text:
[This article was first published on rOpenSci - open tools for open science, and kindly contributed to R-bloggers]. (You can report issue about the content on this page here)
Want to share your content on R-bloggers? click here if you have a blog, or here if you don't.

rOpenSci’s package review system (aka onboarding) is one of our key activities to improve quality and sustainability of scientific R packages. The editorial team are constantly working towards improving the experience for both authors and reviewers. After our first year, we surveyed authors and reviewers who participated in our onboarding process to help us better understand what’s working well and where there is room for improvement. At the end of last year, we did so again, re-designing our survey so as to better track participant opinions year-to-year. In this post we summarize the 45 responses that we received and what we’re doing to address your feedback.

### Section 1: Satisfaction with the review and onboarding process

We’re glad to see high levels of overall satisfaction with the onboarding process. We primarily ask this question to establish a baseline to compare year-to-year, and hope that we can maintain high satisfaction as we continue to grow.

### Section 2: Guidelines for authors and reviewers

We provide packaging and reviewing guides for our authors and reviewers to consult before and during the peer-review process. While both were highly rated, enthusiasm was greater for the packaging guide than the reviewing guide.

For the packaging guide, authors gave feedback that they could be better organized, including a suggestion that the collection of markdown files be converted to a bookdown website for easier navigation. We thought this was a great suggestion, and have begun the process of porting all our documentation for packaging and reviewing, into a common bookdown book. Editor Maëlle Salmon is leading this work, and you can find the in-progress book, rOpenSci Packages: Development, Maintenance, and Peer Review, here.

We also got feedback from authors that they wanted guidance on how best to acknowledge the efforts of reviewers. This was one of our drivers to get reviewers recognized as MARC author type in package DESCRIPTION files, which you can read more about in in this previous blog post.

For the reviewing guide, reviewers suggested greater structure, including separation between code and UI reviewing, more advice and examples on how to approach review, better guidance on expectations for timelines and collaboration between multiple reviewers. We aim to fold all these suggestions into the book. There are great posts about how to tackle a peer review on our blog that we will incorporate.

### Section 3: The review process

This was perhaps the most controversial topic in the responses. Both authors and reviewers were conflicted about the best time to submit for review. Submitting early in development could allow reviewers to provide concrete feedback on software design and usability, but it is generally hard to review incomplete software. On the other extreme, software that is mature and already on CRAN may not be able to implement reviewer suggestions, especially if it would result in breaking changes for users. One suggestion was to consider packages in both stages and provide short (less than one hour of review) for early stage packages and reserve the more detailed reviews for the mature packages.

This has been a perennial tension in the review process. A similar split came up in our previous survey, and Miles McBain recently wrote about it in his blog post about reviewing bowerbird. As we don’t see a strong net pull in either direction, we don’t plan to change submission requirements, and we’re worried that a multi-step process would make too much work for reviewers. One useful suggestion, though, is for authors of early-development packages to open discussions on our forum to solicit early input.

Some respondents noted the challenge of providing an objective review under a fully open system, and one suggestion was to consider a single blinded system in order to allow reviewers (especially more early career folks) to provide critical comments without fear of retaliation.

#### Automated checking

We received specific feedback about our use of the goodpractice package to provide automated checks of some package quality issues. Respondents suggested that authors should pre-run these checks before submitting, and also to provide better guidance as to which outputs can be ignored. We’re adding both suggestions to the new bookdown book.

There were also suggestions for improvement of goodpractice itself. goodpractice was created by Gábor Csárdi and is maintained by Mango Solutions. Thankfully Mango data scientist Hannah Frick has undertaken a project to improve the package and its documentation in recent weeks, and is incorporating suggestions we’ve passed on from the rOpenSci survey.

As goodpractice allows users to ignore some checks and add custom checks, we also aim this year to have a specialized set of rOpenSci checks that will align better align with our packaging guide. Finally, we hope to have these checks run automatically upon submission, reducing effort for both authors and editors.

### Section 4: Value of software review

Package authors are given the option to submit directly to the Journal of Open Source Software (JOSS) for publication.

#### The good and the bad with onboarding

We asked authors to tell us some of the best and worst attributes of participating in onboarding.

Aspects of onboarding that were challenging?

• Deciding when to submit (see above)
• Package review is quite time consuming.
• Keeping up with ongoing revisions to the package during submission.
• Long time lag between review and revision.

Aspects of onboarding that were great

• The opportunity to network and share ideas.
• The openness and transparency of the entire review process.
• Learning about software design, package development, and improving one’s own coding skills.
• The opportunity to have two thorough code reviews which are quite difficult to get elsewhere.
• The collegial and non-adversarial nature of the process.

#### Did blogging post-acceptance have any impact?

At the completion of each review, the editors invite package authors to submit either a long-form blog post aimed at a general reader, or a shorter technical note. Most authors who wrote posts reported greater visibility for their package and some also enjoyed reflecting on the development process.

#### Upcoming improvements to onboarding

We’re grateful to all the authors and reviewers that took this survey. The responses have helped us improve our workflow quite a bit and we are excited to roll out several improvements in the coming weeks to months. Here are a few things to look forward to:

• A much improved (bookdown) guide for both package development and package reviewing.
• Improvements to goodpractice thanks to Mango The Cat’s Hannah Frick and also a goodpractice API down the line.
• A new and improved reviewer sign up form that will make it much easier for editors to match reviewers to submissions.
• Suggestions are always welcome even if you missed this round of the survey. Chime in in the issue tracker of our in-progress guide or in this other issue tracker where we keep track of more general questions & suggestions about onboarding.

We hope to see your package or review expertise on the onboarding repo soon!


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# SNAMP Performance Tips

## Memory calculation

The minimum amount of memory required for single SNAMP process is 512 MB. This amount of memory is enough for launching single gateway and one or two resource connectors.

RAM calculation methodology reflects SNAMP requirements only. So, the total RAM space installed in your hardware must be greater than space required by SNAMP because OS and other daemon processes utilize its own memory. Pay your attention: SNAMP should not swap its memory.

If you have a plan to use up to 10 resource connectors then allocate 1 GB per SNAMP process. Tens of connected resources can be served with 2 GB heap size. In a clustered installation with tens of connected resources we highly recommend to use 4 GB heap size.

## JVM settings

JVM tuning aims to minimize GC pauses. We offer two main strategies on GC tuning:

1. Peak SNAMP performance is the first priority and there are no pause time requirements (or pauses of 1 second or longer are acceptable)
2. SNAMP response time is more important than overall throughput

You may select the most convenient strategy based on your enterprise IT policies.

In most scenarios, number of monitoring & management tools are fixed in time (i.e. number of clients are fixed). If that is your case - choose the second strategy.

According to memory calculation methodology you may specify minimum and maximum Java memory:

• For Linux, go to bin directory (within the SNAMP folder) and open bin/setnv and specify
• export JAVA_MIN_MEM=512m
• export JAVA_MAX_MEM=X, where X is your calculated memory, for example export JAVA_MAX_MEM=2G
• For Windows, go to bin directory (within the SNAMP folder) and open bin/setenv.bat
• SET JAVA_MIN_MEM=512m
• SET JAVA_MAX_MEM=X, where X is your calculated memory, for example SET JAVA_MAX_MEM=2424m

We recommend you not to specify PermGen settings

Additional JVM settings can be specified in EXTRA_JAVA_OPTS environment parameter. The following example demonstrates setup of G1GC garbage collector:

export EXTRA_JAVA_OPTS="-XX:+UseG1GC"

### Memory utilization profile

All the Java objects created during SNAMP execution can be divided by its lifecycle:

• Long-lived objects
• Short-lived objects

Short-lived objects are being created for each attribute request or operation execution from monitoring & management tool, and notification delivery to monitoring & management tool. These objects are never being placed into Old Generation but require a right tunes size of Young Generation Heap (Eden and Survivor spaces). The number of short-lived objects depends on the number of requests per second.

Long-lived objects are being created for each connected managed resource and gateway. So, gateway instance and resource connector instance are long-lived objects. Changing of SNAMP configuration causes releasing long-lived objects in Old Generation Heap (Tenured space). Therefore, if you expect frequent reconfiguring then setup the large OldGen heap (to avoid pauses).

### Throughput first

If peak application performance is the first priority and there are no pause time requirements (or pauses of 1 second or longer are acceptable), then select the parallel collector with -XX:+UseParallelGC. You might also specify -XX:+UseParallelOldGC if you expect frequent SNAMP reconfiguring.

Also, you can specify -XX:GCTimeRatio=<N> JVM option. For example, -XX:GCTimeRatio=19 sets the goal of 1/20 or 5% of the total time in garbage collection.

### Response time first

If response time is more important than overall throughput and garbage collection pauses must be kept shorter than approximately one second, then you may choose one of the following collectors:

• Concurrent Mark & Sweep -XX:+UseConcMarkSweepGC -XX:+CMSClassUnloadingEnabled. Additionally, specify -XX:+CMSIncrementalMode if only two cores are available.
• G1 collector with -XX:+UseG1GC. That requires large heaps - about 6GB or larger, and stable and predictable pause time below 0.5 seconds.

## Number of cores

Each gateway or resource connector uses its own isolated thread pool.
So, recommended number of cores (k) is based on the following metrics:

• λ - expected workload from single monitoring tool, in RPS (requests per second)?
• t - average response time from managed resources (in seconds)?
• P - availability

SNAMP represents multichannel Queuing System so Queuing Theory is applicable for necessary computations:

2. Downtime probability:
3. Availability:

Availability formula contains the required number of cores in indirect form. There is no way to reduce this formula. Therefore, you can use the following simple JS program (use NodeJS or any other JavaScript interpreter) for computing required number of cores:

//source data
var lambda = 50;
var t = 0.1;
var P = 0.999;

//computation
function fact(x){ return x > 1 ? x * fact(x - 1) : 1; }

var rho = lambda * t;

var k = 2;
while(true){
var sum = 0;
for(var i = 0; i <= k; i++){
sum += Math.pow(rho, i) / fact(i);
}
sum = 1 / sum;
sum = sum * Math.pow(rho, k) / fact(k);
sum = 1 - sum;
console.log("Availability %s k %s", sum, k);
if(sum >= P) { console.log("Required number of cores %s", k); return;}
else k += 1;
}

Examples:

1. λ = 50 rps, t = 0.1 seconds and expected availability is P=0,999 (99,9%) then required number of cores k = 14
2. λ = 2 rps, t = 0.3 seconds and expected availability is P=0,99 (99%) then required number of cores k = 4

Many modern CPUs support simultaneous multi-threading (SMT) when one physical CPU core may process two (or more) threads in parallel. In this case, number of cores means number of logical cores.

For example, if workload λ = 50 rps, t = 0.1 seconds and expected availability is P=0,999 (99,9%) then required number of threads is S=14. Therefore, the required number of (logical) cores k = 14/1.5= [9.3]=10. The savings on the number of cores is 28%.

format: plain


text:
# The force applied against an object moving horizontally on a linear path is described by F(x)= x+3 N . By how much does the object's kinetic energy change as the object moves from  x in [ 1 , 2 ]?

Dec 29, 2015

It increases by $4 \text{J}$

#### Explanation:

Work done = force x distance moved in direction of force.

Because force varies with displacement we can say:

$W = {\int}_{1}^{2} F . \mathrm{dx}$

$\therefore W = {\int}_{1}^{2} \left(x + 3\right) . \mathrm{dx}$

$W = {\left[{x}^{2} / 2 + 3 x\right]}_{1}^{2}$

$W = \left[\frac{4}{2} + 6\right] - \left[1 + 3\right]$

$W = 8 - 4 = 4 \text{J}$

This work will appear as kinetic energy.


text:
# Tag Info

## Hot answers tagged momentum

12

First, something we need to get out of the way: Kinetic energy as $\frac{1}{2} m v^2$ is not a precise formula; it is merely a good approximation for anything that is traveling slowly compared to the speed of light. In fact, more precisely, the energy is E = m\, c^2\, \frac{1}{\sqrt{1-v^2/c^2}} \approx mc^2 + \frac{1}{2} m v^2 + ...

6

Your lecturer got the eigenvalue using the fact that the operator $\hat{p}$ is Hermitian so you can do this: \begin{align} \langle p| \hat{p} &= \left( \hat{p}^\dagger |p\rangle\right)^\dagger\\ &= \left( \hat{p} |p\rangle\right)^\dagger\\ &= \left( p |p\rangle\right)^\dagger\\ &= \langle p| p \end{align} I think it ...

5

That formula for momentum is only true for massive particles. Here's what is always true: A particle with a mass $m$ ($\geq 0$) can have an arbitrary momentum $p$ (in some direction, with magnitude $\geq 0$). The energy of such a particle is $$E = \sqrt{m^2c^4 + p^2c^2}$$ The velocity of a particle is equal to $$v = \frac{pc^2}{E}$$ When $m = 0$, $E ... 5 The problem does not mention any radii, but if we did know the radius of each sphere, would it be possible to skip conservation-of–linear momentum calculations altogether The accepted answer has confused you: You can simplify things by considering these conservation laws in the center of mass frame. There the total momentum is zero, therefore ... 4 In both cases and at all times, the force from the (wall/tire) on the hammer equals the force from the hammer on the (wall/tire) : total momentum must be conserved. However, in the first case, the initial energy is dissipated in the wall (as heat and/or damage), so at the end the hammer is stopped. In the second case the initial energy is stored as ... 4 Inertia is an intrinsic characteristic of the object related to its mass. Inertia tells you how much force it will take to cause a particular acceleration on the object. Momentum is a function of an object's mass and velocity. Momentum is a measure of the kinetic energy of the object. A massive object can have any momentum (at least as long as its velocity ... 4 Momentum: The resistance of an object to a change in its state of motion. That sounds like a fishy definition of momentum to me. A slightly better definition, at least at your level, is that momentum represents the "amount of motion" an object has. Granted, "amount of motion" is a very vague term, but it stands to reason that if "amount of motion" were ... 4 A few quick clarifications: a particle cannot just annihilate. It disappears when it interacts with something else. The obvious example of this is an electron and positron annihilating to turn into two photons. Also, the total energy of a particle (this applies to electrons, positrons and photons) is given by: $$E^2 = p^2c^2 + m^2c^4$$ where$p$is the ... 4 Yes,$F=ma$, but also$v=at$. That means that, as you fall for a longer time, your speed will increase. After 1 second, you are going at$9.8 m/s$or$35 km/h$, about the speed of Usain Bolt. After 10 seconds you would reach$98 m/sec$or$350 km/h$. For a free-falling human, the air resistance actually limits you to about$200 km/h$. When you hit the ... 3 You are right. If$q$is a generalized coordinate then$\dot{q}$is the generalized velocity and hence the generalized momentum is $$p = \frac{\partial L}{\partial\dot{q}}$$ Therefore, your sequence looks correct. Further, equations (20) and (21) of the article you have referenced also tell that the$p_{\theta_i}$are indeed ... 3 Find the nearest box-spring mattress. With your hand, execute a slow motion "impact" between your hand and the mattress. You should notice that when your hand is not touching the mattress, there is, of course, no force between your hand and the mattress. As your hand begins to touch the mattress, you should feel a very light force. As your hand presses ... 2 The thrust coming from a rocket engine is exerted on the engine bell, and it is directed along its axis of symmetry. It's not completely clear how you're modelling your ship but it is probably more realistic to apply the force to the "thruster fire" block, whatever that is. It's important to note, though, that if applying the force to the engine bell and ... 2 If the collision is not perfectly along the line connecting the centers of mass of the pucks, they will exert torques on each other as well as forces. The angular momentum of the pair will be conserved, so if the incoming puck was not spinning, the pucks will exit the collision spinning in opposite directions. If the surface they slide on is frictionless, ... 2 Your analysis will need to be a combination of theory and experiment. You are not maximizing impulse alone: you need to think about drag as well. So you need to think about the shape of your rocket as well as the volume of water and initial pressure. The drag on the rocket is probably well modelled by the ram pressure equation ... 2 Force is a concept which describes, and can be used to manipulate, real phenomena that exist regardless of the existence of the human race. Newton's second law (Acceleration = Force / Mass) is a definition of force. Mass certainly exists, as do velocity and its time derivative, acceleration. There is no reason to suppose that force does not exist. Force ... 2 Without defining what a thing is, it makes little sense to discuss the ontology of a thing. Does an apple exist? First, one must say what an apple is; once we agree on that, it's straightforward to show (by example) that apples exist. Given a definition of force, force certainly does exist; we can point to time derivatives of momentum that we observe, and ... 1 This seems like a list question which aren't very good for SE. Here is a proof of your statement:$m_1v_1 = m_2v_2\dfrac{m_1}{m_2} = \dfrac {v_2}{v_1}$Let us assume that$m_1$<$m_2\dfrac{(m_1)^2}{(m_2)^2} = \dfrac {(v_2)^2}{(v_1)^2} < 1v_1$must be greater than$v_2$The only case the energy of a higher velocity is beaten is shown by ... 1 If you consider things classically (for the moment forgetting about virtual particles as mediators of the force) things get more clear. For instantaneous forces (which do not exist in nature), momentum conservation comes from the fact, that the forces in nature fulfil Newtons axiom actio = reactio, meaning, that for two particles, that interact we have the ... 1 I will bring an example from classical electrodynamics. In EM(electromagnetism) you have to consider that the fields(electrical and magnetic) also have energy and momentum. A classical example is to apply the third law of Newton(each action has an equal counter-action) to two moving charges. Then you'll conclude that the third law does not hold-thus the ... 1 You didn't cancel out the mass$m$properly. $$-mv_0\cos a+0=mv_1\cos b+0\\ -v_0\cos a=v_1\cos b\\ \frac{-v_0\cos a}{\cos b}=v_1$$ The incident angle is equal to the exit angle in such collision,$a=b\$. The above reduces to: $$v_1=\frac{-v_0\cos a}{\cos b}=\frac{-v_0\cos b}{\cos b}=-v_0$$ And here you see what you probably expected. In the perpendicular ...

1

Hint: The center of mass of pieces A and B moves in the same path as the intact shell would. (This arises from the conservation of momentum.) Edit: The gravitational force is only acting along the vertical direction. So there is net acceleration only in the vertical direction. Looking at the horizontal one, if we neglect things like air resistance (which ...

1

If we did know the radius of each sphere, would it be possible to skip conservation-of–linear momentum calculations altogether and use a single conservation-of-energy equation... The radius of a ball is always irrelevant to the outcome of a collision, what counts is the ratio of the masses. Two equations are always necessary to determine the outcome ...

1

As CuriousOne inexplicably said in the comments, but not as a formal answer, you should use this equation: $$\vec{L}=m{\vec{r}}\times{\vec{v}}$$ This is the standard equation for angular momentum in vector form. Once you have your angular momentum vector, you can get the individual components. You can see how to take a cross product here. If you need to ...

Only top voted, non community-wiki answers of a minimum length are eligible


text:
• Create Account

### #Actual0Circle0

Posted 19 January 2013 - 06:48 PM

EDIT: Sounds like you want to work with DLL files. Which is not SDL specific but C specific when it comes to programming in SDL.

I always seperated my functions, variables, and DATA files into separate files and put them into *.h headers.

Then in my main.c I'd just do this

#include "data.h"
#include "Variables.h"
#include "Functions.h"

And if you're making a game it is going to be very large to begin with. It is all how you code and how much you are going to put into it. In (WOW)Wrath of the Lich King, there were approximately 5.5 million lines of code in the game, not including database entries and what not. I don't understand what you mean by your game will get big.

### #20Circle0

Posted 19 January 2013 - 06:48 PM

EDIT: Sounds like you want to work with DLL files. Which is not SDL specific but C specific.

I always seperated my functions, variables, and DATA files into separate files and put them into *.h headers.

Then in my main.c I'd just do this

#include "data.h"
#include "Variables.h"
#include "Functions.h"

And if you're making a game it is going to be very large to begin with. It is all how you code and how much you are going to put into it. In (WOW)Wrath of the Lich King, there were approximately 5.5 million lines of code in the game, not including database entries and what not. I don't understand what you mean by your game will get big.

### #10Circle0

Posted 19 January 2013 - 06:45 PM

I always seperated my functions, variables, and DATA files into separate files and put them into *.h headers.

Then in my main.c I'd just do this

#include "data.h"
#include "Variables.h"
#include "Functions.h"

And if you're making a game it is going to be very large to begin with. It is all how you code and how much you are going to put into it. In (WOW)Wrath of the Lich King, there were approximately 5.5 million lines of code in the game, not including database entries and what not. I don't understand what you mean by your game will get big.

PARTNERS


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# chronograph: measure timings of data evaluation

[ bsd3, library, testing ] [ Propose Tags ]

The Chronograph data structure adds a measure field to an existing Haskell expression. This field will be the time necessary to evaluate the expression using an arbitrary evaluation strategy (WHNF by default).

Chronograph preserves laziness, so that the work of performing the evaluation is only done after the result is demanded.

If you want to benchmark your program, you should look to other packages like Criterion that perform statistical analysis of your results so you can determine how reliable they are. Chronograph just takes measurements, leaving the interpretation entirely to you.

[Index]


text:
# ChangeLog: What’s 🆕 at Infinite InkUpdated  2022-September-26

## General News

This section lists recent general news related to Infinite Ink.

Always
• Digital gardening.

2022-September
• Started including lite.qwant.com in the Searching for… sections of some portals. (I’ve also set Qwant Lite as the default search engine in some of my web browsers.)

2022-August
• Started calling this page ChangeLog (rather than News, Updated, or What’s New).

2022-July
2022-June
• For part of June, some Infinite Ink pages wore orange.[1] To learn about this bit of hacktivism, see wearorange.org/about/.

• On some pages, experimenting with previous-next links above the page title. For a long time, previous-next links have been in the footer of most (not all)[2] pages.

2022-May

For more Infinite Ink news, see the rest of this page and Infinite Ink’s To-Do, Doing, and Done Lists👷♀.

2022-June
2022-May

## Endnotes

1. The #activism-hacktivism Portal is still wearing orange thanks to its stylesheet including body{background:#FFDFBF;}. That shade of orange (    ) is sometimes called “orange pale orange.”
2. Infinite Ink’s home page, Portals page, and this ChangeLog page do not have previous-next links.

## Comments 📝 🤔 👎 👍

Your comment might immediately improve this page or help me to (eventually) improve this page. To post or view comments, click this paragraph.


text:
Let f(x) be a function such that,f(0)=f′(0)=0 ,f′′(x)=sec4 x+4 then the function is

# Let f(x) be a function such that,$\mathrm{f}\left(0\right)={\mathrm{f}}^{\mathrm{\prime }}\left(0\right)=0$ , then the function is

1. A

2. B

3. C

4. D

None of the above

Register to Get Free Mock Test and Study Material

+91

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text:
# Summarize raw data in new sheet

Note: This question has a follow up question here.

I have an Excel workbook with two sheets: "Raw data" and "Summary". In the raw data sheet there are several blocks with data structured like the image below:

I want to create a summary of the data below, in the sheet "Summary", looking like this:

The way I've done this is:

Sub TransferAndSumElements()
Dim wb As Workbook
Dim ws1 As Worksheet
Dim ws2 As Worksheet

Set wb = ActiveWorkbook
Set ws1 = Sheets("Raw data")
Set ws2 = Sheets("Summary")

'##############
' Copy / Paste first row
'##############
ws1.Select
Rows("1:1").Activate
Selection.Copy
ws2.Select
Range("A1").Select
ws2.Paste

'##############
' Set Item names in summary
'##############

ws2.Select
Range("A2").Select
ActiveCell.FormulaR1C1 = "Knife"
Range("A3").Select
ActiveCell.FormulaR1C1 = "Fork"
Range("A4").Select
ActiveCell.FormulaR1C1 = "Spoon"
Range("A5").Select
ActiveCell.FormulaR1C1 = "Spork"
Range("A6").Select
ActiveCell.FormulaR1C1 = "Bowl"
Range("A7").Select

'##############
' Find Elements and Sum
'##############

Dim s As Long           ' Sum of elements for each Item
Dim str As String       ' String in first column in "Summary" (Knife, Fork ...)
Dim cellstr As String   ' String in first column in "Raw data" (To be searched through)
Dim DataRange As Range  ' All rows in sheet 1 ("Raw data")

Dim cellA As Range      ' Used to find cellstr = cellA.Value
Dim cellB As Range     ' Used to increment s = s + cellB.Value

ws1.Select              ' Select first worksheet ("This worksheet")
LastRow = ws1.Cells.Find("*", searchorder:=xlByRows, searchdirection:=xlPrevious).Row
Set DataRange = Range("A1:A" & LastRow)

For i = 1 To 5          ' Iterate through Knife, Fork .. in "Summary"...

str = ws2.Cells(i + 1, 1).Value     ' Extract names
s = 0                               ' Initialize sum
ws1.Activate                        ' Activate Worksheet 1 ("Raw data")

For r = 1 To LastRow
Set cellA = Range("A" & r, "A" & r)
cellstr = cellA.Value
If InStr(cellstr, str) > 0 Then
Set cellB = Range("B" & r, "B" & r)
s = s + cellB.Value
End If
Next r

ws2.Cells(i + 1, 2).FormulaR1C1 = s

Next i

ws2.Activate

End Sub

Some stuff that I want to improve, but haven't been able to:

• Automatically fetch the item names from "Raw data". I've tried creating a Collection, but I end up with an empty row in between the others (Knife, Fork, Spoon, Spork, Blank, Bowl).
• I can loop through this list after creating it, and delete the blank row, but I think there should be a way to do this in one go.
• I want to create a list of words ("Knife", "Fork" .. ) and write it to Excel in one command. The way I'm doing it now is to manually write each item name.
• Do the loops make sense, or should I do it in some other way?

And last but not least:

• What about the big picture? Does it make sense to do it like this, or should I do it completely different?

Note: The summary can be sorted alphabetically, by number of elements, or not sorted at all. That doesn't matter.

Tab delimited data dump:

Item    Number
Knife   2
Fork    2
Spoon   1
Spork   2

Item    Number
Spork   2
Fork    3

Item    Number
Bowl    3
Knife   5

• Do you have any power to restructure the raw data? Because that structure desperately needs improving. For more on useful excel data-structures, Joel Spolsky did an excellent introductory tutorial
– Kaz
Jun 16 '16 at 12:01
• I'll check out the link, but first to answer your question: The data above is just a made up format. I chose the format deliberately: Unsorted elements, Variable number of elements in each block, and a header row. I do in fact use a program that outputs data similar to this, only messier. The program is closed, with no API or anything, so I can't alter the formatting of that one. Jun 16 '16 at 12:08

## 1 Answer

You should never, ever, be using Select or Active... in your code.

Unless you explicitly want your sub to click on a cell, or you explicitly want to work off of a user-selection, this is an incredibly slow, inefficient, fragile way of doing things.

This:

Range("A1").Select
Selection.FormulaR1C1 = "Knife"

Or

Range("A1").Select
ActiveCell.FormulaR1C1 = "Knife"

Is much better expressed like so:

Range("A1").FormulaR1C1 = "Knife"

Furthermore, Range("A1") is implicitly calling [ActiveSheet].Range("A1"). You should explicitly qualify your references to avoid bugs down the road. Like so:

ws2.Range("A1").FormulaR1C1 = "Knife"
ws2.Range("A2").FormulaR1C1 = "Fork"
ws2.Range("A3").FormulaR1C1 = "Spoon"

And then you can use a With statement to make it even better:

With ws2
.Range("A1").FormulaR1C1 = "Knife"
.Range("A2").FormulaR1C1 = "Fork"
.Range("A3").FormulaR1C1 = "Spoon"
End With


text:
Article Text

Genetic associations in peripheral joint osteoarthritis and spinal degenerative disease: a systematic review
1. J J Ryder1,
2. K Garrison1,
3. F Song1,
4. L Hooper1,
5. J Skinner1,
6. Y Loke1,
7. J Loughlin2,
8. J P T Higgins3,
9. A J MacGregor1
1. 1
Institute of Health, University of East Anglia, Norwich, UK
2. 2
Institute of Musculoskeletal Sciences, Botnar Research Centre, Nuffield Orthopaedic Centre, University of Oxford, Oxford, UK
3. 3
MRC Biostatistics Unit, Institute of Public Health, University of Cambridge, Cambridge, UK
1. J J Ryder, School of Medicine, Health Policy and Practice, Institute of Health, University of East Anglia, Norwich NR4 7TJ, UK; Jon.Ryder{at}uea.ac.uk

Abstract

We conducted a systematic review of genetic association studies for osteoarthritis of the peripheral joints (OA) and spinal degenerative disease (SDD). Electronic searches were carried out for any English language article reporting on a gene association study for either OA or SDD published up until the end of 2006. A team of seven reviewers used a standardised template to extract data in duplicate. In all, 90 studies fulfilled our inclusion criteria, reporting a total of 94 significant associations from 83 different genes. We found relatively few instances in which a specific gene–disease association had been analysed by more than one study, and there were 14 cases in which significant associations were replicated in independent studies (at joints associated with the AGC1, ASPN, COL9A2, COL9A3, COL11A2, ESR1, FZRB, HFE, IL1A, IL1RN, PTGS2 and VDR genes). Methodological and reporting problems were widespread, including failure to report full results, missing population details, multiple testing, and over-reliance on subgroup analysis. In summary, the complex phenotypes of OA and SDD may have made it difficult for researchers to focus their efforts. The field is dominated by isolated analyses of disparate potential associations, a problem that is amplified by the frequent analysis of different polymorphisms within individual genes. Flaws in study methodology and interpretation undoubtedly increase the risk of publication bias. Closer adherence to published recommendations (in particular those produced by HuGENet) will help to ensure that future studies are well-designed and build on current understanding, rather than simply adding to the growing bank of potential associations.

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Twin and family studies demonstrate that between a half and two-thirds of the occurrences of osteoarthritis (OA) can be attributed to genetic factors.1 This finding has justifiably stimulated the search for specific genes,25 but the task is formidable, as has proven to be the case for other types of complex disease. Genetic factors may vary with the pattern and severity of disease and according to patient characteristics, such as gender and age. OA can also have a range of clinical and radiographic definitions, which further increases the difficulty. For example, while peripheral joint OA and spinal degenerative disease (SDD) share pathological and clinical features, the extent to which the aetiological determinants of these disease processes are similar is not known.

Genetic association studies provide a means of quantifying the effects of specific gene variants on disease occurrence, but interpreting such data can be problematic. Sample sizes, sample selection, population stratification effects, and differences in the pattern of confounding may all have an impact.6 Over the last two decades a large number of gene–OA association studies have been conducted, and it is increasingly recognised that an objective appraisal of such studies can only be approached through systematic review. To date the consistency of the results of all published studies of OA and SDD have not been evaluated in this way.

In this review, therefore, we aimed to retrieve and review all gene–OA and gene–SDD association studies that have been published to date in peer-reviewed, English language journals. Our first objective was to carry out a broad ranging systematic review of genetic association studies. We aimed to document the number of genes that have been screened and to assess the degree to which analyses of individual genes have been replicated by more than one study. Our second objective was to describe common methodological and reporting problems encountered within gene–OA/SDD association studies and to assess the frequency with which these problems occurred. Thus, we aim to help improve the way in which future studies are conducted in this rapidly expanding field.

LITERATURE REVIEW METHODOLOGY

Inclusion and exclusion criteria

A deliberately broad search strategy was used in which gene–OA and gene–SDD association studies of any design, dealing with any gene, were judged suitable for inclusion. We aimed to review associations on a per gene basis: included studies could therefore test for association with regard to one or more genetic polymorphism(s) within a particular gene. We excluded conference abstracts, review papers and non-English language papers.

Search strategy

Electronic searches were performed on Medline and Embase (using the Ovid platform, beginning with 1966 for Medline, and 1980 for Embase) and on the Science Citation Index (SCI; Thompson Web of Knowledge), for all studies matching our search criteria published up until the end of 2006. The general form of the search for Medline is given below. The terms “or” and “and” indicate the standard Boolean operators; $is an Ovid truncation symbol. Lower case terms indicate free-text searches applied to all text fields (title, abstract, keywords, etc). Capitalised words or phrases (eg, Osteoarthritis) indicate MeSH (Medical Subject Heading) terms, which were expanded to include index subheadings (including all available subcategories of disease). The Medline search was adapted for Embase (in which some MeSH terms differ from Medline) and for the SCI. The search was designed to be as broad as possible. We chose to identify a larger number of studies with the initial search and then exclude irrelevant studies manually, rather than risking using an excessively specific (and exclusive) strategy. In addition, gene–disease association studies are sometimes performed alongside other types of analysis (particularly linkage analysis), making a search for related types of study a useful means of identifying smaller association studies. Search terms and structure 1. osteoarthriti$ or osteoarthrosis or osteoarthrotic or spondylosis or osteophytosis

2. (degeneration or degenerative) adj6 (spine or spinal or disc or discs or prolapse)

3. Osteoarthritis or Osteoarthritis (Hip) or Osteoarthritis (Knee) or Spinal Osteophytosis

4. 1 or 2 or 3

5. gene or genes or genetic$or genic or geno$ or chromosom$or allel$ or homozygote or heterozygote or polymorph$or linkage$ or heritab$or inherit$ or mutat$or hered$

6. Heredity or Gene Frequency or Genotype or Gene Dosage or Genetic Predisposition to Disease or Haplotypes or Heterozygote or Homozygote or Inheritance Patterns or Genes (Dominant) or Genes (Recessive) or Multifactorial Inheritance or Quantitative Trait (Heritable) or Linkage (Genetics) or Linkage Disequilibrium or Lod score or Phenotype or Genetic Markers or Penetrance or Variation (Genetics) or Genetic Heterogeneity or Mutation or Polymorphism (Genetic) or Polymorphism (Restriction Fragment Length) or Polymorphism (Single Nucleotide) or Chromosomes or Genome or Genes or Alleles or Major Histocompatibility Complex or Quantitative Trait Loci

7. 5 or 6

8. 4 and 7

9. Limit the above to human studies

The records for each article were checked individually using the title and abstract, and those not dealing specifically with a gene–OA or gene–SDD association were excluded. This initial screen was carried out twice, once each by two members of the review team, to serve as a double check. If it was not possible to exclude articles at the screening stage (for example, if insufficient details were given in the abstract to enable us to judge content), paper or electronic copies were retrieved and re-checked to determine whether or not any further articles should be excluded (also with double-checking within the review team).

Data extraction

We used a simple form to extract the following data from the included studies:

• Study design.

• Type of disease and sites affected. We assigned joint categories on the basis of the primary patient diagnosis of OA (ie, hip, knee or hand) or SDD.

• Method of diagnosis (eg, radiographic).

• Total sample size and number of cases and controls. A “small” sample size was judged to be <150 patients in total (eg, less than 75 cases and 75 controls combined).7

• Proportion of cases male/female.

• Origin of case/control populations and recruitment setting (eg, hospital, with geographic region).

• Gene(s) analysed, with type(s) and number(s) of polymorphism(s) (eg, vitamin D receptor, RFLP).

• Number of alleles per polymorphism (per gene).

• Details of associations and non-significant associations reported.

• Whether or not we judged that a correction for multiple testing may be appropriate, and if so whether or not one had been applied.

• Whether or not any subgroup analyses were used and in what context.

• Missing details: ethnicity, gender, age and details of the recruitment process (eg, origin of the control population not given).

Data extraction was carried out in duplicate by two reviewers. Studies were categorised on the basis of sample size following a recent meta-analysis of gene–disease association studies,7 which found that discrepancies between the results of first and subsequent studies of a particular association were more common when the total sample size of the first study was <150 (“small”) than when it was ⩾150. We judged that correction for multiple testing may be appropriate whenever more than one genetic locus was tested for association, except where the authors put forward an explicit argument as to why correction was considered unnecessary (we made no judgement concerning the validity of any such arguments). This criterion was applied whether the loci under consideration were polymorphic sites within separate genes (ie, a multiple gene study) or discrete polymorphisms within the same gene.

To simplify the results, we grouped together studies that focused on different polymorphic regions within the same gene. This approach necessitates that we qualify our use of the term “replication” in the discussion below. In particular, we stress that “replication” does not necessarily refer to a specific genetic locus, only to the gene within which that locus occurs. Therefore, where a significant association has been reported for a particular gene–disease association in more than one study, we have referred to this as replicated reporting of a significant association, regardless of the specific gene variant.

FINDINGS

Our search identified 5748 articles in total, across all three databases. The number of articles was reduced to 122 by the initial screening process, which excluded irrelevant or unsuitable studies on the basis of the title or abstract. Paper or electronic copies of all 122 articles were retrieved and re-checked. A further 32 articles were excluded after it became apparent that they were not association studies, leaving a total of 90 for data extraction. Table 1 provides an overview of these studies, grouped by joint category.

Table 1 Overview of included studies (n = 90), grouped by joint category, showing genes analysed, numbers of significant associations, and replication across studies

Gene–OA/SDD associations by joint category

Out of the 90 included studies, 68 reported finding one or more significant gene–disease association(s). From these studies, 94 associations were reported from a total of 83 different genes (table 1). Hip OA (HOA) and knee OA (KOA) studies were most prevalent, but we found relatively few instances, either for these or any other joint sites, in which the same gene–disease association had been the focus of more than one study. There were also very few examples of significant associations that had been replicated in separate studies. For example, 31 studies examined gene associations for HOA, analysing a total of 38 genes. Only 15 of these genes were represented by two or more separate studies and there were only two cases in which a significant association was replicated by another study.

Table 2 provides a more detailed summary of the number of significant and non-significant associations for each gene (grouped by joint category).

Table 2 Summary of gene–osteoarthritis (OA)/spinal degenerative disease (SDD) association studies

It is clear that the relatively low incidence of different studies having focussed on the same gene–disease association (which is apparent from table 1) is in part a consequence of the dilution of study effort between different disease phenotypes. For example, we found four studies that tested for an association with the COL1A1 gene, but these were split between cases of HOA, KOA, HOA and KOA combined, and SDD; only one of these separate joint categories was represented by more than one study. A similarly diffuse pattern of research effort is apparent for many of the other genes shown on table 2, with a small number of exceptions (such as the VDR gene, where four of the six joint categories studied are represented by multiple studies).

Study designs, sample sizes and data analysis

The majority of studies had a sample size of ⩾150 (table 3).7 There was no suggestion that the simple categorisation of studies into groups with either < or ⩾150 patients was correlated with a tendency to report a significant association, but caution needs to be exercised here as the number in the “small” category was low.

Table 3 Number of studies reporting one or more significant association grouped by total sample size and disease outcome measure, with additional characteristics of analysis and reporting (number of studies: applying a correction for multiple testing; with reliance on subgroup analysis; and with missing details or results; see text for details)

The majority of studies used some variation of a case–control design to quantify associations, sometimes nesting within a cohort study but more often sampling from a proxy control population, whilst three studies used a family-based design.14 47 93 Most studies analysed the association between a genetic variant(s) and disease prevalence (presence or absence of disease), whilst a smaller proportion focused on quantitative or qualitative measures of disease severity (table 3). Less than a third of all studies that reported a significant association and tested across multiple loci applied any form of correction for a raised type I error rate. Equally striking is the finding that, regardless of sample size, over half of the studies failed to report basic study details. Omissions were evident with regard to one or more of: ethnicity, gender, age, and details of the recruitment process. In many cases these omissions make it difficult to judge whether or not reported associations could potentially be confounded by population stratification, or whether controls have been appropriately selected. Full results of analyses were often not provided and many studies relied on subgroup analyses. Taken together, these problems may create a substantial risk of publication bias.

DISCUSSION

Our results show that there has been much research into gene–OA and gene–SDD associations over the last decade. The 90 studies included in this review reported a total of 94 associations from 83 different genes. Reasonably good evidence is emerging in support of a role for certain genes. We found a number of significant associations that have been replicated within a joint category by two or more independent studies: AGC1, hand;14 15 17 ASPN, hip and knee;1922 COL9A2, spine;41 44 COL9A3, spine;39 45 COL11A2, spine;16 39 ESR1, knee;48 49 FZRB, hip;9 54 55 HFE, hand;57 58 IL1A, knee;67 68 IL1RN, knee;67 68 PTGS2, knee;10 11 and VDR, knee and spine.50 52 8994 We also found several studies where significant associations (FRZB and IL1 gene cluster) were replicated in separate populations.9 53 68 However, in the majority of cases where a significant association had been reported and the same gene–disease association had been tested in another study, the association had not been found again. Therefore, it is unclear to what extent these isolated significant associations are in fact “false positives”. This problem is confounded by the fact that separate studies of the same gene and the same disease phenotype sometimes focus on different polymorphisms within the gene. In such cases, the association may not have been replicated with respect to the same genetic unit (ie, the genetic locus).

Thus, despite the obvious progress that has been made, isolated gene–OA/SDD studies are very common (103 out of a total of 157 by joint category; table 1) and relatively few genes have been the subject of more than two or three studies within any particular joint category (table 2). The complex phenotype of the disease undoubtedly makes it difficult to prioritise research areas and may encourage a somewhat uncoordinated and diffuse approach when the field is viewed as a whole. Indeed, the degree to which different types of OA and SDD genuinely represent discrete forms of disease, with distinct aetiologies, is far from clear. The sheer number of potential gene–OA associations inevitably adds to this problem. However, we suggest that substantial progress could be achieved from focusing on those candidate genes where evidence seems to be growing (eg, VDR). If we continue instead to cast the “net” ever wider in an attempt to detect new associations, unreplicated and untested results will simply continue to accumulate without adding much to our understanding. A more convincing approach may be to carry out smaller numbers of large, high quality, genome-wide association studies (eg, Spector et al).77

A further problem is that relatively simple and avoidable methodological and reporting problems frequently undermine the degree of confidence that can be placed in individual gene–OA/SDD association studies. First, we found that many studies provided insufficient detail to enable the reader to assess whether or not bias could have confounded the analysis (eg, unreported age-biases).6 It should be standard practice to report full case and control population details. Second, many studies tested for associations across multiple loci, but did not apply any form of correction for type I statistical errors. The problem of spurious association may therefore be inflated by the use of inappropriate (permissive) significance levels, presenting a serious challenge to the development of clear, well-supported conclusions. Third, in many cases the main associations reported in a study were derived from some form of subgroup analysis. Whilst this approach may be perfectly valid in the right context, it may be misleading not to include strong caveats. For example, significant associations derived from post hoc subgroup analyses may be emphasised in the abstract of a study over and above the results of the primary analysis of the study, if the latter are non-significant. The omission of appropriate caveats from the abstract of a study can only encourage biased reporting and interpretation. Fourth, many studies failed to report results in full, adding to the familiar concern regarding the degree to which non-significant results are being under-reported.

Huizinga et al have provided recommendations for the conduct of gene–OA association studies; Little et al also recently gave a detailed overview of a wide-range of issues that arise in the appraisal of gene–disease association studies.98100 These studies offer detailed recommendations to aid authors in the preparation of manuscripts and to help readers assess study design and quality. It is undoubtedly too soon to expect these recommendations to have been applied fully in all but the most recent cases (and many studies included here predate these overviews), but our results illustrate why they must be applied in future studies. Adequate replication and attention to study design, methodology and reporting is demonstrably lacking, as is a clear delineation between hypothesis testing vs hypothesis generation. A clearer approach to disease phenotype in the OA/SDD field is also vital. Taken together, these problems present a major obstacle to the systematic review and/or meta-analysis of effect sizes for particular associations. A key challenge for future studies will be to actively minimise the potential for publication bias. For example, the value of published association analyses will be greatly undermined if unpublished studies and datasets are allowed to accumulate beyond the range of the systematic reviewer’s “radar”. The Human Genome Epidemiology Network (HuGENet) has been at the leading edge of calls for registries and investigator networks aimed at overcoming this problem.101

In summary, our primary concern in this review has been to provide an overview of general progress and to highlight areas where it appears that improvements are possible. Whilst it is clear that real progress has been made in the genetic epidemiology of OA and SDD, future association analyses must deal with a range of important problems. Greater focus and coordination between researchers is vital,101 particularly in light of advances in gene-scanning technology, which mean that the potential now exists for the number of reported significant associations to grow at an accelerating rate. As our understanding of the genetic architecture of OA grows, hypothesis testing should become a more realistic and achievable goal. Additionally, basic epidemiological principles must be applied with greater vigour and a clearer standard of reporting maintained, in keeping with the recommendations of HuGENet. We also hope that funding bodies will begin to prioritise research that follows these well-tested principles.

View Abstract

Footnotes

• Funding: This work was supported by an Action Arthritis grant awarded to AJM. JJR was supported by core funding awarded to the Institute of Health at the University of East Anglia, through the Department of Health’s Research Capacity Development Programme.

• Competing interests: None.

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### Home > CC2MN > Chapter 2 > Lesson 2.3.1 > Problem2-117

2-117.

Copy each expression and then simplify it without using a calculator. Be sure to show all steps.

1. $7\frac{5}{8}+2\frac{9}{16}$

This sum can be represented with a diagram: Diagram expression with 2 equal sized rectangles: Left Rectangle + Right Rectangle. Left rectangle, 4 rows of 2 columns, with 5 shaded. Right Rectangle, 4 rows of 4 columns, with 9 shaded.

Find a common denominator to add the two fractions. Changes: additional columns added to left rectangle, to make it 4 rows of 4 columns, with 10 shaded.

Changes: After right rectangle, an equal sign and the fraction 19 sixteenths.

2. $2\frac{7}{8}+6\frac{1}{6}$

See part (a) for strategy.

3. $8-6\frac{2}{5}$

Draw $8$ squares on your paper like this: 8 square tiles.

Cross off $6\frac{2}{5}$. What is left?

$1\frac{3}{5}$


text:
# PSEB 9th Class Science Important Questions Chapter 4 Structure of the Atom

Punjab State Board PSEB 9th Class Science Important Questions Chapter 4 Structure of the Atom Important Questions and Answers.

## PSEB 9th Class Science Important Questions Chapter 4 Structure of the Atom

Long Answer Type Questions:

Question 1.
What is Rutherford’s ∝-ray scattering experiment? Give its observations. How does Rutherford explain this observation?
Answer:
Rutherford’s ∝-ray scattering experiment: In this experiment, Rutherford obtained a thin stream of fast-moving alpha (∝) particles from radioactive source (P0) and was allowed to strike against a thin foil of heavy metal like gold. Alpha particles are also called helium nuclei (He2+ with + 2 unit charge and 4 a.m.u. mass). As a result of the experiment, he made the following observations:

1. Most of the ∝-particles passed through the gold foil without any deflection from their original path.
2. Some of the ∝-particles got deflected by small angles from their original path.
3. A few ∝-particles (approximately 1 out of 10,000) were deflected to a large extent. In some cases, the a-particles came back in the same direction after striking the gold foil.

Explanation for the observations:

We know that the gold foil is made up of atoms of gold which are placed side by side. Since most of the ∝-particles passed through these atoms undeflected they did not come across any obstruction in their path, this shows that most of the space in the atom is light or hollow.

The electrons with negligible mass were supposed to be present in this portion called extra nuclear portion. The a-particies were deflected from their path and a very few came back in the same direction. This shows that they must have met with some obstruction in their path.
The obstruction must be
1. Heavy because a-particles are high speed particles and can remove light obstruction from their path.
2. Small because only a few a-particles got deflected.
3. Positively charged because some a-particles were repelled by this obstruction.

This small, heavy and positively charged portion inside an atom was called nucleus. The protons with mass and positive charge were supposed to be present in the nucleus. Later on, neutrons were also found to be present in the nucleus of the atom.

Question 2.
Discuss in brief Rutherford’s Model of atom.
Answer:
The main points of Rutherford’s model of atom are listed as follows:
1. An atom has two parts. These are nucleus and extra nuclear portion.
2. The nucleus contains protons while electrons are present in the extra nuclear portion.
3. Most of the mass of the atom is due to the nucleus because electrons have negligible mass.

4. The volume occupied by the nucleus is very small as compared to the volume occupied by extra nuclear portion.
5. The electrons in the extra nuclear portion are not stationary. These are revolving around the nucleus. The electrons are called planetary electrons because they can be compared to various planets which revolve around the sun. The nucleus may be compared with the sun.

Question 3.
Discuss the arrangement of the electrons in the different energy levels.
Or
Describe Bohr-Bury scheme for the distribution of electrons in the various orbits or energy levels of an atom.
Answer:
The distribution of electrons in the energy levels is given with the help of Bohr Bury scheme. It is given as follows:
1. The energy levels or energy shells are filled in order of increasing energies. The electrons first enter the K shell (n = 1. which is closest to the nucleus. This is followed by L shell (n = 2), M shell (n = 3) and so on. Here V represents the number of the shell.
2. The maximum number of electrons in any shell is given as 2n2 (n = number of the shell). The distribution in the first four energy shells is given in the table given below:
Distribution of electrons in first four energy shells according to 2n2 rule.

 Symbol of shell Number of shell (n) Maximum number of electrons (2n2) K 1 2 x (1)2 = 2 L 2 2 x (2)2 = 8 M 3 2 x (3)2 = 18 N 4 2 x (4)2 = 32

3. The outermost energy shell cannot have more than 8 electrons. The next inner shell called penultimate shell, cannot have more than 18 electrons.
4. It is not necessary that an energy level or shell is fully filled before the filling in the next energy level starts. In fact, filling of electrons in a new shell starts when any shell contains 8 electrons.
On the basis of the above scheme, let us discus with mass number 24 and atomic number 12.

• Mass no. of atom (A) = 24
• Atomic no. of atom (Z) = 12
• No. of protons in nucleus = 12
• No. of neutrons in nucleus = 12
• No. of electrons to be distributed = 12
• The distribution in the various energy shells: K L M: 2, 8, 2

Question 4.
Give the electronic distribution in the first twenty elements on the basis of Bohr-Bury scheme.
Answer:
On the basis of Bohr-Bury scheme, the distribution of electrons in first twenty elements (Z = 1 to Z = 20) is given below:
Table: Electronic distribution in atoms of first 20 elements

 Element (Symbol) Mass No. (A) Atomic No. (Z) No. of Protons (P) No. of Electrons (e) No. of Neutrons (n) (A-Z) Electronic Distribution K L M N 1. Hydrogen (H) 1 1 1 1 0 1 2. Helium (He) 4 2 2 2 2 2 3. Lithium (Li) 7 3 3 3 4 9 1 4. Beryllium (Be) 9 4 4 4 5 2 2 5. Boron (B) 11 5 5 5 6 2 3 6. Carbon (C) 12 6 6 6 6 2 4 7. Nitrogen (N) 14 7 7 7 7 2 5 8. Oxygen (O) 16 8 8 8 8 2 6 9. Fluorine (F) 19 9 9 9 10 2 7 10. Neon (Ne) 20 10 10 10 10 2 8 11. Sodium (Na) 23 11 11 11 12 2 8 1 12. Magnesium (Mg) 24 12 12 12 12 9 8 2 13. Aluminium (Al) 27 13 13 13 14 ‘ 2 8 3 14. Silicon (Si) 28 14 14 14 14 2 8 4 15. Phosphorus (P) 31 15 15 15 16 2 8 5 16. Sulphur (S) 32 16 16 16 16 2 8 6 17. Chlorine (Cl) 35 17 17 17 18 2 8 7 18. Argon (Ar) 40 18 18 18 22 2 8 8 19. Potassium (K) 39 19 19 19 20 2 8 8 1 20. Calcium (Ca) 40 20 20 20 20 2 8 8 2

Special cases of Potassium and Calcium:

• In case of Potassium (Z = 19), the M shell has only 8 electrons and not 18. Actually, when the M shell gets 8 electrons, the filling of electrons starts in the next shell which is N-shell. When the N-shell gets 2 electrons, the electron are again- filled in the M shell.
• Thus, electronic distribution in potassium (Z = 19) is
K L M N
2 8 8 1 (M shell has 8 electrons and N shell has 1 electron)
• Similarly, the electronic distribution in calcium (Z = 20) is:
K L M N: 2 8 8 2 (M shell has 8 electrons and N shell has 2 electrons)

Question 5.
Define the terms:

1. Atomic number
2. Mass number
3. Atom
4. Ion
5. Element
6. Orbit
7. Nucleons

Answer:

1. Atomic number: It is the number of protons present in the nucleus of an atom of the element.
2. Mass number: It is the sum of number of protons and neutrons present in the nucleus of an atom of the elements.
3. Atom: It is the smallest unit of matter which takes part in chemical reactions and it may or may not exist independently.
4. Ion: It is an atom or group of atoms having charge positive or negative and can exist freely in solution.
5. Element: It is a pure substance which can neither be built up nor broken into two or more still simpler substance by any physical or chemical method and it consists of only one kind of atoms.
6. Orbit: It is the fixed circular path traced by an electron revolving around the
7. Nucleons: The protons and neutrons present in the nucleus of an atom of the element are collectively known as nucleons.u

Short Answer Type Questions:

Question 1.
What do you understand by the structure of atom?
Answer:
According to the Dalton’s atomic theory, the smallest particle of matter is called atom. But later studies have shown that an atom is further made up of sub-atomic particles called electrons, protons and neutrons. Different atoms have different number of such particles and they also differ in the arrangement of these particles. Therefore, the atoms have different properties.

Question 2.
What happens when electric field is applied to the path of cathode rays? What does it show?
Answer:
When electric field is applied to the path of the cathode rays, these are deflected towards the positive plate of the field. This shows that the cathode rays consist of negatively charged particles.

Question 3.
What is the information conveyed by the following observations :

1. Atom is electrically neutral,
2. Mass of the atom is due to the nucleus.

Answer:

1. If an atom is electrically neutral, this means that it will have equal number of protons and electrons. Both protons and electrons have 1 unit charge but with opposite sign.
2. This means that nucleus contains in it both protons and neutrons which have mass.

Question 4.
What is the significance of number of protons found in the atoms in each of the different elements?
Answer:
The number of electrons in atom of the element is equal to the number of protons. By knowing no. of protons, we can calculate no. of electrons and hence its properties.

Question 5.
Write the electronic configuration of the elements A, B, C, D, E with atomic numbers 5, 6, 14, 13 and 15. Which have similar chemical properties?
Answer:

 Element Atomic No. (Z) Electronic configuration K L M A 5 2 3 B 6 2 4 C 14 2 8 4 D 13 ? 8 3 E 15 2 8 5

Elements B and C. have similar chemical properties because both of them have 4 valence elt, trons.
Elements A and D have similar chemical properties because both of them have 3 valence electrons.

Question 6.
Five species P, Q, R, S, T have electrons, protons and neutrons are as follows:

 Species Electrons Protons Neutrons P 4 3 4 Q 8 9 9 R 17 17 20 S 17 17 18 T 18 18 22

Find

1. a cation
2. an anion
3. a noble gas and
4. a pair of isotopes.

Answer:

1. Cation – Q
2. Anion – P
3. A noble gas = T
4. Pair of isotopes = R and S (because they have same number of electrons and protons but different number of neutrons).

Question 7.
Helium has only 2 electrons in the K-shell. But it is called an inert gas element. Why?
Answer:
Helium (He) atom has only one shell (K-shell) which can have a maximum of 2 electrons only. It cannot have more than 2 electrons. As a result, the combining capacity of He is zero. Therefore, its valency is also zero.

Question 8.
Give labelled diagram to show how are cathode rays produced?
Answer:

Question 9.
An atom has mass number A and atomic number Z.
1. How many protons are present in the nucleus?
2. How many electrons are revolving around the nucleus?
3. How many neutrons are present in its nucleus?
Answer:
Mass number = A
Atomic number = Z
1. No. of protons present in the nucleus = Z
2. No. of electrons revolving around the nucleus = Z
3. No. of neutrons present in the nucleus = A – Z.

Question 10.
State three ways by which a proton differs from an electron.
Answer:

 Proton Electron 1. It has one unit of positive charge. 2. It has mass 1.6 x 10-24 g. 3. It is present in the nucleus of an atom. It has one unit of negative charge. It has mass 9.1 x 10″28 g. It is present in the extranuclear part of an atom.

Question 11.
What are isotopes? Give two uses of isotopes. Name the isotopes of hydrogen. Give their structures.
Answer:
Isotopes. The atoms of the same element having the same atomic number but different mass numbers are called isotopes.
Use of isotopes:
1. Isotopes of hydrogen, oxygen, carbon are used to study the mechanisms of organic reactions.
2. Some radioactive isotopes are used for the treatment of cancer and other diseases. The isotopes of hydrogen are:

Question 12.
Explain why elements have fractional atomic masses?
Answer:
The elements have fractional atomic masses due to the existence of isotopes (having different mass numbers) e.g. chlorine has two isotopes $${ }_{17}^{35} \mathrm{Cl}$$ and $${ }_{17}^{37} \mathrm{Cl}$$ and are present in the ratio 3:1.
∴ Average atomic mass of chlorine = $$\frac{35 \times 3+37 \times 1}{4}=\frac{105+37}{4}$$ = 35.5 a.m.u.

Question 13.
What are isobars? Give examples.
Answer:
The atom of different elements having same mass numbers but different atomic numbers are called isobars.
e.g. 1. $${ }_{18}^{40} \mathrm{Ar}$$ and $${ }_{19}^{40} \mathrm{Ar}$$
2. $${ }_{36}^{86} \mathrm{Ar}$$ and $${ }_{38}^{86} \mathrm{Ar}$$

Question 14.
Sulphur has an atomic number 16 and a mass of 32. State the number of protons and neutrons in the nucleus of sulphur. Give a simple diagram to show the arrangement of electrons in an atom of sulphur.
Answer:

Atomic number of S = 16
Mass number of S = 32
∴ No. of protons = 16
No. of neutrons = 32 – 16 = 16
No. of electrons = 16

Question 15.
Write down the electronic configuration of the following:
1. $${ }_{13}^{27} \mathrm{X}$$
2. $${ }_{17}^{35} \mathrm{X}$$
Write down the number of electrons in X and neutrons in Y and the formula of
compound formed by X and Y.
Answer:
1. $${ }_{13}^{27} \mathrm{X}$$
Atomic number of X = 13
Mass number of X =27
No. of protons in X = 13
∴ No. of electrons in X = 13
The electronic configuration of X = 2, 8, 3
Valency of X = 3

2. $${ }_{17}^{35} \mathrm{X}$$
Atomic number of Y = 17
Mass number of Y =35
No. of protons in Y = 17
No. of electrons in Y = 17
No. of neutrons in Y = 35 – 17 = 18
The electronic co0nfiguration of Y = 2, 8, 7
Valency of Y = 1
The formula of the compound formed by X and Y

Question 16.
Give the important properties or characteristics of isotopes.
Answer:

1. The isotopes of an element have same atomic number (Z) i.e. the same number of electrons and same number of protons. Hence, they have identical chemical properties.
2. Isotopes of an element have same number of valence electrons and hence same valency.
3. Isotopes of an element have different mass numbers (or different number of neutrons) and hence they have different physical properties such as mass, density, melting point, boiling point etc.

Question 17.
An atom of the element is represented as 11x?

1. What does the numeral 23 indicate?
2. What does the nemeral 11 indicate?
3. What is the number of protons in X?

Answer:

1. 23 indicates mass number of X.
2. 11 indicates atomic number of X.
3. No. of protons in X = Atomic number = 11.

Question 18.
The atom of an element is made up of 4 protons, 5 neutrons and 4 electrons. what are its atomic number and mass number?
Answer:
No. of proton = 4
Atomic number of the element, Z = 4
No. of neutrons = 5
∴ No. of protons + No. of neutrons =4 + 5 = 9
∴ Mass number of the element, A = 9.

Question 19.
Calculate the number of neutrons in the following elements:
1. $${ }_{1}^{3} \mathrm{Cl}$$
2. $${ }_{17}^{37} \mathrm{Cl}$$
3. $${ }_{20}^{40} \mathrm{Cl}$$
Answer:
1. $${ }_{1}^{3} \mathrm{Cl}$$
Mass number = 3
Atomic number = 1
No.of neutrons = Mass no – Atomic no.
= 3 – 1 = 2

2. $${ }_{17}^{37} \mathrm{Cl}$$
Mass number = 37
Atomic number = 17
No.of neutrons = 37 – 17 = 20

3. $${ }_{20}^{40} \mathrm{Cl}$$
Mass number = 40
Atomic number = 20
No.of neutrons = 40 – 20 = 20

Question 20.
Give two conditions under which cathode rays are produced.
Answer:
Cathode rays are produced when a gas is subjected to the action of
1. High voltage of the order of 5000-10000 V and
2. Under a low pressure of the order of 10-3 atm of Hg.

Question 21.
The electronic configuration of P (Z = 15) is 2, 8, 5. Why is the valency of phosphorus 3 and not 5?
Answer:
The electronic configuration of phosphorus (P) shows that it has 5 valence electrons. Its valency is expected to be 5. But it is not possible for the atom to lose 5 electrons present in the valence shell because the energy required to remove these electrons will be very high. Therefore, phosphorus atom has valency equal to 3 because to gain 3 electrons is easier than losing 5 electrons as it requires less energy.

Question 22.
What information is conveyed by the statement that the mass number of magnesium is 24 and the atomic number is 12?
Answer:
The atomic number (12) of magnesium indicates that its atom has 12 protons in the nucleus and an equal number of electrons in the extra nuclear portion. The mass number (24) points out that there are also 12 neutrons (24 – 12 = 12) present in the nucleus alongwith protons.

Question 23.
Is it possible for an atom to have 12 protons and 13 electrons? Explain.
Answer:
No, it is not possible. An atom must always be electrically neutral. This means that it has no net charge present on it. Now, each proton has one unit positive charge and each electron has one unit negative charge. In a neutral atom, the number of electrons and protons must always be the same. Thus, an atom cannot have 12 protons and 13 electrons.

Question 24.
Give the important isotopes of hydrogen, carbon, oxygen and chlorine. Indicate their mass numbers and atomic numbers.
Answer:

Numerical Problems (Solved):

Question 1.
Chlorine has two isotopes $${ }_{17}^{35} \mathrm{Cl}$$ and $${ }_{17}^{37} \mathrm{Cl}$$ in the ratio 3 : 1. Calculate the average atomic mass of chlorine.
Solution:

Question 2.
Out of 18X and 16Y which atom is chemically more reactive?
Solution:

No. of valence electrons in X =8
No. of valence electrons in Y =6
∴ Y is more reactive.

Question 3.
An element has 12 neutrons and mass number 23. Give the atomic number and symbol of the element.
Solution:

• Mass number(A) = 23
• Number of neutrons = 12
• Number of protons = 23 – 12 = 11
• Atomic Number(Z)= 11
• Symbol of the element = $$\frac{23}{11} \mathrm{Na}$$

Question 4.
Calculate number of protons, neutrons and electrons in $$\frac{235}{92} \mathrm{U}$$ and $$\frac{238}{92} \mathrm{U}$$. How are these atoms related?
Solution:
table
$$\frac{235}{92} \mathrm{U}$$ and $$\frac{238}{92} \mathrm{U}$$ are isotopes.

Question 5.
Calculate the number of electrons, protons and neutrons is $$\frac{19}{9} \mathrm{U}$$. Also calculate its valency.
Solution:

1. Atomic number = 9
2. Mass number = 19
3. No. of protons = 9
4. No. of neutrons = 19 – 9 = 10
5. Electronic configuration = 2 7
6. Valency = 8 – 7 = 1.

Very Short Answer Type Questions

Question 1.
Who discovered canal rays?
Answer:
Goldstein discovered canal rays in 1886.

Question 2.
Which charge is present on canal ray particles?
Answer:
They carry positive charge.

Question 3.
Which charge is present on protons?
Answer:
Positive charge.

Question 4.
Where are protons present in an atom?
Answer:
Protons are present in the nucleus, deep inside the atom.

Question 5.
What were the characteristics of an atom according to Dalton?
Answer:
According to Dalton an atom was indivisible and indestructible.

Question 6.
Which discovery is against Dalton’s theory?
Answer:
Discovering of electrons and protons in an atom.

Question 7.
Who proposed the first model for the structure of atom?
Answer:
J.J. Thomson.

Question 8.
Why did Rutherford used gold foil in a-ray scattering experiment?
Answer:
This is because he wanted as thin a layer as possible. It was about 1000 atoms thick.

Question 9.
What are α-particles?
Answer:
They are doubly charged helium ions (He2+).

Question 10.
What is the charge on α-particles?
Answer:
+2 units.

Question 11.
What is the mass of an α-particle?
Answer:
4 u.

Question 12.
How the defects of Rutherford’s model were removed by Neil Bohr?
Answer:
The electrons revolve around the nucleus in an atom in fixed circular paths called orbits or energy levels and don’t lose energy.

Question 13.
What are energy levels or shells?
Answer:
The definite orbits in where electrons revolve around the nucleus are called energy levels. These are designated as K, L, M, N ……………. 1, 2, 3, 4.

Question 14.
Define mass number of an element.
Answer:
It is the sum of number of neutrons and protons present in the nucleus of the atom of the element.

Question 15.
Who gave the distribution of electrons in the various shells of an atom?
Answer:
Bohr and Bury.

Question 16.
What is the maximum number of electrons present in an electronic shell?
Answer:
2n2 (Where n is the number of shells).

Question 17.
What is the maximum first four shells of an atom?
Answer:
2, 8, 18, 32.

Question 18.
What is the maximum number of electrons in the outermost shell of an atom?
Answer:
8

Question 19.
Which elements are chemically inert?
Answer:
These are the elements which have 8 electrons in their outermost shells.

Question 20.
Define octet.
Answer:
The maximum number of electrons present in the outermost shell of an atom is called octet.

Question 21.
Define valency of an element.
Answer:
It is the combining capacity of an atom of the element and it is equal to the number of electrons lost or gained or contributed for sharing from its valence shell in order to complete its octet.

Question 22.
Define atomic number.
Answer:
It-is the number of protons present in the nucleus of an atom of the element.

Question 23.
Define nucleon.
Answer:
The neutrons and protons present in the nucleus of an atom of the element are collectively known as nucleons.

Question 24.
Where is most of the mass of the element present?
Answer:
In the nucleus of an atom.

Question 25.
Define atomic mass of an element.
Answer:
It is the sum of the number of neutrons and protons present in the nucleus of its atom.

Question 26.
What are isotopes?
Answer:
Atoms of the same element having same atomic number but different mass numbers are called isotopes.

Question 27.
Write two isotopes of chlorine.
Answer:
$${ }_{17}^{35} \mathrm{Cl}$$ and $${ }_{17}^{37} \mathrm{Cl}$$.

Question 28.
Write two isotopes of carbon.
Answer:
$${ }_{6}^{12} \mathrm{Cl}$$ and $${ }_{6}^{14} \mathrm{Cl}$$.

Question 29.
Isotope of which element is used in the treatment of cancer?
Answer:
Isotope of cobalt ($${ }_{27}^{59} \mathrm{Co}$$).

Question 30.
What is the atomic mass of chlorine?
Answer:
35.5

Question 31.
Isotope of which element is used for the treatment of goitre?
Answer:
An isotope of iodine.

Question 32.
Isotope of which element is used in atomic reactor?
Answer:
Uranium.

Question 33.
What are isobars?
Answer:
Atoms of the different elements having same mass number but different atomic numbers are called isobars.

Question 34.
Give one example of isobars?
Answer:
$$\underset{(\text { Calcium })}{\stackrel{40}{20} \mathrm{Ca}}$$ and $$\underset{(\text { Argon })}{\stackrel{40}{18} \mathrm{Ca}}$$.

Question 35.
Write three isotopes of hydrogen.
Answer:

1. Protium ($${ }_{1}^{1} \mathrm{H}$$)
2. Deuterium ($${ }_{2}^{1} \mathrm{H}$$)
3. Tritium ($${ }_{3}^{1} \mathrm{H}$$).

Question 36.
What is the total number of electrons in an atom if its K-and L-shells are fully filled?
Answer:
10 (K-2, L-8).

Question 37.
What electrons in an atom influence its chemical properties?
Answer:
Valence electrons (electrons in the outermost shell).

Question 38.
How will you represent chlorine atom having mass number 35 and atomic number 17?
Answer:
$${ }_{17}^{35} \mathrm{Cl}$$

Question 39.
Name the sub-atomic particle on which size of an atom depends?
Answer:
Electron.

Question 40.
Name the atoms having same atomic number but different mass numbers.
Answer:
Isotopes.

Question 41.
Mg2+ has 10 electrons, what is the number of protons in it?
Answer:
10 + 2 = 12.

Question 42.
Name the element which has no neutrons in the nucleus of its atom.
Answer:
Hydrogen.

Question 43.
Out of electrons, protons and neutrons, which are same in isotopes?
Answer:
Electrons and protons.

Question 44.
There are three isotopes of hydrogen, $${ }_{1}^{1} \mathrm{H},{ }_{1}^{2} \mathrm{H},{ }_{1}^{3} \mathrm{H}$$ all are electrically neutral, why?
Answer:
This is because each isotope has one electron and one proton.

Question 45.
An element has atomic number 16. Give its electronic configuration and number of valence electrons.
Answer:
Electronic configuration 2, 8, 6.

Question 46.
The number of electrons in the valence shell of sodium (Na) is ……………… .
Answer:
1.

Question 47.
Fluorine belongs to …………… family.
Answer:
Halogen.

Question 48.
In an atom, the number of …………… is the same as the number of ………………. .
Answer:
Electrons, protons

Question 49.
Sodium has …………….. electron …………….. than sodium ion.
Answer:
One, more

Question 50.
Isotopes of an element are …………….. because they have ……….. number of electrons.
Answer:
Chemically similar, same

Question 51.
The number of electrons in the valence shell of an atom cannot be ………………..>
Answer:
More than eight.


text:
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# How to Create an Inspirational Vector Political Poster

Difficulty:IntermediateLength:LongLanguages:

Learn how to create a Shepard Fairey Photoshop effect with the help of Illustrator with this poster tutorial.

This tutorial is inspired by Shepard Fairey's famous political poster series for the Obama Democratic Party campaign in the US in 2008 and now being recreated for the Time magazine cover for November 2020, the month of the US election.

We'll be showing you how to create this style of design. We'll start with basic image editing techniques in Adobe Photoshop to get our guide layers set up, and then we'll jump into Adobe Illustrator. You don't need any fancy equipment to do this. I used an older mouse that still has the rubber ball.

Note: This tutorial has been recently updated; however, the stock image and pattern are no longer available for download. Please find the appropriate items to replace them to complete the exact end result. Envato Tuts+ does not endorse any political party in the publishing of this tutorial. Rather, this tutorial is focused on demonstrating a workflow for recreating this iconic poster.

## Create the Hope Photo Effect in Seconds!

Want to save time and get a Photoshop action which will create the Hope photo effect in seconds? Try this Obama Hope Poster Generator Photoshop Action from Envato Elements.

## 1. Crop and Modify Your Stock Image

### Step 1

For this effect, it's best to have a portrait style picture, preferably of a subject that appears to be thinking or looking off into the distance.

### Step 2

Once you have your image, you need to Open it in Photoshop and Crop it appropriately. The top of the image should be cropped to the top of your subject's head, and the bottom should be a bit higher than chest height. The crop should have about a 2:1 ratio.

### Step 3

Now we need to Posterize the image. Go to Image > Adjustments > Posterize. An appropriate posterization level for the look we are going for is 5.

### Step 4

Now we're going to create our guide layers for use in Illustrator. Start by duplicating your posterized layer, and name this new layer "Pattern Guide." Then go to Image > Adjustments > Threshold. We're going to slide our Point until we get something similar to below. We want it fairly dark, but still with a small amount of detail. Also, when using Threshold, it is best to use the Peaks that you see.

### Step 5

Save this newly created layer as "patternguide.psd," or something similar, as you'll need it for Illustrator.

### Step 6

We need to repeat Step 4 and Step 5 three more times. Apply a lower threshold each time, and save each new layer. Below are the settings I used—again, notice the peaks.

## 2. Set Up Your Document and Palettes

### Step 1

Before we start to outline our layers, we need to create the horizontal blue line pattern. To do this, create a new Illustrator document. Give it a Height and Width of 4 px. Then create a rectangle filled with our light blue color (#4F919F), and cover the top half of the canvas.

### Step 2

Cover the bottom half with a rectangle filled with beige. Then Select All (Control-A), and drag this over to our Swatches panel. Congratulations—you've made a pattern!

### Step 3

Now we have to save this Swatch set so we can use it on the poster we're about to make. In the bottom left corner of our Swatches panel, we'll select Save Swatches from the Swatches Library Menu. Name it anything you want; I used "pattern." You can now close this document, as we won't need it anymore.

### Step 4

Here is the color palette we'll be using for this tutorial. There are four colors and a pattern. We'll be using a mix of beige and light blue colors.

## 3. Create Your Portrait

### Step 1

Now that we have our reference images and pattern created, it's time to create a New document in Illustrator. I used a canvas size of 700 x 850 px.

### Step 2

We're going to start with our darkest guide layer—this will be our pattern layer. We're going to outline the layer using the Pen Tool (P). So to start, we're going to go to File > Place and select our first reference layer in Illustrator. If you've been following verbatim, the file to place is called "patternguide.psd".

### Step 3

Let's stop for a moment and discuss the proper use of the Pen Tool (P) in Illustrator. This is one area where Illustrator handily trumps Photoshop, but only if you use it correctly.

For any curve, you basically have three clicks; two are anchor points, and the other is your curve point. You place your curve point at the highest point of your curve. You place the anchors where the curve starts and ends.

One click is all that's needed when you are creating a curve. You can then change the curve points to curves after you have finished the entire shape. To convert them, use the Direct Selection Tool (A) and click on the Convert Selected Anchor Points to Smooth icon.

### Step 4

Before we can outline our first layer, we need to load our pattern into the Swatches Library. To do this, select Other Library from the Swatches Library menu in the Swatches panel.

### Step 5

Okay, so back to outlining our first layer. Create New Layer below the file you just placed into Illustrator. This layer will hold our outline. We place it below because it is easier to see what we're outlining that way.

Now we grab our Pen Tool (P), giving it a stroke of null and a fill of our pattern. Then start outlining all the black areas with the Pen Tool (P). The key is to average it out—don't follow the guide layer too closely. Just roughly outline it, and don't worry about our curves as we'll fix them after we create our outline.

Note: This is the most tedious part of the process.

### Step 6

You might have noticed that there are lighter portions in the middle that we just went right over with outlines. Well, we're going to fix that by outlining those using the same process. Fill those with our beige color, as shown below. We'll do this on a new layer just above the pattern layer. After you've done that, you should have something similar to the following.

### Step 7

Now we smooth out the shapes we just created. We want most of our points to be smooth, but I find it looks better if you leave a few the way they are. View the effect of smoothing both the pattern and the beige layer below.

Note: I made this image out of order and the pattern was wrong—don't worry about the color difference.

### Step 8

Now we just do this same thing for each layer: roughly outlining it, then outlining the highlights with the color immediately below it, and then smoothing out our outlines. The next layer should have solid light blue, followed by red, and then finally our dark blue.

### Step 9

Okay, you should have something like the following. The tedious part is over. Now to make it look like a poster. First, we'll put a beige filled rectangle below all these layers, which will serve as our background.

### Step 10

Now we'll need to put our red and light-blue halves in. We're going to do this just above the beige rectangle layer. Use the Rectangle Tool (M) to place the two rectangles.

### Step 11

Our poster is looking pretty good. Let's create the space needed to add the inspiring text at the bottom. You'll probably have to enlarge your canvas area. To do this, go to File > Document Setup. I added 300 px onto mine.

### Step 12

Now extend your beige rectangle so that it covers the new length of your canvas. Then create a new rectangle filled with our dark blue color. It should be a quarter of the size of the main subject.

### Step 13

Time to add our inspiring word. We'll use light blue for the font color, and we want to change the text settings so that the word fills most of the dark-blue rectangle. To change the spacing between the characters, use the Character panel. This is available to us when we are using the Text Tool (T). I used the font "Myriad Pro".

### Step 14

Almost finished—we just need to clean up our borders a little bit. First, select every layer except your beige layer. Then create a Group (Control-G). Then Vertically and Horizontally Align them to the artboard.

### Step 15

Now we're going to clean up the edges of our poster. Cover the layers that we grouped in a rectangle of any color. Then use the Pathfinder panel, and select Subtract Shape From Area.

## Awesome Work! You're Now Finished!

Way to go, we're done! We managed to create a really awesome effect without needing a whole lot of artistic ability. We also learned how to efficiently use the Pen Tool, and a nice way to use Photoshop and Illustrator in tandem. This is my first tutorial, so I'm sure there will be questions. Feel free to ask in the comments, and I'll try and answer as quickly as I can. Thanks!