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Rankine
Guest or Trecas
St. Venant
Von Mises
Width b 1/M
Width b M 2
Width b 3 M
Width b M
## Theory of Structures For calculating the allowable stress of long columns σ0 = σy/n [1 - a (1/r)²]is the empirical formula, known as
Perry
Rankine
Parabolic formula
Straight line formula
1/4
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2/3
250 mm
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# Posts by: Nicholas Wong
## Awesome New Layout!
Well, it is always good to have a refreshing layout. I spent like 2 days to tweak my website layout. It's really awesome. Useless widgets are removed and background changed.…
## Happy Android Development with Gradle
Recently I have been working a lot on migrating android projects to use gradle building system. Seriously it is much better than the old time. I can easily use libraries…
## Interview Practice Extra 07 – Universal Value Binary Tree
Question Design an algorithm to verify that a tree is a universal value binary tree. Universal value binary tree means all value in that tree is the same. Solution There is…
## Interview Practice Extra 06 – Vending Machine
Question This is an actual question I encountered in an Amazon phone interview in November 2013. You are going to design the money changing algorithm for a vending machine. That…
## Interview Practice Extra 05 – Identify Prime Numbers
Question Write an algorithm to identify prime numbers from a list of numbers ranging 0-100. Solution The main question is actually to write a program to check if a number…
## Interview Practice Extra 04 – Remove Node from Singly Linked List
Question How do you remove a node from a singly linked list, given only that node? Head node is not given. Solution Set next of this node to the next…
## Interview Practice 28 – Counting One in Binary Expression
Question Given a number, find the number of 1 in the number's binary expression. For example, binary express of 10 is 1010. So the number of 1 in it is 2.…
## Interview Practice 26 – String Right-rotation
Question Write a program to right-rotate a string by m characters. Right-rotating a string means moving m characters at the left of string to the right. Is it required the…
## Interview Practice 25 – Longest Consecutive Digits
Question Given a function prototype: int continumax(char *output_string,char *input_string). Implement it to find the longest consecutive digits. This function must return the length of the longest digits. The found longest…
## Interview Practice 24 – Reverse a Linked List
Question Reverse a linked list. Sample [crayon-535820e6548ee248369460/]
## Interview Practice 22 – Card Guessing
Question Consider there are 4 blue and 4 red cards. Host gets 2 cards randomly, no one knows what cards they are. Then he places 2 random cards at the…
## Interview Practice 21 – Summing Combinations
Question Given a integers m and n, generate all combination within 1 to n that would give the sum m. For example, for m=5 and n=5, the combinations are {5}, {4+1},…
## Interview Pizzle 1 – Find the Heavier Marble
Question You have 9 marbles. 8 marbles weigh 1 ounce each, & one marble weighs 1.5 ounces. You are unable to determine which is the heavier marble by looking at them. You…
## Interview Practice 20 – Convert String to Integer
Question Convert the inputted string to an integer. For example, "345" will output 345. Solution Though it looks simple, in fact it is pretty tricky. First we need to make…
## Interview Practice 19 – Fastest Way to Calculate Fibonacci Sequence
Question Construct an algorithm so that the Nth item of Fibonacci Sequence can be found in the shortest time. Definition of Fibonacci Sequence is: [crayon-535820e658569781704183/] Solution There are many solutions for this,…
## Interview Practice 18 – Last Surviving Number in Loop
Consider Consider there is a list containing N numbers, and which formed a ring. That is, item n+1 is item 1. Construct an algorithm such that it traverses through the…
## Interview Practice 17 – Find The String Appeared Once
Question Given a string, write an algorithm to find the character inside that appeared only once. For example, the result for string "abaccdeff" is b. Solution We can use a…
## Interview Practice Extra 03 – Verify Binary Tree with Same Value
Question Verify whether all nodes have the same value in a binary tree. Solution We can traverse the tree with our usual way, like depth-first or breadth-first algorithm. Then pass…
## Interview Practice Extra 02 – Duplicates in List
Question Write an algorithm to remove duplicated node from a linked list. Solution There are many ways to do it. For the first one, as the simplest one, we could…
## Interview Practice Extra 01 – Find Loop in Linked List
Question Validate a linked list whether there is a loop in it. That is, there is a node in a linked list with the next pointer pointing to a node…
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# Applications of Special Relativity
## Contents
#### Problems on the Relativistic Doppler Effect
Problem : A train is moving directly towards you at 2×108 m/s. The (monochromatic) light on the front of the train has a wavelength of 250 nanometers in the frame of the train. What wavelength do you observe?
Using c = we find the frequency of the emitted light to be 1.2×1015 Hz. The observed frequency is given by:
f = f' = 1.2×1015 = ×1.2×1015 = 2.68×1015
Thus the wavelength is λ = c/f = 3.0×108/2.68×1015 = 112 nanometers.
Problem : Light that is assumed to be from the 22.5 cm microwave Hydrogen line is measured at a frequency of 1.2×103 MHz. How fast is the galaxy from which this light was emitted receding from the earth?
This is the famous 'redshift' effect. We know that the ratio = . Because f = c/λ this must be equal to the ratio , where the unprimed symbols denoted the frequencies and wavelengths measured on earth. Thus = , where c/(1.2×109) = 25 . Thus:
1.23 = âá’1.23 - 1.23v/c = 1 + v/câá’0.23 = 2.23v/câá’v = 0.105c
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Three point charges are on the x axis: $q_1=+2/:/mu C/:/:at/:/:x=0$, $q_2=+2/:/mu C/:/:at/:/:x=3/:m$ and$q_3=-4/:/mu C/:/:at/:/:x=6/:m$. Find the potential difference between points A(0, 2) and B(0, 4).
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# How could I prove this?
I've kind of fallen behind in my Algebra class and I really haven't read much about the theory. I'm wondering, how would you go on about proving that if an odd prime, p, does not divide a nor b, but divides the sum of their squares - a^2 + b^2 -, then p = 1 mod 4.
Up to know, I've been considering the special case in which gcd(a,b) = 1. Of course solving for this case solves for the entire problem; also it leaves something to work with since a^2 + b^2 = a + b mod 3 and a^2 + b^2 = 1 mod 4 or a^2 + b^2 = 2 mod 4. From that point all I tried seemingly led to a dead end. Maybe there is a property or something of the kind that I am not aware of and without which the problem becomes very cumbersome?
Last edited:
morphism
Homework Helper
Do you know anything about the ring of Gaussian integers Z?
Nope. Like I said I'm quite weak with the theory right now as I've allowed myself to slip behind. Could you give me some details on this ring so I can read about it?
morphism
Homework Helper
Z = {a+bi : a,b in Z}. Addition and multiplication done like in C turn it into a ring.
Like in Z, there is a notion of "prime number" in Z. The fact I want to exploit is that if p is a prime in Z, and p=3 (mod 4), then p is also a prime in Z (i.e. a Gaussian prime). So maybe you can read up on Gaussian primes.
After that, if we suppose p = 3 (mod 4) in your problem, then p|(a^2+b^2)=(a+bi)(a-bi) would imply that p|(a+bi) or p|(a-bi), which would lead to a contradiction. So p must be = 1 (mod 4).
But there are probably more elementary ways to do this. Number theory isn't my strong suit.
Hummm I don't believe we've gone into such theory quite yet. There must be a more elementary way like you said...
There is no really easy way on that problem, if what you mean is that p = sum of two squares (in only one way). It was stated by Fermat in 1640 and not proven until 1747 by Euler. 1770 proven by Lagrange. 1887 by Dedikand using Gaussian integers. Wikepedia.
However, it is much easier if you only want an X and Y such that X^2+Y^2 ==0 Mod p, because in that case you only need solve $$\frac{X^2}{Y^2}\equiv -1 Mod p$$
Last edited:
morphism
Homework Helper
Isn't the result you're talking about more general, though? That is, it states those primes that are expressible as a sum of two squares are precisely those that are congruent to 1 mod 4 (with the trivial exception of p=2).
Here we only have that p divides a^2 + b^2, and not that it equals it.
morphism: Here we only have that p divides a^2 + b^2, and not that it equals it.
Yes, that is true. By the time I recognized that and got around to it, you were ahead of me. We only need solve: $$\frac{X^2}{Y^2}\equiv -1 Mod p$$
Last edited:
morphism
$$\frac{X^2}{Y^2}\equiv -1 Mod p$$
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# lab3 - Purdue University: ECE438 - Digital Signal...
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Unformatted text preview: Purdue University: ECE438 - Digital Signal Processing with Applications 1 ECE438 - Laboratory 3: Frequency Analysis October 6, 2010 1 Introduction In this experiment, we will use Fourier series and Fourier transforms to analyze continuous- time and discrete-time signals and systems. The Fourier representations of signals involve the decomposition of the signal in terms of complex exponential functions. These decompositions are very important in the analysis of linear time-invariant (LTI) systems, due to the property that the response of an LTI system to a complex exponential input is a complex exponential of the same frequency! Only the amplitude and phase of the input signal are changed. Therefore, studying the frequency response of an LTI system gives complete insight into its behavior. In this experiment and others to follow, we will use the Simulink extension to Matlab. Simulink is an icon-driven dynamic simulation package that allows the user to represent a system or a process by a block diagram. Once the representation is completed, Simulink may be used to digitally simulate the behavior of the continuous or discrete-time system. Simulink inputs can be Matlab variables from the workspace, or waveforms or sequences generated by Simulink itself. These Simulink-generated inputs can represent continuous- time or discrete-time sources. The behavior of the simulated system can be monitored using Simulinks version of common lab instruments, such as scopes, spectrum analyzers and network analyzers. Questions or comments concerning this laboratory should be directed to Prof. Charles A. Bouman, School of Electrical and Computer Engineering, Purdue University, West Lafayette IN 47907; (765) 494- 0340; bouman@ecn.purdue.edu Purdue University: ECE438 - Digital Signal Processing with Applications 2 2 Background Exercises INLAB REPORT: Submit these background exercises with the lab report. 2.1 Synthesis of Periodic Signals Each signal given below represents one period of a periodic signal with period T . 1. Period T = 2. For t [0 , 2]: s ( t ) = rect( t- 1 2 ) 2. Period T = 1. For t [- 1 2 , 1 2 ]: s ( t ) = rect (2 t )- 1 2 For each of these two signals, do the following: i. Compute the Fourier series expansion in the form s ( t ) = a + summationdisplay k =1 A k sin(2 kf t + k ) where f = 1 /T . Hint: You may want to use one of the following references: Sec. 4.1 of Digital Signal Processing, by Proakis and Manolakis, 1996; Sec. 4.2 of Signals and Systems, by A. Oppenheim and A. Willsky, 1983; Sec. 3.3 of Signals and Systems, A. Oppenheim and A. Willsky, 1997. Note that in the expression above, the function in the summation is sin(2 kf t + k ), rather than a complex sinusoid. The formulas in the above references must be modified to accommodate this. You can compute the cos/sin version of the Fourier series, then convert the coefficients....
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# . The temperature of 600 g of cold water rises by 15o C when 300 g of hot water at 50o C is added to it. What was the initial temperature of the cold wa
The temperature of 600 g of cold water rises by 15C when 300 g of hot water at 50C is added to it. What was the initial temperature of the cold water?
Mass of hot water (m1) = 300 g
Temperature (T1) = 500 C
Mass of cold water (m2) = 600 g
Change in temperature of cold water (T – T2) = 150 C
Final temperature = T0 C
The specific heat capacity of water is c
m1c (T1 – T) = m2c (T – T2)
300 (50 – T) = 600 (15)
T = 6000 / 300
T = 200 C
Final temperature = 200 C
Change in temperature = 150 C
Initial temperature of cold water = 200 C – 150 C
= 50 C
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Homework Help: Determine the negation of the expression
1. Sep 1, 2011
ur5pointos2sl
The question states:
Determine the negation of the expression.
p -> q ^ r.
I am having difficulty finding the negation. Below are the steps I have taken in an attempt to come to the solution.
The - symbol will be the negate symbol.
p -> q ^ r
(-p OR q) ^ r
-(-p OR q) ^ r
(p ^ -q) ^ r
-((p ^ -q) ^ r)
-p OR - r OR q
-(p ^ r) OR q
I feel like I am going in circles. I have made truth tables to compare but have had no luck. Please any advise would be appreciated.
2. Sep 1, 2011
ironspud
Re: Logic
If you look at a truth table for a conditional statement such as $P\rightarrow Q$, you should find that the only time this statement is rendered false is if the antecedent (the first part, $P$) is true and the consequent (the second part, $Q$) is false.
So, what does that tell you about your problem? Keep the antecedent as it is, and negate the consequent, $(Q\wedge R)$.
3. Sep 14, 2011
Re: Logic
Thank you!
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# Dashaun Phillips RAS
### Dashaun Phillips RAS
Dashaun Phillips went undrafted for the NFL in 2014.
He was selected by the Dallas Renegades in the XFL Draft.
He recorded a Relative Athletic Score of 8.55, out of a possible 10.0. RAS is a composite metric on a 0 to 10 scale based on the average of all of the percentile for each of the metrics the player completed either at the Combine or pro day.
He had a recorded height of 5113 that season, recorded as XYYZ where X is feet, YY is inches, and Z is eighths of an inch. That correlates to 5 feet, 11 and 3/8 of an inch or 71.375 inches, or 181.2925 centimeters. This correlates to a 6.68 score out of 10.0.
He recorded a weight of 182 in pounds, which is approximately 82 kilograms. This correlates to a 2.24 score out of 10.0.
Based on his weight, he has a projected 40 yard dash time of 4.44. This is calculated by taking 0.00554 multiplied by his weight and then adding 3.433.
This player did not record any measurements at the NFL Combine that we were able to find.
This player did not have a recorded 40 yard dash for the Combine in the RAS database.
At his pro day, he recorded a 40 yard dash of 4.48 seconds. This correlates to a 7.17 score out of 10.0.
The time traveled between the 20 and 40 yard lines is known as the Flying Twenty. As the distance is also known, we can calculate the player’s speed over that distance. The time he traveled the last twenty yards at his pro day was 1.84 seconds. Over 20 yards, we can calculate his speed in yards per second to 10.87. Taking into account the distance in feet (60 feet), we can calculate his speed in feet per second to 32.61. Breaking it down further, we can calculate his speed in inches per second to 391.3. Knowing the feet per second of 32.61, we can calculate the approximate miles per hour by multiplying that value by 0.681818 to give us a calculated MPH of 22.2 in the last 20 yards of his run.
This player did not have a recorded 20 yard split for the Combine in the RAS database.
At his pro day, he recorded a 20 yard split of 2.64 seconds. This correlates to a 5.54 score out of 10.0.
We can calculate the speed traveled over the second ten yards of the 40 yard dash easily, as the distance and time are both known. The time he traveled the second ten yards at his pro day was 1.07 seconds. Over 10 yards, we can calculate his speed in yards per second to 9.35. Taking into account the distance in feet (30 feet), we can calculate his speed in feet per second to 28.04. Breaking it down further, we can calculate his speed in inches per second to 336.45. Knowing the feet per second of 28.04, we can calculate the approximate miles per hour by multiplying that value by 0.681818 to give us a calculated MPH of 19.1 in the second ten yards of his run.
This player did not have a recorded 10 yard split for the Combine in the RAS database.
At his pro day, he recorded a 10 yard split of 1.57 seconds. This correlates to a 7.08 score out of 10.0.
The time he traveled the first ten yards at his pro day was 1.57 seconds. Over 10 yards, we can calculate his speed in yards per second to 6.0. Taking into account the distance in feet (30 feet), we can calculate his speed in feet per second to 19.0. Breaking it down further, we can calculate his speed in inches per second to 229.0. Knowing the feet per second of 19.0, we can calculate the approximate miles per hour by multiplying that value by 0.681818 to give us a calculated MPH of 13.0 in the first ten yards of his run.
This player did not have a recorded bench press for the Combine in the RAS database.
At his pro day, he recorded a bench press of 14 repetitions of 225 pounds. This correlates to a 5.97 score out of 10.0.
This player did not have a recorded vertical jump for the Combine in the RAS database.
At his pro day, he recorded a vertical jump of 39.0 seconds. This correlates to a 8.97 score out of 10.0.
This player did not have a recorded broad jump for the Combine in the RAS database.
At his pro day, he recorded a broad jump of 1003, which is recorded as FII or FFII where F is feet and I is inches. This correlates to a 7.39 score out of 10.0.
This player did not have a recorded 5-10-5 or 20 yard short shuttle for the Combine in the RAS database.
At his pro day, he recorded a 5-10-5 or 20 yard short shuttle of 3.97 seconds. This correlates to a 9.41 score out of 10.0.
This player did not have a recorded 3 cone L drill for the Combine in the RAS database.
At his pro day, he recorded a 3 cone L drill of 6.65 seconds. This correlates to a 9.44 score out of 10.0.
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# Transcendental function
A transcendental function is a function that does not satisfy a polynomial equation whose coefficients are themselves roots of polynomials, in contrast to an algebraic function, which does satisfy such an equation.[1] (The polynomials are sometimes required to have rational coefficients.) In other words, a transcendental function is a function that "transcends" algebra in the sense that it cannot be expressed in terms of a finite sequence of the algebraic operations of addition, multiplication, and root extraction.
Examples of transcendental functions include the exponential function, the logarithm, and the trigonometric functions.
## Definition
Formally, an analytic function ƒ(z) of one real or complex variable z is transcendental if it is algebraically independent of that variable.[2] This can be extended to functions of several variables.
## Examples
The following functions are transcendental:
${\displaystyle f_{1}(x)=x^{\pi }\ }$
${\displaystyle f_{2}(x)=c^{x},\ c\neq 0,1}$
${\displaystyle f_{3}(x)=x^{x}={{^{2}}x}\ }$
${\displaystyle f_{4}(x)=x^{\frac {1}{x}}\ }$
${\displaystyle f_{5}(x)=\log _{c}x,\ c\neq 0,1}$
${\displaystyle f_{6}(x)=\sin {x}}$
Note that in particular for ƒ2 if we set c equal to e, the base of the natural logarithm, then we get that ex is a transcendental function. Similarly, if we set c equal to e in ƒ5, then we get that ln(x), the natural logarithm, is a transcendental function. For more information on the second notation of ƒ3, see tetration.
## Algebraic and transcendental functions
The logarithm and the exponential function are examples of transcendental functions. Transcendental function is a term often used to describe the trigonometric functions (sine, cosine, tangent, their reciprocals cotangent, secant, and cosecant, the now little-used versine, haversine, and coversine, their analogs the hyperbolic functions and so forth).
A function that is not transcendental is said to be algebraic. Examples of algebraic functions are rational functions and the square root function.
The operation of taking the indefinite integral of an algebraic function is a source of transcendental functions. For example, the logarithm function arose from the reciprocal function in an effort to find the area of a hyperbolic sector. Thus the hyperbolic angle and the hyperbolic functions sinh, cosh, and tanh are all transcendental.
Differential algebra examines how integration frequently creates functions that are algebraically independent of some class, such as when one takes polynomials with trigonometric functions as variables.
## Dimensional analysis
In dimensional analysis, transcendental functions are notable because they make sense only when their argument is dimensionless (possibly after algebraic reduction). Because of this, transcendental functions can be an easy-to-spot source of dimensional errors. For example, log(5 meters) is a nonsensical expression, unlike log(5 meters / 3 meters) or log(3) meters. One could attempt to apply a logarithmic identity to get log(10) + log(m), which highlights the problem: applying a non-algebraic operation to a dimension creates meaningless results.
## Exceptional set
If ƒ(z) is an algebraic function and α is an algebraic number then ƒ(α) will also be an algebraic number. The converse is not true: there are entire transcendental functions ƒ(z) such that ƒ(α) is an algebraic number for any algebraic α. In many instances, however, the set of algebraic numbers α where ƒ(α) is algebraic is fairly small. For example, if ƒ is the exponential function, ƒ(z) = ez, then the only algebraic number α where ƒ(α) is also algebraic is α = 0, where ƒ(α) = 1. For a given transcendental function this set of algebraic numbers giving algebraic results is called the exceptional set of the function,[3][4] that is the set
${\displaystyle {\mathcal {E}}(f)=\{\alpha \in {\overline {\mathbf {Q} }}\,:\,f(\alpha )\in {\overline {\mathbf {Q} }}\}.}$
If this set can be calculated then it can often lead to results in transcendence theory. For example, Lindemann proved in 1882 that the exceptional set of the exponential function is just {0}. In particular exp(1) = e is transcendental. Also, since exp(iπ) = -1 is algebraic we know that iπ cannot be algebraic. Since i is algebraic this implies that π is a transcendental number.
In general, finding the exceptional set of a function is a difficult problem, but it has been calculated for some functions:
• A function with empty exceptional set that doesn't require one to assume this conjecture is the function ƒ(x) = exp(1 + πx).
While calculating the exceptional set for a given function is not easy, it is known that given any subset of the algebraic numbers, say A, there is a transcendental function ƒ whose exceptional set is A.[6] The subset does not need to be proper, meaning that A can be the set of algebraic numbers. This directly implies that there exist transcendental functions that produce transcendental numbers only when given transcendental numbers. Alex Wilkie also proved that there exist transcendental functions for which first-order-logic proofs about their transcendence do not exist by providing an exemplary analytic function.[7]
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Diviya
Members
351
1. Pls. suggest books like Scarlet Letter and To Kill a Mockingbird
My 12 year old read Jane Eyre and Pride and Prejudice right after To Kill a Mockingbird and really loved them. She is also currently working her way through the Wizard of Earthsea series. I will say that she has always self censored, and while she used to be a very sensitive reader, she is not now! She has recently read and enjoyed The Book Thief, The Fault in Our Stars, the Hunger Games, the Maze Runner series to give you an idea of how not sensitive she is now :) I'm reading the suggestions in this thread with interest also.
2. Anyone else taking Math Kangaroo today?
It's possible I'm not thinking of this the right way, but I think only the corner small cubes on the big cube have three painted faces. All the other small cubes would have only one or two painted faces. So, eight corner cubes total - they told you none with 3 red faces, so that also means none with three blue faces, so all eight will have both red and blue? Make sense? Can anyone corroborate?
3. Anyone else taking Math Kangaroo today?
There were about 40 kids at our site, an elementary school. There were all levels 1-5 I think, except for my 7th grader. My girls had fun with the test. I also was able to get an answer that stumped both of them - very gratifying! 4ofus, last year, the coordinator at the test site sent us a list of who made the state and national lists. For the actual scores, at some point you can go on the website to get them. It takes an unreasonably long time for the scores to be posted, given it's a scanned answer sheet. I want to say about two months. Well, phew, I think that's it for the math contests for this year. DD12 is taking the SAT on May 2 though - I'll be interested to see how that goes!!
4. Anyone else taking Math Kangaroo today?
We will be in Redwood City - my girls are taking levels 5 and 7. Good luck!
5. Am I setting my daughter up for failure? (AOPS/math questions)
You could absolutely try Algebra first - I did that and it was obvious she wasn't ready for it. But....I really love the Prealgebra book - you may be able to go through some chapters pretty quickly, but I don't think it will feel like needless repetition. For what it's worth, I think the AOPS pretests are ridiculously easy, and to me they seem to have no actual relation to whether your child actually is ready for the course, unless you know you have a super self-motivated child! The post tests are more useful, as Kathy mentioned above. Generally speaking, I would think an 11 year old could manage the online class for Prealgebra, but not the Algebra if it's her first time doing AOPS. Mine took the Algebra online class at 11, but she had already worked through that part of the book on her own the year before. She took the class to review and solidify. I don't think you were thinking of the online classes, but my kids both enjoy them, so it may be worth thinking about. Sorry, probably this was incoherent, but hopefully there is something useful in there.
6. Worried about ds and AoPS
I went through AOPS pre-algebra with my oldest, and now my younger one is taking it online and working with me on it. I would say, in my opinion, you should definitely not make the call on whether it is right for you after a chapter or two. I would say it took a good 5-6 chapters before my kids started "getting" it, and even enjoying it. There were tears, there was a lot of hand holding. We skipped some of the really tough challenge problems. Some chapters, especially early on, were more direct instruction than discovery method - it says right in the front of the book that you can tailor this to the child you have. The important thing is to not be a slave to the curriculum! If some of the problems are ridiculous, skip them! If they need help, work through some problems with them! I have to say, for my own purposes, that sticking with it was the best thing I have done for them so far. My oldest has totally blossomed - she can't get enough AOPS now - it's kind of taken over our lives LOL. And there are still some problems we have to leave behind. And that's ok. She takes all the classes online now, works totally on her own, and she decides when to pursue help on a difficult problem from her classmates or instructor. I totally agree with those above who say AOPS is not the only or best way to get "there." It was the only and best way for US. My only message is to give it a fair shot. Oh, and the pace was painfully slow at first - my younger one actually dropped the class half way through, and then started over a month or so later. But the pace picked up quite a bit as we went through. So you can't really project how long it will take. Hope that helps.
7. There are times...
Is there a side of the house where you can sit and they can't see you? Or a spot where you can open a window and sit next to it? Or maybe you can listen to a book on tape while standing in the garden "pretending" to weed? Or as a last resort, find some particularly tall weeds and hide behind them with your book. Will require taking something to sit on. I am sending hugs.
8. Rat gave birth in my car
Soak some cottonballs in mint extract and leave them in your car (after you've caught mama if you want to save the babies). I have read (and have some anecdotal evidence) that rats can't stand the smell of mint. I'm not promising though... :)
9. math without a curriculum--need ideas-MEP?
Agreeing with Sparkly Unicorn. Can you just go faster through MM? I wouldn't necessarily worry about him forgetting. Mine do that too. We just go over it again when it comes up. MM may actually have too much repetition for him. Get it done, and then move on to the fun stuff. He sounds like someone who would benefit from AOPS pre algebra.
10. Graduating class breakdown at Harvard
Huh. You got me curious so I looked it up. I'm still in favor of legalizing. :)
11. What are you reading this summer?
Huge Georgette Heyer fan here! Which one is your favorite? I have way too many books. Who knows what I will actually end up reading. I'm currently reading Why Don't Students Like School, recommended here. Just finished Robinson Crusoe and How to Fail at Nearly Everything (really enjoyed both). Also working on How to Read a Book, and Teaching as a Subversive Activity. On deck is Angry Conversations with God; The Liberal Arts Tradition; The Up Side of Down; Food and Western Disease, and whatever my dd11 picks out to read next. So many books, so little time...
12. Just in case you have a 7th grader who doesn't particularly enjoy reading...
You could take a look at Readers Odyssey. She has great ideas for how to tailor to a child who doesn't like to read.
13. Doing all the AoPS PreA problems?
In my opinion, you can't really skip problems in AOPS. None of the problems are the same. You can choose to skip the challenge problems for now, and come back to them later. With my older daughter, we would do a section in two days at first. At some point she picked up speed and got more interested, so we could do a section or even two in a day. The summaries typically took 3-5 days. It is totally worth it to spend more time on it in order to get everything out of it. We found it transformative. Hope that helps.
14. Do you ever wonder how much of your kid being ahead is homeschooling?
I'd love to talk about the larger problem (because I think about it all the time), but I'm barely coherent tonight, so I better not. In my opinion, the problem with a younger kid going into a "regular" algebra class has to do with the textbook and the pacing. You would need to look at the number of problems they would be expected to copy and work, the amount of repetition and the amount of review. All of those things would have simply crushed my dd's spirit at an age when she was ready for algebra. ONLY AOPS would work for her, because the pace is right, and the number of problems per section is just right. No review, no repetition - just incremental difficulty. Looking at a page in Dolciani made her hyperventilate. Ok, not sure that was coherent, but hope it helps.
16. Grr. For the first time, we won't finish our curricula..
We don't school year round, but we've never had everything end tidily at the end of the year! We just start new books when the old ones end...I do try to have specific goals of how much I want to cover, but that gets revised a few times over the course of the year as I get a sense for what is going faster and what is going slower. This year is even messier than usual, as we have added some online classes midyear, and shelved some things that weren't working. Overall, progress is definitely being made! I find that each year, we delink a little more from what a traditional school would look like. Haven't been able to give up having the summer off though!
17. Beast Academy 4B is now available
Hooray! I ordered it a couple of hours ago, as soon as I could get to the computer after the email came. What a junkie...
18. Upper elementary and above: Help me cull our books and decide how much to keep (especially history and science)
Why can't they go in the bathroom?
19. Best text for me to re-learn Latin?
I really like Learn to Read Latin, by Andrew Keller and Stephanie Russell. I haven't had time to work through more than a few chapters yet, but it is extremely thorough and well thought out. (There is also Learn to Read Greek). There is a text and workbook, and you can get an answer key from the publisher. Wheelocks would be fine too. I don't think you'd need something before it. For reference, I took through AP Virgil in high school plus an additional semester of literature in college. I've been able to keep up with the kids so far, even without the refresher. The refresher has been working along with them! And my memory is not so great, and my last latin class was over 20 years ago...I would definitely need to study more as they get further along though. Hope that helps!
20. What subjects, if any, will you continue over the summer?
I always plan on working on some things through the summer. But between camps, travel, other activities and my own need for time off, the reality never comes close to the vision. Hoping this year to keep up with math and Greek. I'm thinking about having them do a class online with AOPS for the accountability!
21. When to start on Latin
You can easily wait until at least 4th grade. If your child is excited to start sooner, that's a reason to do it earlier. Otherwise, understand that they will learn virtually nothing useful in those first few years, and probably add unnecessary anxiety to your day. IMHO :)
22. What have you used for Italian? And a Rosetta Stone question..
I was recently researching Italian programs, and decided to purchase Living Language Italian. It had great reviews and is not expensive. There is a book and CDs. We just started using it, so I can't give you any feedback on it, but it does seem very user friendly. I'd be curious to know if anyone else has used this. The company has other languages as well. ETA: while I would normally look for a program with a strong reading component, in this case I was looking primarily for spoken.
23. Most Readable Divine Comedy Translation?
If you search the boards, you may find a thread in which Ester Maria and others talked about the different translations. Here is what I came away with: Dante translators Mandelbaum - best compromise between form and function Esolen - best faithfulness to original Ciardi - most fun Hope that helps!
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13-relative motion 3
# 13-relative motion 3 - PHYS 211 2D and 3D Motion Describe...
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PHYS 211 2/ 13 /09 2D and 3D Motion Describe: Relative Motion Next Week: Chapters 5,6 Force and Motion Newton’s Laws
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Test 1 Results are Posted !"!! !"\$! !"%! !"&! !"’! ("!! ("\$! ( (! () \$’ *+ %& ,, &% +* ’\$ )( (!! (!) ((’ (\$+ (*& (%, (,% (&* (+\$ (’( ()! ()) Score: Points Correct/57 Average: 72 Score Score See syllabus (BlackBoard) for course grade formula
Displacement, Velocity, Acceleration Which one of the following quantities is the change in object’s velocity divided by the elapsed time as the elapsed time becomes very small? a) average velocity b) instantaneous velocity c) average displacement d) average acceleration e) instantaneous acceleration
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Relative Motion - 1D (disp) Reference frames are related to each other. Physics described must be the same, no matter which frame you use. X BA - position of B relative to A X PB - position of P relative to B X PA - position of P relative to A x PA = x PB + x BA Notice order of subscripts inner ones the same outer ones (on right) match left side
Relative Motion - 1D (acc) x PA = x PB + x BA v PA = v PB + v BA For inertial frames , v BA =constant, so a BA =0 a PA = a PB + a BA a PA = a PB same acceleration!
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# lec16 - Math 124A – Viktor Grigoryan 16 Waves with a...
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Unformatted text preview: Math 124A – November 23, 2010 Viktor Grigoryan 16 Waves with a source Consider the inhomogeneous wave equation u tt- c 2 u xx = f ( x,t ) ,-∞ < x < ∞ ,t > , u ( x, 0) = φ ( x ) , u t ( x, 0) = ψ ( x ) , (1) where f ( x,t ), φ ( x ) and ψ ( x ) are arbitrary given functions. Similar to the inhomogeneous heat equation, the right hand side of the equation, f ( x,t ), is called the source term. In the case of the string vibrations this term measures the external force (per unit mass) applied on the string, and the equation again arises from Newton’s second law, in which one now also has a nonzero external force. As was done for the inhomogeneous heat equation, we can use the superposition principle to break problem (1) into two simpler ones: u h tt- c 2 u h xx = 0 , u h ( x, 0) = φ ( x ) , u h t ( x,t ) = ψ ( x ) , (2) and u p tt- c 2 u p xx = f ( x,t ) , u p ( x, 0) = 0 , u p t ( x,t ) = 0 . (3) Obviously, u = u h + u p will solve the original problem (1). u h solves the homogeneous equation, so it is given by d’Alambert’s formula. Thus, we only need to solve the inhomogeneous equation with zero data, i.e. problem (3). We will show that the solution to the original IVP (1) is u ( x,t ) = 1 2 [ φ ( x + ct ) + φ ( x- ct )] + 1 2 c Z x + ct x- ct ψ ( y ) dy + 1 2 c Z t Z x + c ( t- s ) x- c ( t- s ) f ( y,s ) dy ds. (4) The first two terms in the above formula come from d’Alambert’s formula for the homogeneous solution u h , so to prove formula (4), it suffices to show that the solution to the IVP (3) is u p ( x,t ) = 1 2 c Z t Z x + c ( t- s ) x- c ( t- s ) f ( y,s ) dy ds. (5) For simplicity, we will seize specifying the superscript and write u = u p (this corresponds to the as- sumption φ ( x ) ≡ ψ ( x ) ≡ 0, which is the only remaining case to solve). Recall that we have already solved inhomogeneous hyperbolic equations by the method of character- istics, which we will apply to the inhomogeneous wave equation as well. The change of variables into the characteristic variables and back are given by the following formulas ξ = x + ct, η = x- ct, t = ξ- η 2 c , x = ξ +...
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+0
# Binomial Theorem
0
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(a) Simplify (n/k)/(n/(k - 1))
(b) For some positive integer n, the expansion of (1 + x)^n has three consecutive coefficients a, b, c that satisfy 1:8:28. What must n be?
\(\phantom{1:8:40}\)
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GCSE Maths Algebra Sequences
Here we will learn about quadratic sequences including how to recognise, use and find the nth term of a quadratic sequence.
There are also quadratic sequences worksheets based on Edexcel, AQA and OCR exam questions, along with further guidance on where to go next if you’re still stuck
Quadratic sequences are only studied on the higher GCSE Maths pathway.
Quadratic sequences are ordered sets of numbers that follow a rule based on the sequence n2 = 1, 4, 9, 16, 25,… (the square numbers).
Quadratic sequences always include an n2 term.
The difference between each term in a quadratic sequence is not equal, but the second difference between each term in a quadratic sequence is equal.
Here are two examples of quadratic sequences:
4, 7, 12, 19, 28 requires adding to work out that the second difference is +2
and
1, −4, −15, −32, −55 requires subtracting to work out that the second difference is −6
Example:
## How to generate terms of quadratic sequences
1. Substitute the term number that you want to find as n.
2. Calculate.
### Related lessons on sequences
Quadratic sequences is part of our series of lessons to support revision on sequences. You may find it helpful to start with the main sequences lesson for a summary of what to expect, or use the step by step guides below for further detail on individual topics. Other lessons in this series include:
The quadratic sequence formula is: an^{2}+bn+c
Where,
a, b and c are constants (numbers on their own)
n is the term position
We can use the quadratic sequence formula by looking at the general case below:
Let’s use this to work out the n^{th} term of the quadratic sequence,
4, 5, 8, 13, 20, ...
First let’s find the first and second differences of the sequence:
Comparing this to the general case above, we can see that:
The second difference is equal to 2 so,
The first difference is equal to 1 so,
The first term is equal to 4 so,
We can see that a = 1,\; b = -2 and c = 5
So the nth of this quadratic sequence using the quadratic sequences formula is:
n^{2}-2n+5
## Generate terms of quadratic sequences examples
### Example 1: finding terms of a sequence
Find the first three terms of the sequence with nth term 2n2 + 4n + 1.
1. Substitute the term number that you want to find as n.
We want to find terms 1, 2 and 3 so substituting each of these into the formula
2n2 + 4n + 1 we get,
First term:
n = 1: (2 × 12 ) + (4 × 1) + 1
Second term:
n = 2: (2 × 22 ) + (4 × 2) + 1
Third term:
n = 3: (2 × 32 ) + (4 × 3) + 1
2 Calculate.
Remember to apply BIDMAS
n = 1: (2 × 12 ) + (4 × 1) + 1 = 7
n = 2: (2 × 22 ) + (4 × 2) + 1 = 17
n = 3: (2 × 32 ) + (4 × 3) + 1 = 31
The first term is 7, the next term is 13 and the term after that is 31.
### Example 2: finding terms of a sequence
Find the tenth term of the sequence with nth term 5n2 − 6n − 3.
Substitute the term number that you want to find as n.
Calculate.
## How to find the nth term of quadratic sequences
In order to find the nth term (general term) of a quadratic sequence we have to find the second difference. To do this, we calculate the first difference between each term in a quadratic sequence and then calculate the difference between this new sequence.
1 Find the first difference (d1) and second difference (d2) for the sequence.
2 Halve the second difference
$\left ( \frac{d_{2}}{2} =a \right )$
to find a, the coefficient of n2.
3 Subtract an2 from the original sequence.
4 If this produces a linear sequence, find the nth term of it.
5 Add the nth term for the linear sequence to an2 to work out the nth term of the quadratic sequence.
### What are the 5 steps to find the nth term of a sequence?
Step 4 is an additional step dependent on whether there is a remainder in step 3.
• If there is no remainder after subtracting an2 from the original sequence, step 4 is not needed.
• If the same constant remains for each term after the subtraction, we just add it to an2.
• If a sequence is generated after step 3, continue with step 4.
See the table below.
Here,
• Sequence 1 had no remainder and so the sequence is 2n2
• Sequence 2 had a remainder of 1 and so this sequence is 2n2 + 1
• Sequence 3 had another sequence as the remainder and so the nth term of this linear sequence was calculated and added to 2n2 to get 2n2 + n + 2.
## Find the nth term of quadratic sequences examples
### Example 3: find the nth term of a quadratic sequence of the form an2.
Find the nth term of this quadratic sequence:
2, 8, 18, 32, 50,
Find the first difference (d_{1}) and second difference (d_{2}) for the sequence.
Halve the second difference
Subtract an^{2} from the original sequence.
Find the nth term for the linear sequence generated.
Add the nth term for the linear sequence to an^{2} to find the nth term of the quadratic sequence.
### Example 4: find the nth term of a quadratic sequence of the form an2 + c.
Find the nth term rule of the quadratic sequence:
5, 14, 29, 50, 77,
Find the first difference (d_{1}) and second difference (d_{2}) for the sequence.
Halve the second difference.
Subtract an^{2} from the original sequence.
Find the nth term for the linear sequence generated.
Add the nth term for the linear sequence to an^{2} to find the nth term of the quadratic sequence.
### Example 5: find the nth term of a quadratic sequence of the form an2 + bn + c.
Find the nth term of this quadratic sequence:
9, 18, 31, 48, 69, …
Find the first difference (d_{1}) and second difference (d_{2}) for the sequence
Halve the second difference
Subtract an^{2} from the original sequence.
Find the nth term for the linear sequence generated.
Add the nth term for the linear sequence to an^{2} to find the nth term of the quadratic sequence.
### Example 6: find the nth term of a quadratic sequence of the form an2 + bn + c.
Find the nth term of this quadratic sequence:
19, 36, 61, 94, 135,
Find the first difference (d_{1}) and second difference (d_{2}) for the sequence.
Halve the second difference.
Subtract an^{2} from the original sequence.
Find the nth term for the linear sequence generated.
Add the nth term for the linear sequence to an^{2} to find the nth term of the quadratic sequence.
### Example 7: find the nth term of a quadratic sequence of the form an2 + bn + c where terms are generated using a pattern.
In the diagram below, the area of each pool can be represented by a quadratic sequence.
Find the area for a pool of width n.
Find the first difference (d_{1}) and second difference (d_{2}) for the sequence.
Halve the second difference.
Subtract an^{2} from the original sequence.
Find the nth term for the linear sequence generated.
Add the nth term for the linear sequence to an^{2} to find the nth term of the quadratic sequence.
### Example 8: find the nth term of a quadratic sequence of the form an2 + bn + c where a < 0.
Find the nth term of this quadratic sequence:
8, -10, -34, -64, -100,
Find the first difference (d_{1}) and second difference (d_{2}) for the sequence.
Halve the second difference.
Subtract an^{2} from the original sequence.
Find the nth term for the linear sequence generated.
Add the nth term for the linear sequence to an^{2} to find the nth term of the quadratic sequence.
### Common misconceptions
• The value of the second difference is not halved before using it as the coefficient of n2
For example, for the quadratic sequence n2 = 1, 4, 9, 16, 25, … the second difference is equal to 2 and so the nth term is incorrectly written as 2n2.
• The quadratic sequence is answered as if it were an arithmetic sequence
For a quadratic sequence we will have a common second difference.
• When the sequence is decreasing into negative terms, the second difference is written incorrectly as a positive number
For example, the sequence −n2 = −1, −4, −9, −16, … has a second difference of -2 but is written as 2.
• The remainder is subtracted from the an2 term rather than added as it was found by subtraction
For example, the remainder for the sequence n(n + 2) is the sequence 2n but is incorrectly subtracted to get the nth term of n(n − 2).
• The value for n cannot be negative
n represents the term (position) numbers and therefore it can only be positive integers starting from 1 and should also not include 0, n = 1, 2, 3, 4, 5, ….
This is important when finding the term in the sequence given its value as a zero or negative solution for n can be calculated.
• The value for the term is used instead of n
For example, find n when the term has a value of 25 means that the nth term = 25 and not n = 25.
1. Work out the first five terms of the sequence 2n^2-7n
-7, -6, -3, 4, 15
-5, -3, -1, 1, 3
-5, -6, -3, 4, 15
9, 22, 39, 60, 85
2 \times 1^{2} \hspace{1mm}- 7 \times 1 = -5\\ 2 \times 2^{2} \hspace{1mm}- 7 \times 2 = -6\\ 2 \times 3^{2} \hspace{1mm}- 7 \times 3 = -3\\ 2 \times 4^{2} \hspace{1mm}- 7 \times 4 = 4\\ 2 \times 5^{2} \hspace{1mm}- 7 \times 5 = 15\\
2. Find the nth term of the quadratic sequence:
-2, 1, 6, 13, 22
3n+2
n^{2}-3
2n^{2}-3n^{2}
n^{2}+3
The second difference is 2
2 \div 2 = 1
Therefore it is n^{2}
The remainder is 4 for each term therefore our sequence is
n^{2}-3
3. The first five terms of a quadratic sequence are 0, 13, 34, 63 and 100 . By calculating the second difference, find the nth term of the sequence.
4n^{2}+n-5
4n^{2}-5n
8n^{2}-8n
n^{2}+4n-5
The second difference is 8.
8 \div 2 = 4
Therefore it is 4n^{2}
The nth term of -4, -3,-2, -1, 0, … is n-5 so our sequence is
4n^{2}+n-5
4. Find the nth term of the decimal quadratic sequence:
2.1, 4.4, 6.9, 8.6, 10.5
0.1n^{2}+2n
10n^{2}+2n
2n^{2}-10
0.1n^{2}+2
The second difference is 0.2
0.2 \div 2 = 0.1
Therefore it is 0.1n^{2}
The nth term of 2, 4, 6, 8, 10 is 2n so our sequence is
0.1n^{2} +2n
5. Below are the first four terms of a patterned sequence. Calculate the formula to work out the number of tiles in pattern n.
n^{2}-4
4n^{2}
n^{2}+4
2n^{2}+3
Number of tiles: 5, 8, 13, 20
The second difference is 2
2 \div 2 = 1
Therefore it is n^{2}
The remainder is 4 each time so the nth term is
n^{2}+4
6. The first five terms of a quadratic sequence are 5, 2, -3, -10, and -19 . The formula for the nth term of this sequence can be written in the form a+bn^{2} where a and b are integers. Find the values of a and b .
a=-1,\; b=6
a=7,\; b=-2
a=4,\; b=1
a=6,\; b=-1
The second difference is -2
-2 \div 2 = -1
Therefore it is -n^{2}
The nth term is -n^2+6, which can also be written as 6-n^2.
a=6 and b=-1.
1. Work out the first four terms in the sequence with n th term 2n^{2} + 5n – 3
(2 marks)
\begin{aligned} 2 \times 1^{2} + 5 \times 1 – 3 &=4 \\ 2 \times 2^{2} + 5 \times 2 – 3 &=15\\ 2 \times 3^{2} + 5 \times 3 – 3 &=30\\ 2 \times 4^{2} + 5 \times 4 – 3 &=49 \end{aligned}
= 4, 15, 30, 49
(2)
2.
(a) Here is a patterned sequence made from equilateral triangles.
Find the n th term formula for the number of triangles used to form each pattern.
(b) How many triangles would be used in the 20th pattern?
(6 marks)
(a)
Sequence is 1, 3, 6, 10
First differences: 2, 3, 4
Second differences: 1, 1
(1)
0.5n^{2}
(1)
(1)
N th term of 0.5, 1, 1.5, 2 is 0.5n so formula is
0.5n^{2} + 0.5n
(1)
(b)
0.5 \times 20^{2} + 0.5 \times 20
(1)
210
(1)
3. The first four terms in a sequence are:
12, 20, 30, 42, 56
(a) Find the n th term formula for this sequence.
(b) Hence find which term has value 110 .
(7 marks)
(a)
First differences: 8, 10, 12, 14
Second differences: 2, 2, 2
(1)
n^{2}
(1)
(1)
N th term of 11, 16, 21, 26, 31 is 5n+6 so formula is
n^{2}+5n+6
(1)
(b)
n^{2}+5n+6=110
(1)
\begin{aligned} n^{2}+5n-104&=0 \\ (n-8)(n+13)&=0 \end{aligned}
(1)
n = 8, 8th term
(cannot have -13th term)
(1)
## Learning checklist
You have now learned how to:
• recognise and use quadratic sequences
• deduce expressions to calculate the nth term of quadratic sequences.
## Still stuck?
Prepare your KS4 students for maths GCSEs success with Third Space Learning. Weekly online one to one GCSE maths revision lessons delivered by expert maths tutors.
Find out more about our GCSE maths tuition programme.
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# What are Units
In metrology and physics, units tend to be standards regarding measurement of physical quantities that require clear definitions to be appropriate. Reproducibility of trial and error results is main to the scientific approach. A common system of the units allows for this. Scientific systems regarding units are usually a refinement with the concept of measures and weights produced long ago for that commercial purposes.
Medicine, engineering and Science, usually use smaller and larger units of way of measuring than those employed in everyday activity and indicate them much more precisely. The judicious collection of the units of measurement can help researchers within problem solving for instance, dimensional analysis.
In social sciences, you can find no common units of measurement and the practice and theory of the measurement is actually studied in theory of conjoint measurement and the psychometrics.
## Units of Measurement
A unit of measurement is actually a definite magnitude of the physical quantity, adopted and defined through law or by convention, that is utilized like a standard for measurement of the identical physical quantity. Some other value with the physical quantity could be expressed like a simple numerous of the unit of measurement. For instance, length is actually a physical quantity. The metre is actually a unit of length which symbolizes a definite predetermined size. Whenever we say 10 metres (or 10 m), we actually imply 10 times the definite predetermined length known as metre.
The agreement, definition and practical usage of units of measurement have performed a crucial part in human effort from earlier ages up to this day. Various systems of units utilized to be very standard. Presently there is a worldwide standard, the International System of the Units, the modern day form with the metric system.
## Unit of Measure
A unit of measure is designated to almost all inventory items and should be defined prior to a product could be entered. Each and every product may have a sales ordering unit of measure, a purchasing unit of measure and stocking unit of measure.
The unit of measure also can have programmed conversions with other units of measures by means of a conversion factor. The definition and usage of these factors is just used within the product master maintenance IN0100 whenever entering brand new products for that first time. The conversion factor can automatically fill the sales reporting for stock factor and Purchase to Stock factor areas when getting into a new products should the purchasing and selling units of measure differ through the stocking units of measure.
The usage of many different units of measure within the SAP system is beneficial, for instance, for storage, packaging and accounting purposes. As an example, if a crate includes several thousand parts of a certain material, it is much more expedient to buy, package and sell this particular material through the box or crate instead of by the piece. Furthermore, it is essential to use units of measure which result in no field overflows to take place whenever the quantities are increased.
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# How to Calculate a Better FFT by Leveraging the PSD Method
1 Comment
The power spectral density (PSD) function is an excellent tool for analyzing vibration data such as acceleration, velocity, or displacement data because its output is independent of time duration (see our blog that explains more about this). But the units are g2/Hz which can be a little confusing, engineers are expecting to see the units as g which would mimic the output of a typical Fast Fourier transform (FFT)!
What if we could leverage the benefit of the PSD's calculation technique which involves averaging overlapping segments, but scale it so that the output has the expected units of g? What if there was an FFT function that gave an output you expected but gave you control over defining the frequency resolution? At enDAQ in our open-source Python library, we did just that! This blog will explain this calculation methodology and introduce the function you can use on your own data!
I will make the method available to you with our open-source software and show you examples of how it works using both some waveforms I have created and some vibration data from a medical instrument. Some of the plots will be interactive, so have fun zooming around in them! I will discuss the following topics:
In this blog, I will present:
## Video and Demonstration in Python
Follow along with all the calculations in the video below and/or in this Google Colab that contains all the source code and interactive plots!
## Creation of a Test Signal
First, let's generate three signals to simulate vibration data from a hypothetical device. One is random noise in the frequency range of 10 to 50 Hz. The two other signals are pure sinewaves with amplitudes of 1 g, one at 30 Hz and the second one at 80.25 Hz. The time domain composite signal over a 10 s interval is shown in Figure 1.
Figure 1. Time domain plot of the generated signals
Inspection of the waveform does not give you any real useful information other than an indication of the magnitude of the acceleration of the vibration data. We need to determine the frequencies where the vibration is dominant.
## Using the Aggregate FFT Function
Before we explain exactly how the FFT calculation is performed, let's show you the output of our function: endaq.calc.fft.aggregate_fft(). This function allows you explicitly define the frequency resolution (this controls the length of the overlapping segments we'll discuss next), in this example, we set it to 1 Hz.
Figure 2 below is provided as a still image and an interactive plot as well so you can isolate certain signals by clicking in the legend. The magnitude of the 30 Hz sine wave is displayed as 1 g which is what would be expected. The plot of the 80.25 Hz sinewave shows a magnitude of around 0.9 g and not 1 g. This is due to the fact that the frequency, 80.25 Hz, does not align exactly with the center of a bin. The amplitude of the signal has leaked into adjacent bins so its peak at 80 Hz is a little lower than its total amplitude. Despite the minor difference in amplitude, the signal represents high vibration at 80.25 Hz. Then if you look at the random noise spectrum, you'll notice the amplitudes are relatively constant across this range as we'd expect.
Figure 2. Composite spectrum of the noise and the two sinewaves.
## How Time Interval and Bin Width Affect the Aggregate FFT
Now let me show you the Fourier transform data as a function of different time intervals and different bin widths. See Figure 9. Each row of plots represents a different bin width. The plots on the top row have a bin width of 0.5 Hz. The middle row plots were computed with a bin width of 1 Hz, and the lowest row used a 2 Hz bin width. The different color waveforms represent Fourier transforms of different time segments of the time domain data. Time segments include two seconds, three seconds, eight seconds, nine seconds, and 10 seconds.
The different time segments do not have an appreciable effect on the Fourier transform results. You can see that from the overlap in the plots of the sinewaves in the second column. You do see the amplitude of the 80.25 Hz increase to 1 g as the bin width widens and more of the frequency component is consolidated in a single bin. The wave shape also narrows similar to the 30 Hz sinewave which is centered in its bin.
Using the aggregate or averaged Fourier transform of the multiple time segments, you can see that the single frequency signals are properly identified independent of bin width. However, the wideband data represented by the 10 – 50 Hz noise does show an increase in amplitude and a smoothing of the magnitude of the Fourier transform as the bin width widens to include more spectral content per bin.
That results from the spectrum not being a computed power spectral density. The important point is if you are going to use an averaged Fourier transform for comparison between data sets, you need to make sure that all the data sets computing the Fourier transform using identical bin widths.
Figure 9. Plots of the Aggregate FFT with varying time segments and three different bin widths
## How the Aggregate FFT Function Works
Let me tell you how the Aggregate FFT function works. First, you input the parameters required by the function:
• Input overlapping segmented time domain data
• Define the bin width
Figure 10 shows the segments of the time domain data entered into the Aggregate FTT function.
Figure 10. Plot of time domain data. Each color represents a different time segment.
The Aggregate FFT function does the following:
• Applies a boxcar filter, a square box filter, to create the individual data segments. Unlike with a power spectral density function, a Hanning window is not used preventing distortion of the time domain data.
• Computes a Fast Fourier Transform on each time segment (Figure 11)
• Squares and normalizes the Fourier transforms to the bin width
• Averages all squared, normalized Fourier transforms
• Multiplies by the bin width
↑↓ Converts from a PSD calculation to a processed Fourier transform result
• Computes the square root
• Plots the result (Figure 12).
where Xpsd = the power spectral density, Δf = bin width, units are g.
Figure 5 shows plots of individual Fourier transforms for each time domain segment. The final result is the grey line in Figure 6 which shows the resulting averaged and processed composite Fourier transform of the data.
Figure 5. Individual Fourier transforms of each time segment
Figure 6. Plot of the output of the Aggregate FFT waveform, the grey waveshape, overlaying the individual FFTs
## A Real-World Example
Let’s analyze some real-world data. Figure 7 shows 0. 1 s of acceleration data on all three axes from a surgical instrument. Seven peaks in 0.1 s suggest that the aggregate FFT should show a resonance peak at approximately 70 Hz. That is about all that the time domain data tells us.
Figure 7. Time domain plot of 0.1 s of vibration acceleration data from a surgical instrument. The x-axis, y-axis, and z-axis plots are overlayed on each other.
Using the Aggregate FFT function and specifying a 1 Hz bin width yields the results displayed over a 2500 Hz bandwidth shown in Figure 8. The plot shows a low frequency peak and a number of lower magnitude frequency components. If we zoom into the lower frequencies (it's an interactive plot!) we see the expected 73 Hz acceleration peak, and it has a magnitude of 1.15 g. Designers have the data they need to help them identify the source of the vibration and investigate ways to minimize the vibration at 73 Hz.
Figure 8. Frequency domain plots using the Aggregate FFT function. The plot is over a 2500 Hz bandwidth. Zoom in on the interactive plot to see the peak at 73 Hz.
## Conclusion: Multiple Tools for Vibration Analysis
Now you have two tools for analyzing the frequency domain of a vibration signal, or really any signal! One is power spectral density analysis (see my blog on calculating a PSD) and now you have this modified PSD / FFT!
In this blog, I introduced the second tool, our Aggregate FFT function, to enable you to perform frequency domain analysis while maintaining the units you used to acquire your vibration data. You may find the Aggregate FFT function more intuitive to work with.
We have more information for you on using the enDAQ open source Python library for vibration analysis in the frequency domain. Subscribe to our library of videos on the subject. Feel free to take advantage of our open source Python library.
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# Can't See The Forest From The Trees
Geometry Level 5
There is a circular fence of radius $$R=25\text{ m}$$ that encloses an orchard. The trees in the orchard are laid out on the vertices of a square lattice of edge 5 meters. The trees are circular with equal radius, and do not grow where there is a fence. At the very center, instead of a tree, there stands a man.
What is the minimum radius for the trees (in meters) such that the man cannot see the fence?
Hint: think at Minkowski's theorem.
×
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It is currently 23 Jun 2017, 02:17
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# In the figure above, points P
Author Message
VP
Joined: 18 May 2008
Posts: 1261
In the figure above, points P [#permalink]
### Show Tags
13 Nov 2008, 03:35
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Manager
Joined: 08 Aug 2008
Posts: 230
### Show Tags
13 Nov 2008, 08:01
good one really...
at 1st sight, s looks as $$sqrt3$$
now, distance between P and Q=2$$sqrt2$$
now (s+$$sqrt3$$)^2 + (t-1)^2 =8
to solve this equation i need $$sqrt3$$ value for T, hence t=$$sqrt3$$
Hence s=1.
(if u look at the equation closely, it'll be clear....)
Manager
Joined: 23 Jul 2008
Posts: 194
### Show Tags
13 Nov 2008, 12:06
1
KUDOS
prasun84 wrote:
good one really...
at 1st sight, s looks as $$sqrt3$$
now, distance between P and Q=2$$sqrt2$$
now (s+$$sqrt3$$)^2 + (t-1)^2 =8
to solve this equation i need $$sqrt3$$ value for T, hence t=$$sqrt3$$
Hence s=1.
(if u look at the equation closely, it'll be clear....)
it s an interesting method we will have to rely on hit and trial to get the answer
My approach
Starting from the -ve X axix the order of angles formed will be 30,60,30,60
hence applying tan60 in the triangle created in the first quadrant will have sides hypotenuse= 2 base=1 (which is s) perpendicular =root 3(which is t)
hence s is 1
Director
Joined: 14 Aug 2007
Posts: 728
### Show Tags
13 Nov 2008, 20:00
radius = 2 (cant be -ve)
the equation of line passing through point P is,
y = -1/sqrt(3)x [ slope = (1-0)/(-sqrt(3) - 0) and y intercept of the line is 0)
The line passing through point Q is perpendicular to the above line
hence its equation is
y = sqrt(3)x
i.r t = sqrt(3)*s
Now OQ being radius can be expressed as
OQ^2 = s^2 + t^2
2^2 = s^2 + (sqrt(3)*s)^2
4 = S^2 * 4
S^2 = 1
S= 1 (being in Ist quadrant)
VP
Joined: 18 May 2008
Posts: 1261
### Show Tags
13 Nov 2008, 21:42
Thats a very intelligent ans. but shldnt slope of OP be -1/sqrt(3)? (slope=difference in y/difference in x)
alpha_plus_gamma wrote:
radius = 2 (cant be -ve)
the equation of line passing through point P is,
y = -sqrt(3)x [ slope = (-sqrt(3) - 0) / (1-0) and y intercept of the line is 0)
The line passing through point Q is perpendicular to the above line
hence its equation is
y = sqrt(3)x
i.r t = sqrt(3)*s
Now OQ being radius can be expressed as
OQ^2 = s^2 + t^2
2^2 = s^2 + (sqrt(3)*s)^2
4 = S^2 * 4
S^2 = 1
S= 1 (being in Ist quadrant)
VP
Joined: 18 May 2008
Posts: 1261
### Show Tags
13 Nov 2008, 21:48
Wow ! wht a shortcut. Applying angle formula never occured to me.
+1 from me
hibloom wrote:
prasun84 wrote:
good one really...
at 1st sight, s looks as $$sqrt3$$
now, distance between P and Q=2$$sqrt2$$
now (s+$$sqrt3$$)^2 + (t-1)^2 =8
to solve this equation i need $$sqrt3$$ value for T, hence t=$$sqrt3$$
Hence s=1.
(if u look at the equation closely, it'll be clear....)
it s an interesting method we will have to rely on hit and trial to get the answer
My approach
Starting from the -ve X axix the order of angles formed will be 30,60,30,60
hence applying tan60 in the triangle created in the first quadrant will have sides hypotenuse= 2 base=1 (which is s) perpendicular =root 3(which is t)
hence s is 1
Director
Joined: 14 Aug 2007
Posts: 728
### Show Tags
13 Nov 2008, 21:55
ritula wrote:
Thats a very intelligent ans. but shldnt slope of OP be -1/sqrt(3)? (slope=difference in y/difference in x)
Yes you are correct. I have edited my post. thanks
Manager
Joined: 21 Oct 2008
Posts: 131
Schools: Rady School of Management at UC San Diego GO TRITONS
### Show Tags
14 Nov 2008, 12:28
There's a much faster way to do this problem. To find the coordinates of q, all you have to do is switch the x- and y- values of the coordinates of P, then multiply the new y-value (t) by -1.
To understand why, imagine the full circle. Draw any two perpendicular lines through the center. They will intersect the circle at (a, b), (b, -a), (-a, -b), and (-b, a).
Director
Joined: 27 Jun 2008
Posts: 542
WE 1: Investment Banking - 6yrs
### Show Tags
14 Nov 2008, 12:45
7-t71146
Re: GMATPREP- semicircle [#permalink] 14 Nov 2008, 12:45
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# Understanding The Maths Behind Betting Odds Explained
If you see odds of +300, this means you will win \$300 for every \$100 you stake. The American odds will give the bookmaker influence over the point spread without changing the key number of 3. Odds worse than -110 as we discussed can hurt your breakeven percentage and should be taken into account even when betting spreads. On the other hand, underdog plus odds will help your bottom line. A -135 favorite means you must risk \$135 to win \$100 from the sports book. If you do live betting, don’t forget that you can enjoy over/unders, Moneyline, Point Spreads, and props even with in-game betting.
• There is much debate surrounding the actual origins of the game of baseball we see today.
• For big games with a lot of action, sharp bettors often place big wagers just before a game, which can also create optimal betting opportunities.
• Baseball wagering is unique in that there is no traditional point spread set for a ballgame.
• Likewise, you will have a better chance of having your bet pay off.
• However, the patterns can mislead to the approach you come with if they are perceived.
If you wanted to bet on a number with a negative number, the odds represent the amount you need to wager to receive a winning profit of £100. Decimal Odds are particularly popular in the UK with football betting and offer odds in decimal form. One of the most popular forms of handicap betting is handicap league betting, whereby you make a handicap bet on a team’s performance across the course of a season.
## Where Can I Bet On Nfl Games?
There are many ways to bet on sports, and each requires its informative post own set of knowledge. “Outs” are simply the technical term for the number of cards that can make you the winning hand. For example, if you have Ah-Qh against Jc-Jd on a flop of 8h-5h-2c, you can win with any heart, Queen or Ace for a total of fifteen outs.
## How Do You Read American, Fraction And Decimal Odds?
The coin will definitely land on either heads or tails, which taken together provide us with the certain event. To make Betting Directions, Specifications & Previews online keno games No-cost Playing Specifications » Betfair Web site all this information slightly easier to understand, here is a quick example of a bet on a football match. In betting, odds represent the ratio between the amounts staked by parties to a wager or bet.
If you know how to watch lines move and predict where they’re going, money lines can offer a lot of value. For big favorites, it’s often best to wait until 1-2 hours before game time. For big games with a lot of action, sharp bettors often place big wagers just before a game, which can also create optimal betting opportunities. Positive money lines show the profit that would be won from a \$100 bet, not including the original wager. Negative money lines show the wager required to win \$100 in profit. This is what sports bettors refer to when they talk about a betting line on a given contest, although sometimes they use this term to discuss individual wager types within the betting line.
## Virgin Games Bonus Code And Sign
With the amount of money in the market, betting odds fluctuate depending on where the money is being placed. Less popular sports may potentially have less favorable odds AND less money in the market. The use of decimal odds is something that has become more prevalent during recent years. They are most widely used in mainland Europe and Australia, but younger bettors are beginning to favor them more and more with the ever-growing popularity of sports betting online. Example of American odds The way that American odds work for betting on a favorite is that it shows how much money you would have to risk in order to win a certain amount. Imagine for example you were betting on the Campeonato Brasileiro Serie B soccer match between Boa Esporte Clube and Salgueiro AC, as shown in the images.
This feature allows gamblers to remove selections from their accumulator after the bet has been placed and in some instances after the selected event has started. The betting slip is then revised to feature the amended selections and a new potential return amount. A half bet applies only to the score of the first or second half. This can also be applied to a specific quarter in American football or basketball, a fewer number of innings in baseball, or a specific period in hockey.
They will offer betting odds on the chosen outcome which can be accepted and a bet placed. They keep the money from losing bets and pay winners on correct predictions. This is a very common occurrence in sports betting and sportsbooks have the full right to shift the spread or odds for any given match prior to its start. Many factors can influence a change of the spread such as injuries, the number of bets coming in for either team or the weather, to name a few. Depending on the timing of placing the bet, the bettor can also have an advantage or a disadvantage depending on which way the spread has shifted. Some sports betting experts suggest that you shop different online sportsbooks for lines and odds and specific events in order to find a small edge whenever placing bets on games.
## Everything You Need To Know About Esports Betting Odds
This ensures that the sports books are guaranteed a profit because of the 10 percent commission or “vigorish” charged on most sports wagers. If one side on a game is being bet more heavily, the bookie must move the number in order to attract interest on the other side in order to balance action. In this example the Jets are listed as four-point favorites (-4) over the Bills and the 49ers are three-point underdogs (+3) against the Seahawks. So, if you bet \$110 on the favored Jets, they must defeat the Bills by more than four points in order to win \$100.
A negative money line is indicated with a ‘-‘ in front of a number, like -230. The number represents the amount of money that would need to be wagered in order to win \$100 in profit. So, a -230 money line would require a wager of \$230 to win \$100 more, for a total payout of \$330 (\$230 wager + \$100 profit). Of course a \$230 bet is not required, it’s just the basis for calculating the payout.
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## ››Convert kelvin to degree Celsius
kelvin degree Celsius
How many kelvin in 1 degree Celsius? The answer is 1.
We assume you are converting between kelvin and degree Celsius.
You can view more details on each measurement unit:
kelvin or degree Celsius
The SI base unit for temperature is the kelvin.
1 kelvin is equal to 1 degree Celsius.
Note that rounding errors may occur, so always check the results.
Use this page to learn how to convert between kelvins and degrees Celsius.
Type in your own numbers in the form to convert the units!
## ››Want other units?
You can do the reverse unit conversion from degree Celsius to kelvin, or enter any two units below:
## Enter two units to convert
From: To:
## ››Definition: Kelvin
The kelvin (symbol: K) is the SI unit of temperature, and is one of the seven SI base units. It is defined by two facts: zero kelvins is absolute zero (when molecular motion stops), and one kelvin is the fraction 1/273.16 of the thermodynamic temperature of the triple point of water. The Celsius temperature scale is now defined in terms of the kelvin, with 0 °C corresponding to 273.15 kelvins, approximately the melting point of water under ordinary conditions.
## ››Definition: Celsius
The degree Celsius (°C) is a unit of temperature named after the Swedish astronomer Anders Celsius (1701–1744), who first proposed a similar system in 1742. The Celsius temperature scale was designed so that the freezing point of water is 0 degrees, and the boiling point is 100 degrees at standard atmospheric pressure.
## ››Metric conversions and more
ConvertUnits.com provides an online conversion calculator for all types of measurement units. You can find metric conversion tables for SI units, as well as English units, currency, and other data. Type in unit symbols, abbreviations, or full names for units of length, area, mass, pressure, and other types. Examples include mm, inch, 100 kg, US fluid ounce, 6'3", 10 stone 4, cubic cm, metres squared, grams, moles, feet per second, and many more!
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• Print view
## Mie Paramtres
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Topic author
Cambapo
Posts: 23
Joined: 10.01.2011
With us: 10 years
### Mie Paramtres
Hi. I need some help her.
I first made another post because I'm trying to put Mie parametres to Jupiter, Saturn, Uranus and Neptune atmospheres. Jhon Van Vliet helpedme out a lot with it, and now I have Uanus and Neptune just perfect, but the problem is that still i don't undersatnd what wxactly the parametres mean and what they do, Mie, MieAsymetry, Rayleigh (I suppose that's the color) and Absorption, If you could explain em that would appreciate.
About Jupiter and Saturn, I just know that the format is R, G, B, But I dunno wich values to put there, cus' the atmospheres look foggy and colorless, and in saturn the sunset looks blue, when in both planets it's supposed to be red. I would attach some images so you know what I'm talking about. And beforehand, thanks a lot.
Topic author
Cambapo
Posts: 23
Joined: 10.01.2011
With us: 10 years
### Re: Mie Paramtres
The other pics...
PlutonianEmpire
Posts: 1363
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Age: 36
With us: 16 years 4 months
Location: MinneSNOWta
### Re: Mie Paramtres
The "mie" setting influences the density of the atmosphere. The number of zero's after the decimal point in the "rayleigh" and "aborption values must be the same number of zero's after the decimal point as the "mie" parameters.
Example:
Code: Select all
`Mie 0.0112MieAsymmetry -0.26Rayleigh [ 0.0103 0.025 0.061 ]Absorption [ 0.0234375 0.040625 0.01125 ]MieScaleHeight 13.5`
Code: Select all
`Mie 0.00112MieAsymmetry -0.26Rayleigh [ 0.00103 0.0025 0.0061 ]Absorption [ 0.00234375 0.0040625 0.001125 ]MieScaleHeight 13.5`
Code: Select all
`Mie 0.000112MieAsymmetry -0.26Rayleigh [ 0.000103 0.00025 0.00061 ]Absorption [ 0.000234375 0.00040625 0.0001125 ]MieScaleHeight 13.5`
The "rayleigh" numbers are the RGB numbers for the color of the atmosphere, in RGB order.
The "absorption" numbers influence the colors of the sunset. HOWEVER, aborption goes from the top down, instead of bottom up like the other color values do.
For example, the RGB value for lavender, converted into celestia format, is [ 0.71 0.49 0.86 ], which would normaly give you a lavender sky if you used these numbers for the "rayleigh" setting. To get lavender sunsets, you must subtract these numbers from "1", so it would be 1 - 0.71 = 0.29 for the Red, 1 - 0.49 = 0.51 for the green, and 1 - 0.86 = 0.14 for the blue, and use those numbers for the "absorption" values, so [ 0.29 0.51 0.86 ] as the absorption values. Don't forget to add the zero's in between the decimal point and the first number AFTER the decimal point, so in the end, your "absorption" values should look like this [ 0.0029 0.0051 0.0086 ].
The "MieScaleHeight" adjusts how tall the atmosphere is, multiplied by 5. The .ssc definition for Earth's atmosphere default atmosphere height is 60 km. Divide that by 5, and you get 12, which is the MieScaleHeight value given for Earth's atmosphere. Multiply 12 by 5, you get 60, and so forth.
Also, the size of the planet influences how many zero's you should place in between the decimal point and the first number, in the "Mie", "Rayleigh" and "Absorption" settings. The bigger the planet, the more zeroes you need to insert. So for Earth, you get:
Code: Select all
`Mie 0.001MieAsymmetry -0.25Rayleigh [ 0.001 0.0025 0.006 ]MieScaleHeight 12`
However, for planets with radii higher than, say, 20000 km, you need to insert a third zero, as well as increase the MieScaleHeight. So, for Neptune, you get:
Code: Select all
`Mie 0.0001MieAsymmetry -0.25Rayleigh [ 0.0001 0.00025 0.0006 ]MieScaleHeight 120`
For objects significantly smaller than Earth, with, say, a radii below 2000 km, you'd take a way a zero from the Mie, Rayleigh, and Absorption values.
So, for example, the Moon would get:
Code: Select all
`Mie 0.01MieAsymmetry -0.25Rayleigh [ 0.01 0.025 0.06 ]MieScaleHeight 3`
Also, the "MieAssymetry" values, which is from -1.0 to 1.0, shows how light goes through the atmosphere, with 1.0 deflecting light back to the sun, and -1.0 makes the planet's nightside appear to "shine", if you put the planet between you and the light source.
Finally, the "absorption" line is completely optional. You don't have to use it if you don't want to.
Maybe someone with better understanding can explain it better than I just did.
Terraformed Pluto: Now with New Horizons maps! :D
Topic author
Cambapo
Posts: 23
Joined: 10.01.2011
With us: 10 years
### Re: Mie Paramtres
Thanks a lot, now I DID understand everything, I'm a little dumb, I know , but at least I could undersatnd the meaning of everything. Thanks a lot. Mil Gracias XDD
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# math
posted by on .
Two trains heading toward each other are 400 miles apart. One train travels 15 miles per hour faster than the other train. If they arrive at the same station in 5 hours, how fast is each train traveling?
• math - ,
time = distance over speed
since the time is the same for both trains, and their combined distance is 400,
d/s = (400-d)/(s+15) = 5
d = 5s, so
(400-5s)/(s+15) = 5
s = 32.5
d = 162.5
check: 162.5/32.5 = 5
237.5/47.5 = 5
• math - ,
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# Why can a rational function be put in form $P(x)/Q(x)=\sum_{i=1}^n\frac{P(a_i)}{Q'(a_i)(x-a_i)}$?
In preparation for Lagrange's Interpolation Theorem, I'm curious to know why a rational function can be put in the following form.
Suppose $Q$ is a polynomial with distinct roots $a_1,\dots,a_n$, and let $P$ be a polynomial with $\deg(P)<n$. Then why does $$\frac{P(x)}{Q(x)}=\sum_{i=1}^n\frac{P(a_i)}{Q'(a_i)(x-a_i)}?$$ I tried expanding the right hand sum over a common denominator, but it got horribly messy quickly. I also tried induction, assuming $Q$ has just one distinct root, and thus $Q=C(x-a_1)^k$. But then $\deg(P)<1$, so $P$ is constant, say $P=A$. But then $$\frac{P(a_1)}{Q'(a_1)(x-a_1)}=\frac{A}{kC(a_1-a_1)^{k-1}(x-a_i)}$$ which seems problematic if $k>1$. What's the slicker way to conclude this equality? Thanks,
• You are not interpreting "$Q$ is a polynomial with distinct roots" in the intended way. It is intended to mean that the polynomial has degree say $n$, and all of its $n$ complex roots are different from each other. So $Q(x)=(x-1)^2$ is not covered. Indeed, the result is in general false for this $Q$, as you may remember from partial fractions. – André Nicolas Jan 24 '12 at 4:15
• That clears it up, thanks. – Gemma Jan 24 '12 at 4:17
Multiply through by $Q(x)$ and note that both sides are polynomials of degree less than $n$ agreeing at the $n$ points $x=a_i$.
• How can you be sure the right hand side has degree less than $n$? Each summand $\displaystyle\frac{P(a_i)Q(x)}{Q'(a_i)(x-a_1)}$ has degree $\deg(Q)-1$, but what if $\deg(Q)>n$? – Gemma Jan 24 '12 at 3:58
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# 1/2 Tsp To Ml
1/2 Tsp To Ml – This post may contain affiliate links, which means I’ve computerized a bit of what you sell without English. Please see my full policy
Knowing how many milliliters are in a teaspoon is just as important because imperial units of measurement are not comparable!
## 1/2 Tsp To Ml
Although most people think of spoons or spoons as basic tools, they are also specialized pieces of equipment used to measure sound.
### A Colorful Yellow Measuring Spoon Stock Photo
A teaspoon is defined as 3 teaspoons in imperial measurements, but is also defined as 1/16 of a cup or 1/2 of a US cup of water.
A tablespoon is equivalent to 4 fluid plays and half a volume from a Greek drachm.
Consequently, one teaspoon measures 14.7868 milliliters (ml in SI). It is often closer to 15 ml.
Because a teaspoon is a special measuring piece, it’s important to buy the right measuring spoon rather than just any type.
## Clearance! Measuring Spoons Sets Of 6 With Tablespoon, Teaspoon, 3/4 Tsp, 1/2 Tsp 1/4 Tsp 1/8 Tsp, Slim Design Measuring Cups For Dry Spices And Liquid Cooking, Stackable Spoons Easy Storage, S5629
The conversion method is to multiply the number of spoons by 15. This is difficult to do, but it is very easy to multiply first by 5, then by 3.
A teaspoon is one-third of a tablespoon. Therefore, one US teaspoon measures exactly 4.92892 mL, which is rounded up to 5 mL.
When a recipe calls for something dry, like almond flour or coconut flour, it’s to indicate quantity.
While some recipes call for heaping tablespoons or rounded tablespoons, the measurements don’t have to be exact and you don’t need to measure them.
## Mainstays 4 Pieces Stainless Steel Measuring Spoon Set 1/4, 1/2, 1 Tsp 1 Tbsp Ml
ML stands for milliliter, a unit of measurement in SI (International System of Units) defined as 1/1000 of a liter (from the prefix milli-).
Since one liter equals 1000 cubic centimeters, 1ml = 1 cubic cm. That’s why a teaspoon is equal to 15 inches.
No, the abbreviation for tbsp (also available as a capital T) is a teaspoon or 15ml in the US (20ml in Australia).
Hi, I’m Caryn, food blogger, author, recipe developer, published author of several books and founder of Sweet as Honey.
### Amazon.com: Measuring Spoons Stainless Steel Metal Teaspoon Measure Spoon Set Of 9 For Dry And Liquid Ingredients Fits In Spice Jar: Home & Kitchen
I look forward to sharing all my easy and delicious recipes that are delicious and healthy. My background in chemistry and years of following a low-carb keto diet have given me expertise in this area. But since my husband is a vegetarian, I also know how to cook vegetarian and vegan.
My true passion is cooking and baking. In fact, I am sharing only a small portion of my results on Sweet As Honey. Most of them were eaten by my husband and two kids before I could take any pictures!
All my solutions have been tested at least three times to make sure they work and I am happy to keep them as good as possible.
The information, suggestions and articles on this site should not be taken or used as medical advice. The nutritional values given should be used as a guide only. Recipes are calculated using WP Recipe Maker. Net carbs are calculated by subtracting fiber and some carbs.
### Pc Measuring Spoon & Cup Set, Turquoise
You should always calculate the nutritional information yourself instead of relying on statistics. And its contents and content are not intended to cure, prevent, diagnose or treat any disease. cannot be held responsible for any adverse or other consequences arising from the use of recipes or advice found on the website.
I love to cook most of all, you will find many dishes here, but everything you and your loved ones need, even if they are on a diet!
I have answers for everyone, whether regular exercise, keto, gluten-free, vegetarian, vegan, nut-free, Mediterranean, or grain-free (see all answers by diet). How many teaspoons are in a teaspoon? you are not alone This is a common question that we will answer in detail so that you can easily navigate the recipe!
Today we are breaking spoons and spoons. Understanding these basics is important for preparing recipes such as many breakfast, dinner and dessert dishes at TSV. 🧑 🍳
### If It Says To Add 2 Tsp To 10 Gallons How Much Do I Put In 1 Gallon?
So let’s dive into this ‘how many tablespoons in a Tbsp’ question – although the answer is simple: 3 tablespoons make 1 tablespoon. We’ll cover US/Imperial and metric conversions charts, conversion formulas, measurement tips, and more!
Tablespoons and teaspoons both measure amounts in recipes. The main difference between them is their size and capacity:
A small spoon is suitable for measuring small quantities of ingredients such as spices, seasonings or extracts, while large spoons are suitable for measuring large quantities such as oil, butter or sugar.
Using the correct measuring spoon for the measurements specified in the recipe ensures accurate results. You definitely don’t want to use a tablespoon of baking powder when the recipe only calls for one!
#### Stainless Steel Measuring Spoons, Set Of 6 For Measuring Dry And Liquid Ingredients
A teaspoon is abbreviated as ‘tsp’ or ‘t’. Abbreviations are never capitalized to indicate that they are abbreviations. A teaspoon is abbreviated as ‘Tbsp’, ‘tbsp’, ‘tbs’, or ‘T’. A capital T indicates that it is the largest quantity. Smaller text helps save time and makes the recipe easier to read. Although when I write recipes I prefer to use all caps so that there is no mistake between the two!
1 teaspoon contains 3 teaspoons. The conversion can be expressed as 3 tsp = 1 tbsp or 3 t = 1 T. It can also be written as 1 teaspoon = 3 teaspoons.
Another way to look at this variable is: 1 teaspoon is 1/3 of a teaspoon, so 3 teaspoons = 1 teaspoon.
It’s good to know that there are 3 teaspoons in a tablespoon if you double or double it. This allows for more flexibility and flexibility!
#### Deiss Pro 7 Piece Stainless Steel Measuring Spoon Set With Leveler For Cooking & Spices
To convert teaspoons to tablespoons, simply divide the number of tablespoons by 3. This is because 1 tablespoon contains 3 teaspoons.
For example, if you have 9 teaspoons and want to convert them to teaspoons, you would calculate:
To convert tablespoons (tbsp) to tablespoons (tsp), you simply multiply the number of tablespoons by 3. Because 1 tablespoon contains 3 tablespoons.
For example, if you have 4 teaspoons and want to convert them to tablespoons, you would calculate:
### Adjustable Measuring Spoons Set Of 2 Kitchen Utensils Spice
Sometimes you can use recipes that use milliliters (milli) instead of teaspoons (teaspoon) and teaspoons (tbsp). Learning how to convert between the traditional US and metric systems will help make the transition a breeze.
To convert tablespoons (teaspoon) to milliliters (ml), multiply the number of teaspoons by 5. For example, 2 teaspoons = 2 x 5 = 10 milliliters.
To convert teaspoons (tsp) to milliliters (ml), multiply the number of tablespoons by 15. For example, 2 teaspoons = 2 x 15 = 30 milliliters.
Here’s a straightforward chart to help you convert teaspoons (teaspoon) and tablespoon (tsp) to metric units, specifically milliliters (mL):
### Pcs/set Magnetic Measuring Spoons Set With Leveler Stainless Steel Liquid Ingredients Double Sided Measurement Set For Cooking
Here’s a simple conversion table so you can see how different measurements look in different areas.
Using the wrong amount of baking powder in recipes like Vegan Banana Bread or Blueberry Muffins can significantly affect the end result of your baked goods. If you use a lot of baking powder, it can be light and airy on top and have a strong flavor. If you use too much baking powder, it can become dense and heavy and not rise well. A general rule of thumb is to use 1 to 1/4 teaspoon (teaspoon) of baking powder per 1 cup of flour.
There were times when small and large silver spoons were used to measure ingredients. Today, measuring spoons are common, while silver can vary in size. For this reason, it is not recommended to use silverware to measure spices. But if you’re short, and you don’t have standard measuring spoons or teaspoons, you can use a small silver spoon or a large tablespoon as a measuring spoon.
Before standard measuring tools, cooks often used their hands, fingers, or common household tools to measure spices, giving rise to terms like “pinch” and “dash”. Even today, recipes will include a pinch, a dash, or ingredients. Basically all these words mean the same thing.
### Piece Stainless Steel Measuring Cups And Spoons Set In Country Chic
In the US system, 1 teaspoon contains 6 half teaspoons (1/2). Since 1 teaspoon contains 2 teaspoons (1/2) and 1 teaspoon contains 3 teaspoons, the conversion is correct. So half (1/2) teaspoon 3 teaspoons = 6 half (1/2) teaspoons multiplied by one teaspoon.
1 teaspoon in the US system contains 12 – 1/4 teaspoons. The conversion is correct because 1 teaspoon contains 4 (1/4) teaspoons and 1 teaspoon contains 3 teaspoons. Multiply 4 teaspoons – 1/4 teaspoon by 3 = 12 (1/4) teaspoons.
In Australia there are 4 teaspoons in a teaspoon. The conversion is correct because 1 Australian teaspoon is equal
Ml to tsp, 5 ml to tsp, is 1 tsp 5 ml, convert 2 ml to tsp, 1 8 tsp to ml, 1 ml to tsp conversion, 1 1 2 tsp to ml, 1 tsp to ml, ml to tsp conversion chart, convert 1 tsp to ml, 2 5 ml to tsp, 2 ml to tsp
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# SOLUTION: can you please help me with this problem? thank you Hint: Group Factoring 4x^2-x-3 thanks
Algebra -> Algebra -> Test -> SOLUTION: can you please help me with this problem? thank you Hint: Group Factoring 4x^2-x-3 thanks Log On
Ad: Algebrator™ solves your algebra problems and provides step-by-step explanations! Ad: Algebra Solved!™: algebra software solves algebra homework problems with step-by-step help!
Test Calculators and Practice Answers archive Word Problems Lessons In Depth
Hint: Group Factoring
4x^2-x-3
thanks
You can put this solution on YOUR website!
Looking at the expression , we can see that the first coefficient is , the second coefficient is , and the last term is .
Now multiply the first coefficient by the last term to get .
Now the question is: what two whole numbers multiply to (the previous product) and add to the second coefficient ?
To find these two numbers, we need to list all of the factors of (the previous product).
Factors of :
1,2,3,4,6,12
-1,-2,-3,-4,-6,-12
Note: list the negative of each factor. This will allow us to find all possible combinations.
These factors pair up and multiply to .
1*(-12)
2*(-6)
3*(-4)
(-1)*(12)
(-2)*(6)
(-3)*(4)
Now let's add up each pair of factors to see if one pair adds to the middle coefficient :
First NumberSecond NumberSum
1-121+(-12)=-11
2-62+(-6)=-4
3-43+(-4)=-1
-112-1+12=11
-26-2+6=4
-34-3+4=1
From the table, we can see that the two numbers and add to (the middle coefficient).
So the two numbers and both multiply to and add to
Now replace the middle term with . Remember, and add to . So this shows us that .
Replace the second term with .
Group the terms into two pairs.
Factor out the GCF from the first group.
Factor out from the second group. The goal of this step is to make the terms in the second parenthesis equal to the terms in the first parenthesis.
Combine like terms. Or factor out the common term
---------------------------------------------
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# Homework Help: Clausius-Clapeyron equation help
1. Nov 19, 2005
I know that the relationship between pressure and temperature for an ideal gas is linear. The relationship between vapor pressure and temperature for a liquid, however, is exponential. To make it linear we take the natural log and end up with: $$\ln P = -\frac{\Delta H_{vap}}{RT} + b$$. How did we get $$-\frac{\Delta H_{vap}}{RT}$$ to be the slope? The y-axis is $$\ln P$$ and the x-axis is $$\frac{1000}{T}$$.
Thanks
Last edited: Nov 19, 2005
2. Nov 20, 2005
anybody have any ideas?
thanks
3. Nov 20, 2005
### da_willem
I'm not sure what your question really is but if you plot lnP vs 1000/T the slope will be
$$-\frac{\Delta H_{vap}}{1000 R}$$
4. Nov 20, 2005
### ShawnD
The slope will be whatever -Hvap/RT is divided by to get the x axis. If I have y=ab and I plot y vs b, the slope is a. If I plot y vs a, the slope is b. The only (theoretical) way you will get -Hvap/RT as the slope is if you plotted lnP vs 1, but this doesn't make any sense. So in conclusion, you will never get -Hvap/RT as your slope
Which variable are you trying to solve or prove something for?
5. Nov 20, 2005
The slope was actually $$\frac{\Delta H_{vap}}{R}$$. I think it was meant to be written as: $$\ln P = -\frac{\Delta H_{vap}}{R}\frac{1}{T} + b$$
Is this correct?
Thanks
6. Nov 20, 2005
### GCT
yes, and assuming that we're referring to vapor pressure 1 as 1atm, b is
(DHvap/R)1/T(1atm)
Last edited: Nov 20, 2005
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Win a copy of Functional Reactive Programming this week in the Other Languages forum!
# puzzle - water jar
ankur rathi
Ranch Hand
Posts: 3830
You have three jars that hold max 19, 13 and 7 liters of water respectively and 20 liters of water in them (not unlimited). I want two jars with 10 liters of water.
Sandip Sankeshwar
Ranch Hand
Posts: 210
Dude,I used to solve these puzzled in early 90s.Later these kind of puzzles disappeared from scene somehow.I will give it a try in tea break at 4 PM.
Ryan Maurer
Greenhorn
Posts: 5
• 1
Here you go. I did it in 14 steps. I don't know if it can be done in fewer.
I will label the 3 jars j1, j2 and j3.
j1 holds 19 litres
j2 holds 13 litres
j3 holds 7 litres
I am starting off by filling j1 full (19 litres) and putting the remaining litre in j2, so I have this makeup
j1 = 19
j2 = 1
j3 = 0
Step 1 Fill j3
j1 = 12
j2 = 1
j3 = 7
Step 2 put j3 in j2
j1 = 12
j2 = 8
j3 = 0
Step 3 fill j3
j1 = 5
j2 = 8
j3 = 7
Step 4 fill j2 with j3
j1 = 5
j2 = 13
j3 = 2
Step 5 put j2 into j1
j1 = 18
j2 = 0
j3 = 2
Step 6 put j3 into j2
j1 = 18
j2 = 2
j3 = 0
Step 7 fill j3
j1 = 11
j2 = 2
j3 = 7
Step 8 put j3 into j2
j1 = 11
j2 = 9
j3 = 0
Step 9 fill j3
j1 = 4
j2 = 9
j3 = 7
Step 10 fill j2 with j3
j1 = 4
j2 = 13
j3 = 3
Step 11 put j2 in j1
j1 = 17
j2 = 0
j3 = 3
Step 12 put j3 into j2
j1 = 17
j2 = 3
j3 = 0
Step 13 fill j3
j1 = 10
j2 = 3
j3 = 7
Step 14 put j3 into j2
j1 = 10
j2 = 10
j3 = 0
Raghavan Muthu
Ranch Hand
Posts: 3381
Ryan you rock
Rajaraman Sriraman
Greenhorn
Posts: 3
Am trying to make it to 14 steps
j1 = 19
j2 = 1
j3 = 0
-------------
j1 = 12
j2 = 1
j3 = 7
----------------
put j3 in j2
j1 = 12
j2 = 8
j3 = 0
-----------
fill j3
j1 = 5
j2 = 8
j3 = 7
--------------
fill j2 with j3
j1 = 5
j2 = 13
j3 = 2
-----------------
put j2 into j1
j1 = 18
j2 = 0
j3 = 2
-----------
put j3 into j2
j1 = 18
j2 = 2
j3 = 0
--------------
fill j3
j1 = 11
j2 = 2
j3 = 7
--------------
j3 into j2
j1 = 11
j2 = 9
j3 = 0
--------------
fill j3
j1 = 4
j2 = 9
j3 = 7
--------------
fill j2 with j3
j1 = 4
j2 = 13
j3 = 3
--------------
put j2 in j1
j1 = 17
j2 = 0
j3 = 3
--------------
put j3 into j2
j1 = 17
j2 = 3
j3 = 0
--------------
fill j3
j1 = 10
j2 = 3
j3 = 7
--------------
put j3 into j2
j1 = 10
j2 = 10
j3 = 0
--------------
Rammy Rocks
Rajaraman Sriraman
Greenhorn
Posts: 3
j1 = 12
j2 = 1
j3 = 7
Step 2 put j3 in j2
j1 = 12
j2 = 8
j3 = 0
Step 3 fill j3
j1 = 5
j2 = 8
j3 = 7
Step 4 fill j2 with j3
j1 = 5
j2 = 13
j3 = 2
Step 5 put j2 into j1
j1 = 18
j2 = 0
j3 = 2
Step 6 put j3 into j2
j1 = 18
j2 = 2
j3 = 0
Step 7 fill j3
j1 = 11
j2 = 2
j3 = 7
Step 8 put j3 into j2
j1 = 11
j2 = 9
j3 = 0
Step 9 fill j3
j1 = 4
j2 = 9
j3 = 7
Step 10 fill j2 with j3
j1 = 4
j2 = 13
j3 = 3
Step 11 put j2 in j1
j1 = 17
j2 = 0
j3 = 3
Step 12 put j3 into j2
j1 = 17
j2 = 3
j3 = 0
Step 13 fill j3
j1 = 10
j2 = 3
j3 = 7
Step 14 put j3 into j2
j1 = 10
j2 = 10
j3 = 0
Rajaraman Sriraman
Greenhorn
Posts: 3
ok
Campbell Ritchie
Sheriff
Posts: 50204
79
Welcome to JavaRanch and sorry I didn't notice you earlier.
colin shuker
Ranch Hand
Posts: 750
I'm missing something, why not just pour 10 litres in j1, and 10 litres in j2?
Campbell Ritchie
Sheriff
Posts: 50204
79
You have no way of measuring 10. Only 7, 13 and 19.
Ryan McGuire
Ranch Hand
Posts: 1078
4
Since this is the Programming Diversions forum, would anyone care to discuss how you'd go about programming a solution to this type of puzzle?
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## Instructions:
Replace vegetables by digits so that the result of the sum in the bottom line is correct. One row makes of one number. Add two numbers to get the sum. Beware, each type of vegetable is a different digit.
In fact, this is a written addition, with the difference that the individual digits of the numbers or even the sum are hidden behind the vegetables.
+
=
+
=
## Solution
8
0
2
6
+
1
6
7
6
=
9
7
0
2
8026 + 1676 = 9702
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# Tuenti Challenge 9
## Challenge 2 - Help Battlestar Galactica and save humanity
The Galactica spaceship must escape from Cylon persecution. Moreover, Adama (Battlestar Galactica's commander) and Laura Roslin (President of the 12 Colonies) want to reach a new home for humanity. Right now, Galactica is waiting on a planet and wants to go to New Earth (Adama knows where it is). But, to avoid being caught by the Cylon, Adama is planning to visit some other planets before heading towards the final destination. Gaius Baltar is a little busy right now, so you will be the one who saves humanity. And, just to be sure it's worth the trouble, Adama asks you how many different paths can be taken from your current planet to the final destination.
Because of some complicated laws of physics, you can't go everywhere from every planet. And, because you don't want to backtrack, the Battlestar must always go forwards to another planet that’s allowed by the laws of hyperspace jumps, so there can't be any loops. See the chart showing possible situations for an explanation.
In order to explain all above, let’s see this graph, that represents a possible situation:
How many paths are there for the Galactica to reach New Earth? If you count them you can see there are 5 different paths.
• Path 1: Galactica -> A -> E -> New Earth
• Path 2: Galactica -> A -> D -> F -> New Earth
• Path 3: Galactica -> B -> D -> F -> New Earth
• Path 4: Galactica -> C -> D -> F -> New Earth
• Path 5: Galactica -> C -> F -> New Earth
Your goal is to help Adama and Laura by writing a program that will output the number of different paths to reach New Earth given a random map like the one above so they can make a decision.
Assumptions:
• The map has no loops. From any given planet you can only go forward to another planet.
• There is at least one path to every planet, except the planet where the Galactica is waiting (the initial planet).
• All planets, except New Earth which is the final planet, have at least one path to move forwards.
### Input
The first line of the file has the number of cases. Each case starts with a line containing the number P of planets from which you are able to jump. After that there is one line for every planet (except “New Earth”) with a comma separated list of allowed hyperspace jumps for that planet. So, each case will have this format:
```P
Galactica:Destination1,Destination2,..,DestinationM
Destination1:DestinationX,DestinationY
...
DestinationZ:New Earth
```
The first planet will always be named “Galactica” and the final planet will always be named “New Earth”.
### Output
Your output should be an integer with the number of different paths with one line for each case.
1 ≤ P < 200
```1
7
Galactica:A,B,C
A:E,D
B:D
C:D,F
D:F
E:New Earth
F:New Earth
```
### Sample Output
```Case #1: 5
```
### Problem stats
Completion time: min: 0:04:59 h 10th percentile: 0:27:13 h 90th percentile: 24:30:30 h max: 131:42:20 h 10th percentile: 0:24:27 h 90th percentile: 24:06:38 h 10th percentile: 0:00:46 h 90th percentile: 0:12:12 h 382
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Wed May 7 08:50:09 UTC 2008
```On Wed, May 07, 2008 at 06:00:38AM +0200, Jens M Andreasen wrote:
> One thing that I wonder is what the pattern of addition in Fons's
> application really looks like. I assume the fftA * fftB is some windowed
> precalculated impulsresponse and a signal? The addition/accumulate
> suggests that the output fftD has been touched before, implying that
> there could be more variables to work on at once or that the vectors
> would still be in the cache if the order of addition was changed.
It's quite complex. I'll try to build up a picture in four steps.
1. The first step to imagine is just a scalar matrix operating on
a number of input sample streams A [j][t] and producing a number
of output sample streams B [i][t]:
index i = output
index j = input
index t = time
B [i][t] = sum_j (M [i][j] * A [j][t])
2. Now imagine that the matrix is not scalar, but a matrix of FIR
filters:
B [i][t] = sum_j (sum_k (M [i][j][k] * A [j][t- k]))
with k = 0...L-1, for filters of length L. We now need to
store the past L-1 samples of each input as well.
Assume this calculation is done for one value of t (i.e. one
sample) at a time (we'll see why soon). There are six ways to
organise the three nested loops. The one actually used is such
that k is the innermost index.
3. Now imagine that each sample in the above equation is actually a
block of P samples. This is how FFT based partitioned convolution
works. We need a size 2P FFT on each input block (the second half
are zeros) and on each matrix element (the latter is precomputed
of course), a size 2P IFFT on each output block, and each of the
multiplications now becomes a MAC loop over P complex values. The
output size is 2P, i.e. the second half overlaps witht the next
iteration and has to be stored. There are size(A) * size(B) * L
such MAC loops for each block, so it helps to optimize this
operation.
4. The scheme above is FFT-based partitioned convolution with a single
partition size P. For efficiency you want large P, but this also
introduces processing delay as you can only start the computation
when P new input samples are available. To avoid this delay in a
real-time application zita-convolver uses multiple partition sizes,
small at the start of an IR, and larger ones for the later parts.
There can be up to five sizes. Calculations for P == period size
are performed directly in the JACK callback, the longer ones are
performed by lower priority threads. A final optimisation is that
a sparse matrix representation is used in all three dimensions,
so no time or memory is wasted on zero-valued data.
--
FA
Laboratorio di Acustica ed Elettroacustica
Parma, Italia
Lascia la spina, cogli la rosa.
```
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## Test #2 question 4
lauren chung 2f
Posts: 50
Joined: Fri Sep 29, 2017 7:03 am
### Test #2 question 4
On Test 2, it asks to calculate the standard potential for the galvanic cell,
U (s)| U+3 (aq) || V+2 (aq) | V (s)
Since
U+3 + 3e- --> U E= -1.79
V+2 + 2e- --> V E = -1.18
but U is the anode, so I made the equation
U --> U+3 + 3e- E= +1.79
Then I wrote
Ecell = Ecathode - Eanode
= -1.18 - (1.79)
= -2.97 V
Can someone tell me where I went wrong?
Xin He 2L
Posts: 38
Joined: Fri Sep 29, 2017 7:05 am
### Re: Test #2 question 4
You don't flip the signs of the cell potential when you flip the equation.
lauren chung 2f
Posts: 50
Joined: Fri Sep 29, 2017 7:03 am
### Re: Test #2 question 4
Can you explain why you don't flip the signs of the cell potential?
Rachel Wang
Posts: 49
Joined: Fri Sep 29, 2017 7:04 am
### Re: Test #2 question 4
For the equation Ecell = Ecathode - Eanode, it assumes you have not already flipped the sign of the anode. Therefore Ecell = -1.18-(-1.79).
However, you made the equation E=+1.79. In this case you have already flipped the sign and hence the eq is -1.18+1.79.
Cristina Sarmiento 1E
Posts: 52
Joined: Wed Nov 16, 2016 3:02 am
### Re: Test #2 question 4
When you flip the sign because you flip the equation, you must do Ecell = Ecathode + Eanode.
It is also helpful to know that for galvanic cells, the Ecell must always be positive.
Return to “Galvanic/Voltaic Cells, Calculating Standard Cell Potentials, Cell Diagrams”
### Who is online
Users browsing this forum: No registered users and 2 guests
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This activity is designed to be accessible to both GCSE and A-level maths students (Key Stage 4 and Key Stage 5), who can tackle it at different levels.
Detailed teachers' notes for this activity are available on our NRICH website.
##### Stage: 4 and 5 Challenge Level:
One of the challenges in designing an Olympic stadium is to make sure that spectators can see the event without their views being blocked by the spectators in front.
Your task is to design the tiered seating for the stadium. The back of each seat is 80cm behind the back of the seat in front, but raised up so that each row can see over the row in front.
Make some sketches showing what the cross-section (side view) of the seats might look like. Will the seats lie on a straight line, or a curve? How steep?
What factors do you think you will need to take into account when working out how high each seat needs to be?
The Olympic organisers have stipulated that all spectators must be able to see clearly a point 10m in front of the front row of seating:
The spectator in the second row needs to have line of sight to the same point as the spectator in the first row, as seen in the diagram above. Notice that the spectator in the second row needs some extra clearance in order to see comfortably over the first spectator's head.
Here is a zoomed-in diagram:
The first spectator's eye-level is 1.2m above the ground.
There is an extra 0.2m of clearance from his eye-level to the second spectator's line-of-sight. The second spectator is a further 0.8m away from the point on the pitch.
How high above ground level does spectator 2's seat need to be?
Now draw a similar diagram with the dimensions and unknowns for spectators 2 and 3.
How high above ground level does spectator 3's seat need to be?
Finally, imagine there were 40 rows of seating in the stadium.
Can you work out the heights above ground level of each of the 40 rows, and hence plot a side view of the seating?
It is very helpful to use a spreadsheet to perform the repeated calculations and plot the results.
Further information
If you're finding hard to get started, try looking at this hint.
Detailed teachers' notes for this activity are available on our NRICH website.
#### Related resources
You can read more about some of the challenges involved in designing an Olympic venue, including some issues about seating, in our article How the Velodrome found its form, based on an interview with the structural engineers involved in the design team for the London 2012 Velodrome.
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# Count Distinct Elements In Every Window
Posted: 14 Dec, 2020
Difficulty: Moderate
## PROBLEM STATEMENT
#### Given an array of integers ‘arr’ of size ‘n’ and an integer ‘k’. You need to find the count of distinct elements in every sub-array of size ‘k’ in the given array. Return an integer array ‘count’ that consists of n - k + 1 integers where ‘count[i]’ is equal to the number of distinct elements in a sub-array of size ‘k’ starting from index ‘i’.
##### Note:
``````1. The sub-array of an array is a continuous part of the array.
2. Consider ‘0’ based indexing.
3. ‘k’ will always be less than or equal to ‘n’.
3. Don’t print anything, just return the integer array ‘count’.
``````
##### Input format:
``````The first line of input contains an integer ‘T’ denoting the number of test cases.
The next ‘2*T’ lines represent the ‘T’ test cases.
The first line of each test case contains two space-separated integers ‘n’ and ‘k’ respectively representing the size of array ‘arr’ and size of sub-array to be considered.
The second line of the test case contains ‘n’ space-separated integers representing elements of the array ‘arr’.
``````
##### Output format :
``````For each test case, return an integer array ‘count’ that consists of n - k + 1 integers where ‘count[i]’ is equal to the number of distinct elements in a sub-array of size ‘k’ starting from index ‘i’ in array ‘arr’.
Output for each query must be printed in a new line.
``````
##### Constraints:
``````1 <= T <= 50
1 <= n <= 10^4
1 <= k <= n
-10^9 <= arr[i] <= 10^9
Where ‘T’ is the total number of test cases, ‘n’ is the given positive integer, ‘k’ is the size of sub-array to be considered, and arr[i] is the value of the array elements.
Time limit: 1 sec
``````
Approach 1
• Make an integer array ‘count’ of size n-k+1.
• Run a loop where ‘i’ ranges from 0 to n-k and for each ‘i’ count number of distinct elements in sub-array starting from this index, this can be done as follow.
• Initialize an integer variable ‘distinct’:= 0.
• Run a loop where ‘j’ ranges from ‘i’ to ‘i + k -1’ and for each ‘j’ check whether there exist any element in the range [i, j-1] which is equal ‘arr[j]’. If no such element found increment ‘distinct’ by 1.
• Assing count[i] := distinct.
• Return this integer array ‘count’.
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# Untitled
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1. /*
2. Given an array containing n distinct numbers taken from 0, 1, 2, ..., n, find the one that is missing from the array.
3.
4. For example,
5. Given nums = [0, 1, 3] return 2.
6.
7. Note:
8. Your algorithm should run in linear runtime complexity. Could you implement it using only constant extra space complexity?
9.
10. Credits:
11. */
12.
13. /*
14. * 本题要找出 0...n 中缺少的数，使用位运算的效率最高
15. * 使 res 异或 0...n 中的每个数，再异或 nums 向量中的每个数
16. * 异或两次不修改原值，仅异或一次的数被留在 res 中
17. */
18.
19. class Solution {
20. public:
21. int missingNumber(vector<int>& nums) {
22. int i = 0;
23. int res = nums.size();
24. for (auto num : nums)
25. {
26. res ^= num;
27. res ^= (i++);
28. }
29. return res;
30. }
31. };
RAW Paste Data
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# Baccarat Valley
## Some Techniques in the Game of Baccarat
The Game of Baccarat is usually played using eight decks of cards placed in a shoe. Cards that have a value under ten are worth their own value while the tens, jacks, queens and the kings have a value of zero. The ace card has a value of one.
Wagers are placed on the banker, the player or the tie in the game. About two card hands are then given to the banker and the player.
The total for the banker and the player hand is the total of the two cards but the first digit is taken away. For example if a player has a card hand of seven and five, it has a score of two (7 plus 5 = 12 ; you will drop the one). A third card may be given to the players in the game depending on the rules. Some of these rules are: First, if the player or the banker has a score of a nine or eight, they both have to stand.
Second, if the player has a total score of five or less, the player must then hit. The player must then stand if he will not hit. If the player chooses to stand, the banker must then hit five or less, if the player will hit a chart is used as a basis if the banker will stand or hit. The odds in a Baccarat game are that the higher score wins.
The winning wagers on the banker payout in a ratio of 19:20 (cash less of a total of 5% percent in commissions). The Commissions that are needed to be paid are carefully kept track of and cleared when you leave the game so as a player, you must make sure that you have enough money before leaving. Winning wagers of the player pays in a ratio of 1:1.
Winning wagers for the tie usually pay an 8:1 ratio but they also sometimes pay a 9:1 ratio. This is usually considered to be a bad wager as ties happen less often. Players must refrain from wagering on the tie at all costs. As with all casino table games, a lot of players have misconceptions about the game.
Past games are not a good indicator of what will happen on the succeeding games so it is useless. The most basic and optimum strategy is the one-three-two-six technique. This technique is used to increase the chances of winning in the game and minimize the losses that the player could experience while playing.
Players should begin by wagering one unit. If the player wins, the player can add another to the two at the Baccarat table for a total of three on the second wager. If the player wins, he will have a 6 on the table; remove the four so that the player can have a two on the third wager.
If the player wins on the third wager, add 2 to the 4 on the table so that you will have a total of six on the fourth wager. If you lose on the first wager, you will get a loss of one. A win on the first wager which will be followed by a loss on the second wager will just create a loss of two. Wins on the first 2 wager with a loss on the third wager will give the player a total profit of two.
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https://math.stackexchange.com/questions/2909043/why-do-we-use-the-choose-function-here
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# Why do we use the Choose function here?
Assume two gamblers playing a series of poker games, which is independently won with prob. $p$ for player $1$ and $1-p$ for player $2$. The ultimate winner of the series is the first to win k card games.
What is $Pr[\text{a total of 7 games are played}]$, for $k=4$?
I can see that to have 7 games, we need to be tied at 3 wins for each player after 6 games. So the $$Pr[\text{a total of 7 games are played}] = Pr[\text{each player wins 3 games after 6}] = \binom{6}{3}p^3(1-p)^3$$ I just don't understand quietly the use of the choose function to solve this?
What makes this dst different from a neg. Bin. Dst?
• I know that I have 20 ways the 6 games can be won. I actually enumerated them. But I just want to learn the intuition behind the choose function in this case. – Note Sep 7 '18 at 21:17
• Lets call the two gamblers "A" and "B". There are 6 games, 3 won by each player. To find the total number of ways that could happen, you could write down 6 "A"s and then choose three of them to be change to "B". – user247327 Sep 7 '18 at 22:49
• (Note in passing that poker is a rather curious choice of a game to reduce to a binary "win or lose" outcome, since skill in poker is not about how many times you win vs lose, but about how cheaply you can lose when you lose, vs how much you can win when you win). – hmakholm left over Monica Sep 7 '18 at 23:08
## 2 Answers
Once you know that the first player has won three games out of six, there are $\binom 63$ choices/possibilities for which three games they won.
(The second player won the others)
It is not a negative binomial distribution because there are two different stopping conditions: either that the first player has won 4 games or that the second player has won 4 games. That is a more complex situation than the negative binomial distribution is made to model.
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binarizeBASC: Binarization Across Multiple Scales In BiTrinA: Binarization and Trinarization of One-Dimensional Data
Description
Binarizes real-valued data using the multiscale BASC methods.
Usage
```1 2 3 4 5 6``` ```binarize.BASC(vect, method = c("A","B"), tau = 0.01, numberOfSamples = 999, sigma = seq(0.1, 20, by=.1), na.rm=FALSE) ```
Arguments
`method` Chooses the BASC method to use (see details), i.e. either "A" or "B". `vect` A real-valued vector of data to binarize. `tau` This parameter adjusts the sensitivity and the specificity of the statistical testing procedure that rates the quality of the binarization. Defaults to 0.01. `numberOfSamples` The number of samples for the bootstrap test. Defaults to 999. `sigma` If `method="B"`, this specifies a vector of different sigma values for the convolutions with the Bessel function. Ignored for `method="A"`. `na.rm` If set to `TRUE`, `NA` values are removed from the input. Otherwise, binarization will fail in the presence of `NA` values.
Details
The two BASC methods can be subdivided into three steps:
Compute a series of step functions:
An initial step function is obtained by rearranging the original time series measurements in increasing order. Then, step functions with fewer discontinuities are calculated. BASC A calculates these step functions in such a way that each minimizes the Euclidean distance to the initial step function. BASC B obtains step functions from smoothened versions of the input function in a scale-space manner.
Find strongest discontinuity in each step function:
A strong discontinuity is a high jump size (derivative) in combination with a low approximation error.
Estimate location and variation of the strongest discontinuities:
Based on these estimates, data values can be excluded from further analyses.
Value
Returns an object of class `BASCResult`.
References
M. Hopfensitz, C. Müssel, C. Wawra, M. Maucher, M. Kuehl, H. Neumann, and H. A. Kestler. Multiscale Binarization of Gene Expression Data for Reconstructing Boolean Networks. IEEE/ACM Transactions on Computational Biology and Bioinformatics 9(2):487-498, 2012.).
`BinarizationResult`, `BASCResult`
```1 2 3 4 5 6 7 8``` ```par(mfrow=c(2,1)) result <- binarize.BASC(iris[,"Petal.Length"], method="A", tau=0.15) print(result) plot(result) result <- binarize.BASC(iris[,"Petal.Length"], method="B", tau=0.15) print(result) plot(result) ```
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# Summing a multiplicative function
$$f(n)$$ is a multiplicative function, meaning $$f(m\cdot n)=f(m)\cdot f(n)$$.
I want to evaluate the sum: $$(1)\qquad\sum_{k=1}^{n}f(m\cdot k)$$ over a fixed $$m$$. Because $$f$$ is multiplicative, I can rewrite the sum as: $$(2)\qquad\sum_{k=1}^{n}f(m)\cdot f(k)$$ And because $$m$$ is fixed, I thought this sum is equal to: $$(3)\qquad f(m)\cdot \sum_{k=1}^{n} f(k)$$ But it's not, as simple paper & pencil check shows. I can easily evaluate the sum $$\sum_{k=1}^{n} f(k)$$, so I tried to rewrite $$(2)$$ in terms of it.
For example, suppose $$f(n)$$ is Euler's totient function, $$\varphi(n)$$, known as the amount of numbers $$ that are comprime to $$n$$. Suppose $$m=4$$ and I want to evaluate the sum up to $$k=3$$, that is $$\sum_{k=1}^{3}\varphi(4\cdot k)$$. The true sum would be written as $$(1)$$ & $$(2)$$ respectively as: $$\varphi(4)+\varphi(8)+\varphi(12) = \varphi(4)\cdot \varphi(1)+\varphi(4)\cdot \varphi(2)+\varphi(4)\cdot \varphi(3) = 10$$
In my wrong approach, the sum would be:
$$\varphi(m)\cdot \sum_{k=1}^{n} \varphi(k)=\varphi(4)\cdot \biggl(\varphi(1)+\varphi(2)+\varphi(3)\biggl)= 2\cdot \space (1+1+2)=8$$ which is different than the original sum. I've noticed this is also true if $$f(n)$$ is the sum of divisors function too $$\sigma_1(n)$$, and other multiplicative functions as well.
My question is how can I rewrite $$(2)$$ in terms of $$\sum_{k=1}^{n} f(k)$$ or perhaps in another way too, and why is my approach wrong?
Thanks.
• Can you please elaborate on: "But it's not, as simple paper & pencil check shows" Which function $f$ (or which $m$) did you choose? – Viktor Glombik Apr 29 at 12:58
• Is it possible, Matan, that you are comfusing "multiplicative functions" and "completely multiplicative functions"? – Gerry Myerson Apr 29 at 13:11
• I think the functions I am talking about are not completely multiplicative, but rather merely multiplicative. @GerryMyerson – Matan Apr 29 at 13:16
• In that case, you don't always have $f(mn)=f(m)f(n)$ – you're only guaranteed that when $\gcd(m,n)=1$. – Gerry Myerson Apr 29 at 13:18
• (3) and (2) are equivalent, providing $f$ is completely multiplicative (which precisely means $f(mn)=f(m)f(n)$ for all $m,n$). The totient function is multiplicative, but not completely multiplicative. – Gerry Myerson Apr 29 at 13:25
The reason you get 8 is a failure of mind: $$2\cdot(1+2+2)=2\cdot 5=10\neq8$$
but $$\varphi(2)=1$$ , so this sum is completely invalidated.
if not completely multiplicative, as has been talked about, this isn't always going to hold up though. In fact it fails any time $$n\geq (p\mid m)$$
• Thanks, I correct it. But if $f(n)$ is only multiplicative and not completely multiplicative, then how can I perform the summation? What "phrase" would I need to add/subtract from $(3)$ to make it correct? – Matan Apr 29 at 14:31
Let $$f$$ be multiplicative, but not completely multiplicative, and consider $$\sum_{k=2}^2f(2k)$$ This is, of course, $$f(4)$$, and you are asking for a way to modify $$(f(2))^2$$ to get $$f(4)$$, or to write $$f(4)$$ in terms of $$f(2)$$. This is clearly impossible, since no matter what $$f(2)$$ is, $$f(4)$$ could be anything. Multiplicativity is not enough to get an answer to the question you are asking.
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## Solve the system of linear equations by elimination. x+4y=6 −x−2y=−4
Question
Solve the system of linear equations by elimination. x+4y=6 −x−2y=−4
in progress 0
6 months 2021-09-05T06:05:37+00:00 1 Answers 3 views 0
## Answers ( )
(x , y) = (2 , 1)
Step-by-step explanation:
// Solve equation [1] for the variable x
[1] x = -4y + 6
// Plug this in for variable x in equation [2]
[2] -(-4y+6) – 2y = -4
[2] 2y = 2
// Solve equation [2] for the variable y
[2] 2y = 2
[2] y = 1
// By now we know this much :
x = -4y+6
y = 1
// Use the y value to solve for x
x = -4(1)+6 = 2
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# Co-axial Lines Field Analysis MCQ’s
This set of RF & Microwave Circuit Design Multiple Choice Questions & Answers (MCQs) focuses on “Co-axial Lines Field Analysis”.
1. Expression for resistance R of a coaxial transmission line outer radius b and inner radius a is:
a) Rs/2π ( 1/a+1/b)
b) 2πω∈”/ln(b/a)
c) μ/π cos-1(b/a)
d) πϵ’/cosh-1(b/a)
2. If the outer and the inner diameter of a coaxial transmission line are 20 mm and 10 mm respectively, then the inductance /m of the transmission line is:
a) 0.13 µH
b) 0.2 µH
c) 0.3 µH
d) 0.1 µH
3. The expression for conductance G of a coaxial transmission line with outer radius ‘b’ and inner radius ‘a’ is given by:
a) 2πωε”/ (ln b/a)
b) (R/2π)(1/a+1/b)
c) Rb/πa
d) 2Rb/a
4. If the outer circumference and the inner circumference of a transmission line are 40π and 25π units respectively, then the capacitive reactance of the coaxial transmission line is:
a) 0.376 nF
b) 0.45 nF
c) 0.9 nF
d) none of the mentioned
5. Characteristics impedance of a coaxial line with external and inner diameter 5mm is:
a) 40 Ω
b) 41.58 Ω
c) 47.78 Ω
d) 54.87 Ω
6. Flow of power in transmission line takes place through:
a) Electric field and magnetic field
b) Voltage and current
c) Voltage
d) Electric field
7. For a 2 wire transmission line, if the complex part of permittivity is 2.5, then the given distance between the 2 wires is 10mm and operated at a frequency of 1.2 MHz and the radius of the line being 5mm, then the conductance of the transmission line is:
a) 0.2 µH
b) 0.1 µH
c) 0.5 µH
d) 1 µH
8. The characteristic impedance of the transmission line if the outer diameter and inner diameter of the transmission line is 20 mm and 10 mm respectively, given the intrinsic impedance of the medium is 377 Ω, then the characteristic impedance of the transmission line is:
a) 41.58 Ω
b) 50 Ω
c) 377 Ω
d) None of the mentioned
9. When a transmission line is exited by a source, total power supplied is delivered to the load.
a) True
b) False
10. Expression for propagation constant. γ In terms of ω is:
a) √ω2µ∈
b) ω2µ∈
c) – ω2µ∈
d) None of the mentioned
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1. ## Solving ln Equations
I don't fully understand how "ln" works yet. I would really appreciate some help with this problem.
ln(x+31)−ln(x+3)=ln(8)−ln(x).
2. Originally Posted by mmfoxall
I don't fully understand how "ln" works yet. I would really appreciate some help with this problem.
$\ln(x+31) - \ln(x+3) = \ln(8) - \ln(x)$
$\displaystyle \ln\left(\frac{x+31}{x+3}\right) = \ln\left(\frac{8}{x}\right)$
$\displaystyle \frac{x+31}{x+3} = \frac{8}{x}$
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# NoCoug SQL Challenge Entry
Posted in: Technical Track
I love puzzles. So when I heard about the NoCoug SQL Challenge I felt tempted to give it a go.
The Northern California Oracle Users Group (NoCoug) has challenged us to find a good way to calculate the probability of getting different sums for x throws of a n-sided die using only SQL. The probabilities for the faces of a single die are stored in a table and that’s all you need to start playing with the problem. The SQL Challenge rules can be found on the NoCoug website, along with some other relevant information.
After working out my very first solution, I read the rules and found it wasn’t fit for the challenge, as it used non-SQL extensions (SQL*Plus). So I started again, this time using pure SQL. I came up with a few options but wasn’t happy with them from a performance perspective. They needed more sweating.
I find that walking is very good for thinking. Whenever I can, and when weather permits, I walk home from work at the end of the day. The distance between work and home is about 6 km, which takes me around one hour to cover. After you’ve done it a few times the walking becomes automatic and you don’t have to think about it anymore; obstacles, kerbs, corners, street-crossings are all handled in auto-pilot mode. Then, as you don’t have anything else to do, you think.
During the days I was working on the challenge solution, the SQL query used to occupy my thoughts along most of my (almost) daily walk. I decided to use only standard SQL features and wanted to be able to run the solution on both Oracle and SQL Server. While walking, I started thinking on ways to improve my initial solution’s performance without having to resort to non-standard tricks.
### Joins of joins
The most straightforward way to calculate the probabilities for the possible sums in x throws of the die (whose face’s probabilities are stored in the `die` table) is to write a x-way self-cross-join of the table. For example, the probabilities for the sums in 3 throws of the die is given by:
```select
d1.face_value + d1.face_value + d2.face_value total_value,
sum(d1.probability * d2.probability * d3.probability) probability
from
die d1
cross join die d2
cross join die d3
group by d1.face_value + d1.face_value + d2.face_value
```
The problem is that the complexity of the query escalates when a high number of throws is involved; imagine writing and executing a 1024 join of that table. Also, the number of rows generated by the multiple cross-joins increases exponentially every time a new join is added to the query. It’s just not feasible to run the query above for a large number of throws. For a 20-sided die, for example, the query above yields 58 rows after the `GROUP BY` aggregation (totals ranging from 3 to 60). Before the aggregation can be performed, though, all the joins must be calculated, which generates `203` = 8000 combinations.
To improve performance, I needed a way to reduce the number of joins required to calculate the probabilities and the cost of those joins.
### Reusing calculations
Let’s suppose I pre-calculate the probabilities for n throws of the die. To calculate the chances for n+1 throws, I don’t need to start from scratch again—I can start from the probabilities for n throws and perform a Cartesian product (or a “cross-join”, in other words) of that, with the probabilities for one die. Similarly, to calculate the probability for 2n throws, I can use the probabilities for n throws and do a “self-cross-join.”
Using this principle, I thought I could pre-calculate probabilities for some chosen number of throws, and then use those results as building blocks to obtain the probabilities for any other numbers. The most obvious “sizes” for the building blocks would be powers of 2, since it’s easy to find a combination of them to generate any other number using binary arithmetics.
Thus, if the probabilities for 1, 2, 4, 8, … 2n throws of a die are pre-calculated, we can easily combine them to calculate probabilities for any number of throws up to 2(n+1)-1.
### Combining the building blocks
I wanted to use those building blocks together in a single parameterised SQL so I could generate probabilities for any given number of throws without having to rewrite the query every time. To do that, I used some binary arithmetic and logic to manipulate the joins between the building blocks.
Let’s say T1, T2 and T4 are the pre-calculated probabilities for the throw of 1, 2, and 4 dice, respectively. The probability for 7 throws of the die is given by:
```
T1
cross join
T2
cross join
T4
```
Or, alternatively:
``` (select face_value, probability from T1)
cross join
(select face_value, probability from T2)
cross join
(select face_value, probability from T4)
```
The query above, though, is not parameterised and only calculates probabilities for 7 throws of the die. Using a parameter (`&&throws`) we can manipulate which building blocks will be involved in the join:
``` (select face_value, probability from T1
where bitand(1, &&throws) != 0)
cross join
(select face_value, probability from T2
where bitand(2, &&throws) != 0)
cross join
(select face_value, probability from T4
where bitand(4, &&throws) != 0)
```
The query is now parameterised. We have a big problem, however: for `&&throws = 2`, for example, `bitand(1, &&throws) = 0`, which causes the first select to return no rows, which in consequence causes the whole cross-join to return no rows. And this is definitely not what we want.
To avoid this problem, whenever one of the building blocks is not to be used, instead of just suppressing it, we need to replace it with a neutral element—an element that could be used in the cross-join without altering the values. This element is the select shown below:
`select 0 face_value, 1 probability from dual`
The value `0` for `face_value` won’t change the sums for the dice faces, and the value `1` for probability won’t alter the probability products calculate in the cross join. Adding this element to the previous query we then have:
``` (select face_value, probability from T1
where bitand(1, &&throws) != 0
union all
select 0 face_value, 1 probability from dual
where bitand(1, &&throws) = 0)
cross join
(select face_value, probability from T2
where bitand(2, &&throws) != 0
union all
select 0 face_value, 1 probability from dual
where bitand(2, &&throws) = 0)
cross join
(select face_value, probability from T4
where bitand(4, &&throws) != 0
union all
select 0 face_value, 1 probability from dual
where bitand(4, &&throws) = 0)
```
With this query, setting the `&&throws` parameter to any value ranging between 1 and 7 will correctly choose which of the pre-calculated building blocks should be included in the calculation. With one query and one parameter we can then have probabilities calculated for 7 different numbers of throws.
### Reducing cardinality
Another factor affecting the performance of the probabilities calculation is the cardinality of the subsequent joins, which escalate exponentially. With a 20-sided die, for example, after n throws you’ll have 20n rows in the result set. For 3 throws, that means 8000 rows in the resulting set.
For 3 throws of a 20-sided die, though, we can only have sums ranging from 3 (1+1+1) to 60 (20+20+20). That means that, in those 8000 rows we have only 58 distinct summed values. Probabilities for any given sum may appear several times because it may be obtained by summing different combinations of the die’s values. For example, using a 20-sided die, the sum 18 can be obtained in the following ways (among many, many others): 1+1+16, 1+2+15, 1+3+14, etc.
To help performance, I decided to perform an aggregation (`GROUP BY`) after every join to reduce the number of rows before the execution of the subsequent join. The aggregation operation itself has a cost but it pays off, preventing the cardinality of the joins from increasing exponentially.
Applying this idea to the query above we have:
``` (select face_value, probability from T1
where bitand(1, &&throws) != 0
union all
select 0 face_value, 1 probability from dual
where bitand(1, &&throws) = 0)
cross join
(select T2.face_value + T4.face_value face_value,
sum(T2.probability * T4.probability) probability
from (
(select face_value, probability from T2
where bitand(2, &&throws) != 0
union all
select 0 face_value, 1 probability from dual
where bitand(2, &&throws) = 0)
cross join
(select face_value, probability from T4
where bitand(4, &&throws) != 0
union all
select 0 face_value, 1 probability from dual
where bitand(4, &&throws) = 0)
)
group by T2.face_value + T4.face_value
)
```
### The query
The final solution is the combination of the techniques described above. The query was written with standard SQL and makes use of Subquery Factoring (a.k.a. Common Table Expressions, a.k.a. the WITH clause) to calculate the building blocks’ probabilities. The probabilities for a 2n-throws building block is calculated based on the probabilities previously calculated for 2(n-1).
### Limitations
Each written query, like the one below, can calculate probabilities up to a certain number of throws (max # of throws =` 2Q - 1`, where Q is the number of T* subqueries used in the query). This number can be easily increased by adding new subqueries to the query. The maximum number of throws increases exponentially for linear increases in the size of the query.
When testing on Oracle (10.2.0.4 and 11.1.0.7), I found that the maximum Q I could use for a successful execution was 9 (query below). For queries with more than 9 subqueries I got an ORA-600. I couldn’t find an existing solution for the problem in Metalink.
### Code
```-- First International NoCoug SQL Challenge
-- Author: Andre Araujo (araujo@pythian.com)
-- Objective: Calculate the probability of getting certain sums for N throws of a dice.
undefine throws
with
T1 as (
-- Results for the throw of 1 die
select d1.face_value face_value, sum(d1.probability) probability
from die d1
group by d1.face_value
)
,
T2 as (
-- Results for the throw of 2 dice
select d1.face_value + d2.face_value face_value, sum(d1.probability * d2.probability) probability
from T1 d1
cross join T1 d2
group by d1.face_value + d2.face_value
)
,
T4 as (
-- Results for the throw of 4 dice
select d1.face_value + d2.face_value face_value, sum(d1.probability * d2.probability) probability
from T2 d1
cross join T2 d2
group by d1.face_value + d2.face_value
)
,
T8 as (
-- Results for the throw of 8 dice
select d1.face_value + d2.face_value face_value, sum(d1.probability * d2.probability) probability
from T4 d1
cross join T4 d2
group by d1.face_value + d2.face_value
)
,
T16 as (
-- Results for the throw of 16 dice
select d1.face_value + d2.face_value face_value, sum(d1.probability * d2.probability) probability
from T8 d1
cross join T8 d2
group by d1.face_value + d2.face_value
)
,
T32 as (
-- Results for the throw of 32 dice
select d1.face_value + d2.face_value face_value, sum(d1.probability * d2.probability) probability
from T16 d1
cross join T16 d2
group by d1.face_value + d2.face_value
)
,
T64 as (
-- Results for the throw of 64 dice
select d1.face_value + d2.face_value face_value, sum(d1.probability * d2.probability) probability
from T32 d1
cross join T32 d2
group by d1.face_value + d2.face_value
)
,
T128 as (
-- Results for the throw of 128 dice
select d1.face_value + d2.face_value face_value, sum(d1.probability * d2.probability) probability
from T64 d1
cross join T64 d2
group by d1.face_value + d2.face_value
)
,
T256 as (
-- Results for the throw of 256 dice
select d1.face_value + d2.face_value face_value, sum(d1.probability * d2.probability) probability
from T128 d1
cross join T128 d2
group by d1.face_value + d2.face_value
)
-- The select below combines the results above to achieve the wanted results.
-- The bitand checks ensure only necessary selects are executed.
select T1.face_value + T2.face_value face_value, sum(T1.probability * T2.probability) probability
from
(select face_value, probability from T1 where bitand(1, &&throws) != 0 union all select 0, 1 from dual where bitand(1, &&throws) = 0) T1
cross join
(select T2.face_value + T4.face_value face_value, sum(T2.probability * T4.probability) probability
from (select face_value, probability from T2 where bitand(2, &&throws) != 0 union all select 0, 1 from dual where bitand(2, &&throws) = 0) T2
cross join
(select T4.face_value + T8.face_value face_value, sum(T4.probability * T8.probability) probability
from (select face_value, probability from T4 where bitand(4, &&throws) != 0 union all select 0, 1 from dual where bitand(4, &&throws) = 0) T4
cross join
(select T8.face_value + T16.face_value face_value, sum(T8.probability * T16.probability) probability
from (select face_value, probability from T8 where bitand(8, &&throws) != 0 union all select 0, 1 from dual where bitand(8, &&throws) = 0) T8
cross join
(select T16.face_value + T32.face_value face_value, sum(T16.probability * T32.probability) probability
from (select face_value, probability from T16 where bitand(16, &&throws) != 0 union all select 0, 1 from dual where bitand(16, &&throws) = 0) T16
cross join
(select T32.face_value + T64.face_value face_value, sum(T32.probability * T64.probability) probability
from (select face_value, probability from T32 where bitand(32, &&throws) != 0 union all select 0, 1 from dual where bitand(32, &&throws) = 0) T32
cross join
(select T64.face_value + T128.face_value face_value, sum(T64.probability * T128.probability) probability
from (select face_value, probability from T64 where bitand(64, &&throws) != 0 union all select 0, 1 from dual where bitand(64, &&throws) = 0) T64
cross join
(select T128.face_value + T256.face_value face_value, sum(T128.probability * T256.probability) probability
from (select face_value, probability from T128 where bitand(128, &&throws) != 0 union all select 0, 1 from dual where bitand(128, &&throws) = 0) T128
cross join
(select face_value, probability from T256 where bitand(256, &&throws) != 0 union all select 0, 1 from dual where bitand(256, &&throws) = 0) T256
--
-- Grouping after each join ensures rows with equal sums are combined, reducing the cardinality for the next join.
--
group by T128.face_value + T256.face_value) T128
group by T64.face_value + T128.face_value) T64
group by T32.face_value + T64.face_value) T32
group by T16.face_value + T32.face_value) T16
group by T8.face_value + T16.face_value) T8
group by T4.face_value + T8.face_value) T4
group by T2.face_value + T4.face_value) T2
group by T1.face_value + T2.face_value
order by 1;
```
### Portability
The code above doesn’t use any special, non-standard SQL features, and can be easily ported to other databases with minor changes.
Running it against SQL Server, for example, all you need to do is to replace the `BITAND` function with the bitwise `AND` operator, and replace the use of the substitution variable with a TSQL variable.
### Testing
I ran the tests on an Oracle 10.2.0.4 instance. The results looked quite good and the solution seemed to scale as expected. I compared it to the performance of n straightforward joins of the `die` table (using `GROUP BY`s to reduce cardinality as explained above) and the results were quite similar.
The proposed solution, though, has the advantage of reduced size of the query, and that the same query can be used to calculated the probabilities for a different number of throws.
The results below were run for a 20-sided die. I ran tests for all the number of throws between 1 and 511. The values shown are the elapsed time average over 10 runs of the query for each number of throws. The complete test took 30 hours to complete.
Interested in working with André? Schedule a tech call.
### About the Author
#### André Araújo
DBA since 1998, having worked with Oracle from version 7.3.4 to the latest one. Working at Pythian since 2009.
#### 15 Comments. Leave new
Wow. Very clean and efficient.
Log Buffer #147: a Carnival of the Vanities for DBAs | Pythian Group Blog
May 22, 2009 12:44 pm
[…] Pythian’s André Araujo responds to a different test with his very elegant NoCoug SQL Challenge Entry. […]
Andre Araujo
May 25, 2009 12:10 am
Thanks, Waldar. Looking forward to seeing more solutions submitted. There’s been already some very interesting ones.
Challenges !!! | Waldar's SQLing and Datawarehousing Place
May 27, 2009 7:34 pm
[…] Chen Shapira wrote a summary about this challenge. In the comments i saw a trackback (or pingback, whatever it is) leading to André Aurajo’s solution. […]
Laurent Schneider
June 2, 2009 4:36 am
I love this solution because it is so easy to understand! Really a fantastic piece of code and impressive explanation, congrats !
Andre Araujo
June 2, 2009 6:07 pm
Thanks, Laurent! I appreciate the feedback.
The First International NoCOUG SQL Challenge: Nine Ways To Change a Lightbulb « So Many Oracle Manuals, So Little Time
July 8, 2009 1:17 am
[…] ninth solution was sent in by André Araujo from Australia who used binary arithmetic and common table expressions […]
Who Should Tune SQL: The DBA or The Developer? « So Many Oracle Manuals, So Little Time
July 12, 2009 2:40 pm
[…] ninth solution—by André Araujo from Australia—used binary arithmetic and common table expressions to solve […]
NoCOUG Got Nine, TSQL Got Eleven at Waldar’s SQLing and Datawarehousing Place
July 19, 2009 9:01 pm
[…] I’m the seventh in the list there. As I told before, my favorite solution to this problem is André Aurajo’s one. […]
Iggy Fernandez
July 19, 2009 11:51 pm
The first argument of the ORA-600 error is 15160. The ORA-600 troubleshooting tool in Metalink says that Oracle is “attempting to find a suitable table as the starting point in a complex join operation.” The real problem is that the estimated cost is too high for the optimizer to handle. The workaround is to introduce a Cardinality hint that instructs the query optimizer that the Die table has only one row :-) You will then be able to handle much higher values of the Throws variable.
T1 as (
— Results for the throw of 1 die
select /*+ CARDINALITY(d1 1) */ d1.face_value face_value, sum(d1.probability) probability
from die d1
group by d1.face_value
)
Regards,
Iggy
P.S. The contest is now closed and the winners will be announnced soon.
First International NoCoug SQL Challenge – And the Winner is… « I’m just a simple DBA on a complex production system
July 31, 2009 12:46 am
[…] wins the challenge for his wonderful solution using Discrete Fourier Transforms; the runner-up is André Araujo from Australia, who used binary arithmetic and common table expressions in his […]
Results of the First International NoCOUG SQL Challenge « So Many Oracle Manuals, So Little Time
July 31, 2009 3:45 pm
[…] ninth solution—by André Araujo from Australia—used binary arithmetic and common table […]
First International NoCOUG SQL Challenge Is Over ! at Waldar’s SQLing and Datawarehousing Place
August 1, 2009 9:45 am
[…] think his solution is the one I would have found by myself ! I’m sure the same clever logic could be reused for more problems […]
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# Number 1000222121
### Properties of number 1000222121
Cross Sum:
Factorization:
Divisors:
Count of divisors:
Sum of divisors:
Prime number?
Yes
Fibonacci number?
No
Bell Number?
No
Catalan Number?
No
Base 3 (Ternary):
Base 4 (Quaternary):
Base 5 (Quintal):
Base 8 (Octal):
3b9e2da9
Base 32:
tpsbd9
sin(1000222121)
-0.99982351706121
cos(1000222121)
-0.018786557197886
tan(1000222121)
53.220156654021
ln(1000222121)
20.723487933281
lg(1000222121)
9.0000964552126
sqrt(1000222121)
31626.288448062
Square(1000222121)
### Number Look Up
Look Up
1000222121 which is pronounced (one billion two hundred twenty-two thousand one hundred twenty-one) is a very special number. The cross sum of 1000222121 is 11. If you factorisate the number 1000222121 you will get these result . The figure 1000222121 has 2 divisors ( 1, 1000222121 ) whith a sum of 1000222122. 1000222121 is a prime number. The number 1000222121 is not a fibonacci number. The figure 1000222121 is not a Bell Number. The figure 1000222121 is not a Catalan Number. The convertion of 1000222121 to base 2 (Binary) is 111011100111100010110110101001. The convertion of 1000222121 to base 3 (Ternary) is 2120201002112210202. The convertion of 1000222121 to base 4 (Quaternary) is 323213202312221. The convertion of 1000222121 to base 5 (Quintal) is 4022024101441. The convertion of 1000222121 to base 8 (Octal) is 7347426651. The convertion of 1000222121 to base 16 (Hexadecimal) is 3b9e2da9. The convertion of 1000222121 to base 32 is tpsbd9. The sine of the figure 1000222121 is -0.99982351706121. The cosine of the figure 1000222121 is -0.018786557197886. The tangent of the number 1000222121 is 53.220156654021. The root of 1000222121 is 31626.288448062.
If you square 1000222121 you will get the following result 1000444291337738641. The natural logarithm of 1000222121 is 20.723487933281 and the decimal logarithm is 9.0000964552126. I hope that you now know that 1000222121 is impressive number!
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# RBSE Solutions for Class 8 Maths Chapter 9 Algebraic Expressions Ex 9.2
RBSE Solutions for Class 8 Maths Chapter 9 Algebraic Expressions Ex 9.2 is part of RBSE Solutions for Class 8 Maths. Here we have given Rajasthan Board RBSE Class 8 Maths Chapter 9 Algebraic Expressions Exercise 9.2.
Board RBSE Textbook SIERT, Rajasthan Class Class 8 Subject Maths Chapter Chapter 9 Chapter Name Algebraic Expressions Exercise Exercise 9.2 Number of Questions 3 Category RBSE Solutions
## Rajasthan Board RBSE Class 8 Maths Chapter 9 Algebraic Expressions Ex 9.2
Question 1.
Multiply the given binomials
(i) (2x + 5) and (3x – 7)
(ii) (x – 8) and (3y + 5)
(iii) (1.5p – 0.5q) and (1.5p + 0.5q)
(iv) (a + 3b) and (x + 5)
(v) (21m + 3m²) and (3lm – 5m²)
(vi) ($$\frac { 3 }{ 4 }$$a² + 3b²) and (4a² – $$\frac { 5 }{ 3 }$$b²)
Solution
(i) (2x + 5) and (3x – 7)
= 2x(3x – 7) + 5(3x – 7)
= 2x × 3x – 2x × 7 + 5 × 3x – 5 × 7
= 2 × x × 3 × x – 2 × x × 7 + 5 × 3 × x – 35
= 2 × 3 × x × x – 2 × 7 × x + 15 × x – 35
= 6 × x² – 14 × x + 15x – 35
= 6x² – 14x + 15x – 35
= 6x² + x – 35
(ii) (x – 8) and (3y + 5)
= x(3y + 5) – 8(3y + 5)
= x × 3y + x × 5 – 8 × 3y – 8 × 5
= x × 3 × y + 5x – 8 × 3 × y – 40
= 3 × x × y + 5x – 24 × y – 40
= 3x × y + 5x – 24y – 40
= 3xy + 5x – 24y – 40
(iii) (1.5p – 0.5q) and (1.5p + 0.5q)
= 1.5p (1.5p + 0.5q) – 0.5q (1.5p + 0.5q)
= 1.5p × 1.5p + 1.5p × 0.5q – 0.5q × 1.5p – 0.5q × 0.5q
= 1.5 × p × 1.5 × p + 1.5 × p × 0.5 × q – 0.5 × q × 1.5 × p – 0.5 × q × 0.5 × q
= 1.5 × 1.5 × p × p + 1.5 × 0.5 × p × q – 0.5 × 1.5 × q × p – 0.5 × 0.5 × q × q
= 2.25 × p² + 0.75 × pq – 0.75 × qp – 0.25 × q²
= 2.25p² + 0.75pq – 0.75pq – 0.25q²
= 2.25p² – 0.25q²
(iv) (a + 3b) and (x + 5)
= a (x + 5) + 3b (x + 5)
= a × x + a × 5 + 3b × x + 3b × 5
= ax + 5a + 3bx + 3 × 5 × b
= ax + 5a + 3 bx + 15 b
(v) (2lm + 3m2) and (3lm – 5m2)
= 2lm (3lm – 5m2) + 3m2 (3lm – 5m2)
= 2lm × 3lm – 2lm × 5m2 + 3m2 × 3lm – 3m2 × 5m2
= (2 × 3) l2m2 – (2 × 5) lm3 + (3 × 3) lm3 – (3 × 5) m4
= 6l2m2 – 10lm3 + 9lm3 – 15m4
= 6l2m2 – lm3 – 15m4
(vi) ($$\frac { 3 }{ 4 }$$a² + 3b²) and (4a² – $$\frac { 5 }{ 3 }$$b²)
Question 2.
Find the product
(i) (3x + 8) (5 – 2x)
(ii) (x + 3y) (3x – y)
(iii) (a² + b) (a + b²)
(iv) (p² – q²) (2p + q)
Solution
(i) (3x + 8) (5 – 2x)
= 3x (5 – 2x) + 8(5 – 2x)
= 3x × 5 – 3x × 2x + 8 × 5 – 8 × 2x
= 15x – 6x² + 40 – 16x
= – 6x² + 15x – 16x + 40
= – 6x² – x + 40
(ii) (x + 3y) (3x – y)
= x (3x – y) + 3y (3x – y)
= x × 3x – x × y + 3y × 3x – 3y × y
= 3x² – xy + 9yx – 3y²
= 3x² – xy + 9xy – 3y²
= 3x² + 8xy – 3y²
(iii) (a² + b) (a + b²)
= a² (a + b²) + b (a + b²)
= a² × a + a² × b² + b × a + b × b²
= a3 + a²b² + ba + b3
= a3 + a²b² + ab + b3
(iv) (p² – q²) (2p + q)
= p² (2p + q) – q² (2p + q)
= p² x 2p + p² x q – q² x 2p – q² x q
= 2p3 + p²q – 2q²p – q3
Question 3.
Simplify
(i) (x + 5) (x – 7) + 35
(ii) (a² – 3) (b² + 3) + 5
(iii) (t + s²) (t² – s)
(iv) (a + b) (c – d) + (a – b) (c + d) + 2 (ac + bd)
(v) (a + b) (a² – ab + b²)
(vi) (a + b + c) (a + b – c)
(vii) (a + b) (a – b) – a² + b²
Solution
(i) (x + 5) (x – 7) + 35
= x (x – 7) + 5 (x – 7) + 35
= x² – 7x + 5x – 35 + 35
= x² – 2x
(ii) (a² – 3) (b² + 3) + 5
= a² (b² + 3) – 3 (b² + 3) + 5
= a²b² + 3a² – 3b² – 9 + 5
= a²b² + 3a² – 3b² – 4
(iii) (t + s²) (t² – s)
= t (t² – s) + s² (t² – s)
= t3 – ts + s2t2 – s3
= t3 – s3 + s2t2 – ts
(iv) (a + b) (c – d) + (a – b) (c + d) + 2(ac + bd)
(a + b) (c – d) + (a -b) (c + d) + 2(ac + bd)
= a (c – d) + b(c – d) + a(c + d) – b(c + d) + 2 (ac + bd)
= ac – ad + bc – bd + ac + ad – bc – bd + 2 ac + 2 bd
= ac + ac + 2ac – ad + ad + be – bc – bd – bd – 2 bd
= 4ac
(v) (a + b) (a² – ab + b²)
= a (a² – ab + b²) + b (a² – ab + b²)
= a3 – a²b + ab² + ba² – ab² + b3
= a3 + b3 – a²b + a²b + ab² – ab²
= a3 + b3
(vi) (a + b + c) (a + b – c)
= a (a + b – c) + b (a + b – c) + c (a + b – c)
= a² + ab – ac + ab + b² – bc + ac + bc – c²
= a² + b² – c² + 2ab
(vii) (a + b) (a – b) – a² + b²
= a (a – b) + b (a – b) – a² + b²
= a² – ab + ab – b² – a² + b²
= a² – a² – b² + b² – ab + ab
= 0 + 0 + 0
= 0
We hope the given RBSE Solutions for Class 8 Maths Chapter 9 Algebraic Expressions Ex 9.2 will help you. If you have any query regarding Rajasthan Board RBSE Class 8 Maths Chapter 9 Algebraic Expressions Exercise 9.2, drop a comment below and we will get back to you at the earliest.
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The Rings Between Z and Q
We start with a definition of the rings in question whose first appearance on the scene in [49, 50] dates back to 1994.
Definition 3.6 (A Ring in between) Let S be a set of primes of Q. Let RS be the following subring of Q.
If S = 0, then RS = Z. If S contains all the primes of Q, then RS = Q. If S is finite, we call the ring small. If S is infinite, we call the ring big.
Some of these rings have other (canonical) names: the small rings are also called rings of S-integers, and when S contains all but finitely many primes, the rings are called semi-local subrings of Q. To measure the “size” of big rings we use the natural density of prime sets defined below.
Definition 3.7 (Natural Density) If A is a set of primes, then the natural density of A is equal to the limit below (if it exists):
The big and small rings are not hard to construct.
Example 3.1 (A Small Ring not Equal to Z)
Example 3.2 (A Big Ring not Equal to Q)
Given a big or a small ring R we can now ask the following questions which were raised above with respect to Q:
• • Is HTP solvable over R?
• • Do integers have a Diophantine definition over R?
• • Is there a Diophantine model of integers over R?
Here one could hope that understanding what happens to HTP over a big ring can help to understand HTP over Q.
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## Pre Equilibruim Approach
priyasihota3c
Posts: 11
Joined: Fri Sep 25, 2015 3:00 am
### Pre Equilibruim Approach
What is the difference in the pre equilibrium approach and the steady-state approach?
Katherine Jabba 2H
Posts: 15
Joined: Fri Sep 25, 2015 3:00 am
### Re: Pre Equilibruim Approach
The pre-equilibirum approach uses a psuedo-equilibrium situation where one of the steps of the reaction is said to be at equilibrium and set equal to K, the equilibirum constant. This equation involving K is then manipulated and subsituted into the other reaction rate equation, so that the the reaction rate equation of the steps ends up looking like the observed reaction rate. It is a lot simpler method than the steady-state approach, which involves calculus.
Jeremiah Hutauruk
Posts: 67
Joined: Fri Sep 28, 2018 12:28 am
### Re: Pre Equilibruim Approach
involves k
Vicky Lu 1L
Posts: 60
Joined: Fri Sep 28, 2018 12:18 am
### Re: Pre Equilibruim Approach
The pre-equilibrium approach starts with has a fast elementary step followed by a slow elementary step. With the pre-equilibrium approach, we can't have intermediates between the fast and slow in the rate law. This approach also doesn't factor in account the reverse reaction that is occurring in the fast reaction, therefore, we want to substitute a term in for the intermediate step to have an overall rate law that is based on the slow step and does not include in intermediates.
haleyervin7
Posts: 61
Joined: Fri Sep 28, 2018 12:15 am
### Re: Pre Equilibruim Approach
Do you need the observed rate to be given for this approach?
Destiny Diaz 4D
Posts: 51
Joined: Fri Sep 28, 2018 12:28 am
### Re: Pre Equilibruim Approach
would we be specifically asked to solve using the pre equilibrium approach?
Destiny Diaz 4D
Posts: 51
Joined: Fri Sep 28, 2018 12:28 am
### Re: Pre Equilibruim Approach
Destiny Diaz 4D wrote:would we be specifically asked to solve using the pre equilibrium approach?
I feel like based on the question and what is given you would determine this, but I wanted to ask to be sure?
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# How do you simplify 7x+3y-2+6x-1+y^2?
Mar 14, 2018
See below...
#### Explanation:
Here we must collect the like terms. By like terms, we mean they are either a constant or they have the same variable
e.g: $x + x = 2 x$
Let's group the like terms.
$7 x + 3 y - 2 + 6 x - 1 + {y}^{2}$
$7 x + 6 x - 2 - 1 + 3 y + {y}^{2}$
And now simplify
$13 x - 3 + 3 y + {y}^{2}$
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# Volume to Weight conversion
## based on density
### for 683 common substances and materials
#### weightw = ρ × v
carat 1 087.12 ounce 7.67 gram 217.42 pound 0.48 kilogram 0.22 tonne 0 milligram 217 424.59
#### Density
Ice, solid weigh(s) 0.92 gram per (cubic centimeter) or 0.53 ounce per (cubic inch)
#### volumev = w ⁄ ρ
centimeter³ 236.59 oil barrel 0 foot³ 0.01 US cup 1 Imperial gallon 0.05 US fluid ounce 8 inch³ 14.44 US gallon 0.06 liter 0.24 US pint 0.5 meter³ 0 US quart 0.25 metric cup 0.95 US tablespoon 16 metric tablespoon 15.77 US teaspoon 48 metric teaspoon 47.32
• Calculation Formulas
• weight
w = ρ × v
• volume
v = w ⁄ ρ
• where the Greek letter ρ (rho) stands for density.
• Reference
• N. N. Greenwood, A. Earnshaw.
Chemistry of the Elements.
Butterworth – Heinemann.
Oxford, Amsterdam, Boston, London, New York, Paris, San Diego, San Francisco, Singapore, Sydney, Tokyo.
1997.
• National Center for Biotechnology Information,
United States National Library of Medicine
8600 Rockville Pike
Bethesda MD, 20894 USAhttps://www.ncbi.nlm.nih.gov
#### Foods, Nutrients and Calories
CAMPBELL'S Red and White, Chicken Vegetable Soup, condensed weigh(s) 266.29 gram per (metric cup) or 8.89 ounce per (US cup), and contain(s) 63 calories per 100 grams or ≈3.527 ounces [ calories | weight to volume | volume to weight | price | density ]
#### Gravels and Substrates
CaribSea, Freshwater, African Cichlid Mix, Congo River density is equal to 1313.51 kg/m³ or 82 lb/ft³ with specific gravity of 1.31351 relative to pure water. Calculate how much of this gravel is required to attain a specific depth in a cylinderquarter cylinder or in a rectangular shaped aquarium or pond [ weight to volume | volume to weight | price ]
#### Materials and Substances
Bran weigh(s) 0.256 gram per (cubic centimeter) or 0.148 ounce per (cubic inch) [ weight to volume | volume to weight | price | density ]
#### Weight/Volume at Temperature
Coconut oil weigh(s) 0.91 gram per (cubic centimeter) or 0.526 ounce per (cubic inch) at 39°C or 102.2°F [ weight to volume | volume to weight | price ]
#### What is grain per cubic centimeter?
The grain per cubic centimeter density measurement unit is used to measure volume in cubic centimeters in order to estimate weight or mass in grains
#### What is density measurement?
The objects made of different materials may occupy the same volume but have different masses or weights, e.g. aluminum and steel.
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https://hextobinary.com/unit/conductance/from/gemmho
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# Gemmho Converter
Electrical Conductance
Gemmho
Megasiemens
## How many Megasiemens are in a Gemmho?
The answer is one Gemmho is equal to 1e-12 Megasiemens and that means we can also write it as 1 Gemmho = 1e-12 Megasiemens. Feel free to use our online unit conversion calculator to convert the unit from Gemmho to Megasiemens. Just simply enter value 1 in Gemmho and see the result in Megasiemens.
## How to Convert Gemmho to Megasiemens (gemʊ to MS)
By using our Gemmho to Megasiemens conversion tool, you know that one Gemmho is equivalent to 1e-12 Megasiemens. Hence, to convert Gemmho to Megasiemens, we just need to multiply the number by 1e-12. We are going to use very simple Gemmho to Megasiemens conversion formula for that. Pleas see the calculation example given below.
$$\text{1 Gemmho} = 1 \times 1e-12 = \text{1e-12 Megasiemens}$$
## What is Gemmho Unit of Measure?
Gemmho is a unit of measurement for electric conductance. Gemmho is decimal fraction of electric conductance unit mho. One gemmho is equal to 0.000001 mho.
## What is the symbol of Gemmho?
The symbol of Gemmho is gemʊ. This means you can also write one Gemmho as 1 gemʊ.
## What is Megasiemens Unit of Measure?
Megasiemens is a unit of measurement for electric conductance. Megasiemens is multiple of electric conductance unit siemens. One megasiemens is equal to 1000000 siemens.
## What is the symbol of Megasiemens?
The symbol of Megasiemens is MS. This means you can also write one Megasiemens as 1 MS.
## Gemmho to Megasiemens Conversion Table
Gemmho [gemʊ]Megasiemens [MS]
11e-12
22e-12
33e-12
44e-12
55e-12
66e-12
77e-12
88e-12
99e-12
101e-11
1001e-10
10001e-9
## Gemmho to Other Units Conversion Table
Gemmho [gemʊ]Output
1 gemmho in siemens is equal to0.000001
1 gemmho in megasiemens is equal to1e-12
1 gemmho in kilosiemens is equal to1e-9
1 gemmho in millisiemens is equal to0.001
1 gemmho in microsiemens is equal to1
1 gemmho in ampere/volt is equal to0.000001
1 gemmho in mho is equal to0.000001
1 gemmho in micromho is equal to1
1 gemmho in abmho is equal to1e-15
1 gemmho in statmho is equal to899000.04
## Convert Gemmho to Other Electrical Conductance Units
Disclaimer:We make a great effort in making sure that conversion is as accurate as possible, but we cannot guarantee that. Before using any of the conversion tools or data, you must validate its correctness with an authority.
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https://globalschoolofeconomics.com/classicalassumption2/
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# Econometrics: Classical Assumption 2 – The error term has a zero population mean.
When the intercept term is present in any given regression equation, it forces the average of the error term to be equal to zero. This occurs because the intercept, when included, will account for the fixed portion change of the dependent variable and the explanatory variables will account for the non-fixed portion. Therefore, the error term must be falling in size as both variable and fixed changes are accounted for. Therefore when an intercept is included and the sample size approaches infinity, then the average of the included will approach a zero population mean.
When assessing the stochastic error term and you have a series of possible values for that particular error term, if you were to add them all up and then get the average, this average number should be equal to zero. This probably would not be the case when assessing a small sample but when dealing with a large scale of numbers approaching infinity, this is what should occur.
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https://nl.mathworks.com/matlabcentral/cody/problems/2595-polite-numbers-politeness/solutions/1504279
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Cody
# Problem 2595. Polite numbers. Politeness.
Solution 1504279
Submitted on 25 Apr 2018 by Binbin Qi
This solution is locked. To view this solution, you need to provide a solution of the same size or smaller.
### Test Suite
Test Status Code Input and Output
1 Pass
x = 1; y_correct = 0; assert(isequal(politeness(x),y_correct))
2 Pass
x = 2; y_correct = 0; assert(isequal(politeness(x),y_correct))
3 Pass
x = 3; y_correct = 1; assert(isequal(politeness(x),y_correct))
4 Pass
x = 7; y_correct = 1; assert(isequal(politeness(x),y_correct))
5 Pass
x = 9; y_correct = 2; assert(isequal(politeness(x),y_correct))
6 Pass
x = 15; y_correct = 3; assert(isequal(politeness(x),y_correct))
7 Pass
x = 18; y_correct = 2; assert(isequal(politeness(x),y_correct))
8 Pass
x = 21; y_correct = 3; assert(isequal(politeness(x),y_correct))
9 Pass
x = 1024; y_correct = 0; assert(isequal(politeness(x),y_correct))
10 Pass
x = 1025; y_correct = 5; assert(isequal(politeness(x),y_correct))
11 Pass
x = 25215; y_correct = 11; assert(isequal(politeness(x),y_correct))
12 Pass
x = 62; y_correct = 1; assert(isequal(politeness(x),y_correct))
13 Pass
x = 63; y_correct = 5; assert(isequal(politeness(x),y_correct))
14 Pass
x = 65; y_correct = 3; assert(isequal(politeness(x),y_correct))
15 Pass
% anti-lookup & clue nums=primes(200); pattern=[1 nums([false ~randi([0 25],1,45)])]; x=prod(pattern)*2^randi([0 5]); y_correct=2^numel(pattern)/2-1; assert(isequal(politeness(x),y_correct))
16 Pass
for k=randi(2e4,1,20) assert(isequal(politeness(k*(k-1))+1,(politeness(k)+1)*(politeness(k-1)+1))) end
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latest
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https://mathcabin.com/tag/area-of-the-shaded-region-video-tutorial/
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Home » Posts tagged 'area of the shaded region video tutorial'
# Tag Archives: area of the shaded region video tutorial
## Shaded Area in a Square puzzle example question
In a unit square ABCD, point A is joined to the midpoint of BC, point B is joined to the midpoint of CD, point C is joined to the midpoint of DA, and point D is joined to the midpoint of AB. Find the area of the shaded region.
Solution to this Puzzle practice problem is given in the video below!
## Shaded Area in a Square puzzle example problem
In a unit square ABCD, M is a midpoint of AD, and AC is a diagonal. Find the area of the shaded regions.
Solution to this Puzzle practice problem is given in the video below!
## Shaded Area in a Square puzzle example
In the unit square ABCD, M is the midpoint and AC and BD are diagonals. Find the area of the shaded region.
Solution to this Puzzle practice problem is given in the video below!
## Shaded Area in a Square puzzle example question
In the following unit square ABCD, M is the midpoint and BD is a diagonal. Find the area of the shaded region.
Solution to this Puzzle practice problem is given in the video below!
## Shaded Area in a Square puzzle TRIGONOMETRY example problem
A square with side 1 is rotated around one vertex by an angle , where
and
.
Find the area of the shaded region.
Solution to this Puzzle practice problem is given in the video below!
## Shaded Area in a Rectangle puzzle example
In the following rectangle, an isosceles triangle is drawn. Find the area of the shaded region.
Solution to this Puzzle practice problem is given in the video below!
## Shaded Area in a Rectangle puzzle example question
In the following rectangle below, find the area of the shaded region.
Solution to this Puzzle practice problem is given in the video below!
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https://proofwiki.org/wiki/User:Dfeuer/Intersection_of_Complete_Meet_Subsemilattices
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# User:Dfeuer/Intersection of Complete Meet Subsemilattices
## Theorem
Let $(S, \preceq)$ be an ordered set.
Let $C_i$ be a complete meet subsemilattice (or a meet-complete subsemilattice?) of $S$.
Then $C = \displaystyle \bigcap_{i \in I} C_i$ is a complete meet subsemilattice.
## Proof
Let $D \subseteq C$.
By Intersection is Largest Subset, $D \subseteq C_i$ for each $i \in I$.
Thus $D$ has an infimum in $S$ and $\inf D \in C_i$ for each $i \in I$.
By the definition of intersection, $\inf D \in C$.
$\blacksquare$
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https://math.stackexchange.com/questions/1755700/adjoint-operator-to-the-derivative
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Adjoint Operator to the Derivative
Let $V \subset \Bbb R[X]$ be the Vectorspace of all Polynomials of degree $\le 3$. The inner product on $V$ is defined as follows: $$\langle f,g \rangle:=\int^1_{-1}f(t)g(t)dt$$Let $L:V \to V$ be the derivative $L(p) = p'$
I now have to determine the adjoint Operation $L^*$ so that $$\langle L(f),g \rangle=\langle f,L^*(g) \rangle$$ Any ideas? ( i tried partial Integration but i don't get anywhere.)
• i saw another post with a similar problem but they came to no solution. The problem i have is that even with partial integration L* would be dependent on f but it should of course be independently defined from g and f – DeltaChief Apr 23 '16 at 18:15
Integration by parts $\int_{-1}^1 f'g=[fg]_{-1}^1-\int_{-1}^1 fg'$ is a good idea. But you have to be in a context where the term $[fg]_{-1}^1$ is zero. In order that this condition is always fulfilled, you must restrict your working space to functions that are zero at both bounds $-1$ and $1$ (it is no longer a vector space). In this framework, we have
$$\int_{-1}^1 f'g=\int_{-1}^1 f(-g)'$$
proving that the adjoint of $L$ is its opposite
$$L^*=-L$$
Remark (a discrete analogy): If you are acustomed to the discrete equivalent of the derivation operator $L$, represented by the infinite matrix:
$$\begin{bmatrix}&\cdots&&&&&\\&-1&1&&&&\\&&-1&1&&&\\&&&-1&1&&\\&&&&-1&1&\\&&&&&\cdots&\\&&&&&&\cdots\\\end{bmatrix}$$ its transpose is as well its opposite.
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https://devsenv.com/example/%5Bsolved%5D-sherlock-and-moving-tiles-solution-in-hackerrank-hacerrank-solution-c,-c++,-c-,-java,-js,-python
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## Algorithm
Problem Name: Mathematics - Sherlock and Moving Tiles
In this HackerRank in Mathematics - Sherlock and Moving Tiles solutions,
Sherlock is given 2 square tiles, initially both of whose sides have length l placed in an x - y plane. Initially, the bottom left corners of each square are at the origin and their sides are parallel to the axes.
At t = 0, both squares start moving along line y = x (along the positive x and y) with velocities s1 and s2.
For each querydetermine the time at which the overlapping area of tiles is equal to the query value, queries[i].
Note: Assume all distances are in meters, time in seconds and velocities in meters per second.
Function Description
Complete the movingTiles function in the editor below.
movingTiles has the following parameter(s):
• int l: side length for the two squares
• int s1: velocity of square 1
• int s2: velocity of square 2
• int queries[q]: the array of queries
Returns
• int[n]: an array of answers to the queries, in order. Each answer will be considered correct if it is at most 0.0001 away from the true answer.
Input Format
First line contains integers l,s1,s2.
The next line contains q, the number of queries.
Each of the next q lines consists of one integer queries[i] in one line.
Constraints
1 <= l,s1,s2 <= 109
1 <= q <= 105
1 <= queries[i] <= L2
s1 not= s2
Sample Input
``````10 1 2
2
50
100
``````
Sample Output
``````4.1421
0.0000
``````
Explanation
For the first case, note that the answer is around `4.1421356237...`, so any of the following will be accepted:
``````4.1421356237
4.14214
4.14215000
4.1421
4.1422``````
## Code Examples
### #1 Code Example with C Programming
```Code - C Programming```
``````
#include <stdio.h>
#include <string.h>
#include <math.h>
#include <stdlib.h>
int main() {
double l, s1, s2;
int q;
int i;
scanf("%lf %lf %lf", &l, &s1, &s2);
scanf("%d", &q);
for (i = 0; i < q; i++) {
double qi;
scanf("%lf", &qi);
printf("%.20f\n", sqrt(2) * (l - sqrt(qi)) / abs(s2 - s1));
}
return 0;
}
``````
Copy The Code &
### #2 Code Example with C++ Programming
```Code - C++ Programming```
``````
#include <bits/stdc++.h>
using namespace std;
string ltrim(const string &);
string rtrim(const string &);
vector < string> split(const string &);
/*
* Complete the 'movingTiles' function below.
*
* The function is expected to return a DOUBLE_ARRAY.
* The function accepts following parameters:
* 1. INTEGER l
* 2. INTEGER s1
* 3. INTEGER s2
* 4. INTEGER_ARRAY queries
*/
vector<long double> movingTiles(int l, int s1, int s2, vector<long long> queries) {
int v = abs(s2 -s1);
vector < long double> ans;
for (long long q : queries) {
long double d = sqrt(2)* (long double)l - sqrt(2)*sqrt(q);
ans.push_back(d/v);
}
return ans;
}
int main() {
int l, s1, s2;
cin >> l >> s1 >> s2;
int q;
cin >> q;
vector < long long> queries;
long long query;
for (int i = 0; i < q; i++){
cin >> query;
queries.push_back(query);
}
vector < long double> responses = movingTiles(l, s1, s2, queries);
for (auto time : responses){
cout<< fixed << setprecision(20) << time << endl ;
}
return 0;
}
vector < string> split(const string &str) {
vector<string> tokens;
string::size_type start = 0;
string::size_type end = 0;
while ((end = str.find(" ", start)) != string::npos) {
tokens.push_back(str.substr(start, end - start));
start = end + 1;
}
tokens.push_back(str.substr(start));
}
``````
Copy The Code &
### #3 Code Example with C# Programming
```Code - C# Programming```
``````
using System.CodeDom.Compiler;
using System.Collections.Generic;
using System.Collections;
using System.ComponentModel;
using System.Diagnostics.CodeAnalysis;
using System.Globalization;
using System.IO;
using System.Linq;
using System.Reflection;
using System.Runtime.Serialization;
using System.Text.RegularExpressions;
using System.Text;
using System;
class Result
{
/*
* Complete the 'movingTiles' function below.
*
* The function is expected to return a DOUBLE_ARRAY.
* The function accepts following parameters:
* 1. INTEGER l
* 2. INTEGER s1
* 3. INTEGER s2
* 4. INTEGER_ARRAY queries
*/
public static List < double> movingTiles(int l, int s1, int s2, List<int> queries)
{
}
}
class Solution
{
public static void Main(string[] args)
{
TextWriter textWriter = new StreamWriter(@System.Environment.GetEnvironmentVariable("OUTPUT_PATH"), true);
int l = Convert.ToInt32(firstMultipleInput[0]);
int s1 = Convert.ToInt32(firstMultipleInput[1]);
int s2 = Convert.ToInt32(firstMultipleInput[2]);
List < int> queries = new List<int>();
for (int i = 0; i < queriesCount; i++)
{
}
List < double> result = Result.movingTiles(l, s1, s2, queries);
textWriter.WriteLine(String.Join("\n", result));
textWriter.Flush();
textWriter.Close();
}
}
``````
Copy The Code &
### #4 Code Example with Java Programming
```Code - Java Programming```
``````
import java.io.*;
import java.math.*;
import java.text.*;
import java.util.*;
import java.util.regex.*;
public class Solution {
/*
* Complete the movingTiles function below.
*/
static double[] movingTiles(int l, int s1, int s2, long[] queries) {
/*
*/
int len=queries.length;
double[] time = new double[len];
double root2 = Math.sqrt(2.0);
double diff =Math.abs(1.0*s1-1.0*s2);
for(int i= 0; i < len; i++)
{ time[i]=(l*1.0-Math.sqrt(queries[i]*1.0))*root2/diff;
}
return time;
}
private static final Scanner scanner = new Scanner(System.in);
public static void main(String[] args) throws IOException {
BufferedWriter bufferedWriter = new BufferedWriter(new FileWriter(System.getenv("OUTPUT_PATH")));
String[] lS1S2 = scanner.nextLine().split(" ");
scanner.skip("(\r\n|[\n\r\u2028\u2029\u0085])*");
int l = Integer.parseInt(lS1S2[0]);
int s1 = Integer.parseInt(lS1S2[1]);
int s2 = Integer.parseInt(lS1S2[2]);
int queriesCount = scanner.nextInt();
scanner.skip("(\r\n|[\n\r\u2028\u2029\u0085])*");
long[] queries = new long[queriesCount];
for (int queriesItr = 0; queriesItr < queriesCount; queriesItr++) {
long queriesItem = scanner.nextLong();
scanner.skip("(\r\n|[\n\r\u2028\u2029\u0085])*");
queries[queriesItr] = queriesItem;
}
double[] result = movingTiles(l, s1, s2, queries);
for (int resultItr = 0; resultItr < result.length; resultItr++) {
bufferedWriter.write(String.valueOf(result[resultItr]));
if (resultItr != result.length - 1) {
bufferedWriter.write("\n");
}
}
bufferedWriter.newLine();
bufferedWriter.close();
scanner.close();
}
}
``````
Copy The Code &
### #5 Code Example with Javascript Programming
```Code - Javascript Programming```
``````
'use strict';
const fs = require('fs');
process.stdin.resume();
process.stdin.setEncoding('utf-8');
let inputString = '';
let currentLine = 0;
process.stdin.on('data', function(inputStdin) {
inputString += inputStdin;
});
process.stdin.on('end', function() {
inputString = inputString.split('\n');
main();
});
return inputString[currentLine++];
}
/*
* Complete the 'movingTiles' function below.
*
* The function is expected to return a DOUBLE_ARRAY.
* The function accepts following parameters:
* 1. INTEGER l
* 2. INTEGER s1
* 3. INTEGER s2
* 4. INTEGER_ARRAY queries
*/
function movingTiles(l, s1, s2, queries) {
return queries.map(q =>
Math.abs(Math.sqrt(2) * (Math.sqrt(q) - l) / (s1 - s2)).toFixed(4)
);
}
function main() {
const ws = fs.createWriteStream(process.env.OUTPUT_PATH);
const firstMultipleInput = readLine().replace(/\s+\$/g, '').split(' ');
const l = parseInt(firstMultipleInput[0], 10);
const s1 = parseInt(firstMultipleInput[1], 10);
const s2 = parseInt(firstMultipleInput[2], 10);
let queries = [];
for (let i = 0; i < queriesCount; i++) {
queries.push(queriesItem);
}
const result = movingTiles(l, s1, s2, queries);
ws.write(result.join('\n') + '\n');
ws.end();
}
``````
Copy The Code &
### #6 Code Example with Python Programming
```Code - Python Programming```
``````
#!/bin/python3
import math
import os
import random
import re
import sys
#
# Complete the 'movingTiles' function below.
#
# The function is expected to return a DOUBLE_ARRAY.
# The function accepts following parameters:
# 1. INTEGER l
# 2. INTEGER s1
# 3. INTEGER s2
# 4. INTEGER_ARRAY queries
#
def movingTiles(l, s1, s2, queries):
arr = []
for i in queries:
arr.append(round(math.sqrt(2)*(l-math.sqrt(i))/(abs(s1-s2)),4))
return arr
if __name__ == '__main__':
fptr = open(os.environ['OUTPUT_PATH'], 'w')
first_multiple_input = input().rstrip().split()
l = int(first_multiple_input[0])
s1 = int(first_multiple_input[1])
s2 = int(first_multiple_input[2])
queries_count = int(input().strip())
queries = []
for _ in range(queries_count):
queries_item = int(input().strip())
queries.append(queries_item)
result = movingTiles(l, s1, s2, queries)
fptr.write('\n'.join(map(str, result)))
fptr.write('\n')
fptr.close()
``````
Copy The Code &
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# Search results
1. ### Rotation in space.
We have a rod and keep floating in ISS. If we tap giving force at one end, will it, 1. Rotate about an axis or 2. Has translational and rotational motion. If it rotates, where is the axis? I have interest in yoyo motion too. From YouTube, they spin them while holding. Can we rotate yoyo from...
2. ### Transit of Venus
A transit of Venus across the Sun takes place when thplanet"etVenus passes directly between the Sun and a superior planet, becoming visible against (and hence obscuring a small portion of) the solar disk. During a transit, Venus can be seen from Earth as a small black disk moving across the face...
3. ### Electric Field Intensity. Finite sheet of charge.
problem statement, all variables and given/known data A finite sheet of charge, of density ρ=2x(x2+y2+4)^3/2, lies in the z=0 plane for 0≤x≤2m and 0≤y≤2m.Determine E at (0,0,2)m Ans:(18x10^9)(-16/3ax-4ay+8az) Homework Equations E=kQ/R2 The Attempt at a Solution dE=ρdA / R^2 aR...
4. ### Electric Field Vector
Homework Statement From attachment R=-xaa-yay+3az Homework Equations The Attempt at a Solution I see 4 different values, one for each quadrant. 1st.quadrant R=+xaa-yay+3az 2nd. quadrant R=-xaa+yay+3az 3rd. R=-xaa+yay+3az 4th R=-xaa-yay+3az Where did I go wrong in determine value of R?
5. ### Electric Field Intensity
Homework Statement Show that the electric field E outside a spherical shell of uniform charge density ρs is the same as due to the total charge on the shell located at the centre. Homework Equations Using only Coulomb's Law E=Q/4πε0 ar The Attempt at a Solution If i assumed it as circular...
6. ### Coulomb Forces
Homework Statement Find the force on a point of charge of 30μC at (0,0,5)m due to a 4m square int he z=0 plane between x=±2m and y=±2m with a total of 500μC, distributed uniformly. Homework Equations The Attempt at a Solution R=-x,-y,+5z dQ=ρdydx=500/4 dydx dE=dQ(5az) / (4πε0((x2+y2+25)3/2...
7. ### Spherical coordinates vector
Homework Statement Find the vector directed from (10,3π/4,π/6) to (5, π/4,π), where the endpoints are given in spherical coordinates. Ans -9.660ax, - 3ay. + 10.61az Homework Equations az=rCosΦ The Attempt at a Solution az=10Cos(π/6) +5Cos(π) =13.6 My answer differs. Where did i go wrong?
8. ### Probability of placing marbles in a compartment
Homework Statement 1. (a) A shelf contains 6 separate compartments. In how many ways can 4 distinguishable marbles be placed in the compartments?(b) Work the problem if there are n compartments and r marbles. This type of problem arises in physics in connection with Bose-Einstein statistics...
9. ### Rotational Inertias.
How to derive the Rotational Inertia for a flat plate. I do not see any symmetry about the axis of rotation as in a thin rod. http://imageshack.us/a/img845/1134/53433231.jpg [Broken]
10. ### There is a kingdom where if a person drinks poison he will die. The
There is a kingdom where if a person drinks poison he will die. The only way to counteract the poison is to drink a stronger one. Then the reaction stops. The king decides that he must have the strongest poison available in his possession. So, he sets up a contest between his court adviser and...
11. ### For each real number x, let f(x) be the minimum of the numbers 4x+1,
For each real number x, let f(x) be the minimum of the numbers 4x+1, x+2, and -2x+4. What is the maximum value of f(x)?
12. ### Torque and momentum.
Sorry the title should be : Impulse and Momentum 3000Solved Problems in Physics-Shaum's 9.6: A camper lets fall a heavy mallet of mass M from the height y upon the top of a tent stake of mass m and drives it into the ground a distance d. Find the resistance of the ground, assuming it to...
13. ### Unit of Torque
Why unit of torque is not m.N 1.Torque=rFSinθ =>m.N 2. 1 N.m = 1 Joule but 1 N.m in torque ≠ 1 Joule For the above 2 reasons i reckon it should have unit of m.N.
If we are to answer multiply questions on a single topic, do the step is vital. For example question#5, do we have to get answer for question 5(a) before we are able to do question b(b). Does it mean the first one is the easiest? Thank you.
15. ### Maybe triangle inequalities theorem
Homework Statement Prove: If a2+b2=1, and c2+d2=1, then ac+bd ≤1 Homework Equations Maybe triangle inequalities theorem. The Attempt at a Solution
16. ### Newton's Laws of Motion
At the instant that one-fourth of block 1 remains on block 2, x2+l=x1+(l/16). For days trying to figure out how this equation derived from. Thank You. http://img15.imageshack.us/img15/3926/kotak.jpg [Broken]
17. ### Limit homework issue
Homework Statement If f(x)=x2 prove that \lim_{x \to 2} f(x)= 4 Solution given: We must show that given any ε >0, find δ >0 such that |x2-4|<ε when 0<|x-2|<δ Choose δ≤1 so that <|x-2|<1 ----------------------------------------------- Confuse between the word 'find' and 'choose'.
18. ### Work and Energy
A vertical spring with constant 200N/m has a light platform on its top. When a 500-g mass is set on the platform, the spring compresses 0.0245m. The mass is now pushed down 0.0755m farther and released. How far above this latter position will the mass fly? The answer from the book. If it does...
19. ### Quantum World
I watch a series on quantum world. The host said particles can pass through each others. And then he pushed a wall and said if he keep on pushing for billion of years, there will be possibilities he will pass through the wall. We have more than 7 billions people in this world. I never heard...
20. ### The WMAP objective is to measure the temperature differences in the
The WMAP objective is to measure the temperature differences in the Cosmic Microwave Background (CMB) radiation. My question are, 1.How to distinguish all the radiations of 13 billion years ago from others of the latter ones. 2.How the distance of radiation source is measured. Thanks.
21. ### Amgular impulse
How far above its center should a billiard ball be struck in order it roll without any initial slippage? Denote the ball's radius by R, and assume that the impluse delivered byt the cue is purely horizontal Book Solution: Since any static frictional forced supplied by the table is limited...
22. ### Really couldn't catch the concept on epsilon and delta in limits
Really couldn't catch the concept on epsilon and delta in limits. Let ∂x=x2 - x1 In finding a gradient the value ∂y is taken at certain value. But in finding area using integral, the ∂y is seen to taken as zero. F(x2)=F(x1) Maybe one multiplication and the other is division.
23. ### Depletion region- diode
At depletion layer of p region, there more negative ions thus negatively charged region. Likewise at n region, it is positively charged. Then why in equivalent forward bias the diode behave as a battery of .7v with positive at depletion region of the p region where it is negatively charged?
24. ### Probability- My poor English
Find the probability of scoring a total of 7 points (a) once, (b) at least once, (c) twice, in 2 tosses of a pair of dice. My question. 1. What does scoring a total of 7 points mean? Is it 1+6 or 2+5 or 3+4 or 4+3 or 5+2 or 6+1 2. Is in 2 tosses of a pair of dice means that throwing 2...
25. ### Rotational motion, frictionless.
http://img444.imageshack.us/img444/8148/rollwx.jpg [Broken] A rope is wound around a cylinder of mass 4kg, and I=0.02kg.m2 about the cylinder axis. The frictional force between table and cylinder is negligible. Solution: a=20/4=5.0m/s2 α=ar is not applicable when slippage occurs...
26. ### Dynamic of Ratational Motion
If the ball were rolling uphill, the force of friction would still be directed uphill as in FJg. 1O.19b. Can you see why'? http://img252.imageshack.us/img252/4516/rollingm.jpg [Broken] I always assume the friction is always in opposite direction of the motion.
27. ### Deep freeze
Today i watched Discovery about the fate of the universe. One of the presenters said that finally all the energies used up and the universe will be in deep freeze. My question is where all the energies go? What about conservation of energies? Thank you.
28. ### Inequalities #2
Homework Statement a. Prove: If a≠b≠c are real numbers, then a2+b2+c2>ab+bc+ca b. Prove: If a>0, b>0and a≠b, then a/b+b/a>2 Homework Equations (real numbers)2>0 The Attempt at a Solution a. (a+b+c)2>0 a2+b2+c2>-2(ab+bc+ca) Try to prove -2(ab+bc+ca) > ab+bc+ca but not true, -2.4...
29. ### Centrifugal force
"Centrifugal force" Homework Statement A smooth horizontal tube of length l rotates about a vertical axis. A particle placed at the extreme end of the tube is projected towards O(axis) with a velocity lω while at the same time the tube rotates about the axis with constant angular speed ω...
30. ### Prove the inequality?
Homework Statement Prove: If a>b and c>d, then a+c>b+d Hint: (a-b)+(c-d)=(a+c)-(b+d)>0 Homework Equations The Attempt at a Solution How to use the hint to prove the inequality? My method, not sure it's right. Given c>d, c-d>0 Given a>b => a+(c-d)>b Thus a+c>b+d
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sparseland.blogspot.jp
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## May 6, 2012
### Zero, one, and infinity in compressed sensing
I am very happy I got many comments for my first "technical" blog, where I made a question on relationship between zero and infinity norms in CS (compressed sensing).
Igor pointed out his blog entries (here and here) in his comment. According to the entry, an L-infinity optimization may produce a vector the elements of which are sticked to ±||x||_∞. Applying DFT to such a vector, e.g., x=(1,1,...,1), result in a sparse vector in the Fourier domain. This may be a simple explanation why the infinity norm is related to CS.
Mahesh and Thomas pointed out the infinity norm is also used in the Dantzig selector. This is based on the fact that L-1 (ell-1) is the dual vector space of L-infinity (ell-infinity). This is yet another link between zero and infinity (note that L-1 is related to L-0 by so many studies on CS). This also interests me very much.
1. Masaaki,
You might also be interested in this paper
http://ssg.mit.edu/~venkatc/crpw_lip_preprint10.pdf
In the figure "A summary of the recovery bounds obtained using Gaussian width arguments."
there is this info about +/-1 vector being recoverable with gaussian matrices with p/2 rows.
of related interest:
http://www.lx.it.pt/~mtf/ICASSP%202012%20Tutorial%20[Cevher,%20Figueiredo].pdf
Igor.
1. Igor,
Thanks a lot for linking nice pdf's.
At p.4 of the tutorial slides, I remember Volkan had a lecture at my university, Kyoto univ, just after ICASSP2012. The lecture was nice on high-dimensional statistics, which is different from one you suggested, but he also used the "<S>parsity" notation (<S> for the trademark of Superman). That's why I remember Volkan's lecture from the tutorial slides.
Best regards,
Masaaki
| 456
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| 2.9375
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CC-MAIN-2018-22
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https://security.stackexchange.com/questions/42359/is-there-symmetric-encryption-like-aes-or-camelia-with-keys-longer-than-256-bits
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Is there symmetric encryption like AES or Camelia with keys longer than 256 bits?
I am wondering if is there any (secure!) cipher like AES or Camelia with keys longer than 256 bits (and offering higher security than 256 bits). I have not found anything except http://www.ciphers.de/eng/index.html where they say they have such cipher (but I have not found anything about that cipher).
• 256 bit AES in a EDE chain is trivially 768 bits Sep 13, 2013 at 11:34
• @ratchetfreak That is incorrect. 256-bit AES in EDE would be 512 bits due to the meet-in-the-middle attack. This is also the reason 3DES has a 112 bit security level, not 168 (despite it taking 168 bits of key material). Apr 23, 2018 at 3:32
There is no higher security than 128 bits. There are longer keys, for sure, but we cannot say that they offer "more" security, because they is no security level beyond "meh, cannot break it".
A symmetric encryption algorithm can be generically broken by exhaustive search, i.e. trying out all possible key values; with n-bit keys, there are 2n possible key values. When n grows, this number soon rises way beyond that which is technologically feasible. Threshold depends on current technology, but physics tell us that there are hard limits which seem unavoidable, related to energy consumption (see this answer for details). Right now, all the might of mankind, united for a decade for a single crypto-breaking goal, might pull off an exhaustive search on a 95-bit or so key, but a realistic organization (say, with the combined resources of Apple, Google and Microsoft together) would stop below, around 80 bits. Each additional bit doubles the cost, so 128 bits are more than enough to be in the "cannot break now or in twenty years" zone.
Additional key bits are there for the psychological effect, not for security. We like to go to 128 bits because 128 is a power of 2 and cryptographers just love powers of 2. Beyond that, extra key size is no more than an assertion of manhood. It is a bit like cars. A 128-bit key is a comfortable, fast and powerful BMW car. A 192-bit key is a Hummer with chrome bumpers. A 256-bit is an Aircraft Carrier. When looking for a "more than 256 bits" key, you are actually trying to paint your Aircraft Carrier in bright red.
Now there are a few algorithms which allow for longer keys, not because it helps with anything, but because there was room for it and it would have taken some effort to forbid it. For instance, Blowfish goes to 448 bits. SHACAL (both SHACAL-1 and SHACAL-2) accept keys up to 512 bits. Threefish works with a whooping 1024 bits.
• Ok, I know that 128 bit ciphers are totally sufficient (for now or next thirty years). But I was thinking about a cipher that would be (probably secure for over 50 years even against quantum computers). According to this security.stackexchange.com/questions/14068/…, the quantum computer divides the security by two. So if there is a cipher with complexity about 2^128 it would become 2^64 which might be breakable. That is the reason of my question.
– Pter
Sep 13, 2013 at 15:39
• Then 256-bit keys provide 128-bit security against quantum computers -- if they ever exist. Sep 13, 2013 at 15:44
• @Pter Dividing the keyspace by two would turn 2^128 into 2^127. Apr 23, 2018 at 3:30
• "There is no higher security than 128 bits" - this is nonsense. May 1, 2018 at 10:50
• Bitcoin miners reached ≈2^92 SHA-256 hashes per year in 06 Agust 2019. Also, this answer missing the Grover's algorithm. If ready we must switch to 256-bit. Sep 17, 2019 at 15:20
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June 2008
# It Ain't No Repeated Addition
In my column for September 2007, which was titled What is conceptual understanding? I remarked that I wished schoolteachers would stop telling pupils that multiplication is repeated addition. It was little more than a throwaway line, albeit one that I feel strongly about. I put it in to provide a further illustration for the overall theme of the column, to indicate that there are examples beyond the ones I had focused on. In the intervening months, however, I've received a number of emails from teachers asking for elaboration. Their puzzlement, they make clear, stems from their understanding that multiplication actually is repeated addition.
If ever there were needed a strong argument that professional mathematicians need to interest themselves in K-12 mathematics education and get involved, this example alone should provide it. The teachers who contact me do so because they genuinely want to know what I mean, having been themselves taught, presumably either in schools of education or else from school textbooks, that multiplication is repeated addition.
Let's start with the underlying fact. Multiplication simply is not repeated addition, and telling young pupils it is inevitably leads to problems when they subsequently learn that it is not. Multiplication of natural numbers certainly gives the same result as repeated addition, but that does not make it the same. Riding my bicycle gets me to my office in about the same time as taking my car, but the two processes are very different. Telling students falsehoods on the assumption that they can be corrected later is rarely a good idea. And telling them that multiplication is repeated addition definitely requires undoing later.
How much later? As soon as the child progresses from whole-number multiplication to multiplication by fractions (or arbitrary real numbers). At that point, you have to tell a different story.
"Oh, so multiplication of fractions is a DIFFERENT kind of multiplication, is it?" a bright kid will say, wondering how many more times you are going to switch the rules. No wonder so many people end up thinking mathematics is just a bunch of arbitrary, illogical rules that cannot be figured out but simply have to be learned - only for them to have the rug pulled from under them when the rule they just learned is replaced by some other (seemingly) arbitrary, illogical rule.
Pretending there is just one basic operation on numbers (be they whole numbers fractions, or whatever) will surely lead to pupils assuming that numbers are simply an additive system and nothing more. Why not do it right from the start?
Why not say that there are (at least) two basic things you can do to numbers: you can add them and you can multiply them. (I am discounting subtraction and division here, since they are simply the inverses to addition and multiplication, and thus not "basic" operations. This does not mean that teaching them is not difficult; it is.) Adding and multiplying are just things you do to numbers - they come with the package. We include them because there are lots of useful things we can do when we can add and multiply numbers. For example, adding numbers tells you how many things (or parts of things) you have when you combine collections. Multiplication is useful if you want to know the result of scaling some quantity.
You don't have to use these applications, but both are simple and familiar, and to my mind they are about as good as it gets in terms of appropriateness. (I do think that you need to present simple everyday examples of applications. Teaching a class of elementary school students about axiomatic integral domains is probably not a good idea! This column is not a rant in favor of the "New Math", a term that I use here to denote the popular conception of the log-ago aborted education reform that bears that name.)
Once you have established that there are two distinct (I don't say unconnected) useful operations on numbers, then it is surely self-evident that repeated addition is not multiplication, it is just addition - repeated!
But now, you have set the stage for that wonderful moment when you can tell kids, or even better maybe they can discover for themselves, this wonderful trick that multiplication gives you a super quick way to calculate a repeated addition sum. Why deprive the kids of that wonderful piece of magic?
[Of course, any magic trick loses a lot once you see behind the scenes. In the very early days of the development of the number concept, around 10,000 years ago, there were only whole numbers, and it may be that the earliest precursor of what is now multiplication was indeed repeated addition. But that was all 10,000 years ago, and things have changed a lot since then. We don't try to understand how the iPod works in terms of the abacus, and we should not base our education system on what people knew and did in 8,000 B.C.]
Mathematics is chock full of examples where something that is about A turns out to be useful to do B.
Exponentiation turns out to provide a quick way to do repeated multiplication - wow, it's happened again! Is this math thing cool or what!
Anti-differentiation turns out to be a quick way to calculate an integral. Boy, is that deep!
I can just hear some pupils wondering, "Hey, how many more examples are there like this? This is really, really intriguing. It all seems to fit together. Something deep must be going on here. I've gotta find out more."
I assume the reason for the present state of affairs is that teachers (which really means their instructors or the writers of the textbooks those teachers have to use) feel that children will be unable to cope with the fact that there are two basic operations you can perform on numbers. And so they tell them that there is really only one, and the other is just a variant of it. But do we really believe that two operations is harder to come to terms with than one? The huge leap to abstraction comes in the idea of abstract numbers that you can do things with. Once you have crossed that truly awesome cognitive chasm, it makes little difference whether you can do one abstract thing with numbers or a dozen or more.
Of course, there are not just two basic operations you can do on numbers. I mentioned a third basic operation a moment ago: exponentiation. University professors of mathematics struggle valiantly to rid students of the false belief that exponentiation is "repeated multiplication." Hey, if you can confuse pupils once with a falsehood, why not pull the same stunt again? I'm teasing here. But with the best intentions of drawing attention to something that I think needs to be fixed.
And the way to fix it is to make sure that when we train future teachers, and when authors write, or states adopt, textbooks, we all do it right. We mathematicians bear the ultimate responsibility here. We are the world's credentialed experts in mathematical structures, including the various numbers systems. ("Systems" here includes the operations that can be performed on them.) Our professional predecessors constructed those structures. They are part of our world view, things we mastered so long ago in our educational journey that they are second nature. For too long we have tacitly assumed that our knowledge and understanding of those systems is shared by others. But that isn't the case. I have a file of puzzled emails from qualified teachers that testifies to the gap.
I should end by noting that I have not tried to prescribe how teachers should teach arithmetic. I am not a trained K-12 teacher, nor do I have any first-hand experience to draw on. But the term "mathematics teaching" comprises two words, and I do have expertise in the first. That is my focus here, and I defer to others who have the expertise in teaching. The best way forward, surely, is for the two groups of specialists, the mathematicians and the teachers, to dialog - regularly and often.
In the meantime, teachers, please stop telling your pupils that multiplication is repeated addition.
Devlin's Angle is updated at the beginning of each month.
Mathematician Keith Devlin (email: devlin@csli.stanford.edu) is the Executive Director of the Center for the Study of Language and Information at Stanford University and The Math Guy on NPR's Weekend Edition. Devlin's most recent book, Solving Crimes with Mathematics: THE NUMBERS BEHIND NUMB3RS, is the companion book to the hit television crime series NUMB3RS, and is co-written with Professor Gary Lorden of Caltech, the lead mathematics adviser on the series. It was published last September by Plume.
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# Explain Images formed in Concave Mirror
The position. size and nature of an image formed in a spherical mirror depend on the position of the object placed in front of the mirror. If there is a change in the position of the object, then a corresponding change in the position, size and nature of the image also takes place. Let, MPM’ be a concave mirror. Here, P is the pole, F is the principal focus and C is the centre of curvature of it. AO is the object situated perpendicularly on the principal axis in front of it.
If the object is placed anywhere between infinity and the principal focus, the image thus formed will always be real and inverted Again, in case of an object placed in between the principal focus and the pole, the image will be virtual and erect. The real and virtual images formed by a concave mirror are described below:
Real image: From the point O, a ray OM incident at the point M parallel to the principal axis and reflected through the principal focus along the path ME. Another ray OCIVI1 from the point 0 incident through the centre of curvature C on the mirror and reflects back along the same path. After reflection, the two rays meet at the point I actually. Therefore I am the real image of the point 0. From A. ray incident along the principal axis reflects back along the same path. So. the image of A will be formed on that line. Let us draw normal. IB, from the point I on the principal axis. Now. BI is the real image of the object AO (below figure). The nature of the image is real and inverted.
Fig (1)
Virtual image: In figure 2, the object is situated between the pole and the principal focus. From the point 0, a ray incident parallel to the principal axis and reflected through the principal focus. Another ray incident on the mirror through the centre of curvature reflects back along the same path. After reflection, two rays become mutually diverging. If the two rays are extended backward, they appear to come from the point I. So, the point I am the virtual image of the point O. Normal B3 is drawn on the principal axis from the point I. therefore, BI is the virtual and erect image of the object. The position of the image so developed is behind the mirror, its nature is virtual and erect, and magnified i.e. larger than the size of the object.
Fig (2)
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VDU_Physics_1_011
## Physics 1.- Mechanics: Accelerated Circular Motion
Javier Montenegro Joo
jmj@VirtualDynamicsSoft.com
Accelerated Circular Motion.-
[1] A disk of 2.5 kg and 1 m diameter is rotating at 3000 rpm when it is disconnected from the source of energy that keeps it rotating. For the moment just before disconnection calculate: (a) angular speed of the disc (b) tangential speed of a point on the edge of the disc, (c) centripetal acceleration of a point on the edge of the disc (d) If the disk stops 30 s after disconnection, determine how many turns are completed during that time.
Solution.-
Ballistic pendulum simulator in the Physics Virtual Lab (PVL).
The PVL includes 200+ Physics Simulation Modules
This module displays an application of the conservation of energy and linear momentum during collisions.
To know about the PVL click: http://www.virtualdynamicssoft.com
[2] A 2.50 m diameter disk starts rotating with an angular speed of 1.75 rad/s and a constant acceleration of 1.60 rad/s2 . For a dot at the edge of the disk and for 1.50s after initiated the motion, calculate: (a) Displacement, (b) Angular velocity, (c) Tangential velocity, (d) Magnitude of the acceleration vector, (e) Angle between acceleration and radius.
Solution.-
[3] When the motor of a 34 cm diameter disk which is rotating at 180 rpm, is turned off, the disk gradually comes to rest in 18 s. Determine: (a) angular acceleration of the disk. (b) Its average angular speed (c) Number of turns completed until the disk stops (d) Displacement of a point on the edge of the disk until this comes to stop.
Solution.-
Simulation module dealing with static friction in the Physics Virtual Lab.
The block on the gradually tilted plane remains there while the static friction is not overcome by the component of the weight opposing the friction.
PVL includes 200+ Physics Simulation Modules
[4] A 90-cm diameter wheel is rotating at 250 Rpm when the brakes are applied for 30 seconds and at the end of that time the wheel has reduced its velocity in 40%. Determine: (a) Acceleration experimented by the wheel (b) Number of turns completed during the time the brakes are applied (c) Distance covered by an edge point of the wheel during the time the brakes are applied.
Solution.-
New web page dealing with Algorithmic (Mathematical) Art and Nonlinear Dynamics & Chaos:
www.VirtualDynamics.tech
[5] A disk of 25 kg and 1.80 m in diameter is spinning at 5000 Rpm, when the engine that keeps it rotating is switched off. Before turn off engine, calculate: (a) speed of the disk. (b) Speed of a point on the edge of the disk. (c) The centripetal acceleration of a point on the edge of the disk. (d) If the disk stops 3 min after turning off the engine, calculate how many turns are completed in that time.
Solution.-
[6] A 1.25m diameter disk that is at rest receives a constant acceleration of 0.40 rad/s2 during 5 minutes, and then the acceleration is turned off. Determine: (a) Number of turns completed during the 5 minutes (b) Maximum speed reached during the 5 min. (c) Time to stop after disconnecting the acceleration. (d) Number of turns completed after disconnecting the acceleration.
Solution.-
[7] A 120 gr ball is tied to the end of a string of 3 m which is rotating with its other end fixed to a rotating shaft. The ball describes a circular orbit and moves with an acceleration of 10 m/s2 . When the acceleration vector makes an angle of 25o with rope, calculate (a) radial acceleration (b) tangential acceleration (c) speed of the ball.
Solution.-
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# How far is El Aaiún from Smara?
The distance between Smara (Smara Airport) and El Aaiún (Hassan I Airport) is 99 miles / 159 kilometers / 86 nautical miles.
The driving distance from Smara (SMW) to El Aaiún (EUN) is 136 miles / 219 kilometers, and travel time by car is about 3 hours 25 minutes.
99
Miles
159
Kilometers
86
Nautical miles
## Distance from Smara to El Aaiún
There are several ways to calculate the distance from Smara to El Aaiún. Here are two standard methods:
Vincenty's formula (applied above)
• 99.002 miles
• 159.328 kilometers
• 86.030 nautical miles
Vincenty's formula calculates the distance between latitude/longitude points on the earth's surface using an ellipsoidal model of the planet.
Haversine formula
• 98.868 miles
• 159.113 kilometers
• 85.914 nautical miles
The haversine formula calculates the distance between latitude/longitude points assuming a spherical earth (great-circle distance – the shortest distance between two points).
## How long does it take to fly from Smara to El Aaiún?
The estimated flight time from Smara Airport to Hassan I Airport is 41 minutes.
## Flight carbon footprint between Smara Airport (SMW) and Hassan I Airport (EUN)
On average, flying from Smara to El Aaiún generates about 40 kg of CO2 per passenger, and 40 kilograms equals 87 pounds (lbs). The figures are estimates and include only the CO2 generated by burning jet fuel.
## Map of flight path and driving directions from Smara to El Aaiún
See the map of the shortest flight path between Smara Airport (SMW) and Hassan I Airport (EUN).
## Airport information
Origin Smara Airport
City: Smara
Country: Western Sahara
IATA Code: SMW
ICAO Code: GMMA
Coordinates: 26°43′54″N, 11°41′4″W
Destination Hassan I Airport
City: El Aaiún
Country: Western Sahara
IATA Code: EUN
ICAO Code: GMML
Coordinates: 27°9′6″N, 13°13′9″W
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# Elastic Collisions and Violation Thereof in the Case of Shattering
In Taylor's Classical Mechanics, he includes the following discussion as a follow up to a discussion of energies of interaction and how elastic collisions arise in cases where the force of interaction of the inbound particles is conservative and goes to $$0$$ at large separations (ie. all inbound kinetic energy of the two particles is recovered after the particles separate):
The foregoing discussion may suggest that elastic collisions should be a very com- mon occurence. All that is needed is two particles whose interaction is conservative. In practice, elastic collisions are not as widespread as this seems to imply. The trouble comes from the requirement that it be two particles that enter and leave the collision. For example, if we fire one billiard ball at a second with sufficient energy, the two balls may shatter. Similarly, if we fire an electron with sufficient energy at an atom, the atom may fall apart or, at least, change the internal motion of its constituents. Even in the collision of two genuine particles, such as an electron and a proton, relativity tells us that, with sufficient energy, new particles can be created. Clearly, at high enough energy, the assumption that the two objects entering a collision can be approximated as indivisible particles eventually breaks down, and we cannot assume that collisions will be elastic, even if all the underlying forces are conservative.
Now, I am curious as to why we have the restriction that the particles cannot shatter or else the collision is not elastic. I am assuming that this is because, if there is shattering that goes on, then the otherwise constant energies of interaction between the internal particles of the "macro-particle" that shattered are not constant throughout the period of observation (ie. the rigid-body assumption no longer holds for that shattered object). Is this correct, or am I missing something deeper that Taylor is getting at?
As for the case when constituents shatter-the particle identity has changed: what went in was one particle $$A$$: whats coming out are multilple particles $$A_i$$. The $$A\to A_i$$ reaction may eat up energy* making the collision inelastic.
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Greatest Common Factor
Greatest Common Factor
Remember that a factor is a number that divides evenly into a given number. A number that is a factor of two or more numbers is a common factor of those numbers. The greatest common factor (GCF) is the largest factor that those numbers have in common.
Listing the factors of each number is one way to find the GCF. Another way to find the GCF uses the prime factorization of each number.
Example: Find the GCF of 24 and 30.
We can list out the factors of each number and find the largest one they have in common.
Factors of 24: {1, 2, 3, 4, 6, 8, 12, 24}
Factors of 30: {1, 2, 3, 5, 6, 10, 15, 30}
We see their common factors are 1, 2, and 6. The GCF is the largest one, which is 6.
To find the GCF of larger numbers, you can write out the prime factorization of each number. Next, identify the prime factors that each number has in common. The GCF is the product of these common factors.
Using our example once again:
24: 2 x 12
2 x 2 x 6
2 x 2 x 2 x 3
30: 2 x 15
2 x 3 x 5
So the prime factorizations are
24: 2 x 2 x 2 x 3
30: 2 x 3 x 5
We see that both factorizations have a 2 and a 3, so the GCF of 24 and 30 is 2 x 3 = 6.
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# Jordan Matrix
(redirected from Canonical box matrix)
## Jordan matrix
[zhȯr′dän‚mā·triks]
(mathematics)
A matrix whose elements are equal and nonzero on the principal diagonal, equal to 1 on the diagonal immediately above, and equal to 0 everywhere else.
## Jordan Matrix
(also Jordan form of a matrix, or simple classical matrix). With every square matrix A = ǀǀaikǀǀn1 there is associated a class of similar matrices. This class of matrices always includes a matrix of special normal (canonical) Jordan form, named after M. E. C. Jordan. The Jordan form of a certain eighth-order matrix is given in Figure 1.
Figure 1
Special square blocks, enclosed by broken lines in Figure 1, lie on the principal diagonal. All elements of the matrix that lie outside these blocks are equal to zero. The (complex) numbers forming the principal diagonal of a block are the same; for example, in the first block they are all equal to λ1, and in the second to λ2. The elements of the diagonal immediately above the principal diagonal equal to unity. All other elements in the diagonal blocks are equal to zero. The matrix in Figure 1 contains three diagonal blocks, of which the first is of order 4, and the second and third of order 2. In the general case, the number of blocks and their orders are arbitrary. It is also possible for some of the numbers λ1, λ2, … to be equal. The initial matrix A in this example has the elementary divisors (λ – λ1)4, (λ – λ2)2, and (λ – λ3)2. A Jordan matrix is uniquely determined by its elementary divisors.
If the matrix A has the Jordan form I, then there exists a nonsingular matrix T such that A = TIT–1. Replacement of matrix A by the similar Jordan matrix I is called a reduction of matrix A to normal Jordan form.
To get an idea of the uses of the Jordan form of a matrix, we consider a system of linear differential equations with constant coefficients:
dx1/dt = a11x1 + a12x2 + … + a1nxn
dx2/dt = a21x1 + a22x2 + … + a2nxn
dxn/dt = an1x1 + an2x2 + … + annxn
In matrix notation we have
(1) dx/dt = Ax
Let us introduce new unknown functions y1, y2, …, yn by using a nonsingular matrix T = ǀǀtikǀǀn1 where the tik are numbers (i, k = 1, 2, …, n):
x1 = t11y1 + t12y2 + … + t1nyn
x1 = t21y1 + t22y2 + … + t2nyn
xn = tn1y1 + tn2y2 + … + tnnyn
or in matrix notation
x = Ty
Substituting this expression for x in (1), we obtain
(2) dy/dt = Iy
where matrix I is related to matrix A by the equation
A = TIT–1
The matrix T is usually selected in such a way that matrix A is a Jordan matrix. In this case, the system of equations (2) is much simpler than system (1). For example, if matrix A = ǀǀaikǀǀ81 (n = 8) has the Jordan form given in Figure 1, then system (2) will have the form
dy1/dt = λ1y1 + y2dy5/dt = λ2y5 + y6
dy2/dt = λ1y2 + y3dy6/dt = λ2y6
dy3/dt = λ1y3 + y4dy7/dt = λ3y7 + y8
dy4/dt = λ1y4dy8/dt = λ3y8
Integration of this system reduces to a number of integrations of single differential equations.
### REFERENCES
See references under .
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# Mental And Emotional Health Lesson Plan
According to the concept of "biorhythm," there is a physical cycle of 23 days, a mental cycle of 28 days and an intellectual cycle of 33 days. Wilhelm Fliess established the theory that all living beings, consisting of human beings, go through a series of rhythmic cycles in the Calculation of Biorhythms" in the 19th century. Biorhythm The biorhythm is the regular change of physical and psychological states in a person.
## Biorhythm Calculator
If you experience your life uninformed of your biorhythms the opportunities are you won't completely recognize your moods or why points take place the way they do. It's likely your brain will develop a mental narrative for the feeling or occurrence that exists outside of real influence. A biorhythm calculator can help. An excellent biorhythm calculator will take a note of your birthday and today's date. It will after that create a map of your power centers using information connecting to your physical, emotional, and psychological attributes. Your biorhythm curve will certainly reveal you today problem of your life pressure power, allowing your to make better options.
## How Are Biorhythms Calculated?
Analysts state that between the criticality cycle (day 1) and also the level of sensitivity cycle (day 14) there is a duration of 24 hrs focused in between day 11 (physical cycle) and also day 16 (intellectual cycle). The 23-day cycle is typically referred to as the male cycle as well as the 28-day cycle is usually referred to as the female cycle because both males and women are influenced by the biorhythms. When we checked out the biorhythm chart, we look for ups as well as downs in the biorhythm cycle, and also adverse crucial days (0 days) tell an fascinating tale. If the set value is as well big compared to the moment interval of the residential or commercial property ( as an example, 3 days vs. 1 minute), there is no adverse result on the performance, readability or security of programs that calculate as well as display more biorhythm information. If the collection worths are also small contrasted to the property, 1 minute vs. 3 days, there is a negative impact on the performance and also readability/ stability of the program, which computes as well as plays more "biorhythm" information elements. The program can calculate personal biorhythm predictions for selected data and also individuals with a easy interface, thus figuring out unsafe days and months and carrying out biorhythm estimations with high accuracy. It is likewise possible to calculate Biorhythm for other individuals as well as you can include as several individuals as you such as. According to criminalistic research on biorhythm, some individuals plead guilty when they have a critical day. During a period of three or even more biorhythms that at the same time surpass the zero point, you need to be mindful and mindful and avoid demanding intellectual activities. The theory of biorhythm insists that one's very own life is affected by a balanced organic cycle, and also tries to make predictions regarding the cycle as well as individual ease in performing the associated tasks. Biorhythm concept is a pseudoscientific concept that our daily life is affected by rhythmic cycles over a period of 23, 28 or 33 days (the 23-day physical cycle, the 28-day psychological cycle as well as the 33-day intellectual cycle). You will obtain a total representation of your biorhythm for this cycle or get in touch with the representations for all 3 cycles which are indicated in various shades. Experts state that between the urgency cycle (day 1) as well as the sensitivity cycle (day 14) there is a duration of 24 hours centered in between day 11 (physical cycle) as well as day 16 (intellectual cycle). The 23-day cycle is commonly referred to as the male cycle as well as the 28-day cycle is typically referred to as the female cycle due to the fact that both guys and women are affected by the biorhythms.
## What Is A Biorhythm
The reason why human beings think, act and really feel is due to the fact that we are made from life pressure energy that participates in our bodies before we are born. This life force power inhabits paths in our bodies and discovers centres or centers understood as chakras. There are thousands of chairs in the body, not simply the seven commonly illustrated in publications and in website. Each of these chakras has its very own regularity that engages with the planet as well as the universe. The energy stamina of the chakras differs depending on the activity of the worlds and also different external stimulations. The variant in the chakra power is referred to as your biorhythm. party
## The Biorhythm
Biorhythm Theory services the property that our day-to-days live and futures are affected in large component by the balanced cycles that match to our physical, emotional, as well as intellectual health and wellbeing and direction. According to biorhythm theory a physical cycle is 23 days long, an emotional one is 28 days long, and also an intellectual one is 33 days long. This concept was originally created by Wilhelm Fliess in the late 19th century. In the United States in the 1970s it grew in appeal. Today, biorhythm concept is once more undergoing a renaissance as more and also extra adherents are discovering its individual advantages as well as finding success in their lives. Mental And Emotional Health Lesson Plan
## Biorhythm And Sex
The relationship between a 23-day physical cycle and a 28-day emotional-biorhythmic cycle and free toss precision. Our emotional-borihythmic cycle lasts 28 days, while our intellectual-borihythmic cycle lasts 33 days. Huffman, R. (1977) The relationship between biorhythm cycles and performance in sports throughout a day of competition. The theory of biorhythm asserts that one’s own life is influenced by rhythmic biological cycles, and attempts to make forecasts about these cycles and individual ease in performing the associated jobs. The psychological cycle is considered an idea and reasoning cycle and manages intelligence, logic, mental responses, awareness, sense of orientation, decision-making, judgement, deductivity, memory and ambition.
## Biorhythm Compatibility
When it pertains to love you’ve heard all the cliches. You’ve become aware of true love, concerning people that were ‘indicated’ to be with each other, after that you’ve additionally read about people who are simply incompatible. What does it actually mean? The solution lies in your life pressure energy and your biorhythms. Working with an additional human can be varied. Because you aren’t suitable today doesn’t indicate you will not be extra compatible following week, just. This leaves a lot to possibility, specifically if you go on a date. Your day can be perfect to bit your biorhythms out of positioning. Why not provide it time and utilize a calculator.
## Biorhythm And Health
In one study, there was a considerable association between the portion of hazardous behavior in the psychological cycle and the basic cycle of biorhythms with at least one important day in each cycle, which suggests that the value of the average unsafe behavior on a crucial day is (i.e. (1978) A research study on the influence of three biorhythm cycle on the suggestivity of behavior and self-reported feelings by Garnett, R. E. McKissick, T. L. (1979) Study on the impacts of biorhythm cycles on psychological and physical performance. The study showed that risky driving behavior was associated with the BioRhythm analyzed by the BioRhythm software application, which revealed an association in between risky driving behavior and tracking the vital days of their BioRhythmic cycle. In one research study, there was a substantial association between the portion of hazardous behavior in the emotional cycle and the basic cycle of biorhythms with at least one vital day in each cycle, which indicates that the value of the typical hazardous habits on an important day is (i.e. (1978) A study on the impact of 3 biorhythm cycle on the suggestivity of habits and self-reported sensations by Garnett, R. E. McKissick, T. L. (1979) Study on the effects of biorhythm cycles on physical and mental efficiency. The study showed that unsafe driving habits was associated with the BioRhythm examined by the BioRhythm software application, which showed an association in between hazardous driving habits and tracking the important days of their BioRhythmic cycle. Mental And Emotional Health Lesson Plan
## Biorhythm Chart
Biorhythms are a complicated matter that needs a high degree of ability and also expertise to obtain. That’s why you require a biorhythm calculator and also a biorhythm chart. These devices are specially developed to draw up your present biorhythm frequencies to give you the information you need to make even more productive life options. When you consult your biorhythm chart you will certainly see 3 lines, one is for the physical power, one is for the emotional power, and also one is for your intellectual energy. These lines go up as well as down in waves– it’s a wrong wave– when the lines pass the no factor or weave you need to pay attention. Mental And Emotional Health Lesson Plan
## How To Read A Biorhythm Chart
If you desire your biorhythm calculator and also chart to operate in your favor you require to know exactly how to utilize them. There’s no point in investing in a product you can not read, that’s why we provide you all the info required to recognize the significance of your biorhythm information. The initial point to do is find a dependable biorhythm calculator and also input your data. This is normally a set of dates. After the calculator has actually given you your information it’s time to check out the graph and translate the results. The chart utilizes a line chart layout with coloured lines to show each regularity. Review where these lines converge.
## Intuition Biorhythm
The climax of a physical cycle is an outstanding time to use up a difficulty or set out. The low point of the physical cycle suggests that you have actually spent too long and require to take it simple for a while. Highlights in the emotional cycle show that you are strong, that your outlook of life tends to be finest and that your interactions with others achieve success. The highlights of the cycle are when your intuition is at its greatest and you experience minutes of satisfaction. The biorhythm manifests itself in the physical, psychological and intellectual cycles of our lives. A biorhythm is a crucial point in a private cycle that alternates between low and high periods, and vice versa. As your biorhythm cycle increases and decreases, your capability to perform particular tasks, perform exercise, tension and make reasonable decisions increases. The middle and end point of a biorhythm cycle is when the cycle moves from absolutely no to a point of polarity change. The modification of the day takes place in the middle or at the end of the cycle, and scientists call it half of the periodic day or routine day change point. There have actually been substantial studies and computations in biorhythmic research to identify the precise period of these cycles. Although biorhythms claim to be precise, unchanged durations of body clocks have been observed in the cycle itself, and the durations discovered vary in length due to biological and ecological factors. Physical biorhythms (23-day cycles) are connected with physical strength, endurance, resistance, endurance and courage. Psychological biorhythm (28-day cycle) is associated to emotional stability, sensations of instinct, imagination, receptivity and mood. Studies have actually revealed that risky driving behavior associates with a chauffeur’s biorhythm, and analysis of this information with software application has actually shown a link between unsafe driving behavior and tracking the important days of their bioirrhythmic cycle. Now that we understand the various cycles and their length and results, it is essential to understand how to track and determine one’s own biorhythm. This calculator comprises six cycles and periods of the biorhythm. Calculatornet uses the whole spectrum with a basic two-step method for entering the date of birth and span of the biohythmic diagrams. According to the theory of biorhythm, a person’s life is influenced by a balanced biological cycle that affects his capabilities in various areas such as psychological, psychological and physical activity. The cycle starts at birth and vibrates in an uniform sinusoidal wave form throughout life, and its modelling recommends that an individual’s efficiency level in these locations can be anticipated from day to day. These attributes are determined by sinusoidal curves whose periodicity depends upon the characteristic: the physical cycle is 23 days, the psychological cycle 28 days and the longest intellectual cycle 33 days. The psychological cycle quantifies our imagination, sensitivity, mood and awareness. The visual cycle (43 days) explains the interest in a unified and gorgeous confidence, the visual cycle (48 days) expresses the capability to perceive one’s character and individuality, and the spiritual cycle (53 days) describes inner stability and unwinded posture. Twenty-three days is the length of the physical cycle, which promotes more awareness, time for optimum effort and time for rest and recuperation. We wish to assume that the two greatest biorhythms (23 days and 28 days) represent a kind of improvement or prolongation of the larger Sun-Moon cycle. The middle and end point of a biorhythm cycle is when the cycle moves from no to a point of polarity modification. These qualities are identified by sinusoidal curves whose periodicity depends on the characteristic: the physical cycle is 23 days, the psychological cycle 28 days and the longest intellectual cycle 33 days. The visual cycle (43 days) describes the interest in a beautiful and unified confidence, the aesthetic cycle (48 days) reveals the ability to view one’s personality and individuality, and the spiritual cycle (53 days) describes inner stability and relaxed posture. We would like to presume that the 2 strongest biorhythms (23 days and 28 days) represent a kind of change or prolongation of the bigger Sun-Moon cycle.
## Biorhythm Critical Days
We often get lost in abstract theory and philosophical concepts. But there are practical aspects of biorhythm manifestation technique. According to scientists, researchers, and statisticians, Biorhythms is a specific life path attached to you since birth and as unique to you as your fingerprint. Those who are naturally in harmony with their biorhythms are the one who go on to effortlessly achieve great success in all aspects of life. Those who don’t, experience hurdles and pitfalls. Mental And Emotional Health Lesson Plan
HEALTH:
We reveal a person’s physical rhythm, and this accurately shows someone the best and worst days for exercising. The Biorhythm shows you the days you can really push the limits of your workout and the days you may struggle or should avoid working out altogether. Mental And Emotional Health Lesson Plan
SURVIVAL:
Everyone’s Biorhythm reading will include a few dangerous days each day we call “Critical Days”. These are days you’re most likely to experience physical harm and knowing these days ahead of time could actually be lifesaving.
SELF-HELP:
This offer has truly transformed people’s lives and we’ve even heard we’ve helped prevent potential suicides it’s so powerful. Everyone goes through life pretty blindly with no idea what tomorrow will bring. We help reveal what your days ahead may look like for the first time. When you know what’s coming, you can properly prepare yourself. When you’ve wandered off your lifepath… This is a roadmap with a shortcut to finding your way back. Mental And Emotional Health Lesson Plan
WEALTH:
Go to any casino and you’ll spot that one little old lady who means business… She’s not there to play, she’s there to WIN and win she does! Odds are she knows something no one else in the casino does… Her biorythm! We’ve all heard the expressions “You’re on a lucky streak” and “your lucks run out” Well, this actually helps show people the days they’ll experience a lucky streak and the days they should avoided any gambling. Members love this for casino’s, sports betting, and lottery tickets!
LOVE & DATING:
Will I ever find my soulmate? Is the person I’m with the right one for me? Is my celebrity crush compatible with me? These are a few questions members all have answered when they join and make this an easy sell to the dating and relationship markets. This is the only tool that understands compatibility is a flowing and changing thing.
SPIRITUALITY:
This one is the most obvious and why I saved the best for last! Personal development, law of attraction, manifestation, alternative beliefs, numerology, astrology and so on are our bread and butter for this offer. It converts like crazy because this is something brand new in the space and compliments each of these elements without having to compete with them! party
## The Biorhythm Calculator: Romance Biorhythm And Passion Biorhythm
By default, the Biorhythm Calculator diagram shows 4 primary biorhythms. Biorhythm evaluation in the The Biorhythm app, nevertheless, includes a holistic aspect of your life in terms of physical health, mental health and relationships. In addition to the three main (physical, psychological and psychological) biorhythm waves, there is an extra series of secondary (biorhythm) cycles. By default, the Biorhythm Calculator diagram displays 4 primary biorhythms. Biorhythm review in the The Biorhythm app, however, includes a holistic aspect of your life in terms of physical health, mental health and relationships. No one leaves out the emotional cycle and focuses on the intellectual and physical facets of your biorhythm. In addition to the three primary (physical, mental and emotional) biorhythm waves, there is an additional series of secondary (biorhythm) cycles. The secondary biorhythm cycles are calculated as the average of the two cycles of the first stage.
## The Biorhythm: Biorhythm Luck Chart And Biorhythmic Cycles
Experts say that between the criticality cycle (day 1) and the sensitivity cycle (day 14) there is a period of 24 hours focused in between day 11 (physical cycle) and day 16 (intellectual cycle). These cycles begin at birth and vibrate in a constant sine wave throughout life, and modelling these cycles suggests that an individual's performance level in these areas can be forecasted day by day. The 23-day cycle is typically referred to as the male cycle and the 28-day cycle is often referred to as the female cycle, given that both women and guys are affected by the biorhythm.
## What Is My Biorhythm
As we have laid out over your biorhythm is the regularity of your life force energy located around the factor of energy in your body. These points of power are often called chakras. These chakras are leading as well as effective pressures in your life. They are not just accountable for your health yet likewise for your top quality of life as well as future course. You can review the biorhythmic information that arises from these energy centers making use of one of our biorhythmic calculators and also charts. The information you acquire from these evaluations can aid you make life selections that are extra in tune with truth nature of fact as well as the means things really are. Biorhythms describe the forecast of habits and experience with physical, intellectual and also emotional cycles. There are three classic cycles of biorhythm: endogenous, infradian and balanced . It was ended that the high, low as well as critical settings of the twenty-three day physical, twenty-eight-day psychological, and thirty-thirty-five day intellectual biorhythmic cycles did not impact the academic analysis efficiency of trainees with titles as well as solutions . The theory of biorhythm insists that a person's own life is influenced by a balanced biological cycle, and also tries to make predictions concerning the cycle and also personal convenience in doing the relevant tasks. Fans of the biorhythm believe that the cycle, which impacts all 3 arms, impacts all physical, emotional as well as intellectual abilities. Physical biorhythm is a 23-day cycle pertaining to physical strength, durability, strength, endurance as well as courage. These as well as other discoveries show that individuals can be affected by physiological, emotional, and intellectual rhythms, however the exact connection in between biorhythm cycles is still not completely understood.
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# What is the improved quadratic formula to solve quadratic equations ?
Jan 1, 2018
There is only one quadratic formula, that is $x = \frac{- b \pm \sqrt{{b}^{2} - 4 a c}}{2 a}$.
#### Explanation:
For a general solution of $x$ in $a {x}^{2} + b x + c = 0$, we can derive the quadratic formula $x = \frac{- b \pm \sqrt{{b}^{2} - 4 a c}}{2 a}$.
$a {x}^{2} + b x + c = 0$
$a {x}^{2} + b x = - c$
$4 {a}^{2} {x}^{2} + 4 a b x = - 4 a c$
$4 {a}^{2} {x}^{2} + 4 a b x + {b}^{2} = {b}^{2} - 4 a c$
Now, you can factorize.
${\left(2 a x + b\right)}^{2} = {b}^{2} - 4 a c$
$2 a x + b = \pm \sqrt{{b}^{2} - 4 a c}$
$2 a x = - b \pm \sqrt{{b}^{2} - 4 a c}$
$\therefore x = \frac{- b \pm \sqrt{{b}^{2} - 4 a c}}{2 a}$
Jan 3, 2018
This could refer to...
#### Explanation:
One of the nuisances when using the quadratic formula is that often the square root can be simplified, involving at least one more step than necessary. If the middle coefficient is even, then we can avoid this by using an alternative formulation of the quadratic formula.
Given:
$a {x}^{2} + 2 \mathrm{dx} + c = 0$
The roots are given by the formula:
$x = - \frac{d}{a} \pm \frac{\sqrt{{d}^{2} - a c}}{a}$
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# Tagged Questions
119 views
### Question about ratios and combinatorics
In this question that I posted yesterday (11/15): I am solving a programming puzzle that consists of finding all the possible ways to build a brick wall of $48$" $\times$ $10$" (width $\times$ height ...
112 views
### How many ways to reach $1$ from $n$ by doing $/13$ or $-7$?
How many ways to reach $1$ from $n$ by doing $/13$ or $-7$ ? (i.e., where $n$ is the starting value (positive integer) and $/13$ means division by $13$ and $-7$ means subtracting 7)? Let the number ...
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### Counting $2010^2$-tuples
Here's a problem i invented myself, but i'm not sure about my solution. I'll show it later, so people can enjoy trying to find one: Consider the function $f:\mathbb{R}^2\to\{1,2,...,2012\}$ that ...
### Bijective functions $f(n)=f(f(n-1))=f^n(1)$
How can I find a function (or more) which satisfy $f(n)=f(f(n-1))=f^n(1)$, defined for positive integers n, such that it is a bijection between the positive integers? And which satisfy i) For every ...
I have a sequence of functions $f_k(n)$ defined as follows: $f_1(n)=n^{n-2}$ $f_k(n)=\sum_{i=1}^{n-1}f_{k-1}(i)\cdot(n-i)^{n-i-2}\cdot{n-k \choose n-i-1}$ My goal is to find and prove a closed-form ...
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## Significant figures [ENDORSED]
Moderators: Chem_Mod, Chem_Admin
Alondra Cervantes 1F
Posts: 8
Joined: Fri Sep 25, 2015 3:00 am
### Significant figures
Can someone please further explain how many significant figures to use when writing an answer. I am still a bit confused.
Chem_Mod
Posts: 18400
Joined: Thu Aug 04, 2011 1:53 pm
Has upvoted: 435 times
### Re: Significant figures [ENDORSED]
plz refer to the following doc to check what you need to know about sigfig
https://lavelle.chem.ucla.edu/wp-content/supporting-files/Chem14A/EVERYTHING_YOU_WANTED_TO_KNOW_ABOUT_SF.pdf
Alondra Loera 1A
Posts: 20
Joined: Fri Sep 25, 2015 3:00 am
### Re: Significant figures
Significant figures are the number of digits in an answer relative to the numbers in the problem.
11.2 has 3 significant figures
11.20 has 4 significant figures
0.011 has 2 significant figures
with decimals, the significant figures are counted after the first integer so for 0.011, 1 and anything after counts as a sig. fig.
This Khan Academy video is very helpful as well https://www.khanacademy.org/math/pre-algebra/decimals-pre-alg/sig-figs-pre-alg/v/significant-figures
Sara Juarez 1B
Posts: 8
Joined: Tue Nov 24, 2015 3:00 am
### Re: Significant figures
The number of significant figures will depend on the problem. If I am not mistaken, I believe you look to your original problem and look for the number that has the smallest significant figure and go off of that. Significant figures don't take into account leading 0's. So 0.1 and 0.000001 are both 1 sig fig. However trailing 0's are significant. 0.0102 is 3 sign figures because the first non zero number starts with 1 and than it is followed by two additional numbers.
EmmaSkeie 1F
Posts: 2
Joined: Fri Jun 17, 2016 11:28 am
### Re: Significant figures
Hi,
For quiz 1: If we are given a problem that has 3 different values, each of different number of significant figures (say the least number of SF is 2), do we round off the other numbers to 2 SF for every step, or do we keep the values given and then round off the answer to SF at the very end?
Thank you!
Chem_Mod
Posts: 18400
Joined: Thu Aug 04, 2011 1:53 pm
Has upvoted: 435 times
### Re: Significant figures
The rules for rounding are also described in the link of the endorsed post.
Rob_Carter_1L
Posts: 10
Joined: Wed Sep 21, 2016 2:59 pm
### Trailing zeroes
Just confirming, I know that 0.0100 would have 3 sigfigs due to trailing zeros, but would 1.00 have 3 as well? Is it the same after decimal points?
Kathy Vu 3L
Posts: 24
Joined: Fri Jul 15, 2016 3:00 am
### Re: Trailing zeroes
Rob_Carter_1L wrote:Just confirming, I know that 0.0100 would have 3 sigfigs due to trailing zeros, but would 1.00 have 3 as well? Is it the same after decimal points?
Yes, 1.00 would have 3 significant figures as well since the 2 zeros are trailing zeros behind a decimal point
Kathy Vu 3L
Posts: 24
Joined: Fri Jul 15, 2016 3:00 am
### Re: Significant figures
Is there usually a range of answers we can put on quizzes and exams, or is there only 1 answer that is allowed?
For example what if my answer is 2.5, but others got 2.6 due to a different amount of decimals used during calculations
Chem_Mod
Posts: 18400
Joined: Thu Aug 04, 2011 1:53 pm
Has upvoted: 435 times
### Re: Significant figures
I provide all students with the same constants, equations, and periodic table.
Therefore all students should have the same answer if sig-fig and round-off are done correctly.
Return to “Significant Figures”
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• Research Article
• Open access
• Published:
# A New Conservative Difference Scheme for the General Rosenau-RLW Equation
## Abstract
A new conservative finite difference scheme is presented for an initial-boundary value problem of the general Rosenau-RLW equation. Existence of its difference solutions are proved by Brouwer fixed point theorem. It is proved by the discrete energy method that the scheme is uniquely solvable, unconditionally stable, and second-order convergent. Numerical examples show the efficiency of the scheme.
## 1. Introduction
In this paper, we consider the following initial-boundary value problem of the general Rosenau-RLW equation:
(11)
with an initial condition
(12)
and boundary conditions
(13)
where is a integer and is a known smooth function. When , (1.1) is called as usual Rosenau-RLW equation. When , (1.1) is called as modified Rosenau-RLW (MRosenau-RLW) equation. The initial boundary value problem (1.1)–(1.3) possesses the following conservative quantities:
(14)
(15)
It is known the conservative scheme is better than the nonconservative ones. Zhang et al. [1] point out that the nonconservative scheme may easily show nonlinear blow up. In [2] Li and Vu-Quoc said " in some areas, the ability to preserve some invariant properties of the original differential equation is a criterion to judge the success of a numerical simulation". In [3–11], some conservative finite difference schemes were used for a system of the generalized nonlinear Schrödinger equations, Regularized long wave (RLW) equations, Sine-Gordon equation, Klein-Gordon equation, Zakharov equations, Rosenau equation, respectively. Numerical results of all the schemes are very good. Hence, we propose a new conservative difference scheme for the general Rosenau-RLW equation, which simulates conservative laws (1.4) and (1.5) at the same time. The outline of the paper is as follows. In Section 2, a nonlinear difference scheme is proposed and corresponding convergence and stability of the scheme are proved. In Section 3, some numerical experiments are shown.
## 2. A Nonlinear-Implicit Conservative Scheme
In this section, we propose a nonlinear-implicit conservative scheme for the initial-boundary value problem (1.1)–(1.3) and give its numerical analysis.
### 2.1. The Nonlinear-Implicit Scheme and Its Conservative Law
For convenience, we introduce the following notations
(21)
where and denote the spatial and temporal mesh sizes, , , respectively,
(22)
and in the paper, denotes a general positive constant, which may have different values in different occurrences.
Since , then the finite difference scheme for the problem (1.1)–(1.3) is written as follows:
(23)
(24)
(25)
Lemma 2.1 (see [12]).
For any two mesh functions, , one has
(26)
Furthermore, if , then
(27)
Theorem 2.2.
Suppose that , then scheme (2.3)–(2.5) is conservative in the senses:
(28)
(29)
Proof.
Multiplying (2.3) with , according to boundary condition (2.5), and then summing up for from 1 to , we have
(210)
Let
(211)
Then (2.8) is gotten from (2.10).
Computing the inner product of (2.3) with , according to boundary condition (2.5) and Lemma 2.1, we obtain
(212)
where
(213)
According to
(214)
we have . It follows from (2.12) that
(215)
Let
(216)
Then (2.9) is gotten from (2.15). This completes the proof of Theorem 2.2.
### 2.2. Existence and Prior Estimates of Difference Solution
To show the existence of the approximations for scheme (2.3)–(2.5), we introduce the following Brouwer fixed point theorem [13].
Lemma 2.3.
Let be a finite-dimensional inner product space, be the associated norm, and be continuous. Assume, moreover, that there exist , for all , , . Then, there exists a such that and .
Let , , then have the following.
Theorem 2.4.
There exists which satisfies scheme (2.3)–(2.5).
Proof.
It follows from the original problem (1.1)–(1.3) that satisfies scheme (2.3)–(2.5). Assume there exists which satisfy scheme (2.3)–(2.5), as , now we try to prove that , satisfy scheme (2.3)–(2.5).
We define on as follows:
(217)
where . Computing the inner product of (2.17) with and considering and , we obtain
(218)
Hence, for all , there exists . It follows from Lemma 2.3 that exists which satisfies . Let , then it can be proved that is the solution of scheme (2.3)–(2.5). This completes the proof of Theorem 2.4.
Next we will give some priori estimates of difference solutions. First the following two lemmas [14] are introduced:
Lemma 2.5 (discrete Sobolev's estimate).
For any discrete function on the finite interval , there is the inequality
(219)
where are two constants independent of and step length .
Lemma 2.6 (discrete Gronwall's inequality).
Suppose that the discrete function satisfies the inequality
(220)
where and are nonnegative constants. Then
(221)
where is sufficiently small, such that .
Theorem 2.7.
Suppose that , then the following inequalities
(222)
hold.
Proof.
It is follows from (2.9) that
(223)
According to Lemma 2.5, we obtain
(224)
This completes the proof of Theorem 2.7.
Remark 2.8.
Theorem 2.7 implies that scheme (2.3)–(2.5) is unconditionally stable.
### 2.3. Convergence and Uniqueness of Difference Solution
First, we consider the convergence of scheme (2.3)–(2.5). We define the truncation error as follows:
(225)
then from Taylor's expansion, we obtain the following.
Theorem 2.9.
Suppose that and , then the truncation errors of scheme (2.3)–(2.5) satisfy
(226)
as ,
Theorem 2.10.
Suppose that the conditions of Theorem 2.9 are satisfied, then the solution of scheme (2.3)–(2.5) converges to the solution of problem (1.1)–(1.3) with order in the norm.
Proof.
Subtracting (2.3) from (2.25) letting
(227)
we obtain
(228)
Computing the inner product of (2.28) with , we obtain
(229)
From the conservative property (1.5), it can be proved by Lemma 2.5 that . Then by Theorem 2.7 we can estimate (2.29) as follows:
(230)
According to the following inequality [11]
(231)
Substituting (2.30)–(2.31) into (2.29), we obtain
(232)
Let
(233)
then (2.32) can be rewritten as
(234)
Choosing suitable which is small enough, we obtain by Lemma 2.6 that
(235)
From the discrete initial conditions, we know that is of second-order accuracy, then
(236)
Then we have
(237)
It follows from Lemma 2.5, we have . This completes the proof of Theorem 2.10.
Theorem 2.11.
Scheme (2.3)–(2.5) is uniquely solvable.
Proof.
Assume that and both satisfy scheme (2.3)–(2.5), let , we obtain
(238)
Similarly to the proof of Theorem 2.10, we have
(239)
This completes the proof of Theorem 2.11.
Remark 2.12.
All results above in this paper are correct for initial-boundary value problem of the general Rosenau-RLW equation with finite or infinite boundary.
## 3. Numerical Experiments
In order to test the correction of the numerical analysis in this paper, we consider the following initial-boundary value problems of the general Rosenau-RLW equation:
(31)
with an initial condition
(32)
and boundary conditions
(33)
where . Then the exact solution of the initial value problem (3.1)-(3.2) is
(34)
where is wave velocity.
It follows from (3.4) that the initial-boundary value problem (3.1)–(3.3) is consistent to the boundary value problem (3.3) for . In the following examples, we always choose , .
Tables 1, 2, and 3 give the errors in the sense of -norm and -norm of the numerical solutions under various steps of and at for and . The three tables verify the second-order convergence and good stability of the numerical solutions. Tables 4, 5, and 6 shows the conservative law of discrete mass and discrete energy computed by scheme (2.3)–(2.5) for and .
Figures 1, 2, and 3 plot the exact solutions at and the numerical solutions computed by scheme (2.3)–(2.5) with at , which also show the accuracy of scheme (2.3)–(2.5).
## References
1. Zhang F, Pérez-GarcÃa VM, Vázquez L: Numerical simulation of nonlinear Schrödinger systems: a new conservative scheme. Applied Mathematics and Computation 1995,71(2-3):165-177. 10.1016/0096-3003(94)00152-T
2. Li S, Vu-Quoc L: Finite difference calculus invariant structure of a class of algorithms for the nonlinear Klein-Gordon equation. SIAM Journal on Numerical Analysis 1995,32(6):1839-1875. 10.1137/0732083
3. Chang Q, Xu L: A numerical method for a system of generalized nonlinear Schrödinger equations. Journal of Computational Mathematics 1986,4(3):191-199.
4. Chang Q, Jia E, Sun W: Difference schemes for solving the generalized nonlinear Schrödinger equation. Journal of Computational Physics 1999,148(2):397-415. 10.1006/jcph.1998.6120
5. Wang T-C, Zhang L-M: Analysis of some new conservative schemes for nonlinear Schrödinger equation with wave operator. Applied Mathematics and Computation 2006,182(2):1780-1794. 10.1016/j.amc.2006.06.015
6. Wang T, Guo B, Zhang L: New conservative difference schemes for a coupled nonlinear Schrödinger system. Applied Mathematics and Computation 2010,217(4):1604-1619. 10.1016/j.amc.2009.07.040
7. Zhang L: A finite difference scheme for generalized regularized long-wave equation. Applied Mathematics and Computation 2005,168(2):962-972. 10.1016/j.amc.2004.09.027
8. Zhang F, Vázquez L: Two energy conserving numerical schemes for the sine-Gordon equation. Applied Mathematics and Computation 1991,45(1):17-30. 10.1016/0096-3003(91)90087-4
9. Wong YS, Chang Q, Gong L: An initial-boundary value problem of a nonlinear Klein-Gordon equation. Applied Mathematics and Computation 1997,84(1):77-93. 10.1016/S0096-3003(96)00065-3
10. Chang QS, Guo BL, Jiang H: Finite difference method for generalized Zakharov equations. Mathematics of Computation 1995,64(210):537-553. 10.1090/S0025-5718-1995-1284664-5
11. Hu JS, Zheng KL: Two conservative difference schemes for the generalized Rosenau equation. Boundary Value Problems 2010, 2010:-18.
12. Hu B, Xu Y, Hu J: Crank-Nicolson finite difference scheme for the Rosenau-Burgers equation. Applied Mathematics and Computation 2008,204(1):311-316. 10.1016/j.amc.2008.06.051
13. Browder FE: Existence and uniqueness theorems for solutions of nonlinear boundary value problems. In Proceedings of Symposia in Applied Mathematics, vol. 17. American Mathematical Society, Providence, RI, USA; 1965:24-49.
14. Zhou Y: Applications of Discrete Functional Analysis to the Finite Difference Method. International Academic Publishers, Beijing, China; 1991:vi+260.
## Acknowledgments
The authors would like to express their sincere thanks to the referees for their valuable suggestions and comments. This paper is supported by the National Natural Science Foundation of China (nos. 10871117 and 10571110).
## Author information
Authors
### Corresponding author
Correspondence to Jin-Ming Zuo.
## Rights and permissions
Reprints and permissions
Zuo, JM., Zhang, YM., Zhang, TD. et al. A New Conservative Difference Scheme for the General Rosenau-RLW Equation. Bound Value Probl 2010, 516260 (2010). https://doi.org/10.1155/2010/516260
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MEDIAN OF UNGROUPED DATA EXAMPLES
About "Median of ungrouped data examples"
Median of ungrouped data examples :
Median is defined as the middle value of the data when the data is arranged in ascending or descending order.
Find the median of the following :
40, 50, 30, 60, 80, 70
Arrange the given data in ascending order.
30, 40, 50, 60, 70, 80.
Here the number of terms is 6 which is even. So the third and fourth terms are middle terms. The average value of the these terms is the median.
Median = (50 + 60)/2 = 110/2 = 55
• When the number of observations is odd, the middle number is the median.
• When the number of observations is even, the median is the average of the two middle numbers.
Example 1 :
Find the median of the following data.
17, 15, 9, 13, 21, 7, 32
Solution :
Arrange the data in ascending order.
7, 9, 13, 15, 17, 21, 32
The number of observation is 7 which is odd.
Hence the middle value 15 is the median.
Example 2 :
A cricket player has taken the runs 13, 28, 61, 70, 4, 11, 33, 0, 71, 92. Find the median.
Solution :
Arrange the runs in ascending order as 0, 4, 11, 13, 28, 33, 61, 70, 71, 92.
Here n = 10 (even number).
There are two middle values 28 and 33.
Median = Average of the two middle values
= (28 + 33) / 2 = 61/2 = 30.5
Example 3 :
Find the median of the following data.
83, 66, 86, 30, 81
Solution :
Arrange the data in ascending order.
30, 66, 81, 83, 86
The number of observation is 5 which is odd.
Hence the middle value 81 is the median.
Example 4 :
Find the median of the following data.
3, 4, 5, 3, 6, 7, 2.
Solution :
Arrange the data in ascending order.
2, 3, 3, 4, 5, 6, 7
The number of observation is 7 which is odd.
Hence the middle value 4 is the median.
Example 5 :
Find the median of the data
12, 14, 25, 23, 18, 17, 24, 20.
Solution :
Arrange the data in ascending order
12, 14, 17, 18, 20, 23, 24, 25
The number of observation is 8 which is even.
Median is the average of the two middle terms 18 and 20.
Median = (18 + 20) / 2
= 38/2
= 19
Example 6 :
Find the median of the first 5 prime numbers.
Solution :
The first five prime numbers are 2, 3, 5, 7, 11.
The number of observation is 5 which is odd.
Hence the middle value 5 is the median.
Example 7 :
Find the median of the following data.
6, 14, 5, 13, 11, 7, 8
Solution :
Arrange the data in ascending order
5, 6, 7, 8, 11, 13, 14
The number of observation is 7 which is odd.
Since the number of terms is odd, the middle term is the median.
Hence 8 is the median.
Example 8 :
The weight of 7 chocolate bars in grams are
131, 132, 125, 127, 130, 129, 133
Find the median.
Solution :
Arrange the data in ascending order
125, 127, 129, 130, 131, 132, 133
The number of observation is 7 which is odd.
Since the number of terms is odd, the middle term is the median.
Hence 130 is the median.
After having gone through the stuff given above, we hope that the students would have understood "Median of ungrouped data examples".
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# Foundations for Algebra, Year 2
Welcome to the Foundations for Algebra Parent Guide. The purpose of this guide is to assist you should your child need help with homework or the ideas in the course. We believe all students can be successful in mathematics as long as they are willing to work and ask for help when they need it. We encourage you to contact your child’s teacher if your student has additional questions that this guide does not answer.
The guide is written for use with year 2 of the series. It is arranged by the topical bar graph displayed at the beginning of each chapter. These topics are: number sense, algebra and functions, measurement and geometry, statistics, data, and probability, and mathematical reasoning. Each topic is referenced to the chapter in which the major development of the concept occurs. Detailed examples follow a summary of the concept or skill and include complete solutions. The examples are similar to the work your child has done in class. Additional problems, with answers, are provided for your child to try. Read More...
## Resource Pages
[ Open All | Close All ]
Chapter 1: Getting Organized: Data Interpretation
Chapter 2: The Field Trip: Integer Operations and Graphing Equations
Chapter 3: Mystery of the Dual Spinners: Probability and Fractions
Chapter 4: Going Camping: Algebraic Sentences
Chapter 5: The Math Club: Solving Equations
Chapter 6: Ray Sheo's Pro-Portion Ranch: Ratios and Proportions
Chapter 7: The Class Tournament: Division of Fractions, Percents, and Formulas
Chapter 8: Gold Strike at Dried-Up Creek: The Pythagorean Theorem, Surface Area, and Volume
Chapter 9: College Bound: Slopes and Rates of Change
Chapter 10: Modeling Growth: Exponents, Scientific Notation, and Volume
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Complementarity
Going ‘deep’ in order to understand ‘reality.’
You cannot have an X without an ‘other’ X. Technically, one, and only one, other ‘X.’ Meaning, ‘two’ is the correct explanation for (any, every) one.
Complementarity
This explains complementarity. Which explains ‘reality.’ In a very deep way. If you want to ‘go there…’ (It’s all about the diagram.)
Meaning, a human’s ‘universe’ is virtual. At all times. In all places. Because the human is interacting with, and as, a circle. Which is, technically, a ‘virtual’ event. (The circular relationship between ‘virtual’ and ‘real.’) (Again, it’s all about the diagram.)
Meaning, you cannot have a ‘virtual’ existence without a ‘real’ existence because there is, technically, no such thing as ‘virtual’ or ‘real.’ Except for the circular relationship between virtual and real (technically, pi, in general) (a circle).
Technology (Information in General)
This explains technology. Where the ‘zero’ and the ‘one’ are, more correctly, ‘circumference and diameter.’ (If zero, then one, because if circumference, then diameter.) (The explanation for ‘if… then…’)
So, now we’re forced to ‘realize’ you cannot have ‘zero and one’ without ‘zero or one.’
This means, technically, there is no such thing as complementarity. Or identity. And-Or, paradoxically, it explains both complementarity and identity (the virtual persona).
The Constant Norm
Where, from a human’s point of view, 50–50 (or 50–50–50, if you so desire) is the constant and the norm. Where any string of characters reduces and expands to (50–50) a zero and a one.
But, actually, the whole idea of 50–50 is dependent on 100–1 (99–1) (or 100–0) if you decide you want to ‘go there…’
So, technically, you can say whatever you want about anything (any dichotomy) because the word ‘complementarity’ is, actually, an ‘other’ word for ‘identity.’ Meaning, everything is half-true.
Duplicity (Duality) (Redundancy)
Meaning, you cannot have a ‘one’ without a ‘two.’ Which is the basis for number systems. Linguistic systems. Information systems, in general.
This explains why, and how, an asset is a liability (and, always, vice versa). The concept of ‘vice versa’ in general (circularity in general).
And it removes (solves)the whole idea of ‘paradox.’ Confusion, in general. (Chaos in general.)
Negation (Duplication) (Observation)
Where you cannot have X without X. However, you need to have X without X. In order to have ‘X.’
This explains negation. Duplication. And, most important, observation.
And, that is all there is to, what a human labels, ‘reality.’ (It’s all about the diagram.)
Natural Reality
This takes all of us to a new, and more realistic, understanding of what we all experience as ‘reality.’ (Meaning, technically, as you already know, there is no such thing as ‘reality.’) (Just your own ‘interpretation’ of whatever you are observing (hearing, seeing, sensing)).
Explaining, and overcoming, confrontation (conflict and opposition (disagreement and agreement) in general). (Life and death in general).
Conservation of the circle is the core dynamic in Nature. What a human calls ‘reality.’ (Meaning, pi controls, and explains, everything.)
--
--
More from The Circular Theory
Conservation of the circle is the core dynamic in Nature.
Ilexa Yardley
Author, The Circular Theory
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• # HARRISON CUTTER AUDIO
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## [1] HISTORY
### [1] Maxwell* and Rayleigh*
The origins of most electro-mechanical devices rely on ideas inspired either by Maxwell or Rayleigh, so it is interesting to enquire why the first successful electrical disc cutter and the accompanying WE4A pickup were developed in the USA and not Britain.
At the time of his death on 5th of November 1879, Maxwell's equations had been solved for the behaviour of electromagnetic waves in free space. The solution of two outstanding problems remained:-
[1] The generation of electromagnetic waves from accelerated ( ie oscillating ) charges.
[2] The propagation of electromagnetic disturbances (signals) along cables.
*Maxwell:James Clerk Maxwell: A Treatise on Electricity and Magnetism.
First published: 1st February 1873
*Rayleigh:Baron Rayleigh or John William Strutt: The Theory of Sound
First published: 1877
### [2] Fitzgerald and Heaviside
The first and most difficult problem was solved by George Francis FitzGerald of Trinity College Dublin.
"On the Quantity of Energy Transferred to the Ether by a Variable Electric Current" Transactions of the Royal Dulin Society: 19 November 1883.
The second problem was solved by Oliver Heaviside ( 1850 - 1925 ), a highly eccenctric mathematician and engineer of genius working outside the British establishment. He rediscovered a version of the Laplace Transform which came to be named "Heaviside Operational Calculus" His research into propagation along transmission lines would eventually lead to Harrison's cutting head and and its complement - the WE4A. Heaviside's work showed the importance of inductance in the transmission of signals along a line. An increase in inductance increases the characteristic impedance of the line, so that the the resistance of the conductors becomes less significant and the attenuation lower. High voltages with corresponding lower currents are employed in power transmission for the same reason.
Zo = [L/C]1/2.
Zo = characteristic resistance.
L = inductance per unit length.
C = capacity per unit length.
### [3]Bell Labs.
Sir William Henry Preece (1834 - 1913) was first Electrician and Engineer - in - Chief of the British Post Office.
Sir William did not believe in the existence of self inductance.
In 1888 he described it as "just a lot of bugaboo!"
As a consequence, Heaviside's ideas were not applied to transmission problems in Britain, but were forced across the Atlantic where they were taken up by Professor Michael Pupin at Columbia university and then, eventually, by Bell Labs.
Pupin greatly improved American telephony by winding a thin steel ribbon around cable to increase the self inductance. This was found to be very expensive, and, to decrease the costs, an approximation was made by adding lumped inductance along the line at regular intervals. This worked fine up to a critical frequency where the lumped nature would manifest itself and reject all higher frequencies.
This apparent failure at high frequencies proved the germ for progress, not only in telephonic transmission, but also for disc recording.
It suggested to the fertile minds at Bell that repeated or iterative networks could be made flat or transparent up to a certain frequency and above this frequency to sharply reject all others.
Filter theory was born.
From the clue described above, Campbell, Zobel and Carson working at Bell developed classical filter theory. Networks transparent to any prescribed range of frequencies could be designed at will.
Immersed in such a milieu, Maxfield and Harrison realised that this was the way forward in disc recording, since any mechanical system can be expressed as an analogous electrical network.
For instance:-
Force converts to voltage.
Velocity converts to current.
Mass converts to inductance.
Compliance or springiness converts to capacitance.
Converting the other way, there does not seem to be any mechanical equivalent for mutual inductance.
The route to the Western Electric cutter and the WE4A described above is marked by two classical papers:
(1) " Wave Propagation over Non-Uniform Cables " by M.I.Pupin Trans AIEE,Vol. 17, 1900
(2) " Theory and Design of Uniform and Composite Wave Filters " by Otto J. Zobel B.S.T.J. Vol 2 No1 pp 1-46 1923
Norton, the Norton of Norton's Theorem, had the desk next to Harrison at Bell and his influence on the development of network theory and of electro - mechanical equivalents should never be underestimated.
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## [2] DESIGN
### [1]Design Philosophy
BRIEF DESCRIPTION of THE WE4A
============================
In early pickups the needle usually was coupled to a large iron armature of great mass.
The designer of the WE4A realised that the required signal was already present in the deflection of the centering spring.
All that had to be done was to read this deflection in some way.
In the WE4A the centering spring is made into a circular steel plate - reinforced in the centre so that it bends anti-symmetrically.
As the needle deflects to the right, the bottom of the plate deflects inwards and the top outwards.
The plate is made part of the wall of a sealed enclosure filled with oil.
The plate forms part of a magnetic circuit, so the air gap at the top is increased and the gap at the bottom decreased.
This means that the magnetic flux is diverted either into the top or bottom coil, and so induces a voltage into these coils.
The symmetry reduces even order distortion.
The total flux around the magnetic circuit is constant.
Only the iron under the two coils experiences a change in flux.
The pickup, then, is a true variable reluctance system.
It has a constant velocity response.
The plate forms part of a sealed enclosure filled with oil.
This has two functions:-
(1) The oil against the plate produces mechanical damping.
(2) The oil covering the windings protects them.
For the time of its design (1925 - 1926), the pickup was a lightweight.
The needle point weight on mine is 79.9 gms. Since it uses no rubber in its design, the characteristics remain constant over great lengths of time.
### [2]Equivalent Circuit
#### WE4A Approximate Lumped Equivalent Circuit
Note: Because the WE4A is a magnetic transducer, the following electro-mechanical equivalents have been adopted:-
Velocity (v) --> Voltage (V)
Force (F) -----> Current (I)
*Compliance (U) --> Inductance (L)
Mass (M) --> Capacity C
**Viscous Force Constant (J ) --> Resistance (R)
*Compliance = ( Elastic Deflection )/( Force ) = 1/S where S is the spring constant.
**Viscous Force Constant (J) = ( velocity)/(viscous force)
It is assumed that the pickup is tracking the disc, so that the needle tip velocity (Voltage) is determined entirely by the disc. It can therefore be represented as an independent voltage generator.
APPROXIMATIONS IN DERIVING THE LUMPED EQUIVALENT
The lumped equivalent circuit can, at best, give only a rough approximation to the performance of the WE4A, but it does explain the general behaviour and the dependence of the response on the compliance of the steel needle.
The following complications arise in the development of the equivalent circuit:-
[A] Oil must flow in and out of the magnetic gap, so the viscous damping has a capacitive as well as a resistive component.
[B] The viscous damping is distributed along the retaining spring.
[C] The retaining spring, in the form of a circular plate, probably has significant distributed mass.
[D] The needle clamping screw acts as a mass supported by a short cantilever, and so should be represented as a series tuned circuit across the main moment of inertia Cnh. In the equivalent this is lumped in with the moment of inertia of the needle holder.
SIMPLIFICATION OF THE LUMPED EQUIVALENT
[A] Ln and Lcs can be regarded as an inductive voltage divider to give a Thevenin equivalent input voltage generator of Vin[ Lcs/( Ln + Lcs ) ]
[B] The equivalent output inductance, Lout, of this generator is Ln//Lcs
ie: Lout = (Ln x Lcs)/( Ln + Lcs )
[C] The WE4A equivalent then reduces to the simple resonant system shown in the above diagram.
Ln represents the compliance of the steel needle.
This can be regarded as a cantilever of circular cross section for which the compliance U is given by:-
U = K l3/d4
K = constsnt
l = length of needle
d = diameter of needle
The length of both "loud" and "soft" steel needles is about 16mm.
The diameter of "loud" needles is about 1.5mm: the diameter of "soft" needles is about 0.8mm.
This inplies that the compliance of soft needles is much greater than that of loud needles, so that:-
[1] There is greater attenuation in the inductive divider Ln, Lcs. (Since Ln is now very large)
[2] The output inductance of the equivalent Thevenin generator increases giving rise to high frequency attenuation as shown in the steady state frequency plots above.
If CL = compliance of a loud needle and CS the compliance of a soft needle
then:-
CL/CS = (1.5/0.8)^4 = 12.36 - a large ratio.
In some needles the compliance depended on the direction of loading , so that they could be "loud" or "soft" depending on the angle of mounting in the pickup.
A section of the cylindrical needle was hammered flat into a flange.
If the sharp end of the flange pointed to the turntable spindle, the compliance was low - giving a "loud" needle with good high frequency response. Turning the needle by 90 degrees gave a high compliance and a "soft" needle.
A picture of such a needle is shown on the right.
Picture courtesy of the Vintage Wireless and Gramophone Club of Western Australia.
WARNING
There is more know-how in the manufacture of steel gramophone needles than is often realised.
THEY ARE NOT SIMPLY SHARPENED PIECES OF STEEL.
Large bags of needles were tumbled for many hours. This produced the following qualities:-
[A] A curved edge on the point.
[B] A high polish.
[C] A hardened point.
This process has not been performed on many of the cheap needles sold today, with the consequence that about a third of the way into a 10 inch 78 the point goes blunt with horrendous tracing distortion and poor high frequency response.
The conclusion is that soft needles not only decrease the signal level, but act as a tone control to give high frequency attenuation.
### [3]Needle Tip Force
In the equivalent circuit the needle tip forces are represented by the current in the inductor Ln.
The following general conclusions can be drawn from the equivalent:-
[1] Needle tip forces are at a maximum when the inductors resonate with Cnh to give a peak in the response.
They are limited only by the oil damping resistance.
In acoustic reproducers the horn presents a suitable damping resistance.
In electrical reproducers the external resistive load does not provide sufficient damping, so extra damping has to be introduced. This is usually in the form of lossy rubber.
In the WE4A oil damping is used, and this accounts for the long life and stability of characteristics of the WE4A.
[2] The inductors Ln and Lcs can also resonate with the large capacity representing the mass of the pickup head and arm. This also increases the needle forces at a much lower frequency.
[3] With "soft" needles Ln is large and the current through it decreases. "Soft" needles therefore greatly reduce needle tip forces.
Recording levels have increased since the WE4A was introduced in 1926. It is found the the WE4A tracks modern heavily cut discs with low distortion.
A combination of needle wear and reduced track velocity increases the tracing distortion towards the inside of a disc. To minimise this effect most transcription discs were inside start.
The output of the WE4A on old worn discs is much cleaner than that from a modern light weight cartridge. The effect is purely mechanical.
Initially the steel needles grinds to the shape of the groove, and the weight of the WE4A ensures that the needle stays in the groove and is not buffeted by dirt and steel particles.
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## [3] TEST RESULTS
#### Steady State Amplitude Response from Test Disc Decca Z718
Both curves have been normalised at 1KHz. to emphasise the attenuation of high frequencies with "soft" needles.
The test results follow the general trend predicted by the lumped equivalent circuit.
The pickup was designed for use with "loud" needles, so the 3db peak at 4KHz is intentional.
It was said to render the reproduction more lively.
The output level was high and the output impedance low, so the WE4A can drive a long cable to the amplifier. Optical sound heads with an inbuilt amplifier were designed to produce an output comparable to the WE4A.
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## WE4A AUDIO
The folowing tracks illustrate the quality of reproduction with the WE4A.
WE4A DETAILS
Design Year: 1925
Tracking Weight 79.9 gms.
Transducer: Magnetic - Variable Reluctance.
Acoustics and early electricals were played with a medium compliance needle:
Diameter: 1.32mm.-------Length: 16mm
Electricals with better surfaces were played with a low compliance needle:
Diameter: 1.65mm.-------Length: 16mm.
Playback Equalisation:
Acoustics: Constant velocity.
Electricals: Constant velocity above 250Hz. :: Constant amplitude below 250 Hz.
Filtering:
All filtering 6th order maximally flat.
High pass:
50 Hz. for electricals.
150 Hz. for acoustics.
Low pass:
4.5 KHz for acoustics.
6 KHz for electricals.
DISC DETAILS
[1] IN THE SHADE OF THE OLD APPLE TREE
( Williams and von Alstyne )
Joe Daniels and His Hotshots in "Drumnastics"
Parlophone
Disc recorded using the Blumlein cutter, probably at Abbey Road. 1935
Joe Daniels: born 1908, Zeerust/Transvaal, South Africa.
Drummer with Harry Roy 1931 - 1937
Formed Hotshots in 1935.
This disc is from the ABC record library in Brisbane. Although it carries over 80 years of wear and tear, it probably gives a good idea of the broadcast quality from the ABC in the thirties and forties when the WE4A was the standard studio pickup.
In the late thirties and into the forties many of the national transmitters employed over 30db of envelope feedback, so, overall, the broadcast quality was good.
Note the increase in tracing distortion towards the end of the track as the steel needle wears and the track velocity decreases.
Inside start discs greatly reduced this distortion.
To listen to the track, press In the Shade of the Old Apple Tree
[2] NOLA
( Felix Arndt )
Les Paul
Multitrack Guitar
MX42049 5860
Red Capitol
Les Paul: born Lester William Polsfuss 9 June 1915.
Pioneered the solid body electric guitar and multitrack recording. First disc to disc, then using 8 track tape made by Ampex.
Although designed in 1925 when the recording level was low, the WE4A will track heavily modulated 78 discs of the late era as illustrated below.
To listen to the track, press NOLA
THE 1911 HIT PARADE: Early Acoustics.
==================================
[3] The Bad Girl of The Family
Whit Cunliffe
Sung by Mr. Arthur Osmond
Recorded in London: Reproduced in Linden
Favorite Record
Serial N307a
Matrix 2662 - b 1 - 67165
Recorded 10/04/1911
Monday 10 April 1911
Note: This disc came from a sheep station (ranch) in the West of Queensland, It was probably a hit with the Jackeroos (cowboys) in 1911. It was covered in clay and grit, which had to be hosed off. It is amazing that the WE4A managed to lift a relatively clean signal from such a surface.
The WE4A proved to be a superb device for playing worn acoustic lateral discs.
A modern light weight pickup responds to every obstacle on the disc such as small pieces of grit and steel. Because of its weight, the WE4A simply ploughs through them.
Further, the groove shape and size is highly variable, The steel needle rapidly grinds to fit the groove as was intended with the old discs. To listen to the track, press The Bad Girl of the Family
Roughly thirty three years later on the same theme, but with some improvement in recording quality, we have:
[4] Our Fanny's Gone All Yankee
George Formby with Ukulele and Orch.
Regal Zonophone [red/green]
Recorded November 1944
To listen to the track, press Our Fanny's Gone All Yankee
Here is an example of the WE4A negotiating a heavily worn disc from the thirties.
On discs with a surface which is no longer flat, but has become rippled, the WE4A does introduce some low frequency rumble. This is noticeable on this disc and is very characteristic of the WE4A's performance.
[5] Voices of Spring
Erna Sack
German Opera House Orchestra Berlin
Conducted by Dr. Hans Schmidt - Isserstedt
D 020724 0 E0171
Note: The sustained note at the end of this track is 14 Seconds long and has a measured frequency of 1.455KHz.
The waveform is close to sinusoidal. The third harmonic is 32db below the fundamental.
It is interesting to evaluate which note this is in the musical scale.
In the Western 12 tone system the frequency of any note is given by:-
f = fo x 2^( n/12)
where f is the frequency in Hz.
fo is the reference frequency.
n the number of intervals between them.
So that: n = (12 x log( f / fo ) )/( log( 2 ) )
In the 30s when this disc was recorded :
Middle C = fo = 256 Hz.
f = 1455 Hz
So that:
n = 30.08
Considering the errors in orchestra tuning, the singer's sense of pitch, and the tolerance on recording and reproducing turntable speeds, it seems close to a miracle that n comes out so close to an integer.
In other words Erna has nailed it right in the middle.
Two Cs above middle C occupy 24 intervals, so that the note is 6 intervals above this or F sharp.
To listen to the track, press Voices of Spring
The WE4A is useful in recovering a signal from a disc with extreme wear : Here - The Teddy Bears' Picnic
This 78 is out of the ABC's (The National Broadcaster ) record library in Alice street Brisbane. It may well have been there since 1932. It was extremely popular - hence the wear.
Note the rise in distortion at the end due to wear of the steel needle and also to wear on the disc.
[6] The Teddy Bears' Picnic
The Rhythmic Troubadours : This was the name given to Henry Hall and The BBC Orchestra on Australian Pressings - Reason not known
Vocalist: Val Rosing
Regal Zonophone : Matrix: MA5642
Recording Date: Wednesday 28 September 1932
Recorded at Abbey Road with the Blumlein Moving Coil Cutter.
To listen to the track, press Teddy Bears' Picnic
This is the reverse side of the Teddy Bears' Picnic and so all the remarks made above apply.
The WE4A is useful in recovering a signal from a disc with extreme wear : Here - Hush, Hush. Hush, Here Comes the Bogey Man
This 78 is out of the ABC's (The National Broadcaster ) record library in Alice street Brisbane. It may well have been there since 1932. It was extremely popular - hence the wear.
Note the rise in distortion at the end due to wear of the steel needle and also to wear on the disc.
[7] Hush, Hush, Hush. Here Comes the Bogey Man
The Rhythmic Troubadours : This was the name given to Henry Hall and The BBC Orchestra on Australian Pressings - Reason not known
Vocalist: Val Rosing
Regal Zonophone : Matrix: CA 13048
Recorded at Abbey Road with the Blumlein Moving Coil Cutter.
To listen to the track, press Hush, Hush, Hush, Here Comes the Bogey Man
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## [4] BELL DISC CUTTING HEAD
### [1]Electro-Mechanical Equivalence
This cutter was one of the masterpieces of electro-mechanical design of the 20th century.
It also had a profound effect on the shape of the century with the enormous improvement in the quality of disc recording and the introduction of film sound on disc.
The realisation of the recorder was made possible by:-
[1] The development of classical filter theory at Bell.
[2] The realisation of the dynamic equivalence between electrical and mechanical systems.
ELECTRO-MECHANICAL EQUIVALENCE
Linear differential equations with real constant co-efficients govern the dynamic behaviour of both electrical amd mechanical systems.
If an electrical system is designed to have the same governing differential equation as a mechanical system, then one will be the exact analog of the other.
With a magnetic transducer, force is produced by current and voltage is produced by velocity. It is then natural to equate these variables.The link between the two then becomes an ideal electrical transformer with the transformer ratio determined by the transducer constants.
It is possible to get an equivalent with the variables reversed, but then the transducer is represented by a gyrator with a much indreased difficulty in interpretation.
With a piezo-electric transducer, force is produced by voltage and current is induced by velocity, so it is natural to equate these.
### PIEZO-ELECTRIC TRANDUCER
Put:-
Voltage (V) --> Velocity (v) or angular velocity (w)
Current (I) --> Force (F) or Torque (T)
Put:-
Current (I) --> Velocity (v) or angular velocity (w)
Voltage (V) --> Force (F) or Torque (T)
I = C dV/dt
F = M dv/dt
C = M
Capacity translates to mass or moment of inertia.
V = L dI/dt
F = M dv/dt
L = M
Inductance translates to mass or moment of inertia.
V = L dI/dt
If x is displacement and U is the compliance then:-
x = UF
dx/dt = U dF/dt
v = U dF/dt
L = U
Inductance translates to compliance.
I = C dV/dt
If x is displacement and U is the compliance then:-
x = UF
dx/dt = U dF/dt
v = U dF/dt
C = U
Capacity translates to compliance.
V = R I
If J is Viscous Force Constant then:-
v = J F
R = J
Resistance translates to viscous force constant.
I = G V
If J is Viscous Force Constant then:-
v = J F
G = J
Conductance translates to viscous force constant.
### [2]Equivalent Circuit
The original paper on the cutting head was:-
Methods of High Quality Recording and Reproducing Music and Speech based on Telephone Research.
Bell Telephone Journal: July 1926 pp 493 -- 526.
The recorder had a magnetic transducer, but the equivalent circuit included in the paper uses equivalents appropriate to a piezo-electric transducer. Consequently, much of the equivalent is represented by a nameless block.
It also difficult to relate the circuit to the dynamic behaviour of the mechanical system.
The complete equivalent circuit including the driving amplifier, and using appropriate equivalents, is given below.
A sectioned drawing of the cutter is shown opposite.
Each mechanical element is tagged with its electrical eauivalent.
The required output is the velocity at the tip of the stylus. The maximum displacement for a 78 disc is +(-) 2 thou. of an inch, so, for the small angles involved, angular displacement translates into linear displacement.
The full equivalent circuit using equalities for a magnetic transducer are shown opposite.
If the two inductances and resistance to earth ( -La, Lc and Rc ) are removed, the circuit reduces to a classical low pass constant k filter terminated at the far end.
Over the pass band the filter reflects a constant resistance, Rf, into the primary of the transformer, so that the total primary resistance is Rtot = Ra + Rw + Rf .
If the winding inductance of the transducer, Lw, is not to cause high frequency attenuation, then the resistance must be greater than the inductive reactance at the cutoff frequency fc.
ie: Rtot > 2 Π fc Lw
The filter, then, is driven from a resistive impedance.
We can now observe that :-
[1] The filter is a classical constant k low pass filter.
[2] The filter has three sections: two half L sections at the end and a full T section in the middle.
[3] The filter is terminated at the receiving end and driven from a resistive impedance at the sending end.
We now discuss the action of the three elements which we removed from the equivalent circuit to produce a classical constant k low pass filter.
##### The Resistor Rc
The required output is not taken from the terminating load but from the centre of the middle section.
The voltage appearing across Cc represents the required output, the angular velocity of the cutting stylus.
In lateral recording the dynamic forces at the stylus tip are proportional to velocity.
They therefore appear as a resistive load across Cc.
Note: In vertical recording the forces are highly non-linear.
A good account of this is given in Edison patent No.954,221 July 12 1910.
It is important that the cutting load Rc be much greater than the characteristic resistance of the filter for two reasons:-
[A] It is desirable that the recorded signal be independent of the characteristics of the master disc.
[B] The cutting process produces random forces at the tip of the cutting tool. These excite movement which will be cut back onto the disc as noise.
Keeping Rc high compared with the filter impedance minimises this effect.
##### The Centering Spring Inductor Lc
With constant velocity recording the displacement is inversely proportional to frequency, so that this becomes excessively large at low frequencies. The introduction of a 6db/octave slope in the velocity response at low frequencies results in constant amplitude recording. This is conveniently introduced by the centering spring Lc.
At low frequencies this shunts the signal to earth.
The turnover frequency fc is set by the time constant of the total driving resistor R// and the inductor Lc.
T = Lc/R// :: fc = 1/2 Π T
Note that the turnover frequency is set by the strength of the centering spring: not by a network in the amplifier.
At low frequencies the recorder reduces to a very simple system.
The curerent drive is independent of frequency, and so produces a constant torque applied to the centering spring: giving a constant displacement and so constant amplitude recording.
##### The Negative Compliance -La
Imagine the DC field of the recording head energised with the audio input left open.
If the magnetic vane marked ( Ca La ) is displaced from its central position, magnetic poles are induced on its tip, and these produce a torque in the DC field.
The torque is proportional to angular deflection, but is of opposite sign to that produced by a spring.
It therefore must be represented by a NEGATIVE compliance.
The elements Cc and La are in parallel so the total susceptance y(p) is given by:-
y(p) = pCc - 1/pLa
In the steady state this yields:-
y(i ω) = i ωCc[ 1 + ( ωo/ω )2 ]
where ωo = 1/√LaCc
No data was given to calculate ωo , but it is assumed the squared term is very small at high frequencies approaching cutoff, and so has little effect on the constant k filter action.
Note: There are accounts of the pole piece hitting and sticking to the stationary laminations on very high recording peaks.
The design of electrical wide band resistive loads is difficult: the design of wide band mechanical loads is even more difficult.
Wire wound resistors exhibit inductance and the resistance changes with frequency due to skin effect.
Rubber exhibits histeresis and so resistance along with compliance. The histeresis changes with frequency to mimic the complications of skin effect in wire wound resistors.
This technique is illustrated in the bottom part of the cutter equivalent circuit.
To realise a wideband load of resistance Ro up to a frequency fo:-
[1] Take a constant k section R3 L3 C3
[2] L3 has a series resistive component R3 - a small fraction of Ro
[3] Design L3 C3 to have a characteristic impedance Ro - R3 and a cutoff frequency fo
[4] In the next section L4 C4 has a characteristic impedance of Ro - R3 - R4 and a cutoff frequency of fo
[5] Eventually a termination of 0 ohms will be reached and the output operated into a short circuit.
The technique was used in the fifties to terminate the output lines in distributed amplifiers.
An example used in the Tektronix 545A oscilloscope is given opposite.
It is interesting to note that the Bell engineers ( probably Norton ) were aware of this technique in the twenties and used it first in a mechanical analog form.
The lumped line described above approaches a transmission line as the size of each element is decreased.
The characteristic impedance of a lossless uniform line is purely resistive, and so a line of infinite length would act as an ideal load.
Lines in the real world need not have an infinite length.
The characteristic impedance of a uniform transmission line closely approximates a resistance if the attenuation per wavlength is not too great.
If the line is of sufficient length, so that the reflected wave is small compared with the transmitted wave, then the input impedance will still remain largely resistive.
Further, it does not matter if the trmination is a short or open circuit.
A lossy rubber transmission line replaced the lumped equivalent in the final form of the cutting head.
Its response was more even than the lumped line and it was easier to manufacture.
Unfortunately the properties of rubber are time and temperature dependent.
This was a disadvantage in an otherwise brilliant design.
The application of motional negative feedback by Bell rendered mechanical damping unnecessary and produced a design with an excellent dynamic response.
##### Motional Feedback
Bell were the first to apply motional feedback to a vertical cutting head in the Thirties. Sectioned drawings of the cutter are shown.
Note that no damping material is required in the head: the motional feedback has complete control over the movement of the cutting stylus.
Two of the main difficulties in applying motional feedback are:-
[1] Preventing inductive pickup by the motion sensing coil from the main driving coil.
[2] Ensuring the motion of the pickup coil and the cutting stylus are the same.
Coupling between the two coils is reduced by a thick copper ring shield.
The pickup coil is attched as close as possible to the cutting stylus to ensure they share the same motion.
The mass and compliance of the moving system produce high Q mechanical resonance at about 700 Hz. This produces a transfer function stable under a large degree of feedback.
The feedback flattens the curve over the complete audio spectrum.
The recording amplifier for the motional feedback cutting head is shown on the right.
The transfer function of the forward path is modified by an internal feedback loop with elements B,C,D.
The transfer function of the motional feedback path is modified by the network A.
In a vacuum tube the transfer characteristics of the control, screen, and suppressor grids are different non-linear functions, and it is not good practice to apply the input and feedback signal on different grids.
This general principle is violated twice in this amplifier:-
[1] The internal feedback is applied to the screen of pentode 33.
[2] The input signl is applied to the suppressor of pentode 32.
The circuit technique criticised above comes from a patent by Wiebusch.
Note that when a signal is applied to different grids, a component of plate current is the product of the two - not a desirable state of affairs in a linear system.
### [4]Audio Tracks
The following hill and dale tracks were cut with the feedback cutter.
They were not commercial recordings, but experimental tracks cut by the Bell engineers in the thirties.
To listen to the track, press Hill and Dale Track 1
To listen to the track, press Hill and Dale Track2
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## HARRISON CUTTER AUDIO
The folowing tracks illustrate the quality of early tracks cut with the Western Electric Bell Harrison Cutter.
Initially, Bell decided to limit the use of the recording system using the Harrison cutter to the USA.
The system was licensed to Columbia.
One of the first tasks was to make a demonstration disc.
Here it is.
To listen to the track, press Columbia Demonstration Disc
The British overcame this monopoly by buying the Columbia Company.
One of the first discs cut with the system in Britain was:-
LIONEL MONKTON MEMORIES
[1] Pt. 1 W AX 5154
[2] Pt. 2 W AX 5155
De Broy Somers Band
12" Black Columbia
Recorder: Harrison's WE cutter of 1926
Early English Electrical Recording
DeBroy Somers:
Born Dublin 11 April 1890
Died: Knightsbridge May 1952
Formed The Savoy Hotel Orpheans
Admired by musicians because he could play all the instruments in the band.
EQUALISATION:-
Constant Velocity above 250Hz.:: Constant Amplitude below 250Hz.
FILTERING:-
Maximally Flat 6th order Low Pass Filter: 3db at 6KHz.
Maximally Flat 6th order High Pass Filter: 3db at 50Hz.
WE RECORDING EQUIPMENT DETAIL
Microphone: High quality condenser microphone type 394 designed by Wente of Bell Labs.
Preamp: Type 47A mounted in a cylinder directly under the condenser microphone.
Battery powered miniature triode type 239-A with thoriated tungsten filament. No feedback.
Recording amplifier: Transformer coupled push - pull triodes with thoriated tungsten filament.
Type 205D. No feedback
Designed by Henry Harrison of Bell Labs.
Transducer: Moving Iron. No feedback.
Response: Flat with low frequency turnover built into head.
Constant Velocity - Constant Amplitude turnover adjustable between 250Hz and 300Hz.
To listen to the track, press Lionel Monkton Part1
To listen to the track, press Lionel Monkton Part2
The WE recording equipment was installead in tne Victor recording studios, Camden, between March and April of 1925.
The following brilliant recording "THE WHISTLER AND HIS DOG " was recorded by Victor on Sunday, 11th of October 1925.
Disc Detais:-
THE WHISTLER AND HIS DOG <
ARTHUR PRYOR"S BAND
COMPOSER: ARTHUR PRYOR
WHISTLING BY MARGARET McKEE and BILLY MURRAY
VICTOR HMV VE 19869A
ORTHOPHONIC [ELECTRICAL] RECORDING
To listen to the track, press The Whistler and HIS DOG
VITAPHONE
The WE Harrison recorder gave rise to sound on disc for movie film called "VITAPHONE"
Equipment installation into theatres took some time, so short sound clips were made before the first feature films for demonstration purposes. These usually lasted from eight to ten minutes.
At the end of the run the discs were returned for destruction. However, this did not occur in Queensland, Australia. The end point was usually far removed from a capital city.
Consequently, Vitaphone discs sometimes can be unearthed in remote locations in Queensland.
The following disc is one of them, and, despite the age and wear, gives a good indication of the audio quality of the first "film" sound.
It is believed that the audio quality of "sound on disc" was much better than that of the first "sound on film".
The first very temporary setup for reproducing a Vitaphone disc with the WE4A is shown opposite.
The "drag" from the WE4A on heavily cut passages produces a large torque on the outside of a 16 inch diameter disc. It was found that this was enough to slow the average turntable down, and a professional transcription turntable had to be pressd into service.
The turntable used was Australian designed and made - A Byer Professional 12.
Disc Details:-
Diameter: 16 inches.
Speed: 33 1/3 RPM.
Start: Inside Start.
First Vitaphone Track Details
[1] John Barclay
Impersonator
Matrix: VA-436-2
Rec. 97 Vol. +8
Recording Date: 21-01-27
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## Almost Satisfaction
(Eingestellt am 12. Februar 2022, 07:15 Uhr von Joseph nehme)
This puzzle can be solved without either of the inequality signs. I first posted the puzzle without them but the solving path turned out to be extremely difficult and hard to spot so I added them for a smoother solving path. For a more challenging puzzle solve it without the inequality in box 3 and for a really hard one solve it without both inequalities.
Rules:
- Normal Standard Sudoku rules apply.
- On the marked diagonals digits cannot repeat.
- Normal Little killer Sudoku Rules: Digits on the diagonals sum to the clues outside the grid. Digits can repeat on diagonals.
- The digits in the circles are the sum of the digits along their respective arrows. Digits may repeat along the arrows
- The inequality signs mean that R1C8 is higher than R2C8 and that R9C9 is higher than R8C9
CTC app Link to play below.
As always, your feedback, ratings and comments are highly appreciated :) Enjoy !
Lösungscode: Row 5 followed by column 7 (no spaces and no commas) ie (123456789987654321)
Gelöst von Fred Yamred, Vebby, cdwg2000, Crawlie, SKORP17, purpl, Playmaker6174, PippoForte, Knitabit, spq, wenchang, le bonhomme, Elliott810, CaneloC, Doctor, zlotnleo, argl, AvonD, by81996672, 999ARMEN999
Komplette Liste
### Kommentare
am 11. April 2022, 20:40 Uhr von argl
Broke it once, let it sit, came back to it, somehow managed :D
It was hard, but good
am 31. März 2022, 19:44 Uhr von Hausigel_mod
Labels/Tags changed
Zuletzt geändert am 12. Februar 2022, 20:55 Uhr
am 12. Februar 2022, 19:40 Uhr von Playmaker6174
Very beautiful design, quite challenging solve but I enjoyed the aha moments in this one :)
@Playmaker6174 Thank you my friend :)
Zuletzt geändert am 12. Februar 2022, 12:44 Uhr
am 12. Februar 2022, 12:41 Uhr von Vebby
Very clever geometry! Thanks Joseph :)
@Vebby Thank you my friend :)
Schwierigkeit: Bewertung: 97 % Gelöst: 20 mal Beobachtet: 2 mal ID: 00093O
Lösungscode:
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# Thread: Inverse Cotangent Function
1. ## Inverse Cotangent Function
The question states:
Find the exact value of each real number y. Do not use a calculator.
y = arccot (
-√3)
What's the process for finding the answer???
2. Originally Posted by nee
The question states:
Find the exact value of each real number y. Do not use a calculator.
y = arccot (
-√3)
What's the process for finding the answer???
$\displaystyle arccot=\frac{arccos}{arcsin}\rightarrow \frac{arccos(\frac{-\sqrt{3}}{2})}{arcsin(\frac{1}{2})}$
$\displaystyle arccos\bigg(\frac{-\sqrt{3}}{2}\bigg)=\frac{5\pi}{6}$
or
$\displaystyle arcsin\bigg(\frac{1}{2}\bigg)=\frac{5\pi}{6}$
3. Originally Posted by dwsmith
$\displaystyle arccot=\frac{arccos}{arcsin}\rightarrow \frac{arccos(\frac{-\sqrt{3}}{2})}{arcsin(\frac{1}{2})}$
$\displaystyle arccos\bigg(\frac{-\sqrt{3}}{2}\bigg)=\frac{5\pi}{6}$
or
$\displaystyle arcsin\bigg(\frac{1}{2}\bigg)=\frac{5\pi}{6}$
$\displaystyle \arctan(x){\color{red}\ne}\frac{\arcsin(x)}{\arcco s(x)}$
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# ATAN2
In this comprehensive guide, we will explore everything you need to know about the ATAN2 function in Excel. The ATAN2 function is a trigonometric function that calculates the arctangent of the quotient of two specified numbers, returning the angle in radians between the positive x-axis and the point (x, y) on the plane. This function is particularly useful in various mathematical and engineering applications, such as calculating the angle between two points or determining the direction of a vector.
## ATAN2 Syntax
The syntax for the ATAN2 function in Excel is as follows:
ATAN2(y, x)
Where:
• y is the y-coordinate of the point on the plane.
• x is the x-coordinate of the point on the plane.
Note that the order of the arguments is (y, x) and not (x, y), which is a common source of confusion for users.
## ATAN2 Examples
Let’s take a look at some examples of how to use the ATAN2 function in Excel:
Example 1: Calculate the angle between the positive x-axis and the point (3, 4) on the plane.
=ATAN2(4, 3)
Example 2: Calculate the angle between the positive x-axis and the point (-3, 4) on the plane.
=ATAN2(4, -3)
Example 3: Calculate the angle between two points (2, 3) and (5, 7) on the plane.
=ATAN2(7 – 3, 5 – 2)
## ATAN2 Tips & Tricks
Here are some tips and tricks to help you get the most out of the ATAN2 function in Excel:
• Remember that the ATAN2 function returns the angle in radians. To convert the result to degrees, use the DEGREES function, like this: =DEGREES(ATAN2(y, x)).
• If you need to calculate the arctangent of a single number, use the ATAN function instead of ATAN2.
• Keep in mind that the order of the arguments is (y, x), not (x, y). This is a common source of confusion and errors.
## Common Mistakes When Using ATAN2
Here are some common mistakes users make when using the ATAN2 function in Excel:
• Using the wrong order of arguments: Remember that the correct order is (y, x), not (x, y).
• Forgetting to convert radians to degrees: The ATAN2 function returns the angle in radians. If you need the result in degrees, use the DEGREES function to convert it.
• Using ATAN2 when ATAN is more appropriate: If you only need to calculate the arctangent of a single number, use the ATAN function instead.
## Why Isn’t My ATAN2 Working?
If you’re having trouble with the ATAN2 function in Excel, consider the following possible issues:
• Check the order of the arguments: Make sure you’re using (y, x) and not (x, y).
• Ensure that you’re using the correct function: If you only need to calculate the arctangent of a single number, use the ATAN function instead of ATAN2.
• Verify that your input values are correct: Double-check your coordinates and make sure they’re accurate.
## ATAN2: Related Formulae
Here are some related formulae that you might find useful when working with the ATAN2 function in Excel:
• ATAN: Calculates the arctangent of a single number. Syntax: =ATAN(number).
• SIN: Calculates the sine of an angle in radians. Syntax: =SIN(angle).
• COS: Calculates the cosine of an angle in radians. Syntax: =COS(angle).
• TAN: Calculates the tangent of an angle in radians. Syntax: =TAN(angle).
• DEGREES: Converts an angle in radians to degrees. Syntax: =DEGREES(angle).
By mastering the ATAN2 function and its related formulae, you’ll be well-equipped to tackle a wide range of mathematical and engineering problems in Excel. Happy calculating!
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Calculus: Early Transcendentals 8th Edition
$$\int\tan^2 xdx=\tan x-x+C$$
$$A=\int\tan^2 xdx$$ Remember that $\sec^2 x=1+\tan^2 x$. So, $\tan^2 x=\sec^2 x-1$ $$A=\int(\sec^2x-1)dx$$ Also, $\int\sec^2 x=\tan x+C$. Therefore, $$A=\tan x-x+C$$
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## ››Convert quarter [print] to palm [US, Roman major]
quarter [print] palm [US, Roman major]
Did you mean to convert quarter quarter [cloth] quarter [print] to palm [Britain, Roman minor] palm [Dutch] palm [US, Roman major]
How many quarter [print] in 1 palm [US, Roman major]? The answer is 914.4.
We assume you are converting between quarter [print] and palm [US, Roman major].
You can view more details on each measurement unit:
quarter [print] or palm [US, Roman major]
The SI base unit for length is the metre.
1 metre is equal to 4000 quarter [print], or 4.3744531933508 palm [US, Roman major].
Note that rounding errors may occur, so always check the results.
Use this page to learn how to convert between quarters and palm [US, Roman major].
Type in your own numbers in the form to convert the units!
## ››Quick conversion chart of quarter [print] to palm [US, Roman major]
1 quarter [print] to palm [US, Roman major] = 0.00109 palm [US, Roman major]
10 quarter [print] to palm [US, Roman major] = 0.01094 palm [US, Roman major]
50 quarter [print] to palm [US, Roman major] = 0.05468 palm [US, Roman major]
100 quarter [print] to palm [US, Roman major] = 0.10936 palm [US, Roman major]
200 quarter [print] to palm [US, Roman major] = 0.21872 palm [US, Roman major]
500 quarter [print] to palm [US, Roman major] = 0.54681 palm [US, Roman major]
1000 quarter [print] to palm [US, Roman major] = 1.09361 palm [US, Roman major]
## ››Want other units?
You can do the reverse unit conversion from palm [US, Roman major] to quarter [print], or enter any two units below:
## Enter two units to convert
From: To:
## ››Metric conversions and more
ConvertUnits.com provides an online conversion calculator for all types of measurement units. You can find metric conversion tables for SI units, as well as English units, currency, and other data. Type in unit symbols, abbreviations, or full names for units of length, area, mass, pressure, and other types. Examples include mm, inch, 100 kg, US fluid ounce, 6'3", 10 stone 4, cubic cm, metres squared, grams, moles, feet per second, and many more!
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# Why is the nonlinearity of this Boolean function evaluating to $\frac12$?
I am using the method presented in this paper to find the nonlinearity of the function
$$f: \mathbb{F}^1_2 \to \mathbb{F}^1_2 \\ f(x) = x$$
The truth table is $$f = [0 \space \space 1]$$. Now, I read from the paper by Terry Ritter that
Nonlinearity is the number of bits which must change in the truth table of a Boolean function to reach the closest affine function.
This means the nonlinearity value should be a whole number.
The algorithm to calculate nonlinearity is to first use the Fast Walsh Transform to find the Walsh spectrum, then use the formula
$$Nl(f_k) = 2^{k-1} - \dfrac12 \cdot\max_{a\in\mathbb{F_2^{2^k}}} |W_f(a)|$$
where the Walsh spectrum is calculated by multiplying the truth table of the function by the corresponding Hadamard matrix.
So, since $$k = 1$$, we use the Hadamard matrix of size $$2^1$$ giving the following Walsh spectrum:
$$\begin{bmatrix}0 & 1\end{bmatrix} \begin{bmatrix}1 & 1\\1 & -1\end{bmatrix} = \begin{bmatrix}1 & -1\end{bmatrix} \implies \max_{a\in\mathbb{F_2^{2^k}}} |W_f(a)| = |-1| = 1$$
Therefore
$$Nl(f_{k=1}) = 2^{0} - \dfrac12 \cdot 1 = \dfrac12$$
What am I missing?
1. Calculating Nonlinearity of Boolean Functions with Walsh-Hadamard Transform by Pedro Miguel Sosa
2. Measuring Boolean Function Nonlinearity by Walsh Transform by Terry Ritter
In this formulation you need to convert your function's output range to $$\{-1,+1\}$$ via $$f(x)=(-1)^{f(x)}$$ and apply the Walsh Hadamard to the new function $$f(x)$$. Using the zero one formulation means you are off by a constant depending on the number of variables since
$$(-1)^u=1-2u$$ for $$u\in \{0,1\}.$$
See my answer below on Boolean functions and crypto, it may be useful given your recent questions.
How are boolean functions used in cryptography?
In addition to the answer by kodlu, after carefully re-reading the papers, I was able to figure it out. Key things to note:
1. If we use the Fast Walsh Transform on Boolean functions consisting of $$\{0,1\}$$ then the formula for nonlinearity is
... half the number of bits in the function, less the absolute value of the unexpected distance.
That is $$Nl(f) = \dfrac12 \cdot 2^k - \max_{a\in\mathbb{F}_2^{2^k}} |W_f(a)|\\ = 2^{k-1} - \max_{a\in\mathbb{F}_2^{2^k}} |W_f(a)|$$
Therefore, for the question in the original post we have
$$Nl(f) = 2^{0} - |1| = 0$$
Alternatively, page 20 here (alt link) suggests to proceed as follows: After finding the Fast Walsh transform,
1. Add $$2^{k-1}$$ to each entry in the row except the first entry. This gives us a new row, call it $$FHT'$$
2. If an entry in less than $$2^{k-1}$$ it remains unchanged. Otherwise, if an entry of $$FHT'$$ is greater than $$2^{k-1}$$ then subtract it from $$2^k$$.
3. Finally, the nonlinearity is the smallest of these adjusted elements.
2. If we use the Fast Walsh Transform on Boolean functions consisting of $$\{1,-1\}$$ then the formula for nonlinearity is
$$Nl(f) = 2^{k-1} - \dfrac12 \cdot\max_{a\in\mathbb{F}_2^{2^k}} |W_f(a)|$$
Because
Using real values $$\{1,-1\}$$ doubles the magnitude and changes the sign of the FWT results
Source
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Parallel Parentheses Matching
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# Parallel Parentheses Matching - PowerPoint PPT Presentation
## Parallel Parentheses Matching
- - - - - - - - - - - - - - - - - - - - - - - - - - - E N D - - - - - - - - - - - - - - - - - - - - - - - - - - -
##### Presentation Transcript
1. Parallel Parentheses Matching Plus Some Applications
2. Parentheses Matching Problem Definition: • Given a well-formed sequence of parentheses stored in an array, determine the index of the mate of each parentheses stored in the array
3. Example
4. Sequential Solution • Traditional solution uses a stack • Push left parentheses, pop for a right parenthesis; these are a pair • Can this method be implemented in parallel? Why or why not?
5. Parallel Solution: Divide & Conquer • Lemma 1: The mate of a parenthesis at an odd position in a balanced string lies in an even position (and vice versa). • Lemma 2: If a balanced string has no left parenthesis at an even location (or, equivalently, a right at an odd location), then the mate of each left parenthesis in the string lies immediately to its right.
6. Lemma 2 • Any string that satisfies Lemma 2 is of form ( ) ( ) ( ) ( )….( ) and is referred to as form F.
7. Algorithm Overview • Each left position at an odd position and each right position are marked with a 0. • All others are marked with a 1. These 2 disjoint sets are copied (packed) into a new array. • Repeat for the 2 sets. • Stop when each new substring is of form F.
8. Algorithm Match For i = 1 to log n – 1 do if “(“ & index is odd then mark 0 else mark 1 Use segmented prefix sums to compute new index for each parenthesis Move parentheses to new location Determine if string is now in form F; if not, terminate – unbalanced string Match parentheses and store in original array
9. Example
10. Example – Keep Index
11. Segmented Prefix Sum Problem Definition Given an array containing elements, some marked 0 and some marked 1. Compute the prefix sum of each subset. (For this application the sums will be on values of 1, to number the items.)
12. Segmented Prefix Sum - Example How can this be accomplished with one prefix sums operation?
13. Parentheses Matching on Hypercube • Use the Divide & Conquer strategy • Consider 2 processors • Each PC – assign 0/1 • P0 send 1’s to P1; P1 send 0’s to P0 • Each solve the sub-problem • Does the problem split evenly? • Consider Large problem – P0 & P2 take 0 items, P1 & P3 take 1 items
14. PPM - Hypercube • Overview of Algorithm • 2-Cube: Special case of 2 pc hypercube • 4-Cube: Used to partition large sub-problems consisting of 4 pc cubes • Match: The Driving Algorithm
15. Data Distribution • INPUT array is divided into P equal partitions of size n/p • First n/p items are given to P0, next n/p items to P1, etc. • Final MATCH information for each item is stored in the original PC
16. Data Distribution • Array consists of elements INPUT which holds the parentheses & MATCH which will hold the final matching information • Local Match: if the match is determined by the pc in which the match information is to be stored • Non-local: otherwise
17. Algorithm 2-Cube • Mark left and right parentheses with 0/1 as previously discussed • P0 & P1 exchange entries – P0 contains 0 and P1 contains 1 • Each PC use stack to sequentially match parentheses • Send non-local match operation to appropriate processor
18. Algorithm 2-Cube – Step 1 Mark 0/1 P0 P1
19. Algorithm 2-Cube – Step 2 & 3 Exchange & Match P0 P1
20. Algorithm 2-Cube – Step 4 Send non-local match information P0 P1
21. Algorithm 4-Cube • Basis of Algorithm Match • Insures near-equal data distribution • Overview • Phase 1: local matches determined sequentially & communicated • Phase 2: Unmatched parentheses marked, redistributed; P0 & P1 have 0’s, P2 & P3 have 1’s (half each)
22. Algorithm 4- Cube Phase 1: Sequential Processing 1. Each PC use stack to match 2. Send non-local Match to other PC 3. Count unmatched parentheses; prefix sum to reindex
23. Algorithm 4-Cube Phase 2 1. Mark parentheses with 1/0 2. P0 & P2 exchange: P0=0 & P2=1 Likewise, P1=0 and P3=1 • P0 & P1 exchange number information; likewise for P2 & P3 • P0 & P1 exchange entries; P0 obtains 1st half: likewise for P2 & P3
24. Algorithm 4-Cube Distribution of Data – 64 entries
25. Algorithm Hypercube Match • Each subcube of size 4 executes 4-Cube // Logically P/2 subcubes with independent subproblems • Each subcube (p/2) prefix sums to determine new index • Each subcube from Step 1 recursively repeat 1, 2, 3 until each subcube is of size 2 4. Execute 2-Cube to complete the solution
26. Algorithm Hypercube Match Complexity Analysis • O (log 2 p + n/p log p) • For p=n/log n simplifies to O(log 2 n) Is the Algorithm Optimal?
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04 Civil Engineering Quiz | Water Resource MCQ #mcq.cetjob.com - Multiple choice questions and Objectives
# 04 Civil Engineering Quiz | Water Resource MCQ #mcq.cetjob.com
## 04 Civil Engineering Quiz | Water Resource MCQ
1 >>Pick up the correct statement from the following :
 ?
• (A) The actual prevailing rate of infiltration of water in the soil at any time, is known as infiltration rate
• (B) The actual infiltration rate at any time may be equal to or less than the infiltration capacity
• (C) When rainfall rate is less than the infiltration capacity, the infiltration rate is approximately equal to the rainfall rate
• (D) all
2 >>In India the recording type rain gauge generally used, is
 ?
• (A) tipping type
• (B) float recording type
• (C) weighing type
• (D) none of these.
3 >>Pick up the correct statement from the following :
 ?
• (A) The moisture deficiency will be different at different points
• (B) The path of the water required to bring the soil moisture content of a soil up to its field capacity is called soil moisture deficiency
• (C) The moisture content at which permanent wilting of plants takes place, is called the wilting point
• (D) all
4 >>The deficiency in rain catch due to vertical acceleration of air forced upward over the gauge, is
 ?
• (A) greater for lighter rain
• (B) greater for heavy rain
• (C) lesser for small rain drops.
• (D) greater for large drops
5 >>Infiltration capacity of soil depends upon
 ?
• (A) compaction of the soil particles
• (B) arrangement of soil particles
• (C) shape and size of soil particles
• (D) all
6 >>Pick up the correct statement from the following :
 ?
• (A) Relative humidity is the ratio of the weight of the vapours present per unit volume to the weight of vapours which could be contained at the same temperature when fully saturated
• (B) Relative humidity is the ratio of actual vapour pressure and saturation vapour pressure at the same temperature
• (C) Humidity can be measured by psychrometer
• (D) all
7 >>If a gauge is installed perpendicular to the slope, its measurement is reduced by multiplying
 ?
• (A) tangent of the angle of inclination with vertical
• (B) cosine of the angle of inclination with vertical
• (C) sine of the angle of inclination with vertical
• (D) calibration coefficient of the gauge.
8 >>A recording type rain gauge
 ?
• (A) produces a mass curve of rain fall
• (B) is sometimes called integrating rain gauge or
• (C) records the cumulative rain
• (D) all
9 >>Absolute humidity in air
 ?
• (A) increases at higher altitudes
• (B) decreases at higher altitudes
• (C) remains constant at all altitudes
• (D) none of these.
10 >>Pick up the correct statement from the following :
 ?
• (A) The maximum rate of absorbing water by the soil in any given condition, is known as infiltration capacity
• (B) When rainfall exceeds the interception rainfall, water reaches the ground and infiltration starts
• (C) The difference between the total rainfall and intercepted rainfall, is generally called ground rainfall
• (D) all
11 >>Pick up the correct statement from the following :
 ?
• (A) Water remains in atmosphere as vapours
• (B) Rain water is obtained by evaporation from rivers, lakes and oceans
• (C) Hydrologic cycle is a continuous process of evaporation and precipitation of water in atmosphere
• (D) all
12 >>Precipitation caused by lifting of an air mass due to the pressure difference, is called
 ?
• (A) orographic precipitation
• (B) cyclonic precipitation
• (C) convective precipitation
• (D) none of these.
13 >>In India, rain fall is generally recorded at
 ?
• (A) 12 Noon
• (B) 8 A.M.
• (C) 4 P.M.
• (D) 8 P.M.
14 >>Unit Hydrograph theory was enunciated by
 ?
• (A) Merril Bernard
• (B) Le-Roy K. Shermen
• (C) Robert E. Horten.
• (D) W.W. Horner
15 >>The surface Run-off is the quantity of water
 ?
• (A) intercepted by buildings and vegetative cover
• (B) required to fill surface depressions
• (C) that reaches the stream channels
• (D) absorbed by soil
16 >>The standard height of a standard rain gauge, is
 ?
• (A) 20cm
• (B) 30cm
• (C) 10cm
• (D) 40cm
17 >>Hydrology helps in
 ?
• (A) predicting the effects on the river water level on completion of dams
• (B) deciding the minimum reservoir capacity
• (C) forecasting the availability of quantity of water at reservoir site
• (D) all
18 >>Water contains
 ?
• (A) three hydrogen atoms and two oxygen atoms
• (B) one hydrogen atom and two oxygen atoms
• (C) one hydrogen atom and one oxygen atom
• (D) two hydrogen atoms and one oxygen atom
19 >>The theory of infiltration capacity was given by
 ?
• (A) W.W. Horner
• (B) Le-Roy K. Shermen
• (C) Merrill Bernard
• (D) Robert E. Horten.
20 >>Consumptive use of a crop during growth, is the amount of
 ?
• (A) evaporation
• (B) transpiration
• (C) interception
• (D) all
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https://hackaday.com/2024/09/05/launching-model-airplanes-with-a-custom-linear-induction-motor/
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# Launching Model Airplanes With A Custom Linear Induction Motor
Launching things with electromagnetism is pretty fun, with linear induction motors being a popular design that finds use from everywhere in hobby designs like [Tom Stanton]’s to the electromagnetic launchers on new US and Chinese aircraft carriers. Although the exact design details differ, they use magnetic attraction and repulsion to create a linear motion on the propulsive element, like the sled in [Tom]’s design. Much like the electromagnetic catapults on a Gerald R. Ford-class carrier, electrical power is applied to rapidly move the sled through the channel, akin to a steam piston with a steam catapult.
For [Tom]’s design, permanent magnets are used along both sides of the channel in an alternating north/south pole fashion, with the sled using a single wound coil that uses brushes to contact metal rails along both sides of the channel. Alternating current is then applied to this system, causing the coil to become an electromagnet and propel itself along the channel.
An important consideration here is the number of turns of wire on the sled’s coil, as this controls the current being passed, which is around 90 A for 100 turns. Even so, the fastest sled design only reached a speed of 44 mph (~71 km/h), which is 4 mph faster than [Tom]’s previous design that used coils alongside the channels and a sled featuring a permanent magnet.
One way to increase the speed is to use more coils on the sled, with a two-coil model launching a light-weight model airplane to 10.2 m/s, which is not only a pretty cool way to launch an airplane, but also gives you a sense of appreciation for the engineering challenges involved in making an electromagnetic catapult system work for life-sized airplanes as they’re yeeted off an aircraft carrier and preferably not straight into the drink.
## 18 thoughts on “Launching Model Airplanes With A Custom Linear Induction Motor”
1. Sok Puppette says:
How is it an induction motor?
1. RunnerPack says:
‘Tisn’t. It’s a linear, brush-commutated, DC motor.
2. TG says:
It induces linear. A distinct linear is induced in whatever you strap on top of it
Hi, this is nice and relatively simple. The biggest problem is mechanical contacts. Would you like to consider to use electronically one using Hall sensors and MOS-FET tranzistors replacing fizical contacts. Another ideea is to use a magnets for moving part and fixed coils, switched electronically using the same system with Hall sensors.
1. Miles says:
I agree with this, it would allow a computer to ramp and time the pulses for maximum effect.
However it would mean either electronics and capacitors on the sled with continuous brushed power, or a ribbon capable of acceleration to follow the sled. Possibly ‘noodle wire’ and higher voltages?
Either way, fun!
2. He already did it in a previous vid with a teensy and hall sensors before each coil, the sled used magnets only.
2. Edgee says:
I’ve always wondered looking at linear rails designs is, doesn’t the sled speed up along the channel which means, that the timing when to trigger the poles need to change? Like, using AC on the coil will switch poles at 50/60Hz. this would not necessarily have the correct pole alignment at the correct time for best efficiency.
1. TG says:
Who needs AC? It has brushes that switch the coil at whatever speed it happens to be going
3. The Commenter Formerly Known As Ren says:
“linear induction motor”
Is that another way of saying “large rubber band”?
B^)
1. Mike says:
I thought it was a fancy way of saying “miniature rail gun” 😂
1. Krzysztof says:
Yeah, but this one is miniature coil gun.
4. robert says:
Linear motors are very cool but, 4mph/44mph = 1.1 times as good as the earlier design. I think I’d sooner take the reliability and longevity improvements of the earlier brushless version and tolerate not getting quite as much speed at the output end.
5. Paul says:
138 g airplane at 10.2 m/s is 7.2 J imparted to the load.
His current numbers don’t add up (claims 90 amps at 48 volts, but measures 2 ohms in the maybe different coil).
But assuming a pair of 2 ohm coils in parallel in the final rig, and 3x12V supply, that’s 36 amps x 36 volts. With a 0.1 second duration that’s 130 J from the source, for an efficiency of 5.6%. Right on par with an aircraft carrier steam catapult.
If you count the energy gone into the sled, that actually 11.4J imparted into the accelerated mass, or 8.8% efficiency.
Pretty good for a design with so much obvious room for improvement.
Interestingly, scaling this up to something like a F-14, you find that it would take the simultaneous full capacity output of all eight nuclear reactors at on the USS Enterprise to launch the plane.
6. concretedog says:
Does anyone know what the video tracking software he uses to characterise the speed of the sleds is? Looks very useful.
7. Yet Another Robert Smith says:
Too bad you can’t use a model rocket engine as JATO.
Maybe a CO2 cartridge?
8. Winston says:
A very cool, just to do it (I assume) project. Bungee cords, of course, are cheaper and more practical.
9. David Leclair says:
Freewheeling diodes one each coil will reduce the arcing. The disadvantage is with the stored energy being released back into the coil, the magnetic fields slightly extended.
10. Dork says:
It’s not an induction motor. It’s a linear brushed DC motor.
Please be kind and respectful to help make the comments section excellent. (Comment Policy)
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# Number 111020121
### Properties of number 111020121
Cross Sum:
Factorization:
3 * 3 * 941 * 13109
Divisors:
Count of divisors:
Sum of divisors:
Prime number?
No
Fibonacci number?
No
Bell Number?
No
Catalan Number?
No
Base 2 (Binary):
Base 3 (Ternary):
Base 4 (Quaternary):
Base 5 (Quintal):
Base 8 (Octal):
Base 32:
39s22p
sin(111020121)
0.24753544568268
cos(111020121)
0.96887883820975
tan(111020121)
0.25548648181858
ln(111020121)
18.525222013119
lg(111020121)
8.0454016963304
sqrt(111020121)
10536.608609985
Square(111020121)
### Number Look Up
Look Up
111020121 which is pronounced (one hundred eleven million twenty thousand one hundred twenty-one) is a great figure. The cross sum of 111020121 is 9. If you factorisate 111020121 you will get these result 3 * 3 * 941 * 13109. 111020121 has 12 divisors ( 1, 3, 9, 941, 2823, 8469, 13109, 39327, 117981, 12335569, 37006707, 111020121 ) whith a sum of 160545060. The number 111020121 is not a prime number. 111020121 is not a fibonacci number. The figure 111020121 is not a Bell Number. 111020121 is not a Catalan Number. The convertion of 111020121 to base 2 (Binary) is 110100111100000100001011001. The convertion of 111020121 to base 3 (Ternary) is 21201220101222100. The convertion of 111020121 to base 4 (Quaternary) is 12213200201121. The convertion of 111020121 to base 5 (Quintal) is 211410120441. The convertion of 111020121 to base 8 (Octal) is 647404131. The convertion of 111020121 to base 16 (Hexadecimal) is 69e0859. The convertion of 111020121 to base 32 is 39s22p. The sine of 111020121 is 0.24753544568268. The cosine of the number 111020121 is 0.96887883820975. The tangent of the number 111020121 is 0.25548648181858. The square root of 111020121 is 10536.608609985.
If you square 111020121 you will get the following result 12325467266854641. The natural logarithm of 111020121 is 18.525222013119 and the decimal logarithm is 8.0454016963304. I hope that you now know that 111020121 is very amazing number!
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https://community.gainsight.com/gainsight/topics/aggregate-to-a-median-value-in-rules
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Aggregate to a Median Value in Rules
• Idea
• Updated 2 weeks ago
• Implemented
I would love to be able to aggregate to a median number in the transformation task of a rule. Right now, it's possible to calculate an average, but average is not always the best way to aggregate a series of numbers. Median would allow me to do more of my reporting in Gainsight, rather than using outside software.
• 752 Points
Posted 10 months ago
Srinivas Muthyam, Employee
• 1,294 Points
Hi Dallis,
Thanks for sharing the Idea. We have planned for few statistical formulas in coming releases(Later after winter release). And those statistical formulas includes median!
• 752 Points
Hi - this release just came out this week and I was disappointed to see that Median calculations only exist as Formula Fields and not as an aggregate calculation. I need to calculate Median values across many rows of data which isn't possible in the current configuration. I hope you plan on introducing aggregate median calculations soon as I'm having to work with other software to build a number of graphs at the moment and it's causing a very disjointed workflow for our CSMs
Jitin Mehndiratta, Product Manager
• 3,212 Points
Hi Dallis,
Can you elaborate more on the use case you are trying to solve? It will help us understand and provide you with the best solution.
Regards,
Jitin
• 752 Points
Hi,
One example is if I wanted to calculate the Median number of days that it takes an account to reach "value". I have an MDA table that tracks when an account hits a certain milestone, records that date and then calculates how many days past it's onboarding it takes to reach that. We have a few outlier accounts that really skew the results if I just calculate the average number of days to value across all accounts, so a median value is a better representation.
I want to be able to calculate the median days to value per month to see if changes to our onboarding process are producing the results we're looking for. So in a bionic rule, I want to be able to take a transform step where I Group By the month, and then Show the Month and a Median Aggregate of all of my calculated Days To Value. Then I could build reports off of this data.
We also have a lot of customers who ask us how they compare against other customers as far as product usage goes. I calculated the average number of actions per month across all accounts and built MDA tables to display that number vs an individual account's number. However, again we've found outliers really skew that number. I need to be able to calculate the Median instead of the Average across all our accounts to get a better representation.
• 684 Points
Hi Dallis,
I am running into the same issue and am wondering if you have found any work arounds?
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https://www.zhycasting.com/time-varying-meshing-stiffness-of-hypoid-gears-of-automobile-drive-axle/
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# Time-varying meshing stiffness of hypoid gears of automobile drive axle
The transmission error of gear motion is the main excitation of gear vibration energy. In most gear equipment, it will cause other components to vibrate and radiate outwards in the form of noise, which makes people feel upset. For the hypoid gear system in the main reducer of automobile drive axle, the previous research mainly focused on the simulation of the gear processing process to obtain the high-precision gear outer contour, through the gear geometric contact analysis to optimize the gear geometric parameters and machine processing parameters, and then calculated the gear root stress, load distribution, motion transmission error, etc, However, these results can not fully reflect the vibration and noise response of the gear system in the process of gear meshing. Many undesirable gear meshing noise problems still exist in the vehicle drive axle. For this reason, many researchers have established the dynamic response mathematical model of the main reducer gear system, studied the influence of the stiffness, damping and inertia mass of each component in the gear system on the gear dynamic system, and established the relationship between the static analysis of the gear and the dynamic response of the system through the meshing stiffness. The biggest difference between the hypoid gear and the spur gear in the three kinds of parameters is the meshing stiffness. The main reason is that the direction of meshing force is basically unchanged when the spur involute gear is meshed. There are more mature calculation methods and empirical formulas [8], but the geometry and meshing process of the hypoid gear are complex, The direction of meshing force and the position of meshing point in the process of gear meshing change with the change of gear rotation angle and loading force [9]. Compared with the spur gear, the research on meshing stiffness of this kind of gear is less.
In foreign countries, teik et al. Established a multi degree of freedom model of quasi double faced gear mesh, and used a constant and trigonometric series to approximate the gear mesh stiffness. Mohamadpour et al. Obtained the time-varying meshing stiffness of hypoid gears through the professional finite element software calyx, simplified the meshing stiffness of gears in the form of trigonometric series, and finally obtained the empirical formula of meshing stiffness of gears. In China, Fang Zongde established the dynamic meshing model of hypoid gear according to the static stress state of gear meshing, but this model did not consider the time-varying characteristics of meshing stiffness in the process of gear meshing. Tang Jinyuan et al. Calculated the meshing stiffness of a single gear tooth from the basic theory of gear stiffness, and then obtained the meshing stiffness of multiple gear teeth when meshing at the same time from the superposition of a single gear tooth. This model ignored the influence of the direction of gear meshing force changing with the change of gear meshing position in the process of hypoid meshing.
It can be concluded from the above analysis that the time-varying meshing stiffness of hypoid gears has put forward corresponding calculation methods in foreign countries, and the meshing stiffness has been approximately processed by Fourier series, and the gear meshing stiffness expression is directly output by software, while the actual Fourier series can not fully represent the meshing stiffness characteristics of gears, and the specific calculation details have not been disclosed, However, there are few researches on the meshing stiffness of hypoid gears based on ABAQUS or ANSYS. In China, there are few researches on the real meshing characteristics of hypoid gears, and there are few applications of calyx software. It is difficult to establish accurate geometric model and finite element model of hypoid gears. Therefore, this paper proposes a complete calculation method of time-varying stiffness of hypoid gears, and uses the current mature Numerical Calculation software MATLAB to get the coordinate points of gear tooth surface, Then, the 3D model of hypoid gear is built in CATIA, and the finite element modeling process of hypoid gear is described in detail by using ABAQUS software, and the time-varying meshing stiffness of hypoid gear is obtained by postprocessing the finite element calculation results. This meshing stiffness can be applied to the dynamic analysis of the gear system without any other assumptions, so it can provide a basis for better prediction of the dynamic response of the hypoid gear drive system of the automobile drive axle.
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http://www.physics.emory.edu/faculty/weeks/idl/kit/plot_hist.pro
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; plot_hist ; ; see http://www.physics.emory.edu/~weeks/idl ; for more information and software updates ; ; A user-friendly way to plot roughly normal distributed histograms ; easily with optional gaussian fits. ; The relevant data can be returned with the optional parameter 'data' ; and coff. set number of terms in fit (must be 3 or more) using nterms ; ; original version by John C. Crocker ; slightly revised, Feb '99, Eric R. Weeks; /log added 2-23-99 ; pro plot_hist,d,data,coff,nterms=nterms,fit=fit,xrange=xrange,binsize=binsize,\$ noplot=noplot,oplot=oplot,jitter = jitter,center=center,log=log if keyword_set(fit) then center=1; ERW 2-22-99 if not keyword_set(jitter) then jitter = 0 if keyword_set(nterms) then begin if (nterms lt 3 or nterms gt 6) then begin print,'Resetting nterms to 6. nterms must be 3, 4, 5, or 6! See gaussfit.' nterms = 6 endif endif if not keyword_set(nterms) then nterms = 3 ; don't know why, just gotta do it data = float(d) ; get a little info about the distribution n=n_elements(data) logn=alog10(n) min = min(data,max=max) stdev = stdev(data,mean) min = min([min,mean-(6*stdev)]) max = max([max,mean+(4*stdev)]) if keyword_set(xrange) then begin min = xrange(0) max = xrange(1) w=where(data ge min and data le max,ndata) logn = alog10(ndata) stdev=stdev(data(w)) endif ; calculate a nice binsize if the user doesn't give us one if not keyword_set(binsize) then begin binsize=stdev/logn ; reduce the binsize slightly so that its either 1,2 or 5 times a power of ten logbin=alog10(binsize) fp = (logbin-fix(logbin)) ip = fix(logbin) if (fp lt 0) then begin fp = fp+1 & ip = ip-1 endif coeff = 10^(fp) if (coeff lt 2) then fp = 0 if (coeff ge 2) AND (coeff lt 5) then fp = alog10(2) if (coeff ge 5) then fp = alog10(5) binsize = 10^(float(ip + fp)) endif ; adjust the min value down so that 0 is at the center of a partition minbin = long(min/binsize) if minbin lt 0 then minbin = minbin-1 min = minbin * binsize if (keyword_set(center) and (not keyword_set(xrange))) then xrange=[min,max] ; oh yeah, calculate the histogram hist = histogram(data,binsize=binsize,min=min-(binsize*jitter),max=max+binsize*jitter) ; pad out the hist with a zero so the plot doesn't hang in the air! hist=[hist,[0]] np = n_elements(hist) ; make an 'x' vector for the plot and the fit x = (findgen(np)*binsize) + min - (binsize*jitter) if (not keyword_set(center)) then begin w=where(hist gt 0,nw) minw=w(0) > 1 maxw=w(nw-1) < (n_elements(hist)-2) hist=hist(minw-1:maxw+1) x=x(minw-1:maxw+1) endif ; plot the histogram if not keyword_set(noplot) then begin if keyword_set(oplot) then begin oplot,x,hist,psym=10 endif else begin w=where(hist gt 0) if (keyword_set(xrange)) then begin if (keyword_set(log)) then begin plot,x(w),hist(w),xrange=xrange,psym=4,/ylog endif else begin plot,x,hist,xrange=xrange,psym=10 endelse endif else begin if (keyword_set(log)) then begin plot,x(w),hist(w),psym=4,/ylog endif else begin plot,x,hist,psym=10 endelse endelse endelse endif data = [transpose(x),transpose(hist)] ; do the fit, if desired if keyword_set(fit) then begin ft = gaussfit(x,hist,coff,nterms=nterms) data = [transpose(x),transpose(hist),transpose(ft)] if not keyword_set(noplot) then begin oplot,x,ft,linestyle=3 dy = !y.crange(1)-!y.crange(0) dx = !x.crange(1)-!x.crange(0) yy = !y.crange(0) xx = !x.crange(0) if (keyword_set(log)) then begin xyouts, xx + (dx*0.15), 10^(yy + (dy*.90)), "Height = "+\$ strcompress(string(coff(0),format='(g9.4)')) ,alignment=0.5 xyouts, xx + (dx*0.15), 10^(yy + (dy*.85)), "Offset = "+\$ strcompress(string(coff(1),format='(g9.4)')) ,alignment=0.5 xyouts, xx + (dx*0.15), 10^(yy + (dy*.80)), "Sigma = "+\$ strcompress(string(coff(2),format='(g9.4)')) ,alignment=0.5 endif else begin xyouts, xx + (dx*0.15), yy + (dy*.90), "Height = "+\$ strcompress(string(coff(0),format='(g9.4)')) ,alignment=0.5 xyouts, xx + (dx*0.15), yy + (dy*.85), "Offset = "+\$ strcompress(string(coff(1),format='(g9.4)')) ,alignment=0.5 xyouts, xx + (dx*0.15), yy + (dy*.80), "Sigma = "+\$ strcompress(string(coff(2),format='(g9.4)')) ,alignment=0.5 endelse if (nterms gt 3) then begin if (keyword_set(log)) then begin xyouts, xx + (dx*0.15), 10^(yy + (dy*.75)), "Polynomial:" ,alignment=0.5 xyouts, xx + (dx*0.15), 10^(yy + (dy*.70)), \$ strcompress(string(coff(3),format='(g10.4)')) ,alignment=0.5 endif else begin xyouts, xx + (dx*0.15), yy + (dy*.75), "Polynomial:" ,alignment=0.5 xyouts, xx + (dx*0.15), yy + (dy*.70), \$ strcompress(string(coff(3),format='(g10.4)')) ,alignment=0.5 endelse if (nterms gt 4) then begin if (keyword_set(log)) then begin xyouts, xx + (dx*0.15), 10^(yy + (dy*.65)), "+ "+\$ strcompress(string(coff(4),format='(g9.4)'))+" x" ,alignment=0.5 endif else begin xyouts, xx + (dx*0.15), yy + (dy*.65), "+ "+\$ strcompress(string(coff(4),format='(g9.4)'))+" x" ,alignment=0.5 endelse if (nterms gt 5) then begin if (keyword_set(log)) then begin xyouts, xx + (dx*0.15), 10^(yy + (dy*.60)), "+ "+\$ strcompress(string(coff(5),format='(g10.4)'))+" x^2" ,alignment=0.5 endif else begin xyouts, xx + (dx*0.15), yy + (dy*.60), "+ "+\$ strcompress(string(coff(5),format='(g10.4)'))+" x^2" ,alignment=0.5 endelse endif ; five endif ; four endif ; three endif endif end
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http://www.jiskha.com/members/profile/posts.cgi?name=Bobby
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# Posts by Bobby
Total # Posts: 740
social studies
citizins of the country elect represenatives for the national legislature. Those represenatives then choose a well-respected member of the majority party to be the one who will make sure that laws are properly carried out. What type of democratic system is described om this ...
social studies
what is the difference between a federal form of government and a confederal form of government A. A federal government is democratic while a condederal government is dictatorial B.A federal governtment has a constitution while a confederal government does not C.A federal ...
social studies
citizins of the country elect represenatives for the national legislature. Those represenatives then choose a well-respected member of the majority party to be the one who will make sure that laws are properly carried out. What type of democratic system is described om this ...
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Cody
# Problem 434. Return the Fibonacci Sequence
Solution 3465418
Submitted on 29 Oct 2020 by Jiseon Shin
This solution is locked. To view this solution, you need to provide a solution of the same size or smaller.
### Test Suite
Test Status Code Input and Output
1 Pass
X = fib_seq(34); assert(isequal(X(end),21) && length(X)==8)
2 Pass
X = fib_seq(35); assert(isequal(X(end),34) && length(X)==9)
3 Pass
X = fib_seq(145); assert(isequal(X(end),144) && length(X)==12)
4 Pass
X = fib_seq(4196); assert(isequal(X(end),4181) && length(X)==19)
5 Pass
X = fib_seq(987419996); assert(isequal(X(end),701408733) && length(X)==44)
6 Pass
X = fib_seq(1134903171); assert(isequal(X(end),1134903170) && length(X)==45)
7 Pass
X = fib_seq(98691443031971); assert(isequal(X(end),72723460248141) && length(X)==68)
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# Control of Climatic Load on Insulated Glass Panes of Glass Structures
### Technical Article
001463 07/20/2017
Loading panes of insulating glass due to climatic effects are clearly regulated in DIN 18008. In the case of the corresponding pane geometry, this load type can also be governing for the ultimate limit state design. The FE design on the entire structure with the space between panes represented as the volume of a gas provides exact results for the analysis. However, a plausibility check is also becoming more and more important. This article shows various options of how to perform these checks.
#### System Basis
A vertical glass pane with a height of h = 1,600 mm and a width of b = 400 mm is examined. The pane is supported by restrained supports on four sides for the horizontal loads as well as by singular supports for the vertical loads. The insulated glass pane with double glazing consists of two edge panes of 3.0 mm each, and a space between panes of 16.0 mm. The examined effect is considered as the “Summer” climatic load case in compliance with DIN 18008‑1 [1].
#### Check of Resulting Gas Pressure
The relationship between the deformation and the resulting pressure in the space between panes can be determined using the general gas equation.
$$\frac{{\mathrm t}_1\;\cdot\;{\mathrm V}_1}{{\mathrm T}_1}\;=\;\frac{{\mathrm t}_2\;\cdot\;{\mathrm V}_2}{{\mathrm T}_2}$$
Due to the deformations calculated in the finite element analysis, there is a change in the gas volume. If these are applied to the system, the following values are obtained:
Load case 2 temperature difference: ΔV = 645.13 cm³ Load case 3 atmospheric pressure difference: ΔV = 186.99 cm³ Load case 4 local altitude difference: ΔV = 704.16 cm³
Using the initial volume and the temperature change, we can now calculate the resulting gas pressure. The following values are used:
p1 = 103 kN/m² V1 = 10,240 cm³ T1 = 292 K T2 = 312 K (LC 2) T2 = T1 = 292 K (LC 3 + LC 4)
Thus, the following results are obtained:
Load case 2 p2 = 103.53 kN/m² Load case 3 p2 = 101.15 kN/m² Load case 4 p2 = 96.37 kN/m²
In comparison with the FE analysis performed in RFEM, this gives exactly the same values.
#### Check Using Applied Surface Load
When comparing the applied load on the entire system with a surface system, the most difficult thing is to convert the surface load to be applied according to DIN 18008‑1 to the surface system. However, such cases are documented in technical literature (in [2], for example), so you can always find help there.
Based on the dimensions of the glass pane and the glass layer structure, the insulating glass factor is calculated. Thus, you can determine the load distribution in both panes.
The following parameters are taken into account:
$$\begin{array}{l}\frac{\mathrm a}{\mathrm b}\;=\;0.25\\{\mathrm B}_\mathrm V\;=\;0.07215\\\mathrm a^\ast\;=\;28.9\;\cdot\;\;\sqrt[4]{\frac{{\mathrm d}_\mathrm{SZR}\;\cdot\;\mathrm d_\mathrm a^3\;\cdot\;\mathrm d_\mathrm i^3}{\left(\mathrm d_\mathrm a^3\;+\;\mathrm d_\mathrm i^3\right)\;\cdot\;{\mathrm B}_\mathrm V}}\;=\;213.77\;\mathrm{mm}\\\mathrm\varphi\;=\;\frac1{1\;+\;\left({\displaystyle\frac{\mathrm a}{\mathrm a^\ast}}\right)^4}\;=\;0.0754\end{array}$$
##### Load Case of Temperature Difference
In the climatic load case of the temperature difference (summer), a temperature change of 20°C is applied. The internal and external pressures are 1.03 bar. This results in a load of:
$$\mathrm q\;=\;0.34\;\cdot\;\mathrm{ΔT}\;=\;6.8\;\mathrm{kN}/\mathrm m^2$$
The resulting load on the single pane is:
$$\mathrm q\;=\;6.8\;\cdot\;0.0754\;=\;0.513\;\mathrm{kN}/\mathrm m^2$$
Based on the surface load on the single pane, it is now possible to perform a “manual” design. However, this is not further pursued in this article.
This surface load can be used to determine the relation between the load and the resulting gas pressure:
$${\mathrm p}_{\mathrm{end},\mathrm{in}}\;=\;103.0\;\mathrm{kN}/\mathrm m^2\;+\;0.513\;\mathrm{kN}/\mathrm m^2\;=\;103.513\;\mathrm{kN}/\mathrm m^2$$
##### Load Case of Atmospheric Pressure Difference
The atmospheric pressure difference is specified by a pressure difference of 0.02 bar. This results in the following load on the entire system:
$$\mathrm q\;=\;103.0\;-\;101.0\;=\;2.0\;\mathrm{kN}/\mathrm m^2$$
The load on a single pane with the same dimensions is therefore:
$$\mathrm q\;=\;2.0\;-\;0.0754\;=\;0.151\;\mathrm{kN}/\mathrm m^2$$
The resulting gas pressure in the space between the panes also results from the sum of the final pressure and the applied surface load:
$${\mathrm p}_{\mathrm{end},\mathrm{in}}\;=\;101.0\;\mathrm{kN}/\mathrm m^2\;+\;0.151\;\mathrm{kN}/\mathrm m^2\;=\;101.151\;\mathrm{kN}/\mathrm m^2$$
##### Load Case of Altitude Difference
In the load case of the altitude difference, the difference of the local altitude of 600 m is applied by default. The resulting load is thus calculated as follows:
$$\mathrm q\;=\;0.012\;\cdot\;600\;=\;7.2\;\mathrm{kN}/\mathrm m^2$$
This is converted to the single system as follows:
$$\mathrm q\;=\;7.2\;\cdot\;0.0754\;=\;0.543\;\mathrm{kN}/\mathrm m²$$
Assuming that the atmospheric pressure at the installation site is about 7.2 kN/m² less than at the production site, the resulting gas pressure in the space between the panes can also be calculated as follows:
$${\mathrm p}_{\mathrm{end},\mathrm{in}}\;=\;(103.0\;\mathrm{kN}/\mathrm m^2\;-\;7.2\;\mathrm{kN}/\mathrm m^2)\;+\;0.543\;=\;96.343\;\mathrm{kN}/\mathrm m^2$$
#### Summary
The comparative calculation has shown that the results of the nonlinear FEM calculation are very similar to the calculation using analytical formulas. The described procedure shows a simple verification of the global computer-aided calculation. Furthermore, this article tried to clarify the relations between the loads on the glass pane and the pressure conditions in the space between the panes.
Using the loads calculated above, you can also verify the deformations and stresses. In this case, it should be noted that the computational calculation is usually based on the nonlinear, large deformation analysis where the analytical formulas were developed according to the linear static analysis. Therefore, there might be small differences in the results.
#### Reference
[1] DIN 18008‑1:2010‑12 (2010). Glass in Building - Design and construction rules - Part 1: Terms and general bases. Berlin: DIN Deutsches Institut für Normung e. V. [2] Albert, A. et col. (2016). Schneider - Bautabellen für Ingenieure, (22nd edition). Cologne: Bundesanzeiger. [3] Feldmeier, F. (2006). Klimabelastung und Lastverteilung bei Mehrscheiben-Isolierglas. Stahlbau, 75(6), 467-478.
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# Wide Tv Unit Argos 80cm Corner Hannah Tall And Cabinet St 90cm Low
By Wilde Becker. Corner Tv Stands. At Friday, July 19th 2019, 19:19:52 PM.
Input the equation into your calculator. Use the smaller measurement of the two walls as both the A and B measurements. For example, if the smaller measurement is 20 inches, you want to input 20 squared (or 20 x 20) plus 20 squared (or 20 x 20). Once you have the answer (in this case it is 800), find the square root of the number and you will have the size, in inches, your TV must be smaller than in order to fit onto a TV stand (in this case it is about 28 inches).
Measure the front of your television. Unlike picture size, which is determined by measuring on a diagonal (lower left to upper right), the width of the TV is ascertained by measuring from the lower right corner to the lower left. Write this number down.
Some corner TV stands are angular, while others are rounded. Rounded corner TV stands are much safer in households with pets or children as the sharp edges of more angular designs could be potentially dangerous.
### Wonderful Small Corner Tv Stand’s Adventure
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# N < B < P(N) ?? (logic set and theory)
• Jul 2nd 2011, 04:09 AM
mIXpRo
N < B < P(N) ?? (logic set and theory)
hi ,
as of the title , is there a B that N < B < P(N) ??
N are the complete numbers set.
P(N) is { x | x in N } ??
if there is , would please give an example , else proof that there isn't , thanks a lot ?
• Jul 2nd 2011, 04:33 AM
Also sprach Zarathustra
Re: N < B < P(N) ?? (logic set and theory)
Quote:
Originally Posted by mIXpRo
hi ,
as of the title , is there a B that N < B < P(N) ??
N are the complete numbers set.
P(N) is { x | x in N } ??
if there is , would please give an example , else proof that there isn't , thanks a lot ?
You mean to Continuum hypothesis - Wikipedia, the free encyclopedia?
• Jul 2nd 2011, 04:39 AM
mIXpRo
Re: N < B < P(N) ?? (logic set and theory)
thanks that what i was looking for :D .
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# If I know a given action will result in a collision, should I allow the action to occur anyway?
I've finished programming my game engine, and now that I've been testing it, I've been noticing some graphics problems.
The big one is that when a player tries to push against a wall, their character will "jiggle" against it as it constantly tries to enter the wall's space and the collision detection constantly pushes it back outside.
What is the proper way to handle such a situation?
My first instinct was, if moving in a certain direction causes a collision, disable movement in that direction until the entity's position has changed. I.e., once an object falls onto a platform, disable gravity until the object is no longer above a platform.
Is there a better way to resolve repeated collision detections between the same two objects in the same overlapping space?
About character collisions with a wall, for instance (I'll use some arbitrary values as example): if you are 10 units (e.g. pixels) away from a wall, and a single step would normally move you 20 units in that direction, the correct behavior is for your character to move 10 units and stop there. If you cancelled the action instead, you'd remain 10 units away from the wall which is not what you'd want.
One way to do this is to move the character by the full amount, then detect the collision and its depth i.e. how much the bounding boxes of the character and the wall are overlapping, and move the character back by that same amount. If you do this correctly you shouldn't be getting any "jiggle" behavior from your character, he'll just press against the wall and stop there. If he's jiggle'ing then you're probably moving him back too much (you're probably moving him back to position he was before the collision, instead of clamping it against the wall boundary).
That being said, I usually see gravity being treated as a special case. In most implementations I've seen, the character stores an onFloor flag and gravity is only applied when that flag is false.
Edit: I guess you could expand this last concept and create additional flags such as touchingRightWall or touchingLeftWall, but only when the boundaries of the character and the wall match exactly. In that case you could perhaps bypass movement altogether. But if the boundaries are not matching precisely, you should let the movement take place and let the collision detection system resolve any intersections.
• That is how I'm handling collisions, currently. The "jiggle" looks more like the sprite is vibrating than a detectable amount of forward/backwards. Commented Dec 27, 2011 at 22:12
• Are you calculating the intersection depth correctly? Try to debug the character's position values and the intersection depth against the ones you'd expect to have. I've done this before with no vibrations at all. Could also be some case of subpixel floating point problem? If that's the case, try rounding your values. And finally, check the XNA's platformer sample. Commented Dec 27, 2011 at 22:16
• Honestly, I think it's just because the IO (which moves the entity) happens at a different time than the Collision detection, which means the two entities overlap until the game engine gets to the collision subsystem's update loop. Commented Dec 27, 2011 at 22:44
• But aren't the IO and collision detection subsystems being updated in the same iteration of your update loop? The only way I could see this making a difference is if your rendering loop were completely separate and could somehow be called between your subsystems updates, or if you're stretching this processing between several frames. Commented Dec 27, 2011 at 23:11
• The entire process will usually be [IO -> Collision Detection -> Collision Resolution] -> [Rendering]. Even if the Update and Render portions are called at different rates, you should never be able to see your objects rendered in some intermediate state (i.e. after IO but before Collision Detection, or after Collision Detection but before Collision Resolution). You'll always see your character rendered after all the stages of Update have been applied, so it shouldn't matter that they're processed in different steps or not. Commented Dec 27, 2011 at 23:21
This problem usually occurs when the code clamps positions before detecting a collision. I.e:
MoveCharacter();
ClampPosition();
CheckCollision();
DrawFrame();
MoveCharacer();
CheckCollision();
ClampPosition();
DrawFrame();
The reason it "jiggles" is because the check happens after the character has been clamped to a "safe" location, but because the character is in a non-colliding location at the time of the check the engine allows for the character to continue to move, then the character is drawn and then the next frame the check fails, Clamp is called, causing the character to "snap" back to the safe location.
• I don't really see how this makes sense. How could the clamp ever happen before checking for collisions if you need the results from your collisions to know against which values to clamp in the first place? Commented Dec 27, 2011 at 23:14
• What's really happening is the character is being drawn before being moved to the correct location on the next frame instead of being moved to the correct location before the current frame is drawn. Commented Dec 27, 2011 at 23:33
I've had the same problem recently and fixed it like this:
move the object, checking for a collision, if collision is found, place object outside of colliding object. draw the object.
The player could keep running into the wall without any jiggling because before he is drawn, he will be placed at the side of the object.
Using vector / plane mathematics you can calculate exact point of collision with wall that will happen in next frame. This assumes that your character has velocity which you can use to calculate movement vector that will happen during next frame. After finding point of collision you can calculte sliding vector that will move your character along wall, this vector actually can collide with another wall.
So back to your question, you should calculate the exact position of collision where your character should be located and what actually should happend then. If this is a wall and its FPS game, then let character slide along it.
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# [Numpy-discussion] Fastest way to compute summary statistics for a specific axis
Sebastian Berg sebastian at sipsolutions.net
Mon Mar 16 12:12:56 EDT 2015
```On Mo, 2015-03-16 at 15:53 +0000, Dave Hirschfeld wrote:
> I have a number of large arrays for which I want to compute the mean and
> standard deviation over a particular axis - e.g. I want to compute the
> statistics for axis=1 as if the other axes were combined so that in the
> example below I get two values back
>
> In [1]: a = randn(30, 2, 10000)
>
> For the mean this can be done easily like:
>
> In [2]: a.mean(0).mean(-1)
> Out[2]: array([ 0.0007, -0.0009])
>
If you have numpy 1.7+ (which I guess by now is basically always the
case), you can do a.mean((0, 1)). Though it isn't actually faster in
this example, probably because it has to use buffered iterators and
things, but I would guess the performance should be much more stable
depending on memory order, etc. then any other method.
- Sebastian
>
> ...but this won't work for the std. Using some transformations we can
> come up with something which will work for either:
>
> In [3]: a.transpose(2,0,1).reshape(-1, 2).mean(axis=0)
> Out[3]: array([ 0.0007, -0.0009])
>
> In [4]: a.transpose(1,0,2).reshape(2, -1).mean(axis=-1)
> Out[4]: array([ 0.0007, -0.0009])
>
>
> If we look at the performance of these equivalent methods:
>
> In [5]: %timeit a.transpose(2,0,1).reshape(-1, 2).mean(axis=0)
> 100 loops, best of 3: 14.5 ms per loop
>
> In [6]: %timeit a.transpose(1,0,2).reshape(2, -1).mean(axis=-1)
> 100 loops, best of 3: 5.05 ms per loop
>
>
> we can see that the latter version is a clear winner. Investigating
> further, both methods appear to copy the data so the performance is
> likely down to better cache utilisation.
>
> In [7]: np.may_share_memory(a, a.transpose(2,0,1).reshape(-1, 2))
> Out[7]: False
>
> In [8]: np.may_share_memory(a, a.transpose(1,0,2).reshape(2, -1))
> Out[8]: False
>
>
> Both methods are however significantly slower than the initial attempt:
>
> In [9]: %timeit a.mean(0).mean(-1)
> 1000 loops, best of 3: 1.2 ms per loop
>
> Perhaps because it allocates a smaller temporary?
>
> For those who like a challenge: is there a faster way to achieve what
> I'm after?
>
>
> Cheers,
> Dave
>
>
>
>
>
>
>
>
>
> _______________________________________________
> NumPy-Discussion mailing list
> NumPy-Discussion at scipy.org
> http://mail.scipy.org/mailman/listinfo/numpy-discussion
>
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Home >>Python math Module >Python math.remainder() Method
# Python math.remainder() Method
### Python math.remainder() Method
Python math.remainder() method in python is used to returns the remainder of the first number with respect to the second number in the float of given numbers and , it accepts two numbers (integer or float).
Syntax:
`math.remainder(x, y)`
### Parameter Values
Parameter Description
x It is required a number you want to divide.
y It is required a number you want to divide with.
Here is an Example of Python math.remainder() Method:
``````
import math
print (math.remainder(5, 1))
print (math.remainder(14, 10))
```
```
Output:
0.0
4.0
Example 2:
``````
import math
x = 12
y = 5
print(math.remainder(x,y))
x = 12.4
y = 5.7
print(math.remainder(x,y))
```
```
Output:
2.0
1.0
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# At a large university, the mean amount spent by students for cell phone service is \$38.90 per month with a standard deviation of \$3.64 per month. Consider a group of 44 randomly chosen university students. What is the probability that the mean amount of their monthly cell phone bills differs from the mean for the university by more than \$1?
Question
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At a large university, the mean amount spent by students for cell phone service is \$38.90 per month with a standard deviation of \$3.64 per month. Consider a group of 44 randomly chosen university students. What is the probability that the mean amount of their monthly cell phone bills differs from the mean for the university by more than \$1?
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## What is the side length of a cube with a volume of 64 mm? Cube V=93 1. Substitute the value into the formula: 2. Undo the
Question
What is the side length of a cube with a volume of 64 mm?
Cube V=93
1. Substitute the value into the formula:
2. Undo the cube by applying the cube root
64 = 93
3164 = 3153
What is the side length of the cube?
mm
in progress 0
1 week 2022-01-14T14:58:05+00:00 1 Answer 0 views 0
Step-by-step explanation:
Data
side length = ?
Volume = 93 mm³
To find the answer, look for the formula to calculate the volume of a cube, substitute values, and simplify.
-Formula
Volume of a cube = side³
-Substitution
93 = side³
-Simplification
∛93 = side
-Result
side = 4.53 mm
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1. We went through the Ackermann function before, here, Johan.
http://www.thinbasic.com/community/s...searchid=60624
Maybe it's time to go back to it again, and at least calculate, A(4, 2).
[A(4, 3) = 2^2^65536 - 3.
From within the REPL in DrRacket (using unlimited memory), I tried to execute,
(define a (- (expt 2 (expt 2 65536)) 3)),
and, DrRacket crashed.]
2. Now I know why DrRacket crashed when I tried to evaluate the value of A(4, 3), i.e., 2^2^65536 - 3.
(I think it's easy to become confused about how big a number is, when it is written as an exponent raised to an exponent.)
In the same article,
http://en.wikipedia.org/wiki/Ackermann_function ,
it says that A(4, 3) is approximately equal to,
10^(6.031 * 10^19727).
So, how big is that number?
Let's forget about the factor, 6.031. It doesn't matter, since as we will see, even without it, the number is effectively infinite.
So instead, just consider the number,
10^10^19727.
Let's approach it like this.
10^10^6,
then, that number would have one million and one (1,000,001) (decimal) digits.
10^10^9,
then, that number would have one billion and one (1,000,000,001) digits.
10^10^12,
then, that number would have one trillion and one (1,000,000,000,001) digits.
10^10^303,
then, that number would have one centillion and one (10^303 + 1) digits. "cent", is the largest number prefix I can find, -->
http://en.wikipedia.org/wiki/Names_of_large_numbers .
And, from the same article, the largest named number is, "Googolplex". But it is only equal to,
10^10^100.
Anyway,
10^10^19727,
would have, 10^19727 + 1, digits.
I think, we have no way to imagine how many digits that is, because, there is nothing to compare it to.
In fact, if I am not mistaken, then,
10^10^19727 = 10^10^19627 * Googolplex.
I would be surprised if any computer can ever calculate, A(4, 3).
Interestingly, in the Ackermann article, it also says,
"A(4, 2), which appears as a decimal expansion in several web pages, cannot possibly be computed by simple recursive application of the Ackermann function in any tractable amount of time."
3. I did the calculation in Racket for A(4, 1). It took approximately 48 minutes.
If I tried A(4, 2), I would have to wait forever (if I didn't overflow the stack).
```; code -------------------------------------------------------------
#lang scheme
(define (a m n)
(cond
((= m 0) (+ n 1))
((and (> m 0) (= n 0)) (a (- m 1) 1))
((and (> m 0) (> n 0)) (a (- m 1) (a m (- n 1))))))
(define (ack-time f x y)
(let ([t1 (current-milliseconds)])
(f x y)
(let ([t2 (current-milliseconds)])
(/ (- t2 t1) 1000.0))))
; REPL interaction -------------------------------------------------
Welcome to DrRacket, version 5.1 [3m].
Language: scheme.
> (ack-time a 4 1)
2862.129
>
```
4. Originally Posted by danbaron
In fact, if I am not mistaken, then,
10^10^19727 = 10^10^19627 * Googolplex.
Absolutely correct. A googol, defined as 10^100, is already an astounding number. The estimated number of atoms in the universe is 10^80. The factor between the two is not 20, as many non-mathematically-inclined people think, but 10^20: 100,000,000,000,000,000,000. A googolplex is the number one followed by 10^100 zeroes.
And even that number is dwarfed by A(4,3). Actually, "dwarfed" is the wrong word because that word indicates a certain size relation. A difference in magnitude of 10^10^19627 is incomprehensible.
For interest, extended-precision floating-point values (Ext) have a maximum magnitude of 10^4932. Even that number is basically meaningless.
5. ## Funny Fibonacci numbers
While testing the fixed-point stuff I saw that F(65) has '65' as its last two digits. And of course F(0)=0 and F(1)=1.
```Uses "Console"
Module "BigInt"
Alias String As BigInt
BigInt a
String b,c
BigInt_SetFormat(0,".",FALSE)
a=BigInt_FromInteger(0)
Do
b=BigInt_ToString(BigInt_Fib(a))
c=BigInt_ToString(a)
If RIGHT\$(b,Len(c))=c Then Print c+\$SPC
a=BigInt_Inc(a)
Loop
```
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# Elixir v1.4.4 Float View Source
Functions for working with floating point numbers.
# Link to this section Summary
## Functions
Rounds a float to the smallest integer greater than or equal to `num`
Rounds a float to the largest integer less than or equal to `num`
Parses a binary into a float
Returns a pair of integers whose ratio is exactly equal to the original float and with a positive denominator
Rounds a floating point value to an arbitrary number of fractional digits (between 0 and 15)
Returns a charlist which corresponds to the text representation of the given float
Returns a binary which corresponds to the text representation of the given float
# Link to this section Functions
Link to this function ceil(number, precision \\ 0) View Source
`ceil(float, 0..15) :: float`
Rounds a float to the smallest integer greater than or equal to `num`.
`ceil/2` also accepts a precision to round a floating point value down to an arbitrary number of fractional digits (between 0 and 15).
The operation is performed on the binary floating point, without a conversion to decimal.
The behaviour of `ceil/2` for floats can be surprising. For example:
``````iex> Float.ceil(-12.52, 2)
-12.51``````
One may have expected it to ceil to -12.52. This is not a bug. Most decimal fractions cannot be represented as a binary floating point and therefore the number above is internally represented as -12.51999999, which explains the behaviour above.
This function always returns floats. `Kernel.trunc/1` may be used instead to truncate the result to an integer afterwards.
## Examples
``````iex> Float.ceil(34.25)
35.0
iex> Float.ceil(-56.5)
-56.0
iex> Float.ceil(34.251, 2)
34.26``````
Link to this function floor(number, precision \\ 0) View Source
`floor(float, 0..15) :: float`
Rounds a float to the largest integer less than or equal to `num`.
`floor/2` also accepts a precision to round a floating point value down to an arbitrary number of fractional digits (between 0 and 15). The operation is performed on the binary floating point, without a conversion to decimal.
The behaviour of `floor/2` for floats can be surprising. For example:
``````iex> Float.floor(12.52, 2)
12.51``````
One may have expected it to floor to 12.52. This is not a bug. Most decimal fractions cannot be represented as a binary floating point and therefore the number above is internally represented as 12.51999999, which explains the behaviour above.
This function always returns a float. `Kernel.trunc/1` may be used instead to truncate the result to an integer afterwards.
## Examples
``````iex> Float.floor(34.25)
34.0
iex> Float.floor(-56.5)
-57.0
iex> Float.floor(34.259, 2)
34.25``````
Link to this function parse(binary) View Source
`parse(binary) :: {float, binary} | :error`
Parses a binary into a float.
If successful, returns a tuple in the form of `{float, remainder_of_binary}`; when the binary cannot be coerced into a valid float, the atom `:error` is returned.
If the size of float exceeds the maximum size of `1.7976931348623157e+308`, the `ArgumentError` exception is raised.
If you want to convert a string-formatted float directly to a float, `String.to_float/1` can be used instead.
## Examples
``````iex> Float.parse("34")
{34.0, ""}
iex> Float.parse("34.25")
{34.25, ""}
iex> Float.parse("56.5xyz")
{56.5, "xyz"}
iex> Float.parse("pi")
:error``````
Returns a pair of integers whose ratio is exactly equal to the original float and with a positive denominator.
## Examples
``````iex> Float.ratio(3.14)
{7070651414971679, 2251799813685248}
iex> Float.ratio(-3.14)
{-7070651414971679, 2251799813685248}
iex> Float.ratio(1.5)
{3, 2}
iex> Float.ratio(-1.5)
{-3, 2}
iex> Float.ratio(16.0)
{16, 1}
iex> Float.ratio(-16.0)
{-16, 1}``````
Link to this function round(float, precision \\ 0) View Source
`round(float, 0..15) :: float`
Rounds a floating point value to an arbitrary number of fractional digits (between 0 and 15).
The rounding direction always ties to half up. The operation is performed on the binary floating point, without a conversion to decimal.
This function only accepts floats and always returns a float. Use `Kernel.round/1` if you want a function that accepts both floats and integers and always returns an integer.
The behaviour of `round/2` for floats can be surprising. For example:
``````iex> Float.round(5.5675, 3)
5.567``````
One may have expected it to round to the half up 5.568. This is not a bug. Most decimal fractions cannot be represented as a binary floating point and therefore the number above is internally represented as 5.567499999, which explains the behaviour above. If you want exact rounding for decimals, you must use a decimal library. The behaviour above is also in accordance to reference implementations, such as “Correctly Rounded Binary-Decimal and Decimal-Binary Conversions” by David M. Gay.
## Examples
``````iex> Float.round(12.5)
13.0
iex> Float.round(5.5674, 3)
5.567
iex> Float.round(5.5675, 3)
5.567
iex> Float.round(-5.5674, 3)
-5.567
iex> Float.round(-5.5675)
-6.0
iex> Float.round(12.341444444444441, 15)
12.341444444444441``````
Link to this function to_charlist(float) View Source
`to_charlist(float) :: charlist`
Returns a charlist which corresponds to the text representation of the given float.
It uses the shortest representation according to algorithm described in “Printing Floating-Point Numbers Quickly and Accurately” in Proceedings of the SIGPLAN ‘96 Conference on Programming Language Design and Implementation.
## Examples
``````iex> Float.to_charlist(7.0)
'7.0'``````
Link to this function to_string(float) View Source
`to_string(float) :: String.t`
Returns a binary which corresponds to the text representation of the given float.
It uses the shortest representation according to algorithm described in “Printing Floating-Point Numbers Quickly and Accurately” in Proceedings of the SIGPLAN ‘96 Conference on Programming Language Design and Implementation.
## Examples
``````iex> Float.to_string(7.0)
"7.0"``````
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