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# World Generation

This is a simple world made of chunks of 32³ voxels. The world itself is not static : as you can see on the top of the image, chunks only exist if there is at least one solid voxel in them. In other words, the world is dynamic and can contain as many chunks as the player's computer can handle.

In this particular screenshot, the world is generated with the simple vectorial gradient equation that I invented in high school but which I suppose already existed. Here's the basic equation :

$$\overrightarrow{ \textit{voxel value} } = \frac{ \overrightarrow{\textit{position} } \cdot \overrightarrow{ \textit{gradient}}}{ \overrightarrow{\textit{gradient} } \cdot \overrightarrow{ \textit{gradient}} }$$

That's the equation I came up with and remembered. The gradient * gradient can be simplified for the magnitude (length) of the gradient power squared.

$$\overrightarrow{ \textit{voxel value} } = \frac{ \overrightarrow{\textit{position} } \cdot \overrightarrow{ \textit{gradient}}}{ \left \| \overrightarrow{ \textit{gradient}} \right \| ^{2} }$$

In conclusion, this gives an N dimensional gradient which gives a single decimal number.

# Voxel Traversal Algorithm

As for the voxel traversal algorithm, I decided to go with the most popular one, which was made by John Amanatides and Andrew Woo. As much as I like research papers, I also despise them because they lack simplicity, examples and full source code. That's why I had to google implementations of it and later on remembered that I had actually already implemented this algorithm a few years ago.

## Summary

The simplest way to understand the algorithm is to imagine a line in an 3D world made of blocks. Which blocks does the line touch? Then, in which order are they touched based on the line's start and end positions? The goal is to traverse iteratively the blocks that are touched by the line .

More simply, the logic of the algorithm can be summed making a distinction between the ray's direction's components. Those three define the importance of their axes in terms of how many blocks need to be traversed in what direction. Think of this with integers : two people are running to reach a goal; the fastest runs a 5 km/h, while the slowest runs at 1 km/h. For each time step, i.e. an hour, how many kilometers have each runner traveled? The ratio is 5 : 1, so, to merge the analogy, a ray would traverse each step 5 blocks on the X axis and 1 block on the Y axis. Of course, it's more complicated than that, as there are more parameters to it, especially because of exceptions such as what to do when each component is equal with one another?

## Implementation

The first thing to know about my implementation is that each voxel index is centered within the voxel itself. In other words, the voxel at the position (0, 0, 0) starts at (-0.5, -0.5, -0.5) inclusively and ends at (0.5, 0.5, 0.5) exclusively. This is for a cube extent of 1, naturally. The original research paper doesn't work that way as it starts at the lowest corner, i.e. the voxel spans from (0, 0, 0) to (1, 1, 1). Without any further delay, here is the class for the VoxelRay

import com.cheesygames.colonysimulation.math.MathExt;
import com.cheesygames.colonysimulation.math.vector.Vector3i;
import com.cheesygames.colonysimulation.world.World;
import com.jme3.math.Vector3f;
import com.jme3.scene.plugins.blender.math.Vector3d;

import java.util.function.Function;

/**
* Ray for ray casting inside a voxel world. Each voxel is considered as a cube within this ray. A ray consists of a starting position, a direction and a length. The voxel distance
* is computed once the method {@link #rayCastLocal(double, Function, Vector3i)} or {@link #rayCast(double, Function)} is called.
*/
public class VoxelRay {

private Vector3d m_start;
private Vector3d m_offsettedStart;
private Vector3d m_direction;
private double m_length;
private int m_voxelDistance;
private boolean m_wasStopped;

/**
* Constructs an invalid {@link VoxelRay} as its direction and length are null. The setters must be called after constructing a {@link VoxelRay} with this constructors.
*/
public VoxelRay() {
this.m_start = new Vector3d();
this.m_offsettedStart = new Vector3d();
this.m_direction = new Vector3d();
this.m_length = 0;
}

/**
* Constructs a {@link VoxelRay} from two points : start and end.
*
* @param start The absolute starting position of the ray.
* @param end   The absolute ending position of the ray.
*/
public VoxelRay(Vector3d start, Vector3d end) {
this.m_start = new Vector3d(start);
this.m_offsettedStart = new Vector3d();
this.m_direction = end.subtract(start);
this.m_length = m_direction.length();
this.m_direction.normalizeLocal();
}

/**
* Constructs a {@link VoxelRay} from two points : start and end.
*
* @param start The absolute starting position of the ray.
* @param end   The absolute ending position of the ray.
*/
public VoxelRay(Vector3f start, Vector3f end) {
this.m_start = new Vector3d(start);
this.m_offsettedStart = new Vector3d();
this.m_direction = new Vector3d(end).subtractLocal(m_start);
this.m_length = m_direction.length();
this.m_direction.normalizeLocal();
}

/**
* Constructs a {@link VoxelRay} from a start, a direction and a length.
*
* @param start     The absolute starting position of the ray.
* @param direction The direction of the ray. Must be normalized.
* @param length    The length of the ray.
*/
public VoxelRay(Vector3d start, Vector3d direction, double length) {
this.m_start = new Vector3d(start);
this.m_offsettedStart = new Vector3d();
this.m_direction = new Vector3d(direction);
this.m_length = length;
}

/**
* Constructs a {@link VoxelRay} from a start, a direction and a length.
*
* @param start     The absolute starting position of the ray.
* @param direction The direction of the ray. Must be normalized.
* @param length    The length of the ray.
*/
public VoxelRay(Vector3f start, Vector3f direction, float length) {
this.m_start = new Vector3d(start);
this.m_offsettedStart = new Vector3d();
this.m_direction = new Vector3d(direction);
this.m_length = length;
}

/**
* Casts the ray from its starting position towards its direction whilst keeping in mind its length. A lambda parameter is supplied and called each time a voxel is traversed.
* This allows the lambda to stop anytime the algorithm to continue its loop.
*
* @param onTraversingVoxel The operation to execute when traversing a voxel. This method called the same number of times as the value of {@link #getVoxelDistance()}. The
*                          supplied {@link Vector3i} parameter is not a new instance but a local instance, so it is a reference. The return value {@link Boolean} defines if
*                          the algorithm should stop.
*
*/
public void rayCast(Function<Vector3i, Boolean> onTraversingVoxel) {
rayCastLocal(World.VOXEL_HALF_EXTENT, onTraversingVoxel, new Vector3i());
}

/**
* Casts the ray from its starting position towards its direction whilst keeping in mind its length. A lambda parameter is supplied and called each time a voxel is traversed.
* This allows the lambda to stop anytime the algorithm to continue its loop.
*
* @param voxelHalfExtent   The half extent (radius) of a voxel.
* @param onTraversingVoxel The operation to execute when traversing a voxel. This method called the same number of times as the value of {@link #getVoxelDistance()}. The
*                          supplied {@link Vector3i} parameter is not a new instance but a local instance, so it is a reference. The return value {@link Boolean} defines if
*                          the algorithm should stop.
*
*/
public void rayCast(double voxelHalfExtent, Function<Vector3i, Boolean> onTraversingVoxel) {
rayCastLocal(voxelHalfExtent, onTraversingVoxel, new Vector3i());
}

/**
* Casts the ray from its starting position towards its direction whilst keeping in mind its length. A lambda parameter is supplied and called each time a voxel is traversed.
* This allows the lambda to stop anytime the algorithm to continue its loop.
*

* This method is local because the parameter voxelIndex is locally changed to avoid creating a new instance of {@link Vector3i}.
*
* @param onTraversingVoxel The operation to execute when traversing a voxel. This method called the same number of times as the value of {@link #getVoxelDistance()}. The
*                          supplied {@link Vector3i} parameter is not a new instance but a local instance, so it is a reference. The return value {@link Boolean} defines if
*                          the algorithm should stop.
* @param voxelIndex        The voxel index to locally modify in order to traverse voxels. This parameter exists simply to avoid creating a new {@link Vector3i} instance.
*
*/
public void rayCastLocal(Function<Vector3i, Boolean> onTraversingVoxel, Vector3i voxelIndex) {
rayCastLocal(World.VOXEL_HALF_EXTENT, onTraversingVoxel, voxelIndex);
}

/**
* Casts the ray from its starting position towards its direction whilst keeping in mind its length. A lambda parameter is supplied and called each time a voxel is traversed.
* This allows the lambda to stop anytime the algorithm to continue its loop.
*

* This method is local because the parameter voxelIndex is locally changed to avoid creating a new instance of {@link Vector3i}.
*
* @param voxelHalfExtent   The half extent (radius) of a voxel.
* @param onTraversingVoxel The operation to execute when traversing a voxel. This method called the same number of times as the value of {@link #getVoxelDistance()}. The
*                          supplied {@link Vector3i} parameter is not a new instance but a local instance, so it is a reference. The return value {@link Boolean} defines if
*                          the algorithm should stop.
* @param voxelIndex        The voxel index to locally modify in order to traverse voxels. This parameter exists simply to avoid creating a new {@link Vector3i} instance.
*
*/
public void rayCastLocal(double voxelHalfExtent, Function<Vector3i, Boolean> onTraversingVoxel, Vector3i voxelIndex) {
assert !Double.isNaN(voxelHalfExtent);

assert !Double.isNaN(m_start.x);
assert !Double.isNaN(m_start.y);
assert !Double.isNaN(m_start.z);

assert !Double.isNaN(m_direction.x);
assert !Double.isNaN(m_direction.y);
assert !Double.isNaN(m_direction.z);

assert !Double.isNaN(m_length);

m_wasStopped = false;
final double voxelExtent = voxelHalfExtent * 2;

// This id of the first/current voxel hit by the ray.
VoxelWorldUtils.getVoxelIndexNoOffsetLocal(voxelExtent, m_offsettedStart, voxelIndex);

computeVoxelDistance(voxelExtent, voxelIndex);
assert !Double.isNaN(m_voxelDistance);

// In which direction the voxel ids are incremented.
int stepX = (int) MathExt.getSignZeroPositive(m_direction.x);
int stepY = (int) MathExt.getSignZeroPositive(m_direction.y);
int stepZ = (int) MathExt.getSignZeroPositive(m_direction.z);

// Distance along the ray to the next voxel border from the current position (tMaxX, tMaxY, tMaxZ).
double nextVoxelBoundaryX = (voxelIndex.x + (MathExt.getNegativeSign(stepX) + 1)) * voxelExtent;
double nextVoxelBoundaryY = (voxelIndex.y + (MathExt.getNegativeSign(stepY) + 1)) * voxelExtent;
double nextVoxelBoundaryZ = (voxelIndex.z + (MathExt.getNegativeSign(stepZ) + 1)) * voxelExtent;

// tMaxX, tMaxY, tMaxZ -- distance until next intersection with voxel-border
// the value of t at which the ray crosses the first vertical voxel boundary
double tMaxX = (m_direction.x != 0) ? (nextVoxelBoundaryX - m_offsettedStart.x) / m_direction.x : Double.MAX_VALUE;
double tMaxY = (m_direction.y != 0) ? (nextVoxelBoundaryY - m_offsettedStart.y) / m_direction.y : Double.MAX_VALUE;
double tMaxZ = (m_direction.z != 0) ? (nextVoxelBoundaryZ - m_offsettedStart.z) / m_direction.z : Double.MAX_VALUE;

// tDeltaX, tDeltaY, tDeltaZ --
// how far along the ray we must move for the horizontal component to equal the width of a voxel
// the direction in which we traverse the grid
// can only be FLT_MAX if we never go in that direction
double tDeltaX = (m_direction.x != 0) ? stepX * voxelExtent / m_direction.x : Double.MAX_VALUE;
double tDeltaY = (m_direction.y != 0) ? stepY * voxelExtent / m_direction.y : Double.MAX_VALUE;
double tDeltaZ = (m_direction.z != 0) ? stepZ * voxelExtent / m_direction.z : Double.MAX_VALUE;

if (onTraversingVoxel.apply(voxelIndex)) {
m_wasStopped = true;
return;
}

int traversedVoxelCount = 0;
while (++traversedVoxelCount < m_voxelDistance) {
if (tMaxX < tMaxY && tMaxX < tMaxZ) {
voxelIndex.x += stepX;
tMaxX += tDeltaX;
}
else if (tMaxY < tMaxZ) {
voxelIndex.y += stepY;
tMaxY += tDeltaY;
}
else {
voxelIndex.z += stepZ;
tMaxZ += tDeltaZ;
}

if (onTraversingVoxel.apply(voxelIndex)) {
m_wasStopped = true;
break;
}
}
}

/**
* Computes the voxel distance, a.k.a. the number of voxel to traverse, for the ray cast.
*
* @param voxelExtent The extent of a voxel, which is the equivalent for a cube of a sphere's radius.
* @param startIndex The starting position's index.
*/
private void computeVoxelDistance(double voxelExtent, Vector3i startIndex) {
m_voxelDistance = 1 +
MathExt.abs(VoxelWorldUtils.getVoxelIndexNoOffset(voxelExtent, m_offsettedStart.x + m_direction.x * m_length) - startIndex.x) +
MathExt.abs(VoxelWorldUtils.getVoxelIndexNoOffset(voxelExtent, m_offsettedStart.y + m_direction.y * m_length) - startIndex.y) +
MathExt.abs(VoxelWorldUtils.getVoxelIndexNoOffset(voxelExtent, m_offsettedStart.z + m_direction.z * m_length) - startIndex.z);
}

public Vector3d getStart() {
return m_start;
}

public Vector3d getDirection() {
return m_direction;
}

public double getLength() {
return m_length;
}

public int getVoxelDistance() {
return m_voxelDistance;
}

public void setStart(Vector3d start) {
m_start.set(start);
}

public void setStart(Vector3f start) {
m_start.set(start);
}

/**
* Sets the direction.
*
* @param direction The direction to set to the ray. Must be normalized.
*/
public void setDirection(Vector3d direction) {
m_direction.set(direction);
}

/**
* Sets the direction.
*
* @param direction The direction to set to the ray. Must be normalized.
*/
public void setDirection(Vector3f direction) {
m_direction.set(direction);
}

/**
* Sets the length of the ray.
*
* @param length The new length of the ray. Must be positive.
*/
public void setLength(double length) {
m_length = length;
}

/**
* Sets the end position of the ray, which is not a real variable but a way to set the direction and the length at the same time. The start position does matter for this
* method.
*
* @param end Where the ray ends.
*/
public void setEnd(Vector3d end) {
m_direction.set(end).subtractLocal(m_start);
m_length = m_direction.length();
m_direction.normalizeLocal();
}

/**
* Gets if the voxel ray cast was stopped by the "onTraversingVoxel" method call.
*
* @return True if the voxel ray cast was stopped by the "onTraversingVoxel" method call, false otherwise.
*/
public boolean wasStopped() {
return m_wasStopped;
}
}

Here are the external static methods :

/**
* Gets the voxel index of the specified position. This method suppose that the parameter position is already offsetted with + voxel half extent. This method local because the
* supplied voxel index will be locally modified and returned.
*
* @param voxelExtent The  extent of a voxel, which is the equivalent to a cube of a sphere's diameter.
* @param position    The position to get the voxel index from. Must already be offsetted with + voxel half extent
* @param voxelIndex  Where to store the voxel index.
*
* @return The voxel index parameter that is set to the supplied position's voxel index.
*/
public static Vector3i getVoxelIndexNoOffsetLocal(double voxelExtent, Vector3d position, Vector3i voxelIndex) {
return voxelIndex.set(getVoxelIndexNoOffset(voxelExtent, position.x), getVoxelIndexNoOffset(voxelExtent, position.y), getVoxelIndexNoOffset(voxelExtent, position.z));
}
/**
* Gets the sign of the supplied number. The method being "zero position" means that the sign of zero is 1.
*
* @param number The number to get the sign from.
*
* @return The number's sign.
*/
public static long getSignZeroPositive(double number) {
assert !Double.isNaN(number);
return getNegativeSign(number) | 1;
}
/**
* Gets the negative sign of the supplied number. So, in other words, if the number is negative, -1 is returned but if the number is positive or zero, then zero is returned. It
* does not check if the parameter is NaN.
*
* @param number The number to get its negative sign.
*
* @return -1 if the number is negative, 0 otherwise.
*/
public static long getNegativeSign(double number) {
assert !Double.isNaN(number);
return Double.doubleToRawLongBits(number) >> BIT_COUNT_EXCLUDING_SIGN_64;
}

The important parts to adjust the algorithm to fit my voxel boundaries are the following :

m_offsettedStart.set(m_start).addLocal(voxelHalfExtent, voxelHalfExtent, voxelHalfExtent);

It is mandatory to add the half extent to the starting position.

double nextVoxelBoundaryX = (voxelIndex.x + (MathExt.getNegativeSign(stepX) + 1)) * voxelExtent;
double nextVoxelBoundaryY = (voxelIndex.y + (MathExt.getNegativeSign(stepY) + 1)) * voxelExtent;
double nextVoxelBoundaryZ = (voxelIndex.z + (MathExt.getNegativeSign(stepZ) + 1)) * voxelExtent;

What the MathExt method call does could be programmed as : (stepX >= 0 ? 1 : 0).

I don't know how to express how it is delightful when everything starts to fit and work properly :')

Here are some screenshots

14 minutes ago, swiftcoder said:

They are not scalars but vectors; that's what the arrow means on top of a variable. So it's not scalar multiplications but dot products.

My friend just told me that I can use TeX instead of images for the equations : I'll update that at the same time that I'll upload the video.

Quote

// In which direction the voxel ids are incremented.
double stepX = MathExt.getSignZeroPositive(m_direction.x);

Shouldn't the steps be integers? Actually there is double to int conversion in the inner loop caused by this.

10 hours ago, JoeJ said:

Shouldn't the steps be integers? Actually there is double to int conversion in the inner loop caused by this.

You're right, thank you. I'll fix it.

Very impressive.  Nicely done.  If you don't mind saying, what language did you write this code in?  I can't follow or understand it.  I know English isn't your native language, but would you be willing to offer up a simplified explanation of the logic you used for this blog post?

6 hours ago, Awoken said:

If you don't mind saying, what language did you write this code in?

It's in Java but everything remains the same as other C languages, except the calls to my personal methods, which is kind of egoistic of me. I just haven't tried yet if they are more performant than Java's JDK math methods.

6 hours ago, Awoken said:

ould you be willing to offer up a simplified explanation of the logic you used for this blog post

For the gradient or the algorithm? Well, first of all, for the algorithm, you need to at least read the research paper to get a grasp of what was the intent and the result wanted by the authors, even if you don't understand everything.

Second of all, the simplest way to understand the algorithm is to imagine a line in an 3D world made of blocks. Which blocks does the line touch? Then, in which order are they touched based on the line's start and end positions? The goal is to traverse iteratively the blocks that are touched by the line .

Third of all, the logic of the algorithm can be summed making a distinction between the ray's direction's components. Those three define the importance of their axes in terms of how many blocks need to be traversed in what direction. Think of this with integers : two people are running to reach a goal; the fastest runs a 5 km/h, while the slowest runs at 1 km/h. For each time step, i.e. an hour, how many kilometers have each runner traveled? The ratio is 5 : 1, so, to merge the analogy, a ray would traverse each step 5 blocks on the X axis and 1 block on the Y axis. Of course, it's more complicated than that, as there are more parameters to it, especially because of exceptions such as what to do when each component is equal with one another?

I'll add this explanation to the article ^^

I'm still confused.  Which part of the code you posted actually determines which voxels are hit?

4 hours ago, Awoken said:

I'm still confused.  Which part of the code you posted actually determines which voxels are hit?

        if (onTraversingVoxel.apply(voxelIndex)) {
m_wasStopped = true;
return;
}

int traversedVoxelCount = 0;
while (++traversedVoxelCount < m_voxelDistance) {
if (tMaxX < tMaxY && tMaxX < tMaxZ) {
voxelIndex.x += stepX;
tMaxX += tDeltaX;
}
else if (tMaxY < tMaxZ) {
voxelIndex.y += stepY;
tMaxY += tDeltaY;
}
else {
voxelIndex.z += stepZ;
tMaxZ += tDeltaZ;
}

if (onTraversingVoxel.apply(voxelIndex)) {
m_wasStopped = true;
break;
}
}

Look more specifically at onTraversingVoxel.apply(voxelIndex) : this line of code applies a Java functional interface, which is a lambda. The supplied parameter is the absolute (world) index of the voxel touched.

Posted (edited)

Oh fantastic explanation.  I now understand the logic and it's remarkably simple, which is a good thing.  Great blog post, very educational.

Edited by Awoken

It's interesting that neither this thread nor the linked paper mentions it: I'm convinced this is just a 3D bresenham implementation. (credit where due)
Back in early 2010, I played around with implementing bresenham in a fragment shader working on a volume texture, and it actually did pretty good.
It's sort of a nice way to avoid using polygons, if you're into that sort of thing, but it was a little too much branching for my GPU at the time...

3 hours ago, SuperVGA said:

It's interesting that neither this thread nor the linked paper mentions it: I'm convinced this is just a 3D bresenham implementation. (credit where due)
Back in early 2010, I played around with implementing bresenham in a fragment shader working on a volume texture, and it actually did pretty good.
It's sort of a nice way to avoid using polygons, if you're into that sort of thing, but it was a little too much branching for my GPU at the time...

I didn't know that algorithm. However, the result of the Bresenham's method is the first of two possible choices, which can traverse diagonally. For my algorithm, I need the second choice because I don't want the ray cast to select voxels diagonally, as it might select hidden voxels.

49 minutes ago, thecheeselover said:

I didn't know that algorithm. However, the result of the Bresenham's method is the first of two possible choices, which can traverse diagonally. For my algorithm, I need the second choice because I don't want the ray cast to select voxels diagonally, as it might select hidden voxels.

Ah, right. Yeah, I don't think it's something that requires a whole lot of thought either, so it's fair enough. Mostly I was just puzzled as to why the paper didn't reference it.

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We are pleased to announce the release of Matali Physics 4.4. The latest version introduces comprehensive support for Android 9.0 Pie, iOS 12.x and macOS Mojave (version 10.14.x). Latest version also introduces Matali Render 3.4 add-on with normal mapping and parallax mapping based on the distance from the observer as well as other improvements and fixes.
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JEE Mains Previous Years Questions with Solutions

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AIEEE 2012

MCQ (Single Correct Answer)
If a simple pendulum has significant amplitude (up to a factor of $1/e$ of original ) only in the period between $t = 0s\,\,to\,\,t = \tau \,s,$ then $\tau \,$ may be called the average life of the pendulum When the spherical bob of the pendulum suffers a retardation (due to viscous drag) proportional to its velocity with $b$ as the constant of proportionality, the average life time of the pendulum is (assuming damping is small) in seconds :
A
${{0.693} \over b}$
B
$b$
C
${1 \over b}$
D
${2 \over b}$

Explanation

The equation of motion for the pendulum, suffering retardation

$I\alpha = - mg\left( {\ell \sin \theta } \right) - mbv\left( \ell \right)$ where $I = m{\ell ^2}$

and $\alpha = {d^2}\theta /d{t^2}$

$\therefore$ ${{{d^2}\theta } \over {d{t^2}}} = - {g \over \ell }\tan \theta + {{bv} \over \ell }$

on solving we get $\theta = {\theta _0}\,{e^{{{bt} \over 2}\sin \left( {\omega t + \phi } \right)}}$

According to questions ${{{\theta _0}} \over e} = {\theta _0}{e^{{{ - b\tau } \over 2}}}$

$\therefore$ $\tau = {2 \over b}$
2

AIEEE 2011

MCQ (Single Correct Answer)
Two particles are executing simple harmonic motion of the same amplitude $A$ and frequency $\omega$ along the $x$-axis. Their mean position is separated by distance ${X_0}\left( {{X_0} > A} \right)$. If the maximum separation between them is $\left( {{X_0} + A} \right),$ the phase difference between their motion is:
A
${\pi \over 3}$
B
${\pi \over 4}$
C
${\pi \over 6}$
D
${\pi \over 2}$

Explanation

For ${X_0} + A$ to be the maximum separation $y$ one body is at the mean position, the other should be at the extreme.
3

AIEEE 2011

MCQ (Single Correct Answer)
A mass $M,$ attached to a horizontal spring, executes $S.H.M.$ with amplitude ${A_1}.$ When the mass $M$ passes through its mean position then a smaller mass $m$ is placed over it and both of them move together with amplitude ${A_2}.$ The ratio of $\left( {{{{A_1}} \over {{A_2}}}} \right)$ is :
A
${{M + m} \over M}$
B
${\left( {{M \over {M + m}}} \right)^{{1 \over 2}}}$
C
${\left( {{{M + m} \over M}} \right)^{{1 \over 2}}}$
D
${M \over {M + m}}$

Explanation

The net force becomes zero at the mean point. Therefore, linear momentum must be conserved.

$\therefore$ $M{v_1} = \left( {M + m} \right){v_2}$

$M{A_1}\sqrt {{k \over M}} = \left( {M + m} \right){A_2}\sqrt {{k \over {m + M}}}$

$\therefore$ $\left( {V = A\sqrt {{k \over M}} } \right)$

${A_1}\sqrt M = {A_2}\sqrt {M + m}$
$\therefore$ ${{{A_1}} \over {{A_2}}} = \sqrt {{{m + M} \over M}}$
4

AIEEE 2009

MCQ (Single Correct Answer)
If $x,$ $v$ and $a$ denote the displacement, the velocity and the acceleration of a particle executing simple harmonic motion of time period $T,$ then, which of the following does not change with time?
A
$aT/x$
B
$aT + 2\pi v$
C
$aT/v$
D
${a^2}{T^2} + 4{\pi ^2}{v^2}$

Explanation

For an $SHM,$ the acceleration $a = - {\omega ^2}x$ where ${\omega ^2}$ is a constant. Therefore ${a \over x}$ is a constant. The time period $T$ is also constant. Therefore ${{aT} \over x}$ is a constant.

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text:
# Use the following information for the Exercises below. [The following information applies to the questions displayed...

###### Question:

Use the following information for the Exercises below. [The following information applies to the questions displayed below.] During April, the first production department of a process manufacturing system completed its work on 300,000 units of a product and transferred them to the next department. Of these transferred units, 60,000 were in process in the production department at the beginning of April and 240,000 were started and completed in April. April's beginning inventory units were 60% complete with respect to materials and 40% complete with respect to conversion. At the end of April, 82,000 additional units were in process in the production department and were 80% complete with respect to materials and 30% complete with respect to conversion.

Exercise 16-9 Weighted average: Costs assigned to output and inventories LO C2

The production department had $850,368 of direct materials and$649,296 of conversion costs charged to it during April. Also, its April beginning inventory of $167,066 consists of$118,472 of direct materials cost and $48,594 of conversion costs. 1. Compute the direct materials cost per equivalent unit for April. (Round "Cost per EUP" to 2 decimal places.) 2. Compute the conversion cost per equivalent unit for April. (Round "Cost per EUP" to 2 decimal places.) 3. Using the weighted-average method, assign April’s costs to the department’s output—specifically, its units transferred to the next department and its ending work in process inventory. (Round "Cost per EUP" to 2 decimal places.) Exercise 16-9 Weighted average: Costs assigned to output and inventories LO C2 The production department had$850,368 of direct materials and $649,296 of conversion costs charged to it during April. Also, its April beginning inventory of$167,066 consists of $118,472 of direct materials cost and$48,594 of conversion costs.

1. Compute the direct materials cost per equivalent unit for April. (Round "Cost per EUP" to 2 decimal places.) 2. Compute the conversion cost per equivalent unit for April. (Round "Cost per EUP" to 2 decimal places.) 3. Using the weighted-average method, assign April’s costs to the department’s output—specifically, its units transferred to the next department and its ending work in process inventory. (Round "Cost per EUP" to 2 decimal places.)

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text:
# OpenGL Opengl and Framerate

## Recommended Posts

Opengl it tells it problematic reaches it. The drawing does it if later I turn the screen. Things are being drawn some late. The screen is her strength to other problematic soot and interested is high fps lerde the physics system is blather. How do I straighten not can these problems? Game: http://mosengine.inativa.com/dosya/crazyrace.zip note: sorry . i am speak turkish.This pity the translator is housing estate and wrote it: http://www.1-800-translate.com/machine_trans/free_result.asp

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I'm sorry, but I have no idea what your problem is... [sad]

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Perhaps a turkish speaking member could translate for him instead. That translator doesn't seem to be getting the message through properly

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Turkish:

1. Opengl de cok yuksek fpslerde ekranda rotasyon cok hizli ise ekranda bazi atlamalar yani ekranin ortasinda sonraki frame altinda onceki frame gozukuyor bunun nedeni nedir? Matrix ayalarimda bir problem mi var?

2. Frame rate ile ilgili gene cok yuksek fpslerde fizik sistemim cok hizli calisiyo bunu nasil normale ceviririm ?

oyunda bu:
http://mosengine.inativa.com/dosya/crazyrace.zip

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The perils of machine translation... sigh!

Lets hope someone comes up with a better translation.

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what the hell, i'll give it a try. you'll have to excuse me though, my Turkish is rusty:

Quote:
 Opengl it tells it problematic reaches it.

English: I've told you newbs, you have problems and OpenGL is the answer.

Quote:
 The drawing does it if later I turn the screen.

English: The graphics pwn and make my monitor spin during Doom 3.

Quote:
 Things are being drawn some late.

English: Though sometime in Turkey we must use Direct3D...

Quote:
 The screen is her strength to other problematic soot and interested is high fps lerde the physics system is blather.

English: ...OpenGL pwns the monitor and all the other problematic sh*t, interestinglty high FPS and awesome physics that blather(Turkish for kick ass).

Quote:
 How do I straighten not can these problems?

How do you fix problems you don't have?

Quote:
 San Dimas ruckas

English: OpenGL biatches!

## Create an account

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Before using void glBindImageTexture(    GLuint unit, GLuint texture, GLint level, GLboolean layered, GLint layer, GLenum access, GLenum format), does need to make sure that texture is completeness.
• By cebugdev
hi guys,
are there any books, link online or any other resources that discusses on how to build special effects such as magic, lightning, etc. in OpenGL? i mean, yeah most of them are using particles but im looking for resources specifically on how to manipulate the particles to look like an effect that can be use for games,. i did fire particle before, and I want to learn how to do the other 'magic' as well.
Like are there one book or link(cant find in google) that atleast featured how to make different particle effects in OpenGL (or DirectX)? If there is no one stop shop for it, maybe ill just look for some tips on how to make a particle engine that is flexible enough to enable me to design different effects/magic
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• By dud3
How do we rotate the camera around x axis 360 degrees, without having the strange effect as in my video below?
Mine behaves exactly the same way spherical coordinates would, I'm using euler angles.
Tried googling, but couldn't find a proper answer, guessing I don't know what exactly to google for, googled 'rotate 360 around x axis', got no proper answers.

References:
Code: https://pastebin.com/Hcshj3FQ
The video shows the difference between blender and my rotation:

• By Defend
I've had a Google around for this but haven't yet found some solid advice. There is a lot of "it depends", but I'm not sure on what.
My question is what's a good rule of thumb to follow when it comes to creating/using VBOs & VAOs? As in, when should I use multiple or when should I not? My understanding so far is that if I need a new VBO, then I need a new VAO. So when it comes to rendering multiple objects I can either:
* make lots of VAO/VBO pairs and flip through them to render different objects, or
* make one big VBO and jump around its memory to render different objects.
I also understand that if I need to render objects with different vertex attributes, then a new VAO is necessary in this case.
If that "it depends" really is quite variable, what's best for a beginner with OpenGL, assuming that better approaches can be learnt later with better understanding?

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text:
# Harmonic-analysis definition

The study of functions given by a Fourier series or analogous representations, such as periodic functions and functions on topological groups.
noun
The study of Fourier series.
noun
The act of breaking a periodic function into components, each expressed as a sine or cosine function.
noun
The study of functions given by a Fourier series or analogous representations, such as periodic functions and functions on topological groups.
(analysis) A study of the representation of functions or signals as the superposition of basic waves, involving the notions of harmonic functions, trigonometric series, Fourier series, Fourier transforms, almost periodic functions, and others.
noun


text:
by Rasalhague
 P: 867 I think Wikipedia explained it pretty well in the excerpt you included from Avogadro constant. Avogadro's number is just a number: 6.022×1023 Avogadro's constant has the reciprocal mole (mol^−1) unit attached to the number. I suppose in this case you can see more significance/purpose in Avogadro's number, or giving it a kind of definition, by attaching a unit to it, making it Avogadro's constant: 6.022×1023 of something/mol of something. Someone else could probably explain it with the kind of depth you've been studying it and give a better answer. "Mole" and "Avogadro's number" are interchangeable since each is just a number, actually the same number.
P: 1,402
 Quote by Bohrok I think Wikipedia explained it pretty well in the excerpt you included from Avogadro constant. Avogadro's number is just a number: 6.022×1023 Avogadro's constant has the reciprocal mole (mol^−1) unit attached to the number. I suppose in this case you can see more significance/purpose in Avogadro's number, or giving it a kind of definition, by attaching a unit to it, making it Avogadro's constant: 6.022×1023 of something/mol of something. Someone else could probably explain it with the kind of depth you've been studying it and give a better answer. "Mole" and "Avogadro's number" are interchangeable since each is just a number, actually the same number.
Thanks Bohrok! So was I right to conclude that the Avogadro constant = mol mol-1 = 1? Under what circumstances can 1 be regarded as not dimensionless; or to put it another way, how do we do dimensional analysis on equations involving moles? Do they even follow the usual rules of dimensional analysis? I found a couple more definitions in undergraduate physics textbooks.

"A mole of any substance is the amount of that substance that contains Avogadro's number NA of atoms or molecules defined as the number of carbon atoms in 12 g of carbon-12: NA = 6.022 * 1023. If we have n moles of a substance, then the number of molecules is N = nNA" (Tipler & Mosca: Physics for Scientists and Engineers, 538).

"One mole of a given gas is the amount of gas with an amount in grams equal to the atomic or molecular weight of the gas, e.g. 1 mol of helium gas has mass 4 g. One mole of gas always contains Avogadro's number of molecules: NA = 6.022 * 1023. The total number of molecules in a container, N, can be written in terms of the number of moles, n, as: N = nNA" (Fishbane et al.: Physics for Scientists and Engineers, 468-469).

In these definitions, it seems that NA is dimensionless, in contrast to Wikipedia's notation. In which case, are all of the other quantities in the relation N = nNA being treated as dimensionless by these authors, or should we regard N and n as both having the dimension "amount of substance"? I suspect they're all meant to be dimensionless though, given that the unit of "amount of substance" is the mole, whereas both N and n are numbers of individual molecules.

How would this relation be rewritten, or reinterpreted, using the conventions of the Wikipedia article Avogadro constant

where NA has dimension inverse moles?

The Wikipedia article Mole dismisses criticisms to the concept of the mole being a unit like the meter or the second, including the criticism that "the mole is simply a shorthand way of referring to a large number." Is the article then in disagreement with the definitions I quoted and with your statement that the mole and Avogadro's number are both simply a number? Would the Wikepedia writer, and/or the SI, disagree with those who say you can have "a mole of anything", including a mole of meters, etc.? It continues:

"The second misconception, that the mole is simply a counting aid, has even found its way into elementary chemistry textbooks. These books and others often contend that the mole is defined in terms of the Avogadro constant, rather than the other way around, [...]"

http://en.wikipedia.org/wiki/Mole_%28unit%29

The distinction here seems to be that the mole is defined as the number of atoms of carbon 12 in 12 grams of it (approximately 6.022... * 1023), whatever that number may turn out to be exactly, rather than being defined as exactly 6.022... * 1023. But I'm still struggling to see how uncertainty alone would make a mole qualitatively any less of a number than pi.

 HW Helper PF Gold P: 2,797 Mole = Avogadro's number? Basically it's just a number, much the way "dozen" is a number; the main difference is that the value is different. Dozen means 12; mole means 6.022*1023
P: 1,402
 Quote by symbolipoint Basically it's just a number, much the way "dozen" is a number; the main difference is that the value is different. Dozen means 12; mole means 6.022*1023
So you reckon the Wikipedia writer(s) is making a mountain out of a molehill here? Sorry, I couldn't resist... I'm trying to get my head around how a dimension ("amount of substance") can be defined in terms of a base unit that's just a number. On the one hand, we can't use "the dozen" or "the pi" as a unit of length, mass or time because dozen and pi are dimensionless. On the other, we can multiply the meter, kilogram or second by a dozen (i.e. twelve), or a score (20) or pi (3.141...) without changing their dimension. In these two ways, meters, kilograms and seconds seem different from dozens and scores and pi. So where does the mole belong? What qualities of a base unit does it have? How is mole defined as the measure of amount without that definition also allowing dozen and pi and 1 and 1000 to be dimensional quantities of dimension "amount". Is it correct to say a mole of meters = 6.022*1023 m? Or do we need to balance the dimensional books somehow?
 Other Sci Sci Advisor P: 1,378 You can convert a quantity in moles into a dimensionless quantity by multiplying by Avogadro's number. e.g. 0.5 mol * (6.02x1023 mol-1) = 3.01x1023 In this respect, Avogadro's number is simply a conversion factor to convert moles and dimensionless quantities.
P: 1,402
 Quote by Ygggdrasil You can convert a quantity in moles into a dimensionless quantity by multiplying by Avogadro's number. e.g. 0.5 mol * (6.02x1023 mol-1) = 3.01x1023 In this respect, Avogadro's number is simply a conversion factor to convert moles and dimensionless quantities.
Ah, thanks. So, if I've understood this right, your "Avogadro's number" is equal to Wikipedia's "the Avogadro constant" and Atkins & Jones's "the Avogadro constant", and to mol-1 * "Avogadro's number" as defined by Atkins & Jones, Tipler & Mosca and Fishbane et al.

Atkins & Jones use the symbol NA for what they call the Avogadro constant, as does Wikipedia, whereas Tipler & Mosca and Fishbane et al. use this symbol for what A&J, T&M and F call "Avogadro's number".

In T&M and F's equation N = n NA, are all three quantities dimensionless? That was my first guess. Alternatively, N and n might both have dimension "amount of substance". I don't know. But when A&J present an equation using the same symbols, it seems that NA has dimension inverse amount; N is the number of objects (dimensionless?), and n is the number of moles (dimension "amount of substance").
 P: 1,402 Hmm... So how exactly is "conversion factor" defined. Given that 1 day = 24 hours, would we say that the "conversion factor" for converting days into hours is 24, or would it be 24 day-1?
 Sci Advisor P: 1,866 Yes they're all dimensionless. Note that mole is not a dimension! It's a quantity - just like dozens. A "dozen" isn't a thing or a measure of a thing - it's an amount of a thing. So dozens, thousands, scores and moles are all quantities, whereas inches, pounds and degrees are units. Moles just happen to be a convenient number for dealing with chemistry, because it's the number of things, each having a chemical weight of 1 u, that you need to have 1 gram of that substance. You could also write N = n * ND Where ND is the 'dozen constant', i.e. 12. and n is then the number of dozens you have, and N is the ordinary number. Or, if you wanted to you could do a "unit analysis" (even though they're not properly called units) like: [things] = [dozens]*[things/dozen]
Other Sci
P: 1,378
 Quote by Rasalhague Hmm... So how exactly is "conversion factor" defined. Given that 1 day = 24 hours, would we say that the "conversion factor" for converting days into hours is 24, or would it be 24 day-1?
You have the right idea. Mathematically, if we multiply any number by 1, we still have the same number. Now, since 24 hours = 1 day, we can write 24hrs/day = 1. So, if we want to convert 0.5 days into hours, we can do the following calculation:

0.5 days = 0.5 days * 1 = 0.5 days * (24 hrs / 1 day) = 12 hrs

Similarly, 6.02x1023 (atoms) = 1 mol, so we can use the conversion factor 6.02x1023 (atoms)/mol = 1. (the atoms is in parenthesis because it's an example. It could also refer to the number of ions, molecules, birds, etc. Technically this is the unit, but usually a number of particles is just represented without units as a dimensionless quantity).

Note: I would not pay any attention to the difference between Avogadro's number and Avogadro's constant. Even if they have a different meaning, it is absolutely useless to our understanding of chemistry to know what that meaning is. As long as you get the general idea of what a mole means and why working with particles in units of 6.02x1023 is useful, you're all good.
 P: 233 I think this has been explained thoroughly, but I just wanted to let you guys know I like the "dozen" analogy. I can't believe I haven't heard that one before.
 P: 1,402 If amount of a substance is expressed as some quantity of moles of the substance, then it seems that amount of a substance is a ratio of dimensionless numbers (pure numbers)--the number of items (molecules, etc.) divided by the number of atoms of carbon-12 in 12g of carbon-12. This suggests that "amount of a substance" would also be dimensionless. But even if amount of a substance was defined as a dimension in the SI, wouldn't any quantity of moles still be dimensionless, since it would be a ratio between two quantities which have the same dimension as each other (like an angle measured in radians is the ratio between the length of the arc and the length of the radius)? Could mole, Avogadro's number or the Avogadro constant be added to the table of dimensionless quantities here, or would that be wrong? http://en.wikipedia.org/wiki/Pure_number
 P: 1,402 Thanks for all your replies. Yes, I like the dozens too: I hope that's all there is to it! Some of these definitions, and in particular the Wikipedia articles, give me the impression that there's still something I'm missing though... If the difference between a mole and Avogadro's number was the difference between a dozen and twelve, that's pretty trivial! What's so unique about the mole, as opposed to "the one", "the pi" or "the 3 times 10^8" or any of those dimensionless physical constants--they're all more-or-less handy units of amount? In SI units, if we multiply a number by a meter, the result has dimension length. If we multiply a mole and a meter, the result has dimension length, or does it? That would make moles dimensionless, but isn't it part of what it means to be a base unit that the unit possesses a unique dimension and defines that dimension. If the dimension of a thousand monkeys is amount, and the dimension of a thousand meters is amount times length, and any quantity can be measured in moles, does the definition of the mole as the base unit of amount of substance redefine all dimensionless quantities as quantities of dimension amount of substance? Is "amount of substance" like the identity element of dimensional analysis: amount * length = length? Hmm, but I guess not, or there wouldn't be a distinct concept of inverse amount...
P: 1,866
 Quote by Rasalhague What's so unique about the mole, as opposed to "the one", "the pi" or "the 3 times 10^8" or any of those dimensionless physical constants--they're all more-or-less handy units of amount?
Yup. Look at it from the chemical perspective: Once upon a time, they just didn't know how many atoms there were in any given amount of matter. However - given the theory that stuff was made out of atoms in various combinations, they could figure out how much these different elements weighed, relative each other. If you split water into hydrogen and oxygen gas, you got a weight-proportion of 2:32. (or 1:16)

So given your unit of atomic weight, obviously there had to be an Avogadro's number telling us how many atomic units there were in a gram. Determining that number is a lot more difficult though!

 In SI units, if we multiply a number by a meter, the result has dimension length. If we multiply a mole and a meter, the result has dimension length, or does it?
Yup.

 That would make moles dimensionless
Yup. No more dimensions than 'kilo' or 'milli'.

 but isn't it part of what it means to be a base unit that the unit possesses a unique dimension and defines that dimension.
Well, but moles doesn't. Avogadro's number is just a number. It's a 'unit' in the sense that SI includes it among the units. But it's not really a unit in the sense that it's a measure of anything in particular. It's just a quantity. A mole of meters is a distance, a mole of bicycles are still bicycles.
P: 1,402
Before I read alxm's last post, I made some guesses which I think must be partly mistaken... I wrote:

Is it perhaps that a dimensionless quantity (any dimensionless quantity?) can be given the dimension "amount of substance" by dividing the quantity by the number of atoms in 12g of carbon-12, called variously a mole or Avogadro's number, approximately 6.022 * 1023, and thereby expressing the quantity as a number of moles? And the quantity has dimension "amount of substance" when expressed in terms of some number of moles. But if expressed as a number of atoms, e.g. ten atoms or a dozen atoms, or in any units other than a mole, then the same quantity is considered dimensionless?

And to convert a quantity expressed in terms of moles (an SI "amount of substance"), into a quantity expressed in terms of the number of individual items that make it up (a pure number, a dimensionless quantity), the number of moles of the substance is multiplied by the conversion factor, Avogadro's constant: 1 mol / 1 mol = 1 = approx. 6.022 * 10^23 mol-1. (Or would it be more correct to say that the number 6.022... * 10^23 is the conversion factor?)

But from what you say, it seems more like the mole and any quantity measured in moles or inverse moles, has no dimension, in spite of the textbooks that list its dimension as "amount of a substance" and the strong wording of the Wikipedia articles which insist that "amount of substance" is a true "physical quantity" or "independent dimension of measurement", somehow qualitatively different from absolute number--althought I don't understand their arguments.

http://en.wikipedia.org/wiki/Mole_%28unit%29
http://en.wikipedia.org/wiki/Amount_of_substance

 Quote by alxm Avogadro's number is just a number. It's a 'unit' in the sense that SI includes it among the units. But it's not really a unit in the sense that it's a measure of anything in particular. It's just a quantity. A mole of meters is a distance, a mole of bicycles are still bicycles.
Do the objects that make up the substance have to all be of the same kind or defined in the same way; could we, say, have a mole of "cycles and cyclists", or a mole of "hydrogen atoms and helium atoms", or a mole of "hydrogen atoms and hydrogen molecules, H2"? What does the official definition mean when it says: "Note that this definition specifies at the same time the nature of the quantity whose unit is the mole"; does it perhaps mean that only quantities of a physical substance should be expressed in moles (atoms and bicycles, as opposed to meters and desires), or does it mean that the quantity must be made up of discrete items (material or immaterial, but not continuous)?

 P: 1 I think you are putting way too much into this (not necessarily a bad thing). When used in dimensional analysis it is simply a number that can be applied to something. Like if I wanted to convert 36 grams of water to molecules of water:  (1 mole H20) (6.022×10^23 molecH20) (36gH2O) × ------------- × ------------------------ = 1.2044 × 1024 molecH2O (18gH20) (1 mole H20) In this case it is simply a number. Almost always it is just a number. I have not come across a case yet where, in practice, Avogadro's Number has a dimension. You have the understanding of it. Now just do it.


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# Definition:Euler's Number/Limit of Sequence

## Definition

The sequence $\left \langle {x_n} \right \rangle$ defined as $x_n = \left({1 + \dfrac 1 n}\right)^n$ converges to a limit as $n$ increases without bound.

That limit is called Euler's Number and is denoted $e$.

### Decimal Expansion

The decimal expansion of Euler's number $e$ starts:

$2 \cdotp 71828 \, 18284 \, 59045 \, 23536 \, 02874 \, 71352 \, 66249 \, 77572 \, 47093 \, 69995 \ldots$

## Source of Name

This entry was named for Leonhard Paul Euler.


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# Viewpoint: Particle Decays Point to an Arrow of Time

• Michael Zeller, Department of Physics, Yale University, New Haven, CT 06520, USA
Physics 5, 129
An experiment studying B meson decays makes a direct observation of time-reversal violation without relying on assumed relationships with other fundamental symmetries.

Time moves irrevocably in one direction. Things get old, decay, and fall apart, but they rarely ever reassemble and grow young. But at the particle level, time’s arrow is not so clearly defined. Most collisions and other particle interactions look the same whether run forwards or backwards. Physicists have, however, identified a few reactions that appear to change when time is reversed, but the reasoning has assumed certain relations between fundamental symmetries of particle physics. The BaBar collaboration has now observed time-reversal violation directly and unambiguously in decays of B mesons. The measured asymmetry, reported in Physical Review Letters [1], is statistically significant and consistent with indirect observations.

In trying to understand the nature of particle interactions, observing the behavior of those interactions under different symmetry transformations has proven invaluable in formulating and verifying the fundamental theory. It is well known, and has been experimentally shown, that the strong and electromagnetic interactions are unchanged when viewed in a mirror world, in which particle positions are reflected ( r to r ). In contrast, experiments in 1956 [2] demonstrated that the weak interaction is not invariant under such parity inversion (P). A decade later, researchers found evidence in K meson decays [3] that weak interactions may also violate a combination of parity inversion with charge conjugation (C), where particles are interchanged for antiparticles. Physicists continue to study CP violation, in part to explore whether it can explain the dominance of matter over antimatter in the universe. But a related symmetry, time inversion (T), has been more elusive. It involves running an experiment backwards ( t to t) and converting initial states to final states. Weak interactions may violate time-reversal symmetry, but observing this directly with high enough precision proves difficult.

It is, however, possible to infer time-reversal violation (TRV) through observations of CP violation in K and B meson decays using the CPT theorem. The CPT theorem states that all local Lorentz invariant quantum field theories are invariant under the simultaneous operation of charge conjugation, parity reversal, and time reversal. Thus an experiment that measures a violation of CP might infer a corresponding lack of invariance under time reversal in order to maintain CPT invariance. Attempts to measure TRV without reference to CP violation have been controversial. Experiments at CPLEAR (measuring the difference in rates between K0 transitions to K¯¯¯0 and K¯¯¯0 transitions to K0 [4]) and at Fermilab (observing an asymmetry seen in the decay K0Lπ+πe+e [5]) claim to detect TRV directly, but some researchers have doubted this interpretation because of complicating factors in the K meson decays.

Using a method suggested previously [6], Lees et al. of the BaBar collaboration have now observed an explicit time-reversal violation in the decay of B mesons [1]. The experiment, performed at the Stanford Linear Accelerator Center (SLAC) in California, takes advantage of entangled B0 and B¯¯¯0 mesons in the Υ(4s) resonance produced in positron-electron ( e+e) collisions at SLAC. This allows measurement of an asymmetry that can only come about through a T inversion, and not by a CP transformation.

Each of the entangled B0 and B¯¯¯0 mesons resulting from the Υ(4s) can decay into either a CP eigenstate (e.g., J/ψKL, called B+ for CP even, and J/ψKs , called B for CP odd), or a state that identifies the flavor of the meson (e.g., lX for B¯¯¯0 and l+X for B0, where l± is a lepton and X represents the other particles in the decay). To study T inversion, the experimenters selected events where one meson decayed into a flavor state and the other decayed into a CP eigenstate. The time between these two decays was measured, and the rate of decay of the second with respect to the first was determined. As an example, let’s consider events in which the first meson decays at time t1 into l+X for B0. Due to entanglement, the surviving meson is tagged as “ B¯¯¯0” without being measured. At a later time t2 , this second meson decays into J/ψKs, implying a B (see Fig. 1). The researchers compared these “ B¯¯¯0 B events to the time-reversed pair: “ B B¯¯¯0, which are those events identified by both reversing the order of decays and exchanging initial and final states. In this example, the time-reversed events are the ones with a J/ψKL decay followed by a lX decay.

One of the keys to identifying the various B meson combinations is the energetic, asymmetric e+e collider of the PEP-II at SLAC. In the asymmetric collider, the beams are tuned to a center-of-mass energy of 10.58GeV, corresponding to the Υ(4s) mass. At that setting, 9GeV electrons collide with positrons at 3.1GeV, giving a boost to the resulting Υ(4s) and thus to its B and B¯¯¯ decay products. This extra momentum makes it possible for the BaBar detector to determine the species of each final state meson. The detector—built inside the flux return of a superconducting solenoid magnet—surrounds the e+e collision point with cylindrical trackers followed by cylindrical calorimeters [7]. Those detectors both track the outgoing trajectories and identify the particle species of the collision products, with reconstruction of that information yielding the identities of the B mesons. The flavor identity of neutral B mesons was made on the basis of the charges of prompt leptons, kaons, pions from D* mesons, and high-momentum charged particles, while the B states were reconstructed from several CP-odd states in addition to the J/ψKs mode. K0L mesons passed through the trackers undetected, so J/ψK0


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# Constant angular acceleration

## Homework Equations

? constant acceleration equations I guess

## The Attempt at a Solution

They didn't get to this part in class today, yet it's expected on our online homework due tonight. No idea how to even get started.

## Answers and Replies

Related Introductory Physics Homework Help News on Phys.org
rock.freak667
Homework Helper
For these questions you can consider conservation of energy.

PE + KE = constant.

For these questions you can consider conservation of energy.

PE + KE = constant.
jtw2e: Your post title makes my think you might not yet have been introduced to rotational kinetic energy. For an alternate approach: consider the relationship between how fast the block lands and how fast the disk rotates. Do you see that you can work out the torque acting on the disk and find the rotational speed?

jtw2e: Your post title makes my think you might not yet have been introduced to rotational kinetic energy. For an alternate approach: consider the relationship between how fast the block lands and how fast the disk rotates. Do you see that you can work out the torque acting on the disk and find the rotational speed?
Thank you. As of today we began rotational kinetic energy but did not get very much covered in the topic. We are supposed to begin torque on Friday.

For these questions you can consider conservation of energy.

PE + KE = constant.
While I would like to use CoE, we're supposed to find our answers with rotational constant acceleration approach. I've actually been doing my other homework with CoE just to get the answers turned in. I don't know how to find them with this rotational stuff.

jhae2.718
Gold Member
If you don't go with the energy approach, you'll want to find the angular acceleration of the pulley system.

You can use these equations:
\begin{align} \vec{\tau} &= \vec{r} \times \vec{F} \\ \vec{\tau}_{net} &= I\vec{\alpha}_{net} \end{align}
In these equations, $\vec{\tau}$ is torque, I is moment of inertia, $\vec{\alpha}$ is angular acceleration, and r is the distance vector from the axis of rotation to the force F.

These are the rotational analogues to force and Newton's second law. You can use Eq. (2) on the pulley system to find the net angular acceleration, which will give the system's acceleration. To find the linear acceleration of the block, $\vec{a} = r\vec{\alpha}$.


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# Decimals 1

Level pending

The smallest rational number with which $$\frac{9072}{56000}$$ should be multiplied so that its decimal expansion terminated after four decimal places is

×


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# A Boundary conditions on the Euclidean Schwarzschild black hole

1. Jun 15, 2017

### Afonso Campos

This question is based on page 71 of Thomas Hartman's notes on Quantum Gravity and Black Holes (http://www.hartmanhep.net/topics2015/gravity-lectures.pdf).

The Euclidean Schwarzschild black hole

$$ds^{2} = \left(1-\frac{2M}{r}\right)d\tau^{2} + \frac{dr^{2}}{1-\frac{2M}{r}} + r^{2}d\Omega_{2}^{2}$$

is obtained from the Lorentzian Schwarzschild black hole via Wick rotation $t \to -i\tau$.

Why does the fact that the coordinates must be regular at the origin imply that the angular coordinate must be identified as

$$\phi \sim \phi + 8\pi M?$$

Last edited: Jun 15, 2017
2. Jun 15, 2017

### Staff: Mentor

It isn't the angular coordinate $\phi$ that's identified, it's the "angular" coordinate $\tau$--the one that is derived via $t \rightarrow - i \tau$. The "origin" is $r = 2M$, and the solution has no interior, so the only range covered is $r \ge 2M$.

3. Jun 16, 2017

### Afonso Campos

How can we see explicitly that the range $r < 2M$ in the Lorentzian Schwarzschild black hole is not covered by the $\tau$ coordinate in the Euclidean Schwarzschild black hole?

4. Jun 16, 2017

### Staff: Mentor

The question doesn't make sense. To see whether the range $r < 2M$ is covered or not, you look at the behavior of the $r$ coordinate, not any other coordinate.


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# Hit and run. Think Bayes!

At the R in Insurance conference Arthur Charpentier gave a great keynote talk on Bayesian modelling in R. Bayes’ theorem on conditional probabilities is strikingly simple, yet incredibly thought provoking. Here is an example from Daniel Kahneman to test your intuition. But first I have to start with Bayes’ theorem.

### Bayes’ theorem

Bayes’ theorem states that given two events $D$ and $H$, the probability of $D$ and $H$ happening at the same time is the same as the probability of $D$ occurring, given $H$, weighted by the probability that $H$ occurs; or the other way round. As a formula it can be written as:

$$P(H \cap D) = P(H|D) \, P(D) = P(D|H) \, P(H)$$

Or if I rearrange it:

$$P(H|D) = \dfrac{P(D|H) \, P(H)}{P(D)}$$

Imagine $H$ is short for hypothesis and $D$ is short for data, or evidence. Then Bayes’ theorem states that the probability of a hypothesis given data is the same as the likelihood that we observe the data given the hypothesis, weighted by the prior belief of the hypothesis, normalised by the probability that we observe the data regardless of the hypothesis.

The tricky bit in real life is often to figure out what the hypothesis and data are.

### Hit and run accident

This example is taken from Daniel Kahneman’s book Thinking, fast and slow [1].

A cab was involved in a hit and run accident at night. Two cab companies, the Green and the Blue, operate in the city. 85% of the cabs in the city are Green and 15% are Blue. A witness identified the cab as Blue. The court tested the reliability of the witness under the same circumstances that existed on the night of the accident and concluded that the witness correctly identified each one of the two colours 80% of the time and failed 20% of the time.

What is the probability that the cab involved in the accident was Blue rather than Green knowing that this witness identified it as Blue?

What is here the data and what is here the hypothesis? Intuitively you may think that the proportion of Blue and Green cabs is the data at hand and the witness accusation that a Blue cab was involved in the accident is the hypothesis. However, after some thought I found the following assignment much more helpful, as then $P(H|D)$ matches the above question:

$H =$ Accident caused by Blue cab. $D =$ Witness said the cab was Blue.

With this it is straightforward to get the probabilities of $P(H)=15\%$ and $P(D|H)=80\%$. But what is $P(D)$? Well, when would the witness say that the cab was Blue? Either, when the cab was Blue and so the witness is right, or when the cab was actually Green and the witness is incorrect. Thus, following the law of total probability:

\begin{align} P(D) & = P(D|H) P(H) + P(D | \bar{H}) P(\bar{H})\\ & = 0.8 \cdot 0.15 + 0.2 \cdot 0.85 = 0.29 \end{align}

Therefore I get $P(H|D)=41\%$. Thus, even if the witness states that the cab involved in the accident was Blue, the probability of this being true is only $41\%$.

An alternative way to think about this problem is via a Bayesian Network. The colour of the cab will influence the statement of the witness. In R I can specify such a network using the gRain package [2], which I discussed in an earlier post. Here I provide the distribution of the cabs and the conditional probabilities of the witness as an input. After I compile the network, I can again read off the probabilities that a Blue cab was involved, when the witness said so.

### References

[1] Daniel Kahneman. (2011). Thinking, Fast and Slow. New York : Farrar, Straus and Giroux.

[2] Søren Højsgaard (2012). Graphical Independence Networks with the gRain Package for R. Journal of Statistical Software, 46(10), 1-26. URL http://www.jstatsoft.org/v46/i10/

### Session Info

R version 3.1.1 (2014-07-10)
Platform: x86_64-apple-darwin13.1.0 (64-bit)

locale:
[1] en_GB.UTF-8/en_GB.UTF-8/en_GB.UTF-8/C/en_GB.UTF-8/en_GB.UTF-8

attached base packages:
[1] grid stats graphics grDevices utils datasets
[7] methods base

other attached packages:
[1] Rgraphviz_2.8.1 gRain_1.2-3 gRbase_1.7-0.1 graph_1.42.0

loaded via a namespace (and not attached):
[1] BiocGenerics_0.10.0 igraph_0.7.1 lattice_0.20-29
[4] Matrix_1.1-4 parallel_3.1.1 RBGL_1.40.0
[7] Rcpp_0.11.2 stats4_3.1.1 tools_3.1.1

### Related

comments powered by Disqus


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# How To Weed out long paths easily ahead of a OneDrive for Business migration

You know exactly what’s happened as soon as you see it. That pesky error. One of your users has managed to create a directory structure so long courtesy of a sneaky mapped drive and now you’re unable to do anything with it; least of all get it synchronising with OneDrive for Business. If only there was a way to weed out those long paths in advance and fix them before people encounter the error?

Helpfully, there are some options to make this easier. I went scouring the web for something free, simple and easy to find all files where the path length would exceed the limit for OneDrive for Business (256 chars). A user by the handle “deadlydog” had posted some PowerShell script to Stack Overflow that looked like it might do the job.

In the spirit of building on the great ideas of others, I’ve taken deadlydog’s initial solution and extended it slightly to include a character limit. I figured someone running this script is more interested in the files longer than the limit rather than every file path length. I’ve also altered it to output to a CSV file since that might be more useful when it comes to manipulating the data in Excel or Power BI should you have large numbers of files that exceed the limit.

## Sample PowerShell Script

$pathToScan = "C:\Some\Directory" # The path to scan and the the lengths for (sub-directories will be scanned as well).$outputFilePath = "C:\temp\PathLengths.csv" # This must be a file in a directory that exists and does not require admin rights to write to.
$writeToConsoleAsWell =$false   # Writing to the console will be much slower.
$maxLength = 256 #Max path length. Script will record paths equal to, or longer than this. # Open a new file stream (nice and fast) and write all the paths and their lengths to it.$outputFileDirectory = Split-Path $outputFilePath -Parent if (!(Test-Path$outputFileDirectory)) { New-Item $outputFileDirectory -ItemType Directory }$stream = New-Object System.IO.StreamWriter($outputFilePath,$false)
Get-ChildItem -Path $pathToScan -Recurse -Force | Select-Object -Property FullName, @{Name="FullNameLength";Expression={($_.FullName.Length)}} | Sort-Object -Property FullNameLength -Descending | ForEach-Object {
$filePath =$_.FullName
$length =$_.FullNameLength
$string = "$length,$filePath" # Write to the Console. if ($writeToConsoleAsWell) { Write-Host $string } #Write to the file. if ($length -ge $maxLength) {$stream.WriteLine($string)} }$stream.Close()

By running this with the appropriate permissions you should generate a CSV with a line per file broken out by path length and full path. This will make it easier to identify potential problems when synchronising with OneDrive for Business or SharePoint Online.

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• 全部四级六级高考考研Why not instead think of learning as if it were food—something so valuable to humans that they have evolved to experience it as a pleasure?Joy should not be trained out of ...
全部四级六级高考考研Why not instead think of learning as if it were food—something so valuable to humans that they have evolved to experience it as a pleasure?Joy should not be trained out of children or left for after-school programs.出自-2015年12月阅读原文The author's mother taught him a valuable lesson by pointing out lots of flaws in his seemingly perfect essay.出自-2015年12月阅读原文A look at what goes on in most classrooms these days makes it abundantly clear that when people think about education, they are not thinking about what it feels like to be a child, or what makes childhood an important and valuable stage of life in its own right.出自-2015年12月阅读原文They're willing to do the hard work of identifying their valuable skills; be creative about where and how to look; learn how to present themselves to potential employers; and keep going, even after repeated rejections.出自-2012年12月阅读原文Plus, importantly, the power source is rechargeable, so this does not involve using any valuable resources.出自-2013年6月听力原文It's not the knowledge that is valuable, it's the practice of learning new things and understanding how things work.2019年6月四级真题(第三套)阅读 Section BOptions are more valuable the longer they can be held.2019年6月四级真题(第一套)阅读 Section BPeople need to realize that bees are valuable and they should be looked after.2017年6月四级真题(第二套)听力 Section ASo if you believe you will live longer, then options become more valuable, and early commitment becomes less attractive.2019年6月四级真题(第一套)阅读 Section BThat thing that you make is valuable to you and you can share it with others.2019年6月四级真题(第三套)阅读 Section BThe lessons prove valuable outside the classroom as well.2019年6月四级真题(第一套)听力 Section CUniversities posted research data on the Internet, so students could find valuable information without leaving their dormitories.2017年12月四级真题(第三套)阅读 Section AThey found the antiques there more valuable.出自-2017年6月阅读原文They argue that when a company like Monsanto goes to great expense to create a valuable new genetically modified seed, it must be able to protect its property interests.出自-2013年12月阅读原文People who do not have access to a good listener may not only be denied the opportunity to heighten their self-awareness, but they are also denied valuable feedback as to the validity and acceptability of their inner thoughts and feelings.出自-2012年12月阅读原文The author takes walks and reads and loses himself in the stillness of the hermitage so that he can bring his wife and bosses and friends around him anthing valuable出自-2012年12月阅读原文One problem is that time becomes more valuable and time becomes more worth money.出自-2014年6月听力原文At a time when natural history was a valuable tool for discovery, Merian discovered facts about plants and Insects that were not previously known.2018年12月六级真题(第二套)阅读 Section BEgo aside, a third advantage is that the emotional neutrality of indifferent relationships has been found to enhance critical evaluation, to strengthen one's focus on task resolution, and to gain greater access to valuable information.2019年6月六级真题(第三套)阅读 Section CFor example, when you own something, you feel it as more valuable than when you don't own it.2019年6月六级真题(第二套)听力 Section CIn her work, she revealed a side of nature so exotic, dramatic and valuable to Europeans of the time that she received much acclaim.2018年12月六级真题(第二套)阅读 Section BIn the other comer is the world's most valuable company, whose chief executive, timothy Cook, has said he will appeal the court's order.2017年12月六级真题(第三套)阅读 Section BNot surprisingly, much of Nokia's most valuable design and programming talent left as well.2015年12月六级真题(第三套)阅读 Section AThe knowledge she collected over decades didn't just satisfy those curious about nature, but also provided valuable insights into medicine and science.2018年12月六级真题(第二套)阅读 Section BAlthough skills and talents bring great success, the most valuable asset one can hold is the heart.2016年高考英语天津卷 完形填空 原文Being able to communicate ideas in a meaningful way is a valuable skill.2014年高考英语全国卷1 阅读理解 七选五 选项Depending on which role you choose, you could help to increase a community's access to safe drinking water, or help to protect valuable local cultures.2019年高考英语北京卷 阅读理解 阅读A 原文Electronic devices contain valuable metals such as gold and silver.2015年高考英语江苏卷 阅读理解 阅读B 原文In the context of this discussion, it seems valuable to first find out the cause of change in the case of parents' involvement with their grown children.2015年高考英语北京卷 阅读理解 阅读D 原文Many airlines now allow passengers to print their boarding passes online to save their valuable time.2017年高考英语北京卷 单项填空 原文Of course, you want to make sure that you become an accepted and valuable part of your new neighborhood.2018年高考英语浙江卷 阅读理解 七选五 原文The information they collected could become more valuable in future.2016年高考英语上海卷 阅读理解 阅读C 选项The Internet has made data abundant, all-present and far more valuable, changing the nature of data and competition.2017年高考英语江苏卷 阅读理解 阅读C 原文The most valuable firms are Google, Amazon, Facebook and Microsoft.2017年高考英语江苏卷 阅读理解 阅读C 原文The work on health is the most valuable experience.2016年高考英语四川卷 阅读理解 阅读B 选项The Yellowstone wolf project has been a valuable experiment to help biologists decide whether to reintroduce wolves to other parts of the country as well.2017年高考英语全国卷3 阅读理解 阅读C 原文They've learned life's most valuable lesson.2019年高考英语天津卷 阅读理解 阅读D 原文We have all heard how time is more valuable than money, but is it possible to have too much time?2018年高考英语浙江卷 完形填空 原文We went 10-1 and I was named most valuable player, but I often had crazy dreams in which I was to blame for Miller's accident.2016年高考英语全国卷3 完形填空 原文At puberty, however, the brain shuts down half of that capacity, preserving only those modes of thought that have seemed most valuable during the first decade or so of life.出自-2009年考研阅读原文Moreover, some have been found to have a valuable function in building up soil fertility.出自-2010年考研翻译原文Besides generating income, the presence of other marketers makes the site seem objective, gives companies opportunities to learn valuable information about the appeal of other companies’ marketing, and may help expand user traffic for all companies concerned.出自-2011年考研阅读原文"The valuable examples which they furnish of the power of self-help, of patient purpose, resolute working and steadfast integrity, issuing in the formation of truly noble and manly character, exhibit," wrote Smiles, " what it is in the power of each to ac2012年考研真题(英语二)阅读理解 Section ⅡAt the end of adolescence, however, the brain shuts down half of that capacity, preserving only those modes of thought that have seemed most valuable during the first decade or so of life.2009年考研真题(英语一)阅读理解 Section ⅡBesides generating income, the presence of other marketers makes the site seem objective, gives companies opportunities to learn valuable information about the appeal of other companies' marketing, and may help expand user traffic for all companies concer2011年考研真题(英语一)阅读理解 Section ⅡHelping forests flourish as valuable "carbon sinks" long into the future may require reducing their capacity to absorb carbon now.2019年考研真题(英语二)阅读理解 Section ⅡThese purchases often become more valuable with time — as stories or memories — particularly if they involve feeling more connected to others.2014年考研真题(英语二)阅读理解 Section Ⅱ
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• 使确立牢固的地位 (cause to) establish a secure position for oneselfdig inThey dug in and waited for the enemy's ...The soldiers were ordered to dig in.士兵们奉命挖掘壕沟。They were attacked before the...
使确立牢固的地位 (cause to) establish a secure position for oneselfdig inThey dug in and waited for the enemy's attack.他们挖好壕沟,等候敌人进攻。The soldiers were ordered to dig in.士兵们奉命挖掘壕沟。They were attacked before they had time to dig in.他们还来不及挖战壕就受到了攻击。The enemy scarcely had time to dig in before our assault was on.敌人还没有来得及挖好掩体,我们的进攻就开始了。dig oneself inOur soldiers have landed and are now digging themselves in.我们的战士已经登陆,现在正在挖战壕把自己隐蔽起来。As soon as they had seized the enemy position, the infantry dug themselves in.步兵一拿下敌军阵地,立即掘壕固守。The fighters steadily dug themselves in.战士们沉着地挖掘战壕。dig sth ⇔ inDig this chemical in and your crops will grow faster.把这种化学剂埋到地里庄稼就会长得更快。The chemical fertilizer should be well dug in.化学肥料应该均匀地混入土壤中。dig inJohn dug in and finished his homework quickly.约翰开始认真地学习并很快就把家庭作业做完了。Come on! Dig in before the food goes cold.来吧! 趁热快吃。The boys dug in ravenously.男孩们开始狼吞虎咽地吃起来。Don't stand on ceremony — dig in!别客气,开始吃吧!Mother set the food on the table and told the children to dig in.妈妈将食品摆在餐桌上告诉孩子开始吃。dig in to sthThey dug in to the meal.他们大吃了一顿。He sat down and at once dug in to his meal.他坐下来,立刻开始吃饭。He dug in to a pie.他津津有味地吃馅饼。dig inThe more the boss tries to get rid of her, the more she digs in.老板越是想除掉她,她就越是拼命保住自己的地位。dig oneself inIf anyone is to be dismissed when the company falls on hard times, it won't be Jack; he's dug himself in too well.如果公司碰到困难时刻要裁员的话,那是裁不到杰克头上来的。他在公司里的地位是稳固的。
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• 全部Methods Three different concentrations of the n - butanol extract of MOH ( BEM ) : 0.038 g · L ~ ( - 1 ), 0.11 g · L ~ ( - 1 ) and 0.34 g · L ~ ( - 1 ) were prepared.巴戟天醇提取物 ( BEM ) 配成...
全部Methods Three different concentrations of the n - butanol extract of MOH ( BEM ) : 0.038 g · L ~ ( - 1 ), 0.11 g · L ~ ( - 1 ) and 0.34 g · L ~ ( - 1 ) were prepared.巴戟天醇提取物 ( BEM ) 配成含生药0.038g·L~ ( -1)、0.11g·L~ ( -1 ) 、 0.34g·L~ ( -1 ) 三种浓度药液.互联网BEM triangulation mesh generation method was mainly investigated and further proposed generation algorithm through restriction points.重点研究了BEM中三角网格剖分方法,提出了过约束点的剖分算法.互联网Two - dimensional stress intensity factors are solved by BEM with crack closure integrals.应用边界元法结合裂纹闭合积分求解二 维 裂纹应力强度因子.互联网Interaction between foundation and superstructure has been analysised concretely with BEM theory.利用边界元法对地基与上部结构相互作用做了较为具体的理论分析.互联网Fast multipole BEM is a new algorithm developed for BEM these years.快速多极边界元法(fastmultipoleBEM)是近几年发展起来的边界元新型算法.互联网We compile the second - order element algorithm for BEM analysis calculation software.编制了三维静电场边界元法的二阶等参元有限元单元的计算软件.互联网Depois de iniciar o projecto , primeiramente faz bem o trabalho da prepara & ccedil ; & atilde ; o.工程开工后, 首先做好施工准备工作.互联网The boundary integral in BEM Boundary Element Method effects the precision and the speed method.边界元中的边界积分计算影响计算精度和计算速度.互联网Based on the original scheme, a new version fast multipole BEM is presented for two - dimensional elasticity.在初始格式的基础上, 进一步提出适用于二维弹性力学问题的新型快速多极边界元格式.互联网Eight - nodded curvilinear quadrilateral isoparametric shell element is used both in the FEM and in the BEM.在有限元方法和边界元法中均采用了八结点曲面四边形等参单元.互联网A finite element method with boundary element method ( FEM - BEM ) is presented for computing electromagnetic induction.提出了用有限元 - 边界元法计算三维电磁场感应问题.互联网The advantage of boundary element method ( BEM ) is that one dimension is reduced for solving problems.边界 元法在势流计算中具有使问题降低一维的优点.互联网And the applications of solving the two mathematical models with the domain discretization BEM are studied.研究了区域剖分边界元方法在求解这两类数学模型时的应用.互联网The difficulty of treatment of the nearly singular integrals hindered applications of Boundary Element Method ( BEM ).几乎奇异积分的计算困难阻碍了 边界 元法的工程应用.互联网The study on BEM has been a challenging topic for the scientists, the engineers and technicians.故对边界元法的研究一直很受学者和工程人员的重视.互联网
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• 全部香槟酒Her father picked up the tab for all the champagne...《简明英汉词典》There were two glasses of champagne on the tray.托盘里有两杯香槟酒.《简明英汉词典》Other sparkling wines are often consid...
全部香槟酒Her father picked up the tab for all the champagne at the wedding.婚礼上所有的香槟酒都是他父亲买单.《简明英汉词典》There were two glasses of champagne on the tray.托盘里有两杯香槟酒.《简明英汉词典》Other sparkling wines are often considered the poor relations of champagne.其他起泡的葡萄酒通常被认为较香槟酒为次.《简明英汉词典》Try some of this sparkling white wine — the poor man's champagne.尝尝这种白葡萄汽酒吧,这是穷人的香槟酒.《简明英汉词典》They sat there swilling champagne.他们坐在那里大喝香槟酒.《简明英汉词典》It was a party flowing with the best champagne.那是一次人们可痛饮上等香槟酒的宴会.《简明英汉词典》I'm no connoisseur but I know a good champagne when I taste one.我不是行家,但我一尝就知道是不是好香槟.《简明英汉词典》We celebrated by getting loaded on champagne.我们喝香槟酒庆祝.《现代英汉综合大词典》Jon's arrival was a cue for more champagne.乔恩一来就意味着要喝更多的香槟酒了。《牛津高阶英汉双解词典》I'm being very naughty—I've ordered champagne!我今天放肆一回—我要了香槟了！《牛津高阶英汉双解词典》She has a penchant for champagne.她酷爱香槟酒。《牛津高阶英汉双解词典》The show bubbles like pink champagne with pretty sets, exquisite costumes and enchanting dance routines.这个节目满眼都是包装漂亮的粉红香槟、华美的服装和迷人的舞蹈。柯林斯例句Robin opened some champagne.罗宾开了一些香槟。柯林斯例句The champagne sloshed and spilt.香槟酒晃洒了。柯林斯例句The champagne was sweet and too gassy.这种香槟味道很甜，但气太多了。柯林斯例句He untwisted the wire off the champagne bottle, and the cork popped and shot to the ceiling.他拧掉香槟瓶上的铁丝封口，瓶塞砰的一声直冲天花板。柯林斯例句They toasted her in champagne.他们以香槟为她干杯。柯林斯例句The owner knew about the engagement and brought them glasses of champagne on the house.店主知道了订婚一事，于是免费请他们喝香槟。柯林斯例句If you can't run to champagne, buy sparkling wine.如果买不起香槟酒，就买汽酒好了。柯林斯例句Steve uncorked bottles of champagne to toast the achievement.史蒂夫打开几瓶香槟酒庆功。柯林斯例句He had reached over to turn round a bottle of champagne so that the label didn't show.他伸手把香槟转了一下，这样它的商标就不会露出来。柯林斯例句Berlin people drank champagne, set off fireworks and tooted their car horns.柏林市民喝香槟、放焰火、鸣车笛。柯林斯例句Belvoir Farms produce a delicious elderflower tea. Talking of elderflower, you might wish to try Elderflower Champagne.贝尔沃农场生产一种非常好喝的接骨木花茶。说到接骨木花，你兴许想尝尝接骨木花香槟酒。柯林斯例句The waitress brings over some congratulatory glasses of champagne, courtesy of the restaurant.女侍者端来几杯饭店赠送的用来庆祝的香槟。柯林斯例句
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• 全部四级六级高考考研But this and other price shocks were event-driven—drought in the Soviet Union, crop-shrinking heat in the U.S.Corn Belt.出自-2016年6月阅读原文Since a union representative visited ...
• 全部麦片粥稀饭The father bear said, " Who has eaten some of my porridge? "熊爸爸说: “ 谁吃了我的一些粥? ”期刊摘选Mother has always ...期刊摘选During the day, he did odd jobs that were just enoug...
• ## Enough 用法

千次阅读 2008-10-04 22:16:00
enough一词在英语中既可放在被修饰的词之前，...请看下面几个例句： 1. He is not old enough to go to school. 他年龄太小，不能去上学。 2. But there were not enough people to pick them. 但是没有足够的人采摘它
• 同位语从句用法详解(例句丰富) 一、同位语从句的引导词 引导同位语从句的词语通常有连词that，whether，连接代词和连接副词等。 1. 由that引导 We heard the news that our team had won. 我们听到消息说我们...
• 一、详细释义：v.保证，担保；保护，使安全，确保 [T]例句：Close cooperation will ensure better progress in ...例句：The doctors were zealous in their efforts to ensure the survival of premature babies...
• Be 动词的八种形式 1.be 2.is 3.am 4.are 5.was 6.were 7.being 8.been be 是动词原形。 is,am,are 是一般现在时时态的be动词。 was ,were 时过去时的be动词....例句 1.The man is back. 那个男人回来了。 （...
• 抽签draw lots抽签，抽签决定lots of production生产批量drawing lots抽签双语例句用作名词(n.)He wrote lots of poems in his life.他一生中写了许多诗。Lots of people were thus denied the right to vote.就...
• 与…一致双语例句用作名词(n.)The queen consort has come together with the king.王后也随同国王前来。用作动词(v.)Don't consort with criminals.不要和犯罪分子结交。If Mary and Kate were friends, then to ...
• 分离例句用法:1. I detached my watch from the chain. 我从表链上解下手表。2. A number of men were detached to guard the right flank. 派遣了一些士兵守卫右翼.3. A number of soldiers were d
• ## slide

2010-06-07 11:55:00
幻灯片，滑动，滑坡，雪崩 例句用法： 1.House values may begin to slide. 房价可能开始逐渐下降了。 2.The children were taking turns on the slides in the playground. 孩子们在操场上轮流滑滑梯。3.The t...
• 使负义务,强制,预留同义词：oblige例句用法:1. We were obligated to attend the opening ceremony. 我们有必要参加开幕式.2. He felt obligated to help. 他觉得有责任予以帮助. equivalent:
• ## idle

2010-06-21 16:15:00
i·dle['aidl]i·dleadj.... (机器)空转例句用法：1.The cowboy lives an idle life.牛仔过着悠闲的生活。 2.His words were just idle threats: he can't hurt us.他的话只不过是无聊的吓唬人罢了，他是伤害...