OpenLLM-France/wikimedia huggingface dataset:
# Chip's Challenge

This is a walkthrough for the 1989 computer game Chip's Challenge, written for the by Chuck Sommerville and others. It shows how to complete every level. Note that the Microsoft Entertainment Pack version (the most well-known) differs from the original in many respects. Some of these differences are bugs which allow shortcuts to be taken. When these differences occur, the walkthrough will (sometimes) explain both ways.
This walkthrough frequently uses directions such as "north" and "west". The direction north means up. A notation such as N12 means to move north twelve squares. An intersection is where you have a choice of directions to take.
Chip's Challenge:
General hints | Levels 1-20 | Levels 21-40 | Levels 41-60 | Levels 61-80 | Levels 81-100 | Levels 101-120 | Levels 121-140 | Levels 141-149 | Passwords
Level Pack 2:
Levels 1-20 | Levels 21-40 | Levels 41-60 | Levels 61-80 | Levels 81-100 | Levels 101-120 | Levels 121-140 | Levels 141-149 | Passwords
# Chinese Tractor Maintenance/KAMA 454 554
## Operation

(Starting, Break-in, Normal Operation, Operating Limits)

## Fluid Capacities

(Engine oil, coolant, differential, power steering, hydraulic system) Engine 495 T2 - 12 quarts of 15W-40 Diesel rated oil Rear Axle and transmission - 17 quarts of 75W-90 gear oil - Royal Purple synthetic makes shifting much better. Front axle - 9.5 quarts of 75W-90W Fuel injection - 7 ounces of motor oil 15W-40W

## Maintenance Schedule

(pre-startup,50,100,...hour maintenance

## 295T Engine

(May want to link to a page specific to the engine model so that information is not repeated).

### Block
#### Lubrication System
#### Timing Chain
## Drive Train
### PTO
## Brake System
## Trouble Shooting

This section contains trouble shooting information that is specific to the these models of tractors. A good place to start for general trouble shooting is the Chinese Tractor Maintenance/Trouble Shooting page.

### Cooling System
## Owner Modifications/Enhancements

There are many modifications described in Chinese Tractor Maintenance#General Modifications that apply to these KAMA models.
# Chinese Tractor Maintenance/Modified Cooling System

This page contains information on the modifications that Chinese tractor owners have made to their tractors to improve the cooling system.
This page is part of the Chinese Tractor Maintenance book. Go to the Chinese_Tractor_Maintenance book.

## Radiator Screen

On some Chinese tractors the radiator screen that is supposed to protect the radiator fins has holes that are too large, allowing small debri to fill the radiator fins.
* Several tractor owners have replaced their screens with inexpensive cardboard framed furnace filters. Buy a filter that is larger than the original screen and cut it to size to fit inside the metal screen rails. These filters are easier to replace then try to clean when they get dirty. They can be removed and reinstall 2 or 3 times before they begin to deteriorate.
* Another option is to buy a stiffer and hardier furnace filter that is biodegradable. It is moister resistance and be clean and resinstalled more than 3 times before deterioration.
* You can keep your original screen a wrap it with aluminum window screen that can be obtained at hardware store. This has the advantage of keepin the original screen to protect the radiator from being damaged by objects, etc.
Regardless of the method used to reduce the size of debri entering the raditor fins, it is still necessory from time to time to remove dust from the radiator fins.

## Thermostat

The thermostat need not be adjusted but just going to the supplier for a new one.
# Drag force

The actual force of air resistance is quite complicated. Experiment by moving a book through the air with the face of the book and then the side of the book forward, you will agree that the area of the book makes a difference as to how much you must work in order to move the book at the same speed in both cases. This is why racing cars are slim-lined in design, and not shaped like a big box!
Get a plastic container lid (or anything waterproof) swing it around in air and then try to swing it around under water. The density of the water is much larger than the air, making you have to work harder at swinging the lid in water. This is why boats and submarines are a lot slower than aeroplanes!
So we know that density, area and speed all play a role in the drag force.The expression we use for drag force is
* ${\begin{matrix}D={\frac {1}{2}}C\rho Av^{2}\end{matrix}}$
* (9.25)
where C is a constant which depends on the object and fluid interactions, $\rho$ is the density, A is the area and v is the velocity.
# Mass and Weight

Weight is a force which is measured in Newtons, it is the force on an object due to gravity. People are always asking other people "What is your weight?" when a physicist might ask the same question as "What is your mass?".
Mass is measured in kilograms (kg) and is the amount of matter in an object, it doesn't change unless you add or remove matter from the object (if you continue to study physics through to university level, you will find that Einstein's theory of relativity means that mass can change when you travel as fast as light does, but you don't need to worry about that right now). There are 1000 g in 1 kg and 1000 kg in a tonne.
To change mass into weight we use Newton's 2nd Law which is $F=ma$. The weight is the force and gravity the acceleration, it can be rewritten as:
| ${\begin{matrix}W=mg\end{matrix}}$ | (9.2) |
W is the weight, measured in newtons. m is the mass, measured in kilograms and g is the acceleration due to gravity, measured in metres per second squared; it is equal to about 10 on the Earth. The exact value of Earths equatorial surface gravity is 9.7801 $m.s^{-2}$.

## Examples

1. A bag of sugar has a mass of 1 kg, what is its weight? (Acceleration due to gravity = 10 $m.s^{-2}$)
Step 1: Always write out the equation, it helps you to understand the question, and you will get marks as well.
* ${\begin{matrix}W=mg\end{matrix}}$
* (9.7)
Step 2: Fill in all the values you know. (remember to make sure the mass is in kilograms and NOT in grams or tonnes!)
* ${\begin{matrix}W=1\times 10\end{matrix}}$
* $W=10$
* (9.8)
Step 3: Write out the answer remembering to include the units! You will lose marks if you don't
* ${\begin{matrix}W=10\ newtons\end{matrix}}$
* (9.9)
2. A space-man has a mass of 90 kg, what is his weight (a) on the earth? (b) on the moon? (c) in outer space? (The acceleration due to gravity on the earth is 10 $m.s^{-2}$, on the moon gravity is about 1/6 of the gravity on earth).
1:
* ${\begin{matrix}W=mg\end{matrix}}$
* ${\begin{matrix}W=90\times 10=900\end{matrix}}$
* $W=900\ newtons$
* (9.10)
2:
* $W=M\ g$
* ${\begin{matrix}W=90\times \ 10\times {\frac {1}{6}}=150\end{matrix}}$
* $W=150\ newtons$
* (9.10)
3: Weightless in outer space because g = 0.
So now when somebody asks you your weight, you know to reply ``Anything!! But my mass is a different matter!"
# Properties

Gravitational attraction is a force and therefore must be described by a vector - so remember magnitude and direction. The force due to gravity acts between any two objects with mass. To determine the magnitude of the force we use the following equation:
| ${\begin{matrix}F={\frac {Gm_{1}m_{2}}{r^{2}}}\end{matrix}}$ | (9.1) |
This equation describes the force between two bodies, one of mass m$_1$, the other of mass m$_2$ (both have units of kilograms, or kg for short). The G is Newton's`Gravitational Constant' (6.673×10$^{−11}$ N·m$^2$·kg$^{−2}$) and r is the straight line distance between the two bodies in meters.
This means the bigger the masses, the greater the force between them. Simply put, big things matter big with gravity. The 1/r$^2$ factor (or you may prefer to say r$^{−2}$) tells us that the distance between the two bodies plays a role as well. The closer two bodies are, the stronger the gravitational force between them is. We feel the gravitational attraction of the Earth most at the surface since that is the closest we can get to it, but if we were in outer-space, we would barely even know the Earth's gravity existed!
Remember that
| ${\begin{matrix}F=ma\end{matrix}}$ | (9.2) |
which means that every object on the earth feels the same gravitational acceleration! That means whether you drop a pen or a book (from the same height), they will both take the same length of time to hit the ground... in fact they will be head to head for the entire fall if you drop them at the same time. We can show this easily by using the two equations above (9.1 and 9.2). The force between the Earth (which has the mass m$_e$) and an object of mass m$_o$ is
| ${\begin{matrix}F={\frac {Gm_{o}m_{e}}{r^{2}}}\end{matrix}}$ | (9.3) |
and the acceleration of an object of mass m$_o$ (in terms of the force acting on it) is
| ${\begin{matrix}a_{o}={\frac {F}{m_{o}}}\end{matrix}}$ | (9.4) |
So we substitute equation (9.3) into equation (9.4), and we find that
| ${\begin{matrix}a_{o}={\frac {Gm_{e}}{r^{2}}}\end{matrix}}$ | (9.5) |
Since it doesn't matter what m$_o$ is, this tells us that the acceleration on a body (due to the Earth's gravity) does not depend on the mass of the body. Thus all objects feel the same gravitational acceleration. The force on different bodies will be different but the acceleration will be the same. Due to the fact that this acceleration caused by gravity is the same on all objects we label it differently, instead of using a we use g which we call the gravitational acceleration.
# Newtonian Gravitation

All objects on Earth are pulled downward, towards the ground. This phenomenon is called gravity. Every object falls just as fast as any other object (unless the air slows it down like a feather, or pushes it up like a balloon), as first shown by Galileo. In 1687 Isaac Newton stated that gravity is not restricted to the Earth, but instead, there is gravity everywhere in the universe. Newton explained that planets, moons, and comets move in orbits because of the effect of gravity.
# Falling bodies

Objects on the earth fall because there is a gravitation force between them and the earth - which results in an acceleration - as we saw above. So if you hold something in front of you and let it go - it will fall.
It falls because of an acceleration toward the centre of the earth which results from the gravitational force between the two.
These bodies move in a straight line from the point where they start to the centre of the earth. This means we can reuse everything we learnt in rectilinear motion. the only thing that needs thinking about is the directions we are talking about.
We need to choose either up or down as positive just like we had to choose a positive direction in standard rectilinear motion problems. this is the hardest part. If you can do rectilinear motion you can do falling body problems. Just remember the acceleration they feel is constant and because of gravity - but once you have chosen your directions you can forget that gravity has anything to do with the problem - all you have is a rectilinear motion problem with a constant acceleration!!
What is the weight of a free falling body ? Would you be surprised to know that it is zero ?
Weight is measured using a spring balance. (Mass is measured using a physical balance). When a body is falling, the spring balance to which it is attached (the spring balance needs to fall along with the body too !) would show zero weight. When the body stops falling, then it would exert a force on the spring (equivalent to the force exerted by earth on itself) and hence it would show a reading.
# Comparative problems

Here always work with multiplicative factors to find something new in terms of something old.

## Worked Example 57 Comparative Problem 1

Question: On Earth a man weighs 700 N. Now if the same man was instantaneously beamed to the planet Zirgon, which has the same size as the Earth but twice the mass, what would he weigh?
Answer:
Step 1 :
We start with the situation on earth:
* ${\begin{matrix}W=mg=G{\frac {m_{E}m}{r^{2}}}\end{matrix}}$
* (9.13)
Step 2 :
Now we consider the provje
* ${\begin{matrix}W_{Z}=mg_{Z}=G{\frac {m_{Z}m}{r_{Z}^{2}}}\end{matrix}}$
* (9.14)
but we know that $m_{Z}=2m_{E}$ and we know that r$_Z$ = r so we could write the equation again and substitute these relationships in:
Step 3 :
* ${\begin{matrix}W_{Z}=mg_{Z}=G{\frac {(m_{Z})m}{(r_{Z})^{2}}}\end{matrix}}$
* (9.15)
Step 4 :
* ${\begin{matrix}W_{Z}=mg_{Z}=G{\frac {(2m_{E})m}{(r)^{2}}}\end{matrix}}$
* (9.16)
* ${\begin{matrix}W_{Z}=2(G{\frac {(m_{E})m}{(r)^{2}}})\end{matrix}}$
* (9.17)
Step 5 :
* ${\begin{matrix}W_{Z}=2(W)\end{matrix}}$
* (9.18)
so on Zirgon he weighs 1400 N.

## Principles

* Write out first case
* Write out all relationships between variable from first and second case
* Write out second case
* Substitute all first case variables into second case
* Write second case in terms of first case
| Mercury | 0.376 |
| Venus | 0.903 |
| Earth | 1 |
| Mars | 0.38 |
| Jupiter | 2.34 |
| Saturn | 1.16 |
| Uranus | 1.15 |
| Neptune | 1.19 |
| Pluto | 0.066 |

## Worked Example 58 Comparative Problem 2

Question: On Earth a man weighs 70 kg. On the planet Beeble how much will he weigh if Beeble has mass half of that of the Earth and a radius one quarter that of the earth.
Answer:
Step 1 :
We start with the situation on earth:
* ${\begin{matrix}W=mg=G{\frac {m_{E}m}{r^{2}}}\end{matrix}}$
* (9.19)
Step 2 :
Now we consider the provje
* ${\begin{matrix}W_{B}=mg_{B}=G{\frac {m_{B}m}{r_{B}^{2}}}\end{matrix}}$
* (9.20)
but we know that $m_{Z}={\frac {1}{2}}m_{E}$ and we know that $r_{Z}={\frac {1}{4}}r$ so we could write the equation again and substitute these relationships in:
Step 3 :
* ${\begin{matrix}W_{Z}=mg_{Z}=G{\frac {(m_{Z})m}{(r_{Z})^{2}}}\end{matrix}}$
* (9.21)
Step 4 :
* ${\begin{matrix}W_{Z}=mg_{Z}=G{\frac {({\frac {1}{2}}m_{E})m}{({\frac {1}{4}}r)^{2}}}\end{matrix}}$
* (9.22)
* ${\begin{matrix}W_{Z}=8(G{\frac {(m_{E})m}{(r)^{2}}})\end{matrix}}$
* (9.23)
Step 5 :
${\begin{matrix}W_{Z}=8(W)\end{matrix}}$
But although the man exerts 8 times as much force due to gravity, he still weighs 70 kg on Beeble!
* Did you know that the largest telescope in the Southern Hemisphere is the South African Large Telescope (SALT) which came online in 2004 outside Sutherland in the Karoo.
# Important Equations and Quantities

Table 9.1: Units used in Newtonian Gravitation :
| Units ||||||
| Quantity | Symbol | Unit | SI base units | SI base units | Direction |
| mass | m | -- | kg or -- | kg or -- | -- |
| velocity | ${\overrightarrow {v}}$ | -- | ${\frac {\mbox{m}}{\mbox{s}}}$ or $\ \ {\mbox{m}}\cdot {\mbox{s}}^{-1}$ | ${\frac {\mbox{m}}{\mbox{s}}}$ or $\ \ {\mbox{m}}\cdot {\mbox{s}}^{-1}$ | yes |
| force | ${\overrightarrow {F}}$ | N | ${\frac {{\mbox{kg}}\cdot {\mbox{m}}}{{\mbox{s}}^{2}}}$ or $\ \ {\mbox{kg}}\cdot {\mbox{m}}\cdot {\mbox{s}}^{-2}$ | ${\frac {{\mbox{kg}}\cdot {\mbox{m}}}{{\mbox{s}}^{2}}}$ or $\ \ {\mbox{kg}}\cdot {\mbox{m}}\cdot {\mbox{s}}^{-2}$ | yes |
| energy | E | J | ${\frac {{\mbox{kg}}\cdot {\mbox{m}}^{2}}{{\mbox{s}}^{2}}}$ or $\ \ {\mbox{kg}}\cdot {\mbox{m}}^{2}\cdot {\mbox{s}}^{-2}$ | ${\frac {{\mbox{kg}}\cdot {\mbox{m}}^{2}}{{\mbox{s}}^{2}}}$ or $\ \ {\mbox{kg}}\cdot {\mbox{m}}^{2}\cdot {\mbox{s}}^{-2}$ | -- |
# Terminal Velocity

Physics is all about being simple - all we do is look at the world around us and notice how it really works. It is the one thing everyone is qualified to do - we spend most of our time when we are really young experimenting to find out how things work.
Take a book - wave it in the air - change the angle and direction. what happens of course there is resistance. different angles make it greater - the faster the book moves the greater it is. The bigger the area of the book moving in the direction of motion the greater the force.
So we know that air resistance exists! it is a force. So what happens when an object falls? of course there is air resistance - or drag as it is normally called. There is an approximate formula for the drag force as well.
The important thing to realise is that when the drag force and the gravitational force are equal for a falling body there is no net force acting on it - which means no net acceleration. That does not mean it does not move - but it means that its speed does not change.
It falls at a constant velocity! This velocity is called terminal velocity.
# FHSST Physics/Pressure

Essay 3: Pressure and Forces
Author: Asogan Moodaly
Asogan Moodaly received his Bachelor of Science degree (with honours) in Mechanical Engineering from the University of Natal, Durban in South Africa. For his final year design project he worked on a 3-axis filament winding machine for composite (Glass re-enforced plastic in this case) piping. He worked in Vereeniging, Gauteng at Mine Support Products (a subsidiary of Dorbyl Heavy Engineering) as the design engineer once he graduated. He currently lives in the Vaal Triangle area and is working for Sasol Technology Engineering as a mechanical engineer, ensuring the safety and integrity of equipment installed during projects.

# Pressure and Forces

In the mining industry, the roof (hangingwall) tends to drop as the face of the tunnel (stope) is excavated for rock containing gold.
As one can imagine, a roof falling on one's head is not a nice prospect! Therefore the roof needs to be supported.
The roof is not one big uniform chunk of rock. Rather it is broken up into smaller chunks. It is assumed that the biggest chunk of rock in the roof has a mass of less than 20 000 kg therefore each support has to be designed to resist a force related to that mass. The strength of the material (either wood or steel) making up the support is taken into account when working out the minimum required size and thickness of the parts to withstand the force of the roof.
Sometimes the design of the support is such that the support needs to withstand the rock mass without the force breaking the roof..
Therefore hydraulic supports (hydro = water) use the principles of force and pressure such that as a force is exerted on the support, the water pressure increases. A pressure relief valve then squirts out water when the pressure (and thus the force) gets too large. Imagine a very large, modified doctor's syringe.
In the petrochemical industry, there are many vessels and pipes that are under high pressures. A vessel is a containment unit (Imagine a pot without handles, that has the lid welded to the pot that would be a small vessel) where chemicals mix and react to form other chemicals, amongst other uses.
The end product chemicals are sold to companies that use these chemicals to make shampoo, dishwashing liquid, plastic containers, fertilizer, etc. Anyway, some of these chemical reactions require high temperatures and pressures in order to work. These pressures result in forces being applied to the insides of the vessels and pipes. Therefore the minimum thickness of the pipe and vessels walls must be determined using calculations, to withstand these forces. These calculations take into account the strength of the material (typically steel, plastic or composite), the diameter and of course the pressure inside the equipment.
# Diagnostic Radiology/Musculoskeletal Imaging/Tumors Basic/Periosteal Osteosarcoma

Periosteal osteosarcoma are more rare than conventional or parosteal osteosarcoma and arises from the surface of the bone, most commonly in diaphyses of femur and tibia. Age of presentation is typically 10–20 years of age. Scalloping of the adjacent cortex is common and there are often spicules of bone emanating from the lesion in a sunburst pattern. Also seen commonly is periosteal reaction. Intramedullary invasion and extension is uncommon. MRI signal is typically not helpful (low T1 and high T2.) Spicules of bone emanating in sunburst pattern tend to be low T2 in signal. Differential includes juxtacortical chondroma and high-grade surface osteosarcoma.

## References

* Osteosarcoma by Todd Kooy, M.D., University of Washington Department of Radiology
* Treatment of Osteosarcoma (cancer.gov)
* Wikipedia Listing for Osteosarcoma
* Murphey MD, Robbin MR, MacRae GA, Flemming DJ, Temple HT, Kransdorf MJ. The many faces of osteosarcoma. RadioGraphics 1997;17(5).
* Osteosarcoma (Emedicine)
# Diagnostic Radiology/Musculoskeletal Imaging/Tumors Basic/Telangiectatic Osteosarcoma

Telangiectatic osteosarcoma This is a rare variant that present as an entirely lytic bone lesion with a much less aggressive appearance. Typical age is 10–25 years of age and lesions commonly occur in the metadiaphysis of long bones, similar to conventional osteosarcoma. Lesions are more geographic but portions of the mass can have a wide zone of transition and cortical breakthrough with a soft tissue mass. Lesions are highly vascular, partly necrotic and often have large pools of blood within them with soft tissue peripherally and along septa. Leading to an MRI appearance with internal septa, fluid fluid levels, peripheral heterogeneity. Telangiectatic osteosarcoma commonly enhances. Differential includes aneurysmal bone cyst and giant cell tumor.

## References

* Osteosarcoma by Todd Kooy, M.D., University of Washington Department of Radiology
* Treatment of Osteosarcoma (cancer.gov)
* Wikipedia Listing for Osteosarcoma
* Murphey MD, Robbin MR, MacRae GA, Flemming DJ, Temple HT, Kransdorf MJ. The many faces of osteosarcoma. RadioGraphics 1997;17(5).
* Osteosarcoma (Emedicine)
# Diagnostic Radiology/Neuroradiological Anatomy

Neuroradiological anatomy is the study of the structures and components of the central nervous system (CNS), including the brain and spinal cord, using radiological imaging techniques. Radiological imaging plays a crucial role in diagnosing, monitoring, and researching neurological disorders and conditions. Here's an overview of key structures and imaging modalities used in neuroradiological anatomy:
Neuroradiological anatomy is crucial for diagnosing and understanding neurological disorders such as strokes, tumors, vascular malformations, neurodegenerative diseases, and traumatic brain injuries. Radiologists, neurologists, neurosurgeons, and other healthcare professionals use neuroradiological imaging to make accurate diagnoses and guide treatment decisions.

## Brain and Coverings

* Skull, sutures
* Major apertures
* Hemispheres, cortex, gyri, sulci
* Major fissures
* Major cisterns
* Basal ganglia, thalamus, hypothalamus, pituitary gland, pineal gland
* Pons, cerebellum, cerebellopontine angle
* Ventricles, choroid plexus
* Vessels and major branches
** External carotid artery
** Common carotid artery
** Internal carotid artery
** Circle of Willis
** Anterior cerebral artery
** Middle cerebral artery
** Vertebral, basilar artery
** Supratentorial veins, dural sinuses
** Infratentorial veins, dural sinuses

## Head and Neck

* Buccal space
* Sublingual space
* Submandibular space
* Parotid space
* Parapharyngeal space
* Poststyloid parapharyngeal (carotid) space
* Masticator space
* Retropharyngeal space
* Perivertebral space
* Posterior cervical space
* Pharyngeal mucosal space
** Oral cavity
** Oropharynx and nasopharynx
** Mandible and dental structures
* Visceral space
** Larynx and hypopharynx
** Thyroid
* Orbit
* Paranasal sinuses
* Skull base
** Anterior
** Central
** Posterior
* Temporal bone

## Spine

* Vertebral bodies
* Facet joints and transverse processes
* Lamina and spinous processes
* Support ligaments
* Specific characteristics of cervical, thoracic, and lumbar segments
* Cranio-vertebral and lumbo-sacral junctions
* Normal stability and motion