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History of video games/Platforms/Slow Games

History
Slow Games is a 2014 art console made by designer Ishac Bertran of New York City. The system attracted media attention for having games that only allowed one move per day, and for it's minimalist aesthetics.

The game consoles were not intended for commercial production.

Technology
Three models of the device were made supporting one game each, with each device featuring slightly different controls on top of the console as a result.


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Using an Abacus/Addition and subtraction

Introduction
As has already been stated in the introduction to this book, addition and subtraction are the only two operations that can be carried out on the abacus; everything else must be reduced to a sequence of addition and subtraction, so learning these two operations is the most fundamental step in the study of the abacus.

Learning to add and subtract with the abacus is another case of psychomotor learning, similar to learning to dance, ride a bicycle, drive or learn a musical instrument.
 * In a first phase you will need a continuous cognitive effort trying to determine what is the next movement you have to do.
 * Later, you will notice that progressively you have to think less while the movements arise in an increasingly automatic way.
 * Finally, the movements will emerge spontaneously from you, you will have them definitely hardwired in your motor cortex and you will not have to think about them again. Although you will have a lifetime to perfect them.

Yes, it is like learning a musical instrument, but learning the abacus is much easier and faster than learning to play the viola and you will be sensitive to your progress from day to day.

In what follows we will deal with addition and subtraction together; It would be very difficult to separate the learning of one of these operations from the other since, as we will see, when we are adding we spend half the time subtracting complementary numbers and vice versa, when we are subtracting we spend half the time adding complementary numbers.

It has also been anticipated in the introduction of this book that it is not necessary to know how to add and subtract to use an abacus, only to know how to manipulate the beads or counters. In fact, for centuries the abacus was taught to people who had no previous knowledge of arithmetic and that the only knowledge they would have of it throughout their lives was going to be the use of the abacus itself. They learned to add and subtract by memorizing a long series of verses, rhymes, or rules intended to be sung as they were used. For example, taken from Xú Xīnlǔ's Pánzhū Suànfǎ, the first book entirely devoted to the abacus published in 1573 (late Ming Dynasty), and liberally translated from Chinese:

Which, obviously, inform us of which beads we have to move in order to add or subtract a digit. For example, the third line of the addition table contains three rules to try to add a 3: which apply, for example, to the following cases:
 * 3 activate 3, i.e. just activate three lower beads.
 * 3 activate 5 deactivate 2, i.e. activate one upper bead and deactivate two lower ones.
 * 3 subtract 7 carry 1, i.e. subtract 7 and add 1 to the left column.

You will understand these rules better later on, but don't worry anyway, you won't have to follow these 48 rules as you will go an easier path by memorizing just six rules that can be summed up into just three.

1-digit Addition and subtraction
The first step in learning addition and subtraction with an abacus is learning to add or subtract one of the 9 digits 1, 2, ..., 9 to / from any other 0, 1, 2, ..., 9; in total 180 cases that we will go through in our daily practice until we have them integrated into our motor memory. After this, adding or subtracting multi-digit numbers will be as simple as iterating this process in an orderly fashion for all the digits of the addend or subtrahend.

What do you need to know
To deal with the 180 cases mentioned above without memorizing the 48 rules of Panzhu Suanfa we need to memorize some almost trivial data:
 * the beads necessary to form a digit.
 * the complements to 5 of the digits 1, 2, 3, 4 and 5
 * the complements to 10 of the digits 1, 2, ..., 10

Beads needed to form a digit
Remember what was said in the Introduction of this book: "Addition is simulated by gathering the sets of counters representing the two addends, while subtraction is simulated by removing from the set of counters representing the minuend a set of counters representing the subtrahend". So we need to know the beads that make up each digit to be able to add or subtract them, but we already know this from the figure:

or in table form:

Complementary numbers
We also need to memorize two types of digit pairing: These are the digit pairs that together add up to 5 or 10. We can always find them mentally with our knowledge of addition and subtraction, but with practice they will end up solidly installed in our memory without the need to mentally "calculate" them. They are the basis of the mechanics of the abacus.
 * 5-complements
 * 10-complements

At a later stage, to deal with negative numbers, you will also need to handle 9-complements:

But for now you can safely forget about them.

The rules to use
The mechanics of addition and subtraction are based on three rules to be tried in sequence with the following protocol:
 * Only if a rule fails (because we do not have the necessary beads to complete the operation) we proceed to try the next rule.
 * The second of the rules only works for digits 1, 2, 3 and 4.
 * The third rule decomposes the operation into two other "simpler" ones: a carry to the column directly to the left or a borrow from that column plus an operation of the opposite type (i.e. a subtraction if we are adding or a sum if we are subtracting). This case raises the following points to take into account:
 * The first operation (carry or borrow) is trivial most of the time.
 * The second operation (the opposite of the starting one) is guaranteed to culminate using rules 1 or 2 (never 3) of the opposite operation.
 * We need to decide in what order we do these two operations.

Addition rules
These are the rules for the addition of a digit:

Subtraction rules
And these are the rules for subtraction:

Joint rules for addition and subtraction
The above rules for addition and subtraction are of identical structure so we can merge them into:

and we have only three rules to memorize!

Order of operation
Before moving on to some preliminary examples we have to decide what order of operation to use in case we reach rule 3, which will happen half the time. This rule leads us to a split of the original problem into two hopefully simpler ones: a carry or borrow and a operation of the opposite type to the one we are performing. What do we do first?...(sakidama, atodama...)

The Japanese standard method currently taught since the end of the 19th century proposes to first carry out the borrow and then the addition of the complementary number in the case of subtraction (sakidama 先珠), while in the case of addition, the subtraction of the complementary number is done first and then carry to the left column (atodama 後珠). This seems inspired by the structure of Chinese rules / verses / rhymes used for teaching the abacus since ancient times, but there does not seem to be any compelling logical reason to do so and not everyone agrees.

As we will see, with the abacus one works from left to right during the addition and subtraction of multi-digit numbers, so it seems natural to try to respect this movement from left to right of the hand without disturbing it with continuous comings and goings to the column of the left. Always using sakidama (carries and borrows first) seems the most natural thing to do.

It goes without saying that if you have a teacher or coach you should scrupulously follow their directions, but if you are self-taught feel free to experiment until you find your way.

By the way, in some Asian countries it is taught to use the left hand for carries and borrows.

Some preliminary examples
Ex: Enter 1 to a column of your abacus and add 3 to it:
 * 1) Do we have at our disposal (inactive) the necessary beads (3 lower ones) to add them to the 1 in our abacus? Yes!
 * 2) *then we activate them and we have completed the operation with the first rule.

Ex: Enter 3 to a column of your abacus and add 3 to it:
 * 1) Do we have at our disposal (inactive) the necessary beads (3 lower ones) to add them to the 3 in our abacus? No!
 * 2) * then we go to the second rule.
 * 3) As the addend 3 is less than 5 we can try the second rule: Do we have at our disposal (inactive) an upper bead? Yes!
 * 4) * then we apply the second rule: we activate the upper bead and retire two lower beads (the 5-complement of the addend 3).

Ex: Enter 9 to a column of your abacus and add 3 to it:
 * 1) Do we have at our disposal (inactive) the necessary beads (3 lower ones) to add them to the 3 in our abacus? No!
 * 2) * then we go to the second rule.
 * 3) As the addend 3 is less than 5 we can try the second rule: Do we have at our disposal (inactive) an upper bead? no!
 * 4) * the we proceed to the third rule:
 * 5) Carry one to A and subtract 7 (the 10-complement of 3) from B
 * 6) Do we have at our disposal (active, we are subtracting now!) the necessary beads (one upper bead and 2 lower ones) to retire them from the 9 on our abacus? Yes!
 * 7) *then retire them and we have completed this part of the operation with the first rule.

As you can see, the rules used here are the same that appeared in Xu Xinlu's Panzhu Suanfa, but condensed into only three rules thanks to the concept of complementary numbers!

Let us see now the reverse mouvements for subtraction:

Ex: Enter 4 to a column of your abacus and subtract 3 from it:
 * 1) Do we have at our disposal (active) the necessary beads (3 lower ones) to retire them from the 4 in our abacus? Yes!
 * 2) *then we deactivate them and we have completed the operation with the first rule.

Ex: Enter 6 to a column of your abacus and subtract 3 from it:
 * 1) Do we have at our disposal (active) the necessary beads (3 lower ones) to subtract them from the 6 in our abacus? No!
 * 2) * then we go to the second rule.
 * 3) As the subtrahend 3 is less than 5 we can try the second rule: Do we have at our disposal (active) an upper bead? Yes!
 * 4) * then we apply the second rule: we deactivate the upper bead and add two lower beads (the 5-complement of the subtrahend 3).

Ex: Enter 12 to a pair of columns of your abacus (AB) and subtract 3 from B:
 * 1) Do we have at our disposal (active) the necessary beads (3 lower ones) to retire them from B? No!
 * 2) * then we go to the second rule.
 * 3) As the subtrahend 3 is less than 5 we can try the second rule: Do we have at our disposal (active) an upper bead? no!
 * 4) * the we proceed to the third rule:
 * 5) Borrow 1 from A and ADD 7 (the 10-complement of 3) to B
 * 6) Do we have at our disposal (inactive, we are adding now!) the necessary beads (one upper bead and 2 lower ones) to add them to the 2 on B? Yes!
 * 7) *then activate them and we have completed this part of the operation with the first rule.

Types of one-digit addition and subtraction
The following table shows for each of the 180 elementary operations of addition and subtraction which rule solves the problem. It can be useful during your first practice, to choose which digits to add or subtract.

As you can see, the tables for addition and subtraction are mirror images of each other. Also note how half the cases correspond to rule three, that is, they require carries and borrows and from them, those marked in bold, end with an opposite type 2 operation. Also check how rule 2 only affects the addition / subtraction of digits less than 5.

Examples
Ex: Enter 5 to a column of your abacus and add 7 to it:

In this example, which requires a carry, the subtraction of the complementary number in turn requires the use of rule 2, affecting the upper bead.

Ex: Enter 95 into your abacus and add 7 to it:

Now the carry leads to another type 3 operation, requiring in turn a new carry. Three columns of the abacus are affected in this operation.

Ex: Enter 999995 into your abacus and add 7 to it:

This is an extreme situation, extrapolation from the previous case, which you should study carefully. The carry spreads or runs through the columns to the left until it finds a hole to lodge!

Note that, if we had at our disposal a lower fifth bead, in the case of the traditional abacus, we could have avoided this "carry run" at least temporarily.

For details on the use of the lower fifth ball you can consult the book: Traditional Abacus and Bead Arithmetic once you have acquired practice in addition and subtraction.

Ex: Enter 50 into your abacus and subtract 3 from it it: In this case of a type 3 operation, the borrow in turn requires a type 2 operation (affecting the upper bead).

Ex: Enter 10006 into your abacus and subtract 7 from it it:

Finally, this is a case of "borrow run" where we have to travel far to the left to find something to subtract from! Study this case carefully as well.

Two advices
So far our theoretical or intellectual explanations about the abacus. Now you know what the eastern abacus "is" and you are on your way. This intellectual knowledge will be your guide during your first steps, but with practice the movements of the beads will become second nature to you and you will never think about all these rules again (at least, until you write your first book on the abacus). To achieve this you will need to practice and practice and we offer you a couple of important tips to help you complete the path you are taking now.

 * Never read intermediate results. This is a bad habit that does not lead to anything, only to wasting time and wasting mental energy, and what you want is to acquire speed and comfort in the use of the abacus. Your abacus is there, and you have paid for it, just to keep your numbers safe for you without you having to worry about them. You only have to "react" to the arrangement of the beads without having to be aware of what number they represent.
 * Forget the addition and subtraction tables, except what we have extracted from them in the form of numbers complementary to five and ten. In particular, never think: "I have to add 7 + 8, this gives 15, then a fifteen has to appear on the abacus." If you do this, you will be "thinking" while adding and subtracting and that will tire you out and slow you down. If you have to think about something, think about the rules of movement of beads and not numbers, until you are able to add and subtract mechanically while thinking about anything else.

If you do not follow these tips, you will develop a bad habit that can be very difficult to correct later, as with the bad habits that are acquired when studying a musical instrument.

And now the practice
Your first exercises should be as simple as possible and nothing seems easier than randomly choosing two digits, for example: 6 and 8, and trying to add or subtract them (perhaps adding a one in front of the first digit if when subtracting you will need to borrow). You can use the table of types of operations explained earlier in this chapter to know in advance the type of operations to be carried out.

Subsequently, you should proceed to a systematic practice of all 180 cases of addition and subtraction of a single digit, for which the following exercise is proposed, which will also serve as an introduction to the addition and subtraction of multi-digit numbers.

Start with the abacus in the following state and add the same digit to each of the nine columns B-J proceeding from left to right

For example, to add 1 to each digit of 123456789 follow the steps indicated in the following table

and you should arrive at the result: 234567900; that is, 123456789+111111111. The following table shows the results of adding 111111111, 222222222, ... 999999999 to 1234568789.

For subtraction, add an additional 1 to column A for future borrows:

and proceed similarly

and the result is 1012345678 = 1123456789-111111111. For the rest of digits, the following table shows the results of subtracting 111111111, 222222222, ... 999999999 from 1123456789.

During a time you should practice these exercises daily until you notice that little by little you are replacing your intellectual work (thinking about the rules to use) by an instinctive mechanical response. Then you can say that you have started to learn the abacus.

Always work from left to right
In English, numbers are stated starting with the highest power of ten; 327 is "three hundred twenty-seven" and not "seven-twenty-three hundred". This is the case for many other languages, including Chinese and Japanese, but not for others such as those of the Semitic family. This is the main reason why in the abacus the sum of multi-digit numbers is worked from left to right; everything will be much easier for you, whether you have to read the numbers on a list or if someone else is dictating them to you.

For instance, let us obtain 44+78. Start with a cleared abacus and introduce the first addend 44 anywhere on the abacus (aligned to a unit rod mark if you wish, this is convenient but not essential)

then enter 4 (40) in B then enter 4 in C now add 7 (70) to B and, finally, add 8 to C The result: 122 appears on ABC.

In a more synthetic form:

Another example. Suppose we have to obtain the total of these amounts in euros: Start by clearing your abacus and enter the first number (from left to right). Align it with some of the unit dots markers if you wish.

With exercise sheets
You should start your practice by adding and subtracting short series of small integers; for example, 3 to 5 numbers of 2 or 3 digits. For instance:

Progressively increase the size of these series until you reach 10 numbers and, from here, progressively increase the size of the numbers to be added / subtracted to 5 or 6 digits. For instance:

You will therefore need collections of problems of this type that you can generate with some free utilities on the internet .

With this type of practice you will develop two different skills The latter is essential to, for example, make use of the abacus in accounting.
 * efficiently add and subtract with the abacus.
 * read numbers at a glance and keep them in memory long enough to work on the abacus.

The 123456789 exercise
It was common in ancient books on the abacus to demonstrate addition and subtraction using a well-known exercise that consists of adding the number 123456789 nine times to a cleared abacus until the number 1111111101 is reached, and then erase it again by subtracting the same number nine times. This is a convenient exercise because it uses many of the 1-digit addition and subtraction cases (but not all) and allows you to practice addition and subtraction without a paper worksheet, but it is not an elementary exercise given its length. You will need some practice time to complete it without error; consider it rather as a test of your proficiency in addition and subtraction.

Throughout this exercise the following partial results are obtained:

For more details, please refer to the chapter: Extending the 123456789 exercise of the book: Traditional Abacus and Bead Arithmetic.

Addition and subtraction with other abacuses
Everything you learn about the eastern abacus will work well with other types of abacus or at least simplify your learning. Remember that the basic operations of the abacus are addition and subtraction and everything else must be reduced to a sequence of these two operations and to the problem of how to organize such a sequence of operations on the abacus.

Counting rods
Counting rods are another example of a bi-quinary abacus, so the same three rules of addition and subtraction studied here apply to it. You just have to bear in mind that the concepts of activating / deactivating beads translate to placing / removing rods from the table and that "having at our disposal" rods to add does not refer to the pile of rods ready to be used that we have in the box, but to "fit" within the limits of representation of numbers (one 5-rod and 5 1-rods maximum).

Russian abacus
The Russian abacus (Schoty) is not a bi-quinary abacus; so the second of the addition and subtraction rules given here does not work, everything is solved with the help of the first and third rules only.

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


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History of video games/Platforms/Neo Geo Mini

Announcement
In 2018 SNK announced a miniature version of the Neo Geo in an arcade style format in celebration of the SNK 40th anniversary.

Launch
The console was released in Japan by September 2018. A release occurred on September 10th, 2018 in Europe and the United States of America.


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History of video games/Platforms/Capcom Home Arcade

Development
System development was handled by Koch Media under license from Capcom.

Launch
The Capcom Home Arcade was released on November 8th, 2019.

The use of FB Alpha caused some controversy in the emulator community due to the use of a non-commercial emulator in a commercial product.

Compute
The system is based on an Allwinner H3 SOC, which includes a quad core ARM Cortex A7 architecture processor clocked at 1.08 gigahertz and a Mali400 MP2 GPU. It is speculated that the system had 4 gigabytes of RAM, while others list 512 megabytes of DDR3 RAM.

The system has 4 gigabytes of NAND flash storage provided by a Kingston eMMC chip.

Software
The system runs an operating system based on the Linux 5.0 kernel.

The open source FB Alpha emulator is used to run games on the system, which were originally made for the CPS1 and CPS2 platforms.

Connectivity
The system supports Wi-Fi, which is used for leaderboards, and system updates. The Wi-Fi is handled by a Realtek chipset.

A micro USB port is used to provide power to the device, at 5 Volts and 2 Amps. A full size USB port is also present on the system.

An HDMI port is used to output AV. The system outputs at 1080p resolution at 60 frames per second.

Hardware
The system was contained in a fight deck shaped around the Capcom logo. Despite being commonly categorized as a Microconsole, the system is often considered to be quite large. The system could be used on a table, on a lap (Though this was somewhat unoptimal), or a similar surface.

Sanwa sticks and OBSF buttons are used for the control deck.

The system was designed in the United Kingdom, and manufactured in China.

Games

 * Progear
 * Giga Wing
 * Mega Man: The Power Battle
 * Eco Fighters
 * Strider'
 * Armored Warriors
 * 1944 The Loop Master
 * Cyberbots: Full Metal Madness
 * Capcom Sports Club
 * Super Street Fighter II Hyper Fighting'
 * Super Puzzle Fighter II Turbo
 * Ghoul's N Ghosts
 * Alien Vs. Predator
 * Final Fight
 * Darkstalkers: The Night Warriors
 * Captain Commando

External Resources

 * Official Website


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History of video games/Platforms/Roshambo Retro Gaming Kit

History
The system was announced on March 15, 2019.

The kit was sold later that year in 2019 for $33.00 (Without a computer board).

Technology
Kits were sold with either a Rock64 or Rock64Pro board.

The system used cartridges to contain SSDs. The system featured a working eject button for these cartridges.


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History of video games/Platforms/Neo Geo Arcade Stick Pro

Announcement
A new Neo Geo device was hinted at on August 1st, 2019.

The Neo Geo Arcade Stick Pro was announced on September 9th, 2019. In late September 2019 a website was launched for the system.

Launch
The console was released on November 11th, 2019.

In December 2019 system cost roughly $130 USD, and $300 Australian dollars in Australia. The system has an MSRP of $129.99 USD.

Legacy
The console was generally received well.

The good emulation of the console was generally seen as held back by the lackluster controller.

Modes of Operation
The system could be used as an independent console, or as a controller for a different device. The console could also connect with another Neo Geo Arcade Stick Pro console to facilitate multiplayer.

The system supported Neo Geo Mini arcade pads as input devices. Conversely the system itself could be used as a controller for the Neo Geo Mini.

Compute
The system is powered by an ATM7051H SOC, containing an quad core ARM Cortex A9 architecture processor clocked at 1.3 gigahertz. This is complemented by 256 megabytes of DDR3 RAM.

The system contains 1 gigabyte of eMMC flash memory for storage.

Software
The console supported a number of visual filters for display output.

Hardware

The controller arrangement was based on that used by the Neo-Geo CD.

Micro switches are used in the control stick. Button configurations on the device could be changed by the user.

An internal compartment accessible from the bottom of the device could be used to store cabling.

Various optional accessories were made available, including silicon covers for the console, the stick ball, and for external control pads. Other accessories included button stickers, A branded red USB type A to type C power cable, and a branded white USB type A to type C adapter.

Built In

 * Art of Fighting
 * Garou: Mark of the Wolves
 * Fatal Fury Special
 * Fatal Fury 3
 * The King of Fighters '95
 * The King of Fighters '97
 * The King of Fighters '98
 * The King of Fighters '99
 * The King of Fighters 2000
 * The King of Fighters 2002
 * Samurai Shodown II
 * Samurai Shodown III
 * Samurai Shodown IV
 * Samurai Shodown V Special
 * World Heros 2
 * World Heros 2 Jet
 * World Heros Perfect
 * The Last Blade 2
 * Kizuna Encounter
 * Ninja Master's

2019

 * Metal Slug
 * Metal Slug 2

2020

 * Metal Slug 3
 * Metal Slug 4
 * Metal Slug 5
 * Metal Slug X
 * Shock Troopers
 * Shock Troopers 2nd Squad
 * Fatal Fury
 * Real Bout Fatal Fury Special
 * The Last Blade
 * Savage Reign
 * League Bowling
 * The Path of the Warrior Art of Fighting 3
 * The Super Spy
 * Ninja Fury 2
 * Soccer Brawl
 * Super Sidekicks
 * Samurai Shodown

Special Edition Neo Geo Arcade Stick Pro
Sample units were subtly stamped with the word "Sample" on the front.

External Resources

 * Official Website


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History of video games/Platforms/PiCade

History
The Pimoroni Picade was an kit used to turn raspberry pis into mini arcade consoles.

An improved Picade model was released in 2018.

title: History of video games/Platforms/PiCade


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History of video games/Platforms/RetroN Sq

History
The RetroN Sq was formally announced around January 2020, after having been briefly teased in 2017. It is publicly unknown what effect, if any, the COVID-19 pandemic had on the development and release of the video game console.

The Hyperkin RetroN Sq home console was released on March 25th, 2021 for $75. Following launch a number of firmware patches were released to fix severe issues with the framerate and aspect ratio of the video output.

Technology
The system uses software based emulation to run Game Boy and Game Boy Advance games, with Game Boy compatibility being good and Game Boy Advance compatibility being flaky at launch.

The console outputs a 720p signal, using pillarboxing to maintain original aspect ratios.

The console was subject to a California Proposition 65 warning.


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Template:User Oregon

Adds the page to the following category:

title: Template:User Oregon


text: Human brain/Printable version

title: Human brain/Printable version


text: Classical Chinese/Printable version

title: Classical Chinese/Printable version


text: The Legend of Zelda: Ocarina of Time Master Quest/Printable version


text: Super Mario Bros.: The Lost Levels/Printable version


text: Template:StrategyWikiPrint

title: Template:StrategyWikiPrint


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German/Grammar/Imperatives

Imperatives
We now talk about how to make requests, give instructions, issue orders, and other ways of telling someone else what to do.

Terminology
The imperative is a special verb conjugation used to form requests and commands. Grammarians call it a mood rather than a tense since it's not a function of time, but it seems more convenient to place the Imperative in a category of it's own. In this section we'll be talking about the grammatical Imperative. German, like English, has many ways of expressing one's wishes, and many of them do not involve a special form of the verb. For example you can say "Would you bring me some tea?" Grammatically, this is a question though it's intended as a request. You can also say "I'd like you to bring me some tea." Again, this is really a request but this time it has the form of a statement. The grammatical imperative is "Bring me some tea." This sounds rather impolite in English and the German equivalent would sound equally impolite. We'll ignore the forms that are not grammatical imperatives here because their grammar is covered under the corresponding verb and sentence forms. Though it may be impolite depending on the circumstances, the grammatical imperative is still used quite often. There are actually two forms of imperative in German. We'll call the one which most closely resembles the imperative in English the imperative (without a qualifier), and the second one the impersonal imperative.

Forming the imperative
The subject in an imperative sentence is the person you want to do whatever it is you want to be done. This means that not all person/number combinations make sense and the corresponding inflections do not exist. We'll cover each combination which does exist separately since there are slight differences in grammar and shades of meaning associated with each.

Imperative sentences follow the V1 plan like yes/no questions; in most cases you can change a question to an imperative just by changing the verb inflection and the question mark.

Second person singular (familiar)
The subject here is du and this form is used when addressing a single person you are on "du terms" with. To form the imperative here, simply remove the -st from the second person singular of the present tense. So stellen ("to place, set"), which is stellst in the second person singular, becomes stell as an imperative. The du here is optional and usually dropped; it's still there grammatically though since it remains the subject. It would be left in if you wanted to stress the du for some reason. So a more normal phrasing would be: If the verb stem ends with an 's' sound, then you only add -t to form the second person singular, and so you only drop the -t: As seen in the previous example, you would normally soften an imperative with bitte – "please". This can be added as a preamble, postamble, or as an adverb after the verb. Note that you can use bitte equally well with the other person/number combinations. Another softening word used in spoken German is mal, which is difficult to translate directly into English, but lends a certain amount of casualness to the request. Unlike bitte, it follows the verb. (We'll talk about mal again in a later section.)
 * Stell du die Teller auf den Tisch. – "Place the plates on the table."
 * Stell die Teller auf den Tisch. – "Place the plates on the table."
 * Sitz hier, bitte. – "Sit here, please."
 * Bitte stell die Teller auf den Tisch. – "Please place the plates on the table."
 * Stell bitte die Teller auf den Tisch. – "Please place the plates on the table."
 * Stell die Teller auf den Tisch, bitte. – "Place the plates on the table, please."
 * Stell mal die Teller auf den Tisch. – "Place the plates on the table (if you don't mind)."

You can optionally add an -e ending to the verb with no change in meaning. Verbs which add -est in the second person would leave the -e part anyway.
 * Arbeite hart. – "Work hard."

Verbs ending in -ig will always have a final -e as well. Presumably this is because the g in the -ig ending is usually soft in German, and the -e helps to keep the hard g sound.
 * Beleidige sie nicht. – "Don't offend her."

Remember that stem changing verbs have a vowel change in the second person singular. This change is normally carried over to the imperative.
 * Gib mir die Löffel. – "Give me the spoons." (From geben)

However, if the change is to add an umlaut, then it is not carried over to the imperative:
 * Lauf! – "Run!" (From laufen, which is a stem changing verb with du läufst for the second person.)

Separable verbs remain separated in the imperative, with the prefix placed at the end of the sentence as usual:
 * Fang an. – "Begin." (From anfangen.)

Of the irregular verbs we've seen, the only ones that don't follow the above rule are: There are irregular verbs we haven't covered yet, but the imperative isn't applicable to them.
 * haben
 * Hab keine Angst! – "Have no fear!"
 * werden
 * Werd nicht hysterisch! – "Don't become hysterical!"
 * wissen
 * Wisse mehr. – "Know more."
 * sein
 * Sei aufmerksam. – "Be alert."

Second person plural (familiar)
The subject in this case is ihr and this form is used when addressing multiple people you are on "du terms" with. In this case the imperative is the same as the present tense, only the position of the verb tells you that it's the imperative. As with the singular case, the subject ihr is usually dropped but is still there in the sense that it determines the verb conjugation:
 * Stellt die Teller auf den Tisch. – "Place the plates on the table." (To multiple people.)

Keep in mind that even though the English forms are the same, you have to use the correct ending in German to avoid confusion. This might be important; if you spot danger approaching (an angry troll, for example) you need to shout Lauft! to get your companions to run away. Shouting Lauf! may just result in puzzled looks as they try to figure out which one of them you're talking to.

Second person (formal)
The subject in this case is Sie and this form is used when addressing one or more people you are on "Sie terms" with. As with the Sie form of the present tense, this is usually the same as the infinitive. However, the imperative of sein here is seien. Unlike the familiar forms, the subject Sie is never dropped.
 * Stellen Sie die Teller auf den Tisch. – "Place the plates on the table." (Formal form.)
 * Seien Sie aufmerksam. – "Be alert." (Formal form.)

Note that, unlike the infinitive, separable verbs are still separated.
 * Fangen Sie an. – "Begin." (Formal form.)

First person plural
The subject in this case is wir and this form is used when you want to suggest something for a group of people including yourself. The verb form is the same as for the present tense, which for wir is usually the same as the infinitive. As with Sie, the subject is never dropped. The form for sein is again seien.

 * Gehen wir ins Kino. – "Let's go to the movies."
 * Fangen wir an. – "Let's begin."
 * Seien wir klug. – "Let's be smart."

There is a third person form as well, but this is closely related to the special subjunctive and we'll cover it in the corresponding section.

The impersonal imperative
This is a feature of German which does not exist in English, and so it can be tricky to translate. (It's not really even that common in German, but it is used frequently enough that you should be familiar with it.) It's used when no one in particular is being addressed, so, for example, for recipes, instruction booklets, signs, etc. It can also be used in the spoken language, but in that case it has a harsh tone and would only be used in situations where politeness is not an issue. For example it might be used by a military officer giving orders to subordinates. Its form is simply the infinitive, placed at the end of the sentence where the infinitive normally goes. Since there is no finite verb, this might be called the V0 sentence plan, but this is the only time it's used. You can translate this in a number of ways, but none capture the German sense exactly so you must settle for the closest match under the circumstances. For example, you might see this in a recipe for bread: You might use the imperative in English: Or you could use the passive voice to avoid having a person as the subject: Or you could use a vague pronoun like "you" or "one":
 * Den Teig kneten. – "(Literally) To knead the dough."
 * "Knead the dough."
 * "The dough is kneaded."
 * "One kneads the dough."

Just as it's customary in English to place a question mark at the end of an imperative to form computer prompts, German computer prompts are usually formed by placing a question mark at the end of the impersonal imperative:
 * Änderungen speichern? Ja Nein – "Save changes? Yes No"
 * Spiel verlassen? – "Exit game?"

title: German/Grammar/Imperatives


text: Digital Electronics/Printable version

title: Digital Electronics/Printable version


text:
A Beginner's Arduino Guide/Printable version

 * Introduction

 * What is Arduino?

 * Where to get Arduino?

 * Arduino Sketch

 * Arduino Language


text:
History of video games/Platforms/Retro Freak

Background
CYBER Gadget Corporation, a Chiyoda Tokyo based company, was founded on July 19th, 2000. This company would later go on to develop the Retro Freak game console, an emulation based console which played cartridges from older systems.

Launch
The Retro Freak console was was announced around 2015.

By December 2015 the console was being reviewed by western outlets, and was likely released sometime thereabouts.

Sales of the system were temporarily frozen for a software audit regarding the potential use of the not commercially licensed SNES9x emulator.

A limited edition console was released on the 18th of March, 2020.

Compute
The system is based on a Rockchip 3066 system on a chip. This system on a chip contains a dual core ARM Cortex-A9 CPU clocked at an unknown speed, likely around 1.6 gigahertz. This system on a chip also contains a quad core Mali-400MP4 GPU clocked at 250 megahertz.

The system has 512 megabytes of total DDR3 RAM split across dual NANYA 1508 NT5CB128M16FP - DII chips.

Unusually for a console, all computation occurs on a small removable microconsole, which can be separated from the cartridge reading module and used as a smaller standalone device, or as a larger console with the cartridge reading module, giving it the ability to read cartridges directly instead of relying on ROMs.

Storage
Onboard storage consists of 4 gigabytes of NAND flash made by Micron.

The system was able to backup cartridges to a microSD card,. The microSD card can also use to write a file needed to request firmware from the manufacturer, which can then be used to update the system, adding a rather unusual step compared to typical firmware updates.

Software
The system runs Android as an operating system.


text:
Learning anatomy/Oral cavity

Anatomy
The teeth (singular: tooth) are made up of parts shown below.
 * Crown
 * Enamel
 * Dentin
 * Neck
 * Gingiva
 * Pulp cavity
 * Pulp
 * Root
 * Cementum
 * Root canal
 * Periodontal ligament
 * Nerves and blood vessels
 * Alveolar bone
 * Jaw
 * Maxillae
 * Mandible

Function
The teeth function to chew food. The types of teeth are the incisors, canines, premolars, and molars. Incisors cut the food. The canines grip the food. The premolars and molars (from Latin: mola = grindstone) grind the food. A baby has 20 teeth, but they are not visible until they erupt through the gingiva (gums) and an adult has 32 teeth. Maxillary (upper) teeth and their associated periodontal ligament are innervated by the superior alveolar nerves, branches of the maxillary division, termed the posterior superior alveolar nerve, anterior superior alveolar nerve, and the variably present middle superior alveolar nerve. The teeth help us speak.

Anatomy
The parotid, sublingual, and submandibular glands are major salivary glands. The other ones are minor salivary glands.

Function
The salivary glands make saliva. Saliva travels through ducts and then into the mouth.

Saliva
Saliva has alpha-amylase which breaks down starch and lingual lipase which breaks lipids. Saliva also cleans the mouth.

Anatomy
The lingua (tongue) is made up of the base (root), body, and apex.

Function
The lingua helps in speech. Saliva helps taste the food. The chemicals from the food travel from the lingual papillae to the taste buds. The lingua and saliva from the food bolus. It gets sent down the pharynx.

Air
The mouth is also a passage for air.

title: Learning anatomy/Oral cavity


text:
Human brain/Cerebellum

Anatomy
The cerebellum (Latin for little brain) is made up of the cerebrocerebellum, the spinocerebellum, and the vestibulocerebellum.

Function
The cerebellum controls your posture and your balance.

title: Human brain/Cerebellum


text:
Human brain/Brainstem

Anatomy
The brainstem is made up of the midbrain, the pons, and the medulla (medulla oblongata).

Function
The brainstem controls your heart and lungs.

title: Human brain/Brainstem


text: Learning anatomy/Printable version

title: Learning anatomy/Printable version


text:
Template:Book is finished tag/doc

Usage
When book is done:

What you get:

Plan 2:

When book is done:

What you get:

title: Template:Book is finished tag/doc


text:
Template:Proto-Turkic/Header

style="text-align: center; background: #ffffff; border: solid 10px #87CEFA" rules="all" class="wikitable" ! style="background-color:#87CEFA; height: 50px; text-align: center; display: " colspan="" |

title: Template:Proto-Turkic/Header


text:
User:JacobOfChess/sandbox

My sand box

User:JacobOfChess/sandbox/Image test

title: User:JacobOfChess/sandbox


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User talk:Kokomei

Blocked
You cannot edit for indefinite due to disruptive editing. You are welcome to contribute constructively when editing privileges are restored. You may appeal this action with a reasonable argument by adding the text below. —Atcovi (Talk - Contribs) 18:29, 19 September 2021 (UTC)

title: User talk:Kokomei


text:
User:Rolodexter1/sandbox

asdf

title: User:Rolodexter1/sandbox


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User talk:Rolodexter1/sandbox

asdf
asdf Rolodexter1 (discuss • contribs) 18:29, 19 September 2021 (UTC)

title: User talk:Rolodexter1/sandbox


text:
User:Idoghor Melody

I'm Melody Idoghor

title: User:Idoghor Melody


text:
User:Mohmad Abdul sahib

Mohmad Abdul sahib

title: User:Mohmad Abdul sahib


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Minecraft resource gathering/Honey

Acquisition methods
To get honey, first craft a campfire, place it under a bee hive/nest. Then right click the hive/nest with a glass bottle to get honey. If you collect honey without placing a campfire, all nearby bees will attack you.

Use
Honey can remove the poison status effect when drank, but keeps other effects untouched.

You can combine 4 honey bottles into a single honey block. This block may be crafted back into 4 bottles. Honey blocks slow you down, half jump height and reduce fall damage by 80%. Fall speed is decreased if you slide on one, and allow you to stand in the same block as it in the way chests do.

title: Minecraft resource gathering/Honey


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Minecraft resource gathering/Meat

Acquiring meat
To get meat, craft a weapon and kill an animal. The recommended weapon to craft is a sword, but you can also use a bow and arrow if you would like to practice archery. Theoretically you can punch using any tool, item, or even with your bare hand, however a sword is recommended because it deals more damage, meaning you need less hits to kill the animal.

Cooking meat
To cook meat, put it an a furnace or a smoker (for half the exp. points and half the time) with fuel, like coal or charcoal. Cooked meat restores a lot more hunger than raw meat, making it the preferred choice for consumption at a base or settlement, or when preparing for a journey.

It is still possible to eat raw meat in Minecraft for a slight hunger reduction, and this is common when players are on the go and need to eat something immediately (Presuming they are out of superior provisions). This is also fairly common in the early game, when players may not have furnace access. Players are not penalized for eating meat raw, unless that meat is raw chicken, then they will get food poisoning.

Real life inspiration
In real life, any undercooked or raw meat has an increased chance of carrying a food-borne illness. However raw chicken commonly carries the disease Salmonella, and so cooking chicken thoroughly to kill the disease is especially important, which is reflected in Minecraft.

title: Minecraft resource gathering/Meat


text:
Exercise as it relates to Disease/Importance of physical activity and skeletal muscle fat infiltration in older adults

This article is an orginial critical analysis of; Effects of physical activity on strength and skeletal muscle fat infiltration in older adults:a randomized control trial by Bret Goodbaster et al. - Applied physiology - Americal physiological society 2008

What is the background of the research?
The loss of muscle and increase in skeletal muscle fat is highly associated with ageing, and how closely fat interacts with muscle both metabolically and chemically. It's been established in the literature that fat can infiltrate into muscle as we age. Relationships have been reported between accumulation of fat in muscle and the decline of muscle cells. Intermuscular adipose tissue (IMAT) is the term to describe this fatty infiltration in the muscle, referring to where the lipids (fats) are stored in adipocytes located underneath the deep fascia of muscle fibres (also termed as intramuscular fat). This refers to between the muscle fibres (intra) and also between the muscle groups (inter where there is visible storage of lips in the adipocytes). This reduction in muscle mass in the ageing population is a result of the greater fat infiltration into the muscle which is associated with functional decline and increased risk of mobility loss in older men and women. The association between low muscle mass and functional decline seems to be a function of underlying muscle strength. Of course, traditional techniques of doing so involve improving exercise and diet, but the effectiveness on whether physical activity can influence this decline on muscle cells and it's muscle fat infiltration is understudeied method gaining ground with distince advatiages fo declining this aging skeletat muscle fat infiltration.

Where is this research from?
This study was supported by a Cooperative Agreement with the National Institute on Aging (UO1-AG-022376).

This study can be found on Pub med: https://www.ncbi.nlm.nih.gov/pmc/articles/PMC2584841/

The author: Bret Goodpaster

Bret Goodpaster is part of the Department of Medicine, University of Pittsburgh, Pittsburgh, Pennsylvania and has been publishing 231 articles on health disease and physiology since 1993 and actively involved in the research and publication on health and physical activity.

What kid of research was this?
This study was a ranomized control trial which looked at the effectiveness of physical activity on strength and skeletal muscle fat infiltration in older adults. The research involved 52 subjects who were measure for baseline strength and CT scanning and were randomized into either a physical activity (PA) or successful aging health educational control (SA) group.

What did reasearch involve?
The intervention took its effect through different phases: adoption (weeks 1-8) ;transition (weeks 9-24) and maintenance (weeks 25 to end of intervention). The initial intervention in the early stage of phase 1 was primarily centre-based with a shift to home-based activity in the transitiion and mainentance phase. In the adoption phase, participants were given three 40 to 60 min supervise physical activtiy sessions per week which were conducted in the centre. Progressing along to the transition phase of the intervention there was more emphasis on home-based exercise and centre0based exercise was reduced to 2 time per week. The home-based exercises included training of the endurance/strengthening/flexibility. In the maintenance phase participants were encouraged to peform home-based physical activity a minimum 5 days per week. This was all in comparison to the control group ho participated in attention and health education. Participants were slinto into small groups weekly for the first 26 weeks and then monthly on health topics relavent to older adults and physical activity. Physical activity and peformance over the duration of this intervention in both groups were measured via extensor strength, body weight adn axial CT-derived images of midthigh skeletal muscle (CSA) and adipose tissue at base line and 12 months. With the knee extensor strength determined at 60°/s with a Kin-Com dynamometer.

Results
Specific torque was reduced in the control group. This effect was not observed in the physical activity activity group, with a between-group difference in the change in specific torque (Figure 1).

Both groups lost muscle strength as well as a reduction in the CSA although this was not statistically different between the 2 groups (Figure 2 and 3).

IMAT increased in the control group but not in the physical activity. There was an 18% increase of IMAT in the SA controls in this 1 year trial. This gain was nearly completely preventred by physical activity. This gain of fat, however was clearly depot specific; the subcutaneous thigh adipose tissue did not significantly change in either group (Figure 4).

Interpretation and Critical analysis
On of the key findings of the study was that the physical activity group prevented further loss of muscle strength in the older adults who in comparison with the control were on the steep slop of functional decline. The degree of strength loss being ~22% during the 1 year trial in the control group. Another novel finding was the reduced knee extensor strength in both the PA and control groups. Although these results were statistical different these results indicate that even with physical activity program there still a reduction of the maintenance of the muscle strength in the PA group which has to be further looked into. This is a possible direct caveat of the methodology, as the participants were given three 40 to 60 min physical activity supervised and unsupervised sessions focussing on endurance /strengthening/flexibility but the intensity at which the exercises was conducted was not outlined and consistent throughout the 1 year trial. The intensity is very important in physical activity to stress the body in a strenuous amount. In the Australian physical activity guidelines clearly state "all healthy adults aged 18-65 years should participate in moderate intensity aerobic physical activity for a minimum of 30 minutes on 5 days per week or vigorous intensity aerobic activity for a minimum of 20 minutes on three days per week. Another key finding was that the physical activity group at the end of the trial in comparison to the SA group prevented the increase in IMAT associated with ageing. This ~18% increase in IMAT in control group show that these effects on physical activity can be a strategy to compute the age associated decline in muscle mass and strength. This key finding true effect on the population or the clinical significance is possible to be another caveat to the methodology. There was limited information on any pre-existing conditions of these adults. Ageing also comes with a vast number of complications and degenerating changes that may influence to carry out these interventions to it's full potential. For example, patients with multiple conditions including neurocognitive and cardiac disorders would find these as barriers to performing the recommended amounts of exercise. All of which can impact their ability for adequate mobility and participation in the intervention.

Conclusion
Effects of physical activity can be a strategy to combat the age associated increase in muscle fat infiltration and prevent the functional decline associated with it.

Adequate physical activity can have changes in muscle fat infiltration, maintain strength and torque at muscle and prevent loss of muscle mass.

Practical advice
An adequate dose of physical activity intervention including intensity and time has to be given to see clinically significant changes.

Underlying pathologies of the older populations have to be considered of ageing populations and their compliance over the course of the intervention.

Further information/resources

Muscle mass, muscle strength, and muscle fat infiltration as predictors of incident mobility limitations in well-functioning older persons - https://pubmed.ncbi.nlm.nih.gov/15860469/

Sarcopenia: characteristics, mechanisms and functional significance’ -

‘Protein-containing nutrient supplementation following strength training enhances the effect on muscle mass, strength, and bone formation in postmenopausal women’ - https://dx.doi.org/10.1152/japplphysiol.00935.2007


text:
Talk:Guide to Dayi input/Example Codes

Recommend Local Document
I recommend creating a local example, rather then relying on external Google Docs link. Of course, no content should be copied without permission from the original creator. --Mbrickn (discuss • contribs) 08:27, 20 September 2021 (UTC)

title: Talk:Guide to Dayi input/Example Codes


text: Guide to Dayi input/Printable version

title: Guide to Dayi input/Printable version


text:
Talk:Guide to Array input

Recommend Local Documents
I recommend creating local examples, rather then relying on external Google Docs links. Of course, no content should be copied without permission from the original creator. --Mbrickn (discuss • contribs) 08:31, 20 September 2021 (UTC)

title: Talk:Guide to Array input


text:
User talk:Phanbade

If you want to make editing tests, please use the sandbox, or your own sandbox and don't edit main space pages like you did with Using Wikibooks. QuiteUnusual (discuss • contribs) 10:08, 20 September 2021 (UTC)

title: User talk:Phanbade


text:
Carnatic Music/Mandhra sthayi varisai

1. Focus on Sa, touch of ni
 * s______ | s___|____ |
 * g r s n.| s r | g m |

title: Carnatic Music/Mandhra sthayi varisai


text:
Talk:Carnatic Music

Main Book
The main book is an introduction, the subpages teach about Carnatic Music. Faster than Thunder (discuss • contribs) 16:24, 20 September 2021 (UTC)

title: Talk:Carnatic Music


text: Flora of New York/iTax-head

title: Flora of New York/iTax-head


text:
History of wireless telegraphy and broadcasting in Australia/Topical/Biographies/Thomas William Watkins Burgess

Thomas William Watkins Burgess
Burgess, Thomas William Watkins - 1861(NSW)-1929(NSW) - Licences: XDA Wagga Wagga (1913-1914) - Qualifications: Nil yet identified - early wireless experimenter - Electoral Rolls: medical practitioner (Wagga Wagga, 1901-1903)

A comprehensive biography of Thomas William Watkins Burgess has not yet been prepared for this Wikibook, however the following resources have been assembled in preparation:

 * Research - Thomas William Watkins Burgess - Summary of research to date

 * Transcriptions and Notes - Thomas William Watkins Burgess - More than 40 transcriptions

 * Photos, QSL cards and other graphics - Thomas William Watkins Burgess - Nothing uploaded as yet

 * Incubator of Wikipedia Article on Thomas William Watkins Burgess - Coming real soon now!

Pending further progress on the foregoing, the following lovely obituary from the Wagga Wagga Daily Advertiser provides an excellent life summary, though sadly missing any reference to his wireless experiments: "OBITUARY. DR. T. W. W. BURGESS. In a private message received in Wagga on Christmas Day, the news was conveyed of the sudden death of Dr. Thomas William Watkins Burgess, of Vaucluse, who practised in Wagga for a great many years. He was well known throughout Wagga and District, and the news of his death will be learned with very great regret — by a large circle of friends and former patients. Dr. Burgess, who first came to Wagga as assistant chemist to the late Mr. H. J. Williams, as a very young man, afterwards went into business for himself as Chemist, starting in Baylis street in premises now occupied by Mr. C. Perritt as a tyre repair depot. After conducting this business successfully for some years he purchased the site at the corner of Baylis and Morrow streets and there erected the premises now occupied by Mr. L. Anderson. It was while conducting his business in these premises that Dr. Burgess studied for medicine, eventually going to England, where he became attached to Guy's Hospital, London. Taking his degrees in medicine, he returned to Australia and began practicing in Wagga about 30 years ago. Dr. Burgess established a very extensive practice. He met the lady who eventually became his wife in England, and Dr. and Mrs. Burgess lived at 'Wollundry' which he had erected in Morrow street on the site of the original Temperance Hall. There were three sons, Messrs. Whittley, Clifford, and Neale Burgess; the elder is now Dr. Whittley Burgess, and practises at Vaucluse. Dr. Burgess's retirement from Wagga followed a serious breakdown in health, accentuated by an attack of pneumonic influenza contracted during the epidemic of 1919. He afterwards practised at 21 Coolong road, Vaucluse, and in the last few years had made a couple of trips to England for the benefit of his health. At the time that his death occurred at a private hospital in Darlinghurst on Christmas Eve, he was 68 years of age. Throughout the whole period of his practise in Wagga, Dr. Burgess was very successful and accumulated valuable property, both here and in Sydney. He also won a name for his very great kindness to poor families who were unable to pay doctors' fees. For many years he was the Government Medical Officer here, and an honorary doctor of the Wagga District Hospital. He was also one of the oldest members of the Masonic Lodge Harmony, No. 22. He joined the order in Wagga and became Master in 1901, while 22 years ago he was made a grand officer of the order, being appointed Grand District Inspector of Workings. He had held office as trustee of the Wagga Masonic Hall for many years right up to the time of his death. He was always a great worker for the fraternity. The funeral took place yesterday morning at the crematorium at Rookwood."


text: Template:Gffsgeys page

title: Template:Gffsgeys page


text:
Halo/Warthog

The Warthog, also known as the Hog, the M12 LRV, and the M12 Force Application Vehicle, is a small Jeep-like vehicle with a machine gun turret, while also comfortably seating 3. It is humorously known for it’s physics which more than often results in the vehicle flipping over, the power of the machine gun, and it’s presence in another form of media, Red vs Blue.

Usage in Multiplayer
The Warthog is useful for traversing maps that are medium to large sized, as the regular walkspeed is too slow and wastes too much time if you happen to die often. With the ability to seat a passenger and a gunner, bringing the total passenger count to 3 with the driver, you can quickly assemble a small attack force to assault the enemy team’s objectives.

Another aspect is the machine gun mounted on the back of the vehicle. It is also medium and somewhat long ranged, however at longer ranges the crosshair for the gun may not be sufficient for aiming. Instead, make sure to not focus on one target to continually hammer away at with bullets, focus on multiple other targets if possible and try to get someone or yourself to drive the warthog constantly in order to throw off snipers who are aware of your little run and gun tactic.

Since the warthog is a vehicle, another tactic is to attempt to run over enemy players, especially if your warthog has lost all it’s passengers and you are the last man standing, or if close range tactics are the only tactic available due to either a lack of a gunner or the gun cannot hit a target.

If you happen to be on the receiving end of a warthog attack, grenades, sniper rifles(if you can eliminate the driver or other armed targets), and rocket launchers are all effective at eliminating warthogs. If you wish to use plasma grenades, avoid throwing them on the window of the warthog because they simply slide off. Instead, try to throw the grenade on the turret and generally anywhere on the body of the vehicle.

title: Halo/Warthog


text:
Halo/Plasma Grenade

The Plasma Grenade, or the Type-1 plasma grenade, is a thrown grenade seen in Halo and it’s consecutive games. It resembles a small blue orb with orange and green markings on it. The grenade’s function allows it to stick to targets such as players and vehicles.

Usage in Multiplayer
The plasma grenade is not seen too frequently due to it’s power, and as such if an opportunity arises for you to obtain or even use one, it is important to know which situations to use it in.

One type of situation could be that you are engaged in a close range firefight with a few other enemy players in a somewhat small hallway, you could use the plasma grenade and stick one of the opponents to possibly a multi-kill out of it, and you probably won’t miss the throw either, however you will run the risk of killing yourself from the explosion if the hallway is especially close-quarters. In this situation, it is advised to back up as much as possible before sticking the enemy with a grenade.

Another situation is you are standing on a cliff or another ledge and a enemy player runs past below, taking a few potshots at your teammates. They don’t see you, but you can see them, and you have a chance of throwing the grenade onto them at this range. With good enough tracking and timing, you could possibly stick them with the grenade, however you are probably going to miss by a few inches. Instead, get as close enough to stick them, but far enough to not kill or injure yourself, then throw the grenade.

Another situation is a vehicle, say a warthog, or a scorpion, is barreling towards you and your teammates. You happen to have a plasma grenade, so where do you throw it? You will want to target the parts of the vehicle body closest to the driver or the passengers. On a warthog this means the mounted machine gun and the sides of the vehicle, on a scorpion this means the rollcage and the area next to it. Targeting these areas will often allow you to quickly eliminate or wound the passengers.;

title: Halo/Plasma Grenade


text:
Using an Abacus/Modern multiplication

Introduction
The basic concept of multiplication for natural numbers is that of a repeated addition.

$$a\times b = \underbrace{b + \cdots + b}_{a \text{ times}}$$

For example, to multiply 47 by 23 it is only necessary to add 23 47 times or add 47 23 times; we can do it with our abacus:

Where we repeat 23 times the sum of 47 to the IJ columns while we add 1 to C to have a "counter" at our disposal. But this is desperately slow! A more effective way to do the same thing can be the following:

Where this time, after adding 47 three times to IJ (and 1 to C) we have moved one column to the left and we have started adding 47 to columns HI (and 1 to B). Adding 47 in HI is equivalent to adding 470 = 10×47 to HIJ (10 to BC) drastically reducing the number of operations to be carried out, because after doing it twice only we reach 23 in the counter BC and 1081 in GHIJ, the final result. This way of multiplying was the usual one in mechanical calculators that appeared at the end of the 19th century and that continued in use until the 1970s. But this is still excessively slow.

Think that the abacus as we know it now allows adding very quickly, but that before its invention Chinese mathematicians used counting rods which are extraordinarily slow to handle. It is not surprising therefore that Chinese mathematicians, seeking to abbreviate calculations, eventually invented the decimal multiplication table, as we know it, a few centuries before our era.

The multiplication table
This is the decimal multiplication table as we learn it in school:

But living in the computer age, the most likely thing is that we will soon start using an electronic calculator and in adulthood we will do little multiplication by hand. Often many of us, even mathematicians, do not have the multiplication table "fresh" in memory and this can be bad news for you: if you want to multiply (and divide) efficiently with an abacus, you necessarily have to refresh the multiplication table in your memory!

Using the multiplication table we can solve the multiplication problem $$ 47 \times 23 $$ in the form:

$$47\times 23=(40+7)\times (20+3)=$$

$$=40\times 20+40\times 3+7\times 20+7\times 3=$$

$$=(4\times 2)\times 100+(4\times 3)\times 10+(7\times 2)\times 10+(7\times 3)$$

i.e. we only have to retrieve the partial products: $$(4\times 2)=8, (4\times 3)=12, (7\times 2)=14, (7\times 3)=21$$ from the multiplication table and add them in the correct places, as we do with paper and pencil

47 ×23 -   21  12   (×10)  14   (×10) + 8    (×100) - 1081 This is absolutely parallel to the multiplication method that we are going to follow with the abacus.

The modern multiplication method
When we multiply two numbers $$ a $$ and $$ b $$, we call both numbers factors and product to the result $$ a \times b $$, but it is also common to call multiplicand to one of the factors and multiplier to the other. Nevertheless, when it comes to multiplying with the abacus:
 * Multiplicand
 * It is the number that we are going to manipulate on the abacus and that will guide us to obtain the partial products in an orderly manner and to align them correctly for their addition in the correct positions.

 * Multiplier
 * It is the factor that we are not going to manipulate on the abacus. in fact it is not mandatory to even enter it (but it is convenient). It will usually be the factor of the two with the fewest digits.

Multiplication arrangement
There are two ways of entering both factors in the abacus that can be considered practically equivalent; Each of them has its own advantages and disadvantages. The same can be said of the division that we will study in the next chapter. Feel free to experiment with both arrangements.

Traditional Chinese arrangement
The multiplicand is located to the left of the abacus and the multiplier far enough away from the multiplicand. At least as many columns as digits have the multiplier plus two or better three must be left free.
 * Example:

or in table form:

Traditional Japanese arrangement
This is the reverse way. The multiplier is on the left and the multiplicand on the right, leaving at least two empty columns in between. We need to have at least as many free columns to the right of the multiplier as the number of digits in the multiplier plus one.

or in table form:

This is the form that has been most popular in Japan and also ended up being imported to China. It is also the form that we will use in this book.

1-digit × 1-digit multiplication
Of course this is so trivial that we don't need an abacus, but it serves to introduce the rest of the examples. Suppose we have to multiply $$ 7 \times 8 $$, let's take 7 as a multiplier, 8 as multiplicand and adopt the Japanese arrangement just explained; that is, we start from:

Yes, you are right; it is you who did the multiplication, not the abacus. In the following example, the abacus begins to show its usefulness.

1-digit × 2-digit multiplication
Let us multiply $$7\times 83$$, the multiplicand will be 83.

At least, the abacus has served to add the two partial products in FG and EF.

2-digit × 2-digit multiplication
Now, let us multiply $$79\times 83$$.

Multi-digit multiplication
Generalizing what was seen in the previous examples:
 * For each digit of the multiplicand, starting from the right:
 * Multiply the current digit of the multiplicand by the digits of the multiplier (from left to right), adding the first partial product to the two columns to the right of the current digit of the multiplicand, and the rest of the products by successively shifting one column to right every time.
 * Clear the current multiplicand digit.

Let us see it with the following example: $$799\times 835=667165$$:

The unit rod and decimals
Please, review all the examples seen so far and check that, in all cases: This is a general rule for the multiplication of natural numbers following the modern method of multiplication that we are studying. It is convenient to keep this rule in mind since the product could have zeros at the end, as in the case $$ 32 \times 1625 = 52000 $$; which could confuse you. For instance

In the above diagram, the unit rod of multiplicand is column H (signaled with a white dot on the bar). After multiplication, the abacus shows:

You need to know that the unit rod of the result is $$n+1=2+1=3$$ rods to the right of H (i.e. in J) to correctly read the result 52000.

We can extend this rule to decimal numbers:

The following table shows the $$ n $$ values for some multipliers:

Let us multiply $$ 0.0032\times 16.25$$; The unit rod of multiplicand is F.

and for the multiplier $$0.0032$$, we have $$n=-2$$

so that the unit rod of the product is $$n+1=-2+1=-1$$ rods to the right of F, i.e. one rod to its left (E) and the result must be read as $$0.052$$.

External resources
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text:
Overview of Elasticity of Materials/Introducing Strain

This section introduces strain and show tensor symmetry of strain tensor. We also will discuss special subsets of stress and strain including dilatation and deviatoric stresses and strains.

= Average Strain = In this section, we are going to revisit strain to consider it in the infinitesimal limit and to investigate its relationship to displacement using a tensor notation. When we began our discussion, we examined average strain, engineering strain in terms of linear elongation (Figure 1a) and shear deformations (Figure 1b).

= Infinitesimal Strain = While average strain generally looks at the strain of a volume, we will now consider how a point on an elastic body moves and how points near it do also. We will begin in 2D. Say there is a point ($$P$$) on an elastic body that is located at coordinate $$\{x_1, \ x_2\}$$ such as in Figure 2 (this notation will be more convenient when we want to work with tensors). If we deform the body, then $$P$$ is displaced to $$P'$$ which has the coordinates $$\{x_1 + u_1, \ x_2 + u_2 \} $$. We call $$\mathbf{u}$$ the displacement vector.

Looking at Figure 2, there is a point infinitesimally close to $$P$$, called $$Q$$, with coordinates $$\{x_1 + dx_1, \ x_2 + dx_2 \} $$. When $$P$$ is displaced to $$P'$$ by the deformation, $$Q$$ is similarly displaced to $$Q'$$ with coordinates $$\{x_1 + u_1 + dx_1, \ x_2 + u_2 + dx_2 \} $$. Thinking critically, the displacement experienced on a body depends on the position on the body. Therefore $$\mathbf{u} = u (x_1, \ x_2) $$. This allows us to use the chain rule to express infinitesimal displacements. Now define the following terms:

 * $$\begin{align}

\operatorname{d} \! u_1 &= {\partial u_1 \over \partial x_1} \operatorname{d} \! x_1 + {\partial u_1 \over \partial x_2} \operatorname{d} \! x_2 \end{align} $$    [3]
 * $$\begin{align}

\\ \operatorname{d} \! u_2 &= {\partial u_2 \over \partial x_1} \operatorname{d} \! x_1 + {\partial u_2 \over \partial x_2} \operatorname{d} \! x_2 \end{align} $$    [4]

 * $$\begin{matrix}

e_{11} &= {\partial u_1 \over \partial x_1} \qquad e_{12} &= {\partial u_1 \over \partial x_2} \\ e_{21} &= {\partial u_2 \over \partial x_1} \qquad e_{22} &= {\partial u_2 \over \partial x_2} \end{matrix} $$    [5-8] This allows us to write our infinitesimal displacements using Einstein Notation:
 * $$\begin{align}

du_i = e_{ij} \ dx_j \end{align} $$    [9]

Displacement Tensors
What is the physical significance of this? This is easier to see looking in special directions. Considering the points $$P = \{ x_1, \ x_2 \}$$, $$Q_1 = \{ x_1 + dx_1, \ x_2 \}$$ where $$dx_2 = 0$$ and $$Q_2 = \{ x_1, \ x_2 + dx \}$$ where $$dx_1 = 0$$ as seen in Figure 3. Then after the deformation:

 * $$\begin{align}

P' &= \{ x_1 + u_1, \ x_2 + u_2 \} \\ {Q_1}' &= \{ x_1 + u_1 + dx_1 + du_1, \ x_2 + u_2 + du_2\} \\ {Q_2}' &= \{ x_1 + u_1 + du_1, \ x_2 + u_2 + dx_2 + du_2\} \end{align} $$

How do we interpret this? In the case of $$Q_1 $$, we have $$\operatorname{d} \! x_2 = 0$$. Thus, based on Equations 3 & 5 and Equations 4 & 7 expressing the infinitesimal displacements, we can infer that $$\operatorname{d} \! u_1 = e_{11} \operatorname{d} \! x_1 $$, and $$\operatorname{d} \! u_2 = e_{21} \operatorname{d} \! x_1 $$.

This tells us that $$e_{11} $$ is an expression of uniaxial extension in the $$x_1 $$ direction and $$e_{21} $$ is a rotation of $$Q_1 $$ around the point $$P $$. Similarly, in the case of $$Q_2 $$, we can again combine Equation 3 with Equation 6 and Equation 4 with Equation 8 which yields $$\operatorname{d} \! u_1 = e_{12} \ \operatorname{d} \! x_2 $$, and $$\operatorname{d} \! u_2 = e_{22} \ \operatorname{d} \! x_2 $$. Thus, $$e_{22} $$ is a uniaxial extension in the $$x_2 $$ direction, and $$e_{12} $$ is a rotation of $$Q_2 $$ around point $$P $$. These $$\mathbf{e} $$ are our displacement tensors.

The Strain Tensor
Let's return to $$P $$, displacing to $$P' $$. What is the relationship between $$x_i$$, and $${x_i}'$$?

 * $${x_1}' = x_1 + u_1$$    [10]

From Equations 3, 5-8:
 * $$\begin{align}

\operatorname{d} \! u_1 &= e_{11} \operatorname{d} \! x_1 + e_{12} \operatorname{d} \! x_2 \end{align} $$    [11]

Integrating, we have
 * $$\begin{align}

\int_0^{u_1} \operatorname{d} \! u_1 &= e_{11} \int_0^{x_1} \operatorname{d} \! x_1 + e_{12} \int_0^{x_2} \operatorname{d} \! x_2 \\ u_1 &= e_{11} x_1 + e_{12} x_2 \\ {x_1}' &= x_1 + e_{11} x_1 + e_{12} x_2 \end{align} $$

Which can be rewritten as
 * $$\begin{align}

{x_1}' &= \left ( 1+e_{11} \right ) x_1 + e_{12} x_2 \end{align} $$    [12]

In a similar fashion, we can also prove that
 * $$\begin{align} {x_2}' = e_{21} x_1 + (1 + e_{22}) x_2 \end{align}$$    [13]

Looking at our tensor, we can see that our deformations also have translations and rotations. We are not interested in these because they do not tell us about material response such as dilatation (change in volume) or distortion (change in shape). Translations and rotations are a part of the field of mechanics called dynamics. Here we are interested in small scale elastic deformations. Our $$e_{12} \neq e_{21} $$, but we know that our stress tensor is symmetric as $$\sigma_{12} = \sigma_{21} $$. We can therefore rewrite our displacement tensor as a combination of a symmetric and antisymmetric tensor.

 * $$\underbrace{\left ( \begin{matrix}

e_{11} & e_{12} \\ e_{21} & e_{22} \end{matrix} \right )}_{\mathbf{e}} = \underbrace{\left ( \begin{matrix} e_{11} & {1 \over 2} (e_{12} + e_{21}) \\ {1 \over 2} (e_{21} + e_{12}) & e_{22} \end{matrix} \right )}_{\boldsymbol{\varepsilon}} + \underbrace{\left ( \begin{matrix} 0 & {1 \over 2} (e_{12} + e_{21}) \\ {1 \over 2} (e_{21} + e_{12}) & 0 \end{matrix} \right )}_{\boldsymbol{\omega}} $$    [14]

Here, $$\boldsymbol{\varepsilon} $$ is the strain tensor and $$\boldsymbol{\omega} $$ is the rotation tensor. This can be seen schematically in Figure 4. In the scope of this text, we are only interested in $$\boldsymbol{\varepsilon} $$, but it is generally still worth remembering that displacement includes both shear and rotation components:
 * $$\begin{align} u_i = \varepsilon_{ij} x_j + \omega_{ij} x_j \end{align}$$    [15]

If a deformation is irrotational, or in other words, the directions of the principal axes of strain do not change as a result of displacement, then $$w_{ij}=0$$ and

 * $$u_i=\varepsilon_{ij}x_j$$    [16]

The strain tensor maps the irrotational displacement at a point to an imaginary plane, with normal in any direction that cuts through the point. Because the strain tensor $$(\boldsymbol{\varepsilon}) $$ is a tensor, it must transform in the same manner as the stress tensor did in earlier sections of this text. As a reminder:
 * $$\begin{align}

{\sigma_{kl}}' &= a_{ki} \ a_{\ell j} \ \sigma_{ij} \end{align} $$    [17]

 * $$\begin{align}

{\varepsilon_{k\ell}}' &= a_{ki} \ a_{\ell j} \ \varepsilon_{ij} \end{align} $$    [18]

Average Engineering Strain
Note that when we first started to look at this subject we defined shear strain as $$\gamma = \tan \theta $$, which is asymmetric. In terms of our strain tensor, this would be $$\gamma_{ij} = 2 \varepsilon_{ij} $$. (It must be rotated back so each side had an angle of ${1 \over 2} \theta $ .) You may frequently see a matrix written as:
 * $$\left ( \begin{matrix}

\varepsilon_x & \gamma_{xy} \\ \gamma_{yx} & \varepsilon_{y} \end{matrix} \right ) $$

It is sometimes useful to write it this way. However, it is not a tensor, because it does not transform the same as Equations 17 & 18 do, due to its asymmetry. Textbooks generally like using this "average engineering strain" $(\gamma_{ij}) $, but we will not be using this here unless absolutely necessary.

Generalizing 2D to 3D
The results we found for our 2D strain tensor can easily be generalized to 3D by writing them in Einstein Notation and using "3" in the place of "2" in the implicit sums. Equations 3 & 4 are

 * $$\operatorname{d} \! u_j = {\partial u_j \over \partial x_i} \operatorname{d} \! x_i$$    [19]

Which in 3D expresses 3 equations $$j = 1, \ 2, \ 3 $$, and expands to:

 * $$\operatorname{d} \! u_j = {\partial u_j \over \partial x_1} \operatorname{d} \! {x_1}

+ {\partial u_j \over \partial x_2} \operatorname{d} \! {x_2} + {\partial u_j \over \partial x_3} \operatorname{d} \! {x_3} $$

The displacement tensor is:

 * $$e_{ij} = {\partial u_i \over \partial x_j} $$    [20]

The strain tensor is:

 * $$\varepsilon_{ij} = {1 \over 2} \left ( e_{ij} + e_{ji} \right ) $$    [21]

The rotation tensor is:

 * $$\omega_{ij} = {1 \over 2} \left ( e_{ij} - e_{ji} \right ) $$    [22]

Which gives us the displacement:

 * $$u_i = \varepsilon_{ij} x_j + \omega_{ij} x_j = e_{ij} x_j $$     [23]

And the new displaced coordinates are:

 * $$x_j' = x_j + u_j $$    [24]

Now that we have a symmetric strain tensor with properties analogous to stress, we can examine the other properties using similar methods as when we analyzed stress. For small strains where $$\Delta \approx \varepsilon_{11} + \varepsilon_{22} + \varepsilon_{33}=I_1 $$, we define the mean strain as:

 * $$\begin{align}

{\varepsilon_{11} + \varepsilon_{22} + \varepsilon_{33} \over 3} = {\varepsilon_{kk} \over 3} = \varepsilon_m \end{align} $$

Thus,
 * $$ \varepsilon_m \approx {\Delta \over 3} $$    [25]

The total strain tensor can be broken into dilatation and deviatoric components:

 * $$\varepsilon_{ij} = {\varepsilon_{ij}}' + \varepsilon_{m}

= \left ( \varepsilon_{ij} - {\Delta \over 3} \delta_{ij} \right ) + {\Delta \over 3} \delta_{ij} $$    [26]

In a similar fashion, we also have deviatoric stresses and hydrostatic stresses which are analogous to the deviatoric and dilatation strains. The hydrostatic, or mean, stress is:

 * $$\sigma_m = {\sigma_{kk} \over 3} = {\sigma_{11} + \sigma_{22} + \sigma_{33} \over 3}  $$     [27]

Therefore, the deviatoric stress can be deduced because:
 * $$\begin{align}

\sigma_{ij} &= {\sigma_{ij}}' + {1 \over 3} \delta_{ij} \sigma_{kk} \\ {\sigma_{ij}}' &= {\sigma_{ij}} + \sigma_m \delta_{ij} \\ \\ {\sigma_{ij}}' &= \left ( \begin{matrix}    \sigma_{11} - \sigma_m & \sigma_{12} & \sigma_{13}     \\ \sigma_{12} & \sigma_{22} - \sigma_m & \sigma_{23}    \\ \sigma_{13} & \sigma_{23} & \sigma_{33} - \sigma_m \end{matrix} \right ) \end{align} $$    [28-31]

The principal components of $${\sigma_{ij}}' $$ break down to:

 * $${\sigma_{1}}' = {2 \over 3} \left( {\sigma_{1} - \sigma_{2} \over 2}

+ {\sigma_{1} - \sigma_{3} \over 2} \right ) $$     [32]

And we know that these are just the maximum shear stresses:

 * $${\sigma_{1}}' = {2 \over 3} \left( {\sigma_{12}}^{MAX}

+ {\sigma_{13}}^{MAX} \right )  $$     [33]

Keep in mind that we took this from:

 * $$0 = \left ( \sigma' \right )^3 - J_1 \left ( \sigma' \right )^2

- J_2 \sigma' - J_3 $$    [34]

Where:

 * $$\begin{align}

J_1 &= (\sigma_{11} - \sigma_m) + (\sigma_{22} - \sigma_m) + (\sigma_{33} -\sigma_m) \\ J_2 &= {1 \over 6} \left [ \left ( \sigma_{11} - \sigma_{22} \right )^2 + \left ( \sigma_{22} - \sigma_{33} \right )^2 + \left ( \sigma_{33} -\sigma_{11} \right )^2 + 6 \left ( \sigma_{12}^2 + \sigma_{23}^2 + \sigma_{31}^2 \right ) \right ] \\ J_3 &= \det|\boldsymbol{\sigma}'| \end{align} $$    [35-38]


text:
Overview of Elasticity of Materials/Isotropic Response

This section covers linear isotropic response.

We now understand tensor relations and have established two tensors that we need for elasticity; $$\boldsymbol{\sigma}$$ and $$\boldsymbol{\varepsilon}$$. In this chapter we will learn how to relate these to each other. In this particular section we will begin studying elasticity in the case of isotropic elasticity where we assume that all directions have the same material response.

Modulus of Elasticity
In earlier sections we learned that Hooke's Law can relate uniaxial loading to uniaxial strain:

 * $$\sigma_{11} = E \ \varepsilon_{11}$$    [1]

Here, $$E$$ is the modulus of elasticity. Note that we picked the $$x_1$$ direction arbitrarily. It could have easily been $$x_2$$ or $$x_3$$, or any other direction in this crystal, and found the same material response, $$E$$. This is what comes from working with an isotropic solid.

Poisson's Ratio
When we load stress in the $$x_1$$ direction, there is also a natural contraction in the transverse directions, $$x_2$$ and $$x_3$$. This is expressed as Poisson's Ratio, $$\nu$$.

 * $$\varepsilon_{22}

= \varepsilon_{33} = -\nu \varepsilon_{11} = -{\nu \ \sigma_{11} \over E}$$    [2] For a perfectly incompressible material, which means it conserves volume, we will have a Poisson's Ratio of 0.50. However, most real materials such as common metals have a Poisson's Ratio of around 0.33.

For linear elastic, isotropic materials, normal stresses will not induce shear strain and shear stress will not cause normal stress. In linear elasticity, the various contributions to strain can be superimposed. We can use the two above equations for the Modulus of Elasticity and Poisson's Ratio to determine that if a triaxial normal stress is applied, the strain will be:

 * $$\begin{align}

\varepsilon_{11} &= {\sigma_{11} \over E} - {\nu \sigma_{22} \over E} - {\nu \sigma_{33} \over E} \quad \begin{cases} \varepsilon_{11} &= {1 \over E} \left [ \sigma_{11} - \nu \left ( \sigma_{22} \sigma_{33} \right ) \right ] \\ \varepsilon_{22} &= {1 \over E} \left [ \sigma_{22} - \nu \left ( \sigma_{33} \sigma_{11} \right ) \right ] \\ \varepsilon_{33} &= {1 \over E} \left [ \sigma_{33} - \nu \left ( \sigma_{11} \sigma_{22} \right ) \right ] \end{cases} \end{align}$$    [3-5] Moving to shear relations, the shear stress and strain are related by the shear modulus, $$G$$.

 * $$\sigma_{12} = G \gamma_{12} = G 2 \varepsilon_{12}$$    [6]

Again, due to the isotropic nature, this relation holds in all the other directions as well. These three constants of proportionality are sufficient to describe the isotropic linear response. They take typical values (for common engineering metals):

 * $$\begin{align}

50 < \ & E < 200 \quad GPa \\ 25 < \ & G < 75 \quad GPa \\ 0.25 < \ & \nu < 0.35 \end{align}$$

There are many other useful relations that we will summarize here. The Block Modulus, $$K$$, is the ratio of the hydrostatic pressure to the dilatation it produces:
 * $K = {-p \over \Delta} = { \sigma_m \over \Delta} = {1 \over \beta}$     [7]

Where $-p$ is the hydrostatic pressure and $$\beta $$ is the compressibility.

Adding together Equations 3-5 for the triaxial normal strain, we get:

 * $$\begin{align}

\underbrace{\varepsilon_{11} + \varepsilon_{22} + \varepsilon_{33}}_{\Delta} &= {1 - 2\nu \over E} ( \underbrace{\sigma_{11} + \sigma_{22} + \sigma_{33}}_{3 \sigma_m} ) \\ \Delta &= {1 - 2\nu \over E} 3 \sigma_m \end{align}$$    [8]

Applying this to Equations 7 & 8 for the Block Modulus results in:

 * $$K = {\sigma_m \over \Delta} = {E \over 3(1-2\nu)}$$    [9]

These next solutions are given without proof, because several advanced topics are used to show the relationships.

 * $$ G = {E \over 2(1+\nu)} $$    [10]

 * $$ E = {9K \over 1+3K/G} $$    [11]

 * $$ \nu = {1-2G/3K \over 2+2G/3K} $$    [12]

 * $$ G = {3(1-2\nu)K \over 2(1+\nu)} $$    [13]

 * $$ K = {E \over 9-3E/G} $$    [14]

The stress-strain relationships can be expressed in compact tensor notation:

 * $$\varepsilon_{ij} = {1 + \nu \over E} \sigma_{ij} - {\nu \over E} \sigma_{kk} \delta_{ij}$$    [15]

Looking specifically at a normal strain, $$\varepsilon_{22}$$ for example:

 * $$\varepsilon_{22} = {1 + \nu \over E} \sigma_{22}

- {\nu \over E} (\sigma_{11} + \sigma_{22} + \sigma_{33}) \delta_{22}$$

Similarly, for a shear strain, $$\varepsilon_{12}$$, gets us:

 * $$\varepsilon_{12} = {1 + \nu \over E} \sigma_{12}

- {\nu \over E} (\sigma_{11} + \sigma_{22} + \sigma_{33}) \delta_{12}$$

From $G = {E \over 2(1+\nu)}$, we find that:

 * $$\begin{align}

{ 1 + \nu \over E} &= {1 \over 2G} \\ \varepsilon_{12} &= {1 \over 2G} \ \sigma_{12} \\ \sigma_{12} &= 2G \ \varepsilon_{12} \end{align}$$

For a given stress state:
 * $$\boldsymbol{\sigma} = \left (

\begin{matrix} \sigma_{11} & \sigma_{12} & \sigma_{13} \\ \sigma_{21} & \sigma_{22} & \sigma_{23} \\ \sigma_{31} & \sigma_{32} & \sigma_{33} \end{matrix} \right)$$

We can only write the strain tensor as:
 * $$\boldsymbol{\varepsilon} = \left (

\begin{matrix} {1 \over E} (\sigma_{11} - \nu(\sigma_{22} + \sigma_{33})) & {\sigma_{12} \over 2G} & {\sigma_{13} \over 2G} \\ {\sigma_{12} \over 2G} & {1 \over E}(\sigma_{22} - \nu (\sigma_{33} + \sigma_{11})) & {\sigma_{23} \over 2G} \\ {\sigma_{13} \over 2G} & {\sigma_{23} \over 2G} & {1 \over E} (\sigma_{33} - \nu (\sigma_{11} + \sigma_{22})) \end{matrix} \right)$$    [16]

So we have an expression for strain in terms of the applied load, the stress. This can certainly be useful for predicting the deformation to anticipate when loading in application. There is another application we are interested in. Given we deform a part, what stress does it feel? This answers questions such as, "how much can I bend this part before I reach the yield strength?" Basically, we want to take our existing solution and invert it.

To do this, we can take our first triaxial strain equation, Equation 3, and rearrange it:

 * $$\begin{align}

\varepsilon_{11} &= {\sigma_{11} \over E} - {\nu \sigma_{22} \over E} - {\nu \sigma_{33} \over E}       \\ & \quad +0 = {\nu \sigma_{11} \over E} - {\nu \sigma_{11} \over E}    \\ &= {\sigma_{11} \over E} + {\nu \sigma_{11} \over E} - {\nu \sigma_{11} \over E}        - {\nu \sigma_{22} \over E} - {\nu \sigma_{33} \over E} \end{align}$$

And we arrive at:

 * $$\begin{align}

\varepsilon_{11} &= {1+\nu \over E} \sigma_{11} - {\nu \over E} (\sigma_{11} + \sigma_{22} + \sigma_{33}) \end{align}$$    [17]

We can additionally take our earlier equation for $$\Delta$$, Equation 8, and rearrange it to get:

 * $$\begin{align}

\Delta &= {1 - 2\nu \over E} 3\sigma_m \\ \varepsilon_{11} + \varepsilon_{22} + \varepsilon_{33} &= {1-2\nu \over E}(\sigma_{11} + \sigma_{22} + \sigma_{33}) \end{align}$$

And

 * $$ \sigma_{11} + \sigma_{22} + \sigma_{33} = {E \over 1-2\nu}(\varepsilon_{11} + \varepsilon_{22} + \varepsilon_{33})

$$    [18]

Then by substituting Equation 18 into Equation 17, we get:

 * $$\begin{align}

\varepsilon_{11} &= {1+\nu \over E} \sigma_{11} - {\nu \over E} \left (    {E \over 1+2\nu}(\varepsilon_{11} + \varepsilon_{22} + \varepsilon_{33}) \right ) \\ \sigma_{11} &= \varepsilon_{11} \left ( {E \over 1+\nu} \right ) + {E \over 1+\nu} {\nu \over E} {E \over 1-2\nu} (\varepsilon_{11} + \varepsilon_{22} + \varepsilon_{33}) \end{align}$$

Which becomes
 * $$\begin{align}

\sigma_{11} &= \left ( {E \over 1+\nu} \right ) \varepsilon_{11} + {\nu E \over (1+\nu) (1-2 \nu)} (\varepsilon_{11} + \varepsilon_{22} + \varepsilon_{33}) \end{align}$$    [19]

Thinking about shear stress is simpler:

 * $$\begin{align}

\sigma_{12} &= 2G \ \varepsilon_{12} \end{align}$$

Substituting in Equation 10 results in
 * $$\begin{align}

\sigma_{12} &= 2 \ {E \over 2(1+\nu)} \ \varepsilon_{12} \end{align}$$

Which becomes
 * $$\begin{align}

\sigma_{12} &= \left ( {E \over 1+\nu} \right ) \varepsilon_{12} \end{align}$$    [20]

Taking Equations 19 & 20 gives us the tensor expression for stress in terms of strain, which is:

 * $$\sigma_{ij} = {E \over 1+\nu} \varepsilon_{ij}

+ \underbrace{\nu E \over (1+\nu) (1-2\nu)}_{\lambda} \varepsilon_{kk} \delta_{ij}$$    [21]

Here, $$\lambda$$ is the Lamé Constant. We can analyze these results and extract useful equations. We will begin by extracting the deviatoric and hydrostatic contributions to stress.

 * $$\sigma_{ij}' = 2G \ \varepsilon_{ij}'$$    [22]

We can show this is true by considering shear and normal components.

Shear:
 * $$\sigma_{12}' = \sigma_{12} = {E \over (1+\nu)} \varepsilon_{12}

= 2G \ \varepsilon_{12} = 2G \ \varepsilon_{12}'$$

Normal:
 * $$\begin{align}

\sigma_{11}' &= {2 \sigma_{22} - \sigma_{22} - \sigma_{33} \over 3} \\ &= {1 \over 3} \left [ 2 \ (2G\varepsilon_{11}            + \lambda \varepsilon_{11} + \lambda \varepsilon_{22} + \lambda \varepsilon_{33}) - (2G\varepsilon_{22}            + \lambda \varepsilon_{11} + \lambda \varepsilon_{22} + \lambda \varepsilon_{33}) - (2G\varepsilon_{33}            + \lambda \varepsilon_{11} + \lambda \varepsilon_{22} + \lambda \varepsilon_{33}) \right ] \\ &= {1 \over 3} (4 G \varepsilon_{11} - 2G \varepsilon_{22} - 2G \varepsilon_{33}) \\ &= 2G {2 \varepsilon_{11} - \varepsilon_{22} - \varepsilon_{33} \over 3} \\ &= 2G \varepsilon_{11}' \end{align}$$

The relationship between the hydrostatic stress and the mean strain is:
 * $$\sigma_{ii} = {E \over 1-2\nu} \varepsilon_{kk} = 3K \varepsilon_{kk}$$    [23]

Note that we've seen this expression several times before!

Simplified Cases
Plane stress $$\sigma_{33} = 0$$.

Once again looking at our earlier triaxial strain equations, Equations 3 & 4, we will start by adding the first two together:

 * $${\begin{align}

\varepsilon_{11} &= {1 \over E} [\sigma_{11} - \nu \sigma_{22}] = {1 \over E} \sigma_{11} - {\nu \over E} \sigma_{22} \\ + \nu (\varepsilon_{22} &= {1 \over E} [\sigma_{22} - \nu \sigma_{11}]        = -{\nu \over E} \sigma_{11} + {1 \over E} \sigma_{22}) \end{align} \over \varepsilon_{11} + \nu \varepsilon_{22} = \left ( {1 \over E} - {\nu^2 \over E} \right ) \sigma_{11}} \begin{align} \\ \\ \rightarrow \sigma_{11} &= {E \over 1-\nu^2} (\varepsilon_{11} + \nu \varepsilon_{22}) \\ \sigma_{22} &= {E \over 1-\nu^2} (\varepsilon_{22} + \nu \varepsilon_{11}) \end{align}$$

Plane stress encountered either as loaded sheet, or more likely a pressure vessel.

Another simplification is Plane Strain, where $$\varepsilon_{33} = 0$$, typically when one dimension is much greater than the other two so $$\varepsilon_{11}$$ and $$\varepsilon_{22}$$ are much greater than $$\varepsilon_{33}$$, such as a long rod where strain along the length of the rod is constrained. Here, Equation 5 rearranges to:

 * $$\begin{align}

\varepsilon_{33} &= {1 \over E} \left [ \sigma_{33} - \nu (\sigma_{11} + \sigma_{22}) \right ] = 0 \\ \sigma_{33} &= \nu ( \sigma_{11} + \sigma_{22}) \end{align}$$

Note that $$\sigma_{33}$$ does not equal zero just because $$\varepsilon_{33}$$ equals zero. Substituting this $$\sigma_{33}$$ equation into Equations 3-5 yields us:

 * $$\begin{align}

\varepsilon_{11} &= {1 \over E} \left [ (1-\nu^2) \sigma_{22} - \nu(1+\nu) \sigma_{11} \right ] \\ \varepsilon_{22} &= {1 \over E} \left [ (1-\nu^2) \sigma_{22} - \nu (1+\nu) \sigma_{11} \right ] \end{align}$$


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Overview of Elasticity of Materials/Appendix A: Coordinate Notation

This is a discussion of the notational choices used in the book including the coordinate choices.