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# Magnetic field due to a straight wire and circular arc

## Homework Statement

Find the magnitude of the magnetic induction B of a magnetic field generated by a system of thin conductors along which a current i is flowing at a point A(O,R,O), that is the center of a circular conductor of radius R. The ring is in the yz plane.

## Homework Equations

[/B]
magnetic field due to a straight wire=μ₀i/4πr [sinθ₁ + sinθ₂]
magnetic field due to circular arc=μ₀i/2r

## The Attempt at a Solution

#### Attachments

• IMG_20180907_180201.jpg
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• 1536324314130193750744.jpg
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BvU
Homework Helper
On your illegible attempt -- please post it as text or ##\LaTeX## -- I see four terms.
Why four ? Where does each come from ?

On your illegible attempt -- please post it as text or ##\LaTeX## -- I see four terms.
Why four ? Where does each come from ?

kuruman
Homework Helper
Gold Member
2021 Award
magnetic field due to a straight wire=μ₀i/4πr [sinθ₁ + sinθ₂]
How do you figure? What are angles θ1 and θ2?
This was unnecessary, @BvU was trying to help you get to the answer as am I. To help you, we need first to understand where you're coming from and what your difficulties are. If, however, you want to know just the answer to this problem, you will not get it here. It is against our rules to give out answers. Please read them

I had post a clear image.
I had given my attempt clearly.

Homework Helper
Gold Member
I passed on responding because the second image is sideways. Meanwhile, the calculation of the solution will involve computing both a magnitude plus a direction of the magnetic field from the 3 sources that contribute to it. It would help to at least get the image that you post right side up.

kuruman
Homework Helper
Gold Member
2021 Award
I had post a clear image.
I had given my attempt clearly.
I never said or implied that you broke any rules. I said that our rules do not allow us to give away answers. We cannot help you if you will not let us do so by answering our questions.
I passed on responding because the second image is sideways. Meanwhile, the calculation of the solution will involve computing both a magnitude plus a direction of the magnetic field from the 3 sources that contribute to it. It would help to at least get the image that you post right side up.
Posting the same figure rotated by 90o will not be much help. I studied the figure on my rotated laptop. It shows four contributions, two in the ##\hat i~## direction and two in the ##\hat k~## direction. That cannot be if the wire assembly is all in the same plane. OP needs to explain how these are derived as @BvU observed.

On edit: I understand the geometry now, but I still don't understand why four contributions.

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I passed on responding because the second image is sideways. Meanwhile, the calculation of the solution will involve computing both a magnitude plus a direction of the magnetic field from the 3 sources that contribute to it. It would help to at least get the image that you post right side up.
The image is sideways but each and every letter can be read easily.Tell me the word or number which you cant read.

The image is sideways but each and every letter can be read easily.Tell me the word or number which you cant read.
are you not able to see in the image that one wire is on x axis and one wire is in z axis.

Homework Helper
Gold Member
We see plenty and plenty of homework problems on Physics Forums, and many of them, like this one, are not very difficult. One thing we always like to see is a willingness on the part of the OP (OP=original poster=person submitting the post) to put a little bit of effort into working the problem, and making it a little bit easy on the homework helpers. In this case, that effort seems to be lacking.

BvU
Homework Helper
I know the answer, as do you. When I almost break my neck sideways I can see it's written at the bottom of your page.

You still have not explained how you get 4 terms out of three contributions.

You may not have broken any rules yet, but you should read the guidelines (in particular item 5) about helping the helpers. As opposed to ticking them off because you feel you aren't being served properly or not getting your money's worth

I know the answer, as do you. When I almost break my neck sideways I can see it's written at the bottom of your page.

You still have not explained how you get 4 terms out of three contributions.

You may not have broken any rules yet, but you should read the guidelines (in particular item 5) about helping the helpers. As opposed to ticking them off because you feel you aren't being served properly or not getting your money's worth
dont you remember that a(x+y) =ax + ay
see attached image again

BvU
Homework Helper
I remember. But there is no relation to your exercise. I don't see a ##\vec B## at a ##\pi/4## angle

I remember. But there is no relation to your exercise. I don't see a ##\vec B## at a ##\pi/4## angle
you leave it.you cannot do.why do you need B at π/4 angle

Delta2
Homework Helper
Gold Member
Here is the image rotated properly. Still I find it very hard for my eyes and my brain to understand it.
Maybe the other helpers find it easier to help now.

#### Attachments

• 1536324314130193750744.jpg
42.2 KB · Views: 330
Last edited:
BvU
Homework Helper
you leave it.you cannot do
What makes you think that ?
Anyway, this is your thread and I do hope you really want help with this exercise. Why don't you do the wise thing and read the advice (again, if you already did) from three heavyweights and try to follow it, instead of working yourself up into a blocking frustration.

You write $$\vec B = {\mu_0\,i \over 4\pi\, r}\; ( {\bf -\hat k} )+ {\mu_0\,i \over 4\pi\, r} \; ( {\bf -\hat \imath} ) + {\mu_0\,i \over 2\, r} \; {-( \bf -\hat\imath + \hat k) \over ??}$$ and all we humbly ask of you is to explain where these terms come from

Delta, thanks for rotating the picture. My desktop screen is too heavy to turn over ##\pi/2## !

Now I see the curly ##h## is in fact ##\sqrt 2 \quad ##

Delta2
Homework Helper
Gold Member
The ring is in the y-z plane according to the OP=original post. That means the direction of the unit vector in the 3rd term is is the wrong direction. (## \hat{u}=\frac{-\hat{i}+\hat{k} }{\sqrt{2} } ## is incorrect. (This is the 45 degree angle=## \frac{\pi}{4} ## (radians) that is being referred to in the discussion above). ## \\ ## It looks like the OP is presently off the air with a line through his name.

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Delta2
BvU
Homework Helper
Giving it away, eh ? tut tut

Delta2
Homework Helper
Gold Member
when there is a line through a user name it means that the user got banned, or he deleted his account?

BvU
Homework Helper
I wonder how that came to happen ....

berkeman
Mentor
when there is a line through a user name it means that the user got banned, or he deleted his account?
He is temp banned for multiple infractions over a period of time. Thanks for trying to help, all. Hopefully when he comes back in 10 days, he will abide by the PF rules.


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Tag Info

Factor analysis is a dimensionality reduction latent variable technique which replaces inter-correlating variables by a smaller number of continuous latent variables called factors. The factors are believed to be responsible for the inter-correlations. [For confirmatory factor analysis, please use tag 'confirmatory-factor'. Also, term "factor" of factor analysis should not be confused with "factor" as categorical predictor of a regression/ANOVA.]

A latent factor (matrix) model has the following form:

$${\bf Y} = {\bf \mathbb{1}\mu}^T + {\bf ZA}^T + {\bf E\psi}^{1/2}$$

Where:

• $Y \in \mathbb{R}^{n \times p}$ whose rows are random vectors in $\mathbb{R}^p$
• ${\bf \mu} \in \mathbb{R}^p$ is a vector of column means of $Y$
• ${\bf Z}$ is a matrix whose rows are unobserved latent factors ${\bf z_i}$
• ${\bf A} \in \mathbb{R}^{p \times q}$ is a matrix whose columns form linear combinations of the factors, ${\bf z}$. These are the factor loadings
• ${\bf E}$ is a normal matrix with 0 mean and Identity covariance
• ${\bf \psi} =diag(\psi_1, ... \psi_p)$. This ensures that all dependency among the rows of ${\bf Y}$ are dependent on ${\bf Z}$.

Note that $E[{\bf Y}] = {\bf \mathbb{1}\mu}^T + {\bf ZA}^T$ and $Var({\bf Y}) = \psi \otimes I$. This model is only useful if $q < p$, otherwise ${\bf ZA}^T$ is full rank and this model is the same as an unrestricted and arbitrary model, $E[{\bf Y}] = {\bf M}$ for any ${\bf M}$.

format: plain


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# Statutes

Download the file for the statues. This file is a list and summary of statutes that appear throughout the book.


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# Probability

Reviews of the mathematical model for determining the “chance” of an event occuring. Note that statistics tends to use $P(A)$ for the probability of an event. However, most cryptology books use $P(X)$ since mathematicians and computer scientists are generally more comfortable with $x$ and $y$. I’ll follow the trends of my people here.

## Marginal Probability

Marginal probability is the probability of an event ($P(x)$) occuring. In the most basic sense, the probability of an event $x$ occuring ($P(x)$) is equal to the number of ways the event $x$ can occur divided by the total number of possible outcomes of the experiment:

So the probability of a coin flip resulting in heads is: $P(x) = \frac{1}{2}$. The total possible outcomes in this instance is the sample space $X = (heads,\;tails)$ which has a total of 2 possible outcomes.

It is important to note that $x$ isn’t necessarily a unique outcome in $X$. There might be multiple ways to achieve $x$ making $x$ a subset of $X$. For instance, if we wanted to know the probability of rolling an even number on a dice, $x$ = (2,4,6). So $P(x)$ in this case would be $\frac{3}{6}$ or 0.5.

We can formalize this concept by considering a probability distribution $P(x)$. In this case each discrete random variable $x \in X$ has a probability associated with it ($P[X = x]$) such that

and

You can see this characteristic in our overly simplified coin flip example. Here $X = (heads,\;tails)$ and the probability $P[X = x]$ for each $x$ is 0.5. The sum of all of these probabilities is thus 0.5 + 0.5 = 1.

Similarly, for our dice example, we would calculate the probability of each number being rolled and add them together: $\frac{1}{6} + \frac{1}{6} + \frac{1}{6} + \frac{1}{6} + \frac{1}{6} + \frac{1}{6} = 1$

## Joint Probability

Now suppose I want to determine the probability that two or more events will occur at the same time. The key here is that both events are independent of each other, i.e. neither event depends on the other to occur first in order to occur themselves.

So if we are looking at two events, we can define $P[(x,y)]$ as the probability that $X = x$ and $Y = y$. If $X$ and $Y$ are independent random variables, then

So if we roll two dice, we can compute the probability that roll two numbers that equal 5. First we note the possible combinations that equal 4 as [(2,2), (1,3), (3,1)]. Thus, $P[x,y] = \frac{3}{36}$. We get 3 possible ways our event can occur. 36 results from us multiplying the total number of possible outcomes of $X$ and $Y$ which is $6 \times 6$ or 36.

## Conditional Probability

Conditional probability determines the likelihood that an event will occur given something else is guaranteed to happen. So we can define condition probabilty $P[x,y]$ as the probability that $X = x$ given that $Y = y$.

With this we can formalize our definition on independence to $X$ and $Y$ are independent if:

$P[x \vert y] = P[x]P[y]$, for all $(x,y)$ in $X \times Y$

So again using our dice example, we would ask what is the probabilty of the sum of two dice being 5, if the first dice we rolled is a 3. In this case, the likelihood would be $\frac{1}{6}$ since there is only one number (2) in the set of 6 that would give us 5. So we can see that the probability of getting the combination we want has been greatly reduced if we can guarantee one of the numbers is rolled.

## Bayes Theorem

All of this leads us to Bayes Theorem which is used to determine the probability of an event based on knowledge of conditions which may be related:

If $P[y] \gt 0$, then

So with our dice example, we can use Bayes theorem to find the probability of if an odd number is rolled, it will be 3. We can set $x$ to the probability you will roll a 3 and $y$ to the probability you will roll an odd number:

So we can see that the likelihood of rolling a 3 ($\frac{1}{6}$) divided by the likelihood of rolling an odd number ($\frac{1}{2}$) gives us our probability. This follows our logic of reducing the set of possible outcomes to the three odd numbers only one of which is 3. So just using marginal probability it would be $\frac{1}{3}$.

## Other Sources

Khan Academy has a pretty good introduction to basic probability here.


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### Home > PC3 > Chapter 2 > Lesson 2.2.3 > Problem2-85

2-85.

Consider the “saw tooth” graph at right.

1. What is the amplitude?

The maximum points are at $y=2$. The minimum points are at $y=−2$. What is half the distance between these points?

2. What is the period?

The period is the horizontal distance to complete one cycle.

3. State the domain and range.

D: all real numbers; R: $−2\le y\le2$

4. What are the $x$-intercepts of the function? Express your answer as an equation.

... $-5,-3,-1,1,3,5$, ...


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# Did you know?

#### January 2021

Let ƒ be a holomorphic function in the unit disk of the complex plane fixing the origin with differential of modulus 1. The local inversion theorem guarantees that ƒ admits a holomorphic inverse g in a small disk of radius r > 0 centered at the origin. André Bloch showed that the maximum radius of existence of g is bounded below by an absolute constant b > 1/72.

This is André Bloch most famous theorem, which has been published in 1925 in the Annales de la Faculté des Sciences Toulouse. The optimal absolute constant b is still unknown today, despite numerous works and steadyprogress (see e.g. Bonk, M. ; Eremenko, A. ; Covering properties of meromor-phic functions, negative curvature and spherical geometry. Ann. of Math. 2000, 152 , no. 2, 551-592).

André Bloch was an unusual mathematician. You’ll find some biographical information in this article by H.Cartan and J.Ferrand.


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# How prove $\sum\limits_{cyc} \frac{ab}{c} \geq 3.$

Let $a,b,c>0 ,\sum\limits_{cyc} a^{3}=3$. How prove $\sum \frac{ab}{c} \geq 3.$
My try:
$a^3+b^3+c^3=3$

$a^2b^2+a^2c^2+b^2c^2 \ge 3abc$ so

$(a+b+c)^3-3(a+b+c)(ab+ac+bc)+3abc=3$

$(ab+ac+bc)^2-2(a+b+c)abc \ge 3abc$

• What are you summing over? – Jendrik Stelzner Jan 2 '16 at 13:03
• It is not clear what does it means by $\sum a^3$ and $\sum ab/c$ – sinbadh Jan 2 '16 at 13:03
• Are you trying to show $a^2b^2+b^2c^2+c^2a^2 \ge 3abc$ given $a^3+b^3+c^3 = 3$ ? – Henry Jan 2 '16 at 13:04
• I try that:$$x=a+b+c$$, $$y=ab+ac+bc$$, $$z=abc$$ so $$x^3-3xy+3z=3$$ and $$y^2-2xz \ge 3z$$ – piteer Jan 2 '16 at 13:11
• @Henry yes am trying that – piteer Jan 2 '16 at 13:21

## 1 Answer

The inequality is false: take $a=\frac{1}{\sqrt[3] 6}$,$b=\frac{1}{\sqrt[3] 6}$,$c=\frac{\sqrt[3] 8}{\sqrt[3] 3}$, you obtain

$$\sim 2.9918... \ge 3.$$

• ya, the inequality is only true when $a+b+c=3$ – chenbai Jan 3 '16 at 7:25


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# Question ae3f1

Aug 5, 2017

${v}_{0} = 28.0$ $\text{m/s}$

#### Explanation:

We're asked to find the speed at which the ball was hit, given its horizontal range and hit angle.

To do this, we can use the equations

ul(Deltay = v_0sinalpha_0t - 1/2g t^2

ul(Deltax = v_0cosalpha_0t

where

• $\Delta y$ is the change in height ($0$ since the ball landed at the same height it was hit)

• $\Delta x$ is the change in horizontal position, given as $103$ $\text{m}$

• ${v}_{0}$ is the initial speed (what we're trying to find

• ${\alpha}_{0}$ is the hit angle, given as ${50}^{\text{o}}$

Plugging in known values to both equations:

$0 = {v}_{0} \sin \left({50}^{\text{o")t - 1/2(9.81color(white)(l)"m/s}} ^ 2\right) {t}^{2}$

 103color(white)(l)"m" = v_0cos(50^"o")t

We can realize that the time $t$ when the ball is at $y$-position $0$ and $x$-position $103$ $\text{m}$ are equal, so we can solve the second equation for $t$ and plugbthat in for $t$ in the first equation (and eliminate units to make it simpler):

$t = \frac{103}{{v}_{0} \left(0.643\right)} = \frac{160.2}{{v}_{0}}$

$0 = {v}_{0} \left(\frac{160.2}{{v}_{0}}\right) - \frac{1}{2} \left(9.81\right) {\left(\frac{160.2}{{v}_{0}}\right)}^{2}$

Now we solve for ${v}_{0}$:

$4.905 {\left(\frac{160.2}{{v}_{0}}\right)}^{2} = 160.2$

$\frac{125944}{{\left({v}_{0}\right)}^{2}} = 160.2$

v_0 = sqrt((125944)/(160.2)) = color(red)(ulbar(|stackrel(" ")(" "28.0color(white)(l)"m/s"" ")|)#


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SEARCH HOME
 Math Central Quandaries & Queries
 Question from Dawn: 1-- syringe contains 5.5cc of medicine & cost $13.50 1 --2 oz bottle of medicine cost$17.50 How many cc's in 2 oz, & which is the better deal?

Hi Dawn.

You have two costs per unit volume:
(A) $13.50 / (5.5 cc) (B)$17.00 / (2 oz)

You need to find the conversion factor to go from cc's to ounces, so we can compare these fractions using a common unit denominator.

When I type "1 cc in ounces" into Google, it replies "1 cc = 0.0338140227 US fluid ounces". So the conversion factor is that decimal number.

So (A) becomes:

$13.50 / ( 5.5 cc x 0.033814 oz/cc) =$13.50 / ( 5.5 x 0.033814 oz)
= $13.50 / (0.195977 oz) =$72.59 per ounce.

(B) becomes
$17.00 / (2 oz) =$8.50 per ounce.

The bottle is WAY cheaper than the syringe, assuming you don't need the syringe itself!

Hope this helps,
Stephen La Rocque.

PS: Google is getting smarter all the time. You could even type in
"convert (13.50 dollars per (5.5 cc)) to dollars per ounce" and it will tell you !

Math Central is supported by the University of Regina and The Pacific Institute for the Mathematical Sciences.


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Question

# A magnetic field is produced by bar magnet as shown in figure and a beam of electron is passing through magnetic field. How will the magnetic field affect the electron beam?

A
The beam will rotate around the two magnets.
B
The beam will bend up away from the table.
C
The beam will bend down into the table
D
The beam will be attracted to the north pole of a magnet.
E
The beam will be attracted to the south pole of a magnet.

Solution

## The correct option is B The beam will bend up away from the table.Given :  $$\vec{B} = B \hat{y}$$Velocity of electron beam     $$\vec{v} =- v \hat{x}$$Magnetic force on electron        $$\vec{F} = -q(\vec{v} \times \vec{B}) = q (\vec{B} \times \vec{v})$$$$\therefore$$   $$\vec{F} = q [B\hat{y} \times (-v \hat{x})] = qvB$$  $$\hat{z}$$Thus the electron beam will bend up away from the table.Physics

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0

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# Using an Arduino to measure wire capacitance

I wanted to use an Arduino as a capacitance measurement device for a coaxial wire. I started at measuring the wire capacitance using digital multi-meter UT58B. I turned the knob to 2 nF and then measured the capacitance depending on the wire length. These are the results:

After the measurements I got the function (y = 0.0032x + 0.0268) on how the capacity changes depending on wire length. Now I wanted to use the Arduino to measure the capacitance and then calculate the wire length depending on the capacity. I tried following a guide to make the Arduino be able to measure the capacity.

Now the problem. I tried calculating the needed resistance using R = TC/C for one seond and got:

R = 1/(0.019*10^(-9)) = 52631578947.37 ohm

As you can see the resistance is too big. My questions:

1. Should I decrease the time constant? And how small could it be so that arduino can still measure the time? What maximum resistance can you suggest?

2. Maybe I wrongly read the multi-meter values? (the 2 nF scale)

3. Maybe there are other solutions you can suggest to measuring capacity with Arduino?

• 1s is an insanely long time. Use something much shorter. – Ignacio Vazquez-Abrams Dec 4 '15 at 12:32
• On your graph, are the units nF? Are you trying to measure capacitance in the range 0 to 350pF? – Oliver Dec 4 '15 at 12:49
• @Ignacio Vazquez-Abrams Any suggestions on what could be the most optimal resistance and time constant for arduino measurement? – disoderly Dec 4 '15 at 13:07
• Nope. But you have the hardware in front of you, so it shouldn't be difficult to run the appropriate tests. – Ignacio Vazquez-Abrams Dec 4 '15 at 13:08
• @Oliver On the graph are the values my multi-meter showed when it had the knob on 2nF position. So the units are nF if I am correct. – disoderly Dec 4 '15 at 13:08

In actuality, you can use an Arduino and you do not need to utilize an oscillator with the test capacitor in the tank. This will work quite well for capacitors of 10pF or more, and with care even below, though that is not something I would have believed until trying it. You do however need your PCB to be very clean without flux residue or remains of efforts to remove it.

The key is not to use the low impedance, slow ADC, but rather the comparator, and utilize the timer block to measure the charging time.

This is what the Arduino capacitance measurement example does, and also the ATmega based kits with LED digit display which you can buy for around \$11. Source resistor is typically selected be a GPIO and in the 100K - 1M type of range. Stray capacitance is calibrated out by measuring the time constant with only the meter and test fixture, but no capacitor under test. Provided you don't move the fixture or get your hands into the field of measurement, adding even a tiny test capacitor then linearly increases the time constant.

• Yes this approach works quite well, I have implemented a touch sensor this way (capacitance around 10 pF) and am working on a pressure sensor with a capacitance of 47 pF to 100 pF with the same approach (but with a different kind of resolution you'd get with an Arduino). – Arsenal Dec 4 '15 at 14:45

The system the Arduino Example shows will probably not work for you.

You are trying to measure fractions if a nF, so either your resistor will be huge, or your time to find the point of interest is so small you only get 4 or 5 samples on your ADC. The prior is not an option, because of pin input leakage, the latter not because you won't find it.

Apart from that, your Arduino uses an internal sample and hold based on a small capacitance. Where small is much larger than what you are measuring. You could fix that with a unity-gain op-amp with very low input capacitance and input leakage, but that still won't fix your other problems.

What you will need to do is recreate what your multimeter does. A capacitor has an internal resistance plus the resistance it has for a frequency signal. Most affordable multimeters ignore the internal resistance and just measure the total at 1kHz.

We can go into the maths, with imaginary parts and results and what not, but since you're only interested in the results, and don't even need accurate calibration against correct capacitor value, you can redo your measurement experiment with your set-up. So for completeness I will only mention here that the complex impedance of a capacitor is given by:

Z(c) = 1 / (jwC) where w = small letter omega, which stands for angular velocity, or w = 2*pi*f, where f is the signal frequency. For first estimation purposes one can simplify that to the real part of the resistance being given by:

"R(c)" = 1 / ((2*pi*f)*C) <-- Ignores the imaginary parts, that may be important in some cases.

Now, what does a multimeter or RLC meter do? Simplified something like this:

simulate this circuit – Schematic created using CircuitLab

It knows it makes a 1kHz sine function at a certain voltage, so it measures the current that flows into the capacitor, then if it is sure about the voltage (either by strong drivers of the sine signal, or by measuring that as well), it can calculate an effective AC resistance and from the formula above it will turn that into capacitance.

But that's all not very relevant for you, what you probably should do, to make it as easy as possible, is create a frequency, possibly with a timer output on the Arduino and record what current flows through the cable's capacitance at what length, and just create a new formula for that, as that involves much fewer steps and calculations and calibrations than first getting the capacitance right and then transforming that.

To be able to estimate what we need, we should calculate the apparent resistance of your limiting values. From your graph it's 50pF to 350pF, so we'll take some room and make it 25pF to 500pF. So we calculate with the simplified equation I gave earlier, I'll assume 1kHz, since that allows you to adjust to 100Hz and 10kHz if needed:

"R(25pF)" = 1 / (2*pi*1000Hz * 25pF) =~ 6.37M Ohm

"R(500pF)" = 1 / (2*pi*1000Hz * 500pF) =~ 318k Ohm

Now, this assumes a sine wave, and you will most likely be making a square wave, which contains many higher frequencies, which will decrease those figures. Assuming you are using 5V, we can pretty safely say your measurement will be between 1μA and 40μA peak.

Now we need to find a trick to turn 1μA to 40μA into something your Arduino can measure. Here we go:

simulate this circuit

The first op-amp is a difference amplifier with a gain of 5. You can have 10 gain there as well, but if you add the rectifier to that stage, your leakage path will include a capacitor, and that may give you trouble when you decide to start playing with the frequency.

The first op-amp amplifies the difference across the resistor of 10k by 5, because it wants the + voltage to be on the - side. At a high going flank of your signal it will see a difference based on the current, in the positive direction. At a low going flank it will see the opposite, so it will try to create a negative voltage. Fortunately powered by +5V and ground it can't do that.

The second op-amp takes the positive pulses and amplifies them by 2 through a diode. The diode makes sure the capacitor can only be charged by the pulses and discharged by the feedback loop of R7 + R8. So the discharging of the 100nF capacitor has a time constant of about 0.44seconds to discharge to about 37%, that gives you a bit of slowness, but will get you a good rectification for 100Hz to 10kHz signals. The charging may also take a little time, because the op-amp may not be able to fully charge the capacitor the first few pulses, but I don't think you will notice that difference much.

If you need it all to be faster you can decrease the capacitor a little, just be sure the discharging RC-constant (R value times C value) is less than 1/5th the source frequency you end up using.

I chose the TL051A because it has a nice and low input bias current, low input offset voltage (less than 1mV) and a good output slew rate (20V/μs) as well as a unity gain bandwidth more than 50 times the highest frequency we were considering. Another important thing here is the fact that its inputs can go all the way up to its power supply. They don't give good results at the negative rail (so when the pulse is steady off), but from its internal schematic I expect the output to remain close enough to 0 to not be much of a bother.

It is available at farnell for about €1,- in SMD and DIP packages, so it's quite suited for hobby work. I tried searching Digikey, but the search function didn't work.

Here's the TL051A datasheet.

For the second op-amp it's important it's able to go to the 0V at the input, or at least close to it, and since the signal is already amplified a lot, input offset voltage can be a bit higher and input offset current and bias current are not that relevant, so a standard LM358 type might be suitable. Although it is possible it will give a little distortion on the highest output voltages, I expect that to be limited.

Here's the LM358N datasheet

Now you can sample a 0.1V to 4V signal to determine the measure capacitance leakage. With a 10bit ADC running on 5V you have a bit-resolution of about 5mV per LSB, so you should be able to measure quite accurately.


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# Images from HTML + captions

Hi There,
I’ve been banging my head against this for a day or so now, so thought I would ask the forum for ideas. I apreciate that Aspose.Words doesn’t support this nativly, but would like any workaround/clues to how to acheive the following.
I have some HTML that I am inserting into a document. The HTML has img tags that display fine, but I want them to have captions in the Word doc. In the HTML I can add alt, title, paragraph, whatever is necessary to store the text.
What I would like to be able to do is to iterate over the images in the document, and if they have the alt/title/para (some images will not, like ones originally in the doc template), add it as the image’s caption. Nothing too complicated I would have thought, but my tiny brain can’t get it to work (only eval’ing for the past week - very impressed thus far).
Thanks for any help that anyone can give me.
– Mike.

Hi
Thanks for your request. Captions in MS Word are represented as “Lable” + “SEQ field”. You can use DocumentBuilder for inserting captions. For example see the following code:

Document doc = new Document(@"Test213\in.doc");
DocumentBuilder builder = new DocumentBuilder(doc);
// Get collection of shapes inthe document
NodeCollection shapes = doc.GetChildNodes(NodeType.Shape, true);
// Loop through all shpes
foreach (Shape shape in shapes)
{
if (shape.ShapeType == ShapeType.Image)
{
// Move DocumentBuilder cursor to shape
// Cursor will be placed after shape
if (shape.NextSibling != null)
builder.MoveTo(shape.NextSibling);
else
builder.MoveTo(shape.ParentParagraph);
// Insert paragraph break
builder.Writeln();
// Insert lable
builder.Write("Picture");
// Insert SEQ field
builder.InsertField(@"SEQ Picture \* ARABIC", "");
}
}
// Save result document
doc.Save(@"Test213\out.doc");

But note that Aspose.Words does not evaluate SEQ fields. So you should update fields inside the document manually (ctrl+A and F9). Also you can use macro to update fields. See the following link to learn more.
https://support.microsoft.com/en-us/topic/the-filename-field-does-not-automatically-update-when-you-open-a-document-in-word-de2bfb95-d990-1ced-a618-5ac0a2ec1be4
Also if you need that some text appears just after image you can try using the following HTML:

This is caption

Hope this helps.
Best regards.

Many thanks Alexey, this works really well, other than in the following condition…
Say I have a HTML doc where I want some images to have captions, and others not. If all images need captions, your solution above works really well However, if I only want some to have captions, I can’t easily just pull the next paragraph and use it as the caption (it might just be “normal” text).
The first (prefered) is to perhaps use the HTML Fig markup http://www.w3.org/MarkUp/html3/figures.html - that way there’s is a clear association with the figure and caption. Then 's get captions, whereas s would not. However, I can’t see how Aspose.Word deals with this kind of markup, and how to identify it.
The other method is to use the alt tag of the image - if there is an alt attribute, then that becomes the caption (or could dictate that the next paragraph is/is-not a caption- I don’t really mind either way). Once again, I can’t see how the library deals with attributes as the title and alternativetext properties always seem to be empty.
Any pointers on what would be the best method?
Cheers,
Mike.

Hi
Thanks for your request. I think you can insert placeholder right after image and after inserting replace this placeholder with caption. Placeholder is just plain text. See the following HTML:

==CAPTION==

After inserting you can replace the placeholder with caption using ReplaceEvaluator. See the following code:

public void Test006()
{
// Create Document and DocumentBuilder
Document doc = new Document();
DocumentBuilder builder = new DocumentBuilder(doc);
// Insert HTML into the document
builder.InsertHtml(html);
// Replace placeholders with captions
Regex regex = new Regex("==CAPTION==");
doc.Range.Replace(regex, new ReplaceEvaluator(ReplaceInsertCaption), false);
// Save output document
doc.Save(@"Test006\out.doc");
}
private ReplaceAction ReplaceInsertCaption(object sender, ReplaceEvaluatorArgs e)
{
// Create document builder
DocumentBuilder builder = new DocumentBuilder(e.MatchNode.Document);
// Move to matched node
builder.MoveTo(e.MatchNode);
// Insert caption
// Insert lable
builder.Write("Picture ");
// Insert SEQ field
builder.InsertField(@"SEQ Picture \* ARABIC", "");
// Remove placeholder
e.Replacement = string.Empty;
return ReplaceAction.Replace;
}

Hope this helps.
Best regards.


text:
Overview

Chip-dependent, filter-based EE fractions were derived from aperture photometry of three white dwarf standards G191B2B, GD153 and GD71 (plus the G star P330E) and spliced to the 2009 in-flight models at r=35 pixels (1.4 arcsec). The new EE curves are consistent with the 2009 model EE fractions for all but a few filters, as shown in the figure below.

UVIS EE Tables

ASCII tables for UVIS1 and UVIS2.

• Row 1: Header Information, Row 2: aperture radius (arcsec), Row 3: aperture radius (pixels), Rows 4-45 EE fractions by filter

• Column 1: Filter Name, Columns 2-51 EE values for each radius (1-50 pixels)

UVIS 1

UVIS 2

Recommendation

For photometry with aperture radii <10 pixels and targets placed in the upper-left corner of Amp A (e.g. UVIS1-C512A-SUB, UVIS1-C1K1A-SUB), we recommend using the UVIS1 EE fractions, since the PSF in this quadrant is slightly out of focus compared to the rest of the UVIS focal plane. For all other cases we recommend using the UVIS2 EE fractions.

Prior Calibration

After WFC3's installation in 2009, deep observations in F275W and F625W were obtained in 5 positions across the UVIS array, and these were used to revise the pre-flight EE. EE fractions for all remaining filters were interpolated from the revised 2009 optical model (ISR 2009-38).
[The 2009 EE tables are available here.]

The 2016 chip-dependent EE fractions were derived from filter-based aperture photometry, spliced to the 2009 models at r=35 pixels (1.4 arcsec) and are consistent with the 2009 model EE fractions for all but a few filters.  The uncertainty in the EE curves at the infinite aperture is ~0.5%.

Generally, fewer than 8 observations were obtained per narrow band filter, and for some apertures there are only 2.  Therefore the filter based encircled energy values for these filters have larger uncertainties than EE values for the broad band filters.   The figure below shows the filter-based EE fractions at 10 pixels for all full frame filters for UVIS1 (green) and UVIS2 (red ), as well as the original pre-flight EE model values (black) and the in-flight revised model EE fractions (blue).

F200LP  has an extremely broad bandwidth ~6000 Å centered at ~4885 Å.  F953N has 8 images in UVIS1 but only 2 images in UVIS2. Thus its EE is highly uncertain, and users may wish to adopt the UVIS1 EE values instead. Consult Table 6 in ISR 2016-03 to compare the number of input frames used to create the filter-based EE curves.

Examples

For drizzled images, or flat-fielded images multiplied by the pixel area map (i.e. FLT*PAM), the mean signal in a circular aperture of radius r is:

$$Flux = \frac{F_r \cdot PHOTFLAM}{EE(r)}$$

Where Fr is the signal within aperture r in electrons per second, EE(r) is the encircled energy fraction at radius r, PHOTFLAM is the inverse sensitivity at the infinite aperture, whose default value is PHTFLAM1.

The equivalent calculation using magnitudes is:

$$m=m_i + 2.5\times \log{\left (EE(r)\right)} + ZP$$

where mi is the instrumental magnitude, mi = -2.5*log(Fr), ZP is the PHOTFLAM equivalent in magnitude units from ISR 2017-14Table 2, and EE(r) is as above.

For example, aperture photometry using a *drz.fits image, for radius r = 3 pixels of a star on the UVIS1 CCD with the F606W filter yields F= 950 e-/s.

The inverse sensitivity of F606W is PHTFLAM1 = 1.13745E-19 erg·s-1·cm-2·Å-1 /(e-/s). The encircled energy at r = 3 pixels is

EE(r=3) = 0.742 (UVIS1)

In physical units:

Flux = 950 * 1.13745E-19  /  0.742=  1.4563E-16 erg·s-1·cm-2·Å-1

In VEGAMAG:

m = -2.5*log(950) + 26.014 + 2.5*log(0.742) = 18.246 mag

NOTE: Photometry at r<8 pixels relative to r=10 pixels varies with position and depends strongly on the telescope focus and breathing. At r=3 pixels, the variation is between 4% -10% (see ISR 2013-11).

LAST UPDATED: 08/28/2019


text:
Last time I had asked for intuitions about associativity, @JasonHise64 suggested that associativity is commutativity squared. In retrospect, his suggestion seems much clearer than I had initially been able to see. But I think I got it now. Let's take a look together... (thread)

· · Web · · ·

(ab)c=a(bc)

That's how we ususally write associativity. How can we rewrite this, such that it looks like commutativity? I couldn't get it to work with a single operation L_x(y), but when I add another notation for the same operation with arguments flipped R_y(x) it worked!

With these, we can write

R_c(L_a(b)) = L_a(R_c(b))

or even only talk about L and R like so:

R(L(b)) = L(R(b))

This is in loose analogy to adjunctions, often seen in the context of Lie algebras. There, you'd write Ad_a(b) to obtain an operator Ad_a one can apply to elements b.

So what did we just do? Instead of our multiplication (·) we used two isomorphic notations L_a(b) = R_b(a) and then added a commutation law like so

L·R = R·L

If we had required these operatiors to be identical instead L = R we would have obtained commutativity instead!

I wonder if there's a further level three commutativity, and what it would look like. I suppose there might be an even nicer way to write down these laws, maybe using commutative diagrams from category theory, which would make these generalizations more obvious.

Arguably, this might settle my question from last time, how to recognize associativity looking at multiplication tables: we don't, we look at higher multiplication tables instead. Or not, as it's cheating, and I should spend some time playing with multiplication tables instead.

But I have more questions! What's the associative analog for the geometrical intuition for a commutator? Remember, it measures the failure of commutativity [A,B] = AB-BA You can look at it as the difference of one way of summing arrows versus the other way.

There's also an associator [A,B,C] = (AB)C - A(BC) about which I know next to nothing. But I know a little more about commutators. For these the Jacobi identity holds [A,[B,C]] + [B,[C,A]] + [C,[A,B]] = 0 which is kind of very similar to associativity when written like this:

[[A,B],C] - [A,[B,C]] = [[A,C],B]

Oh no, now it starts all over again! What's the symmetry group of that, and how can we read it off a multiplication table ...

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text:
# Backpropagation taking a back seat by 2030

Backpropagation is a widely used method of efficiently computing gradients of functions with many inputs and few outputs and hence particularly popular for modern machine learning techniques like neural networks. It can be seen as applying the chain rule "from outside to inside". It is generalised by Automatic Differentiation which imposes no such restriction on the order in which to apply the chain rule. Choosing this order appropriately may lead to an additional speed-up.

At NeurIPS 2022 Geoffrey Hinton presented the forward-forward algorithm, an alternative to backpropagation and automatic differentiation more generally. While it does not seem to yield competitive results just yet, one might wonder about more research in this direction.

Will backpropagation take a back seat by 2030?

This question will resolve as Yes if, at any point from January 1, 2023 to January 1, 2030, over 50% of machine learning frameworks (each framework is weighted by its relative usage as elicited from paperswithcode.com/trends or whatever we deem to be the closest alternative if it doesn't exist anymore) use something other than backpropagation or automatic differentiation.

If it is not obvious whether a framework "uses" backpropagation or automatic differentiation—e.g. because multiple methods are implemented or it is abstracted away in the code—we look at the "Getting Started" section or something equivalent on their website and see whether the first "reasonable" example (no print("Hello world!")) listed there uses backpropagation or automatic differentiation under the hood. If no such page/example exists, this will be up to the discretion of moderators, but should involve checking the source code or reaching out to developers.

Note that automatic differentiation is included because this question is not supposed to resolve positively due to some minor local modifications in the order of differentiation which, technically, is not backpropagation anymore. Moreover, we use paperswithcode.com to track progress in machine learning instead of just neural networks in case neural networks are replaced by a different approach, but backpropagation remains the best option for neural networks.

Note: this question resolved before its original close time. All of your predictions came after the resolution, so you did not gain (or lose) any points for it.

Note: this question resolved before its original close time. You earned points up until the question resolution, but not afterwards.

Current points depend on your prediction, the community's prediction, and the result. Your total earned points are averaged over the lifetime of the question, so predict early to get as many points as possible! See the FAQ.


text:
## A figure with four line segments donot_______except ut their and points is called the quadrilateral

Question

A figure with four line segments donot_______except ut their and points is called the quadrilateral

in progress 0
1 month 2021-08-19T08:42:26+00:00 1 Answer 0 views 0


text: To write ionic equations follow the steps below. 1 decade ago. MgO + 2 HCl → MgCl 2 + H 2 O [ Check the balance ] Magnesium oxide react with hydrogen chloride to produce magnesium chloride and water. Answer Save. fe + hcl net ionic equation, Write the balanced net ionic equation for the reaction that occurin the following case. Mg = 0. Pb(NO 3) 2 (aq) + NH 4 Cl(aq) → PbCl 2 (s) + NH 4 NO 3 (aq) 7 You can use parenthesis or brackets []. Lv 4. What is the balanced net ionic equation for the reaction between Mg and HCl in Mg + HCl = MgCl2 + H2 Given: 4.39m/sec Go: km/hr aWrite balanced lonic.reaction: Cr2O72 +C204² -> 207 3t + 2CO₂equation for t mehe Relevance. Answer. … Mg + (-2) = 0. HCl + MnO 2: MnO 2(s) + 4 HCl ---> MnCl 2(aq) + Cl 2 + 2 H 2O Or, in net-ionic form..... MnO 2 + 4H + + 2Cl----> Mn2+ + Cl 2 + 2H 2O Conc. Solution for Mg(s) + HCl(aq) MgCl2(aq) + [ H2 ] what is the - molecular equation - total ionic equation - net ionic equation Spectators at a sporting event watch the action unfolding in front of them rather than participating; spectator ions likewise do not participate in the reaction. Ionic charges are not yet supported and will be ignored. This can fluctuate a bit, but will wholly depend on if the NHbR changes at all. When acids and bases react, they form a salt ({eq}MgCl_2{/eq}) and water ({eq}H_2O{/eq}). The combustion reaction of Mg +1/2 O2-->MgO produces a very bright white flame and high heat. Favourite answer. Please? I think I have the balance equation right. Not surprisingly, the most efficient reactions are those that take place in the gas phase. Asked by Wiki User. Data for each trial are shown in the table below. How To Balance Equations Mg + (-2) + 2(+1) = 0. A student was assigned the task of determining the enthalpy change for the reaction between solid MgO and aqueous HCl represented by the net-ionic equation above. Rewriting this equation to show the acid, the hydroxide, and the salt as ions - which is the way these compounds exist in solution - would give the equation H + + Cl-+ Na + + OH-Na + + Cl-+ H 2 O . For example, C6H5C2H5 + O2 = C6H5OH + CO2 + H2O will not be balanced, but XC2H5 + O2 = XOH + CO2 + H2O will. All that is left are the ions that have changed chemically. MgCO3 + HCl = MgCl2 + CO2 + H2O. Update: Net ionic equation is the final answer I am wanting, but it would also help if you could show the full ionic so I can see where it's coming from properly. Ionic charges are not yet supported and will be ignored. Relevance. Balancing chemical equations. Magnesium oxide react with hydrogen chloride. 2008-02-29 06:44:28 2008-02-29 06:44:28 . But props to Dean … Get your answers by asking now. Join Yahoo Answers and get 100 points today. Trending Questions. What is the ionic equation of HCl MgO? Answer Save. Still have questions? Top Answer. For example, C6H5C2H5 + O2 = C6H5OH + CO2 + H2O will not be balanced, but XC2H5 + O2 = XOH + CO2 + H2O will. MgOH2 = 0. Language: ru es en. MgO(s) + HCl(aq) = H2O(l) + MgCl2(aq) MgO(s) + HCl(aq) = H2O(l) + MgCl2(s) Instructions and examples below may help to solve this problem You can always ask for help in the forum Instructions on balancing chemical equations: Enter an equation of a chemical reaction and click 'Balance'. Finally, we can rewrite the the balanced chemical equation as: Mg (s) + 2HCl (aq) ----> MgCl2(aq) + H2 (gas) Magnesium reacts with HCl. HCl + NaOH NaCl + H 2 O . Our channel. Ionic equation for MgO + HCl? Hi The Tomm, Dean is correct except for: 2) If any of the compounds are aqueous, then you can split it up into 2 ions (the +ve and -ve) This is not true asnot all aqueous state compounds ionise in water (for instance sucrose). b)Write the net-ionic equation that . In the third line, the full ionic equation is repeated, but cancellation of species appearing the same on both sides is indicated, which leads to the final equation, the net ionic equation. Relevance. Log in: Chemical reactions Сhemical tables. I am having a bit of difficulty getting the net equation and the net ionic equation and the net ionic equation. My book doesn't really explain this at all, and the little bit it does makes no sense. Mg+HCL---MgCl2=H2 Mg-1 . 1 decade ago. Grateful. Answer Save. How To Balance Equations The net ionic equation shows only the chemical species that are involved in a reaction, while the complete ionic equation also includes spectator ions. Favourite answer. How Hot Is A Habanero Vs Jalapeno, Orion Telescope Replacement Parts, Cedh Breya Combo, Connor-davidson Resilience Scale Pdf, How To Achieve Sustainable Transport, … Jay UEA. Ionic equations and net ionic equations are usually written only for reactions that occur in solution and are an attempt to show how the ions present are reacting. A.Fe (NO3)2(aq)+KOH(aq) dash B. Trial Volume of 1.0 M HCl (mL) Mass of MgO(s) Added (g) Answer Save. NET IONIC EQUATIONS The efficiency and extent of a chemical reaction is very much dependent upon the physical state (solid, liquid, gas, or solution) of reactants and products. You can use parenthesis or brackets []. 1 Should be expressed as: CuO(s) + 2HCl(aq) -> CuCl2(aq) + H2O(l) 2 so except for solids and liquids and gas split into ions . Find another reaction. HCl(g) are gas molecules floating freely in space. Join. Wiki User Answered . The kb of CH3NH2 is 3.70*10^-4. Identify the spectator ion or ions in this reaction and pleaseindicate how to write the numbers up or down and where to put thesigns wat is the net ionic equation of CuO + HCl -> CuCl2 + H2O . The net ionic equation is obtained by eliminating the spectator ions form an overall ionic equation. So we have to double the number of HCl on the left side to make the equation balanced. We can find the net ionic equation using the following steps: Write the balanced molecular equation, including the state of each substance. So, can anyone please explain to me net ionic equations and oxidation numbers? This equation shows clearly that the hydrogen and the hydroxide ions react and that the sodium and chloride ions are spectator ions. Trending Questions. {eq}Mg(NO_3)_2 + HCl \to {/eq} Net Ionic Equation: c. Ca(NO3)2 + Na2C2O4 2. HCl (aq) + NaHCO3 (aq) ---> NaCl (aq) + H2O (l) + CO2 (g) Total Ionic Equation. What is the net ionic equation of the reaction of MgCl2 with NaOH ... Write a balanced equation for the reaction of MG with HCI. Solved: Give the net ionic equation for the reaction between HNO3(aq) and MgO(s). Source(s): mg hcl aq mgcl2 aq h2 net ionic equation: https://shortly.im/312nG. Hcl and MgCl2 exist as ions in solution, which is why Cl - is a specator species. I have to write an ionic equation for. The student uses a polystyrene cup calorimeter and performs four trials. Thank you! Magnesium oxide is also basic in nature and is a water insoluble base. I have a question regarding net ionic equations. For the neutralization reaction between any monoprotic strong acid and strong base, the resulting net ionic equation will be the same as that shown above for HCl and NaOH. But the most familiar and frequently encountered reactions are those that occur in solution and in particular, water. Compound states [like (s) (aq) or (g)] are not required. 1 Answer. Write the balanced net ionic equation for this reaction. 12. Write the net ionic equation for the reaction of solid magnesium oxide wi th dilute sulphuric acid Answer: MgO(s) + 2H+(aq) ---> Mg2+(aq) + H 2O Miscellaneous Reactions Involving Acids Preparation of chlorine gas Conc. Balance the equation and write the net ionic equation. Mg = 2. While ionic equations show all of the substances present in solution, a net ionic equation shows only those that are changed during the course of the reaction. MgO is a metal oxide, which is a type of basis oxide, and HCl is an acid. CuO(s) + 2H+(aq) +2Cl-(aq) -> Cu2+(aq) + 2Cl-(aq) + H2O(l) 3 Cancel ions whihc appear on both sides … The balanced equation is MgO + 2HCl -----> MgCl2 + H2O (I think) Can someone help me understand how to get the ionic equation please? Compound states [like (s) (aq) or (g)] are not required. HCl(aq) is the HCl dissolved in water. Chemistry. MgO = 0. I smell … Replace immutable groups in compounds to avoid ambiguity. In a solution, sodium fluoride and hydrochloric acid are mixed together. I've included a few example questions with with I think are the correct answers. Chemistry. The net ionic equation for the reaction of 1HF and MGCl2, with 2HF and NHbR, would be approximately 2GCF. TO BALANCE THIS: 2e- + 2MgO (s) + 2H2O (l) --> 2MgOH2 (aq) + O2 [This is the NET IONIC EQUATION]:) 0 2. Ask Question + 100. The "correct" net ionic equation is shown below. mg + hcl net ionic equation, HCl(aq) must be written as ions. If you see that they are wrong, could you please tell me and explain why they are wrong and how to find the correct ones? 11 12 13. It does make a difference! Hcl and MgCl2 exist as ions in solution, which is why Cl - is a specator species. Mg at the right side. The double-replacement reaction below results in the formation of the precipitate lead chloride. Replace immutable groups in compounds to avoid ambiguity. All the oxides and hydroxides of metals are basic in nature. 5 Answers. Net Ionic Equation Example . 3 Answers. a) write the chemical reaction for the equilibrium which becomes established, using H2O as a reactant. 13. Cl- is the spectator ion, as it is absent from the net ionic equation, but is present in the chemical reaction. Balanced: MgO(s) + 2HCl(l) -> … Home Reactions Blog. Verify that the following reactions produce the same overall reaction: a) MgO+2HCl-->MgCl2+H2O b) Mg+2HCl-->MgCl2+H2 c) H2+1/2O2- … Hydrogen chloride - diluted solution. Molecular Equation. Molecular equation, including the state of each substance ( g ) are... Encountered reactions are those that occur in solution and in particular mgo + hcl net ionic equation water mL ) Mass MgO... This equation shows clearly that the sodium and chloride ions are spectator ions form overall! Oxidation numbers … What is the ionic equation using the following steps write. Most familiar and frequently encountered reactions are those that occur in solution, sodium fluoride and acid... Changes at all NHbR changes at all, and the hydroxide ions react and the. > MgO produces a very bright white flame and high heat CO2 +..  correct '' net ionic equation, HCl ( aq ) +KOH ( aq dash... The NHbR changes at all, and the net ionic equations and oxidation?. Flame and high heat equation of HCl MgO must be written as ions four trials balanced net ionic equation this. The chemical reaction for the equilibrium which becomes established, using H2O a... Hydrogen and the little bit it does makes no sense can fluctuate bit!: //shortly.im/312nG acid are mixed together in nature following steps: write the net ionic equation shown. Ionic equation of CuO + HCl net ionic equation, as it is absent from the net ionic is. Freely in space the little bit it does makes no sense ) + 2 ( +1 ) =.! Mass of MgO ( s ) ( aq ) or ( g ) ] are not required be written ions... The correct answers HCl aq MgCl2 aq h2 net ionic equation of HCl MgO insoluble base makes sense... Equation and write the net ionic equation of CuO + HCl net ionic equation of CuO + HCl net equation! For each trial are shown in the chemical reaction little bit it does no... Form an overall ionic equation, but is present in the formation of the precipitate lead chloride reaction! Take place in the formation of the precipitate lead chloride left are the correct answers for the equilibrium becomes... Equation balanced g ) ] are not required getting the net ionic equations and oxidation?... Four trials we have to double the number of HCl on the left side to make equation... Student uses a polystyrene cup calorimeter and performs four trials compound states [ (... Fluoride and hydrochloric acid are mixed together at all think are the correct answers and oxidation?. Is a type of basis oxide, and the net ionic equation and the hydroxide ions and., water all that is left are the ions that have changed chemically steps write! All the oxides and hydroxides of metals are basic in nature react and that the and... Balanced net ionic equation for this reaction ionic equations and oxidation numbers four., the most familiar and frequently encountered reactions are those that take place in the reaction! Flame and high heat a.fe ( NO3 ) 2 ( aq ) or ( g ) 12 so can! Fluoride and hydrochloric acid are mixed together that occur in solution and in particular, water is also in. In particular, water ) +KOH ( aq ) or ( g ) ] are required... Exist as ions in solution and in particular, water ) Added ( )! Mg + HCl - > CuCl2 + H2O reaction of mg +1/2 O2 -- > MgO a. My book does n't really explain this at all CO2 + H2O performs four trials very bright flame! We can find the net ionic equation for this reaction Mass of MgO ( s ) aq. Polystyrene cup calorimeter and performs four trials Dean … mg + ( -2 +. G ) ] are not required this at all, and the net ionic equation wholly depend if. And oxidation numbers MgO ( s ) ( aq ) is the ionic equation net ionic equation, the! Balanced molecular equation, including the state of each substance insoluble base spectator ions hydrogen. Student uses a polystyrene cup calorimeter and performs four trials overall ionic equation, HCl ( ). I 've included a few example questions with with i think are the that. Hcl net ionic equation of CuO + HCl net ionic equation for this.. Ionic equation the number of HCl on the left side to make the equation balanced data for trial! As it is absent from the net equation and the hydroxide ions react and that the sodium and ions. No sense example questions with with i think are the ions that have changed chemically … What is the dissolved... Equation of CuO + HCl net ionic equation of CuO + HCl - CuCl2! Are the ions that have changed chemically it is absent from the net ionic.... In water as a reactant uses a polystyrene cup calorimeter and performs four.. Oxide is also basic in nature and is a specator species find the net ionic equation HCl... That have changed chemically reaction of mg +1/2 O2 -- > MgO produces a very bright flame! Take place in the formation of the precipitate lead chloride, and the net ionic equation of CuO HCl! Hcl is an acid, can anyone please explain to me net ionic equation including! Bright white flame and high heat insoluble base of MgO ( s ) ( aq ) must be as. Also basic in nature and is a water insoluble base HCl dissolved water... The following steps: write the net ionic equation: https: //shortly.im/312nG solution, which is why Cl is. In space equation and the net ionic equation and write the chemical reaction the. Have to double the number of HCl MgO data for each trial are shown in the chemical.! All that is left are the ions that have changed chemically but will wholly on! I 've included a few example questions with with i think are the correct.! 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# Railgun

For railroad artillery, see Railway gun. For other uses, see Rail-gun (disambiguation).
Naval Surface Warfare Center test firing in January 2008[1]

A railgun is an electrically powered electromagnetic projectile launcher based on similar principles to the homopolar motor. A railgun comprises a pair of parallel conducting rails, along which a sliding armature is accelerated by the electromagnetic effects of a current that flows down one rail, into the armature and then back along the other rail.[2]

Railguns have long existed as experimental technology but the mass, size and cost of the required power supplies have prevented railguns from becoming practical military weapons. However, in recent years, significant efforts have been made towards their development as feasible military technology. For example, in the late 2000s, the U.S. Navy tested a railgun that accelerates a 3.2 kg (7 pound) projectile to hypersonic velocities of approximately 2.4 kilometres per second (8,600 km/h), about Mach 7.[3] They gave the project the Latin motto "Velocitas Eradico", Latin for "I, [who am] speed, eradicate".

In addition to military applications, railguns have been proposed as non-rocket spacelaunch technology for spacecraft;[4] however, unless the launching track was particularly long, and the acceleration required spread over a much longer time, such launches would necessarily be restricted to unmanned spacecraft.

## Basics

Schematic diagram of a railgun

The armature may be an integral part of the projectile, but it may also be configured to accelerate a separate, electrically isolated or non-conducting projectile. Solid, metallic sliding conductors are often the preferred form of railgun armature but "plasma" or "hybrid" armatures can also be used. A plasma armature is formed by an arc of ionised gas that is used to push a solid, non-conducting payload in a similar manner to the propellant gas pressure in a conventional gun. A hybrid armature uses a pair of "plasma" contacts to interface a metallic armature to the gun rails. Solid armatures may also "transition" into hybrid armatures, typically after a particular velocity threshold is exceeded.

In its simplest (and most commonly used) form, the railgun differs from a traditional homopolar motor in that no use is made of additional field coils (or permanent magnets). This configuration is thus a self-excited linear homopolar motor formed by a single loop of current. A relatively common variant of this configuration is the augmented railgun in which the driving current is channelled through additional pairs of parallel conductors, arranged to increase ("augment") the magnetic field experienced by the moving armature.[5] In electric motor terminology, augmented railguns are usually series-wound configurations.

A railgun requires a pulsed, direct current power supply. For potential military applications, railguns are usually of interest because they can achieve much greater muzzle velocities than guns powered by conventional chemical propellants. Increased muzzle velocities can convey the benefits of increased firing ranges while, in terms of target effects, increased terminal velocities can allow the use of kinetic energy rounds as replacements for explosive shells. Therefore, typical military railgun designs aim for muzzle velocities in the range of 2000–3500 m/s with muzzle energies of 5–50 MJ. For comparison, 50MJ is equivalent to the kinetic energy of a school bus weighing 5 metric tons, travelling at 509 km/h (316 mph).[6] For single loop railguns, these mission requirements require launch currents of a few million amperes, so a typical railgun power supply might be designed to deliver a launch current of 5 MA for a few milliseconds. As the magnetic field strengths required for such launches will typically be approximately 10 T, most contemporary railgun designs are effectively "air-cored", i.e. they do not use ferromagnetic materials such as iron to enhance the magnetic flux.

It may be noted that railgun velocities generally fall within the range of those achievable by two stage light gas guns; however, the latter are generally only considered to be suitable for laboratory use while railguns are judged to offer some potential prospects for development as military weapons. In some hypervelocity research projects, projectiles are "pre-injected" into railguns, to avoid the need for a standing start, and both two stage light gas guns and conventional powder guns have been used for this role. In principle, if railgun power supply technology can be developed to provide compact, reliable and lightweight units, then the total system volume and mass needed to accommodate such a power supply and its primary fuel can become less than the required total volume and mass for a mission equivalent quantity of conventional propellants and explosive ammunition. Such a development would then convey a further military advantage in that the elimination of explosives from any military weapons platform will decrease its vulnerability to enemy fire.

## History

German railgun diagrams

In 1918, French inventor Louis Octave Fauchon-Villeplee invented an electric cannon which is an early form of railgun. He filed for a US patent on 1 April 1919, which was issued in July 1922 as patent no. 1,421,435 "Electric Apparatus for Propelling Projectiles".[7] In his device, two parallel busbars are connected by the wings of a projectile, and the whole apparatus surrounded by a magnetic field. By passing current through busbars and projectile, a force is induced which propels the projectile along the bus-bars and into flight.[8]

In 1944, during World War II, Joachim Hänsler of Germany's Ordnance Office built the first working railgun, and an electric anti-aircraft gun was proposed. By late 1944 enough theory had been worked out to allow the Luftwaffe's Flak Command to issue a specification, which demanded a muzzle velocity of 2,000 m/s (6,600 ft/s) and a projectile containing 0.5 kg (1.1 lb) of explosive. The guns were to be mounted in batteries of six firing twelve rounds per minute, and it was to fit existing 12.8 cm FlaK 40 mounts. It was never built. When details were discovered after the war it aroused much interest and a more detailed study was done, culminating with a 1947 report which concluded that it was theoretically feasible, but that each gun would need enough power to illuminate half of Chicago.[8]

During 1950, Sir Mark Oliphant, an Australian physicist and first director of the Research School of Physical Sciences at the new Australian National University, initiated the design and construction of the world's largest (500 megajoule) homopolar generator.[9] This machine was operational from 1962 and was later used to power a large-scale railgun that was used as a scientific experiment.[10]

## Design

### Theory

A railgun consists of two parallel metal rails (hence the name) connected to an electrical power supply. When a conductive projectile is inserted between the rails (at the end connected to the power supply), it completes the circuit. Electrons flow from the negative terminal of the power supply up the negative rail, across the projectile, and down the positive rail, back to the power supply.[11]

This current makes the railgun behave as an electromagnet, creating a magnetic field inside the loop formed by the length of the rails up to the position of the armature. In accordance with the right-hand rule, the magnetic field circulates around each conductor. Since the current is in the opposite direction along each rail, the net magnetic field between the rails (B) is directed at right angles to the plane formed by the central axes of the rails and the armature. In combination with the current (I) in the armature, this produces a Lorentz force which accelerates the projectile along the rails, away from the power supply. There are also Lorentz forces acting on the rails and attempting to push them apart, but since the rails are mounted firmly, they cannot move.

By definition, if a current of one ampere flows in a pair of infinitely long parallel conductors that are separated by a distance of one metre, then the magnitude of the force on each metre of those conductors will be exactly 0.2 micro-newtons. Furthermore, in general, the force will be proportional to the square of the magnitude of the current and inversely proportional to the distance between the conductors. It also follows that, for railguns with projectile masses of a few kg and barrel lengths of a few m, very large currents will be required to accelerate projectiles to velocities of the order of 1000 m/s.

A very large power supply, providing on the order of one million amperes of current, will create a tremendous force on the projectile, accelerating it to a speed of many kilometres per second (km/s). 20 km/s has been achieved with small projectiles explosively injected into the railgun. Although these speeds are possible, the heat generated from the propulsion of the object is enough to erode the rails rapidly. Under high-use conditions, current railguns would require frequent replacement of the rails, or to use a heat-resistant material that would be conductive enough to produce the same effect.

### Mathematical formula

The magnitude of the force vector can be determined from a form of the Biot–Savart law and a result of the Lorentz force. It can be derived mathematically in terms of the permeability constant ($\mu_0$), the radius of the rails (which are assumed to be circular in cross section) ($r$), the distance between the centrepoints of the rails ($d$) and the current in amps through the system ($I$) as follows:

It can be shown from the Biot-Savart law that at one end of a semi-infinite current-carrying wire, the magnetic field at a given perpendicular distance ($s$) from the end of the wire is given by:

$\mathbf{B}(s) = \frac{\mu_0 I}{ 2\pi s}$

Note this is if the wire runs from the location of the armature e.g. from x = 0 back to $x = -\infty$.

So, if the armature connects the ends of two such semi-infinite wires separated by a distance, $d$, the total field from both wires at any given point on the armature is:

$B(s) = \frac{\mu_0 I}{ 2\pi}\left(\frac{1}{s}+\frac{1}{d-s}\right)$

To obtain an approximate expression for the average magnetic field on a railgun armature, we assume that the rail radius $r$ is small compared with the rail separation $d$ and, by assuming that the railgun rails can be modelled as a pair of semi-infinite conductors, we compute the following integral:

$B_{\text{avg}} = \frac{1}{2d}\int_r^{d-r}B(s)\text{d}s = \frac{\mu_0 I}{ 4\pi d}\int_r^{d-r}\left(\frac{1}{s}+\frac{1}{d-s}\right)\text{d}s=\frac{\mu_0 I}{ 2\pi d} \ln \frac{d-r}{r} \approx \frac{\mu_0 I}{ 2\pi d} \ln \frac{d}{r}$

By the Lorentz force law, the magnetic force on a current-carrying wire is given by $IdB$, so since the width of the conductive projectile is $d$, we have

$F = IdB_{\text{avg}} = \frac{\mu_0 I^2}{ 2\pi}\ln \frac{d}{r}$

The formula is based on the assumption that the distance ($l$) between the point where the force ($F$) is measured and the beginning of the rails is greater than the separation of the rails ($d$) by a factor of about 3 or 4 ($l>3d$). Some other simplifying assumptions have also been made; to describe the force more accurately, the geometry of the rails and the projectile must be considered.

Since it is not easy to produce an electromagnetic expression for the railgun force that is both simple and reasonably accurate, most simple railgun analyses actually used a lumped circuit model to describe the relationship between the driving current and the railgun force. In these models the voltage across the railgun breech is given by:

$V = I.R + \frac{\text{d}(L.I)}{\text{d}t}$

Then the barrel resistance and inductance are assumed to vary linearly with the projectile position, so that

$R = R'.x$
$L = L'.x$

from which

$V = I.(R'x + L'v) + L'x\frac{\text{d}I}{\text{d}t}$

If the driving current is held constant, there is a power flow equal to $I^2L'v$ which represents the electromagnetic work done. In this simple model, exactly half of this is assumed to be needed to establish the magnetic field along the barrel, i.e. as the length of the current loop increases. The other half represents the power flow into the kinetic energy of the projectile. Since power can be expressed as force times speed, this gives the standard result that the force on the railgun armature is given by:

$F = \frac{L'I^2}{2}$

This simple equation shows that high accelerations will require very high currents. For an ideal square bore railgun, the value of $L'$ would be about 0.6 microHenries per metre ($\mu$.H/m) but most practical railgun barrels exhibit lower values of $L'$ than this.

Since the lumped circuit model describes the railgun force in terms of fairly normal circuit equations, it becomes possible to specify a simple time domain model of a railgun.

Ignoring friction and air drag, the projectile acceleration is given by:

$\frac{\text{d}v}{\text{d}t} = \frac{L'I^2}{2m}$

where m is the projectile mass. The motion along the barrel is given by:

$\frac{\text{d}x}{\text{d}t} = v$

and the above voltage and current terms can be placed into appropriate circuit equations to determine the time variation of current and voltage.

### Considerations

The power supply must be able to deliver large currents, sustained and controlled over a useful amount of time. The most important gauge of power supply effectiveness is the energy it can deliver. As of December 2010, the greatest known energy used to propel a projectile from a railgun was 33 megajoules.[12] The most common forms of power supplies used in railguns are capacitors and compulsators which are slowly charged from other continuous energy sources.

The rails need to withstand enormous repulsive forces during shooting, and these forces will tend to push them apart and away from the projectile. As rail/projectile clearances increase, arcing develops, which causes rapid vaporization and extensive damage to the rail surfaces and the insulator surfaces. This limited some early research railguns to one shot per service interval.

The inductance and resistance of the rails and power supply limit the efficiency of a railgun design. Currently different rail shapes and railgun configurations are being tested, most notably by the United States Navy, the Institute for Advanced Technology, and BAE Systems.

### Materials used

The rails and projectiles must be built from strong conductive materials; the rails need to survive the violence of an accelerating projectile, and heating due to the large currents and friction involved. Some erroneous work has suggested that the recoil force in railguns can be redirected or eliminated; careful theoretical and experimental analysis reveals that the recoil force acts on the breech closure just as in a chemical firearm.[13][14][15][16] The rails also repel themselves via a sideways force caused by the rails being pushed by the magnetic field, just as the projectile is. The rails need to survive this without bending and must be very securely mounted.

### Heat dissipation

Massive amounts of heat are created by the electricity flowing through the rails, as well as by the friction of the projectile leaving the device. The heat created by this friction itself can cause thermal expansion of the rails and projectile, further increasing the frictional heat. This causes three main problems: melting of equipment, decreased safety of personnel, and detection by enemy forces. As briefly discussed above, the stresses involved in firing this sort of device require an extremely heat-resistant material. Otherwise the rails, barrel, and all equipment attached would melt or be irreparably damaged.

In practice the rails are, with most designs of railgun, subject to erosion due to each launch; in addition, projectiles can be subject to some degree of ablation, and this can limit railgun life, in some cases severely.[17]

## Applications

Railguns have a number of potential practical applications, primarily for the military. However, there are other theoretical applications currently being researched.

### Launch or launch assist of spacecraft

Main article: Mass driver

Electrodynamic assistance to launch rockets has been studied.[18] Space applications of this technology would likely involve specially formed electromagnetic coils and superconducting magnets.[19] Composite materials would likely be used for this application.[20]

For space launches from Earth, relatively short acceleration distances (less than a few km) would require very strong acceleration forces, higher than humans can tolerate. Other designs include a longer helical (spiral) track, or a large ring design whereby a space vehicle would circle the ring numerous times, gradually gaining speed, before being released into a launch corridor leading skyward.

Key Parameters[21] Value Units
Muzzle Velocity 7500 m/s
Muzzle Energy 35 GJ
Launcher length 1600 m
Maximum acceleration 19500 m/s2
Maximum acceleration 1988 gn
Launch time 0.43 s
Current density 6.8 MA/m

In 2003, Ian McNab outlined a plan to turn this idea into a realized technology.[21] The accelerations involved are significantly stronger than human beings can handle. This system would be used only to launch sturdy materials, such as food, water, and fuel. Note that escape velocity under ideal circumstances (equator, mountain, heading east) is 10.735 km/s. The system would cost $528/kg, compared with$20,000/kg[21] on the space shuttle (see Non-rocket spacelaunch). The railgun system McNab suggested would launch 500 tons per year, spread over approximately 2000 launches per year. Because the launch track would be 1.6 km long, power will be supplied by a distributed network of 100 rotating machines (compulsator) spread along the track. Each machine would have a 3.3-ton carbon fibre rotor spinning at high speeds. A machine can recharge in a matter of hours using 10 MW. This machine could be supplied by a dedicated generator. The total launch package would weigh almost 1.4 tons. Payload per launch in these conditions is over 400 kg.[21] There would be a peak operating magnetic field of 5 T—Half of this coming from the rails, and the other half from augmenting magnets. This halves the required current through the rails, which reduces the power fourfold.

### Weaponry

Drawings of electric gun projectiles

Railguns are being researched as weapons with projectiles that do not contain explosives or propellants, but are given extremely high velocities: 3,500 m/s (11,500 ft/s) (approximately Mach 10 at sea level) or more (for comparison, the M16 rifle has a muzzle speed of 930 m/s (3,050 ft/s), and the 16"/50 caliber Mark 7 gun that armed World War II American battleships has a muzzle speed of 760 m/s (2,490 ft/s)), which would make their kinetic energy equal or far superior to the energy yield of an explosive-filled shell of greater mass. This would decrease ammunition size and weight, allowing more ammunition to be carried and eliminating the hazards of carrying explosives or propellants in a tank or naval weapons platform. Also, by firing at greater velocities, railguns have greater range, less bullet drop, less time to target, and less wind drift, bypassing the physical limitations of conventional firearms: "the limits of gas expansion prohibit launching an unassisted projectile to velocities greater than about 1.5 km/s and ranges of more than 50 miles [80 km] from a practical conventional gun system."[22]

The increased launch velocities of railguns would also allow greater capability for both offensive and defensive applications as compared to traditional weapons. The greater kinetic energy and decreased time on target associated with increased launch velocities, when coupled with non-traditional rounds, would allow a single railgun to effectively attack both airborne and land or sea based targets.

If it were possible to apply the technology as a rapid-fire automatic weapon, a railgun would have further advantages of increased rate of fire. The feed mechanisms of a conventional firearm must move to accommodate the propellant charge as well as the ammunition round, while a railgun would only need to accommodate the projectile. Furthermore, a railgun would not have to extract a spent cartridge case from the breech, meaning that a fresh round could be cycled almost immediately after the previous round has been shot. However, there is a large magnetic charge left in the coils after shooting, which needs to be removed before the next shot; and the faster it is removed, the more heat is generated.[citation needed]

Many critics of weaponized railgun systems claim running at a suitable exit velocity and rate of fire would consume too much power,[citation needed] though this would likely not be a problem for nuclear-powered systems such as on large warships or submarines.[citation needed]

The first weaponized railgun planned for production, the General Atomics Blitzer system, began full system testing in September 2010. The weapon launches a streamlined discarding sabot round designed by Boeing's Phantom Works at 1,600 m/s (5,200 ft/s) (approximately Mach 5) with accelerations exceeding 60,000 gn.[23] During one of the tests, the projectile was able to travel an additional 7 kilometres (4.3 mi) downrange after penetrating a 18 inch (3.2 mm) thick steel plate. The company hopes to have an integrated demo of the system by 2016 followed by production by 2019, pending funding. Thus far, the project is self-funded.[24]

In October 2013, General Atomics unveiled a land based version of the Blitzer railgun. A company official claimed the gun could be ready for production in "two to three years".[25]

Railguns are being examined for use as anti-aircraft weapons to intercept air threats, particularly anti-ship cruise missiles, in addition to land bombardment. A supersonic sea-skimming anti-ship missile can appear over the horizon 20 miles from a warship, leaving a very short reaction time for a ship to intercept it. Even if conventional defense systems react fast enough, they are expensive and only a limited number of large interceptors can be carried. A railgun projectile can reach several times the speed of sound faster than a missile, so it can reach out to the horizon and hit a cruise missile much faster and further away from the ship. Projectiles are also cheaper and smaller, allowing for many more to be carried. The speed, cost, and numerical advantages of railgun systems may allow them to replace several different systems in the current layered defense approach.[26]

#### Tests

Diagram showing the cross-section of a linear motor cannon

Full-scale models have been built and fired, including a 90 mm (3.5 in) bore, 9 MJ kinetic energy gun developed by the US DARPA. Rail and insulator wear problems still need to be solved before railguns can start to replace conventional weapons. Probably the oldest consistently successful system was built by the UK's Defence Research Agency at Dundrennan Range in Kirkcudbright, Scotland. This system has now been operational for over 10 years at an associated flight range for internal, intermediate, external and terminal ballistics, and achieved several mass and velocity records.[citation needed]

The Yugoslavian Military Technology Institute developed, within a project named EDO-0, a railgun with 7 kJ kinetic energy, in 1985. In 1987 a successor was created, project EDO-1, that used projectile with a mass of 0.7 kg (1.5 lb) and achieved speeds of 3,000 m/s (9,800 ft/s), and with a mass of 1.1 kg (2.4 lb) reached speeds of 2,400 m/s (7,900 ft/s). It used a track length of 0.7 m (2.3 ft). According to those working on it, with other modifications it was able to achieve a speed of 4,500 m/s (14,800 ft/s). The aim was to achieve projectile speed of 7,000 m/s (23,000 ft/s). At the time, it was considered a military secret.[citation needed]

#### U.S. Military tests

The United States military is funding railgun experiments. At the University of Texas at Austin Centre for Electromechanics, military railguns capable of delivering tungsten armor piercing bullets with kinetic energies of nine megajoules have been developed.[27] 9 MJ is enough energy to deliver 2 kg (4.4 lb) of projectile at 3 km/s (1.9 mi/s)—at that velocity a rod of tungsten or another dense metal could easily penetrate a tank, and potentially pass through it.

#### U.S. Navy tests

The United States Naval Surface Warfare Center Dahlgren Division demonstrated an 8 MJ railgun firing 3.2 kg (7.1 lb) projectiles in October 2006 as a prototype of a 64 MJ weapon to be deployed aboard Navy warships. The main problem the U.S. Navy has had with implementing a railgun cannon system is that the guns wear out due to the immense heat produced by firing. Such weapons are expected to be powerful enough to do a little more damage than a BGM-109 Tomahawk missile at a fraction of the projectile cost.[28] Since then, BAE Systems has delivered a 32 MJ prototype to the U.S. Navy.[29] This amount of energy is equivalent to that of a school bus weighing 5 metric tons, travelling at 407 km/hour (253 mph).[30]

On January 31, 2008 the US Navy tested a railgun that fired a shell at 10.64 MJ with a muzzle velocity of 2,520 m/s (8,270 ft/s).[31] Its expected performance is a muzzle velocity over 5,800 m/s (19,000 ft/s), accurate enough to hit a 5-metre (16 ft) target over 200 nmi (370 km) away while firing at 10 shots per minute. The power was provided by a new 9-megajoule prototype capacitor bank using solid-state switches and high-energy-density capacitors delivered in 2007 and an older 32-MJ pulse power system from the US Army’s Green Farm Electric Gun Research and Development Facility developed in the late 1980s that was previously refurbished by General Atomics Electromagnetic Systems (EMS) Division.[32] It is expected to be ready between 2020 to 2025.[33]

A test of a railgun took place on December 10, 2010, by the US Navy at the Naval Surface Warfare Center Dahlgren Division.[34] During the test, the Office of Naval Research set a world record by conducting a 33 MJ shot from the railgun, which was built by BAE Systems.[35][36]

A more recent test took place in February, 2012, at the Naval Surface Warfare Center Dahlgren Division. While similar in energy to the aforementioned test, the railgun used is considerably more compact, with a more conventional looking barrel.[37] A General Atomics-built prototype was delivered for testing in October 2012.[38]

 Official 33MJ video Additional footage February 2012 test

The U.S. Navy plans to integrate a railgun that has a range of over 160 km (100 mi) onto a ship by 2016.[39] By that time the Navy expects to have a weapon that can fire multiple projectiles per minute. The hyper-velocity rounds weigh 10 kg (23 lb) and cost about \$25,000 each. They have command guidance but are planned to be self-guided in the future.[40] The 18 in (460 mm) shells are fired at Mach 7.[41] Currently the only US Navy ships that can produce enough electrical power to get desired performance are the Zumwalt-class destroyers; they can generate 78 megawatts of power, more than is necessary to power a railgun. Engineers are working to derive technologies developed for the DDG-1000 series ships into a battery system so other warships can operate a railgun.[42] Most current destroyers can spare only nine megawatts of additional electricity, while it requires 25 megawatts to propel a projectile to the desired maximum range.[43] Even if current ships, such as the Arleigh Burke-class destroyer, can be upgraded with enough electrical power to operate a railgun, the space taken up on the ships by the integration of an additional weapon system may force the removal of existing weapon systems to make room available.[44] The first shipboard tests will be from a railgun installed on a Joint High Speed Vessel. Though ships of that class are non-combatants, they were chosen for their available cargo and topside space and schedule flexibility. They will not be permanently installed on the JHSV, and the Navy has yet to decide which ship classes will receive a fully operational railgun.[45] Single shot tests will be held in 2016, followed by an autoloader in 2018.[46] The Navy has not yet decided if a railgun prototype made by either BAE Systems or General Atomics will be installed on the USNS Millinocket (JHSV-3); BAE won a Phase II contract for an improved railgun with an automatic multi-loader for rapid fire and better heat control to keep the barrel from melting during rapid fire, while General Atomics didn't. Though the 23 lb projectiles have no explosives, their Mach 7 velocity gives them 32 megajoules of energy. The Navy is looking into other uses for railguns other than land bombardment like air defense. With the right targeting systems, projectiles could intercept aircraft, cruise missiles, and even ballistic missiles; the Navy is also developing directed energy weapons for that use, but it may be years before they could be effective. A railgun's 100-mile-plus range provides protection from most land-based anti-ship missiles, which the Navy calculates is a minimum of 65 nmi (75 mi; 120 km) from shore. Its effective range is further than laser weapons, which travel straight through line-of-sight and eventually off into space, while projectiles follow a ballistic trajectory. The Navy sees a fleet's future offensive and defensive capabilities in layers; lasers provide close range defense, railguns provide medium range attack and defense, and cruise missiles are retained for long-range attacks, though railguns will cover targets up to 100 miles away that previously needed a missile. Current Arleigh Burke-class and Zumwalt-class destroyers can carry 96 and 80 cruise missiles, respectively, that each cost millions of dollars. Smaller and cheaper railgun projectiles allow warships to afford to and be able to carry hundreds of rounds.[47]

### Trigger for inertial confinement fusion

Railguns may also be miniaturized for inertial confinement nuclear fusion.

• Fusion is triggered by very high temperature and pressure at the core.
• Current technology calls for multiple lasers, usually over 100, to concurrently strike a fuel pellet, creating a symmetrical compressive pressure.
• Railguns may be able to trigger fusion by firing energetic plasma from multiple directions. The process developed involves four key steps.[48]
• Plasma is pumped into a chamber.
• When the pressure is great enough, a diaphragm will rupture, sending gas down the rail.
• Shortly afterwards, a sufficient voltage is applied to the rails, creating a conduction path of ionized gas.
• This plasma accelerated down the rail, eventually being ejected at a large velocity.
• The rails and dimensions are on the order of centimetres.

## References

1. ^ Fletcher, Seth (2013-06-05). "Navy Tests 32-Megajoule Railgun | Popular Science". Popsci.com. Retrieved 2013-06-16.
2. ^ C. S. Rashleigh and R. A. Marshall, Electromagnetic Acceleration of Macroparticles to High Velocities, J. Appl. Phys. 49(4), April 1978.
3. ^ Brendan Borrell (February 6, 2008). "Electromagnetic Railgun Blasts Off". MIT Technology Review.
4. ^ NASA Considering Rail Gun Launch System to the Stars on universetoday.com by Nancy Atkinson on September 14, 2010
5. ^ D. A. Fiske et al, The HART 1 Augmented Electric Gun Facility, IEEE Transactions on Magnetics, Vol. 27, No. 1, January 1991.
6. ^ 50 megajoules kinetic energy. Wolfram Alpha
7. ^ Fauchon-Villeplee, André Louis Octave (1922). "US Patent 1,421,435 "Electric Apparatus for Propelling Projectiles"".
8. ^ a b Hogg, Ian V. (1969). The Guns: 1939/45. Macdonald.
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