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Observational astronomy
Observational astronomy
In the observational astronomy project, astronomers can gain an understanding of objects in the universe by studying them with a telescope. Observational astronomy is the process of collecting and analyzing data to learn about these objects. In this learning project a computer program is used to view and study images taken with a telescope to learn basic concepts in astronomy. This is part of the astronomy project.
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# Boundary Value Problems/BVP-Exercise 1-1:Problems and Solutions

Review Ordinary-Differential-Equations
# Boundary Value Problems/Bessel's equation

Bessel's Equation is a differential equation that occurs quite often in science and engineering. The Bessel's equation will occur in the solution of BVPs and is part of the classical solution process.Therefore is handled here as a topic of special interest. Using power series as a form for a solution of the differential equation results in solutions of a polynomial form called Bessel polynomials
Return To Main Page of Boundary Value Problems
# Boundary Value Problems/BVP-ODE-1st-order

Return to main course page Boundary Value Problems
Video of solving first order (click on the PLAY BAR) ODE
# Boundary Value Problems/Finite Difference Time Domain

Numerical Methods
This method is for solving the wave propagation specifically for electromagnetic waves. See detailed information at FDTD.

## 1D FDTD

The
# Boundary Value Problems/Examples

A few examples of physical problems that can be posed as Boundary Value problems listed below.
* Hanging Cable
* Temperature in a Rod
* Vibrating String

## Readings

* Examples of boundary value problems at Wikipedia
* Boundary value problem at Wikipedia
# Boundary Value Problems/Introduction to BVPs/IVP-student-1

Return to Introduction_to_BVPs
Problem: Differential equation: $y'=xy$ with initial condition: $y(0)=3$
Start Solution:
$\int {\frac {1}{y}}dy=\int xdx$
$ln(|y|)={\frac {x^{2}}{2}}+C$
$e^{ln(|y|)}=e^{{\frac {x^{2}}{2}}+C}$
$|y|=e^{\frac {x^{2}}{2}}e^{C}$
$|y|=e^{\frac {x^{2}}{2}}C_{1}$
$|y|=C_{1}e^{\frac {x^{2}}{2}}$
Use the initial condition $y(0)=3$ to solve for the constant $C_{1}$.
$|y(0)|=3=C_{1}e^{0}$
$C_{1}=3$
The particular solution to the initial value problem is:
$y=3e^{\frac {x^{2}}{2}}$ Why did I drop the "absolute value" operation?
End Solution:
Return to Introduction_to_BVPs
# Boundary Value Problems/3-D Spherical/Legendre's equation

http://en.wikipedia.org/wiki/Legendre_polynomialsw:Legendre_polynomials
# Boundary Value Problems/Introduction

An example of a boundary value problem in one dimension is the second order linear differential equation:
$ay''+by'+cy=0$ with the end conditions of $y(0)=0$ and $y(L)=0$.
For a simple problem let a=1, b=0, c=1 and $L=\pi$. The resulting differential equation is $y''+y=0$ with boundary conditions $y(0)=0$ and $y(\pi )=0$. A solution is $y(x)=sin({{\pi x} \over L})$ .
A plot of this solution is shown below. Note that the solution satisfies the boundary conditions.
For more information about ordinary differential equations and methods for solving them, use the following link ODE
# Boundary Value Problems/Introduction to BVPs
## Objective

Introduce Boundary value problems for a single independent variable.

## Approach

* What is a Boundary Value problem?
* Solution of a Boundary Value Problem is directly related to solution of an Initial Value Problem. So let's review the material on IVPs first and then make the connection to BVPs.
* Details of solving a two point BVP.

## Initial Value Problems

For a single independent variable $x$ in an interval $I:a<x<b$, an initial value problem consists of an ordinary differential equation including one or more derivatives of the dependent variable, $y$,
$y^{(n)}+p_{n-1}(x)y^{(n-1)}+...+p_{1}(x)y'(x)+p_{0}y(x)=f(x)$
and $n$ additional equations specifying conditions on the solution and the derivatives at a point $x_{0}\in I$
$y^{(n-1)}(x_{0})=b_{n-1}$, ..., $y'(x_{0})=b_{1}$, $y(x_{0})=b_{0}$
Example:
The differential equation is $y'=x$ (First order differential equation.) and the initial condition at $x=0$ is given as $y(0)=1$ .
Solution:
$\int y'dx=\int xdx$
$y={\frac {x^{2}}{2}}+C$.
When,$x=0$ $1=C$ and $y={\frac {x^{2}}{2}}+1$
Get out a piece of paper and try to solve the following IVP in a manner similar to the preceding example:
$y'=xy$ and the initial condition at $x=0$ is given as $y(0)=3$ .
Once you have an answer (or are stuck) check your solution here. Click here for the solution: IVP-student-1
A second order ODE example:
The differential equation is $y''+5y'+4y=0$ (Second order differential equation.) and the two initial conditions at $x=0$ given as $y(0)=1,y'(0)=-2$ .
Solution:
Assume the solution has the form $y=e^{rx}$
$y'=re^{rx},y''=r^{2}e^{rx}$
$y''+5y'+4y=r^{2}e^{rx}+5re^{rx}+4e^{rx}$
$0=r^{2}e^{rx}+5re^{rx}+4e^{rx}$
$0=r^{2}+5r+4$ The characteristic polynomial. Solve for "r".
${\begin{array}{c}r_{1}=4\\r_{2}=1\end{array}}$
See the Wikipedia link for more on Initial Value Problems

## Two point BVPs for an ODE

Begin with second order DEs, $x''=f(t,x,x')$, with conditions on the solution at $t=a$ and $t=b$.
$\scriptstyle {\frac {d^{2}x}{dt^{2}}}+p(t){\frac {dx}{dt}}+q(t)x(t)=f(t)$ with $\scriptstyle a_{0}x(a)+a_{1}x'(a)=g$ and $\scriptstyle b_{0}x(b)+b_{1}x'(b)=h$ on the interval $\scriptstyle I_{ab}=\{x|a\leq t\leq b\}$

## Example

$\scriptstyle {\frac {d^{2}x}{dt^{2}}}+4{\frac {dx}{dt}}+2x(t)=f(t)$ with $\scriptstyle x(0)=0$ and $\scriptstyle x(1)=0$ on the interval $\scriptstyle I_{ab}=\{x|0\leq t\leq 1\}$
See the wikipedia topic
Boundary Value Problems
# Biblical Studies/Revision

Welcome to the The Wikiversity Department of Biblical Studies, part of the School of Theology. According to Wikipedia, "Biblical studies is the academic study of the Judeo-Christian Bible and related texts"; hence this department focuses on the study of the Bible, and of exegetic and hermeneutic techniques.

## Department Centers

* Center of Biblical Hermeneutics
* Center of Biblical Languages
* Center of Biblical Translation
* Center of Biblical Survey

## Department Projects

At present, the primary focus of the department is content development. There are a number of pages that need to be created and/or fleshed out, and we need editors (like you :-) who are willing to contribute time and energy to the cause.
A department sub-page has been created to list some of the projects that need to be done. If you don't already have a project in mind, you can go to that page, pick a project that sounds interesting, and begin working on it. Then if you intend to stick around, you can place your username in the "Active Participants" section below. (if you want to help, but don't know how to edit wikiversity, see wiki 101 for an introduction to editing.)

## Active participants

The histories of Wikiversity pages indicate who the active participants are. If you are an active participant in this Department, you may list your name here. This could help the Department grow and the participants better communicate.
Active Participants:
* OpenSourceJunkie
* Saz909
# Elasticity/Compatibility example 1
## Example 1

Given:
The compatibility equations in terms of the strains are (in index notation)
    $e_{ikr}~e_{jls}~\varepsilon _{ij,kl}=0$
The stress-strain relations for a linear elastic material are
    $\varepsilon _{ij}={\frac {1}{E}}\left[(1+\nu )~\sigma _{ij}-\nu ~\sigma _{mm}~\delta _{ij}\right]$
Show:
Substituting the stress-strain relations into the compatibility equations, show that the compatibility equation of stress can be expressed as
    $\sigma _{ii,jj}~\delta _{rs}-\sigma _{ii,rs}-(1+\nu )(\sigma _{ij,ij}~\delta _{rs}+\sigma _{rs,ii}-\sigma _{is,ir}-\sigma _{ir,is})=0$

### Solution

Substituting the stress-strain relations into the left hand side of the compatibility equations and multiplying both sides by $E$, we get,
    ${\begin{aligned}E~e_{ikr}~e_{jls}~\varepsilon _{ij,kl}&=e_{ikr}~e_{jls}~\left[(1+\nu )~\sigma _{ij,kl}-\nu ~\sigma _{mm,kl}~\delta _{ij}\right]\\&=(1+\nu )~e_{ikr}~e_{jls}~\sigma _{ij,kl}-\nu ~e_{ikr}~e_{jls}~\delta _{ij}~\sigma _{mm,kl}\\&=(1+\nu )~e_{ikr}~e_{jls}~\sigma _{ij,kl}-\nu ~e_{nkr}~e_{nls}~\sigma _{mm,kl}\\&=(1+\nu )~e_{ikr}~e_{jls}~\sigma _{ij,kl}-\nu ~e_{krn}~e_{lsn}~\sigma _{mm,kl}\end{aligned}}$
Now, the $\delta -e$ rule states that
    $e_{ijk}~e_{pqk}=\delta _{ip}\delta _{jq}-\delta _{iq}\delta _{jp}$
Therefore,
    ${\begin{aligned}E~e_{ikr}~e_{jls}~\varepsilon _{ij,kl}&=(1+\nu )~e_{ikr}~e_{jls}~\sigma _{ij,kl}-\nu ~\left(\delta _{kl}\delta _{rs}-\delta _{ks}\delta _{rl}\right)~\sigma _{mm,kl}\\&=(1+\nu )~e_{ikr}~e_{jls}~\sigma _{ij,kl}-\nu ~\left(\sigma _{mm,nn}\delta _{rs}-\sigma _{mm,sr}\right)\end{aligned}}$
Recall that,
    ${\begin{aligned}e_{ikr}~e_{jls}&={\text{det}}{\begin{bmatrix}\delta _{ij}&\delta _{il}&\delta _{is}\\\delta _{kj}&\delta _{kl}&\delta _{ks}\\\delta _{rj}&\delta _{rl}&\delta _{rs}\end{bmatrix}}\\&=\delta _{ij}\delta _{kl}\delta _{rs}-\delta _{ij}\delta _{ks}\delta _{rl}-\delta _{il}\delta _{kj}\delta _{rs}+\delta _{il}\delta _{ks}\delta _{rj}+\delta _{is}\delta _{kj}\delta _{rl}-\delta _{is}\delta _{kl}\delta _{rj}\end{aligned}}$
Therefore,
    ${\begin{aligned}\delta _{ij}\delta _{kl}\delta _{rs}\sigma _{ij,kl}&=\sigma _{ii,jj}\delta _{rs}\\\delta _{ij}\delta _{ks}\delta _{rl}\sigma _{ij,kl}&=\sigma _{ii,rs}\\\delta _{il}\delta _{kj}\delta _{rs}\sigma _{ij,kl}&=\sigma _{ij,ij}\delta _{rs}\\\delta _{il}\delta _{ks}\delta _{rj}\sigma _{ij,kl}&=\sigma _{ir,is}\\\delta _{is}\delta _{kj}\delta _{rl}\sigma _{ij,kl}&=\sigma _{is,ir}\\\delta _{is}\delta _{kl}\delta _{rj}\sigma _{ij,kl}&=\sigma _{sr,jj}\end{aligned}}$
Hence,
    ${\begin{aligned}E~e_{ikr}~e_{jls}~\varepsilon _{ij,kl}&=(1+\nu )\left(\sigma _{ii,jj}\delta _{rs}-\sigma _{ii,rs}-\sigma _{ij,ij}\delta _{rs}+\sigma _{ir,is}+\sigma _{is,ir}-\sigma _{sr,jj}\right)-\nu ~\left(\sigma _{ii,jj}\delta _{rs}-\sigma _{ii,rs}\right)\\&=\sigma _{ii,jj}\delta _{rs}-\sigma _{ii,rs}-(1+\nu )\left(\sigma _{ij,ij}\delta _{rs}+\sigma _{rs,ii}-\sigma _{is,ir}-\sigma _{ir,is}\right)=0\end{aligned}}$
Hence shown.
# Elasticity/Beam bending example 1
## Example 1

Given:
A long rectangular beam with cross section $ab$
Find:
A solution for the displacement and stress fields, using strong boundary conditions on the edges $x_{2}=0$ and $x_{2}=b$.
[Hint : Assume that the displacement can be expressed as a second degree polynomial (using the Pascal's triangle to determine the terms) $u(x,y)=Ax^{2}+By^{2}+Cxy+Dx+Ey+F$]

### Solution

Step 1: Boundary conditions
    ${\begin{aligned}{\text{at}}~x_{1}&=0~;~~\sigma _{13}=0\\{\text{at}}~x_{1}&=a~;~~u_{3}=0\\{\text{at}}~x_{2}&=0~;~~\sigma _{23}=0\\{\text{at}}~x_{2}&=b~;~~\sigma _{23}=S\end{aligned}}$
Step 2: Assume a solution
Let us assume antiplane strain
    $u_{3}(x_{1},x_{2})=Ax_{1}^{2}+Bx_{2}^{2}+Cx_{1}x_{2}+Dx_{1}+Ex_{2}+F~;~~u_{1}=u_{2}=0~.$
Step 3: Calculate the stresses
The stresses are given by $\sigma _{\alpha 3}=\mu u_{3,\alpha }$, and $\sigma _{11}=\sigma _{22}=\sigma _{33}=\sigma _{12}=0$. Therefore,
    ${\begin{aligned}\sigma _{13}&=\mu u_{3,1}=\mu (2Ax_{1}+Cx_{2}+D)\\\sigma _{23}&=\mu u_{3,2}=\mu (2Bx_{2}+Cx_{1}+E)\end{aligned}}$
Step 4: Satisfy stress BCs
Thus we have,
    ${\begin{aligned}0&=\mu (Cx_{2}+D)\\0&=\mu (Cx_{1}+E)\\S&=\mu (2bB+Cx_{1}+E)\end{aligned}}$
Since $x_{1}$ and $x_{2}$ can be arbitrary, $C=D=E=0$.
Hence, $B=S/2\mu b$ which gives us
    ${\begin{aligned}u_{3}&=Ax_{1}^{2}+{\frac {S}{2\mu b}}x_{2}^{2}+F\\\sigma _{13}&=\mu (2Ax_{1})\\\sigma _{23}&=\mu (2{\frac {S}{2\mu b}}x_{2})\end{aligned}}$
Assume that the body force is zero. Then the equilibrium condition is $\nabla ^{2}{u_{3}}=0$. Therefore,
    ${\begin{aligned}&u_{3,11}+u_{3,22}=0\\{\text{or,}}\quad &2A+2{\frac {S}{2\mu b}}=0\\{\text{or,}}\quad &A=-{\frac {S}{2\mu b}}\end{aligned}}$
Therefore, the stresses are given by
    ${\sigma _{13}=-{\frac {S}{b}}x_{1}~;~~\sigma _{23}={\frac {S}{b}}x_{2}}$
Step 5: Satisfy displacement BCs
The displacement is given by
    $u_{3}=-{\frac {S}{2\mu b}}x_{1}^{2}+{\frac {S}{2\mu b}}x_{2}^{2}+F$
If we substitute $x_{1}=a$, we cannot determine the constant $F$ uniquely.
Hence the displacement boundary conditions have to be applied in a weak sense,
    ${\begin{aligned}&\int _{0}^{b}u_{3}(a,x_{2})dx_{2}=0\\{\text{or,}}\quad &\int _{0}^{b}\left(-{\frac {S}{2\mu b}}a^{2}+{\frac {S}{2\mu b}}x_{2}^{2}+F\right)dx_{2}=0\\{\text{or,}}\quad &\left.\left[{\frac {S}{2\mu b}}\left(-a^{2}x_{2}+{\frac {x_{2}^{3}}{3}}\right)+Fx_{2}\right]\right|_{0}^{b}=0\\{\text{or,}}\quad &{\frac {S}{2\mu b}}\left(-a^{2}b+{\frac {b^{3}}{3}}\right)+Fb=0\\{\text{or,}}\quad &{\frac {S}{2\mu }}\left(-a^{2}+{\frac {b^{2}}{3}}\right)+Fb=0\\{\text{or,}}\quad &{\frac {S}{2\mu }}\left(-{\frac {a^{2}}{b}}+{\frac {b}{3}}\right)+F=0\\{\text{or,}}\quad &F={\frac {S}{2\mu b}}\left(a^{2}-{\frac {b^{2}}{3}}\right)\end{aligned}}$
Therefore,
    ${u_{3}={\frac {S}{2\mu b}}\left(x_{2}^{2}-x_{1}^{2}+a^{2}-{\frac {b^{2}}{3}}\right)}$
# Elasticity/Airy example 1
## Example 1 - Beltrami solution

Given:
Beltrami's solution for the equations of equilibrium states that if
    ${\boldsymbol {\sigma }}={\boldsymbol {\nabla }}\times {\boldsymbol {\nabla }}\times \mathbf {A}$
where $\mathbf {A}$ is a stress function, then
    ${\boldsymbol {\nabla }}\bullet {\boldsymbol {\sigma }}=0~;~~{\boldsymbol {\sigma }}={\boldsymbol {\sigma }}^{T}$
Airy's stress function is a special form of $\mathbf {A}$, given by (in 3$\times$3 matrix notation)
    $\left[A\right]={\begin{bmatrix}0&0&0\\0&0&0\\0&0&\varphi \end{bmatrix}}$
Show:
Verify that the stresses when expressed in terms of Airy's stress function satisfy equilibrium.

### Solution

In index notation, Beltrami's solution can be written as
    $\sigma _{ij}=e_{imn}~e_{jpq}~A_{mp,~nq}$
For the Airy's stress function, the only non-zero terms of $A_{mp,~nq}\,$ are $A_{33,~nq}=\varphi _{,nq}\,$ which can have nine values. Therefore,
    ${\begin{aligned}\sigma _{11}&=e_{13n}~e_{13q}~\varphi _{,~nq}\\\sigma _{22}&=e_{23n}~e_{23q}~\varphi _{,~nq}\\\sigma _{33}&=e_{33n}~e_{33q}~\varphi _{,~nq}\\\sigma _{23}&=e_{23n}~e_{33q}~\varphi _{,~nq}\\\sigma _{31}&=e_{33n}~e_{13q}~\varphi _{,~nq}\\\sigma _{12}&=e_{13n}~e_{23q}~\varphi _{,~nq}\end{aligned}}$
Since $e_{33k}=0\,$ for $k=1,2,3$, the above set of equations reduces to
    ${\begin{aligned}\sigma _{11}&=e_{13n}~e_{13q}~\varphi _{,~nq}\\\sigma _{22}&=e_{23n}~e_{23q}~\varphi _{,~nq}\\\sigma _{33}&=0\\\sigma _{23}&=0\\\sigma _{31}&=0\\\sigma _{12}&=e_{13n}~e_{23q}~\varphi _{,~nq}\end{aligned}}$
Now, $e_{13k}\,$ is non-zero only if $k=2$, and $e_{23k}\,$ is non-zero only if $k=1$. Therefore, the above equations further reduce to
    ${\begin{aligned}\sigma _{11}&=e_{132}~e_{132}~\varphi _{,~22}\\\sigma _{22}&=e_{231}~e_{231}~\varphi _{,~11}\\\sigma _{33}&=0\\\sigma _{23}&=0\\\sigma _{31}&=0\\\sigma _{12}&=e_{132}~e_{231}~\varphi _{,~21}\end{aligned}}$
Therefore, (using the values of $e_{132}\,$, $e_{231}\,$ and the fact that the order of differentiation does not change the final result), we get
    ${\begin{aligned}\sigma _{11}&=\varphi _{,~22}\\\sigma _{22}&=\varphi _{,~11}\\\sigma _{33}&=0\\\sigma _{23}&=0\\\sigma _{31}&=0\\\sigma _{12}&=-\varphi _{,~12}\end{aligned}}$
The equations of equilibrium (in the absence of body forces) are given by
    $\sigma _{ji,j}=0$
or,
    ${\begin{aligned}\sigma _{11,1}+\sigma _{21,2}+\sigma _{31,3}&=0\\\sigma _{12,1}+\sigma _{22,2}+\sigma _{32,3}&=0\\\sigma _{13,1}+\sigma _{23,2}+\sigma _{33,3}&=0\end{aligned}}$
Plugging the stresses in terms of $\varphi$ into the above equations gives,
    ${\begin{aligned}\varphi _{,221}-\varphi _{,122}+0&=0\\-\varphi _{,121}+\varphi _{,112}+0&=0\\0+0+0&=0\end{aligned}}$
Noting that the order of differentiation is irrelevant, we see that equilibrium is satisfied by the Airy stress function.
# Elasticity/Antiplane shear

-- Return to Introduction to Elasticity --
Antiplane shear (or antiplane strain) is the state of strain that is obtained when the displacement field is of the form
    $u_{1}=u_{2}=0,~~u_{3}=u_{3}(x_{1},x_{2})$
There is only an out of plane displacement.
Since the strains are given by
    $\varepsilon _{ij}={\cfrac {1}{2}}(u_{i,j}+u_{j,i})={\cfrac {1}{2}}\left({\cfrac {\partial u_{i}}{\partial x_{j}}}+{\cfrac {\partial u_{j}}{\partial x_{i}}}\right)$
we have
    ${\begin{aligned}\varepsilon _{11}&={\cfrac {\partial u_{1}}{\partial x_{1}}}=0\\\varepsilon _{22}&={\cfrac {\partial u_{2}}{\partial x_{2}}}=0\\\varepsilon _{33}&={\cfrac {\partial u_{3}}{\partial x_{3}}}=0\\\varepsilon _{23}&={\cfrac {1}{2}}\left({\cfrac {\partial u_{2}}{\partial x_{3}}}+{\cfrac {\partial u_{3}}{\partial x_{2}}}\right)={\cfrac {1}{2}}~u_{3,2}\\\varepsilon _{31}&={\cfrac {1}{2}}\left({\cfrac {\partial u_{3}}{\partial x_{1}}}+{\cfrac {\partial u_{1}}{\partial x_{3}}}\right)={\cfrac {1}{2}}~u_{3,1}\\\varepsilon _{12}&={\cfrac {1}{2}}\left({\cfrac {\partial u_{1}}{\partial x_{2}}}+{\cfrac {\partial u_{2}}{\partial x_{1}}}\right)=0\end{aligned}}$
Therefore, for antiplane shear, the only nonzero strains are the out-of-plane shear strains
    $\varepsilon _{23}={\cfrac {1}{2}}{\cfrac {\partial u_{3}}{\partial x_{2}}}~;~~\varepsilon _{13}={\cfrac {1}{2}}{\cfrac {\partial u_{3}}{\partial x_{1}}}$
# English-Chinese/Eating oat bran can lower cholesterol levels

Eating oat bran or whole oatmeal combined with a low-fat diet can lower cholesterol levels and reduce the risk of heart disease.
吃燕麦麸或全燕麦片与低脂饮食相结合可以降低胆固醇水平，减少患心脏病的风险。
# English-Chinese/Effective leaders have to take account of everyone's feelings

Effective leaders have to take account of everyone's feelings.
能干的领导者必须考虑到每个人的感受。
# English-Chinese/Eating more fruits, vegetables, and whole grains can prevent hemorrhoids

Eating more fruits, vegetables, and whole grains can prevent hemorrhoids.
多吃水果，蔬菜和粗粮可以预防痔疮。
# English-Chinese/Echocardiography, myocardial perfusion imaging, and cardiac stress testing is not recommended

Echocardiography, myocardial perfusion imaging, and cardiac stress testing is not recommended in those at low risk who do not have symptoms.
对于 没有症状的 低危人群，不推荐 超声心动图，心肌灌注显像 和 心脏压力测试。
# English-Chinese/Economics

Economics is the social science that studies the production, distribution, and consumption of goods and services.
经济学 是 研究 商品 和 服务的 生产，分配 和 消费的 社会科学。
# English-Chinese/Education

Education is about teaching, learning skills and knowledge. Education also means helping people to learn how to do things and encouraging them to think about what they learn. It is also important for educators to teach ways to find and use information.
教育是关于教学，学习技能和知识。教育还意味着帮助人们学习如何做事，并鼓励他们思考他们学到的东西。教育工作者教授查找和使用信息的方法也很重要。
# English-Chinese/Effectiveness in hypertension treatment

Even small decreases in blood pressure can have a large effect on a person's health. For example, decreasing blood pressure by 5 mmHg (for example, from 150/100 to 145/95 mmHg) can decrease the risk of stroke by 34%. It can also decrease the risk of heart disease by 21%.
即使 血压 下降 很小 也 会 对 人体 健康 产生 很大 影响。例如，将 血压 降低 5 mmHg（例如，从 150/100 至 145/95 mmHg）可以 将 中风 风险 降低 34％。它 还 可以 将 心脏病的 风险 降低 21％。
# English-Chinese/Effects of a sedentary work life or lifestyle can be either direct or indirect

Effects of a sedentary work life or lifestyle can be either direct or indirect.
久坐不动的工作生活或生活方式的影响可以是直接的，也可以是间接的。
# English-Chinese/Eating whole-grain barley, as well as other grains with lots of fiber, improves regulation of blood sugar (i.e., reduces blood glucose response to a meal)

Eating whole-grain barley, as well as other grains with lots of fiber, improves regulation of blood sugar (i.e., reduces blood glucose response to a meal).
吃全麦大麦以及其他含有大量纤维的谷物可以改善血糖的调节（即降低血糖对膳食的反应）。
# English-Chinese/Edible fruits

Edible fruits, in particular, have propagated with the movements of humans and animals in a symbiotic relationship as a means for seed dispersal and nutrition; in fact, humans and many animals have become dependent on fruits as a source of food.
特别是食用果实随着人类和动物的移动以共生关系繁殖，作为种子传播和营养的手段; 事实上，人类和许多动物已经变得依赖果实作为食物来源。
# Catalogue 22/Descending the Landscape

Part of Catalogue 22, Catalogue 22 on Google Maps
Designer: Natalie Petricca Degree: B. Architecture School: Carleton University Year: 2012 Location of Project: Rovaniemi, Finland Project Status: Unbuilt
‘Descending the Landscape’ is a hotel, sauna and restaurant located in Rovaniemi Finland and designed with a sensitivity towards light, especially the Northern Lights, which became a guiding principle of the design. Each space was carefully designed so that from a specific angle the sky could be observed.
Site orientation allows visitors to encounter the most public space, the restaurant, first and then descend slowly down the landscape, with the spaces becoming more and more private. In the restaurant, each table has their own window that frames the natural landscape surrounding the site and the celestial phenomena. In the sauna, the Northern Lights can be observed while sitting in a reclined position, halfway through the sauna cycle. In the hotel room, the most private of all the spaces, the Northern Lights can be observed while lying in bed, with the beds being oriented so that the guests face North, where the most amount of activity happens. While the roofs are used to frame the Northern Lights, the walls are used to frame day light. In the shower in the hotel bathrooms, a small slit of light is positioned so that light cascades over the users body in the morning hours. In the sauna, meditation is separated into three distinct areas, with the most meditative space featuring small slits of light, the intermediate area having a view up to the Northern Lights i a reclined position and the most social space having a full view out to the surrounding landscape and out over the hotel rooms below.
# Catalogue 22/Dinner By the River

Part of Catalogue 22, Catalogue 22 on Google Maps
Designer: Jessica Hickling Degree: B. Architecture School: Carleton University Year: 2009 Location of Project: Rideau River at Carleton University Project Status: Built
“Sugar Spice and Everything Ice” a group design build project collaboration to design a temporary pavilion in order to serve food and drink to our critics. The sense of temporary was emphasized by our bold choice of material. Almost half of the structure was made out of ice illuminated with candles and battery powered lights. The central element is the fire cooking surface with an ice ring around it where the caramelized apples were prepared then served on ice cream in a tule cup. Seating was constructed by stacking and clamping newspapers together to create a recycled and quite comfortable seating arrangement. The use of light, smoke, ice and fog combined allowed us to create a pavilion which connected to the river site and the temporary nature of the project.