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text:
# MSL - Mean sea level

The mean sea level is defined in IEC-61400-3-1 (2019) as the "average level of the sea over a period of time long enough to remove variations due to waves, tides and storm surges".

format: plain


text: dc.contributor.author Stoer, J.R.A. Tapia 2018-06-18T17:23:57Z 2018-06-18T17:23:57Z 1984-07 Stoer, J. and R.A. Tapia. "On the Characterization of Q-superlinear Convergence of Quasi-Newton Methods for Constrained Optimization." (1984) https://hdl.handle.net/1911/101574. https://hdl.handle.net/1911/101574 In this paper we present a short, straightforward and self-contained derivation of the Boggs-Tolle-Wang characterization of those quasi-Newton methods for equality constrained optimization which produce iterates which are q-superlinearly convergent. 10 pp On the Characterization of Q-superlinear Convergence of Quasi-Newton Methods for Constrained Optimization Technical report July 1984 TR84-02 Text


text:
# How to change the position of underset?

In the command:

\underset{t \rightarrow x}{\limsup} f(t) = \underset{t \rightarrow x}{\liminf} f(t)

I get the {t \rightarrow x} lower under the limsup then the liminf (due to the "p" in limsup).

How can I change the position of the underset?

-
A tip: If you indent lines by 4 spaces, then they are marked as a code sample. You can also highlight the code and click the "code" button (with "101010" on it). –  Caramdir Mar 10 '11 at 19:18
Shouldn't that be written as \limsup_{t \rightarrow x} f(t) = \liminf_{t \rightarrow x} f(t)? Doesn't solve the problem though. –  Caramdir Mar 10 '11 at 19:23
I tried that first, but it placed the {t \rightarrow x} to the right of limsup. –  milcak Mar 10 '11 at 19:24
That is usually by design in inline mode. You can override that behavior with \limsup\limits_{t \rightarrow x} f(t). –  Caramdir Mar 10 '11 at 19:27
Very much related: Supremum of the infimum (possible duplicate). –  Hendrik Vogt Mar 11 '11 at 7:45

A phantom could be useful:

\documentclass{article}
\usepackage{amsmath}

\begin{document}

$\underset{t \rightarrow x}{\limsup} f(t) = \underset{t \rightarrow x}{\liminf\vphantom{p}} f(t)$

\end{document}

-
Yes that works thank you! Is there a general way of adjusting the position though? –  milcak Mar 10 '11 at 19:17
you can also use a \mathstrut in lieu of the phantom, nothing wrong with your solution either. –  Yiannis Lazarides Mar 10 '11 at 19:22
How does the \mathstrut work? –  milcak Mar 10 '11 at 19:23
@Yiannis: yes. That will also work. –  Gonzalo Medina Mar 10 '11 at 19:30


text:
# Estes knock-offs

## Are You Gonna Be at NERRF3

• ### What's a NERRF?

Results are only viewable after voting.

#### shrox

##### Well-Known Member
I have noticed allot of knock-offs of old Estes stuff being sold by various vendors. I have been thinking about it, and to me it seems less than proper.

When I got back into rocketry in 1997 or so, I was dismayed at the selection available. Little more than the same base 3FNC's with different stickers and colored plastic. Most of the classics where not available. I cloned a few from memory, and then I designed my own. People said "Hey cool, where can I get one?" Then I began selling my own designs.

Making a clone of a long time OOP kit for personal fun and flight is fine I think. Taking something that is known to be a design of a well known company is a little crass, especially if you call it the same thing! (A scale model is something different, using parts from a known company is acceptable too, so long as it is fairly generic, ogive cones, conical cones, etc...)

Honestly, it is an homage to build and fly a clone of a favorite kit, it is in bad taste to make and sell them.

shrox

#### Chilly

##### Well-Known Member
That's an interesting point, but there's nothing wrong with selling clones. Seems to me that if the manufacturer is clear about it, there's a market for it, and (most important) Estes doesn't make a legal challenge, then have at it. Personally, I'm thinking real hard about treating myself to a TP Mars Lander for my birthday...

Bottom line is that Estes apparently hasn't seen fit to sic the lawyers on the Cloners. So if they don't have a problem with it, why should anyone else? I wish they'd pay attention to the obvious demand out there and start issuing some of those classics. Interceptors and Saturn IB's immediately come to mind.

#### wwattles

##### Well-Known Member
I'm of the opinion that kits should fall into the realms of pseudo-copyrights. If Estes (or any other manufacturer, for that matter) has deemed it unprofitable to keep producing a kit, for whatever reason, they have the right to do so. If I was looking to get into the business of making clones, I'd probably wait a respectful time for those remnants to clear the shelves at stores (2-3 years or so) before starting to clone. Otherwise I'd feel like I was somehow robbing the originators of business that they have earned.
I'd also probably contact the original makers and let them know that I intend to start cloning, and maybe even buy up some of their kit-specific parts (odd-shaped nose cones come to mind).

WW

#### wyldbill

##### Well-Known Member
Originally posted by shrox
When I got back into rocketry in 1997 or so, I was dismayed at the selection available. Little more than the same base 3FNC's with different stickers and colored plastic. Most of the classics where not available. I cloned a few from memory, and then I designed my own.
A lot of other people are in the same boat. They remember great kits, and there just aren't many left. This seems to be changing. Estes is putting out some interesting stuff and we've got Fliskits and some others who are concentrating on originals. And we support them (big time). There's still a pretty serious vaccuum tho'. Folks want to build the rocket that they couldn't afford or didn't have the skill to build when they were young.

Originally posted by shrox
Making a clone of a long time OOP kit for personal fun and flight is fine I think. Taking something that is known to be a design of a well known company is a little crass, especially if you call it the same thing!.....Honestly, it is an homage to build and fly a clone of a favorite kit, it is in bad taste to make and sell them.
Funny thing here. None of the cloners I'm aware of are presenting clones as THEIR designs. Everyone knows they're Estes remakes. Estes is not losing any business as they don't offer anything comparable. When they do reissue popular kits, they sell out. Tried to find an ESTES OT lately? They're selling for over sticker on the 'bay. We support Estes as well, I've bough multiples of the new Renegade and Outlander. When I buy a clone kit, I'm paying for convenience. I could go to BMS for cones, TT for tube, TP for decals, here for rings, there for hooks, and end up paying through the nose for shipping unless I've planned for multiple clones and bought in lots. The cloner buys in bulk for me and sells everything I need in a package. The kit isn't any less "real" or less difficult to build. The only work I don't have to do is track down and order the parts (and cut tubes in some cases). Clones are more expensive than the actual parts, but I'm willing to pay for the convenience. I really don't see a difference between buying a kit and cloning it myself, the outcome is the same. The cloner isn't making money by ripping off Estes, they're providing a service that some are willing to pay for.

#### bsexton

##### Well-Known Member
This seems to have stirred up some emotional opinions. Simply stated I believe if you are cloning a kit no longer in production for your personal use it is OK. If you are a vendor doing it for profit ask for permission. It doesn't seem that hard to do, especially since Estes is still in business.

#### GlennW

##### Well-Known Member
I am one of those that responded that it is wrong to sell a clone. However, I took that to mean a currently produced kit. I see no problem with selling a clone kit of an OOP model as long as you have permission. As a matter of fact, one of the posters on the board has been talking of making a new Gemini Titan kit, which I would eagerly buy.

Glenn

#### sandman

##### Well-Known Member
Well, now scale jobs...those really wouldn't apply as clones would they?

Unless it was exactly the same as the original but would it???

hmmm...this is an interesting twist.

Does NASA own the copyrights on the Little Joe II? I don't think so.

Interesting twist on the cloning issue...scale that is.

sandman

##### Well-Known Member
Originally posted by sandman
Well, now scale jobs...those really wouldn't apply as clones would they?

Unless it was exactly the same as the original but would it???

hmmm...this is an interesting twist.

Does NASA own the copyrights on the Little Joe II? I don't think so.

Interesting twist on the cloning issue...scale that is.

sandman
So guess we should call them...SCONES!

#### jflis

##### Well-Known Member
scale models wouldn't apply.

While the internal workings and such may (probably *are*) protected and/or classified, the shape of the rocket is public domain (heck, YOU paid for the darned thing! )

The only risk is if you were foolish enough to just photo copy the Estes instruction set and/or cover art. Other than that, you're cool.

Interesting point here... ...Ever notice that when Estes (or anyone else for that matter) produced a scale kit (like the Saturn V), they would put the TM symbol on the name... ...hard to trademark a public domain term, 'specially one already in use by the originating entity...

#### loopy

##### Well-Known Member
I selected Shrox is a Rocket Deity in the Dark Art of Rocketry. His powers are great...

This is kind of an unfair/loaded question because there is a distinction between cloning OOP kits with permission (seems to be the bulk of what we see) versus cloning something in production with no permission, and submitting it for direct competition with the creating entity. I don't think I'd want to have to compete against my own intellectual property...

That being said, it seems as though proper steps have been taken (at least by the entities on this board) to ensure proper credit is given, and they are not setting up a competitive environment.

Oh, and Shrox is a Rocket Deity in the Dark Art of Rocketry. His powers are great...yes, they are...lol Shrox - I'm glad you joined our little microcosm of rocketry fanatics - you certainly fit in here...

#### cydermaster

##### Well-Known Member
Originally posted by Loopy
Shrox - I'm glad you joined our little microcosm of rocketry fanatics - you certainly fit in here...
Amen Brother Loopy!


text:
# Simplify 4c+3(2+c) giving step-wise details?

Feb 24, 2017

#### Explanation:

$4 c + 3 \left(2 + c\right)$

Step (1) = $4 c + \underline{6 + 3 c}$ - Distributive Property

Step (2) = $4 c + \underline{3 c + 6}$ - Commutative Property of addition

Step (3) = $\left(4 c + 3 c\right) + 6$ - regrouping similar monomials

Step (4) = $\left(4 + 3\right) c + 6$ - Distributive property

Step (5) = $7 c + 6$ - Simple addition


text:
# If two dice are thrown $5$ times, what is the probability of obtaining the same number on both dice $2$ times?

How do I go about solving this question on probability?

Two dice are thrown 5 times. Calculate the probability to obtain the same number on both dice 2 times.

One thing i know that to obtain any number on one die is 1/6. Need help in moving ahead.

• Can you solve: "Two dice are thrown 1 time. Calculate the probability to obtain the same number on both dice"?
– Did
Apr 16, 2017 at 9:08
• Then, can you solve: "The same experiment, with probability of success p, is performed 5 times. Calculate the probability that it is successful 2 times"?
– Did
Apr 16, 2017 at 9:10

$\frac 6 {36}$ of these rolls have the dice the same number, $\{11,22,33,44,55,66\}$, and $\frac {30} {36}$ don't have the same number.
Now think about tossing two dice 5 times. How can you get matching dice exactly two of the five tosses? First and second toss of the two dice or first and third toss or...or fourth and fifth toss. There are $\binom 5 2=10$ ways to choose two of five tosses. The rest of the tosses (three of them) will have non-matching dice. Each of the two tosses with matching dice will give a probability of $\frac 6 {36}$. Each of the three tosses with different dice will give $\frac {30} {36}$.
We end up with our probability. $\binom 5 2(\frac 6 {36})^2(\frac {30} {36})^3$


text:
# A bullet-knot-block system with momentum

1. Sep 15, 2014

### B3NR4Y

1. The problem statement, all variables and given/known data
You shoot a 0.0050-kg bullet into a 2.0-kg wooden block at rest on a horizontal surface (Figure 1) . After hitting dead center on a hard knot that runs through the block horizontally, the bullet pushes out the knot. It takes the bullet 1.0 ms to travel through the block, and as it does so, it experiences an x component of acceleration of -4.5 × 105 m/s2. After the bullet pushes the knot out, the knot and bullet together have an x component of velocity of +10 m/s. The knot carries 10% of the original inertia of the block.
Using conservation of momentum, compute the final velocity of the block after the collision.
2. Relevant equations
$p=mv$
$p_{T1} = p_{T2}$

3. The attempt at a solution

The first part of the problem had me calculate the initial velocity of the bullet, which I did and got a correct answer of 460 m/s. Now I am not sure what to do with the problem because all of the problems we had before had instantaneous collisions, not collisions that happen over time, which I assume has integrals involved, so would I integrate over the time given? I assume the step is a pretty easy and I am just missing what to do. I don't know if I should calculate the momentum the moment to block hits and after the block hits and then find out the velocity of the block, or if since there is a time where the bullet is inside of the block and pushing the "knot" if that causes any significant change in the velocity.

2. Sep 15, 2014

### collinsmark

That looks correct to me.

Very nice.

Conservation of momentum applies whether the collision is instantaneous or not (and of course, assuming there are no external forces or torques). The fact that this collision is not instantaneous doesn't matter. Conservation of momentum still applies.

Just take the system's total momentum before the collision [Edit: i.e., before the bulled strikes the block], and set that equal to the system's total momentum after the collision [Edit: i.e., after the collision is completely finished]. There is only one unknown variable: the velocity of the remainder of the block. Solve for that.

[Edit: btw, no integration is necessary.]

Last edited: Sep 15, 2014
3. Sep 15, 2014

### BvU

Hi B,

If it says "Using conservation of momentum, it probably means "Using conservation of momentum", so why not try that ? Write down an equation for conservation of momentum.

Done? OK, let's continue then:
For the momentum of the bullet before the collision you have all you need.
For the knot too.
For the rest of the block also.
Personally, I don't understand why you don't show that intermediate result.

Now comes the tough question: what information is really needed to "compute the final velocity of the block after the collision" ?

 boy, am I a slow typist...

4. Sep 15, 2014

### B3NR4Y

Okay so since momentum is an extensive quantity, $p_{total, i} = p_{block, i} + p_{bullet, i}$ for the initial momentum, and for the final momentum $p_{total, f} = p_{block, f} + p_{bullet+block \, piece}$ and using the definition of momentum, and knowing that hte piece of the block and bullet's masses are (10% of 4 kg) + 0.0050 kg, or 0.405 kg, and I get the mass of the block to be 3.6 kg, and my equations become:

$0.0050 (kg) \, * \, 460 (m/s) = 3.6 (kg) * v_{f} + 4.05 kg*m/s$

and solving for vf I get

$(2.3 kg*m/s \, - \, 4.05 kg*m/s)/3.6(kg) = v_{f}$

This throws me off cause it is negative, I might be thinking about the problem wrong, but a negative answer doesn't make sense to me. I got an answer of ~-0.49 [edit: which I entered into the program and is wrong]

Last edited: Sep 15, 2014
5. Sep 15, 2014

### BvU

What was the mass of the block ?

 kudos for smelling something wrong, though!

6. Sep 15, 2014

### B3NR4Y

I feel incredibly dumb. 2 kilograms, not 4

[edit: I learned a valuable lesson in reading problems today]

7. Sep 15, 2014

### BvU

Happens all the time. You're in good company. Finding your own errors and fixing them makes you a good scientist in the long run. (ahem...)

8. Sep 15, 2014

### B3NR4Y

Yes, thanks for your help! I'm quickly starting to learn in this physics course that if an answer seems wrong, it probably is.

9. Sep 21, 2015

### Molly Swanson

How do you find initial velocity?

10. Sep 22, 2015

### BvU

For the block it is a given, so I suppose you mean for the bullet ...
Write out some conservation of momentum equations and then work backwards.

11. Sep 22, 2015

### B3NR4Y

Thanks for your help! But this is a year old, this homework is long over lol

12. Sep 23, 2015


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# Is the speed of all waves constant?

1. Dec 21, 2015

### mrsmitten

the equation for wave velocity is velocity = wavelength * frequency. now by using the Doppler equation for change in frequency and the Doppler equation for change in wavelength. if you multiply those two equations together you get. frequency' *wavelength' = original frequency * original wavelength.

2. Dec 21, 2015

### Staff: Mentor

Welcome to the PF. Is there a question you would like to ask?

3. Dec 21, 2015

### mrsmitten

based on the derived equation does it mean that all waves have a constant velocity for all reference frames.

4. Dec 21, 2015

### Staff: Mentor

In the relativistic Doppler effect for light (and other electromagnetic radiation) in a vacuum, what you wrote in the first post is true. For other kinds of waves (mechanical waves, sound waves,...) it is not true. In fact, for light traveling through a medium it is not true. This was measured experimentally by Fizeau as far back as 1851:

https://en.wikipedia.org/wiki/Fizeau_experiment

5. Dec 22, 2015

### mrsmitten

so I should be able to take any sound and measure its frequency and wavelength to determine if the source is moving toward or away from me. I also have no previous knowledge of the sources original frequency. because if the source is moving the wave velocity will be different then if it was stationary.

6. Dec 22, 2015

### alw34

no, the frequency will be different.

If you ran toward a sound source, would that change it's velocity through air?

An electromagnetic wave always travels at 'c' in a vacuum. A sound wave always travels about 1126 ft per second in dry air regardless of the source speed. [says Wikipedia] The frequency varies. See the following.

Electromagnetic wave:
"A Doppler radar is a specialized radar that uses the Doppler effect to produce velocity data about objects at a distance. It does this by bouncing a microwave signal off a desired target and analyzing how the object's motion has altered the frequency of the returned signal.

Sound wave:
"The Doppler effect (or Doppler shift), named after Austrian physicist Christian Doppler who proposed it in 1842, is the difference between the observed frequency and the emitted frequency of a wave for an observer moving relative to the source of the waves. It is commonly heard when a vehicle sounding a siren approaches, passes and recedes from an observer. The received frequency is higher (compared to the emitted frequency) during the approach.."

7. Dec 22, 2015

### Staff: Mentor

There is one really important difference between these two statements. The speed of sound in dry air is relative to the speed of the air, so if the receiver is moving relative to the air, the sound wave will not be moving at 1126 fps relative to them. However, the speed of light in a vacuum is c for all observers regardless of their motion (it kind of has to be because there's no such thing as a speed relative to vacuum - there's nothing in the vacuum to be relative to).

8. Dec 22, 2015

### Ibix

Mechanical waves move at a constant speed with respect to the medium in which they propagate (all things being equal). So sound travels at 330m/s (give or take - it depends on temperature, among other things) relative to the air. If you run through the air, the speed of sound relative to you can change. If you are doing 10m/s relative to the air, the speed of a sound wave relative to you could be anywhere between 320m/s and 340m/s. You can even go supersonic - travel faster through the medium than the wave can propagate.

Light is different. It has no medium, and it travels at 3x108m/s regardless of your state of motion. If you are moving towards the source, light passes you at 3x108m/s. If you are moving away from the source, light passes you at 3x108m/s.

Edit: Beaten to it by Nugatory, appropriately enough.

9. Dec 22, 2015

### mrsmitten

The thing is that the velocity of waves is measured with the equation $\nu=f*\lambda$. It is not measured with $v= \nu \pm v_{o}$. Correct me if i am wrong the Doppler equations are
$f'=f \frac{\nu \pm v_{o}}{\nu \pm v_{s}}$
and
$\lambda '=\lambda \frac{\nu \pm v_{s}}{\nu \pm v_{o}}$
To find the wave velocity you would use the equation
$v'=f'*\lambda'$
Which if you multiply $f'*\lambda'$ it goes to.
$f'* \lambda' = f* \lambda$

Let’s say that you here a frequency sound in the distance. Which equation would you use to determine the wave velocity? Could you be able to tell if the source was moving?

10. Dec 23, 2015

### Ibix

Not without extra information. Your velocity with respect to the medium and the original frequency, off the top of my head.

Where did you get your expressions for the Doppler effect (link, or your working)? They aren't trivially related to the expressions I derive, and they don't match (for example) https://en.wikipedia.org/wiki/Relat...#Systematic_derivation_for_inertial_observers. Also, they have the implausible property you describe, that the speed of sound is invariant in all frames. If that were the case, an aircraft travelling at Mach 1 would be paradoxical.

Last edited: Dec 23, 2015
11. Dec 23, 2015

### mrsmitten

Exactly there is no difference between a 440Hz sound with both observer and source stationary and a 440Hz sound that is produced by relative motion. According to what I have read hear you would not need any extra information because just from the sound you can measure the frequency and wavelength. If the wave velocity $v=f*\lambda$ is different from a source that is stationary then the source is moving. Don't take the last sentences out of the context of this paragraph.

In any fundamentals of physics book that derives the Doppler effect you can get the equations from there. the standard equations that is used to convert from frequency shift to wavelength shift is $f=\frac{\nu}{\lambda}$ and $f'=\frac{\nu}{\lambda'}$. That is how the wavelength equation is derived for the relativistic equation. here is a link that describes Doppler wavelength change http://hyperphysics.phy-astr.gsu.edu/hbase/sound/dopp.html#c2. Under the Doppler wavelength change section there are these two equations.
Source receding
$\lambda'=\frac{\nu +v_{s}}{\nu }*\lambda_{s}$
Source approaching
$\lambda'=\frac{\nu -v_{s}}{\nu }*\lambda_{s}$
Which can be combined to say if only the source is moving
$\lambda'=\frac{\nu \pm v_{s}}{\nu }*\lambda_{s}$
The equation is not complete because they did not address a moving observer.
The definition of wave speed is $\v=\lambda*f$ which applies to all waves.
so the equation for speed of wave with source approaching the observer is
$v=\lambda' * f' = \frac{\nu -v_{s}}{\nu }*\lambda_{s} *\frac{\nu}{\nu-v_{s}}*f_{s} = \lambda{s} * f_{s}$
Which means that the source would measure the same wave speed as the observer.
As for the derivation for inertial observers it proves my point to.
$f'=\gamma (1-\frac{v_o}{\nu })*f$
$\lambda'=\frac{1}{\gamma} (\frac{1}{1-\frac{v_o}{\nu }})*\lambda$
$f'*\lambda'=f*\lambda$
Waves do not have mass. Waves transport energy not mass.
Can i see the expressions that you derive. Why would an aircraft travelling at Mach 1 be paradoxical

Last edited: Dec 23, 2015
12. Dec 23, 2015

### Ibix

Sorry - I assumed you'd messed up the relativistic formula for Doppler, but you're using non-relativistic physics.

Everybody agrees on the wavelength, but they disagree on the frequency. This gets you a different wave speed for observers in relative motion. Think of the wave crests travelling along. If they're 1m apart, they're 1m apart. You or me moving doesn't change that. It does change how long it takes for each subsequent crest to overhaul us, which is where the frequency change comes in.

The expression for Doppler wavelength change is the change in the wavelength emitted by a particular source when it starts moving relative to the medium. The expression for Doppler frequency change is the change in the frequency received when the source and receiver are in relative motion. These are not quite the same thing. Simply dividing the one by the other is implicitly assuming that the receiver is at rest with respect to the medium.

The reason that I described an aircraft doing Mach 1 as paradoxical is that it is travelling at the same speed as sound. That is, the speed of sound is zero from this perspective. Yet according to your formula, it should measure the speed of sound to be 330m/s. That's a paradox - and that's one reason you can't travel at the speed of light, which genuinely is invariant.

13. Dec 25, 2015

### Staff: Mentor

Here is a good page that explains how to unify the Doppler shift for light and for sound.

http://mathpages.com/rr/s2-04/2-04.htm

You can use $v=f*\lambda$ if you like, but the corresponding expression for $\lambda$ is not very simple in any reference frame except for the rest frame of the medium. In other frames $\lambda$ depends on the velocity of the medium as well as the direction of propagation.


text:
My Social Security Card Number

I have forgotten my social security card number. All I remember is that it is the largest integer with the property that the block of any two of its digits that are adjacent is either a two-digit prime number or a two-digit perfect square, and that these are all different.

What is it?

• That's a very interesting thing to remember about your social security card number ... – Rand al'Thor Nov 3 '18 at 16:38
• Do we know how many digits is the number? – Rand al'Thor Nov 3 '18 at 16:39
• @Randal'Thor: No, I cannot remember that either. Quite a few... – Bernardo Recamán Santos Nov 3 '18 at 16:43
• Do the 2-digit blocks overlap? Can a 2-digit number have a leading 0? – Weather Vane Nov 3 '18 at 17:42
• For no particular reason... What is your mother's maiden name and what was the name of your first/current/favorite pet? :) – Chowzen Nov 3 '18 at 18:18

253797364171613119

Proof:

We cannot use zeros; a two digit number ending in zero is not prime and 00 is the only square, so zeros can only appear at the beginning and can therefore be ignored.

Let's list the squares we can use: 16, 25, 36, 49, 64, 81
and the primes: 11, 13, 17, 19, 23, 29, 31, 37, 41, 43, 47, 53, 59, 61, 67, 71, 73, 79, 83, 89, 97

It's a bit like:

domino: to maximize the total number, we first need to maximize the number of pairs we can make.

See the following table:

. start end match
1 5 6 5
2 3 0 0
3 3 6 3
4 4 1 1
5 2 1 1
6 3 2 2
7 3 5 3
8 3 0 0
9 1 6 1

So we can make

16 matches, for a 18 digit Social Security Card Number.

In theory, we want

the largest numbers to appear first, but the only pair ending in 5 is 25.


text:
Figure 1: Plot of the blackbody fraction vs. λT product.

It is often necessary or useful to find the fraction of blackbody spectral emissive power that lies within a particular range of wavelength. This is most conveniently done through use of the following equation:

$\frac{{{E_{\lambda b}}}}{{{T^5}}} = \frac{{2\pi {C_1}}}{{{{\left( {\lambda T} \right)}^5}\left( {{e^{{C_2}/\lambda T}} - 1} \right)}}$

from Planck distribution. Integration of that equation over the range $0 \le \left( {\lambda T} \right)^{*} \le \lambda T$ results in the fraction of blackbody emission in that range as

$\begin{array}{l} {F_{0 - \lambda T}} = \int_{\left( {\lambda T} \right)^{*} = 0}^{\lambda T} {\frac{{{E_{\lambda b}}}}{{{T^5}}}} d(\lambda T)^{*}/\int_{\left( {\lambda T} \right)^{*} = 0}^\infty {\frac{{{E_{\lambda b}}}}{{{T^5}}}} d(\lambda T)^{*} \\ = \frac{1}{\sigma }\int_{\left( {\lambda T} \right)^{*} = 0}^{\lambda T} {\frac{{2\pi {C_1}}}{{\left( {\lambda T} \right)^{*}{^5}\left( {{e^{{C_2}/\left( {\lambda T} \right)^{*}}} - 1} \right)}}d\left( {\lambda T} \right)^{*}} \\ \end{array} \qquad \qquad(1)$

This can be integrated by parts using the same change of variables as for

${E_b} = \int_{\lambda = 0}^\infty {\frac{{2\pi {C_1}}}{{{\lambda ^5}\left( {{e^{{C_2}/\lambda T}} - 1} \right)}}d\lambda } = \frac{{2\pi {C_1}{T^4}}}{{C_2^4}}\int_{\xi = 0}^\infty {\frac{{{\xi ^3}}}{{{e^\xi } - 1}}} d\xi = \sigma {T^4}$

from Planck distribution and its consequences (Chang and Rhee, 1984)) to give the analytical relation

${F_{0 - \lambda T}} = \frac{{15}}{{{\pi ^4}}}\sum\limits_{n = 1}^\infty {\left[ {\frac{{{e^{ - n\xi }}}}{n}\left( {{\xi ^3} + \frac{{3{\xi ^2}}}{n} + \frac{{6\xi }}{{{n^2}}} + \frac{6}{{{n^3}}}} \right)} \right]} \qquad \qquad(2)$

where$\xi \equiv {C_2}/\lambda T$. Equation (2) converges rapidly, generally requiring only a few terms for accurate evaluation. Keeping only five terms in the summation gives results accurate to five significant figures up to F0-λT of at least 0.99812 (at λT = 100,000), and four terms is accurate to five significant figures up to λT = 6,000, where F0-λT = 0.73778. Figure 1 is a plot of eq. (1) or (2).

Given eq. (1), the blackbody emissive power in any interval (λT)1 to (λT)2 can be found from

$\begin{array}{l} {F_{{{\left( {\lambda T} \right)}_2} - {{\left( {\lambda T} \right)}_1}}} = \frac{1}{\sigma }\int_{\left( {\lambda T} \right)^{*} = {{\left( {\lambda T} \right)}_1}}^{{{\left( {\lambda T} \right)}_2}} {\frac{{2\pi {C_1}}}{{\left( {\lambda T} \right){*^5}\left( {{e^{{C_2}/\left( {\lambda T} \right)^{*}}} - 1} \right)}}d\left( {\lambda T} \right)^{*}} \\ = \frac{1}{\sigma }\left[ {\int_{(\lambda T)^{*} = 0}^{{{(\lambda T)}_2}} {\frac{{2\pi {C_1}}}{{(\lambda T)^{*}{^5}[{e^{{C_2}/(\lambda T)^{*}}} - 1]}}d(\lambda T)^{*}} - \int_{(\lambda T)^{*} = 0}^{{{(\lambda T)}_1}} {\frac{{2\pi {C_1}}}{{(\lambda T)^{*}{^5}[{e^{{C_2}/(\lambda T)^{*}}} - 1]}}d(\lambda T)^{*}} } \right] \\ = {F_{0 - {{(\lambda T)}_2}}} - {F_{0 - {{(\lambda T)}_1}}} \\ \end{array} \qquad \qquad(3)$

Interpreting eq. (3) graphically, Fig. 2 shows the difference in the blackbody normalized emissive power (the cross-hatched area) in the range

${{\lambda T}_1} \le \lambda T \le {{\lambda T}_2}$

If this is divided by the area under the entire curve (which has a value of σ), then the areas are the F0-λT values on the right-hand-side of eq. (3).

## References

Chang, S.L. and Rhee, K.T., 1984, ‘’Blackbody Radiation Functions,” Int. Comm. Heat and Mass Transfer, Vol. 11, pp. 451-455.

Faghri, A., Zhang, Y., and Howell, J. R., 2010, Advanced Heat and Mass Transfer, Global Digital Press, Columbia, MO.


text:
101 Sequence Detector Using Moore Machine

In the theory of computation, a Moore machine is a finite-state machine whose output values are determined only by its current state. This is in contrast to a Mealy machine, whose (Mealy) output values are determined both by its current state and by the values of its inputs. The Moore machine is named after Edward F. Moore, who presented the concept in a 1956 paper, “Gedanken-experiments on Sequential Machines.”[1]

Sequence Detector Moore AIM: Design a controller that detects the overlapping sequence “0X01” in a bit stream using moore machine. DESIGN Verilog Program- Sequence Detector 0x01 Moore implementation `timescale 1ns / 1ps ///// // Company: TMP. A sequence detector is a sequential state machine. In a Moore machine, output depends only on the present state and not dependent on the input (x). Hence in the diagram, the output is written with the states. The state diagram of a moore machine for a 101 detector is.

Formal definition

A Moore machine can be defined as a 6-tuple${displaystyle (Q,q_{0},Sigma ,O,delta ,G)}$ consisting of the following:

• A finite set of states${displaystyle Q}$
• A start state (also called initial state) ${displaystyle q_{0}}$ which is an element of ${displaystyle Q}$
• A finite set called the input alphabet${displaystyle Sigma }$
• A finite set called the output alphabet${displaystyle O}$
• A transition function${displaystyle delta :Qtimes Sigma rightarrow Q}$ mapping a state and the input alphabet to the next state
• An output function ${displaystyle G:Qrightarrow O}$ mapping each state to the output alphabet

A Moore machine can be regarded as a restricted type of finite-state transducer.

Visual representation

Table

State transition table is a table showing relation between an input and a state.[clarification needed]

Diagram

The state diagram for a Moore machine or Moore diagram is a diagram that associates an output value with each state.Moore machine is an output producer.

Relationship with Mealy machines

As Moore and Mealy machines are both types of finite-state machines, they are equally expressive: either type can be used to parse a regular language.

The difference between Moore machines and Mealy machines is that in the latter, the output of a transition is determined by the combination of current state and current input (${displaystyle Stimes Sigma }$ as the input to ${displaystyle G}$), as opposed to just the current state (${displaystyle S}$ as the input to ${displaystyle G}$). When represented as a state diagram,

• for a Moore machine, each node (state) is labeled with an output value;
• for a Mealy machine, each arc (transition) is labeled with an output value.

Every Moore machine ${displaystyle M}$ is equivalent to the Mealy machine with the same states and transitions and the output function ${displaystyle G(s,sigma )rightarrow G_{M}(delta _{M}(s,sigma ))}$, which takes each state-input pair ${displaystyle (s,sigma )}$ and yields ${displaystyle G_{M}(delta _{M}(s,sigma ))}$, where ${displaystyle G_{M}}$ is ${displaystyle M}$'s output function and ${displaystyle delta _{M}}$ is ${displaystyle M}$'s transition function.

However, not every Mealy machine can be converted to an equivalent Moore machine. Some can be converted only to an almost equivalent Moore machine, with outputs shifted in time. This is due to the way that state labels are paired with transition labels to form the input/output pairs. Consider a transition ${displaystyle s_{i}rightarrow s_{j}}$ from state ${displaystyle s_{i}}$ to state ${displaystyle s_{j}}$. The input causing the transition ${displaystyle s_{i}rightarrow s_{j}}$ labels the edge ${displaystyle (s_{i},s_{j})}$. The output corresponding to that input, is the label of state ${displaystyle s_{i}}$.[2] Notice that this is the source state of the transition. So for each input, the output is already fixed before the input is received, and depends solely on the present state. This is the original definition by E. Moore. It is a common mistake to use the label of state ${displaystyle s_{j}}$ as output for the transition ${displaystyle s_{i}rightarrow s_{j}}$.

Examples

Types according to number of inputs/outputs.

Simple

Simple Moore machines have one input and one output:

• edge detector using XOR
• clocked sequential systems (a restricted form of Moore machine where the state changes only when the global clock signal changes)

Most digital electronic systems are designed as clocked sequential systems. Clocked sequential systems are a restricted form of Moore machine where the state changes only when the global clock signal changes. Typically the current state is stored in flip-flops, and a global clock signal is connected to the 'clock' input of the flip-flops. Clocked sequential systems are one way to solve metastability problems. A typical electronic Moore machine includes a combinational logic chain to decode the current state into the outputs (lambda). The instant the current state changes, those changes ripple through that chain, and almost instantaneously the output gets updated. There are design techniques to ensure that no glitches occur on the outputs during that brief period while those changes are rippling through the chain, but most systems are designed so that glitches during that brief transition time are ignored or are irrelevant. The outputs then stay the same indefinitely (LEDs stay bright, power stays connected to the motors, solenoids stay energized, etc.), until the Moore machine changes state again.

101 Sequence Detector

A sequential network has one input and one output. The output becomes 1 and remains 1 thereafter when at least two 0's and two 1's have occurred as inputs.

Example moore machine

A moore machine with nine states for the above description is shown on the right. The initial state is state A, and the final state is state I. The state table for this example is as follows:

Current stateInputNext stateOutput
A0D0
1B
B0E0
1C
C0F0
1C
D0G0
1E
E0H0
1F
F0I0
1F
G0G0
1H
H0H0
1I
I0I1
1I

Complex

More complex Moore machines can have multiple inputs as well as multiple outputs.

Gedanken-experiments

In Moore's paper 'Gedanken-experiments on Sequential Machines',[1] the ${displaystyle (n;m;p)}$ automata (or machines) ${displaystyle S}$ are defined as having ${displaystyle n}$ states, ${displaystyle m}$ input symbols and ${displaystyle p}$ output symbols. Nine theorems are proved about the structure of ${displaystyle S}$, and experiments with ${displaystyle S}$. Later, '${displaystyle S}$ machines' became known as 'Moore machines'.

At the end of the paper, in Section 'Further problems', the following task is stated:

Another directly following problem is the improvement of the bounds given at the theorems 8 and 9.

Moore's Theorem 8 is formulated as:

Given an arbitrary ${displaystyle (n;m;p)}$ machine ${displaystyle S}$, such that every two of its states are distinguishable from one another, then there exists an experiment of length ${displaystyle {tfrac {n(n-1)}{2}}}$ which determines the state of ${displaystyle S}$ at the end of the experiment.

In 1957, A. A. Karatsuba proved the following two theorems, which completely solved Moore's problem on the improvement of the bounds of the experiment length of his 'Theorem 8'.

Theorem A. If ${displaystyle S}$ is an ${displaystyle (n;m;p)}$ machine, such that every two of its states are distinguishable from one another, then there exists a branched experiment of length at most ${displaystyle {tfrac {(n-1)(n-2)}{2}}+1}$ through which one may determine the state of ${displaystyle S}$ at the end of the experiment.

Theorem B. There exists an ${displaystyle (n;m;p)}$ machine, every two states of which are distinguishable from one another, such that the length of the shortest experiments establishing the state of the machine at the end of the experiment is equal to ${displaystyle {tfrac {(n-1)(n-2)}{2}}+1}$.

Theorems A and B were used for the basis of the course work of a student of the fourth year, A. A. Karatsuba, 'On a problem from the automata theory', which was distinguished by testimonial reference at the competition of student works of the faculty of mechanics and mathematics of Moscow Lomonosow State University in 1958. The paper by Karatsuba was given to the journal Uspekhi Mat. Nauk on 17 December 1958 and was published there in June 1960.[3]

Until the present day (2011), Karatsuba's result on the length of experiments is the only exact nonlinear result, both in automata theory, and in similar problems of computational complexity theory.

References

1. ^ abMoore, Edward F (1956). 'Gedanken-experiments on Sequential Machines'. Automata Studies, Annals of Mathematical Studies. Princeton, N.J.: Princeton University Press (34): 129–153.
2. ^Lee, Edward Ashford; Seshia, Sanjit Arunkumar (2013). Introduction to Embedded Systems (1.08 ed.). UC Berkeley: Lulu.com. ISBN9780557708574. Retrieved 1 July 2014.CS1 maint: discouraged parameter (link)
3. ^Karatsuba, A. A. (1960). 'Solution of one problem from the theory of finite automata'. Uspekhi Mat. Nauk (15:3): 157–159.

• Conway, J.H. (1971). Regular algebra and finite machines. London: Chapman and Hall. ISBN0-412-10620-5. Zbl0231.94041.CS1 maint: discouraged parameter (link)
• Moore E. F. Gedanken-experiments on Sequential Machines. Automata Studies, Annals of Mathematical Studies, 34, 129–153. Princeton University Press, Princeton, N.J.(1956).
• Karatsuba A. A. Solution of one problem from the theory of finite automata. Usp. Mat. Nauk, 15:3, 157–159 (1960).
• Karatsuba A. A. Experimente mit Automaten (German) Elektron. Informationsverarb. Kybernetik, 11, 611–612 (1975).
• Karatsuba A. A. List of research works.

• Media related to Moore machine at Wikimedia Commons

This VHDL project presents a full VHDL code for Moore FSM Sequence Detector. A VHDL Testbench is also provided for simulation. The sequence to be detected is '1001'.

The Moore FSM state diagram for the sequence detector is shown in the following figure.

VHDL Testbench for Sequence Detector using Moore FSM:

Simulation Waveform for Moore FSM Sequence Detector in VHDL:

As shown in the simulation waveform of the VHDL Moore FSM sequence detector, the detector output only goes high when the sequence '1001' is detected.
Verilog code for Moore FSM Sequence Detector: here.
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