and the Elite (from Video 1 & 3) appears (JUMP CUT).

PRESENTER

Hi [name]. I see you’ve brought your own map.

ELITE

Of course.

PRESENTER (to [name])

That’s great, because I thought you might be able to help me out deciding on a route choice to my first control.

ELITE

Sure.

CLOSE UP of her map.

PRESENTER

Which way would you go?

ELITE (ad-lib, along the lines of…)

Well, I’d go this way…

(points to route choice – CLOSE UP)

But if I wasn’t very confident, I’d go this way.

(points to safer route choice – CLOSE UP).

PRESENTER

Why is that?

ELITE (ad-lib…)

When you’re just starting out, it pays to pick the safest route choice, which means the route choice that will make it easiest for you to find the control.

PRESENTER

But not necessarily the fastest?

ELITE (ad-lib…)

No, but you often can’t tell that anyway until you’ve run the leg and compared your time to other people.

PRESENTER

Which you can do because of the electronic timing at each control?

ELITE (ad-lib…)

Right. But since we’re using control cards today, we could set off together, do our own route choice and meet at the control.

PRESENTER

Sounds like a good idea. But I hope you won’t be waiting too long!

ELITE

I’ll take it easy. See you at the control.

He speeds off, leaving the Presenter staring after him for a moment before setting off at a jog on her own route.

PRESENTER (V/O)

That doesn’t look like taking it easy…

INTERCUT Elite running his route choice, the Presenter running hers.

CUT TO Presenter jogging up to the control, where the Elite is waiting.

PRESENTER

You look like you’ve been here for hours.

ELITE

Only three minutes.

PRESENTER

So you your route choice was three minutes quicker than mine?

ELITE (ad-lib…)

Not taking into account running speed, yes. But you found the control without making a navigational mistake, and that’s saves you the most time in the long run.

PRESENTER

And what happens if I do make a navigational mistake?

ELITE

(laughs)

Depends on how lost you get!

PRESENTER

Shall we do the next control?

MID SHOT of Presenter and Elite planning the route choice to the second control (no sound), then running off in separate directions -

Presenter (V/O)

Making a navigational mistake or – worse - getting lost is something I really don’t want to think about. But I have a feeling that it’s that I’m going to have to confront sooner rather than later.

title: Orienteering/Videos/Route choice


text:
Orienteering/Videos/Relocation

The Presenter runs through the forest, looking at her map. She slows down, looks around then finally comes to a stop. Turning around slowly, she scratches her head.

PRESENTER (V/O)

I’m competing in my second orienteering event, on a course that requires moderate navigational skills. I’ve enlisted the help of [name], one of Australia’s top orienteers, to help me choose the best route choice, but I am beginning to suspect that I might have just come unstuck.

CLOSE UP of Presenter, peering at the map.

PRESENTER (to camera)

I’m on my way from control number three (points to control on map) to control number four. I thought I left the track at this bend here (points to the bend) and now I’m at the place I thought my control should be. But there’s nothing here.

PAN around the forest, where there is indeed no control to be seen.

PRESENTER (to camera)

I think I’m going to need a little help on this one.

She clicks her fingers, and the Novice Instructor (from scripts 1, 4) appears (JUMP CUT)

INSTRUCTOR

Need some help?

PRESENTER

I do, as a matter of fact. I’ve just come to where I think my control ought to be, and it’s not here. And if it’s not here, I haven’t got a clue where I am.

INSTRUCTOR

Can I have a look at your map?

PRESENTER

Sure.

She shows the Instructor her map, and the two of them study it, looking around, pointing etc. (no sound)

SCREEN SHOT of a map and route, with a route choice line going off on a tangent half way to the next control.

PRESENTER (V/O)

Working out where you are after you’ve made a navigational mistake is called relocating. There are many different ways to relocate. For example, you can stay where you are -

(line stops, ‘blinks’ in one place)

While you try and recognize some features in the surrounding forest that you can match with the symbols on the map. Or, you can go back to the last place where you definitely knew you were -

(line goes back to track bend, or other major feature)

Or you can run around like a headless chook.

(line goes around and around in circles)

But that’s not really recommended.

Back on Presenter and Instructor.

PRESENTER

What if I’m really, really lost?

INSTRUCTOR (ad-lib…)

If you’re so lost that you are pretty sure you can’t relocate, you should take a safety bearing. This means heading in a set direction – such as due north – until you hit something major, like a road. The direction of the safety bearing will usually be written on your map, or it may be on the control description.

PRESENTER

Right. Well I don’t think I’m totally lost. But I’m still not sure where I am. Where do you think we are?

INSTRUCTOR

Where do you think we are?

PRESENTER (laughs)

Well, I thought I left the track at this bend here, and followed my compass a hundred meters into the forest. My control is on a boulder [or whatever] but there’s not one here!

INSTRUCTOR (looking around)

So what can you see here?

PRESENTER

Well I can see what looks like a fence corner [or whatever] over there.

INSTRUCTOR

And can you see something like that on the map?

CLOSE UP of map.

PRESENTER

I can see one here. But that’s further back towards my last control.

INSTRUCTOR

So what do you think might have happened?

PRESENTER

Hmm. Well I was looking for a track bend. Maybe I came off one track bend too early.

INSTRUCTOR

Shall we go back and have a look?

They head back out towards the track.

SCREEN SHOT of GPS footage from an elite race showing big navigational error/s.

PRESENTER (V/O)

Navigational mistakes happen to everyone – even top orienteers. Obviously, the best thing you can do is avoid a mistake altogether. But the next best thing you can do is relocate quickly.

Back to Presenter and Instructor, who are now standing on the track.

PRESENTER

Oh I see! So if I just go along this track another hundred meters, there’ll be another bend, and that’s the one I have to go in from. Great!

She starts to run off down the track.

PRESENTER (back over her shoulder)

Thanks!

INSTRUCTOR

You’re welcome.

S/he clicks fingers and is gone.

CUT TO Presenter following her map carefully in the forest. She looks up and sees the control.

PRESENTER

Brilliant!

She jogs over and punches the control. Stops, looks at her map…as another orienteer runs in, punches the control, and runs off again. The presenter watches her go, looks down at her map and sets off in the same direction.

PRESENTER (V/O)

Hmm. Now that I’ve learned to relocate, maybe it’s time l learned a thing or two about moving a little more quickly through the controls.

title: Orienteering/Videos/Relocation


text:
Orienteering/Videos/Etiquette

The Presenter approaches a large feature (eg. boulder, termie) slowing down to check her control description. Without stopping, she rounds the feature, confident that the flag will be on the other side.

PRESENTER (V/O)

I’ve nearly finished my second orienteering competition. For the last few controls I’ve been practicing my control flow – approaching the control smoothly and leaving in the right direction, while starting to plan my next route.

On the other side of the feature, she punches and quickly heads off toward her last control.

PRESENTER (V/O)

At my first competition, I learned that GPS navigational aids – like my iPhone - are not allowed in orienteering competitions. But there are a number of other basic rules that should also be observed.

CUT TO Presenter arriving at a marked fence crossing. She stops.

PRESENTER (to camera)

Orienteering competitions are often held on private land, and organizers need the permission of landowners in order to gain access. It’s important to use marked fence crossings, such as this (CLOSE UP on map) that help to protect the landowners’ property.

She crosses the fence and heads off.

SCREEN SHOT of a map showing an out of bounds area.

PRESENTER (V/O)

Areas that are out of bounds on the map are shown with this symbol.

(circle)

Failing to observe an out-of-bounds area could jeopardize future use of the map, and anyone found to have crossed an out-of-bounds area will be disqualified from the competition.

CUT TO Presenter jogging to her final control, running up the finish chute and dibbing on the finish line, where she gets a few cheers from both the Elite and Instructor.

PRESENTER (to camera)

My second orienteering event - finished! And it’s not even dark yet.

She heads to the tent to download.

CUT TO Presenter shaking hands with the Elite and the Instructor at the drinks, who hand her a cup of water.

PRESENTER

So I have to ask you, are there any other rules that I should know about?

ELITE

You should never grab someone’s map!

PRESENTER

Does that really happen?

ELITE (ad-lib…)

In bigger competitions, yes, when someone comes to a control that’s not theirs, and you’re punching, they might grab your map to work out where they are.

PRESENTER

I’d be tempted to punch them in that case.

INSTRUCTOR

(laughs)

That’s not allowed either.

PRESENTER

So no physical violence of any sort?

ELITE

Unfortunately, no.

PRESENTER

Anything else?

INSTRUCTOR

Well, you’re not supposed to talk to other competitors or ask for help, unless you’re seriously injured.

PRESENTER

So strictly speaking I should be disqualified?

INSTRUCTOR (ad-lib…)

Well, no, there are different rules for novice orienteers and young kids often have their parents or another adult shadow them on a course, so if they do come unstuck there’s a safety net.

PRESENTER

And why if I am seriously injured?

INSTRUCTOR (ad-lib…)

We strongly recommend that you carry a whistle in your pocket, which you can blow on to attract attention. Anyone who hears a whistle blast – or a call for help – is required to abandon their course and assist you. And vice versa, of course.

PRESENTER

Right. Anything else?

INSTRUCTOR

No dogs.

PRESENTER

Really? I can’t bring my dog along for a run?

INSTRUCTOR (ad-lib…)

Generally no, particularly on private land or State forests. Very rarely dogs are OK, but only if it’s some kind of public park that allows dog, and the event information will generally specify that.

PRESENTER

OK, got it.

ELITE

No standing at the control and shouting ‘it’s over here!’

PRESENTER

People do that?

ELITE

More often that you’d think.

They continue to chat, their voices fading out over a long pan (or a montage) around the event, picking up competitors of all ages finishing, discussing courses, kids on a strong course etc.

title: Orienteering/Videos/Etiquette


text:
User:វ័ណថារិទ្ធ

 * [[Image:Internet-group-chat.svg|left|25px]] សូមផ្ញើសាររបស់អ្នក នៅទីនេះ ! Please leave your message here!

Greeting from Cambodia!

This is Vantharith. I am a Khmer Wikipedian, currently working on some other Khmer Wiki such as Wiktionary and Wikibook. I am also an user in English Wiki. Here is my user page link on Khmer Wikipedia វ័ណថារិទ្ធ.

Currently, I am hoping to extend Khmer Wiki into Wikisource as there are many works in Khmer language should be preserved and made available on the internet. As part of our work with some like-minded Cambodian Wikipedians, we have made some online presence as below:
 * Khmer Wikipedia Blog (in Khmer language only)
 * Khmer Wikipedia Twitter
 * Khmer Wikipedia and other sister projects' Fan Page on Facebook

As a result, we have increased awareness among Cambodian users and form them into volunteers teams to work on subject they are interested. This includes few WikiProjects as below:
 * WikiProject History
 * WikiProject Buddhism
 * WikiProject Computing

We still have a long journey to go, one step at a time! We are also looking for support from Wikimedia Foundation and other chapters/communities to training our members so that they can contribute more in creating a online, free encyclopedia in their own mother tongue. Thank you!

Kind regards,

title: User:វ័ណថារិទ្ធ


text:
User:Truongdat120891/enes100/codeing

int switchPin = 2; int led1Pin = 3; int led2Pin = 5; int led3Pin = 6;
 * 1) define FADESPEED 5

int val; int val2; int buttonState; int lightMode = 0;

void setup { pinMode(switchPin, INPUT);

pinMode(led1Pin, OUTPUT); pinMode(led2Pin, OUTPUT); pinMode(led3Pin, OUTPUT); Serial.begin(9600); buttonState = digitalRead(switchPin); }

void loop{ int r, g, b; val = digitalRead(switchPin); delay(10); val2 = digitalRead(switchPin); if (val == val2) { if (val != buttonState) { if (val == LOW) { if (lightMode == 0) { lightMode = 1; } else { if (lightMode == 1) { lightMode = 2; } else { if (lightMode == 2) { lightMode = 3; } else { if (lightMode == 3) { lightMode = 0; }			}         }        }      }    }    buttonState = val; }

if (lightMode == 0) { digitalWrite(led1Pin, LOW); digitalWrite(led2Pin, LOW); digitalWrite(led3Pin, LOW); }

if (lightMode == 1) { delay(500);

{ for (r = 0; r < 256; r++) { analogWrite(led1Pin, r); delay(FADESPEED); }

for (b = 255; b > 0; b--) { analogWrite(led2Pin, b); delay(FADESPEED); }

for (g = 0; g < 256; g++) { analogWrite(led3Pin, g); delay(FADESPEED); }

for (r = 255; r > 0; r--) { analogWrite(led1Pin, r); delay(FADESPEED); }

for (b = 0; b < 256; b++) { analogWrite(led2Pin, b); delay(FADESPEED); }

for (g = 255; g > 0; g--) { analogWrite(led3Pin, g); delay(FADESPEED); } }

if (lightMode == 2)

{    delay(500); digitalWrite(led1Pin, LOW); digitalWrite(led2Pin, HIGH); digitalWrite(led3Pin, HIGH); delay(500); digitalWrite(led1Pin, LOW); digitalWrite(led2Pin, HIGH); digitalWrite(led3Pin, LOW); delay(500); digitalWrite(led1Pin, HIGH); digitalWrite(led2Pin, HIGH); digitalWrite(led3Pin, LOW); delay(500); digitalWrite(led1Pin, HIGH); digitalWrite(led2Pin, LOW); digitalWrite(led3Pin, LOW); delay(500); digitalWrite(led1Pin, LOW); digitalWrite(led2Pin, LOW); digitalWrite(led3Pin, HIGH); delay(500); digitalWrite(led1Pin, HIGH); digitalWrite(led2Pin, LOW); digitalWrite(led3Pin, HIGH); } if (lightMode == 3) { delay(100); digitalWrite(led1Pin, LOW); digitalWrite(led2Pin, HIGH); digitalWrite(led3Pin, HIGH); delay(100); digitalWrite(led1Pin, LOW); digitalWrite(led2Pin, HIGH); digitalWrite(led3Pin, LOW); delay(100); digitalWrite(led1Pin, HIGH); digitalWrite(led2Pin, HIGH); digitalWrite(led3Pin, LOW); delay(100); digitalWrite(led1Pin, HIGH); digitalWrite(led2Pin, LOW); digitalWrite(led3Pin, LOW); delay(100); digitalWrite(led1Pin, LOW); digitalWrite(led2Pin, LOW); digitalWrite(led3Pin, HIGH); delay(100); digitalWrite(led1Pin, HIGH); digitalWrite(led2Pin, LOW); digitalWrite(led3Pin, HIGH); }  } }

title: User:Truongdat120891/enes100/codeing


text:
User:Truongdat120891/enes100/code2

int switchPin = 2; int led1Pin = 3; int led2Pin = 5; int led3Pin = 6;
 * 1) define FADESPEED 5

int val; int val2; int buttonState; int lightMode = 0;

void setup { pinMode(switchPin, INPUT);

pinMode(led1Pin, OUTPUT); pinMode(led2Pin, OUTPUT); pinMode(led3Pin, OUTPUT); Serial.begin(9600); buttonState = digitalRead(switchPin); }

void loop{ int r, g, b; val = digitalRead(switchPin); delay(10); val2 = digitalRead(switchPin); if (val == val2) { if (val != buttonState) { if (val == LOW) { if (lightMode == 0) { lightMode = 1; } else { if (lightMode == 1) { lightMode = 2; } else { if (lightMode == 2) { lightMode = 3; } else { if (lightMode == 3) { lightMode = 0; }			}         }        }      }    }    buttonState = val; }

if (lightMode == 0) { digitalWrite(led1Pin, LOW); digitalWrite(led2Pin, LOW); digitalWrite(led3Pin, LOW); }

if (lightMode == 1) { delay(100); digitalWrite(led1Pin, LOW); digitalWrite(led2Pin, HIGH); digitalWrite(led3Pin, HIGH); delay(100); digitalWrite(led1Pin, LOW); digitalWrite(led2Pin, HIGH); digitalWrite(led3Pin, LOW); delay(100); digitalWrite(led1Pin, HIGH); digitalWrite(led2Pin, HIGH); digitalWrite(led3Pin, LOW); delay(100); digitalWrite(led1Pin, HIGH); digitalWrite(led2Pin, LOW); digitalWrite(led3Pin, LOW); delay(100); digitalWrite(led1Pin, LOW); digitalWrite(led2Pin, LOW); digitalWrite(led3Pin, HIGH); delay(100); digitalWrite(led1Pin, HIGH); digitalWrite(led2Pin, LOW); digitalWrite(led3Pin, HIGH); }

if (lightMode == 2)

{    delay(25); digitalWrite(led1Pin, LOW); digitalWrite(led2Pin, HIGH); digitalWrite(led3Pin, HIGH); delay(25); digitalWrite(led1Pin, LOW); digitalWrite(led2Pin, HIGH); digitalWrite(led3Pin, LOW); delay(25); digitalWrite(led1Pin, HIGH); digitalWrite(led2Pin, HIGH); digitalWrite(led3Pin, LOW); delay(25); digitalWrite(led1Pin, HIGH); digitalWrite(led2Pin, LOW); digitalWrite(led3Pin, LOW); delay(25); digitalWrite(led1Pin, LOW); digitalWrite(led2Pin, LOW); digitalWrite(led3Pin, HIGH); delay(25); digitalWrite(led1Pin, HIGH); digitalWrite(led2Pin, LOW); digitalWrite(led3Pin, HIGH); } if (lightMode == 3) { delay(100); digitalWrite(led1Pin, LOW); digitalWrite(led2Pin, HIGH); digitalWrite(led3Pin, HIGH); delay(100); digitalWrite(led1Pin, LOW); digitalWrite(led2Pin, HIGH); digitalWrite(led3Pin, LOW); delay(100); digitalWrite(led1Pin, HIGH); digitalWrite(led2Pin, HIGH); digitalWrite(led3Pin, LOW); delay(100); digitalWrite(led1Pin, HIGH); digitalWrite(led2Pin, LOW); digitalWrite(led3Pin, LOW); delay(100); digitalWrite(led1Pin, LOW); digitalWrite(led2Pin, LOW); digitalWrite(led3Pin, HIGH); delay(100); digitalWrite(led1Pin, HIGH); digitalWrite(led2Pin, LOW); digitalWrite(led3Pin, HIGH); }  }

title: User:Truongdat120891/enes100/code2


text:
User:U3036219/Poker Machine Reform – The Effects on the Landscape of Australian Sport

Not Yet Completed - From Meeting Friday 1:00pm /BPS2011 Essay/

Title
Poker Machine Reform – The Effects on the Landscape of Australian Sport

Outline
This essay will be focussing on the Poker machine reforms that have been introduced to Parliament and how they will affect the sporting landscape within Australia if they are passed. It will look at the reforms and how they will affect sport at a National, State and Local level. The introduction of the reforms looks set to affect many sporting teams and individuals by lowering the amount of revenue that a club will be able to receive through its parent clubs, where less profits will be made which will in turn, mean that less funding will be available to support local and junior sport in different regions around Australia. This will have a detrimental effect on the communities that these sporting clubs are a part of, as clubs may need to pull out of their involvement in sport competitions and reduce the number of teams that they provide with funding and equipment. The paper will look at the issues at a National level, as well as an insight into the local Canberra region, where sport is a very strong part in the culture of the city. The essay will also incorporate how business and politics have interacted with the sporting landscape of Australia and how they may have affected the outlook on sport in Australia.

Introduction
The Australian Sporting Landscape has always offered a rich and diverse range of sports and events that cater to many different playing levels and age groups. From professional athletes right the way down to grass-roots juniors who are beginning to understand the fun and enjoyment that can be developed from sport, it has always provided Australians with a sense of identity to be successful. Australians thrive on the traditions and values that have been instilled into our culture from sport throughout the ages. From Bradman and Phar Lap during the Depression, through to the success of today’s current sporting teams and individuals, Australia has been recognised globally as a nation who performs strongly at a global scale. This dominance and success, however, could be affected in the future if the laws on poker machines are passed by the Federal Government. The ramifications could be widespread, through the national sporting organisations, filtering down into damaging sport in local communities.

Business and politics form strong bond and associations with sport, whether that is through funding and sponsorship, or through the establishment of the Australian Sports Commission (ASC) to help fund each sport. Without governance or structure of sporting bodies, then they would not be financially successful or viable, so strong administration of business is needed to compliment the success.

Issues at a National Level
The issues that will be defined at a national level will have a significant effect on sport in Australia. The business of sport is intrinsically intertwined in the culture of Australians, and to allow the reforms to be implemented will have a serious effect on sport competitions that run at a National level. Leagues such as the AFL and NRL will be directly affected, because revenues will decrease as a result of potentially less patrons using Leagues Club facilities to enjoy watching their team play. Dupree, (2011), found that a club in Adelaide believes that if the reforms are passed, that the club will lose up between 30-40% of its revenues that they would use on ‘development, facilities, sponsor support and work with charities’. The Club also strongly believes that the community that they support is the main focus of why the club is running, and the development of local kids to be active and be involved in sport. SANFL football operations manager Darren Chandler also stated in the article that he believed up to around 300 clubs throughout the state of South Australia relied on the revenue that pokies bring to their individual clubs.

The Courier-Mail (2011), also found that poker machines have a strong influence on the NRL. An interview with David Gallop, the CEO of the NRL, found that clubs contributed about $25 million dollars each year to the clubs they are associated with, but also found that the contribution to junior sport was heavily supported, with clubs working strongly to try and attract kids and their families to join up at the club at a junior level, and make them more interested in the sport. This is a vital point when referring to the business of the sport, as attracting juniors and retaining them throughout their growth is an important step in gaining revenue. Families soon become involved in the dealings with the club, and may eventually become members of a club. This is a boost to the club, but also a boost for the sport as more revenue is created to help support the teams that they are involved with. Gallop also stated that ‘over 1100 clubs were supported by revenues from clubs in 2010, with $40 million dollars being passed through to the junior clubs’ (Courier-Mail 2011).

A spokesperson for the NRL also stated in The Age (2011), that clubs faced losses of ‘up to $71 million to alter machines in accordance with the reforms’. They also said that clubs would record losses in revenue of an estimated $144 million a year. These 2 figures are worrying thoughts that clubs would have to face if the reforms are passed through parliament. Politics is an important part of what is happening. Corporate politics is working strongly to ensure that these reforms do not become law, by running campaigns to ensure that the public and communities who are involved with clubs are aware of the ramifications that the changes will have.

Issues at a Local Level
The reforms will also provide a significant negative effect on the sport landscape at a Local Level, with many sporting teams across varying sports are reliant on clubs and the revenue that is generated to operate and run smoothly. These teams use the clubs to enhance their image, or use the fields and equipment that is provided to ensure that young juniors from the Canberra region are able to participate in an active lifestyle.

Many local sporting clubs are dependent on the support that clubs provide financially to continue operating successfully. AFL, Football (soccer), rugby league and union, all are involved in different playing levels within the Canberra region that have teams that are funded through clubs. These clubs use the revenue that they receive from poker machines and other forms of gambling and entertainment to sponsor these football sides. Ainslie Football Club, for example, have many junior teams playing in different ages ranges in the local AFL competition. Without the support and backing from the club, these young children would not be able to play for the club with their mates, or it may mean that parents are required to pay more for registration fees and uniform fees, because the revenue from the club has deteriorated.

These reforms will be negative to the growth and maturity of young children in playing sport as they will be deterred from becoming involved in sport because of the cost that is involved.

Conclusion
It is important to understand that sport is an integral part of the Australian ethos and has a strong contribution to the society that we live in today. It is vital than, that the business of sport be fully recognised so that it can be a viable industry for the future and to help increase the success of sport. These reforms will damage that spirit as they have the potential to drive young children away from playing sport, as clubs will not be financially able to support the community through sport and recreation. The business of sport is something that everyone should be able to have the opportunity of cashing in on, whether that be through financial means, or through a more personal meaning, such as healthy lifestyle, of enjoyment. It is important that this is instilled in the future generations as they mature, so that they can continue the traditions that have been before them. These reforms will take away many of those opportunities as clubs will no longer be able to provide financial stability and support to the sports industry.


title: File:CORDIC.VHDL.2.A.20111107.pdf


title: File:CORDIC.VHDL.1.A.20111107.pdf