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User:Eas4200c.f08.carbon.orear/HW3/Lecture Notes

=Open Thin Walled Cross Section=

Homework

 * Prove that the area of the triangle is $$A = \frac{1}{2} bh$$ of whether point C is at point D or not.

 * BC x BE|| = ||BC|| * ||BE|| * sin (θ)

h = ||BE|| * sin(θ)

b = ||BC||

Theta is between the two vectors.

Area = .5 * ||BC x BE|| (can be obtained from cross product definition)

Thus, Area = .5 * bh.

For an open thin walled cross section:

$$T = 2 q \bar{A}$$

$$R e = T = 2 q \bar{A}$$ (Resultant force (R) times distance from origin (e)) {7.50}

=Uniform bar with circular cross section=

Warping vs no warping: (warping = deformation of cross section by axial displacement of a point). A uniform bar with circular cross section exhibits a non-warping case; the cross sections behave as rigid disks.

$$T = \iint\limits_A r \tau dA$$

$$\tau = G \gamma$$ (Hooke's Law)

$$\frac{da}{dx} =: \theta $$ rate of twist

$$T = \int\limits_A r G (r \theta) dA$$

G and theta are independent for (y and z).

$$dA = dydz = rdrdw$$

$$=G \theta (\int\limits_A r^2 da)$$

$$\int\limits_A r^2 da = J $$ – 2nd polar area moment of inertia

a = radius of circular cross section

Homework

 * Prove $$J = \frac{1}{2} \pi a^4$$

When integrating to solve for J in polar form (see external link ) we integrate about a full circle with respect to φ and from the inner to outer radial distance with respect to ρ. For our solid cross section this is from 0 to r (or also called, 0 to a).

$$J = \int_0^{2\pi} \int_0^a \rho^2 \rho\, d\rho\, d\phi = \frac{\pi a^4}{2}$$

We know these bound because we want to integrate about a full 360° (2π radians) and from the inner surface (at the origin in this case) to the outer radial distance.

=Solid vs Hollow Cross Section= Solid circle cross section Hollow circle cross section Thin-walled (t << a) $$r_i = a$$ (inner radius) $$r_0 = b$$ (outer radius)

$$J = \frac{1}{2} \pi (b^4 - a^4)$$

$$J = \frac{1}{2} \pi (b - a)(b + a)(b^2 - a^2)$$

$$(b - a) = t$$

$$(b - a) = 2 \bar{r}$$ (Average Radius)

$$\bar{r} = \frac{(b + a)}{2}$$

$$b^2 \approx \bar{r}^2$$

$$a^2 \approx \bar{r}^2$$

Homework

 * Show this Approximation Rigorously

 * $$\bar{r} = \frac{(b + a)}{2}$$

 * $$\displaystyle a = b - t$$

 * $$\displaystyle b = a + t$$

 * $$\bar{r} = \frac{(b + b - t)}{2}$$

 * $$\bar{r} = \frac{(a + t + a)}{2}$$

 * $$\bar{r} = \frac{(2 * b - t)}{2}$$

 * $$\bar{r} = \frac{(2 * a + t)}{2}$$

 * $$\bar{r}^2 = \frac{(2 * b - t)^2}{2^2}$$

 * $$\bar{r}^2 = \frac{(2 * a + t)^2}{2^2}$$

 * $$\bar{r}^2 = \frac{(2 * b)^2 - t^2}{2^2}$$

 * $$\bar{r}^2 = \frac{(2 * a)^2 + t^2}{2^2}$$

 * $$\displaystyle t^2 << a^2$$

 * $$\displaystyle t^2 << b^2$$

 * $$\bar{r}^2 \approx \frac{(2 * b)^2 - 0^2}{2^2}$$

 * $$\bar{r}^2 \approx \frac{(2 * a)^2 + 0^2}{2^2}$$

 * $$\bar{r}^2 \approx \frac{2^2 * b^2}{2^2}$$

 * $$\bar{r}^2 \approx \frac{2^2 * a^2}{2^2}$$

 * $$\bar{r}^2 \approx b^2$$

 * $$\bar{r}^2 \approx a^2$$

$$J = 2 \pi (t) (\bar{r}^3)$$

$$J = 2 \pi^\frac{-1}{2} (t) (\pi \bar{r}^2)^\frac{3}{2}$$

$$\pi \bar{r}^2 = \bar{A}$$

$$J$$ is proportional to $$\bar{A}^\frac{-1}{2}$$ with $$2 \pi^\frac{-1}{2} t$$ being a proportionality factor.

Homework

 * Comparing a Solid Circular Cross section to a hollow thin walled cross section we note:

 * Material Close to the core doesn't work as hard

 * Push the material to the maxium efficiency! Reduce weight by eliminating core material that doesn't contribute to our inertia as greatly!

 * Pg: 1.8 in the book

 * (a)	Soild circ cross section     $$ r_o ^{(a)} = 1 cm$$
 * (b)	Thin walled (hollow) cross section thickness = $$0.1 cm r_i ^{(b)} = 5 cm$$

 * Compute area for case a and b

 * $$ A^{(a)} = \pi r_o ^{(a) 2} = \pi 0.01^{2} = 0.000314159265 m^3$$

 * $$ A^{(b)} = \pi r_o ^{(b) 2} - \pi r_i ^{(b) 2}= \pi 0.05^{2} - \pi 0.049^{2}= 0.000311017673 m^3$$

 * Compute polor moment of interal (J) for a and b

 * $$ J^{(a)} = 0.5 * \pi r_o ^{(a) 4} = 0.5 * \pi 0.01^4 = 1.57079633 * 10^{-8} m^3$$

 * $$ J^{(b)} = 0.5 * \pi (r_o ^{(b) 4} - r_i ^{(b) 4}) = 0.5 * \pi (0.05^4 - 0.049^4 = 7.62148807 * 10^{-7} m^3$$

 * Compute the ratio of $$J^{(a)} / J^{(b)}$$

 * $$\frac{1.57079633 * 10^{-8} m^3}{7.62148807 * 10^{-7} m^3} = 0.0206101002$$

 * Find $$r_i^{(c)}$$ with $$t = 0.02 r_i ^{(c)}$$ such that  $$\displaystyle J^{(c)} = J^{(a)}$$, compute $$\frac{A^{(a)}}{A^{(c)}}$$

 * $$J^{(c)} = 0.5 * \pi ( (r_i + 0.02 r_i)^{(c) 4} - r_i^{(b) 4}) = J^{(a)} = 1.57079633 * 10^{-8} m^3$$

 * $$r_i^{(c)} = .0187 m$$

=Warping=

Torsion relates the rate of twist of an object and the displacement of any point along a non-centerline axis to an applied torque.

θ=α/x=rate of twist.

The following are mathematical definitions for the warping along the x,y, and z-axes.

1: y-displacement of a point on the cross section, horizontal

→ uy=-rαsinβ=-αy=-θxz

2: z-displacement of a point on the cross section, vertical

→ uz=v=rαcosβ=+(PP')cosβ=+αyp=+θxy, where PP'=OPα, α=θx, and yp=y because the variable is general for all coordinates

3: x-displacement, along centerline

→ ux=θψ(y,z), where this value is proportional to the rate of twist (θ)

These three equations compose the kinematic assumptions of warping.

 * {| class="toccolours collapsible collapsed" width="60%" style="text-align:left"

!Contribution Physical Representation of Torsion, from team Carbon
 * Torsion relates the rate of twist of an object and the displacement of any point along a non-centerline axis to an applied torque.
 * Torsion relates the rate of twist of an object and the displacement of any point along a non-centerline axis to an applied torque.

Contribution by carbon.w
 * }

Roadmap for Torsional Analysis of Aircraft Wing
Below is a picture of a cross section of a wing with a two-cell stringer-skin-web section with a thin, variable wall.

A. Kinematic Assumptions described above with the three equations.

B. Strain-displacement relationship:

εxx = εyy = εzz = γxy = 0

σxx = σyy = σzz = τxy = 0

$$\gamma _{xz} = \frac{\partial w}{\partial x} + \frac{\partial u}{\partial z}$$

$$\gamma _{yz} = \frac{\partial w}{\partial y} + \frac{\partial v}{\partial z}$$

C. Equilibrium Equation for Stresses:

$$\frac{\partial \tau _{xz}}{\partial x} + \frac{\partial \tau_{yz}}{\partial y} = 0$$

$$\frac{\partial \sigma _{xx}}{\partial x} + \frac{\partial \tau _{yx}}{\partial y} + \frac{\partial \tau _{zx}}{\partial z} = 0$$

$$\frac{\partial \tau _{xy}}{\partial x} + \frac{\partial \sigma _{yy}}{\partial y} + \frac{\partial \tau _{zy}}{\partial z} = 0 $$

$$\frac{\partial \tau _{xz}}{\partial x} + \frac{\partial \tau _{yz}}{\partial y} + \frac{\partial \sigma _{zz}}{\partial z} = 0 $$

$$\tau _{xy} = \tau _{yx}$$

$$\tau _{yz} = \tau _{zy}$$

$$\tau _{xz} = \tau _{zx}$$

D. Prandtl Stress Funtion, $$\phi $$

$$\tau _{xz} = \frac{\partial \phi }{\partial y}$$

$$\tau _{yz} = \frac{\partial \phi }{\partial x}$$

E. Strain Compatibility

$$\frac{\partial \gamma _{yz}}{\partial x} - \frac{\partial \gamma _{xz}}{\partial y} = 2\theta $$

F. Equation for $$\phi $$

$$\frac{\partial ^{2}\phi }{\partial x^{2}} = -2G\theta$$

G. Boundary Conditions for $$\phi$$

$$\frac{\partial \phi }{\partial s} = 0$$ or $$\phi $$ = constant (on the lateral surface)

If cross section is solid and thus has only a single contour boundary, then the constant can be chosen arbitrarily as zero: $$\phi$$ = 0 on lateral surface of the bar

H. Torque, $$T = 2\int \int_{A}^{}{\phi dA}$$

→ $$T = GJ\theta $$, where using the above equation and the equation for $$\phi $$ we obtain $$J = -\frac{4}{\bigtriangledown^{2}\phi }\int \int_{A}^{}{\phi dxdy}$$ and $$(\phi dxdy = dA)$$

I. Thin Walled Cross Section

$$\tau = \frac{T}{2abt}$$ Ad-hoc assumption of shear flow

$$T = \oint_{}^{}{\rho qds} = \int \int_{\bar{A}}^{}{2qdA} = 2q\bar{A}$$

J. Twist Angle, θ: Method 1

$$\theta = \frac{1}{2G\bar{A}}\oint_{}^{}{\tau ds} = \frac{1}{2G\bar{A}}\oint_{}^{}{\frac{q}{t}ds}$$

K. Multi-cell Section
 * cell $$i=1,...,n_{cell}$$

For convenience, abbreviate $$n_{cell}$$ as $$n_c$$

$$T = 2\sum_{i=1}^{n_c} q_i\bar {A_i}$$ ( See Eq. 3.62 pg 93) where, $$q_i = $$ shear flow in cell i    $$\bar {A_i} = $$ "average" area in cell i

Define: $$T_i = 2q_i\bar{A_i}$$ to be the torque generated by one cell

$$=> T = \sum_{i=1}^{n_c} T_i$$

K.2: Shape of an airfoil is "rigid" in the plane (y,z), although it can warp out of the plane.

$$\theta = \theta_1 = ... =\theta_{HC}$$

$$\theta_i = \frac{1}{2G_i\bar{A_i}} \oint \frac{q_i}{t_i} ds$$ where, $$G_i = $$ the shear modulus of cell i    $$t_i = $$ the thickness in cell i     *Note: $$t_i(s)$$, i.e. thickness can be a function of circular coordinate along the cell wall

Remember Problem 1.1, which was for the more general case of a rectangular single cell section.

Example

$$ A = \frac{1}{2}\pi R^2 = \frac{1}{2}\pi (\frac{b}{2})^2 $$ -area for half a circle $$ A = \frac{1}{2}ba$$ -area for the triangle

$$\bar {A_i}$$ is found to be:

$$\bar {A_i} = \frac{\pi}{2}(\frac{b}{2})^2 + \frac{1}{2}ba$$

Remember shear flow is given by: $$T = 2q\bar{A} => q = \frac{T}{2\bar{A}}$$
 * Note: T can be variable

And, the twist angle is given by: $$\theta = \frac{1}{2G\bar{A}} \sum_{j=1}^3 \frac{q_jl_j}{t_j}$$ where $$ j = 1,2,3 $$ index for segment number

Summations
This is supplementary material reference in class. Team Carbon has provided links and definitions to those who wish to explore these topics in greater detail.

Integration and summation are two different ways to find an area.

The integration symbol, $$\int$$, is an elongated s standing for countinuous summation.

Riemann Sum, $$\sum$$, is a discrete sum (noncontinuous). Integrals can be broken down into Riemann sums.

Quadrature

Quadrature- numerical integration using quadrangles (rectangles) to find the area for more information on quadrature visit this wikipedia article Numerical Integration

Quadrangle - shape with 4 sides (quadrilateral)
 * ex. squares, rectangles, trapizoids, etc.

cubature - when you subdivide a volume into cubes to find the volume of a given shape

Wikipedia article on Squaring of a Circle

=NACA 4-Digit Airfoil Series=

Let's begin with what the digits in a NACA 4-digit airfoil designation mean physically. When we look at the cross-section of the airfoil we can measure the following values.

The first digit describes the maximum camber as a percentage of the chord length.

The second digit describes the distance to the maximum camber from the airfoil's leading edge in tens of percents of the total chord length.

Two final digits describe the maximum thickness of the airfoil as a percentage of the chord length.

We can note patterns: a symmetric airfoil begins with 00--.

Goal: We will be introducing a [MATLAB] code in Homework 3 which will plot a NACA 4-digit airfoil, compute the centroid, and compute the area Ā.

We can discretize the y-axis by a number of segments that the airfoil is broken up into.

Begin with the cross-product:

r x PQ

We can substitute the length PQ with a force if we scale by the shear flow, q:

dT = r x dF = r x q (PQ)

Examining the equation:

qr x PQ

we can see that r x PQ can be replaced with 2dA î. In homework 4, we will see why dA can be relevant as a vector quantity pointing into our out of the page, but here it is fine to write it as a scalar.

=Torsion of a Non-circular Bar=

We will now examine how torsion of a non-circular bar is different from a circular bar. What we notice most is that warping of a cross section occurs. Warping, again, is axial displacement of a point on the rotated cross section. This means that when we apply torsion to rotate the bar, the point moves both axially and about the axis of rotation. If all points in a cross-section plane are coplanar before torsion, we can easily see that they will not be all coplanar after. This warping effect can be dramatic in our study of structural components.

Question: Why is the vector from P to P' straight, and not circular (an arc)? In other words, why is the P-P' vector perpendicular to line segment OP.

Answer: Simply, because α is very small, the displacement can be drawn in this manner. But why is that the case?

uz = L sin (α) is approximately = Lα

uy = L (1-cos(α)) is approximately = 0

by small angle approximations. This gives L and α a linear relationship seen in the picture.


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User talk:Zetachiick09

== Siwell Middle School Band ==

Siwell Middle School's "Sounds of Distinction" marching band was esatblished in 1940 along with the school. They are located in Jackson Mississippi. The current band direcors name is Mr. Christopher Little.

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User:JoJo

I am the same as on the English wiki, my home wiki. My husband is.

I have the global account (SUL) JoJo.

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Educational Development of Cambodia

The Kingdom of Cambodia is a country in South East Asia with a population of over 3 million people. Cambodia is the successor state of the once powerful Hindu and Buddhist Khmer Empire, which ruled most of the Indochinese Peninsula between the eleventh and fourteenth centuries. Today Cambodia is regarded as one of developing countries, therefore it has many problems, including problems related with educational development.

Cambodian Brief Profile

Cambodia has an area of 181,035 square kilometers, and it borders Thailand to its west and northwest, Laos to its northeast, and Vietnam to its east and southeast. Cambodia gained independence from France on November 9, 1953. Cambodia's main industries are garments, tourism, and construction. As for its language, more than 90% of its population is of Khmer origin therefore speaks the Khmer language, the country's official language. The remainder includes Chinese, Vietnamese, and Indians. As for the religion, most Cambodians are Buddhists of Khmer extraction, but the country also has a substantial number of Muslim Cham, as well as small animist tribes. In addition, according the report of The World Bank, “Cambodia's economy is showing resilience in spite of the challenging international economic environment. Economic growth in 2001 was estimated at 6.3 percent, driven by an expanding tourism sector and robust garment exports. In 2001, there was nearly zero inflation. In 2002 inflation continued to be low (World Bank)”. That is, Cambodia’s economic growth has been moderately strong over the past decade. However, this growth has been concentrated in urban areas (The World Bank). Therefore the rural-urban income gap is widening these days. And around 40 percent of the population still remains under the national poverty line.

History
1863 - Cambodia becomes a protectorate of France. French colonial rule lasts for 90 years.

1941 - Prince Norodom Sihanouk becomes king. Cambodia is occupied by Japan during World War II.

1945 - The Japanese occupation ends.

1946 - France re-imposes its protectorate. Communist guerrillas begin an armed campaign against the French.

1953 - Cambodia wins its independence from France. Under King Sihanouk, it becomes the Kingdom of Cambodia.

Early 1970s - Cambodian army faces two enemies: the North Vietnamese and communist Khmer Rouge. Gradually, the army loses territory.

1975 - Lon Nol is overthrown as the Khmer Rouge led by Pol Pot occupied Phnom Penh. Sihanouk briefly becomes head of state, the country is re-named Kampuchea. Basic freedoms are curtailed and religion is banned. Hundreds of thousands of the educated middle-classes are tortured and executed in special centres. Others starve, or die from disease or exhaustion.

1976 - The country is re-named Democratic Kampuchea. Sihanouk resigns, Khieu Samphan becomes head of state, Pol Pot is prime minister.

1977 - Fighting breaks out with Vietnam.

1979 January - The Vietnamese take Phnom Penh. Pol Pot and Khmer Rouge forces flee to the border region with Thailand. The People's Republic of Kampuchea is established. Many elements of life before the Khmer Rouge take-over started to be re-established.

1981 - The pro-Vietnamese Kampuchean People's Revolutionary Party wins the elections to the National Assembly.

1989 - Vietnamese troops withdraw. The country is re-named the State of Cambodia. Buddhism is re-established as the state religion.

1991 - A peace agreement is signed in Paris. A UN transitional authority shares power temporarily with representatives of the various factions in Cambodia. Sihanouk becomes head of state.

1993 - General election sees the royalist Funcinpec party win the most seats followed by Hun Sen's Cambodian People's Party (CPP). A three-party coalition is formed with Funcinpec's Prince Norodom Ranariddh as prime minister and Hun Sen as deputy prime minister. The monarchy is restored, Sihanouk becomes king again. The country is re-named the Kingdom of Cambodia.

1994 - Thousands of Khmer Rouge guerrillas surrender in government amnesty.

1997 - Hun Sen stages a coup against the prime ministar, Prince Ranariddh, and replaces him with Ung Huot. The coup attracts international condemnation and Cambodia's membership of the Association of Southeast Asian Nations (ASEAN) is delayed. The Khmer Rouge put Pol Pot on trial and sentence him to life imprisonment.

1998 - On April 15, Pol Pot dies in his jungle hideout. Elections in July are won by Hun Sen's CPP, amid allegations of harassment. A coalition is formed between the CPP and Funcinpec. Hun Sen becomes prime minister, Ranariddh is president of the National Assembly.

2001 - Senate approves a law to create a tribunal to bring genocide charges against Khmer Rouge leaders.

2001 June - International donors, encouraged by Cambodia's reform efforts, pledge $560 million in aid at a donor conference in Tokyo.

2001 June - Five Cambodians, three of them US citizens, are sentenced to life for involvement in armed attack by US-based Cambodian Freedom Fighters (CFF) on government buildings in Phnom Penh in 2000. CFF says it will continue campaign to overthrow Prime Minister Hun Sen.

2001 December - First bridge across the Mekong River opens, linking the east and west of the country. The 1.36km bridge cost $56 million.

2002 February - First multi-party local elections; ruling Cambodian People's Party wins in all but 23 out of 1,620 communes.

2004 August - Parliament ratifies kingdom's entry into World Trade Organization (WTO).

2004 October - King Sihanouk abdicates and is succeeded by his son Norodom Sihamoni.

2005 April - Tribunal to try surviving Khmer Rouge leaders gets green light from UN after years of debate about funding.

2005 October - Prime minister signs a controversial border agreement with Vietnam. Legal action is taken against some critics of the deal, prompting international concern.

2006 July - Ta Mok, one of the top leaders of the brutal Khmer Rouge regime, dies aged 80.

2007 July - UN-backed court tribunals begin questioning suspects about allegations of genocide by the Khmer Rouge.

2007 September - Most senior surviving member of Khmer Rouge, Nuon Chea - "Brother Number Two" - is arrested and charged with crimes against humanity.

2008 February - Cambodian court sentences 20 member of small Cambodian Freedom Fighters group to prison for attack on prime minister's office in November 2000.

2008 March - Genocide tribunal refuses bail plea of Nuon Chea.

2008 April - US court convicts Cambodian-born Cambodian Freedom Fighters leader Chhun Yasith of masterminding 2000 attack.

2008 July - PM Hun Sen's ruling Cambodian People's Party (CPP) claims victory in parliamentary elections. EU monitors say the vote fell short of international standards.

2008 July - Cambodia and Thailand move troops to disputed land near ancient Preah Vihear temple after decision to list it as UN World Heritage Site fans nationalist emotions on both sides. Officials from both states start talks to resolve standoff. (BBC News 2008)

This part would focus on the Khmer Rouge, because it has affected the educational system in Cambodia. On 17 April 1975, after five years of civil war, Cambodia’s capital Phnom Penh fell to the forces of the Communist Party of Kampuchea (CPK), popularly known as the Khmer Rouge (Linton). Pol Pot was the leader of the Khmer Rouge. “The country was cut off from the outside world, its cities emptied, its economy militarized, and its Buddhist and folk cultures destroyed. The leadership’s ‘vision’ of a better Cambodia led to extraordinary horrors on Cambodian people, who had to endure the forced movement of their urban populations, starvation, unlawful killings, forced labor, torture, and cruel, inhumane and degrading treatment and persecution on a massive scale”(Linton). Mao-think that Pol Pot studied in China, the ethnocentric policy, and anti-Vietnam policy became the theoretical support which created the regime of terror by the Khmer Rouge (Miyamoto). Pol Pot sanctioned the use of massive violence because he believed it was the only “practical” way to achieve and protect his utopian vision for a better society in Cambodia (Valentino). By the hands of the Khmer Rouge, mainly educated people (such as teachers and respected local leaders and so on) had been killed, since they were likely to be rebel groups. Furthermore, the Khmer Rouge had denied the value of education at all points. Therefore, it can be said that Cambodia had experienced a systematic destruction of education by the Khmer Rouge, until Heng Samrin supported by Vietnamese robbed the Khmer Rouge of its power in 1979. According to Linton, an estimated 20% of the 1975’s population of 7.7million people lost their lives. After the Khmer Rouge was over, the United Nations peacekeeping force, the United Nations Transitional Authority in Cambodia (UNTAC), began monitoring Cambodia in 1992. “The goal of UNTAC was disarmament of the troops, and national elections for a constituent assembly which took place in 1993. (…) Contrary to the fears of many, the elections were peaceful. Over 90 percent of the registered voters went to the polls in Cambodian freest, fairest, and most secret election since the colonial era. (…) For the first time in Cambodian history, a majority of Cambodians had voted against an armed government such as the Khmer Rouge” (Chandler). Therefore it can be said: Cambodia was able to gradually recover from many years of violence by the Khmer Rouge regime, in the wake of those elections, although not all problems were solved.

Educational Development Strategy and Challenge of Cambodia
Although all educational systems of Cambodia were abolished under the regime of terror by the Khmer Rouge, according to the World Bank, there have been significant improvements in the past five years in the education sector in Cambodia, particularly in terms of primary net enrollment gains (it’s because the government of Cambodia has put more efforts on the primary education, according to Komai), the introduction of program based-budgeting and the development of a sound pro-poor policy framework. But in Cambodia, several challenges remain. For example, most Cambodian children attend some schooling, but a large share complete only a few grades-with 85 percent of 15 to 19 year olds completing just grade 1, while only 27 percent complete grade 7 (The World Bank). It is partly because parents (especially those in rural areas) are blind to the importance of education. There are also disparities in education participation rates by different regions. Moreover, inefficiency and poor quality in education service delivery at primary, secondary and tertiary levels are other major challenges. For instance, the number and the quality of teachers in Cambodia are insufficient. It is because almost all of experienced old teachers had been killed under the regime of the Khmer Rouge. In fact, one thirds of teachers in the primary school have not finished even lower secondary education (Komai). And the budget for education is extremely restricted, compared to neighboring nations (Komai). Thus, the salary of teachers is too low, which causes teachers’ corrupt practices. And the time for studying in the school is 15 hours per week, which is extremely insufficient, too (Komai). Another challenge is related to child prostitution. “In Cambodia, child prostitution has become a worsening problem since 1990’s. Many Cambodian children have been deceived by words that there are “good” sources of employment in urban areas and have left their home, and have been forced to work in the house of prostitution. In Cambodia, there are more than 8.000～15.000 children who had been damaged by child prostitution” (Social Ecoo). Furthermore, expanding of child prostitution causes spread of HIV/AIDS among Cambodian children. HIV/AIDS is one of educational challenge, since children who are infected with the HIV virus cannot get an education well.

The Recommendation for Ways to Approach Educational Development of Cambodia
First of all, those involved in educational development in Cambodia should make improvements of Cambodian education quality. It would lead to Cambodian students’ higher promotion and achievement rates. Moreover, if education quality gets improved, even parents in rural areas would come to understand the value of education, and would come to encourage their children to complete higher school grades. Furthermore, the budget for education should be increased. The government of Cambodia has gradually made an increase in the budget for education since 2006, and it is hoped that this trend would be maintained longer. Moreover, the teacher training is also needed, because there are little experienced old teachers left due to the Khmer Rouge’s regime of terror.In addition, vocational education for Cambodian children engaging in child prostitution is necessary. It is because children would not have to sell their body, if there are any other sources of employment. And it would lead to cut down the number of children, who are infected with the HIV virus and cannot go to school though they want to do.

The External Actor Involved in Cambodian Educational Development
Cambodia has been aided by a number of more developed nations like France, West-Germany, Canada, Australia, the United States, and Japan. The United States has contributed up to $5 million through international relief agencies to children within Cambodia (Sutter). As for Japan, many Japanese organizations such as JICA have supported Cambodian educational development. For example JICA has tried to revive Cambodian educational system, by forming projects including “the project for improvement of science and mathematics education (2000-2004)”, which is aimed to enable teachers engaged in science and mathematics education to acquire skills more than cheap tricks. In addition, The World Bank, UNICEF, UNESCO, Asian Development Bank (ADB), and some NGOs (such as World Vision in the United States, Save the Children in the UK, and so on) have also supported educational development in Cambodia. Among interesting projects formed by those organizations is the World Bank’s “Education Quality Improvement Project (1999 - 2004)”. This project aimed to develop and put into practice a participatory approach to school quality improvement and performance-based resource management. The goal was to have participating schools eventually show improvements in student enrollment, attendance and achievement (The World Bank). The result was that Students in more than 1,000 participating schools in Cambodia achieved higher test scores in literacy, had lower drop-out and higher promotion rates. In addition, this project could show that health and vocational training was the important factor in reducing student drop-out, and in improving promotion rates.

Works Cited

BBC News. 2008. “Timeline: Cambodia” ＜http://news.bbc.co.uk/1/hi/world/asia-pacific/country_profiles/1244006.stm＞

“Cambodian Information Center”. 2008. ＜http://www.cambodia.org/＞

Chandler, David. 2008. “A History of Cambodia”. Westview Press: Colorado, America.

“Japan International Cooperation Agency”. 1995. ＜http://www.jica.go.jp/Index-j.html＞

Kennsuke, Miyamoto. 1993. “Tounan Asia no Gennzai”. 〔The Present of Southeast Asia〕. Horupu Publication: Tokyo, Japan.

Komai, Hiroshi. 2001. “Shinnsei Cambodia”. 〔New Birth of Cambodia〕. Akashi Publication: Tokyo, Japan.

Linton, Suzannah. 2004. “Reconciliation in Cambodia”. Documentation Center of Cambodia: Phnom Penh, Cambodia.

“Social Ecoo”. 2007. ＜http://www.socialecoo.jp/series/archives/2007/07/03/entry135.html＞

Sutter, Robert G. 1991. “The Cambodian Crisis and U.S. Policy Dilemmas”. Westview Press: Colorado, America.

The World Bank Group. 2008. “The World Bank”. ＜http://www.worldbank.org/＞

Valentino, Benjamin A. 2004. “Final Solutions: Mass Killing and Genocide in the Twentieth Century”. Cornell University Press: New York, America.

title: Educational Development of Cambodia


text:
User talk:Dhamal

 Hello Dhamal, and welcome to Wikiversity! If you need help, feel free to visit my talk page, or contact us and ask questions. After you leave a comment on a talk page, remember to sign and date; it helps everyone follow the threads of the discussion. The signature icon in the edit window makes it simple. To get started, you may

 * Take a guided tour and learn to edit.
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 * Browse everything on Wikiversity, or visit a portal corresponding to your educational level: pre-school, primary, secondary, tertiary, non-formal education.
 * Find out about research activities on Wikiversity.

 * Read an introduction for teachers and find out how to write an educational resource for Wikiversity.
 * Give feedback about your initial observations
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 * Chat with other Wikiversitans on #wikiversity-en.

And don't forget to explore Wikiversity with the links to your left. Be bold to contribute and to experiment with the sandbox or your userpage, and see you around Wikiversity! --Thuvack | talk 06:21, 2 October 2008 (UTC)

 Woof! My name is Jack(Russell). I am a dog and a Wikiversity mascot. I am pretty new around Wikiversity. Perhaps we can learn together! Tail wag...

Electric Circuit Analysis
Thank you for the brilliant edits you did on electric circuit analysis. If you are enclined towards Electrical/Electronic Engineering, you might like to visit the following pages: You might see other courses you can help out on or even start one! --Thuvack | talk 06:21, 2 October 2008 (UTC)
 * School of Electrical Engineering.
 * School of Electronic Engineering.

title: User talk:Dhamal


title: File:Histogram.1.A.pdf


title: File:Geometry.1.A.pdf


text:
Wikiversity:EasyTimeline

EasyTimeline is an extension of MediaWiki which helps editors to produce colourful histograms ("timelines").

How to use

 * Add suggestions and examples here

Examples
The following example was taken from the extension manual at

User:Jtneill/Sandbox/EasyTimeline is trying to create a small "horizontal bar" for indicating page size

title: Wikiversity:EasyTimeline


text:
Enterprise Resource Planning/Presales and Business Analysis

Programme
This is a module part of the Open Source ERP/Executive Masters course conducted by a private university.

Subject
ERP Presales and Business Analysis "Open Source itself is still a new concept to the public and even within IT Departments there is skepticsm and confusion as what embarking on Open Source means. For ERP, there are its own further issues such as application maturity and gaps resolution. Business owners has a checklist of pains and wants that anyone with Open Source ERP has to grasp and respond to."

Objectives
The objectives of this subject are:
 * To understand the issues facing a prospective customer of an ERP
 * To be familiar with the pre-sales cycle and proposal making
 * To understand the prospect's evaluation process in considering an ERP proposal
 * To grasp the methodology and practice of business analysis for an ERP

Learning Outcomes
On completion of this subject the students shall be able to:
 * Devise a strategy to sell his/her services in the market
 * Go through a pre-sales cycle and identify the issues and address them
 * Introduce a methodology to capture the prospects needs
 * Conduct a scoping and pre-qualification analysis
 * Present a proposal set to the prospect as well as the internal implementing team

Synopsis
This subject brings to the student important business context in which ERP Systems are evaluated before implementation. Starting from the pre-sales process right up to proposal delivery and closing the deal. The prospect has many issues and need to be confident that the would-be implementor can address them.

Selling ERP
(20 hrs)
 * ^1 - ERP Players in The Market
 * ^2 - Suspect or Prospect? - The prequalifying stage
 * ^3 - People Are Not Free - Selling Services
 * ^4 - Knowing the Territory - and Selling Yourself
 * ^5 - Users Want Functionality First

Pre-Sales Cycle
(20 hrs)
 * ^6 - Setting The Stage - The Call
 * ^7 - Various Roles in The Team - from Sales to Engineering Support
 * ^8 - Initial Meeting - What to expect and prepare for
 * ^9 - Handling opposition and competition
 * ^10 - Understanding the Client

Business Analysis
(30 hrs)
 * ^11 - Preview demo ERP via online or Virtual Machine
 * ^12 - Gathering Preliminary Information
 * ^13 - Interviewing the key personnel
 * ^14 - Drawing up Business and Operatonal Flows
 * ^15 - Calculating Return on Investment

Proposal Making
(30 hrs)
 * ^16 - Setting the costs
 * ^17 - Margin for Error
 * ^18 - Pricing Strategy
 * ^19 - Scoping Phases

OS Sales Strategy
(20 hrs)
 * ^20 - OS Services Pricing Model
 * Prelim charges for scoping exercise
 * ^21 - Packaging Services
 * Rate card for standard costs and discounts
 * ^22 - Aligning With Friends
 * Level of consultancy and market needs
 * ^23 - Sub Contractor / Consultant

Download Resource

 * Technical Movies


text:
User:Jtneill/Sandbox/EasyTimeline

Feel free to participate in the evolution of a small, horizontal bar for indicating page size.

1-10 mini-horizontal graph prototype
''Trying to to reduce this to, say, 100px width ... but no luck so far''

title: User:Jtneill/Sandbox/EasyTimeline


text:
Template:SPP/Participant/Size graph

Usage

 * Part of SPP/Participant
 * Currently it plots (user page + user talk page) size relative to a specified largest user size
 * It also selects the colour for the bar based on the combined size of a user's page and their talk page:
 * If under 3660 bytes, red bar
 * If 3660 bytes or over, blue bar

title: Template:SPP/Participant/Size graph


text:
Template:SPP/Participant/Email

Usage

 * This is designed to provide a clickable email icon to send wiki-email to a user whose name is specified by a parameter
 * It currently doesn't work, seemingly due to bug 8835 which is resolved in r40684 / r40685 of mw:Extension:ImageMap, however Wikiversity currently has r35980 (see Special:Version) - how to get upgraded? Report as bug?
 * This is what it should look like/do
 * Now nicely fixed by Darklama

title: Template:SPP/Participant/Email


text:
User:Mankash/monobook.css

img.thumbborder { border: 1px solid #c0c0c0; } div.thumb { margin-bottom: .5em; border-style: solid; border-color: white; width: auto; } td.diff-deletedline span.diffchange { font-weight: bold; color: black !important; background-color: #FFFF80 !important; } td.diff-addedline span.diffchange { font-weight: bold; color: black !important; background-color: #40D040 !important; } td.diff-addedline { background-color: #A0FFA0; } td.diff-deletedline { background-color: #FFC0C0; } td.diff-context { background-color: #FFFFA0; }   background: none !important; color: #008000 !important; padding: 0 !important; } ol.references { font-size: 90%; }
 * 1) bodyContent a.external {

title: User:Mankash/monobook.css


text:
User:Eml4500.f08.qwiki.bishop/Lecture 16

=Lecture 16 - Wednesday, October 1, 2008=

Closing the Loop: continued

 * The infinitesimal displacement of point global node 2 (point A to point D) is related to the concept of virtual displacement from statics.
 * The figure below shows the displacement of global node 1:
 * The displacement in the y direction, yD, is equivalent to the displacement, d4
 * The displacement in the x direction, xD, is equivalent to the displacement, d3
 * The vector summation of these two displacement provide the displacement vector $$\vec{AD}$$
 * To determine the axial forces, P, we must first find the axial displacements, q
 * $$q^{(1)}_{2}=l*d^{(1)}_{3}+m*d^{(1)}_{4}=cos(30)*4.352+sin(30)*6.1871=6.8625$$
 * $$q^{(2)}_{1}=l*d^{(2)}_{1}+m*d^{(2)}_{2}=cos(-45)*4.352+sin(-45)*6.1871=-1.2976$$
 * Then the axial forces
 * $$P^{(1)}_{2}=k^{(1)}*q^{(1)}_{2}=(0.75)*(6.8625)=5.145$$
 * $$P^{(2)}_{1}=k^{(2)}*q^{(2)}_{1}=(5)*(-1.2976)=-6.488$$

 * The length of line AC can be found by:
 * $$AC=\frac{\left|P^{(1)}_2 \right|}{k^{(1)}}=\frac{5.145}{.75}=6.8625$$
 * The length of line AB can be found by:
 * $$AB=\frac{\left|P^{(2)}_1 \right|}{k^{(2)}}=\frac{-6.488}{5}=-1.2976$$
 * Note 1: The negative value the length of AB is due to the assigned local axes of element 2.
 * Note 2: The lengths of AB and AC match the displacements for Elements 2 and 1 respectively.
 * So the x and y coordinates for points B and C are (if point A is the origin):
 * $$\left(x_{B},y_{B} \right)=(-.92,.92)$$
 * $$\left(x_{C},y_{C} \right)=(5.94,3.43)$$

 * So, we still have the 2 unknowns (xD, yD)
 * The figure below shows the xy axes and also the $$\tilde{x}\tilde{y}$$ axes
 * Let's analyze the vector PQ
 * The equation derived from this figure is: $$y-y_{P}=tan\theta (x-x_{P})$$
 * And the equation for the line perpendicular to this line is: $$y-y_{P}=tan(\theta +\frac{\Pi }{2}) (x-x_{P})$$
 * So, by applying some trigonometry and algebra $$(x_D, y_D)=(4.352, 6.187)$$
 * And the vector AD by definition is:
 * $$\vec{AD}=d_{3}\vec{i}+d_{4}\vec{j}$$

3-Bar Truss System

 * The force, P, has a value of 50
 * Make sure the local node numbering is selected so that it is convenient to assemble the stiffness matrix, K
 * The image below shows each Element and its local nodes as well as the global nodes


text: Template:SPP/Participant/Size graph/Colour


text:
User talk:Kieran2112

 Hello Kieran2112, and welcome to Wikiversity! If you need help, feel free to visit my talk page, or contact us and ask questions. After you leave a comment on a talk page, remember to sign and date; it helps everyone follow the threads of the discussion. The signature icon in the edit window makes it simple. To get started, you may

 * Take a guided tour and learn to edit.
 * Visit a (kind of) random random project.
 * Browse everything on Wikiversity, or visit a portal corresponding to your educational level: pre-school, primary, secondary, tertiary, non-formal education.
 * Find out about research activities on Wikiversity.

 * Read an introduction for teachers and find out how to write an educational resource for Wikiversity.
 * Give feedback about your initial observations
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 * Chat with other Wikiversitans on #wikiversity-en.

And don't forget to explore Wikiversity with the links to your left. Be bold to contribute and to experiment with the sandbox or your userpage, and see you around Wikiversity! --Gbaor 17:26, 2 October 2008 (UTC)

 Hi! Our name is Vicky-Verity. We are twins and are Wikiversity mascots. We think Wikiversity is a great place to learn and make friends (when we aren't sleeping). Leave us a message!

title: User talk:Kieran2112


text:
User talk:Zappernapper

 Hello Zappernapper, and welcome to Wikiversity! If you need help, feel free to visit my talk page, or contact us and ask questions. After you leave a comment on a talk page, remember to sign and date; it helps everyone follow the threads of the discussion. The signature icon in the edit window makes it simple. To get started, you may

 * Take a guided tour and learn to edit.
 * Visit a (kind of) random random project.
 * Browse everything on Wikiversity, or visit a portal corresponding to your educational level: pre-school, primary, secondary, tertiary, non-formal education.
 * Find out about research activities on Wikiversity.

 * Read an introduction for teachers and find out how to write an educational resource for Wikiversity.
 * Give feedback about your initial observations
 * Discuss Wikiversity issues or ask questions at the colloquium.
 * Chat with other Wikiversitans on #wikiversity-en.

And don't forget to explore Wikiversity with the links to your left. Be bold to contribute and to experiment with the sandbox or your userpage, and see you around Wikiversity! --Gbaor 17:31, 2 October 2008 (UTC)

 Hi! Our name is Vicky-Verity. We are twins and are Wikiversity mascots. We think Wikiversity is a great place to learn and make friends (when we aren't sleeping). Leave us a message!

title: User talk:Zappernapper


text:
User:Soly~enwikiversity

hi, it's me ^_^ Celeste:-P
SOLY

in Soly.

"La vida es bella, disfrutala!" ^_^

" Life is beautiful, enjoy it!" ^_^         @_@..........*_^

TeReRe

Don't worry be happy
Anime-bowling ^_^

Simpsons-anime

= Funny Computer Jokes =

 * My son is so lazy he hates emptying the trash in the recycle bin on his computer.

 * How a computer geek drinks jajaja

YOU KNOW YOU'RE ADDICTED TO THE INTERNET WHEN...
• Your bookmarks takes 15 minutes to scroll from top to bottom.

• You step out of your room and realize that your parents have moved and you don't have a clue when it happened.

• All of your friends have an @ in their names.

• You've already visited all the links at Yahoo and you're halfway through Excite.

• You check your mail. It says "no new messages." So you check it again.

• Your phone bill is delivered in a box.

• You wake up at 3 a.m. to go to the bathroom and stop and check your e-mail on the way back to bed.

• The last girl you picked up was only a jpeg.

• You forget what year it is.

• You begin to wonder how on earth your service provider is allowed to call 200 hours per month "unlimited."

if only life could be like a computer
If you messed up your life, you could press "Ctrl, Alt, Delete" and start all over!

To get your daily exercise, just click on "run"! If you needed a break from life, click on suspend.

Hit "any key" to continue life when ready.

To get even with the neighbors, turn up the sound blaster.

To add/remove someone in your life, click settings and control panel.

To improve your appearance, just adjust the display settings.

If life gets too noisy, turn off the speakers.

When you loose your car keys, click on find.

"Help" with the chores is just a click away.

Auto insurance wouldn't be necessary. You would use your diskette to recover from a crash.

10 Commandments For Computer Ethics
THE TEN COMMANDMENTS FOR COMPUTER ETHICS

1. Thou shalt not use a computer to harm other people.

2. Thou shalt not interfere with other people's computer work.

3. Thou shalt not snoop around in other people's files.

4. Thou shalt not use a computer to steal.

5. Thou shalt not use a computer to bear false witness.

6. Thou shalt not use or copy software for which you have not paid.

7. Thou shalt not use other people's computer resources without authorization.

8. Thou shalt not appropriate other people's intellectual output.

9. Thou shalt think about the social consequences of the program you write.

10. Thou shalt use a computer in ways that show consideration and respect.

Funny pics
200 dollar fine

scape

addicted

title: User:Soly~enwikiversity


text: German vocabulary

title: German vocabulary


text:
Category:German vocabulary

fr:Catégorie:Vocabulaire/Allemand

title: Category:German vocabulary


text: German vocabulary/Time

title: German vocabulary/Time


text: German vocabulary/Sports

title: German vocabulary/Sports


text:
German vocabulary/Verbs

Important verbs
Verb tables below
 * Sein
 * Haben
 * /Modal verbs/
 * /Regular verbs/

Other Verbs

 * gehen to go
 * machen to do
 * brauchen- to need
 * essen- to eat
 * heissen- to be called (as in, "her name is"; "sie heisst Claudia...")
 * wissen- to know (knowledge, information)
 * kennen- to know (acquaintance)
 * spielen- to play
 * besuchen- to visit
 * laufen- to run
 * radfahren- to ride a bike (Separable prefix. Ex. Ich fahre rad. (I ride a bike.))
 * lachen = to laugh
 * sehen = to see (Stem-changer... for du, er, sie, and es change the first 'e' to 'ie'): Ex. Siehst du mich? (Do you see me?)
 * möchten = to would like
 * fliegen = to fly
 * lieben = to love
 * schlafen = to sleep (Stem-changer. Change 'a' to umlauted 'a'.)
 * weinen = to cry
 * mögen = to like (Stem changer, change umlauted 'o' to 'a' in du, er, sie, and es forms. (Ex. Ich mag Katzen: I like cats.))
 * schnell= to hurry

title: German vocabulary/Verbs


text:
User:Straight Edge PXK

Hi, I am Jordan Payne. For main account, see here.

Current Courses:

Finished Courses:

title: User:Straight Edge PXK


text:
Category:Language learning streams

This category contains pages with list of people willing to learn or teach a specific language on Wikiversity.

title: Category:Language learning streams


text:
User talk:Straight Edge PXK

 Hello Straight Edge PXK, and welcome to Wikiversity! If you need help, feel free to visit my talk page, or contact us and ask questions. After you leave a comment on a talk page, remember to sign and date; it helps everyone follow the threads of the discussion. The signature icon in the edit window makes it simple. To get started, you may

 * Take a guided tour and learn to edit.
 * Visit a (kind of) random random project.
 * Browse everything on Wikiversity, or visit a portal corresponding to your educational level: pre-school, primary, secondary, tertiary, non-formal education.
 * Find out about research activities on Wikiversity.

 * Read an introduction for teachers and find out how to write an educational resource for Wikiversity.
 * Give feedback about your initial observations
 * Discuss Wikiversity issues or ask questions at the colloquium.
 * Chat with other Wikiversitans on #wikiversity-en.

And don't forget to explore Wikiversity with the links to your left. Be bold to contribute and to experiment with the sandbox or your userpage, and see you around Wikiversity! --Gbaor 20:51, 2 October 2008 (UTC)

 Hello , we are the Baah and Bahh twins and we'd both like to welcome you to Wikiversity. While you sit back and explore Wikiversity, enjoy some of our fresh goat milk. You can help us learn by creating new learning resources or help us by improving the guided tours and other exploration tools. We look forward to seeing you around. Baaaahbye for now.

movie
I saw your question. One of the reasons for the movie project is to help us learn how to create video. If you create and share some video then it helps people like me learn about the software tools that exist for creating video. Another reason for participating in the movie project is to help tell the story of Wikiversity. "an animation of people in a school" <-- not sure where you are going with this idea, but I often have images of this flash through my mind when I compare conventional schools to Wikiversity. I'd love to learn more about Endorphin --JWSchmidt 15:44, 8 November 2008 (UTC)

title: User talk:Straight Edge PXK


text:
User:Eas4200c.f08.aeris.guan/hw3

Torque, Shear Flow, and Polar Moment of Inertia
In order to calculate torque on a member induced by shear flow, it is necessary to analyze the average area in which that shear flow acts. In the collapsible box below, it is proved that a triangle's area is half the product of its base length and its height even if the height is measured outside of the triangle.

The torque on an arbitrarily shaped shear panel can be expressed in terms of shear flow, and the average area. The shear flow is the force per unit length the shear panel experiences and the average area is the area bounded by the contour of the shear panel relative to the designated origin of the coordinate system in which the shear panel is described. The math process of obtaining this equation is shown below. $$\displaystyle T= \oint{\rho qds}= \int\int_{ \bar{A}}{2qdA}=2q \bar{A}$$ The below picture describes the forces, shear flow, average area and coordinate system of an open thin-walled cross section.  PQ is the shear panel contour, A is the average area enclosed by the panel relative to the origin, e is the perpendicular vector arm from the origin to the force F (which is calculated by multiplying the shear flow with the length in with the shear flow acts). F is the essentially the equivalent force of the shear flow and is used to calculate the torque. Another equation can be written to accommodate for F and the vector e. $$\displaystyle Fe=T=2Q\bar{A}$$ Now, let's consider a uniform bar with circular cross section subject to torque and resulting torsion. This is not a case in which warping of the member is a problem and thus the cross section behaves like rigid disks with only angular displacement rather than lateral displacement (along the length of the bar). The picture below represents an infinitesimally small part of the circlar bar where dA is the infinitesimally small area. This area will be displaced in the (y,z) plane an angel gamma when the object is torqued.  The derivation of an equation to find the torque in terms of the bar's shear modulus, the polar moment of inertia of the cross section, and the rate of twist (gamma). $$\displaystyle T=\int\int_{A}{r \tau dA}=G \gamma \int\int_{A}{rdA}$$ where $$\tau =G \gamma $$ is obtained from Hooke's Law The below picture describes the deformation after torque when the vector r (vector from the origin to dA) is displaced by a small angle alpha to a new vector (r') in the (y,z) plane. $$\displaystyle \gamma= \frac{rd \alpha}{dx}$$ where $$\displaystyle \frac{d \alpha}{dx}=: \theta$$ (rate of twist) so $$\displaystyle T= \int_A{rG(r \theta)dA}$$ where dA= dydz, and G and $$\theta$$ are independent of y and z. $$\displaystyle = G \theta \int_A{r^2dA}=G \theta J$$ G is the shear modulus, theta is the rate of twist, and J is the polar moment of inertia. If a is the radius of the circular cross section then $$ J=\frac{1}{2} \pi a^4$$.

If it is a hollow, thin-walled (t<$$J=\frac{1}{2}\pi(b^4-a^4)=\frac{1}{2}\pi(b-a)(b+a)(b^2+a^2)=\frac{1}{2}\pi t(2 \bar{r})(b^2+a^2)$$ where $$\bar{r}$$ is the average radius ((a+b)/2) $$\displaystyle b^2 \cong \bar{r}^2$$ and $$a^2 \cong \bar{r}^2$$ so $$J=2\pi t \bar{r}^3$$

$$\displaystyle J=(2\pi^{1/2}t)(\pi \bar{r}^2)^{3/2}$$ where $$\displaystyle \bar{a}^=\pi \bar{r}^2$$ This means that J is proportional to the average area to the 1.5 power with a proportional factor of $$2\pi^{1/2}t$$

title: User:Eas4200c.f08.aeris.guan/hw3


text: File:Triareaderiv.JPG

title: File:Triareaderiv.JPG


title: File:Shearpan1.JPG


text: File:Circxbartorsion.JPG

title: File:Circxbartorsion.JPG


text: File:Infiniangledia.JPG

title: File:Infiniangledia.JPG


title: File:Warpwing1.JPG


text:
User:Eas4200c.f08.radsam/HW3

=Homework #3=

Class Notes
 September $$19^{\mbox{th}}$$ - Engineering Shear Strain  Engineering shear strain, $$\gamma\,$$, is measurement of the total deformation of a material under a shear stress. It is the counterpart of normal strain, $$\epsilon\,$$.

Engineering shear strain can be defined as:

$$\gamma = \frac{du}{dy} + \frac{dv}{dx} = \frac{du_{x}}{dy} + \frac{du_{y}}{dx}\,$$

Which is shown in Figure 1.

Where:

$$u = u_{x}\,$$ = the displacement along the x-direction

$$v = u_{y}\,$$ = the displacement along the y-direction

Contributed by radsam.a and .m.

 September $$22^{\mbox{nd}}$$ - Curved Panels  Figure 1 shows a thin curved panel being exposed to a shear stress. With the help of this diagram, we may derive the following two equations of the book:

$$V_y = \tau\,t a\,$$ (1.4)

$$V_z = \tau\,t b\,$$ (1.5)

where:

$$V_y\,$$ is the shear stress in the y direction.

$$V_z\,$$ is the shear stress in the z direction.

A side view of the curved panel is shown in Figure 2. In this diagram we may see how the force $$d\vec{F}\,$$ acts on a small section $$dl\,$$ at and angle $$\theta\,$$. This force may be decomposed in two components along $$Y\,$$ and $$Z\,$$. Also, knowing that the force is equal to the shear flow $$q\,$$ times the length, we obtain the following relation:

$$q = \tau\, t\,$$

So,

$$d\vec{F} = q dl = q (dl_y + dl_z)\,$$

Following the geometry on Figure 2, we know:

$$dl_y = dl cos \theta\,$$

$$dl_z = dl sin \theta\,$$

Therefore:

$$d\vec{F} = q (dl cos\theta\, + dl sin\theta\,)\,$$

$$d\vec{F} = q (dy + dz)\,$$

A detailed drawing of the shear force being applied on the small element $$dl\,$$ is shown in Figure 3. We may clearly see how $$d\vec{F}\,$$ is out an an angle and how it may be decomposed into components. To obtain the total value of $$\vec{F}\,$$, we integrate $$d\vec{F}\,$$ from point $$A\,$$ to $$B\,$$. The resultant force is computed as follows: $$\vec{F} = \iint_{A}^{B} d\vec{F}\, = q ((\iint_{A}^{B} dy)\hat{j}\, + (\iint_{A}^{B} dz)\hat{k}\,)\,$$

If we look at Figure 2 again, one may see that the horizontal distance from $$A\,$$ to $$B\,$$ is equal to $$a\,$$, and that the vertical distance between $$A\,$$ and $$B\,$$ is equal to $$b\,$$. Therefore whe may solve the two integrals above and obtain an expression for $$\vec{F}\,$$. $$\vec{F} = q (a\hat{j} + b\hat{k})\,$$

This last equation corresponds to equations 1.4 and 1.5, because we know that:  $$q = \tau\, t$$ The expression for F shown above is still decomposed for 'y' and 'z' coordinates, but one may compute the total resultant force in the actual direction of stress:

$$|| \vec{F} || = (F_y^2 + F_z^2)^{(1/2)}\,$$

$$|| \vec{F} || = (q (a^2 + b^2)^{(1/2)})\,$$ Where:

$$d = (a^2 + b^2)^{(1/2)}\,$$ -- Length of a straight line.

So we conclude:

$$\vec{F} = q d\,$$ -- Which is essentially Eq. 3.49a

Content by radsam.d Drawings by radsam.a

 September $$22^{\mbox{nd}}$$ - Proof of equation 3.48  Figure 1 shows a constant flow on a closed thin-walled section. Equation 3.48 implies that the torque 'T' is equal to 2 times the shear flow and the enclosed area. It is written as follows:  $$T = 2 q \bar A\,$$ Where $$\bar A\,$$ is the enclosed area and $$q\,$$ is the shear flow. We know that the torque at a point on the thin wall is the radius times the force being applied. Also, the force is equal to the shear flow times the length of the element.  $$dT = \vec r\ x d \vec F\,$$ $$dT = \rho\, dF\, = \rho\, (q dl)\,$$ Figure 2 shows a closer view of the element $$dl\,$$ and the force $$dF$$. Due to the irregular shape of the cross-section, the radius may not always be perpendicular to the wall, so we draw a vector $$\rho\,$$ perpendicular to $$dF\,$$. This forms the triangle shown in Figure 2, so the area of the triangle $$dA$$ may be computed:  $$dA = \frac{1}{2} \rho\, dl\,$$ The total torque $$T\,$$ applied to the thin-walled is the integral of $$dT\,$$

 $$T = \oint dT\, = q \oint \rho\, dl\,$$ But, $$\rho\, dl\, = 2~dA\,$$

Therefore:

$$T = 2q \int_{A} dA =  2q \bar A\,\,$$

Content by radsam.d Drawings by radsam.a

 September $$26^{\mbox{th}}$$ - Thin-Walled Cross Sections

First, the torque of an open thin-walled cross section (Figure 1) is examined.

From the torque equations applied in the last lecture:

$$T=2q\bar{A}$$

The total torque applied on this cross section generates a resultant force, which can be expressed as:

$$Re=T=2q\bar{A}$$

Secondly, the cross section of a uniform bar is examined. Ultimately we want to compare a thin-walled cross section to a solid cross section and determine which performs better when placed under a torsional load.

The equation for the torsion in a cross sectional area (as shown in Figure 2) can be found by taking the integral of the area:

$$T=\int \int_{A}^{}{}r\tau dA=\int \int_{A}^{}{}rG\gamma dA$$ (from Hooke's Law)

The angle gamma can be defined as: $$\gamma = \frac{r d\alpha }{dx}$$

Where: $$\frac{d\alpha }{dx} \equiv \theta$$ (rate of twist)

Thus, torsion is: $$T = \int_{A}^{}{rG(r\theta )dA}=G\theta \int_{A}^{}{r^{2}dA}$$

Where: $$\int_{A}^{}{r^{2}dA} = J$$ = 2nd polar area moment of inertia.

 * See homework section below for the examination of J for a solid circular cross-section: Proof of Moment of Inertia for a solid circular cross-section

Now, we need to consider how all this can apply to aircraft structures. Typically, the fuselage of an airplane is NOT a solid cylinder, but rather a hollow thin-walled circular cross section. In order to express the torsion for such a case, our equation for J needs to be revised somewhat.

 * First, we must note that a thin wall can be defined in terms of an inner radius and an outer radius. Observing Figure 6 to the right: rinner = a    router = b  and  b-a=t where t is the thickness.

 * Notice that because the structure is a thin wall, we can assume t<<a and t<<b. Also, the average radius will be defined as:

 * $$\bar{r}=r_{avg}=\frac{a+b}{2}$$

 * Using our expression for a solid cylinder: $$J=\frac{1}{2}\pi a^{4}$$

 * We can re-express the equation for a thin-walled cylinder (as shown in Figure 3): $$J=\frac{1}{2}\pi (b^{4}-a^{4})=\frac{1}{2}\pi ((b^{2})^{2}-(a^{2})^2)=\frac{1}{2}\pi (b-a) (b+a) (b^2 + a^2)$$

 * Notice that $$\left(b-a \right)=t$$

 * $$(b+a)=2\bar{r}$$

 * and $$(b^2 + a^2) = 2\bar{r}^2$$  This last expression is proven rigorously in the homework section below titled "Proof of Moment of Inertia for a hollow circular cross-section"

Therefore:

$$J=2\pi t\bar{r}^{3}$$

It is also interesting to note that this equation can be expressed as $$J=(2\pi ^{-\frac{1}{2}}t)(\pi \bar{r}^{2})^{3/2}$$. Noticing that $$(\pi \bar{r}^{2})$$ is an expression for the average area of the cross-section, it can be deduced that

J is proportional to $$\bar{A}^{3/2}$$ with a $$\left( 2\pi ^{-1/2}t\right)$$ proportionality factor.

 September $$29^{\mbox{th}}$$ - NACA Airfoil

ns = # of segments (as depicted in Figure 2) to discretize the y-axis.

As shown in Figure 1 and Figure 3:

$$dT = \vec r\ x ~d \vec F\,$$

$$ = \vec r\ x ~\vec {PQ}\,$$

$$ = q\vec r\ x ~\vec {PQ}\,$$ Torsion of uniform, non-circular bars (leads to warping of cross-section) Warping = Axial displacement x-axis (i.e, along the bar length) of a point on the deformed (rotated) cross-section.

Figure 4 and Figure 5 depict: $$Vy ~= ~y~-~component ~of ~displacement~vector~\vec {PP'}\,$$ $$Vz ~= ~z~-~component ~of ~displacement ~vector~\vec {PP'}\,$$ $$~\vec {OP}\ = ~\vec {PP'}\ = R,$$ $$Vz = Rsin \theta\, = R \alpha\, (small~\alpha\,),$$ $$Vy = R(1-cos \alpha\,) =~ 0,$$

 October $$1^{\mbox{st}}$$ - Torsion and Angle of Twist

Torque is a type of load typically found in aircraft structures, most typically in beams and bars that serve as supports in aircraft. Torque occurs in the longitudinal axis of a shaft or beam.

To further analyze the torsion and rotation of a bar undergoing torsion (Figure 1), we shall use Saint-Venant's displacement theorem, which is based on assumptions of the displacement field.

If we observe at the figure, we can see that the rate of twist is defined as...

$$\theta=\frac{\alpha}{z}\,$$ Where $$\alpha\,$$ is equal to the total angle of rotation at $$z\,$$ in relation to the end where $$z\,$$ is equal to 0.

"Saint-Venant assumed that during torsional deformation, plane sections warp, but their projections on the x-y plane rotate as a rigid body" From this assumption we can conclude that u and v, the in-plane displacement components, follow a rigid body rotation as well. For example, if we refer back to our figure above, point P is displaced through the angle alpha to P'. Then we can see, if the rotation angle, alpha is small, that the displacement components are...

$$u=-r\alpha sin\beta = -\alpha y = -\theta zy\,$$

$$v=r\alpha cos\beta = \alpha x = \theta zx\,$$

Where $$r\,$$ = distance from origin to point $$P\,$$

The displacement in the z-direction is assumed to be independent of z:

$$w(x,y) = \theta \psi (x,y)\,$$

Where $$\psi (x,y)\,$$ is the warping function.
 * Note: The rate of twist is independent of z

Road Map for Torsional Analysis of Aircraft Wing:

 * (A) Kinematic Assumption
 * (B) Strain Displacement
 * (C) Equilibrium Equation (Stresses)
 * (D) Prandtl Stress Function $$\phi\,$$
 * (E) Strain Compatibility Equation
 * (F) Equilibrium
 * (G) Boundary Condition for $$\phi\,$$
 * (H) $$T=2\iint_{A} \phi \,dA\,$$

$$T=GJ\theta\,$$

$$J=\frac{-4}{\nabla^2 \phi}\,$$

 * (I) Thin-Walled cross section (Problem 1.1 Ad Hoc assumption on shear flow)

$$T=2q\bar B\,$$

 * (J)Twist angle $$\theta\,$$ : Method 1

$$\theta=\frac{1}{2G\bar A} \oint\frac{q}{t} ds\,$$

 October $$3^{\mbox{rd}}$$ - Multicell Section
 * (K) Multicell Section

$$all~i= 1,2,3,....,n_c$$

 * (K1)   For a discrete sum:

$$T = 2 \sum_{i=1}^{n_c} q_i \bar A_i$$

$$\displaystyle q_i$$ = shear flow in all i

$$\displaystyle A_i$$ = "average" area in all i

Define: $$\displaystyle T_i =2 q_i A_i$$ torque generated by one cell

$$T = \sum_{i=1}^{n_c} T_i $$

 * (K2) Shape of airfoil is "rigid" in the plane (y,z) but can warp out of plane.

$$\theta = \theta_1 = \theta_2 =~...~= \theta_{n_c}$$

$$\theta_i=\frac{1}{2G_i\bar A_i} \oint\frac{q_i}{t_i} ds\,$$

$$\displaystyle G_i$$ = shear modulus in all i

$$\displaystyle t_i$$ = thickness in all i

$$\displaystyle T =2 q A$$ So shear flow is: $$\displaystyle q =\frac{T}{2\bar A}$$

where $$\displaystyle T $$ is the variable.

Twist Angle $$\displaystyle \theta $$:

$$\theta = \frac{1}{2 G \bar A} \sum_{i=1}^{3} \frac{q_i l_i}{t_i}$$

HW 3 Contributors
Eas4200c.f08.radsam.a. 22:06, 7 October 2008 (UTC) Eas4200c.f08.radsam.z 04:16, 8 October 2008 (UTC) Eas4200c.f08.radsam.m 16:55, 8 October 2008 (UTC) Eas4200c.f08.radsam.s 17:50, 8 October 2008 (UTC) Eas4200c.f08.radsam.d 18:50, 7 October 2008 (UTC) Eas4200c.f08.radsam.r 18:40, 8 October 2008 (UTC)

title: User:Eas4200c.f08.radsam/HW3


text:
Enterprise Resource Planning/Software and Development

Programme
This is a module part of the Open Source ERP/Executive Diploma course conducted by a private university.

Subject
ERP Software Development - Just Enough Java

Objectives
The objectives of this subject are:
 * To know how the software operates in running the ERP System
 * To learn just enough Java, SQL and DB to troubleshoot the ERP Software
 * To use the community to gain experience and obtain apprenticeship
 * To develop further and extend the present ERP functionalities

Learning Outcomes
On completion of this subject the students shall be able to:
 * Setup an Interactive Development Environment (IDE) such as Eclipse
 * Handle the sourcecode, modify and recompile succesfully for deployment
 * Work with others in the community to shorten the learning curve
 * Debug any fault in the system and and resolve the bugs
 * Use SourceForge trackers to conduct software maintenance professionally

Sypnosis
This subject covers the software development environment using popular open source tools. This involves debugging and modifying the software to resolve many common needs. The approach is to jump into some hands-on experience by using the SourceForge trackers to pick up real bugs and resolve them collectively with the community.

Using the IDE
(15hrs)
 * Installing IDE
 * Checking Out The Source
 * Setting Classpath Properties & Other Errors
 * Using the Perspectives & Views

Using SourceForge
(15 hrs)
 * Bug Tracker
 * Features Tracker
 * Tracker Activity
 * Reading Maillist

Just Enough Java
(20hrs)
 * Basics of Java
 * ERP Java
 * Stepping through Code
 * Latest Progress

Just Enough SQL
(15hrs)
 * Basics of SQL
 * Using DB Editor
 * Understanding PL/SQL
 * Java SQL

Solving A Bug
(20 hrs)
 * Choosing a Bug
 * Research the context
 * Communicate the ideas
 * Submitting for Review

Just Enough Database
(15hrs)
 * Importing the Initial DB
 * Exporting DB and Versioning Rules
 * Restoring DB
 * Handling DB Instances

New Feature
(20hrs)
 * Create or Extending Module
 * ModelValidator
 * Design Stage
 * Development Stage
 * Testing Stage
 * Documenting Stage

Main Reference

 * Adempiere Technical Training
 * Project Wiki


text:
User:Eml4500.f08.team.foskey.ckf/HW3

Note and Acknowledgements
The following MATLAB code was based on example programs written by X. G. Tan and Team Meshworks. By reviewing their earlier work, the basic methodology of creating MATLAB plots was found, and understood. The program for the two bar truss problem borrows heavily from the work of X. G. Tan.

The programs by X. G. Tan and Team Meshworks are provided near the end of this article.

Creating A MATLAB Program For The Two Bar Truss
This MATLAB program shows the original unloaded truss, as well as the truss that has deformed as a result of the load P. Essentially, the program calculated the coordinates of the elemental nodes for the loaded and unloaded trusses. From there it assigns those points to each element, and then gives the elemental nodes their corresponding global nodes. The program then connects the global nodes. In the figure, the dotted lines represent the elements of the unloaded truss, and the solid lines are the deformed elements from the loaded truss.

Matlab Program Code
% Matlab Script %*********************************************************************  % filename: two_bar_truss_prob.m   % % PURPOSE: %   Example of plotting with Matlab %   Input nodes and connectivity information %   Plot the two bar truss structure %  % AUTHOR: C. Foskey %  %   % Created on :  Sept. 29, 2008 %  %   %*********************************************************************   clear; close; % model with 2-D beam elements with 2 dofs per node dof = 2;        %  dof per node % 1 : axial disp x,  2 : disp y    % input the nodal coordinates n_node = 4; % total number of nodes (2 fixed point nodes,                        %   1 unfixed-undeformed, 1 unfixed-deformed) n_elemUND = 2;            % total number of elements(undeformed) n_elemD=2;              %total number of elements (deformed) total_dof = 2 * n_node; % total dofs of system Ax=4*cos(30*pi/180); %x-cord of node 2 Ay=4*sin(30*pi/180); %y-cord of node 2 Fx=2*cos(45*pi/180); %x-cord of node 3(called point F)   % must be added to Ax Fy=-2*sin(45*pi/180);  %x-cord of node 3(callled point F)  % must be added to Ay %%undeformend points position(:, 1) = [ 0; 0; 0]; % location of node 1 (origin) position(:, 2) = [ Ax; Ay; 0];   % location of node 2 position(:, 3) = [ Ax+Fx; Ay+Fy; 0]; %location of node 3 %%calculating deformed points p1=5.124*cos(30*pi/180); p2=6.276*sin(45*pi/180); k1=3/4; k2=5; q1=p1/k1; % length AC   q2=p2/k2;   %lenght AB   Bx=-q2*cos(45*pi/180)+Ax;   %x-cordinate of point B   By = q2*sin(45*pi/180)+Ay;   %y-cordinate of point B   Cx=q1*cos(30*pi/180)+Ax;   %x-cordinate of point C   Cy=q1*sin(45*pi/180)+Ay;   %y-cordinate of point C    %find intersect xval = 7.7844912; yAB = (-(tan(-45*pi/180))^-1)*(xval-Bx) +By; yAC = (-(tan(30*pi/180))^-1)*(xval-Cx) + Cy; Dx=xval; Dy=yAB; %%deformed points position(:, 4) = [ Dx; Dy; 0]; %location of node 4 %node 4 = deformed node 3 % nodes 1, 2 remain same for deformed because they are fixed % set up the nodal coordinate arrays x, y  for i = 1 : n_node x(i) = position(1,i); y(i) = position(2,i); end % set up the element connectivity array node_connect, defined as % follows: % node_connect(local node number, element number) = global node number node_connect(1, 1) = 1;  % element 1 node_connect(2, 1) = 2; node_connect(1, 2) = 2;  % element 2 node_connect(2, 2) = 3; node_connect(1, 3) = 1;  % element 3= deformed element 1 node_connect(2, 3) = 4; node_connect(1, 4) = 4;  % element 4=deformed element 2 node_connect(2, 4) = 3; % plot the whole truss system % loop over the elements for i = 1 : n_elemUND % for element i, do the following: % node_1 = global node number corresponding to the local node 1 node_1 = node_connect(1,i); % node_2 = global node number corresponding to the local node 1 node_2 = node_connect(2,i); % xx : 1x2 array containing x coordinates of node_1 and node_2 xx = [x(node_1),x(node_2)]; % yy : 1x2 array containing y coordinates of node_1 and node_2 yy = [y(node_1),y(node_2)]; % plot the element i in 2-D using command plot % use axis command to avoid having the plot window too tight axis([-1 10 -1 10]) % dotted line plot(xx,yy,':') % hold the current figure for the next element hold on  end %%loop for deformed elements for i = (1+2) : (n_elemD+2) % for element i, do the following: % node_1 = global node number corresponding to the local node 1 node_1 = node_connect(1,i); % node_2 = global node number corresponding to the local node 1 node_2 = node_connect(2,i); % xx : 1x2 array containing x coordinates of node_1 and node_2 xx = [x(node_1),x(node_2)]; % yy : 1x2 array containing y coordinates of node_1 and node_2 yy = [y(node_1),y(node_2)]; % plot the element i in 2-D using command plot % use axis command to avoid having the plot window too tight axis([-2 10 -2 10]) % solid line plot(xx,yy,'-') % hold the current figure for the next element hold on  end % label nodes, elements text(-.5,-.5,0, 'Node 1') text(Ax, Ay+.5, 0 ,'Node 2') text(Fx+Ax+.5, Fy+Ay-.5,0, 'Node 3') text(Dx-2, Dy+.5,0, 'Displaced Node 2') text(.5*Ax-.5, .5*Ay, 0, 'Element 1') text(Ax+.5*(Fx)-1, Ay+.5*(Fy),0,'Element 2') text(.5*Dx-2, .5*Dy, 0,'Deformed Element 1') text(Dx-.5*(Dx-Fx), Dy-.5*(Dy-Fy), 0,'Deformed Element 2') % put the title on the figure title('Two bar truss') % label x axis xlabel('x') % label y axis ylabel('y') % label z axis zlabel('z') view([0 0 1])     % xy plane view

The Resulting Figure
The above MATLAB program generates the following figure:

In the figure, the dotted lines represent the undeformed, unloaded elements of the two-bar truss. The solid lines represent the deformed elements that result from the load being applied at global node 2.

The example given in class used the global node 2 of the undeformed truss as the origin. In this figure, the origin is located at global node 1. This was done for two reasons. The first being that global node 1 is a fixed point, and remains the same for both the unloaded and the loaded trusses. The second reason was for node position calculation simplicity. By choosing the origin to be at node 1, all of the node positions are positive, and the math required to calculate their positions was generally more intuitive.

The Example MATLAB Programs
The following programs were provided as examples to study prior to the creation of the program of the two bar truss.

The Crab-Leg Structure (matlab_plot_example2.m)
% Matlab Script %*********************************************************************  % filename: matlab_plot_example2.m   % % PURPOSE: %   Example of plotting with Matlab %   Input nodes and connectivity information %   Plot the crab-leg structure %  % AUTHOR: X.G. Tan %  % Modified on: Tue, 23 Sep 2003, 16:08:58 EDT % Created on : Tue, 09 Feb 1999, 09:40:47 EST %  % DEPENDENCIES: %  call: %  % REMARKS: %  %*********************************************************************   clear; close; % model with 3-D beam elements with 6 dofs per node dof = 6;        %  dof per node % 1 : axial disp x,  2 : disp y  , 3 : disp z                    %  4 : torsion x    ,  5 : bend y  , 6 : bend z  % % input the nodal coordinates %  n_node = 17;             % total number of nodes n_elem = 20;            % total number of elements total_dof = 6 * n_node; % total dofs of system l_h_1 = 200.e-6;  % horizontal beam length at center l_h_2 = 50.e-6;   % horizontal beam length at corner l_v_1 = 200.e-6;  % vertical beam length at center l_v_2 = 100.e-6;  % vertical beam length at corner position(:, 9) = [     0; 0; 0]; position(:, 8) = [- l_h_1; 0; 0]; position(:,10) = [ l_h_1; 0; 0]; position(:, 4) = [     0;  l_v_1; 0]; position(:, 3) = [- l_h_1; l_v_1; 0]; position(:, 5) = [ l_h_1;  l_v_1; 0]; position(:,14) = [     0; - l_v_1; 0]; position(:,13) = [- l_h_1; - l_v_1; 0]; position(:,15) = [ l_h_1; - l_v_1; 0]; position(:, 2) = [- l_h_1 - l_h_2;  l_v_1; 0]; position(:, 6) = [ l_h_1 + l_h_2;   l_v_1; 0]; position(:,12) = [- l_h_1 - l_h_2; - l_v_1; 0]; position(:,16) = [ l_h_1 + l_h_2; - l_v_1; 0]; position(:, 1) = [- l_h_1 - l_h_2;  l_v_1 + l_v_2; 0]; position(:, 7) = [ l_h_1 + l_h_2;   l_v_1 + l_v_2; 0]; position(:,11) = [- l_h_1 - l_h_2; - l_v_1 - l_v_2; 0]; position(:,17) = [ l_h_1 + l_h_2; - l_v_1 - l_v_2; 0]; % set up the nodal coordinate arrays x, y, x  for i = 1 : n_node x(i) = position(1,i); y(i) = position(2,i); z(i) = position(3,i); end % set up the element connectivity array node_connect, defined as % follows: % node_connect(local node number, element number) = global node number node_connect(1, 1) = 1;  % element 1 node_connect(2, 1) = 2; node_connect(1, 2) = 2;  % element 2 node_connect(2, 2) = 3; node_connect(1, 3) = 3;  % element 3 node_connect(2, 3) = 4; node_connect(1, 4) = 4;  % element 4 node_connect(2, 4) = 5; node_connect(1, 5) = 5; node_connect(2, 5) = 6; node_connect(1, 6) = 6; node_connect(2, 6) = 7; node_connect(1, 7) = 3; node_connect(2, 7) = 8; node_connect(1, 8) = 4; node_connect(2, 8) = 9; node_connect(1, 9) = 5; node_connect(2, 9) = 10; node_connect(1,10) = 8; node_connect(2,10) = 9; node_connect(1,11) = 9; node_connect(2,11) = 10; node_connect(1,12) = 8; node_connect(2,12) = 13; node_connect(1,13) = 9; node_connect(2,13) = 14; node_connect(1,14) = 10; node_connect(2,14) = 15; node_connect(1,15) = 11; node_connect(2,15) = 12; node_connect(1,16) = 12; node_connect(2,16) = 13; node_connect(1,17) = 13; node_connect(2,17) = 14; node_connect(1,18) = 14; node_connect(2,18) = 15; node_connect(1,19) = 15; node_connect(2,19) = 16; node_connect(1,20) = 16; node_connect(2,20) = 17; % plot the whole truss system % loop over the elements for i = 1 : n_elem % for element i, do the following: % node_1 = global node number corresponding to the local node 1 node_1 = node_connect(1,i); % node_2 = global node number corresponding to the local node 1 node_2 = node_connect(2,i); % xx : 1x2 array containing x coordinates of node_1 and node_2 xx = [x(node_1),x(node_2)]; % yy : 1x2 array containing y coordinates of node_1 and node_2 yy = [y(node_1),y(node_2)]; % zz : 1x2 array containing z coordinates of node_1 and node_2 zz = [z(node_1),z(node_2)]; % plot the element i in 3-D using command plot3 % use axis command to avoid having the plot window too tight axis([-3e-4 3e-4 -4.5e-4 4.5e-4]) % solid line plot3(xx,yy,zz,'-') % hold the current figure for the next element hold on  end % put the title on the figure title('Crab-leg structure w/ 3-D beam elements') % label x axis xlabel('x') % label y axis ylabel('y') % label z axis zlabel('z') % attempt to view the plotted 3-D truss structure from different % viewpoint. % plot viewed from z axis, i.e., projection on xy plane view([0 0 1])     % xy plane view % plot viewed from y axis, i.e., projection on xz plane % view([0 1 0])    % xz plane view % plot viewed from x axis, i.e., projection on yz plane % view([1 0 0])    % yz plane view % produce PostScript (PS) file for printing on a PS printer print -dps crab_leg.ps

The Electric Pylon (electric_pylon.m)
% Matlab Script %*********************************************************************  % filename: electric_pylon.m   % % PURPOSE: %  Plot an electric pylon ("superman") structure with matlab %  Prepare the data for FE analysis %  % AUTHOR: %   Team MeshWorks %   EAS 4200C Aerospace Structures 2, Spring 2000. %  % Modified on: vql, Fri, 03 Oct 2003, 08:37:46 EDT % Created on : Spring 2000 %  % DEPENDENCIES: %  call: %  % REMARKS: %  %*********************************************************************   % Plot the "superman" structure %  clear; close; % model with 3-D beam elements dof = 2;         %  dof per node:  1= axial disp x,  2= disp y  % obtain the coordinates of all nodes %  n_node = 46;             % number of nodes n_elem = 91;            % number of elements total_dof = 2 * n_node; % total dof of system position(:, 1) = [ 1.5; 0; 0]; position(:, 2) = [ 4.53; 0; 0]; position(:, 3) = [ 1.88; 1.18; 0]; position(:, 4) = [ 3.03; 1.18; 0]; position(:, 5) = [ 4.17; 1.18; 0]; position(:, 6) = [ 2.2; 2.181; 0]; position(:, 7) = [ 3.03; 2.181; 0]; position(:, 8) = [ 3.85; 2.181; 0]; position(:, 9) = [ 2.53; 3.181; 0]; position(:, 10) = [ 3.53; 3.181; 0]; position(:, 11) = [ 2.34; 3.681; 0]; position(:, 12) = [ 3.03; 3.681; 0]; position(:, 13) = [ 3.71; 3.681; 0]; position(:, 14) = [ 2.16; 4.181; 0]; position(:, 15) = [ 2.62; 4.181; 0]; position(:, 16) = [ 3.44; 4.181; 0]; position(:, 17) = [ 3.89; 4.181; 0]; position(:, 18) = [ 2.36; 4.5; 0]; position(:, 19) = [ 3.7; 4.5; 0]; position(:, 20) = [ 1.8; 5.172; 0]; position(:, 21) = [ 4.25; 5.172; 0]; position(:, 22) = [ 0; 6; 0]; position(:, 23) = [ 1.06; 6; 0]; position(:, 24) = [ 1.5; 6; 0]; position(:, 25) = [ 1.82; 6; 0]; position(:, 26) = [ 2.22; 6; 0]; position(:, 27) = [ 2.71; 6; 0]; position(:, 28) = [ 3.34; 6; 0]; position(:, 29) = [ 3.84; 6; 0]; position(:, 30) = [ 4.23; 6; 0]; position(:, 31) = [ 4.55; 6; 0]; position(:, 32) = [ 5; 6; 0]; position(:, 33) = [ 6.05; 6; 0]; position(:, 34) = [ .8; 6.23; 0]; position(:, 35) = [ 5.25; 6.23; 0]; position(:, 36) = [ 1.24; 6.355; 0]; position(:, 37) = [ 4.82; 6.355; 0]; position(:, 38) = [ 1.5; 6.43; 0]; position(:, 39) = [ 2; 6.43; 0]; position(:, 40) = [ 2.48; 6.43; 0]; position(:, 41) = [ 3.03; 6.43; 0]; position(:, 42) = [ 3.58; 6.43; 0]; position(:, 43) = [ 4.05; 6.43; 0]; position(:, 44) = [ 4.55; 6.43; 0]; position(:, 45) = [ 1.5; 7.43; 0]; position(:, 46) = [ 4.55; 7.43; 0]; % print the node coord. for i = 1 : n_node; x(i) = position(1,i); y(i) = position(2,i); z(i) = position(3,i); end node_connect(1, 1) = 1;  % element 1 node_connect(2, 1) = 3; node_connect(1, 2) = 1;  % element 2 node_connect(2, 2) = 4; node_connect(1, 3) = 3;  % element 5 node_connect(2, 3) = 4; node_connect(1, 4) = 4;  % element 3 node_connect(2, 4) = 2; node_connect(1, 5) = 4;  % element 6 node_connect(2, 5) = 5; node_connect(1, 6) = 2;  % element 4 node_connect(2, 6) = 5; node_connect(1, 7) = 3;  % element 8 node_connect(2, 7) = 7; node_connect(1, 8) = 5;  % element 9 node_connect(2, 8) = 7; node_connect(1, 9) = 3;  % element 7 node_connect(2, 9) = 6; node_connect(1, 10) = 5;  % element 10 node_connect(2, 10) = 8; node_connect(1, 11) = 6;  % element 11 node_connect(2, 11) = 7; node_connect(1, 12) = 7;  % element 12 node_connect(2, 12) = 8; node_connect(1, 13) = 6;  % element 13 node_connect(2, 13) = 9; node_connect(1, 14) = 7;  % element 14 node_connect(2, 14) = 9; node_connect(1, 15) = 7;  % element 15 node_connect(2, 15) = 10; node_connect(1, 16) = 8;  % element 16 node_connect(2, 16) = 10; node_connect(1, 17) = 9;  % element 17 node_connect(2, 17) = 10; node_connect(1, 18) = 9;  % element 18 node_connect(2, 18) = 11; node_connect(1, 19) = 9;  % element 19 node_connect(2, 19) = 12; node_connect(1, 20) = 10;  % element 20 node_connect(2, 20) = 12; node_connect(1, 21) = 10;  % element 21 node_connect(2, 21) = 13; node_connect(1, 22) = 11;  % element 22 node_connect(2, 22) = 12; node_connect(1, 23) = 12;  % element 23 node_connect(2, 23) = 13; node_connect(1, 24) = 11;  % element 24 node_connect(2, 24) = 14; node_connect(1, 25) = 11;  % element 25 node_connect(2, 25) = 15; node_connect(1, 26) = 12;  % element 26 node_connect(2, 26) = 15; node_connect(1, 27) = 12;  % element 27 node_connect(2, 27) = 16; node_connect(1, 28) = 13;  % element 28 node_connect(2, 28) = 16; node_connect(1, 29) = 13;  % element 29 node_connect(2, 29) = 17; node_connect(1, 30) = 14;  % element 30 node_connect(2, 30) = 15; node_connect(1, 31) = 15;  % element 31 node_connect(2, 31) = 16; node_connect(1, 32) = 16;  % element 32 node_connect(2, 32) = 17; node_connect(1, 33) = 14;  % element 33 node_connect(2, 33) = 20; node_connect(1, 34) = 14;  % element 34 node_connect(2, 34) = 18; node_connect(1, 35) = 15;  % element 35 node_connect(2, 35) = 18; node_connect(1, 36) = 16;  % element 36 node_connect(2, 36) = 19; node_connect(1, 37) = 17;  % element 37 node_connect(2, 37) = 19; node_connect(1, 38) = 17;  % element 38 node_connect(2, 38) = 21; node_connect(1, 39) = 18;  % element 39 node_connect(2, 39) = 20; node_connect(1, 40) = 19;  % element 40 node_connect(2, 40) = 21; node_connect(1, 41) = 20;  % element 41 node_connect(2, 41) = 24; node_connect(1, 42) = 20;  % element 42 node_connect(2, 42) = 26; node_connect(1, 43) = 21;  % element 43 node_connect(2, 43) = 29; node_connect(1, 44) = 21;  % element 44 node_connect(2, 44) = 31; node_connect(1, 45) = 22;  % element 45 node_connect(2, 45) = 23; node_connect(1, 46) = 23;  % element 46 node_connect(2, 46) = 24; node_connect(1, 47) = 24;  % element 47 node_connect(2, 47) = 25; node_connect(1, 48) = 25;  % element 48 node_connect(2, 48) = 26; node_connect(1, 49) = 26;  % element 49 node_connect(2, 49) = 27; node_connect(1, 50) = 27;  % element 50 node_connect(2, 50) = 28; node_connect(1, 51) = 28;  % element 51 node_connect(2, 51) = 29; node_connect(1, 52) = 29;  % element 52 node_connect(2, 52) = 30; node_connect(1, 53) = 30;  % element 53 node_connect(2, 53) = 31; node_connect(1, 54) = 31;  % element 54 node_connect(2, 54) = 32; node_connect(1, 55) = 32;  % element 55 node_connect(2, 55) = 33; node_connect(1, 56) = 22;  % element 56 node_connect(2, 56) = 34; node_connect(1, 57) = 23;  % element 57 node_connect(2, 57) = 34; node_connect(1, 58) = 23;  % element 58 node_connect(2, 58) = 36; node_connect(1, 59) = 24;  % element 59 node_connect(2, 59) = 36; node_connect(1, 60) = 24;  % element 60 node_connect(2, 60) = 38; node_connect(1, 61) = 25;  % element 61 node_connect(2, 61) = 38; node_connect(1, 62) = 25;  % element 62 node_connect(2, 62) = 39; node_connect(1, 63) = 26;  % element 63 node_connect(2, 63) = 39; node_connect(1, 64) = 26;  % element 64 node_connect(2, 64) = 40; node_connect(1, 65) = 27;  % element 65 node_connect(2, 65) = 40; node_connect(1, 66) = 27;  % element 66 node_connect(2, 66) = 41; node_connect(1, 67) = 28;  % element 67 node_connect(2, 67) = 41; node_connect(1, 68) = 28;  % element 68 node_connect(2, 68) = 42; node_connect(1, 69) = 29;  % element 69 node_connect(2, 69) = 42; node_connect(1, 70) = 29;  % element 70 node_connect(2, 70) = 43; node_connect(1, 71) = 30;  % element 71 node_connect(2, 71) = 43; node_connect(1, 72) = 30;  % element 72 node_connect(2, 72) = 44; node_connect(1, 73) = 31;  % element 73 node_connect(2, 73) = 44; node_connect(1, 74) = 31;  % element 74 node_connect(2, 74) = 37; node_connect(1, 75) = 32;  % element 75 node_connect(2, 75) = 37; node_connect(1, 76) = 32;  % element 76 node_connect(2, 76) = 35; node_connect(1, 77) = 33;  % element 77 node_connect(2, 77) = 35; node_connect(1, 78) = 34;  % element 78 node_connect(2, 78) = 36; node_connect(1, 79) = 35;  % element 79 node_connect(2, 79) = 37; node_connect(1, 80) = 36;  % element 80 node_connect(2, 80) = 38; node_connect(1, 81) = 37;  % element 81 node_connect(2, 81) = 44; node_connect(1, 82) = 38;  % element 82 node_connect(2, 82) = 39; node_connect(1, 83) = 39;  % element 83 node_connect(2, 83) = 40; node_connect(1, 84) = 40;  % element 84 node_connect(2, 84) = 41; node_connect(1, 85) = 41;  % element 85 node_connect(2, 85) = 42; node_connect(1, 86) = 42;  % element 86 node_connect(2, 86) = 43; node_connect(1, 87) = 43;  % element 87 node_connect(2, 87) = 44; node_connect(1, 88) = 38;  % element 88 node_connect(2, 88) = 45; node_connect(1, 89) = 39;  % element 89 node_connect(2, 89) = 45; node_connect(1, 90) = 43;  % element 90 node_connect(2, 90) = 46; node_connect(1, 91) = 44;  % element 91 node_connect(2, 91) = 46; % connect all beam elements by connectivity array for i = 1 : n_elem; node_1 = node_connect(1,i); node_2 = node_connect(2,i); xx = [x(node_1),x(node_2)]; yy = [y(node_1),y(node_2)]; zz = [z(node_1),z(node_2)]; plot3(xx,yy,zz,'-') hold on  end title('Electric Pylon') xlabel('x') ylabel('y') zlabel('z') view([0 0 1])     % xy plane view % view([0 1 0])    % xz plane view %view([1 0 0])    % yz plane view % produce postscript file for later printing % print -dps electric_pylon.ps  print -deps electric_pylon.ps

title: User:Eml4500.f08.team.foskey.ckf/HW3


text: Category:The Crafting Freedom Project

title: Category:The Crafting Freedom Project


text:
ERP Project Implementation

Programme
This is a module part of the Open Source ERP/Executive Masters course conducted by a private university.

Subject
ERP Project Implementation - Modules Workshops "A capable and professional ERP implementor is often a person with real grunt work exposure with business applications. There are various differentiated applications of ERP that are sometimes diverse and unconnected to one another. Learning about such differences and how to map them as common components is a plus."

Objectives
The objectives of this subject are:
 * To work hands-on various functional module workshops
 * To go through some Packaging Tools to Import/Export Modules
 * To gain experience in various ERP disciplines

Learning Outcomes
On completion of this subject the students shall be able to:
 * Make use of a Packaging tool to import a real sub project
 * Recompile the projects and deploy it successfully
 * Gather more experience and exposure among the community
 * Contribute Learning Points back to the Wiki and Forums

Sypnosis
This subject covers the important key modules and tools in an Open Source ERP project. Pratical workshops for the students to gain experience using online resources and the community to achieve a satisfactory self-sufficient capability in making various modules work. The experience and confidence obtained can be applied to other technical challenges in a future career.

Full Case Setup
(20hrs)
 * Sample Importer/Exporter of Packages
 * Using 2Pack
 * Using ADCK
 * Importing EDI Package

Group Project I - POS
(20 hrs)
 * Project Workshop
 * Learning Points

Group Project II - Replication
(20 hrs)
 * Project Workshop
 * Learning Points

Group Project III - Human Resource
(20 hrs)
 * Project Workshop
 * Learning Points

Group Project IV - Manufacturing
(20 hrs)
 * Project Workshop
 * Learning Points

Project Documentation
(20 hrs)
 * Reviewing Wiki
 * Proposing New materials
 * Reorganising Present Materials
 * Submitting for Merit/Credit

title: ERP Project Implementation


text:
Number Theory/Diophantine Analysis

Welcome to the Lesson of Diophantine Analysis Part of the School of Olympiads

This lesson is about Diophantine Equations or indeterminate polynomial equations that allows the variables to be integers only (or in some cases fractions). They have fewer equations than unknown variables and involve finding integers that work correctly for all equations. In more technical language, they define an algebraic curve, algebraic surface or more general object, and ask about the lattice points on it. While individual equations present a kind of puzzle and have been considered throughout history, the formulation of general theories for solving them was an achievement of the twentieth century.

The questions asked in Diophantine analysis include: These traditional problems often lay unsolved for centuries, and mathematicians gradually came to understand their depth (in some cases), rather than treat them as puzzles.
 * Are there any solutions?
 * Are there any solutions beyond some that are easily found by inspection?
 * Are there finitely or infinitely many solutions?
 * Can all solutions be found, in theory?
 * Can one in practice compute a full list of solutions?

Some widely used techniques are
 * Factor Decomposition Method
 * Bounding by Inequalities {especially Discriminant Inequality in Quadratic Equations}
 * Parametrization
 * Modular Arithmetic
 * Mathematical Induction
 * Fermat's Infinite Descent
 * Reduction to Pell's and Continued Fractions
 * Positional Numeral Systems
 * Elliptic Curves

=Theorems=

Simple Systems

 * $$\ x + y = a, x - y = b, xy = c, x^2 + y^2 = d, x^2 - y^2 = e, x^3 + y^3 = f, x^3 - y^3 = g.$$

The problem is, given any two of a, b, c, d, e, f, and g, find x and y. For x:

Bezout's Identity [ax+by=d]
In number theory, Bézout's identity or Bézout's lemma is a linear diophantine equation. It states that if a and b are nonzero integers with greatest common divisor d, then there exist infinitely many integers x and y (called Bézout numbers or Bézout coefficients) such that
 * $$ ax+by=d. \, $$

Additionally, d is the least positive integer for which there are integer solutions x and y for the preceding equation. The Bézout numbers x and y as above can be determined with the Extended Euclidean algorithm. However, they are not unique. If one solution is given by (x, y), then there are infinitely many solutions. These are given by
 * $$ \left\{ \left(x+\frac{kb}{\gcd(a,b)},\ y-\frac{ka}{\gcd(a,b)}\right) \mid k \in \mathbb{Z} \right\}. $$

Extended Euclidean Algorithm
The extended Euclidean algorithm is an extension to the Euclidean algorithm for finding the greatest common divisor (GCD) of integers a and b: it also finds the integers x and y in Bézout's identity
 * $$ax + by = \gcd(a, b). \,$$

(Typically either x or y is negative). The extended Euclidean algorithm is particularly useful when a and b are coprime, since x is the modular multiplicative inverse of a modulo b.

Pythagorean Triples
Pythagorean triplets are sets of three natural numbers that satisfy Pythagoras' theorem;$$x^2+y^2=z^2$$ which describes the relationships between the sides of a right angled triangle. It has been shown that there are an infinite number of them by the following proof. The square numbers; 1, 4, 9, 16 etc can be seen to be separated by the odd numbers 3, 5, 7 etc. This is because $$(n+1)^2=n^2+2n+1$$ As an infinite number of these odd numbers are squares ( as an odd number squared results in an odd number) there must be an infinite number of pythagorean triplets.

Fermat's Last Theorem
Fermats last theorem is a theorem about an equation that is similar to pythagorus' theorem. It is $$x^n+y^n=z^n$$. Fermats last theorem states that there are no integer solutions to this equation for x,y,z does not equal 0 and $$n>2$$. It is particularly famous because Fermat stated that he had a proof. The first case of this to be proved was n=4 which was proved by infinite descent.

Exponential Equations
=Problems=

p/x + q/y = 1
Problem Let p and q be prime numbers. Find the number of pairs of positive numbers x,y that satisfy the the equation: $$p/x + q/y = 1\,\!$$

=Resources=

Textbooks

 * Wikibook on Number Theory
 * Wikibook on High School Mathematics Extensions
 * A Computational Introduction to Number Theory and Algebra
 * Elementary Number Theory

Practice Questions

 * from Math Links Everyone

Related WebApps

 * Online Bezout Calculator

Active participants
The histories of Wikiversity pages indicate who the active participants are. If you are an active participant in this division, you can list your name here (this can help small divisions grow and the participants communicate better; for large divisions a list of active participants is not needed).

title: Number Theory/Diophantine Analysis


text:
User:Jermaine~enwikiversity

Full Name : Tichaona Jermaine Gaza

Date of Birth : 28 December 1987

Sex : Male

Nationality : Zimbabwean

Occupation : Currently a student at Africa University

title: User:Jermaine~enwikiversity


text:
User talk:Jermaine~enwikiversity

 Hello Jermaine, and welcome to Wikiversity! If you need help, feel free to visit my talk page, or contact us and ask questions. After you leave a comment on a talk page, remember to sign and date; it helps everyone follow the threads of the discussion. The signature icon in the edit window makes it simple. To get started, you may

 * Take a guided tour and learn to edit.
 * Visit a (kind of) random random project.
 * Browse everything on Wikiversity, or visit a portal corresponding to your educational level: pre-school, primary, secondary, tertiary, non-formal education.
 * Find out about research activities on Wikiversity.

 * Read an introduction for teachers and find out how to write an educational resource for Wikiversity.
 * Give feedback about your initial observations
 * Discuss Wikiversity issues or ask questions at the colloquium.
 * Chat with other Wikiversitans on #wikiversity-en.

And don't forget to explore Wikiversity with the links to your left. Be bold to contribute and to experiment with the sandbox or your userpage, and see you around Wikiversity! --Cormaggio talk 11:47, 3 October 2008 (UTC)

Your account will be renamed
Hello,

The developer team at Wikimedia is making some changes to how accounts work, as part of our on-going efforts to provide new and better tools for our users like cross-wiki notifications. These changes will mean you have the same account name everywhere. This will let us give you new features that will help you edit and discuss better, and allow more flexible user permissions for tools. One of the side-effects of this is that user accounts will now have to be unique across all 900 Wikimedia wikis. See the announcement for more information.

Unfortunately, your account clashes with another account also called Jermaine. To make sure that both of you can use all Wikimedia projects in future, we have reserved the name Jermaine~enwikiversity that only you will have. If you like it, you don't have to do anything. If you do not like it, you can pick out a different name.

Your account will still work as before, and you will be credited for all your edits made so far, but you will have to use the new account name when you log in.

Sorry for the inconvenience.

Yours, Keegan Peterzell Community Liaison, Wikimedia Foundation 23:34, 17 March 2015 (UTC)

Renamed
 This account has been renamed as part of single-user login finalisation. If you own this account you can |log in using your previous username and password for more information. If you do not like this account's new name, you can choose your own using this form after logging in: . -- Keegan (WMF) (talk) 06:12, 19 April 2015 (UTC)

title: User talk:Jermaine~enwikiversity


text:
User talk:Saeed.Veradi

 Hello Saeed.Veradi, and welcome to Wikiversity! If you need help, feel free to visit my talk page, or contact us and ask questions. After you leave a comment on a talk page, remember to sign and date; it helps everyone follow the threads of the discussion. The signature icon in the edit window makes it simple. To get started, you may

 * Take a guided tour and learn to edit.
 * Visit a (kind of) random random project.
 * Browse everything on Wikiversity, or visit a portal corresponding to your educational level: pre-school, primary, secondary, tertiary, non-formal education.
 * Find out about research activities on Wikiversity.

 * Read an introduction for teachers and find out how to write an educational resource for Wikiversity.
 * Give feedback about your initial observations
 * Discuss Wikiversity issues or ask questions at the colloquium.
 * Chat with other Wikiversitans on #wikiversity-en.

And don't forget to explore Wikiversity with the links to your left. Be bold to contribute and to experiment with the sandbox or your userpage, and see you around Wikiversity! --Cormaggio talk 11:50, 3 October 2008 (UTC)

title: User talk:Saeed.Veradi


text:
User:Tomherrell

I am a motivational speaker.

title: User:Tomherrell


text:
User talk:Tomherrell

 Hello Tomherrell, and welcome to Wikiversity! If you need help, feel free to visit my talk page, or contact us and ask questions. After you leave a comment on a talk page, remember to sign and date; it helps everyone follow the threads of the discussion. The signature icon in the edit window makes it simple. To get started, you may

 * Take a guided tour and learn to edit.
 * Visit a (kind of) random random project.
 * Browse everything on Wikiversity, or visit a portal corresponding to your educational level: pre-school, primary, secondary, tertiary, non-formal education.
 * Find out about research activities on Wikiversity.

 * Read an introduction for teachers and find out how to write an educational resource for Wikiversity.
 * Give feedback about your initial observations
 * Discuss Wikiversity issues or ask questions at the colloquium.
 * Chat with other Wikiversitans on #wikiversity-en.

And don't forget to explore Wikiversity with the links to your left. Be bold to contribute and to experiment with the sandbox or your userpage, and see you around Wikiversity! --Sunstar NW XP 12:53, 3 October 2008 (UTC)

--Sunstar NW XP 12:53, 3 October 2008 (UTC)

Your articles
Thanks for contributing to Wikiversity. However, please feel free to ask if you need any help around here. Your articles have been listed at Requests for Deletion, where you are welcome to participate. Thanks, AC --Sunstar NW XP 12:55, 3 October 2008 (UTC)

title: User talk:Tomherrell


text: Template:Hoax

title: Template:Hoax


text:
User:Toad~enwikiversity

Hi,

I am a professional translator. I mainly do books, novels and the like.

My other passion is computers and I am somewhat of a Linux freak. Unfortunately, I cannot program, although this would help me a lot when it comes to writing programmes ;)

So I registered with Wikiversity to see whether I can learn how to program with python AND see whether there are others out there who have similar interests...

Over and out

toad

title: User:Toad~enwikiversity


text: Template:Atp/PD

title: Template:Atp/PD


text: Template:Atp/GFDL+CC

title: Template:Atp/GFDL+CC


text: File:Element disp vector.jpg

title: File:Element disp vector.jpg