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# All Science Fair Projects
## Science Fair Project Encyclopedia for Schools!
Search Browse Forum Coach Links Editor Help Tell-a-Friend Encyclopedia Dictionary
# Science Fair Project Encyclopedia
For information on any area of science that interests you,
enter a keyword (eg. scientific method, molecule, cloud, carbohydrate etc.).
Or else, you can start by choosing any of the categories below.
# Cauchy principal value
In mathematics, the Cauchy principal value of certain improper integrals is defined as either
• the finite number
$\lim_{\varepsilon\rightarrow 0+} \left(\int_a^{b-\varepsilon} f(x)\,dx+\int_{b+\varepsilon}^c f(x)\,dx\right)$
where b is a point at which the behavior of the function f is such that
$\int_a^b f(x)\,dx=\pm\infty$
for any a < b and
$\int_b^c f(x)\,dx=\mp\infty$
for any c > b (one sign is "+" and the other is "−").
or
• the finite number
$\lim_{a\rightarrow\infty}\int_{-a}^a f(x)\,dx$
where
$\int_{-\infty}^0 f(x)\,dx=\pm\infty$
and
$\int_0^\infty f(x)\,dx=\mp\infty$
(again, one sign is "+" and the other is "−").
In some cases it is necessary to deal simultaneously with singularities both at a finite number b and at infinity. This is usually done by a limit of the form
$\lim_{\varepsilon \rightarrow 0+}\int_{b-1/\varepsilon}^{b-\varepsilon} f(x)\,dx+\int_{b+\varepsilon}^{b+1/\varepsilon}f(x)\,dx.$
## Nomenclature
The Cauchy principal value of a function f can take on several nomenclatures, varying for different authors. These include (but are not limited to): $PV \int f(x)dx$, P, P.V., $\mathcal{P}$, Pv, (CPV) and V.P..
## Examples
Consider the difference in values of two limits:
$\lim_{a\rightarrow 0+}\left(\int_{-1}^{-a}\frac{dx}{x}+\int_a^1\frac{dx}{x}\right)=0,$
$\lim_{a\rightarrow 0+}\left(\int_{-1}^{-a}\frac{dx}{x}+\int_{2a}^1\frac{dx}{x}\right)=-\log_e 2.$
The former is the Cauchy principal value of the otherwise ill-defined expression
$\int_{-1}^1\frac{dx}{x}{\ } \left(\mbox{which}\ \mbox{gives}\ -\infty+\infty\right).$
Similarly, we have
$\lim_{a\rightarrow\infty}\int_{-a}^a\frac{2x\,dx}{x^2+1}=0,$
but
$\lim_{a\rightarrow\infty}\int_{-2a}^a\frac{2x\,dx}{x^2+1}=-\log_e 4.$
The former is the principal value of the otherwise ill-defined expression
$\int_{-\infty}^\infty\frac{2x\,dx}{x^2+1}{\ } \left(\mbox{which}\ \mbox{gives}\ -\infty+\infty\right).$
These pathologies do not afflict Lebesgue-integrable functions, that is, functions the integrals of whose absolute values are finite.
03-10-2013 05:06:04
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# 765907
## 765,907 is a prime number. Like all primes greater than two, it is odd and has no factors apart from itself and one.
What does the number 765907 look like?
As a prime, it is not composed of any other numbers and has no internal structure.
765907 is a prime number. Like all primes (except two), it is an odd number.
## Prime factorization of 765907:
### 765907
See below for interesting mathematical facts about the number 765907 from the Numbermatics database.
### Names of 765907
• Cardinal: 765907 can be written as Seven hundred sixty-five thousand, nine hundred seven.
### Scientific notation
• Scientific notation: 7.65907 × 105
### Factors of 765907
• Number of distinct prime factors ω(n): 1
• Total number of prime factors Ω(n): 1
• Sum of prime factors: 765907
### Divisors of 765907
• Number of divisors d(n): 2
• Complete list of divisors:
• Sum of all divisors σ(n): 765908
• Sum of proper divisors (its aliquot sum) s(n): 1
• 765907 is a deficient number, because the sum of its proper divisors (1) is less than itself. Its deficiency is 765906
### Bases of 765907
• Binary: 101110101111110100112
• Base-36: GEZ7
### Squares and roots of 765907
• 765907 squared (7659072) is 586613532649
• 765907 cubed (7659073) is 449291410950597643
• The square root of 765907 is 875.1611280217
• The cube root of 765907 is 91.4938732001
### Scales and comparisons
How big is 765907?
• 765,907 seconds is equal to 1 week, 1 day, 20 hours, 45 minutes, 7 seconds.
• To count from 1 to 765,907 would take you about twenty hours!
This is a very rough estimate, based on a speaking rate of half a second every third order of magnitude. If you speak quickly, you could probably say any randomly-chosen number between one and a thousand in around half a second. Very big numbers obviously take longer to say, so we add half a second for every extra x1000. (We do not count involuntary pauses, bathroom breaks or the necessity of sleep in our calculation!)
• A cube with a volume of 765907 cubic inches would be around 7.6 feet tall.
### Recreational maths with 765907
• 765907 backwards is 709567
• The number of decimal digits it has is: 6
• The sum of 765907's digits is 34
• More coming soon!
#### Copy this link to share with anyone:
MLA style:
"Number 765907 - Facts about the integer". Numbermatics.com. 2024. Web. 16 June 2024.
APA style:
Numbermatics. (2024). Number 765907 - Facts about the integer. Retrieved 16 June 2024, from https://numbermatics.com/n/765907/
Chicago style:
Numbermatics. 2024. "Number 765907 - Facts about the integer". https://numbermatics.com/n/765907/
The information we have on file for 765907 includes mathematical data and numerical statistics calculated using standard algorithms and methods. We are adding more all the time. If there are any features you would like to see, please contact us. Information provided for educational use, intellectual curiosity and fun!
Keywords: Divisors of 765907, math, Factors of 765907, curriculum, school, college, exams, university, Prime factorization of 765907, STEM, science, technology, engineering, physics, economics, calculator, seven hundred sixty-five thousand, nine hundred seven.
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## Warm up- introduction of matrices.pdf - Section 1: Warm Up
Warm up- introduction of matrices.pdf
Warm up- introduction of matrices.pdf
# Organizing and Calculating Data with Matrices
Unit 9: Statistics
Lesson 1 of 10
## Big Idea: To understand the vocabulary of Matrices including rows, columns, and addresses to set up a Real World Problem. As well as perform Multi-step Operations to that problem.
Print Lesson
2 teachers like this lesson
Standards:
Subject(s):
Statistics, Math, Real world problems, Matrices, Adding Matrices, Subtracting Matrices, Scalar Multiplication, Find the percent of
50 minutes
### Rhonda Leichliter
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Last year, a certain company began manufacturing product X : GMAT Data Sufficiency (DS)
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# Last year, a certain company began manufacturing product X
Author Message
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### Hide Tags
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Joined: 20 May 2012
Posts: 20
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Kudos [?]: 26 [0], given: 9
Last year, a certain company began manufacturing product X [#permalink]
### Show Tags
05 Jun 2012, 14:09
1
This post was
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Question Stats:
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### HideShow timer Statistics
Last year, a certain company began manufacturing product X and sold every unit of product X that it produced. Last year the company's total expenses for manufacturing product X were equal to $100,000 plus 5% of the company's total revenue from all units of product X sold. If the company made a profit on product X last year, did the company sell more than 21000 units of product X? (1) The company's total revenue from the sale of product X last year was greater than$110,000
(2) For each unit of product X sold last year, the company's revenue was $5 [Reveal] Spoiler: OA Kellogg MMM ThreadMaster Joined: 28 Mar 2012 Posts: 314 Location: India GMAT 1: 640 Q50 V26 GMAT 2: 660 Q50 V28 GMAT 3: 730 Q50 V38 Followers: 29 Kudos [?]: 399 [2] , given: 23 Re: Yes/No data DS [#permalink] ### Show Tags 05 Jun 2012, 21:09 2 This post received KUDOS Hi, It is given that the total expense for the company = 100,000 + 5% of total revenue (let say, x) Also, the project made profit. Profit = Total revenue - expenses = x - (100000 + 0.05x) = 0.95x - 100000 Since profit > 0 therefore, 0.95x - 100000 > 0 ...(a) using (1) The company's total revenue from the sale of product X last year was greater than$110,000,
we can only check whether 0.95x - 100000 is positive or not. Insufficient.
using (2) For each unit of product X sold last year, the company's revenue was \$5,
lets find the minimum number of units,
number of units = total revenue/5 = x/5
or 0.95(x/5) - 100000 >0 ....(from (a))
or x > 21052, thus x is greater than 21000. Sufficient.
Regards,
Re: Yes/No data DS [#permalink] 05 Jun 2012, 21:09
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# Trouble with Multiplexers
Thread Starter
#### Davey
Joined Oct 3, 2005
3
I'm working on my DSD Homework here and I'm having a little trouble with the following problem:
Originally posted by Problem 4-18 Logic and Computer Design Fundamentals (Mano 3rd ED)
Construct a quad 9-to-1 line multiplexer with four single 8-to-1 line multiplexers and one quadruple 2-to-1 multiplexer. The multiplexers should be interconnected and inputs labeled so that the selection codes 0000 through 1000 can be directly applied to the multiplexer selection inputs without added logic.
I know how to for example make a 32-to-1 Multiplexer out of two 16-to-1 Multiplexers. That's easy enough, but then again that's a very clear statement. I'm having trouble understanding what is meant by "quad" 9-to-1 multiplexer or "quadruple" 2-to-1 multiplexers. Could I have just have easily said that a 32-to-1 mux could be made from a "dual" 16-to-1 mux?
The book is very vague on the description of adding terms like "dual" "quad" etc to the beginning of multiplexers...in fact I can't figure out the meaning at all. My only bright ideas are that they're talking about IC Chips (a quad 2-to-1 package meaning there's four 2-to-1 mux's on the IC with a common select). Either that they could be telling me to construct "four" 9-to-1 multiplexers with one 8-to-1 line mux's and four 2-to-1 line mux's. I can do that I think.
Does anyone have any hints or explinations on where I should start? Could someone clarify what the question is asking for? I think I could solve it from there.
Thank you,
Davey
#### nomurphy
Joined Aug 8, 2005
567
It means for you to design four 9-1 muxes, using four 8-1's and four 2-1's.
Should be simple enough, the output of each 8-1 is muxed with the 9th line by the 2-1's, just pay attention to the adressing.
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Energy stores accumulate working
• Ee04PNnugget01—see this in its SPT context (PN thread)
The power in a pathway determines the quantity of energy shifted to a store each second
Connect up a circuit and leave it running in a steady state – lamps glow and resistors warm. How much they glow or warm depends both on the electrical current and on the electrical potential difference.
So much for the physical description; now to re-describe it in terms of pathways.
Electrical working at a constant rate occurs in an electrical loop in which there is a constant electrical current. Every device in this loop that has a potential difference across it shifts energy. The power in the electrical pathway tells you how much energy is shifted every second by that device. This power is set by the potential difference across the device and the current through the device.
Power in a pathway here can be equal to power in a pathway over there
The power in this pathway shifts energy to or from a store, adding to or subtracting from the energy already in the store. So a constant power in a pathway leads to a steady accumulation in a store. This accumulation can be either positive or negative.
As a starting example, use a simple large-scale electrical loop that does something useful: a hydro-electric generator, deep in the Welsh countryside, lights a domestic lamp in the West Midlands. Model the circuit, simplifying as much as possible, with wires of negligible (as near to zero as makes no difference) resistance. Then there are only two devices in the circuit – places where pathways are switched. The system is designed to optimise the following switches: at the generator, from the mechanical pathway to the electrical pathway; at the lamp, from the electrical pathway to the heating by radiation pathway (again, visible radiations only – that is our particular interest because of how we evolved).
In both locations the device will not be perfect – there will be some switching to other pathways, chiefly heating by particles as the water is warmed by churning through the turbine, and the wires in the generator are warmed by the current driven through them, as the air surrounding the lamp is warmed.
Describing the process in terms of energy shifted, you end up with an even more abstract picture, as the gravitational store is depleted and the thermal store augmented. This is the view that was emphasised in the earlier SPT: Energy and SPT: Electric circuits topics.
Constant power implies steady rate of accumulation
A larger constant power in any pathway accumulates energy in a store at a larger steady rate. This accumulation can be either positive or negative, thereby augmenting or depleting the energy in the store.
For the electrical pathway, you choose how to make the power larger: set the character of the pathway by using either a larger current or a larger potential difference, or both.
Accumulating over more time, at the same constant rate, also leads to larger quantities of energy being shifted to or from stores.
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Payal Tandon
Co-founder, e-GMAT
Welcome to e-GMAT Support!
I am Payal, Co-Founder of e-GMAT.
Feel free to ask any Query.
We will be contacting you soon on
## OGQR 2020: Question No. 77
If half the result obtained when 2 is subtracted from 5x is equal to the sum of 10 and 3x, what is the value of x?
Source OGQR 2020 Type Problem Solving Topic Algebra Sub-Topic Linear Equation Difficulty Easy-Medium
### Solution
#### Given
In this question, we are given
• 1/2 the result obtained when 2 is subtracted from 5x is equal to the sum of 10 and 3x.
#### To Find
We need to determine
• The value of x.
#### Approach & Working
1/2 the result obtained when 2 is subtracted from 5x = 1/2(5x – 2)
Sum of 10 and 3x = (10 + 3x)
Therefore, as per the question,
• 1/2 (5x – 2) = (10 + 3x)
Or, 5x – 2 = 20 + 6x
Or, x = -2 – 20 = -22
Hence, the correct answer is option A.
If you are planning to take the GMAT, we can help you with a personalized study plan and give you access to quality online content to prepare. Write to us at acethegmat@e-gmat.com. We are the most reviewed GMAT prep company on gmatclub with more than 2400 reviews and are the only prep company that has delivered more than 700+ scores than any other GMAT club partner. Why don’t you take a free trial and judge for yourself?
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Let k be an algebraically closed field (for example C). We are going to make kn into a topological space. In the case of k=C there is already a natural way to do this (since it is a metric space) but the topology we give here, called the Zariski topology is coarser than the usual topology, it has fewer open sets. The advantage of this approach is that it works for all k, not just C. Once we have this topology on kn then it becomes possible to talk about algebraic geometry.
We will be using the polynomial ring k[x1,...,xn]. If f(x)=f(x1,..., xn) in k[x1,...,xn] is a polynomial then we can evaluate it at a point a=(a1,..an) in kn to give an element of k denoted by f(a) or f(a1,...,an). If S is a nonempty set of polynomials then define
Z(S) = { a in kn : f(a)=0 for all f in S }
For example, if n=2 and S={x-y} then Z(S) is the set { (a,a) : a in k }. This is just the line y=x. Note that to get an idea of what's going on you can sketch the real points of of closed sets in C2 or C3. Here's another example for n=2 if we take S={ xy } then this time Z(S) is the union of the x-axis and the y-axis. For some interesting examples take a look at plane algebraic curve. One final example. If we take S={x,y} then Z(S) is a single point, namely the the origin {(0,0)}
Definition Any subset of kn of the form Z(S) is called an affine algebraic variety.
In fact we are going to show that the sets of this form are the closed sets for a topology on kn called the Zariski topology.
First notice that if I is the ideal of k[x1,...,xn] generated by S then we have Z(I)=Z(S). For since S<=I (S is a subset of I) we obviously have Z(I)<=Z(S). Suppose that a is in Z(S) and let f in I. Then
f=f1g1 + ... + fmgm
for some fi in S and polynomials gi. Evidently then f(a)=0 and we have Z(I)=Z(S). Since k[x1,...,xn] is Noetherian by The Hilbert Basis Theorem we know that any ideal has finitely many generators. This means that in the definition of affine varieties set we only need to consider finite sets of polynomials S.
Theorem The affine algebraic varieties in kn are the closed sets of a topology.
Proof: We just have to check the axioms for a topology
1. The empty set is a variety. For this is equals to Z(k[x1,...,xn]) as no point is a zero of the constant polynomial 1.
2. The whole set kn is a variety. For this equals Z({0}), by definition.
3. A finite union of closed sets is closed. By induction we just have to think about two sets Z(S) and Z(T). As above we may suppose that the subsets S,T are ideals. Consider Z(ST). Since ST is a subset of S and T we have Z(S) U Z(T) is a subset of Z(ST). Let's show the reverse inclusion. Suppose that a is in Z(ST) and that WLOG it is not in Z(S). Since a is killed by ST then it is killed by every product fg with f in S and g in T. But if f(a)g(a)=0 then either f(a)=0 or g(a)=0. Since a is not in Z(S) there exists f in S such that f(a) is nonzero. Thus a is in Z(T), as required.
4. An arbitrary intersection of closed sets is closed. For suppose that we have a family Z(Si) of closed sets. As usual we can assume that each Si is an ideal. Let S be the sum of the ideals Si, itself an ideal. Then I claim that Z(S) is the intersection of the Z(Si). Since each Si is a subset of S we see that Z(S) is contained in the intersection of the Z(Si). On the other hand, suppose that a is in the intersection. Then a is killed by the polynomials in each Si so it is killed by any finite linear combination of such polynomials, that is a is in Z(S).
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CC-MAIN-2018-30
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http://www.gradesaver.com/textbooks/math/algebra/intermediate-algebra-12th-edition/chapter-7-section-7-4-adding-and-subtracting-radical-expressions-7-4-exercises-page-466/31
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Intermediate Algebra (12th Edition)
$19\sqrt[4]{2}$
$\bf{\text{Solution Outline:}}$ To simplify the given radical expression, $5\sqrt[4]{32}+3\sqrt[4]{162} ,$ simplify first each term by expressing the radicand as a factor that is a perfect power of the index. Then, extract the root. Finally, combine the like radicals. $\bf{\text{Solution Details:}}$ Expressing the radicand as an expression that contains a factor that is a perfect power of the index results to \begin{array}{l}\require{cancel} 5\sqrt[4]{16\cdot2}+3\sqrt[4]{81\cdot2} \\\\= 5\sqrt[4]{(2)^4\cdot2}+3\sqrt[4]{(3)^4\cdot2} .\end{array} Extracting the roots of the factor that is a perfect power of the index results to \begin{array}{l}\require{cancel} 5(2)\sqrt[4]{2}+3(3)\sqrt[4]{2} \\\\= 10\sqrt[4]{2}+9\sqrt[4]{2} .\end{array} By combining like radicals, the expression above is equivalent to \begin{array}{l}\require{cancel} (10+9)\sqrt[4]{2} \\\\= 19\sqrt[4]{2} .\end{array}
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https://codedump.io/share/o0zdEN5vpDpO/1/d-is-less-efficient-than-0-9
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weston - 1 month ago 14
C# Question
# \d is less efficient than [0-9]
[0123456789]
in a regular expression rather than
[0-9]
or
\d
. I said it was probably more efficient to use a range or digit specifier than a character set.
I decided to test that out today and found out to my surprise that (in the C# regex engine at least)
\d
appears to be less efficient than either of the other two which don't seem to differ much. Here is my test output over 10000 random strings of 1000 random characters with 5077 actually containing a digit:
Regular expression \d took 00:00:00.2141226 result: 5077/10000
Regular expression [0-9] took 00:00:00.1357972 result: 5077/10000 63.42 % of first
Regular expression [0123456789] took 00:00:00.1388997 result: 5077/10000 64.87 % of first
It's a surprise to me for two reasons:
1. I would have thought the range would be implemented much more efficiently than the set.
2. I can't understand why
\d
is worse than
[0-9]
. Is there more to
\d
than simply shorthand for
[0-9]
?
Here is the test code:
using System;
using System.Collections.Generic;
using System.Linq;
using System.Text;
using System.Diagnostics;
using System.Text.RegularExpressions;
namespace SO_RegexPerformance
{
class Program
{
static void Main(string[] args)
{
var rand = new Random(1234);
var strings = new List<string>();
//10K random strings
for (var i = 0; i < 10000; i++)
{
//Generate random string
var sb = new StringBuilder();
for (var c = 0; c < 1000; c++)
{
sb.Append((char)('a' + rand.Next(26)));
}
//In roughly 50% of them, put a digit
if (rand.Next(2) == 0)
{
//Replace one character with a digit, 0-9
sb[rand.Next(sb.Length)] = (char)('0' + rand.Next(10));
}
}
var baseTime = testPerfomance(strings, @"\d");
Console.WriteLine();
var testTime = testPerfomance(strings, "[0-9]");
Console.WriteLine(" {0:P2} of first", testTime.TotalMilliseconds / baseTime.TotalMilliseconds);
testTime = testPerfomance(strings, "[0123456789]");
Console.WriteLine(" {0:P2} of first", testTime.TotalMilliseconds / baseTime.TotalMilliseconds);
}
private static TimeSpan testPerfomance(List<string> strings, string regex)
{
var sw = new Stopwatch();
int successes = 0;
var rex = new Regex(regex);
sw.Start();
foreach (var str in strings)
{
if (rex.Match(str).Success)
{
successes++;
}
}
sw.Stop();
Console.Write("Regex {0,-12} took {1} result: {2}/{3}", regex, sw.Elapsed, successes, strings.Count);
return sw.Elapsed;
}
}
}
\d checks all Unicode digits, while [0-9] is limited to these 10 characters. For example, Persian digits, ۱۲۳۴۵۶۷۸۹, are an example of Unicode digits which are matched with \d, but not [0-9].
You can generate a list of all such characters using the following code:
var sb = new StringBuilder();
for(UInt16 i = 0; i < UInt16.MaxValue; i++)
{
string str = Convert.ToChar(i).ToString();
if (Regex.IsMatch(str, @"\d"))
sb.Append(str);
}
Console.WriteLine(sb.ToString());
Which generates:
0123456789٠١٢٣٤٥٦٧٨٩۰۱۲۳۴۵۶۷۸۹߀߁߂߃߄߅߆߇߈߉०१२३४५६७८९০১২৩৪৫৬৭৮৯੦੧੨੩੪੫੬੭੮੯૦૧૨૩૪૫૬૭૮૯୦୧୨୩୪୫୬୭୮୯௦௧௨௩௪௫௬௭௮௯౦౧౨౩౪౫౬౭౮౯೦೧೨೩೪೫೬೭೮೯൦൧൨൩൪൫൬൭൮൯๐๑๒๓๔๕๖๗๘๙໐໑໒໓໔໕໖໗໘໙༠༡༢༣༤༥༦༧༨༩၀၁၂၃၄၅၆၇၈၉႐႑႒႓႔႕႖႗႘႙០១២៣៤៥៦៧៨៩᠐᠑᠒᠓᠔᠕᠖᠗᠘᠙᥆᥇᥈᥉᥊᥋᥌᥍᥎᥏᧐᧑᧒᧓᧔᧕᧖᧗᧘᧙᭐᭑᭒᭓᭔᭕᭖᭗᭘᭙᮰᮱᮲᮳᮴᮵᮶᮷᮸᮹᱀᱁᱂᱃᱄᱅᱆᱇᱈᱉᱐᱑᱒᱓᱔᱕᱖᱗᱘᱙꘠꘡꘢꘣꘤꘥꘦꘧꘨꘩꣐꣑꣒꣓꣔꣕꣖꣗꣘꣙꤀꤁꤂꤃꤄꤅꤆꤇꤈꤉꩐꩑꩒꩓꩔꩕꩖꩗꩘꩙０１２３４５６７８９
Source (Stackoverflow)
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| 2.59375
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longest
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https://savvycalculator.com/how-to-calculate-rounds-per-minute/
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# How to Calculate Rounds Per Minute
Understanding the speed at which a machine or firearm operates is crucial for enthusiasts, professionals, and anyone curious about the mechanics involved. In this comprehensive guide, we will delve into the intricacies of calculating rounds per minute, demystifying the process and empowering you with the knowledge to gauge speed effectively.
## Exploring the Basics
## Defining Rounds Per Minute Embark on your journey by grasping the concept of rounds per minute (RPM). This fundamental aspect forms the backbone of understanding the operational speed of various mechanisms.
## Importance of Knowing RPM Delve into the significance of RPM awareness in different domains. Whether you’re a firearm aficionado or an engineer working with machinery, comprehending RPM is pivotal for optimal performance and safety.
## Navigating the Formulae
## Formula for Calculating Rounds Per Minute Unravel the mathematical equation behind determining RPM. Mastering the formula equips you with the ability to make precise calculations effortlessly.
## Application in Firearms Explore how calculating rounds per minute is specifically applied in the realm of firearms. Gain insights into the correlation between RPM and firearm efficiency, helping you make informed decisions and optimizations.
## Understanding the Variables
## Factors Influencing RPM Dive into the variables affecting RPM calculations. From the intricacies of machinery to the characteristics of ammunition, understanding these factors ensures accurate and context-specific results.
## Tips for Accurate Measurements Unlock expert tips for ensuring your RPM calculations are spot-on. From calibration techniques to avoiding common pitfalls, these insights guarantee precision in your assessments.
## Real-world Applications
## Industrial Machinery Discover the applications of RPM calculations in the industrial landscape. Uncover how this knowledge is harnessed to enhance productivity, reduce wear and tear, and optimize machinery performance.
## Sports and Entertainment Explore the surprising connection between RPM and sports or entertainment. From sports equipment to audiovisual equipment, witness how RPM plays a role in enhancing experiences.
## How to Calculate Rounds Per Minute: A Step-by-Step Guide
## Step 1: Gather Essential Data Initiate your calculation journey by collecting the necessary data, including the number of rounds fired and the time taken.
## Step 2: Plug into the Formula Follow a simplified step-by-step guide to inputting your data into the RPM formula. Our user-friendly approach ensures seamless calculations.
## Step 3: Interpret the Results Decode the results of your calculation effectively. Understand how the derived RPM figure impacts the overall performance or efficiency of the subject.
## Frequently Asked Questions
How does RPM differ from revolutions per minute (RPM) in machinery? RPM in machinery refers to the rotational speed, while rounds per minute specifically relates to the number of rounds or cycles in firearms or similar devices.
Is there a universal formula for calculating RPM, or does it vary by industry? While the basic formula remains constant, industry-specific nuances may exist. It’s crucial to adapt calculations based on the context, considering factors like ammunition type or machine specifications.
Can RPM calculations be done manually, or is specialized software required? While manual calculations are possible, specialized software can streamline the process, especially in scenarios involving large datasets or complex machinery.
Are there safety considerations when calculating RPM for firearms? Absolutely. Understanding the RPM of a firearm is vital for safety. Always adhere to safety guidelines and consult experts to ensure responsible use and maintenance.
What role does RPM play in the maintenance of industrial machinery? RPM serves as a diagnostic tool for assessing the health of industrial machinery. Monitoring RPM aids in identifying irregularities, facilitating timely maintenance and preventing breakdowns.
How can RPM calculations be applied in the automotive industry? In the automotive sector, RPM calculations are integral for engine performance assessment. They assist in optimizing fuel efficiency, power output, and overall vehicle performance.
## Conclusion
Equipped with a newfound understanding of how to calculate rounds per minute, you are now poised to navigate the intricate world of machinery and firearms with confidence. Embrace the power of precision, and let your knowledge elevate your endeavors.
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https://beta.mapleprimes.com/users/666%20jvbasha/questions?page=1
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javid basha jv
6 years, 53 days
## "unable to compute solution for t>HFloat...
Maple 18
Hi maple users
I am working with the PDE solver.
i am receiving a following error
"unable to compute solution for t>HFloat(0.0):"
Kindly do the needful how to rectify this error.
dumm.mw
## How to rectify the "Error, invalid subsc...
Maple 18
Hi, Maple users
I hope you are doing well.
Here I solved one of the ode problems by dsolve.
But I am getting "Error, invalid subscript selector" error.
Kindly do the needful to find a solution.
Thank you.
JVB.mw
## How to integrate the indefinite integral...
Maple 18
Dear maple users,
A fine day wishes to all.
I have solved the PDE via PDsolve. Here I need to calculate the Psi function. How to calculate the indefinite integral and how to find the constant-coefficient (C1).
Here Psi=0 at x=0
int_c.mw
> restart:
> with(PDEtools):
> with(plots):
> fcns := {f(x,t),theta(x,t)};
(1)
> d:=0.5:xi:=0.1:
> R:=z->piecewise(d<=z and z<=d+1,1-2*xi*(cos((2*3.14)*((z-d)*(1/2))-1/4)-(7/100)*cos((32*3.14)*(z-d-1/2))),1);
(2)
> PDE1 :=(diff(f(x,t),t))=1+(1-2*theta((x,t)))*(1/(R(z)^2))*((diff(f(x,t),x,x))+(1/x)*diff(f(x,t),x))+theta((x,t));
(3)
> PDE2 :=2*(diff(theta(x,t),t))=(1/(R(z)^2))*((diff(theta(x,t),x,x))+(1/x)*diff(theta(x,t),x));
(4)
> IBC := {D[1](f)(0,t)=0,f(1,t)=0,f(x,0)=0,D[1](theta)(0,t)=0,theta(1,t)=1,theta(x,0)=0};
(5)
> z:=0.98:
>
> sol:=pdsolve(eval([PDE1,PDE2]),IBC ,numeric, time = t): sol:-value(f(x,t), output=listprocedure); fN:=eval( f(x,t), sol:-value(f(x,t), output=listprocedure)):
(6)
> t := 1;
(7)
> A1:=x*R(z)*R(z)*(fN)(x, t);
(8)
> A2:=eval(int(A1, x))+C1;
(9)
> W11:=eval(subs(x=0,A2));
> Find_c1:=solve(W11,C1);
(10)
>
Here u is fN(x,t) and t=1.
## How to integrate the computed values ...
Maple 18
Dear maple users,
A fine day wishes to all.
Here, we have computed the fN(x,t) value by using pdsolve.
We have to integrate the computed value and need to find the values with the sequence of x.
A1:=int(fN(x,1.12),x)
A2:=seq(A1,x=0..1,0.1)
How to integrate the computed values
And, How to find the values for the sequence of x.
> restart:
> with(PDEtools):
> with(plots):
> fcns := {f(x,t)};
(1)
> b1:=1.41:d:=0.5/1:xi:=0.1:ea:=0.5:ra:=2:
> L:=z->piecewise(d<=z and z<=d+1, 1-2*xi*(cos((2*3.14)*((z-d)*(1/2))-1/4)-(7/100)*cos((32*3.14)*(z-d-1/2))),1):
> PDE1 :=ra*(diff(f(x,t),t))=+b1*(1+ea*cos(t))+(1/(L(z)^2))*((diff(f(x,t),x,x))+(1/x)*diff(f(x,t),x));
(2)
> IBC := {D[1](f)(0,t)=0,f(1,t)=0,f(x,0)=0};
(3)
> z:=0.5;
(4)
>
> sol:=pdsolve(eval([PDE1]),IBC ,numeric, time = t,spacestep = 0.025, timestep=0.0001): sol:-value(f(x,t), output=listprocedure);
(5)
> fN:=eval( f(x,t), sol:-value(f(x,t), output=listprocedure)):
>
> A1:= int(fN(x,t),x);
(6)
> A2 := seq(A1(x), x = 0.1 .. 1, 0.1);
>
## How to execute the piecewise condition i...
Maple 18
Dear maple users
A fine day wishes to all
In my problem, L(z) is a piecewise condition.
L(z):
I have to calculate the f(x,t) value at x=0.71,t=1.12 and z=0.71 for L(z)=0..1.
How to calculate the f(x,t) value.JVB1.mw
> restart:
> with(PDEtools):
> with(plots):
> fcns := {f(x,t)};
(1)
> b1:=1.41:d:=0.5/1:xi:=0.1:ea:=0.5:ra:=2:
> L:=z->piecewise(d<=z,1-2*xi*(cos((2*3.14)*((z-d)*(1/2))-1/4)-(7/100)*cos((32*3.14)*(z-d-1/2))),z<=d+1,1-2*xi*(cos((2*3.14)*((z-d)*(1/2))-1/4)-(7/100)*cos((32*3.14)*(z-d-1/2))),1);
(2)
> PDE1 :=ra*(diff(f(x,t),t))=+b1*(1+ea*cos(t))+(1/(L(z)^2))*((diff(f(x,t),x,x))+(1/x)*diff(f(x,t),x));
(3)
> IBC := {D[1](f)(0,t)=0,f(1,t)=0,f(x,0)=0};
(4)
> z:=0.71;
(5)
> sol:=pdsolve(eval([PDE1]),IBC ,numeric, time = t,spacestep = 0.025, timestep=0.0001): sol:-value(f(x,t), output=listprocedure);
(6)
> fN:=eval( f(x,t), sol:-value(f(x,t), output=listprocedure)):
>
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https://inst.eecs.berkeley.edu/~cs194-26/fa18/upload/files/proj4/cs194-26-acq/
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# Project 4: Fun with Frequencies and Gradients!
## Part 1: Introduction
The goal of this project was to perform transformations between faces, using morphing as well as using population means. Each method requires defining correspondences manually on an input face, and using affine transformations to transform one shape derived from the corresponding points to another shape. The resulting image would then take weighed averages between the pixels in the original image and pixels in a reference image.
## Part 2: Defining Correspondences
Correspondences were defined manually by picking points on two images such that they corresponded to equivalent features. Input images, and the points defined are as shown
Input Image 1 Input Image 2 Key points of Image 1 Key points of Image 2
## Computing the midway face
From the 2 correspoding points, a "midway" shape was defined by simply averaging the 2 corresponding points from the same locations. A delaunay triangulation was defined on the midway shape, and both faces were morphed into that shape. Finally, the pixel values from the 2 morphed images were averaged to compute the midway face.
Delaunay Triangulation of midpts and morphed Image 1 Delaunay Triangulation of midpts and morphed Image 2
## Part 3: Morph Sequence
The morph sequence was generated in a similar manner to the midway face. For each frame 0-45, a warp and dissolve fraction was defined to be proportional to the frame number. For this case, the two are equivalent. Then, instead of taking an equal ratio of both images for both the shape and the pixel values, a weighed average of the shape and pixels were taken corresponding to the warp and dissolve fractions respectively. When each frame is put together, we get a nice morph sequence
## The "Mean Face" of a population
To generate the "mean face" of a population, we first received a collection of face images, each annotated with corresponding points. Source: M. B. Stegmann, B. K. Ersbøll, and R. Larsen. FAME – a flexible appearance modelling environment. IEEE Trans. on Medical Imaging, 22(10):1319–1331, 2003
An average face of the subpopulation consisting of faces with neutral expressions was computed by first averaging the shapes of each face, and morphing each face into that shape. Then, each morphed face is average together.
A few morphed faces The average face. (computed in grayscale because some input images were in grayscale)
In addition to computing the average face, I decided to annotate my own face corresponding to the face dataset, and morphed my face to the average face, and the average face to my face. The results don't look as good because of manual error in my own annotations.
The average face morphed to my face shape My face morphed to the average face shape
## Caricatures: Extrapolating from the mean
A caricature of my face was generated by extrapolating my face's shape from the population mean. My face was then morphed into the extrapolated shape. The results look wacky due to the difficulty in accurately annotating the points. However, we can see a couple of caricatured elements, including larger lips, larger teeth, and larger nose
## Bells and Whistles, Unsmiling a face
To unsmile myself, I took subpopulation means of smiling people and not smiling people, and added the difference between unsmiling and smiling people to my face in different ways. I first added just the difference in shape, then the difference in pixel colors, then added both the difference in shape and color.
Input Image Adding the difference in color. Interesting to note how the color of the teeth change. Sad people don't show teeth! Adding the difference in shape Adding both the difference in color and shape
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# coset
(redirected from Coset representative)
Also found in: Encyclopedia.
## coset
(ˈkəʊˌsɛt)
n
one of several sets that form a larger set
References in periodicals archive ?
In general, for a subcoset Hy [subset or equal to] Gx (using the notation Hy even if it may be empty), when generators for H and a coset representative are not necessarily available at hand, if we are given a polynomial-time procedure to determine, for any given z [member of] Gx, whether z [member of] Hy, then we say Hy is (polynomial-time) recognizable.
If I is the order ideal encircled by the solid line, then [phi](I) is the coset representative encircled by the solid line, and [MATHEMATICAL EXPRESSION NOT REPRODUCIBLE IN ASCII] is labeled by [S.
I] [member of] W is the unique coset representative of minimum (affine) length.
When we wish to refer to the Poincare polynomial of a minimal length coset representative w for the induced order on either of the quotients [sup.
In our case, the base window of the minimal length coset representative of an element is obtained by ordering the entries that appear in the base window increasingly.
1, every coatom w is a maximal-length coset representative of the subgroup [W.
The authors cover convex preorders and cuspidal systems, imaginary representations, imaginary Schur-Weyl duality, imaginary Howe and Ringel dualities, Gelfand-Graev words and representations, partitions and compositions, Coset representatives, Schur algebras, and a wide-ranging variety of other related mathematical subjects.
i]} be a distinct set of coset representatives of [A.
A set K formed by a collection of such representatives is called a transversal, and by definition K consists of one and just one point from each distinct coset H + a = [a] (the transversal is also called a complete set of coset representatives but we shall stick to the shorter name).
Furthermore, in order to resolve the scaling problems of Wyler's expression, Gilmore showed why it is essential to use dimensionless volumes by setting the throat sizes of the Anti de Sitter hyperboloids to r = 1, because this is the only choice for r where all elements in the bounded domains are also coset representatives, and therefore, amount to honest group operations.
The minimal length left coset representatives [omega] [member of] [[?
J] the set of minimum-length coset representatives in W/[W.
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# How do you find turnover number?
## How do you find turnover number?
The units of Turn over number (kcat) are kcat = (moles of product/sec)/ (moles of enzyme) or sec-1….Turnover Number
1. kcat = Turnover number,
2. Vmax = Maximum rate of reaction when all the enzyme catalytic sites are saturated with substrate and.
3. Et =Total enzyme concentration or concentration of total enzyme catalytic sites.
## What is the turnover rate of an enzyme?
Turnover number is defined as the number of substrate molecules transformed per minute by a single enzyme molecule when the enzyme is the rate-limiting factor.
What is the relationship between turnover number and k2?
At this point we define the turnover number: the number of substrate molecules catalyzed per second per enzyme molecule when the enzyme is saturated with substrate (Nelson and Cox, 2005; Becker et al., 2006). Hence, when the enzyme is saturated with substrate the turnover number (k2) becomes rate limiting.
### What is the turnover number for this enzyme and substrate?
The turnover number of an enzyme, is the number of substrate molecules converted into product by an enzyme molecule in a unit time when the enzyme is fully saturated with substrate. The turnover numbers of most enzymes with their physiological substrates fall in the range from 1 to 10^4 per second (Table link).
### Which enzyme has maximum turnover number?
Catalase
Catalase has the highest turnover numbers of all enzymes. One molecule of catalase can convert over 2.8 million molecules of hydrogen peroxide to water and oxygen per second.
How do you calculate a catalyst turnover number?
Turnover number (TON) = number of moles of reactant consumed/(mole of catalyst). Turnover frequency (TOF) = TON/time of reaction.
#### What is the fastest enzyme?
carbonic anhydrase
In fact, carbonic anhydrase is one of the fastest enzymes known. Each enzyme molecule can hydrate 106 molecules of CO2 per second. This catalyzed reaction is 107 times as fast as the uncatalyzed one.
What is the turnover number in Michaelis Menten equation?
The turnover number represents the number of substrate molecules that can be converted into product per unit time by a single enzyme.
## How is catalase used to study H 2 O 2?
Decomposition of hydrogen peroxide (H 2 O 2 ) at physiological levels was studied in human erythrocytes by means of a recently developed sensitive H 2 O 2 assay. The exponential decay of H 2 O 2 in the presence of purified erythrocyte catalase was followed down to 10 −9 mol/L H 2 O 2 at pH 7.4.
## What is the V max value of catalase?
Kinetic studies on the activity of catalase conducted using a pressure gauge indicates that the enzyme has a V max value of 0.0144, and K mvalue of 0.00275. The catalase appears to be affected by fluctuating pH values, and operates most ideally at pH 9.
How many molecules of hydrogen peroxide can catalase break down?
Catalase is remarkably efficient, and one catalase enzyme can convert 40 million molecules of hydrogen peroxide to oxygen and water per second. This enzyme is necessary as a condition for survival, as the catalase breaks down hydrogen peroxide from accumulating to dangerous levels.
### Are there reagents that show prevalence of catalase?
These direct measurements showed a clear prevalence of catalase even at very low H 2 O 2 concentrations. Reagents. Luminol, catalase, GPO, glutathione reductase, reduced glutathione (GSH), hydrogen peroxide, aminotriazole, NADPH, sodium hypochlorite, and sodium azide were from Sigma (Deisenhofen, Germany).
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Richard Nakka's Experimental Rocketry Web Site
Technical Notepad #5 -- KNER Ideal Performance Calculations
Note 1
Potassium Nitrate - Erythritol (KNER) propellant
65/35 O/F ratio @ 1000 psia chamber pressure
From PROPEP results, for 100 grams mixture:
The effective Molecular Weight is given by dividing the number GAS moles into the system mass. Since the system mass is 100 grams:
g/mole
Note that this is the proper molecular weight to use in the thermodynamic equations.
The mass fraction of condensed phase is given by the mass of the condensed phase (K2CO3) divided by the system mass
The MW of K2CO3 = 138.21 g/mole, thus
KNER 65/35 O/F ratio @ 1000 psia chamber pressure
Mole fractions and mass fractions for each combustion product are calculated in the table below:
The table below shows the computation of k, the ratio of specific heats:
The values for Cp and Cs are taken from the JANAF Thermochemical Tables and NIST Chemistry WebBook.
Note that the highlighted range (yellow) is applicable for interpolation of the values at 1608 K, the chamber combustion temperature under consideration.
The Cp for the gas only products and mixture (gas+condensed) is given by
where ni is the number of moles of gas component i , ns the number of moles of condensed component, n the total number of gas moles. The ratio of specific heats for the mixture, for the gas-only, and for two-phase flow is given by
where = 8.314 J/mol-K (universal gas constant).
where y = X /(1-X).
Note that k for two-phase (gas+condensed) flow is a modified form of the gas-only k'. This is the correct form of k to use in the thermodynamic equations involving products with a significant fraction of condensed-phase particles. The value of k given in the PROPEP output (Cp/Cv) is for the mixture.
Note 3
Characteristic exhaust velocity is given by
with
To = 1608 K
M = 38.78 kg/kmol
k = 1.1390 Note: k for the mixture is the proper value to use, as c* represents a static condition
= 8314 J/kmol-K
this gives c* = 923 m/s (3027 ft/s).
Note 4
The propellant specific impulse is given by the effective exhaust velocity divided by g.
with
To = 1608 K
M = 38.78 kg/kmol
k = 1.0426 Note: k for 2-phase flow is the proper value to use, as Isp involves two-phase flow.
Thus, Isp = 167 sec.
for standard conditions of Po = 68 atm. (1000 psia) and Pe = 1 atm., and g = 9.806 m/s
(maximum theoretical, assumes frozen equilibrium, and no particle velocity lag or thermal lag).
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# Convert Nanohertz to Radians Per Hour
### Kyle's Converter > Frequency > Nanohertz > Nanohertz to Radians Per Hour
Nanohertz (nHz) Radians Per Hour (rad/h)* Precision: 0 1 2 3 4 5 6 7 8 9 12 15 18
Reverse conversion?
(or just enter a value in the "to" field)
Please share if you found this tool useful:
Unit Descriptions
1 Nanohertz:
1 Nanohertz is exactly one billionth of a Hertz. 1 yHz = 1 x 10-9 Hz. 1 nHz = 0.000000001 Hz.
1 Radian per hour is comparative to 1/7200π Hertz. Radians per hour is a measure of angular frequency, it can be compared to Hertz or other angular units. Formula uses an approximation of PI. 1 rad/h is approximately 0.000044209706414415 Hz.
Conversions Table
1 Nanohertz to Radians Per Hour = 070 Nanohertz to Radians Per Hour = 0.0016
2 Nanohertz to Radians Per Hour = 080 Nanohertz to Radians Per Hour = 0.0018
3 Nanohertz to Radians Per Hour = 0.000190 Nanohertz to Radians Per Hour = 0.002
4 Nanohertz to Radians Per Hour = 0.0001100 Nanohertz to Radians Per Hour = 0.0023
5 Nanohertz to Radians Per Hour = 0.0001200 Nanohertz to Radians Per Hour = 0.0045
6 Nanohertz to Radians Per Hour = 0.0001300 Nanohertz to Radians Per Hour = 0.0068
7 Nanohertz to Radians Per Hour = 0.0002400 Nanohertz to Radians Per Hour = 0.009
8 Nanohertz to Radians Per Hour = 0.0002500 Nanohertz to Radians Per Hour = 0.0113
9 Nanohertz to Radians Per Hour = 0.0002600 Nanohertz to Radians Per Hour = 0.0136
10 Nanohertz to Radians Per Hour = 0.0002800 Nanohertz to Radians Per Hour = 0.0181
20 Nanohertz to Radians Per Hour = 0.0005900 Nanohertz to Radians Per Hour = 0.0204
30 Nanohertz to Radians Per Hour = 0.00071,000 Nanohertz to Radians Per Hour = 0.0226
40 Nanohertz to Radians Per Hour = 0.000910,000 Nanohertz to Radians Per Hour = 0.2262
50 Nanohertz to Radians Per Hour = 0.0011100,000 Nanohertz to Radians Per Hour = 2.2619
60 Nanohertz to Radians Per Hour = 0.00141,000,000 Nanohertz to Radians Per Hour = 22.6195
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# How can I perform tensor contractions without first storing an unnecessarily large tensor?
I'm performing a calculation which requires me to start with a collection of tensors and contract out their indices according to a pattern that I've written down. Right now, if I have a pair of two index tensors $A_{ij}$ and $B_{nm}$, and I want to contract out the second index of $A$ with the first index of $B$, I would write
TensorContract[TensorProduct[A,B], {2,3}]
This works acceptably whenever the tensors are of reasonable size, but unfortunately, my arrays quickly become quite large. In particular, on the last step of my calculation, I have a pair of 28-index objects (each with two values, so in total $2^{28}$ elements each) that I need to tensor product together. I then want to contract out 16 pairs of indices, so that I'm left with a 24-index object. Unfortunately, my computer's memory can't seem to handle the step where I need to create the intermediate 56-index object. Hence my question: is there a smarter way to perform tensor contractions that doesn't require the creation of an intermediate object with a higher rank? It seems wasteful to build the intermediate object, but I don't know of any (straightforward) workarounds. Is there a simple way to address this problem?
I should also mention that these arrays are quite sparse, and I have made all of my tensors into sparse arrays. In particular, for my 28-index object with $2^{28}$ entries, only $2^{13}$ of them are nonzero. But even with sparcity, this is killing my computer. So any advice is very welcome!
• A similar question: mathematica.stackexchange.com/questions/111467/… .
– jose
Commented Oct 10, 2016 at 19:19
• Ah, you're my hero! Inactive is exactly what I needed. Thanks friend. Commented Oct 10, 2016 at 19:27
• Glad to help. Inactive is indeed a very powerful simple idea. Here is another question whose answers contain several tricks for efficient contraction of arrays: mathematica.stackexchange.com/questions/17758/… .
– jose
Commented Oct 10, 2016 at 20:23
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## A boat travels with velocity vector (25, 25 StartRoot 3 EndRoot). What is the directional bearing of the boat? N 30° E E 30° S E 30° N N 30°
Question
A boat travels with velocity vector (25, 25 StartRoot 3 EndRoot). What is the directional bearing of the boat? N 30° E E 30° S E 30° N N 30° W
in progress 0
3 weeks 2021-08-26T18:11:39+00:00 2 Answers 0 views 0
## Answers ( )
1. Answer:
a
Step-by-step explanation:
2. Answer:
The directional bearing of the boat is N 30º E
Step-by-step explanation:
Let , where is the vector velocity. Given that such vector is represented in rectangular, a positive value in the first component is the value of the vector in the east direction, whereas a positive value in the second component is in the north direction. The directional bearing of the boat (), measured in sexagesimal degrees, is determined by trigonometrical means:
(1)
If we know that and , then the directional bearing of the boat is:
In consequence, we conclude that the direction bearing of the boat is 30 degrees to the East from the North (N 30º E).
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You are on page 1of 60
Rheology
and
Polymer Characterization
Asst Prof Anongnat Somwangthanaroj
Anongnat.s@chula.ac.th 20 Sep 2010
http://pioneer.netserv.chula.ac.th/~sanongn1/course.html
Fundamentals:
Why Rheology ? Fundamental Rheology Concepts and Parameters Fundamental Rheometry Concepts Viscosity, Viscoelasticiy and the Storage Modulus The Linear Viscoelastic Region (LVR)
AGENDA
Why Rheology ? Fundamental Rheology Concepts and Parameters Fundamental Rheometry Concepts Viscosity, Viscoelasticiy and the Storage Modulus The Linear Viscoelastic Region (LVR)
Sometimes it doesnt ____
BECAUSE
If a material is pumped, sprayed, extended, extruded, molded, coated, mixed, chewed, swallowed, rubbed, transported, stored, heated, cooled, aged RHEOLOGY is important .!!
AGENDA
Why Rheology ? Fundamental Rheology Concepts and Parameters Fundamental Rheometry Concepts Viscosity, Viscoelasticiy and the Storage Modulus The Linear Viscoelastic Region (LVR)
(everything flows )
Heraclito de Samos (500 A.C.)
Time Scale in Rheology
Deborah Number
De = / texp
Judges 5:5
Definition of Rheology
Rheology is the science of ____?____ and ____?___ of matter under controlled testing conditions .
flow deformation
Definition of Rheology
Rheology is the science of deformation and flow of matter under controlled testing conditions .
Simple Shear Deformation and Shear Flow
Shear Deformation F = A
y
x(t)
y0
A
x
. = t
Range of Rheological Material Behavior
Rheology: The study of deformation and flow of matter at specified conditions. Range of material behavior
Solid Like
(Ideal Solid
---------------------
Liquid Like
Ideal Fluid)
Classical Extremes
Classical Extremes: Elasticity
1678: Robert Hooke develops his True Theory of Elasticity
The power of any spring is in the same proportion with the tension thereof. Hookes Law:
=G
or
(Stress = G x Strain)
1000
Linear Region G is constant
Non-Linear Region
100.0
G = f()
osc. stress (Pa)
100.0
G' (Pa)
10.00
Classical Extremes: Viscosity
1687: Isaac Newton addresses liquids and steady simple shearing flow in his Principia The resistance which arises from the lack of slipperiness of the parts of the liquid, other things being equal, is proportional to the velocity with which the parts of the liquid are separated from one another.
Newtons Law: =
where is the Coefficient of Viscosity
1.000E5
Newtonian Region Independent of
Non-Newtonian Region
1.000E5
= f()
10000
1000
(Pa.s)
100.0
10.00
0.1000
1.000 1.000
(Pa)
PARAMETERS for Rheological Properties
Classical Extremes Ideal Solid
STEEL Strong Structure Rigidity Deformation Retains/recovers form Stores Energy
(Purely Elastic R. Hooke, 1678) [Energy]
-- [External Force] --
Ideal Fluid
WATER Weak Structure Fluidity Flow Losses form Dissipates Energy
(Purely Viscous I. Newton, 1687)
REAL Behavior Apparent Solid
[Energy + time]
Apparent Fluid
- viscoelastic materials -
Types of non-Newtonian fluids
Deformation rate dependent viscosity Yield Stress (plasticity) Elasticity Thixotropy Transient behaviour
apparent viscosity as a function of time
AGENDA
Why Rheology ? Fundamental Rheology Concepts and Parameters Fundamental Rheometry Concepts Viscosity, Viscoelasticiy and the Storage Modulus The Linear Viscoelastic Region (LVR)
Viscometer vs. Rheometer
Viscometer: instrument that measures the viscosity of a
fluid over a limited shear rate range
Rheometer: instrument that measures:
Viscosity over a wide range of shear rates, and Viscoelasticity of fluids, semi-solids and solids
Frame of Reference Recognize that a rheometer is a highly sensitive device used to quantify viscoelastic properties of the molecular structure of materials. A rheometer can not always mimic the conditions of a process, application or use. Rheometers determine apparent properties under a wide range of testing conditions.
The apparent behavior can be used as a finger print or benchmark of the material.
Constitutive Relations
Stress = Modulus Strain
Stress = Viscosity Shear rate
Measuring Systems - Geometries
Concentric Cylinders Cone and Plate Parallel Plates Rectangular Torsion
Soft to Rigid Solids
Decane
Water
Steel
AGENDA
Why Rheology ? Fundamental Rheology Concepts and Parameters Fundamental Rheometry Concepts Viscosity, Viscoelasticiy and the Storage Modulus The Linear Viscoelastic Region (LVR)
Dynamic Testing
Deformation
An oscillatory (sinusoidal) deformation (stress or strain) is applied to a sample. The material response (strain or stress) is measured. The phase angle , or phase shift, between the deformation and response is measured.
Response
Phase angle
Phase angle Stress
0 < < 90
Strain
Viscoelastic Parameters
The Complex Modulus: Measure of materials overall resistance to deformation. The Elastic (Storage) Modulus: Measure of elasticity of material. The ability of the material to store energy. The Viscous (loss) Modulus: The ability of the material to dissipate energy. Energy lost as heat. Tan Delta: Measure of material damping - such as vibration or sound damping.
Tan = G"/G'
Stress or Strain
Deformation
Time
The material response is monitored at a constant frequency, amplitude and temperature.
USES Time dependent Thixotropy Cure Studies Stability against thermal degradation Solvent evaporation/drying
AGENDA
Why Rheology ? Fundamental Rheology Concepts and Parameters Fundamental Rheometry Concepts Viscosity, Viscoelasticiy and the Storage Modulus The Linear Viscoelastic Region (LVR)
Dynamic Stress or Strain Sweep (Torque Ramp)
Stress or Strain
Deformation
Time
The material response to increasing deformation amplitude (stress or strain) is monitored at a constant frequency and temperature.
USES Identify Linear Viscoelastic Region Strength of dispersion structure - settling stability Resilience
1000
Linear Region G is constant
Non-Linear Region
100.0
G = f()
10.00
1.000 0.010000 0.10000
Critical Strain c
1.0000 10.000 % strain 100.00 0.01000 1000.0
osc. stress (Pa)
100.0
G' (Pa)
Frequency Sweep
Stress or Strain
Deformation
Time
The material response to increasing frequency (rate of deformation) is monitored at a constant amplitude (stress or strain) and temperature.
USES Viscosity Information - Zero Shear , shear thinning Elasticity (reversible deformation) in materials MW & MWD differences Polymer Melts and Polymer solutions. Finding Yield in gelled dispersions High and Low Rate (short and long time) modulus properties. Extend time or frequency range with TTS
Time Sweep on Latex
Structural Recovery after Preshear
100.0
80.00
G' (Pa)
60.00
40.00
20.00
0 0
25.00
50.00
75.00
100.0
125.0
150.0
175.0
200.0
225.0
time (s)
1000
Linear Region G is constant
Non-Linear Region
100.0
G = f()
10.00
1.000 0.010000 0.10000
Critical Strain c
1.0000 10.000 % strain 100.00 0.01000 1000.0
100.0
G' (Pa)
Oscillation Model Fitting for Classic Polymer Data [Polyacrylamide Soln.]
Polyacrylamide Solution 20 C
1000 100.0 10.00 1.000 0.1000 0.01000 1.000E-3 1.000E-4 0.01000
TA Instruments
100.0 1000 100.0
Slope = 0.97 G"
4 Element Maxwell Fit
10.00
G' (Pa)
n' Slope = 1.96
Straight Line Fit to Terminal Region of Data
0.1000 1.000 10.00 ang. frequency (rad/sec) 100.0
n' (Pa.s)
G'' (Pa)
G'
Defining Shear Rate Ranges
Situation
Sedimentation of fine powders in liquids
Shear Rate Range 10-6 to 10-3 10-2 to 10-1 10-1 to 101 100 to 102 101 to 102 101 to 102 101 to 103 100 to 103 103 to 104 104 to 105 104 to 106 105 to 106 103 to 107
Examples
Leveling due to surface tension Draining off surfaces under gravity Extruders Chewing and Swallowing Dip coating Mixing and stirring Pipe Flow Brushing Rubbing High-speed coating Spraying Lubrication
Paints, Printing inks Toilet bleaches, paints, coatings Polymers, foods Foods Confectionery, paints Liquids manufacturing Pumping liquids, blood flow Painting Skin creams, lotions Paper manufacture Atomization, spray drying Bearings, engines
Strain
0
time
Response of Classical Extremes
Hookean Solid Newtonian Fluid
Stress
Stress
0
time
time
Stress Relaxation Experiment (contd)
Response of
Stress decreases with time starting at some high value and decreasing to zero.
Stress
0
Material
time
For small deformations (strains within the linear region) the ratio of stress to strain is a function of time only. This function is a material property known as the STRESS RELAXATION MODULUS, G(t) G(t) = s(t)/
Stress
t1 time t2 Response of Classical Extremes
Stain for t>t1 is constant Strain for t >t2 is 0
Strain
Strain
Stain rate for t>t1 is constant Strain for t>t1 increase with time Strain rate for t >t2 is 0
t1
time t2
t1 time t2
Creep Recovery Experiment: Response of Viscoelastic Material
Creep t1> 0 Recovery t 2= 0 (after steady state)
/
Strain
Recoverable Strain
t2
time
Strain rate decreases with time in the creep zone, until finally reaching a steady state.
In the recovery zone, the viscoelastic fluid recoils, eventually reaching a equilibrium at some small total strain relative to the strain at unloading.
Reference: Mark, J., et.al., Physical Properties of Polymers ,American Chemical Society, 1984, p. 102.
Polyethylene Rheology @ 150 C
1000000
HDPE
100000
viscosity (Pa.s)
LLDPE
10000
LDPE
1000 1.000E-4 1.00E-3 0.01000 0.1000 1.000 10.00
100000
Creep or Equilibrium Flow
viscosity (Pa.s)
10000
. () = ()
PDMS.05F-Flow step PDMS.08F-Flow step
Dynamic Frequency Sweep
1000
Dynamic data gives high shear rates unattainable in flow
100.0 1.000E-5
1.000E-3
10.00
1000
Dynamic Temperature Ramp or Step and Hold: Material Response
Glassy Region Transition Region
Terminal Region
Temperature
Molecular Structure - Effect of Molecular Weight
Glassy Region Transition Region Rubbery Plateau Region
Low MW
High Med. MW MW
Temperature
Effect of Heating Rate on Temperature of Cold Crystallization in PET
Heating Rate After Quench Cooling
Tg melt
PET Bottle Resin Cold Crystallization
1.000E10 0.4000 0.3500
Temperature Ramp at 3C/min. Frequency = 1 Hz Strain = 0.025%
1.000E10
G
1.000E9 0.3000
- transition Tg = 88.0C
1.000E9
0.2500 tan(delta)
G' (Pa)
Cold Crystallization
G'' (Pa)
1.000E8
0.2000
1.000E8
0.1500
1.000E7
0.1000
- transition -56.62C
1.000E7
0.05000
tan
-150.0 -100.0 -50.0 0 50.0 100.0 150.0 200.0 1.000E6 250.0
1.000E6 0 -200.0
temperature (Deg C)
PET Bottle Resin - Repeat Run After Cold Crystallization
1.000E10 2.250
G
2.000 1.000E9 1.750
1.000E10
Melt Tm = 240C
1.000E9
1.000E8
1.500
G
tan(delta)
1.000E8
G'' (Pa)
1.000E7
1.000E6
0.7500
1.000E6
0.5000 1.000E5 0.2500
tan
- transition Tg = 103C
1.000E5
10000 0 -200.0
-150.0
-100.0
-50.0
50.0
100.0
150.0
200.0
10000 250.0
temperature (Deg C)
1.000E10
1.000E9
G' (Pa) 1.000E8
1.000E7
Temperature Ramp at 3C/min. Frequency = 1 Hz Strain = 0.025%
1.000E6 -150.0 -100.0 -50.0 0 50.0 temperature (Deg C) 100.0
Cold Crystallization
150.0
200.0
250.0
BECAUSE Typical DSC Transitions
Oxidation or Decomposition
Heat Flow
Temperature
Quantitative Description of Consistency (structure) ?
BECAUSE
Thermal Analysis describes thermal transitions
NEED to quantify ...
Physical Properties of Structure Strength or weakness of the Structure
and because
Rheology
Shear Flexure
Tension
Compression
Creep
Creep
Stress relaxation
Creep
Stress relaxation
Constant stressing rate
Creep
Stress relaxation
Constant straining rate
Acknowledgements
Abel Gaspar-Rosas, Ph.D.
TA Instruments Waters, Inc For graphs and figures
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Unary operators in C/C++
Unary operator: are operators that act upon a single operand to produce a new value.
Types of unary operators:
1. unary minus(-)
2. increment(++)
3. decrement(- -)
4. NOT(!)
5. Addressof operstor(&)
6. sizeof()
1. unary minus
The minus operator changes the sign of its argument. A positive number becomes negative, and a negative number becomes positive.
``` int a = 10;
int b = -a; // b = -10
```
unary minus is different from subtraction operator, as subtraction requires two operands.
2. increment
It is used to increment the value of the variable by 1. The increment can be done in two ways:
1. prefix increment
In this method, the operator preceeds the operand (e.g., ++a). The value of operand will be altered before it is used.
``` int a = 1;
int b = ++a; // b = 2
```
2. postfix increment
In this method, the operator follows the operand (e.g., a++). The value operand will be altered after it is used.
``` int a = 1;
int b = a++; // b = 1
int c = a; // c = 2
```
3. decrement
It is used to decrement the value of the variable by 1. The increment can be done in two ways:
1. prefix decrement
In this method, the operator preceeds the operand (e.g., ++a). The value of operand will be altered before it is used.
``` int a = 1;
int b = --a; // b = 0
```
2. posfix decrement
In this method, the operator follows the operand (e.g., a- -). The value of operand will be altered after it is used.
``` int a = 1;
int b = a--; // b = 1
int c = a; // c = 0
```
C++ program for combination of prefix and postfix operations:
```// C++ program to demonstrate working of unary increment
// and decrement operators
#include <iostream>
using namespace std;
int main()
{
// Post increment
int a = 1;
cout << "a value: " << a << endl;
int b = a++;
cout << "b value after a++ : " << b << endl;
cout << "a value after a++ : " << a << endl;
// Pre increment
a = 1;
cout << "a value:" << a << endl;
b = ++a;
cout << "b value after ++a : " << b << endl;
cout << "a value after ++a : "<< a << endl;
// Post decrement
a = 5;
cout << "a value before decrement: " << a << endl;
b = a--;
cout << "b value after a-- : " << b << endl;
cout << "a value after a-- : " << a << endl;
// Pre decrement
a = 5;
cout << "a value: "<< a<<endl;
b = --a;
cout << "b value after --a : " << b << endl;
cout << "a value after --a : " << a << endl;
return 0;
}
```
Output:
```a value: 1
b value after a++ : 1
a value after a++ : 2
a value:1
b value after ++a : 2
a value after ++a : 2
a value before decrement: 5
b value after a-- : 5
a value after a-- : 4
a value: 5
b value after --a : 4
a value after --a : 4
```
The above program shows how the postfix and prefix works.
4. NOT(!): It is used to reverse the logical state of its operand. If a condition is true, then Logical NOT operator will make it false.
``` If x is true, then !x is false
If x is false, then !x is true
```
5. Addressof operator(&): It gives an address of a variable. It is used to return the memory address of a variable. These addresses returned by the address-of operator are known as pointers because they “point” to the variable in memory.
```& gives an address on variable n
int a;
int *ptr;
ptr = &a; // address of a is copied to the location ptr.
```
6. sizeof(): This operator returns the size of its operand, in bytes. The sizeof operator always precedes its operand.The operand is an expression, or it may be a cast.
```#include <iostream>
using namespace std;
int main()
{
float n = 0;
cout << "size of n: " << sizeof(n);
return 1;
}```
Output:
```size of n: 4
```
7. This article is contributed by I. HARISH KUMAR. If you like GeeksforGeeks and would like to contribute, you can also write an article using contribute.geeksforgeeks.org or mail your article to contribute@geeksforgeeks.org. See your article appearing on the GeeksforGeeks main page and help other Geeks.
Please write comments if you find anything incorrect, or you want to share more information about the topic discussed above.
GATE CS Corner Company Wise Coding Practice
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1.5 Average Difficulty : 1.5/5.0
Based on 10 vote(s)
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# Search by Topic
#### Resources tagged with Interactivities similar to Turning Triangles:
Filter by: Content type:
Stage:
Challenge level:
### There are 152 results
Broad Topics > Information and Communications Technology > Interactivities
### Khun Phaen Escapes to Freedom
##### Stage: 3 Challenge Level:
Slide the pieces to move Khun Phaen past all the guards into the position on the right from which he can escape to freedom.
### Cogs
##### Stage: 3 Challenge Level:
A and B are two interlocking cogwheels having p teeth and q teeth respectively. One tooth on B is painted red. Find the values of p and q for which the red tooth on B contacts every gap on the. . . .
### When Will You Pay Me? Say the Bells of Old Bailey
##### Stage: 3 Challenge Level:
Use the interactivity to play two of the bells in a pattern. How do you know when it is your turn to ring, and how do you know which bell to ring?
### Rollin' Rollin' Rollin'
##### Stage: 3 Challenge Level:
Two circles of equal radius touch at P. One circle is fixed whilst the other moves, rolling without slipping, all the way round. How many times does the moving coin revolve before returning to P?
### Rolling Around
##### Stage: 3 Challenge Level:
A circle rolls around the outside edge of a square so that its circumference always touches the edge of the square. Can you describe the locus of the centre of the circle?
### Diagonal Dodge
##### Stage: 2 and 3 Challenge Level:
A game for 2 players. Can be played online. One player has 1 red counter, the other has 4 blue. The red counter needs to reach the other side, and the blue needs to trap the red.
### Conway's Chequerboard Army
##### Stage: 3 Challenge Level:
Here is a solitaire type environment for you to experiment with. Which targets can you reach?
### Picturing Triangle Numbers
##### Stage: 3 Challenge Level:
Triangle numbers can be represented by a triangular array of squares. What do you notice about the sum of identical triangle numbers?
### You Owe Me Five Farthings, Say the Bells of St Martin's
##### Stage: 3 Challenge Level:
Use the interactivity to listen to the bells ringing a pattern. Now it's your turn! Play one of the bells yourself. How do you know when it is your turn to ring?
### Drips
##### Stage: 2 and 3 Challenge Level:
An animation that helps you understand the game of Nim.
### Inside Out
##### Stage: 4 Challenge Level:
There are 27 small cubes in a 3 x 3 x 3 cube, 54 faces being visible at any one time. Is it possible to reorganise these cubes so that by dipping the large cube into a pot of paint three times you. . . .
### Attractive Tablecloths
##### Stage: 4 Challenge Level:
Charlie likes tablecloths that use as many colours as possible, but insists that his tablecloths have some symmetry. Can you work out how many colours he needs for different tablecloth designs?
### Online
##### Stage: 2 and 3 Challenge Level:
A game for 2 players that can be played online. Players take it in turns to select a word from the 9 words given. The aim is to select all the occurrences of the same letter.
### Volume of a Pyramid and a Cone
##### Stage: 3
These formulae are often quoted, but rarely proved. In this article, we derive the formulae for the volumes of a square-based pyramid and a cone, using relatively simple mathematical concepts.
### Muggles Magic
##### Stage: 3 Challenge Level:
You can move the 4 pieces of the jigsaw and fit them into both outlines. Explain what has happened to the missing one unit of area.
### Cubic Rotations
##### Stage: 4 Challenge Level:
There are thirteen axes of rotational symmetry of a unit cube. Describe them all. What is the average length of the parts of the axes of symmetry which lie inside the cube?
### Disappearing Square
##### Stage: 3 Challenge Level:
Do you know how to find the area of a triangle? You can count the squares. What happens if we turn the triangle on end? Press the button and see. Try counting the number of units in the triangle now. . . .
### Factor Lines
##### Stage: 2 and 3 Challenge Level:
Arrange the four number cards on the grid, according to the rules, to make a diagonal, vertical or horizontal line.
### Tracking Points
##### Stage: 4 Challenge Level:
Can you give the coordinates of the vertices of the fifth point in the patterm on this 3D grid?
### Lost
##### Stage: 3 Challenge Level:
Can you locate the lost giraffe? Input coordinates to help you search and find the giraffe in the fewest guesses.
### Diamond Mine
##### Stage: 3 Challenge Level:
Practise your diamond mining skills and your x,y coordination in this homage to Pacman.
### A Tilted Square
##### Stage: 4 Challenge Level:
The opposite vertices of a square have coordinates (a,b) and (c,d). What are the coordinates of the other vertices?
### Excel Interactive Resource: Fraction Multiplication
##### Stage: 3 and 4 Challenge Level:
Use Excel to explore multiplication of fractions.
### Nine Colours
##### Stage: 3 Challenge Level:
Can you use small coloured cubes to make a 3 by 3 by 3 cube so that each face of the bigger cube contains one of each colour?
### Isosceles Triangles
##### Stage: 3 Challenge Level:
Draw some isosceles triangles with an area of $9$cm$^2$ and a vertex at (20,20). If all the vertices must have whole number coordinates, how many is it possible to draw?
### Subtended Angles
##### Stage: 3 Challenge Level:
What is the relationship between the angle at the centre and the angles at the circumference, for angles which stand on the same arc? Can you prove it?
### Shuffles Tutorials
##### Stage: 3 Challenge Level:
Learn how to use the Shuffles interactivity by running through these tutorial demonstrations.
### Teddy Town
##### Stage: 1, 2 and 3 Challenge Level:
There are nine teddies in Teddy Town - three red, three blue and three yellow. There are also nine houses, three of each colour. Can you put them on the map of Teddy Town according to the rules?
### Instant Insanity
##### Stage: 3, 4 and 5 Challenge Level:
Given the nets of 4 cubes with the faces coloured in 4 colours, build a tower so that on each vertical wall no colour is repeated, that is all 4 colours appear.
### Square Coordinates
##### Stage: 3 Challenge Level:
A tilted square is a square with no horizontal sides. Can you devise a general instruction for the construction of a square when you are given just one of its sides?
### Jam
##### Stage: 4 Challenge Level:
To avoid losing think of another very well known game where the patterns of play are similar.
### Shuffle Shriek
##### Stage: 3 Challenge Level:
Can you find all the 4-ball shuffles?
### An Unhappy End
##### Stage: 3 Challenge Level:
Two engines, at opposite ends of a single track railway line, set off towards one another just as a fly, sitting on the front of one of the engines, sets off flying along the railway line...
### More Number Pyramids
##### Stage: 3 Challenge Level:
When number pyramids have a sequence on the bottom layer, some interesting patterns emerge...
### Number Pyramids
##### Stage: 3 Challenge Level:
Try entering different sets of numbers in the number pyramids. How does the total at the top change?
### Fractions and Percentages Card Game
##### Stage: 3 and 4 Challenge Level:
Match the cards of the same value.
### Loci Resources
##### Stage: 4 Challenge Level:
This set of resources for teachers offers interactive environments to support work on loci at Key Stage 4.
### Countdown
##### Stage: 2 and 3 Challenge Level:
Here is a chance to play a version of the classic Countdown Game.
### See the Light
##### Stage: 2 and 3 Challenge Level:
Work out how to light up the single light. What's the rule?
### Balancing 1
##### Stage: 3 Challenge Level:
Meg and Mo need to hang their marbles so that they balance. Use the interactivity to experiment and find out what they need to do.
### Shear Magic
##### Stage: 3 Challenge Level:
What are the areas of these triangles? What do you notice? Can you generalise to other "families" of triangles?
### Changing Places
##### Stage: 4 Challenge Level:
Place a red counter in the top left corner of a 4x4 array, which is covered by 14 other smaller counters, leaving a gap in the bottom right hand corner (HOME). What is the smallest number of moves. . . .
### Icosian Game
##### Stage: 3 Challenge Level:
This problem is about investigating whether it is possible to start at one vertex of a platonic solid and visit every other vertex once only returning to the vertex you started at.
### Interactive Spinners
##### Stage: 3 Challenge Level:
This interactivity invites you to make conjectures and explore probabilities of outcomes related to two independent events.
### Fifteen
##### Stage: 2 and 3 Challenge Level:
Can you spot the similarities between this game and other games you know? The aim is to choose 3 numbers that total 15.
### Overlap
##### Stage: 3 Challenge Level:
A red square and a blue square overlap so that the corner of the red square rests on the centre of the blue square. Show that, whatever the orientation of the red square, it covers a quarter of the. . . .
### Sliding Puzzle
##### Stage: 1, 2, 3 and 4 Challenge Level:
The aim of the game is to slide the green square from the top right hand corner to the bottom left hand corner in the least number of moves.
### Partitioning Revisited
##### Stage: 3 Challenge Level:
We can show that (x + 1)² = x² + 2x + 1 by considering the area of an (x + 1) by (x + 1) square. Show in a similar way that (x + 2)² = x² + 4x + 4
### Konigsberg Plus
##### Stage: 3 Challenge Level:
Euler discussed whether or not it was possible to stroll around Koenigsberg crossing each of its seven bridges exactly once. Experiment with different numbers of islands and bridges.
### Bow Tie
##### Stage: 3 Challenge Level:
Show how this pentagonal tile can be used to tile the plane and describe the transformations which map this pentagon to its images in the tiling.
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# [petsc-users] Unstructured mesh
Karthikeyan Chockalingam - STFC UKRI karthikeyan.chockalingam at stfc.ac.uk
Fri Dec 10 06:42:11 CST 2021
```Hi Matt,
I intend to perform a scaling study – I have a few more questions from ex56
3D, tri-quadratic hexahedra (Q1), displacement finite element formulation.
i) What makes the problem non-linear? I believe SNES are used to solve non-linear problems.
ii) 2,2,1 does it present the number of elements used in each direction?
iii) What makes the problem unstructured? I believe the geometry is a cube or cuboid – is it because it uses DMPlex?
iv) Do any external FEM package with an unstructured problem domain has to use DMPlex mat type?
v) What about -mat_type (not _dm_mat_type) ajicusparse – will it work with unstructured FEM discretised domains?
I tried to run two problems of ex56 with two different domain size ( - attached you find the log_view outputs of both on gpus) using _pc_type asm:
./ex56 -cells 2,2,1 -max_conv_its 2 -lx 1. -alpha .01 -petscspace_degree 1 -ksp_type cg -ksp_monitor -ksp_rtol 1.e-8 -pc_type asm -snes_monitor -use_mat_nearnullspace true -snes_rtol 1.e-10 > output_221.txt
./ex56 -cells 10,10,5 -max_conv_its 2 -lx 1. -alpha .01 -petscspace_degree 1 -ksp_type cg -ksp_monitor -ksp_rtol 1.e-8 -pc_type asm -snes_monitor -use_mat_nearnullspace true -snes_rtol 1.e-10 > output_221.txt
Below is the SNES iteration for problem with 2,2,1 cells which converges after two non-linear iterations:
0 SNES Function norm 0.000000000000e+00
0 SNES Function norm 7.529825940191e+01
1 SNES Function norm 4.734810707002e-08
2 SNES Function norm 1.382827243108e-14
Below is the SNES iteration for problem with 10,10,5 cells– why does it first decrease and then increase to 0 SNES Function norm 1.085975028558e+01 and finally converge?
0 SNES Function norm 2.892801019593e+01
1 SNES Function norm 5.361683383932e-07
2 SNES Function norm 1.726814199132e-14
0 SNES Function norm 1.085975028558e+01
1 SNES Function norm 2.294074693590e-07
2 SNES Function norm 2.491900236077e-14
Kind regards,
Karthik.
From: Matthew Knepley <knepley at gmail.com>
Date: Thursday, 2 December 2021 at 10:57
To: "Chockalingam, Karthikeyan (STFC,DL,HC)" <karthikeyan.chockalingam at stfc.ac.uk>
Cc: "petsc-users at mcs.anl.gov" <petsc-users at mcs.anl.gov>
Subject: Re: [petsc-users] Unstructured mesh
On Thu, Dec 2, 2021 at 3:33 AM Karthikeyan Chockalingam - STFC UKRI <karthikeyan.chockalingam at stfc.ac.uk<mailto:karthikeyan.chockalingam at stfc.ac.uk>> wrote:
Hello,
Are there example tutorials on unstructured mesh in ksp? Can some of them run on gpus?
There are many unstructured grid examples, e.g. SNES ex13, ex17, ex56. The solver can run on the GPU, but the vector/matrix
FEM assembly does not. I am working on that now.
Thanks,
Matt
Kind regards,
Karthik.
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--
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## Strange error message using plot3d...
Asked by:
When trying to perform the following:
p1 := proc (x, y) if x^2+y^2 <= 1 then x*y-y^2 else 0 end if end proc;
plot3d(p1, x = -1 .. 1, y = -1 .. 1);
Error, (in plot3d) expected ranges but received x = -1 .. 1 and y = -1 .. 1
I get this strange error message. To the best of my knowledge x and y ARE provided as ranges. What am I missing/not understanding?
If I omit the ranges in plot3d Maple returns a correct plot, but the default range (-10 .. 10) does not display sufficient details
## Range of function ...
Asked by:
How can i calculate Range (Max. and Min.) of $f(x) = \sqrt{x^2+x+1}-\sqrt{x^2-x+1}$
## Multiplier in a specific range...
Asked by:
Dear all,
I am creating an animation, and I was wondering if I can add a multiplier to my equation in a specific range. So starting from z:=0.2 add a multiplier (z+1). The code I have so far is added. Does anyone know a code for this?
Kind regards
restart;
with(plots);
a := -1/2; b := 1/2; c := -2; d := 2; n := 20;
g := proc (x) options operator, arrow; value(Int(sigma(t), t = 0 .. x)) end proc;
sigma := proc (z) options operator, arrow; 2*sqrt(2*h^2-4*z^2)*z/h^2 end proc;
h := i/n;
for i to n do
an2[i] := plot(sigma(z), z = -(1/2)*h .. (1/2)*h, view = [a .. b, c .. d], color = AQUAMARINE);
an3[i] := plot(2*g(x), x = 0 .. (1/2)*h, view = [a .. b, c .. d], color = RED)
end do;
p := plots[display]([seq(an2[i], i = 1 .. n)], insequence = true);
q := plots[display]([seq(an3[i], i = 1 .. n)], insequence = true); display(p, q)
## nested ode programs...
Asked by:
> restart;
> with(plots);
> n := [0, .5, 1, 5]; pr := .71; p := 0; l := [1, 2, 3]; b := 0; s := 0; L := [green, blue, black, gold];
[green, blue, black, gold]
> R1 := 2*n/(n+1);
2 [0, 0.5, 1, 5]
------------------
[0, 0.5, 1, 5] + 1
for j from 1 to nops(l) do; for j from 1 to nops(n) do R1 := 2*n[j]/(1+n[j]); R2 := 2*p/(1+n[j]); sol1 := dsolve([diff(diff(diff(f(eta),eta),eta),eta)+f(eta)*diff(diff(f(eta),eta),eta)+R1*(1-diff(f(eta),eta)^2) = 0, (1/pr)*diff(diff(theta(eta),eta),eta)+f(eta)*diff(theta(eta),eta)-R2*diff(f(eta),eta)*theta(eta) = 0, f(0) = 0, (D(f))(0) = l+b*((D@@2)(f))(0), (D(f))(-2) =1, theta(0) = 1+s*(D(theta))(0), theta(-2) = 0], numeric, method = bvp); fplt[j]:= plots[odeplot](sol1,[eta,diff(diff(f(eta),eta),eta)],color=["blue","black","orange"]); tplt[j]:= plots[odeplot](sol1, [eta,theta(eta)],color=L[j]); fplt[j]:= plots[odeplot](sol1,[eta,diff(f(eta),eta)],color=L[j]); od:od:
Error, (in dsolve/numeric/bvp) unable to store 'Limit([eta, 2*eta, 3*eta]+eta^2*[.250000000000000, .500000000000000, .750000000000000]-.250000000000000*eta^2, eta = -2., left)' when datatype=float[8]
> plots:-display([seq(fplt[j], j = 1 .. nops(n))], color = [green, red], [seq(fplt[j], j = 1 .. nops(l))]);
> sol1(0);
Dear sir
In the above problem i tried to write a nested program but its not executing and showing the error as Error, (in dsolve/numeric/bvp) unable to store 'Limit([eta, 2*eta, 3*eta]+eta^2*, i want the plot range from -2 to 2 but taking only 0 to -2 ,and -2.5 to 3 but taking only 0 to 1
## How to specify solution range for solve...
Asked by:
Hey,
I want to solve this equation and looking at the plot there are at least 3 solutions. I want the greatest/smallest negative solution. Unfortunately using solve with assumptions produces no results and solve without assumptions only finds two solutions.
Can you please help me?
> #select greatest negative value from solution
> restart:
> expr:= ax*cos(lambda)+ay*sin(lambda)-(a+b*lambda)
(1)
> ax:=1:ay:=2:a:=0.5:b:=0.25: #examplanatory values
> plot(expr)
>
> assume(-2*Pi
>
> sol_lambda:=[solve(expr=0,lambda, useassumptions)];# returns empty list even though without assumption one solution is found
(2)
> sol_lambda:=[solve(expr=0,lambda)]; #returns only two solutions even though looking at the plot 3 are there
(3)
> sol_l_v:=evalb~(sol_lambda<~0); #dirty workaraound
(4)
> sol_l_add:=[ListTools:-SearchAll(true,sol_l_v)] ; #this seems overly complicated
>
(5)
> lambda:=sol_lambda[sol_l_add[-1]]; #to select the last entry
>
(6)
> expr; #test
>
(7)
>
Download select_solution.mw
Thanks!
Honigmelone
## Why are ranges exceeded?...
Asked by:
eq1 := z = y*log(x): eq2 := z = y+x*log(x):
DispIntersecting := implicitplot3d([eq1, eq2], x = 0 .. 10, y = -30 .. 30, z = -40 .. 40, color = [blue, green]):
solve({eq1, eq2}, [x, y, z]);
assign(%):
DispIntersection := spacecurve([x, y, z], x = 0.1e-2 .. 10, color = red, view = [0 .. 10, -30 .. 30, -40 .. 40]):
display(DispIntersecting, DispIntersection, axes = boxed, scaling = constrained);
## Need to restrict range on a graph...
Asked by:
Hello,
I have a procedure which plots a graph. I need the x-axes, which in this case is theta, to range between -3 and +3. However, I am not sure as to how I can create this restricted range. Any help is greatly appreciated! Thank you in advance!
Kind regards,
Gambia Man
>
>
>
>
>
>
>
Download Poincare_section_Boyd_plot.mw
>
>
>
>
>
>
>
Download Poincare_section_Boyd_plot.mw
## Solving a pde system...
Asked by:
restart;
with(DETools, diff_table);
kB := 0.138064852e-22;
R := 287.058;
T := 293;
p := 101325;
rho := 0.1e-2*p/(R*T);
vr := diff_table(v_r(r, z));
vz := diff_table(v_z(r, z));
eq_r := 0 = 0;
eq_p := (vr[z]-vz[r])*vr[] = (vr[]*(vr[r, z]-vz[r, r])+vz[]*(vr[z, z]-vz[z, r]))*r;
eq_z := 0 = 0;
eq_m := r*vr[r]+r*vz[z]+vr[] = 0;
pde := {eq_m, eq_p};
IBC := {v_r(1, z) = 0, v_r(r, 0) = 0, v_z(1, z) = 0, v_z(r, 0) = r^2-1};
sol := pdsolve(pde, IBC, numeric, time = z, range = 0 .. 1);
what am I doing wrong?
it's telling me: Error, (in pdsolve/numeric/par_hyp) Incorrect number of boundary conditions, expected 3, got 2
but i did just as in the example :-/
## extract range from solution...
Asked by:
Is there any simple way to extract the feasible range for the variables from solve result ?
For example, when I do solve({x-y = 10, x+y < 100}) I get {x = y + 10, y < 45}
From here, I need -inf<y<45 and -inf<x<55. Is it possible ? (I tried few methods, but it is not working for all cases.)
## Maple error for NonlinearFit...
Asked by:
I have been trying to fit a function to experimental data. To do this i was using
.
The data is of the type
.
When I use initialvalues, I get a result, that fits the data well, but is clearly not the desired minimum. Maple delivers g always bigger than 10000 which is nowhere near -7, where it has to be for physical reasons. When using parameterranges I get the error
Warning, no iterations performed as initial point satisfies first-order conditions.
He stopps computing and simply prints out my initialvalues or the first value that is in the parameterrange with a huge RSS.
How can I use initialvalues and parameterranges together for my data?
## How to adjust color hue...
Asked by:
In a 3d coordinate system I have a circular spacecurve with z-minimum -4 and z-maximum +4. In the same 3d coordinate system I have a 3d surface plot with z-minimum -0.5 and z-maximum +1.3 . When I choose the color option "Z(Hue)" in order to color-code the z-values on the 3d surface and make the topography more clear, I mostly get a totally green 3d surface. It seems that the color scaling is coupled with the spacecurve with z-values of +-4 . How can I uncouple the color scaling from the spacecurve and couple it with the z-range of the surface, while the color-limits shall be at +-1.3 ?
## Plotting a function over a range...
Asked by:
I've got a piece-wise function(for which I've made the procedure) f defined over x<=-1, -1<x<1 and x>=1 which I am trying to plot over the range of (-2,2). I've tried using plot(f,-2..2) but it doesn't show any curve. Should I add a few more parameters to plot()?
## How to fix: 'Error, (in int) integration range' ??...
Asked by:
int(int(y, 0 <= y, x^2+y^2+z^2 <= 1));
Error, (in int) integration range or variable must be specified in the second argument, got 0 <= y
int(int(y, y = 0, x^2+y^2+z^2 = 1));
Error, (in int) integration range or variable must be specified in the second argument, got y = 0
## How to limit a function to an interval...
Asked by:
Hello,
I have a function, lets say g(x,y,...), that depends on many other functions. But I don´t want the results that are inside certain intervals, and I need to receive the results of those functions as something like NULL when asking for a result that is inside any of those intervals. That way, g(x,y,...) would also have to result in something like NULL if any of the lesser functions are NULL fora any given values.
I tryied using the piecewise command, and for the intervals that I wanted, it worked, but for those I wanted to be NULL, they were understood as 0, and so G(x,y,..) continued to exist but with a very different value.
To clarify what I need, I will try an exemple:
Imagine I have the function f(x)=x
I want to disconsider the results for x<2 and x>6, in a way that if I try the command 'f(1)', I will receive something like NULL and know that it is outside the range.
In the same way, I need the plot of this function f(X) to show the function only from 2 to 6, but not existing for the delimited intervals.
Ad if I continue and make g(x)=f(x)+10 , I don´t want g(x) to exist if f(x) doesn´t exist, and same for the g(x) plot, which shouldn´t be shown in the intervals where f(x) don´t exist.
Thank you very much for your atention!
## How to create command...
Asked by:
Hi,
I want to calculate a range.
eg. j=1...n-1 will use this equation
P(j,n):={((m/(2-1))(m/(2))^(j-1))/(2(m/(2))^(n)-1))}
j=n will use other equation.
How to create the command? I keep receive errors.
Thank you.
Lina
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# Reasoning Questions: Puzzles Set 18
AffairsCloud YouTube Channel - Click Here
AffairsCloud APP Click Here
Hello Aspirants. Welcome to Online Reasoning Section in AffairsCloud.com. Here we are creating question sample in Puzzles set, which is common for all the IBPS,SBI exam and other competitive exams. We have included Some questions that are repeatedly asked in bank exams !!!
I.Study the following information carefully to answer the questions given below.
Eight persons A,B,C,D,E,F,H and I are going to three different destinations Gujarat, Amritsar, Hyderabad in three different cars- Chevrolet, Ford, Hyundai. Out of these three are females and also one in each car. At Least two persons are there in each car.
• E, a male, is travelling with only I and they are not going to Amritsar.
• A is travelling in Chevrolet and and is going to Hyderabad. C is not travelling with B and H.
• C and F are travelling together. H is not going to Amritsar. D is the sister of A and is travelling by Hyundai.
1. Members of which cars are going to Amritsar?
1)Ford
2)Can’t be determined
3)Hyundai
4)Chevrolet
Answer – 3)Hyundai
Explanation :
2. In which car are four members travelling?
1)None
2)Hyundai
3)Ford
4)Chevrolet
Answer – 1)None
3. Which of the following combinations represents the three female members?
1)ABC
2)CID
3)CFE
4)Can’t be determined
Answer – 4)Can’t be determined
4. Who is travelling with H?
1)CF
2)SI
3)AB
4)Can’t be determined
Answer – 3)AB
5. Members of which of the following are travelling in Chevrolet?
1)ACH
2)ABI
3)HAB
4)None of these
Answer – 3)HAB
II.Study the following information carefully to answer the questions given below.
• Five friends A,B,C,D and E wore shirts of Green, Yellow, Pink, Red and Blue colors and shorts of Black, White, Grey, Blue and Green colors.
• Nobody wore shirt and short of same color.
• D wore Blue shirt and C wore Green Short.
• The one who wore Green shirt, wore Black short and the one who wore Blue short wore Red shirt
• A wore White short and Pink shirt.
• E did not wear Red shirt.
1. Which color shirt did C wear?
1)White
2)Green
3)Yellow
4)None
Answer – 3)Yellow
Explanation :
2. Who wore Black Short?
1)B
2)E
3)A
4)C
Answer – 2)E
3. Who wore White short?
1)C
2)E
3)D
4)A
Answer – 4)A
4. Which color short did B wear?
1)Grey
2)Black
3)Green
4)Blue
Answer – 4)Blue
5. If C wore Green shirt and E wore Yellow shirt and if C wore Pink shirt then what color shirt did A wear?
1)Green
2)Blue
3)Yellow
4)None of these
Answer – 3)Yellow
Note : Dear Friends if u know an alternate methods or shortcuts related to any chapter, you can share here.
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# Differential Equations Forum
Differential Equations Help Forum
1. ### DE Tutorial - Part IV: Laplace Transforms
• Replies: 1
• Views: 3,667
Feb 9th 2011, 12:56 AM
2. ### DE Tutorial - Part I: First Order Equations and Homogeneous Second Order Equations
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Mar 15th 2010, 01:44 PM
3. ### DE Tutorial - Part III: Systems of Differential Equations
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4. ### DE Tutorial - Part II: Nonhomogenous Second Order Equations and Their Applications
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• Views: 6,223
Dec 27th 2008, 09:56 AM
1. ### Euler's Method
• Replies: 2
• Views: 3,134
Nov 9th 2009, 08:20 AM
2. ### Boundary Value Problem
• Replies: 0
• Views: 603
Nov 9th 2009, 08:17 AM
3. ### second order differential equation
• Replies: 4
• Views: 581
Nov 9th 2009, 03:36 AM
4. ### Method of Undetermined Coeffiicients Problem
• Replies: 1
• Views: 392
Nov 8th 2009, 06:56 PM
5. ### Damped Spring Question
• Replies: 0
• Views: 744
Nov 8th 2009, 02:23 PM
6. ### PDE Question...
• Replies: 2
• Views: 743
Nov 8th 2009, 06:42 AM
7. ### find the maximum rate of change of x
• Replies: 6
• Views: 1,151
Nov 8th 2009, 05:41 AM
8. ### how to determine separable & linear?
• Replies: 3
• Views: 827
Nov 8th 2009, 03:45 AM
9. ### Existence
• Replies: 2
• Views: 777
Nov 7th 2009, 08:31 PM
10. ### Differetial Equation with Sine Integral
• Replies: 2
• Views: 768
Nov 7th 2009, 08:54 AM
11. ### A Differential Application - Force/Resistance
• Replies: 1
• Views: 452
Nov 7th 2009, 12:43 AM
12. ### Intro to PDE's, Schrodinger's Equation
• Replies: 1
• Views: 897
Nov 6th 2009, 08:50 AM
13. ### Diffy Q word problem
• Replies: 2
• Views: 882
Nov 5th 2009, 07:53 PM
14. ### Springs
• Replies: 3
• Views: 752
Nov 5th 2009, 03:29 PM
15. ### Poisson Equation, Energy Function (need help)
• Replies: 1
• Views: 1,311
Nov 5th 2009, 08:21 AM
16. ### [SOLVED] Differential equation
• Replies: 1
• Views: 715
Nov 5th 2009, 05:44 AM
17. ### Laplace transform
• Replies: 2
• Views: 436
Nov 5th 2009, 04:18 AM
18. ### Wronskian and Linear Independence
• Replies: 0
• Views: 512
Nov 5th 2009, 12:51 AM
19. ### dP/dt=sq. Pt
• Replies: 11
• Views: 4,429
Nov 4th 2009, 10:47 PM
20. ### [SOLVED] Logistic Equation
• Replies: 0
• Views: 613
Nov 4th 2009, 07:29 PM
21. ### Laplace transforms (shift thereom)
• Replies: 1
• Views: 437
Nov 4th 2009, 05:57 PM
22. ### Laplace transform of delta(t - a)
• Replies: 1
• Views: 725
Nov 4th 2009, 05:38 PM
23. ### ODE finding constants...
• Replies: 3
• Views: 557
Nov 4th 2009, 01:28 PM
24. ### Differential Equation
• Replies: 2
• Views: 382
Nov 4th 2009, 03:31 AM
25. ### differential equations #2
• Replies: 1
• Views: 5,614
Nov 4th 2009, 12:22 AM
,
,
,
,
,
,
### forum on what are the applied differential equation topic
Click on a term to search for related topics.
Use this control to limit the display of threads to those newer than the specified time frame.
Allows you to choose the data by which the thread list will be sorted.
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# Questions tagged [trust-region-policy-optimization]
For questions about the Trust Region Policy Optimization (TRPO) algorithm, which was introduced in the paper "Trust Region Policy Optimization" (2015) by J. Schulman et al.
9 questions
Filter by
Sorted by
Tagged with
18 views
### Why does PPO lead to a worse performance than TRPO in the same task?
I am training an agent with an Actor-Critic network and update it with TRPO so far. Now, I tried out PPO and the results are drastically different and bad. I only changed from TRPO to PPO, the rest of ...
71 views
### How can I implement the reward function for an 8-DOF robot arm with TRPO?
I need to get an 8-DOF (degrees of freedom) robot arm to move a specified point. I need to implement the TRPO RL code using OpenAI gym. I already have the gazebo environment. But I am unsure of how to ...
91 views
### Understanding proof of lemma 1 (policy improvement bound) of the “Trust Region Policy Optimization” paper
In the Trust Region Policy Optimization paper, in Lemma 1 of Appendix A, I did not quite understand the transition from (21) from (20). In going from (20) to (21), $A^\pi(s_t, a_t)$ is substituted ...
114 views
### Are these two TRPO objective functions equivalent?
In the TRPO paper, the objective to maximize is (equation 14) $$\mathbb{E}_{s\sim\rho_{\theta_\text{old}},a\sim q}\left[\frac{\pi_\theta(a|s)}{q(a|s)} Q_{\theta_\text{old}}(s,a) \right]$$ which ...
43 views
### How does the TRPO surrogate loss account for the error in the policy?
In the Trust Region Policy Optimization (TRPO) paper, on page 10, it is stated An informal overview is as follows. Our proof relies on the notion of coupling, where we jointly define the ...
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In the Trust Region Policy Optimization paper, in Lemma 2 of Appendix A, I did not quite understand deriving inequality (31) from (30), which is: $$\bar{A}(s) = P(a \neq \tilde{a} | s) \mathbb{E}_{(a,... 2answers 735 views ### Why is the log probability replaced with the importance sampling in the loss function? In the Trust-Region Policy Optimisation (TRPO) algorithm (and subsequently in PPO also), I do not understand the motivation behind replacing the log probability term from standard policy gradients$$L^...
I happened to discover that the v1 (19 Feb 2015) and the v5 (20 Apr 2017) versions of TRPO papers have two different conclusions. The Equation (15) in v1 is $\min_\theta$ while the Equation (14) in v2 ...
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# Homopolar designs and equations?
1. Feb 17, 2009
### confusedone
Homopolar designs and equations???
I have been experimenting with Homopolar generation for some time now and have some experimental observations I would like to share with this community. I am looking for people who have had some exposure to this method of producing electricity and anyone else who may have some constructive ideas about whats going on and ideas about how significant these observations of mine may be.
The part that intrigues me about this method of producing electricity is that you can fix the induction magnet to the induction plate and produce the same current as you would if the the two where seperated. (I will attach illustrations of the homopolar design I am refereing to) The above statement is not assumption but experimentally observed. Its interesting to note that by physically attaching the two components the typical backward torque on a system designed this way is elliminated. The current induced in this system, produces eddy currents in the conductor and consequently random magnetic fields that would interact with the induction magnet and in turn work to slow the motion of the conductor. I am sure that these eddy currents are still present when the two are fixed totheger but the method used to fix the two components together converts this into internal resistance on the mechanical structure instead of the magnetic structure. (I am sure as well that this mechanical resistance is being converted to heat in some small way)
I have further experimented with this principle and found that the conductor can be sectioned radially (pie sections) and each section will produce its own seperate current. This fact is interesting in that, if treated like individual power supplies they can be manipulated the same way and be connected in parallel or series to produce more practical currents. The experiment I conducted showed a significant disproportion between voltage (substantially higher) vs. Amperage (substantially lower). The experimental piece consisted of 3" diameter 1/8" thick neodymium magnet insulated and fixed to eight radially sectioned pieces of copper cut from a copper sheet. The radial dimention for the copper was also 3" and each section was insulated from one another and seperation was confirmed through continuity tests between all components...
Having done some research on this subject (till my eyes hurt) the typical result from such a method as this produces very HIGH currents which result in most experimenters just writing of the whole phenominon (spell check) as being impractical for any use at all. My experiments point to a different result and the potential for practical energy production with unheard of efficiency....
Your thoughts and questions are welcome.
One more thing too... I have not found any reliable math out there that can be used to predict the results of this type of electrical production... If you have some insight on this subjet I am all eyes....
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# GCSE Calculator Questions 1 (2013/Higher Tier)
This worksheet contains a sample of practice questions based on the 2013 GCSE Higher Level syllabus. Calculators may be used.
Key stage: KS 4
Curriculum topic: GCSE Practice Papers
Curriculum subtopic: Selection of Topics for Calculator Practice
Difficulty level:
### QUESTION 1 of 10
Formula Sheet
Higher Tier
Area of trapezium = ½ (a + b) h
Volume of prism = area of cross-section x length
In any triangle ABC:
Area of triangle = ½ab sinC
Sine rule
a = b = c sin A sin B sin C
Cosine rule
a2 = b2 + c2 - 2bc cos A
The solutions of ax2 + bx + c = 0, where a ≠ 0, are given by
Write the following fraction as a decimal:
10 32
Write the following recurring decimal as a fraction in its lowest terms.
Calculate the value of the following to 3 decimal places:
5.962 - 2.145 1.986
Express 350 as a product of prime factors.
2×52 ×73
2×42 ×72
2×52 ×74
2×52 ×7
A triangle has a height of 7x and a base length of 3x.
The area of the triangle is 147 cm2.
Select the equation which shows this information.
3x + 7x = 147
10x² = 147
21x² = 147
21x² = 294
Charlie invests £3200 for 3 years in a savings account that gives 5% interest per year.
How much more interest will Charlie earn if he invests the money at compound interest rather than at simple interest?
Look at the diagram below which shows the sector of a circle of radius 4 cm.
Calculate the arc length AB in cm to 3 sig. figs.
(Just write the number)
Solve the quadratic equation by selecting all the correct solutions below.
8x2 + 2x - 3 = 0
x = ¾
x = -¾
x = -½
x = ½
x = -2
Sam measures the angle of elevation of a point at the top of the Tower of Pisa to be 28°.
Use the information shown in the diagram below to calculate the height of the Tower of Pisa to the nearest metre.
(Just write the number)
In triangle PQR, which is not drawn to scale,
PQ = 4.5 cm, QR = 10 cm, and PR = 12 cm.
Use the cosine rule to find ∠PQR to the nearest degree.
(Just write the number)
In triangle PQR, which is not drawn to scale,
PQ = 4.5 cm, QR = 10 cm, and PR = 12 cm.
Calculate ∠PQR to the nearest degree.
(Just write the number)
Here is a formula invented by Einstein.
E = mc2
Calculate E when m = 6.2 x 10-5 and c = 3 x 108.
5.58 × 103
5.58 × 10-12
5.58 × 1012
1.86 × 104
1.86 × 1012
150 cm of copper wire has a mass of 10.4 grams.
Calculate the mass of 270 cm of the copper wire in grams to 2 dps.
Calculate the length of the line segment shown in the diagram below to the nearest tenth of a unit.
Ben is a keen photographer.
He goes on holiday for 14 days and takes roughly 350 photographs per day. Each photograph uses 12.5 MB of disk space and his camera has a new 64 GB card at the start of his holiday.
What percentage of disk space remains on the card at the end of his holiday?
(1GB = 1000 MB)
4.3%
95.7%
4.2%
95.8%
There are 405 people in a school.
The ratio of children to adults is 25:2.
How many children are at the school?
Find the length of the cuboid's diagonal BH of this 4 cm x 5 cm x 12 cm cuboid in cm to 3 sig. figs.
(Just write the number)
One solution to the following equation lies between x = 4 and x = 5.
2x3 + 2x2 + x = 200
Using trial and improvement, find this solution to 1 d.p.
(Just write the number)
Solve the following equation for a:
7a - 10 = 8a - 5 5 7
(Just write the value of a)
There are 16 boys and 10 girls in a class.
The teacher choses two children at random.
What is the probability of choosing a boy and a girl?
In the diagram below, C is the centre of the circle and TU is a tangent.
Calculate the size of angle CRS.
(Just write the number)
A sequence begins:
12, 19, 26, 33, 40 ...
Write down the nth term of the sequence.
Charlie is experimenting with seedlings and measures the height of his seedling plants after two weeks. The results are recorded in this frequency table.
Estimate the mean height of the seedlings in mm to the nearest whole number.
(Just write the number)
Calculate the area of the triangle PQR in cm2 to 1 decimal place.
All lengths are shown in cm.
The diagram shows a solid cylinder, made from metal with a density of 7.4 grams per cm3.
Calculate the mass of the cylinder in grams to 3 significant figures.
(Just write the number)
• Question 1
Write the following fraction as a decimal:
10 32
0.3125
EDDIE SAYS
10 ÷ 32 = 0.3125
• Question 2
Write the following recurring decimal as a fraction in its lowest terms.
5/11
EDDIE SAYS
45/99 reduces to 5/11
• Question 3
Calculate the value of the following to 3 decimal places:
5.962 - 2.145 1.986
16.806
EDDIE SAYS
On the calculator, type:
5.96 x² - 2.145 = ÷ 1.986 =
• Question 4
Express 350 as a product of prime factors.
2×52 ×7
EDDIE SAYS
350 = 10 × 35
350 = 2 × 5 × 5 × 7
• Question 5
A triangle has a height of 7x and a base length of 3x.
The area of the triangle is 147 cm2.
Select the equation which shows this information.
21x² = 294
EDDIE SAYS
Area of triangle = Base × Height ÷ 2
147 = 3x × 7x ÷ 2
147 × 2 = 3x × 7x
294 = 21x²
• Question 6
Charlie invests £3200 for 3 years in a savings account that gives 5% interest per year.
How much more interest will Charlie earn if he invests the money at compound interest rather than at simple interest?
£24.40
£ 24.40
EDDIE SAYS
Simple interest will earn him 0.05 × 3200 × 3 = £480
Compound interest will earn him 1.05³ × 3200 - 3200 = £504.40
£504.40 - £480 = £24.40
• Question 7
Look at the diagram below which shows the sector of a circle of radius 4 cm.
Calculate the arc length AB in cm to 3 sig. figs.
(Just write the number)
2.09
EDDIE SAYS
Circumference = 8π
Arc Length = 30/360 × 8π = 2.094395.....
• Question 8
Solve the quadratic equation by selecting all the correct solutions below.
8x2 + 2x - 3 = 0
x = -¾
x = ½
EDDIE SAYS
This factorises to give the equation (2x-1)(4x+3)= 0
• Question 9
Sam measures the angle of elevation of a point at the top of the Tower of Pisa to be 28°.
Use the information shown in the diagram below to calculate the height of the Tower of Pisa to the nearest metre.
(Just write the number)
57
EDDIE SAYS
Let height above Sam's eye be h.
Using trig, h = 104 tan 28° = 55.3 m
Height of tower from ground = 55.3 + 1.6 = 56.9 ≅ 57 m
• Question 10
In triangle PQR, which is not drawn to scale,
PQ = 4.5 cm, QR = 10 cm, and PR = 12 cm.
Use the cosine rule to find ∠PQR to the nearest degree.
(Just write the number)
22
EDDIE SAYS
12² = 10² + 4.5² - 2 × 10 × 4.5 × cosθ°
• Question 11
In triangle PQR, which is not drawn to scale,
PQ = 4.5 cm, QR = 10 cm, and PR = 12 cm.
Calculate ∠PQR to the nearest degree.
(Just write the number)
22
EDDIE SAYS
12² = 10² + 4.5² - 2 × 10 × 4.5 × cosθ°
• Question 12
Here is a formula invented by Einstein.
E = mc2
Calculate E when m = 6.2 x 10-5 and c = 3 x 108.
5.58 × 1012
EDDIE SAYS
E = 6.2 × 10-5 × 3 × 108 × 3 × 108
• Question 13
150 cm of copper wire has a mass of 10.4 grams.
Calculate the mass of 270 cm of the copper wire in grams to 2 dps.
18.72
EDDIE SAYS
10.4 ÷ 150 × 270 = 18.72
• Question 14
Calculate the length of the line segment shown in the diagram below to the nearest tenth of a unit.
9.2
EDDIE SAYS
Build a right-angled triangle using the end-points of the line segment.
The triangle has side lengths of 7 and 6.
Use Pythagoras' Theorem to calculate the length as
√(6² + 7²) = √85
• Question 15
Ben is a keen photographer.
He goes on holiday for 14 days and takes roughly 350 photographs per day. Each photograph uses 12.5 MB of disk space and his camera has a new 64 GB card at the start of his holiday.
What percentage of disk space remains on the card at the end of his holiday?
(1GB = 1000 MB)
4.3%
EDDIE SAYS
Disk space used = 14 × 350 × 12.5 = 61250 MB
Percentage remaining = (64000 - 61250) ÷ 64000 × 100 = 4.2968... %
• Question 16
There are 405 people in a school.
The ratio of children to adults is 25:2.
How many children are at the school?
375
EDDIE SAYS
Number of children = 25 ÷ (25 + 2) × 405 = 375
• Question 17
Find the length of the cuboid's diagonal BH of this 4 cm x 5 cm x 12 cm cuboid in cm to 3 sig. figs.
(Just write the number)
13.6
EDDIE SAYS
Use Pythagoras' Theorem to get:
BH² = 4² + 5² + 12² = 16 + 25 + 144 = 185
• Question 18
One solution to the following equation lies between x = 4 and x = 5.
2x3 + 2x2 + x = 200
Using trial and improvement, find this solution to 1 d.p.
(Just write the number)
4.3
EDDIE SAYS
2 × 4.25³ + 2 × 4.25² + 4.25 = 193.90625 (just too small)
2 × 4.3³ + 2 × 4.3² + 4.3 = 200.294 (just too big)
• Question 19
Solve the following equation for a:
7a - 10 = 8a - 5 5 7
(Just write the value of a)
5
EDDIE SAYS
Multiply both sides by 35 to get:
7(7a - 10) = 5(8a - 5)
• Question 20
There are 16 boys and 10 girls in a class.
The teacher choses two children at random.
What is the probability of choosing a boy and a girl?
32/65
EDDIE SAYS
16/26 × 10/25 + 10/26 × 16/25
• Question 21
In the diagram below, C is the centre of the circle and TU is a tangent.
Calculate the size of angle CRS.
(Just write the number)
45
EDDIE SAYS
∠CTP = 90 - 65 = 25°
∠TSR = ½ of 140 = 70°
reflex ∠TCR = 360 - 140 = 220°
∠CRS = 360 - 220 - 70 - 25 = 45°
• Question 22
A sequence begins:
12, 19, 26, 33, 40 ...
Write down the nth term of the sequence.
7n+5
EDDIE SAYS
The terms increase by 7, so the nth term is related to 7n.
Compare the sequence to 7, 14, 21, 28, 35 and we realise that we must add 5 to get the given sequence.
• Question 23
Charlie is experimenting with seedlings and measures the height of his seedling plants after two weeks. The results are recorded in this frequency table.
Estimate the mean height of the seedlings in mm to the nearest whole number.
(Just write the number)
16
EDDIE SAYS
We use the midpoint of each class, i.e. 2.5mm, 7.5mm, 12.5 mm etc.. to work out the total of all the seedling heights as:
4 × 2.5 + 12 × 7.5 + 22 × 12.5 + 29 × 17.5 + 15 × 22.5 + 11 × 27.5 = 1522.5
There are 93 seedlings if we add up the numbers in the frequency column.
Mean = 1522.5 ÷ 93 = 16.37...
• Question 24
Calculate the area of the triangle PQR in cm2 to 1 decimal place.
All lengths are shown in cm.
27.4
EDDIE SAYS
Area = ½ × 5 × 11 × sin95° = 27.395...cm²
• Question 25
The diagram shows a solid cylinder, made from metal with a density of 7.4 grams per cm3.
Calculate the mass of the cylinder in grams to 3 significant figures.
(Just write the number)
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| 4.1875
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http://functions.wolfram.com/Bessel-TypeFunctions/BesselJ/23/01/0005/
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html, body, form { margin: 0; padding: 0; width: 100%; } #calculate { position: relative; width: 177px; height: 110px; background: transparent url(/images/alphabox/embed_functions_inside.gif) no-repeat scroll 0 0; } #i { position: relative; left: 18px; top: 44px; width: 133px; border: 0 none; outline: 0; font-size: 11px; } #eq { width: 9px; height: 10px; background: transparent; position: absolute; top: 47px; right: 18px; cursor: pointer; }
BesselJ
http://functions.wolfram.com/03.01.23.0005.01
Input Form
Sum[BesselJ[k, x] t^k, {k, -Infinity, Infinity}] == E^((1/2) x (t - 1/t))
Standard Form
Cell[BoxData[RowBox[List[RowBox[List[UnderoverscriptBox["\[Sum]", RowBox[List["k", "=", RowBox[List["-", "\[Infinity]"]]]], "\[Infinity]"], RowBox[List[RowBox[List["BesselJ", "[", RowBox[List["k", ",", "x"]], "]"]], " ", SuperscriptBox["t", "k"]]]]], "\[Equal]", SuperscriptBox["\[ExponentialE]", RowBox[List[FractionBox["1", "2"], " ", "x", " ", RowBox[List["(", RowBox[List["t", "-", FractionBox["1", "t"]]], ")"]]]]]]]]]
MathML Form
k = - J k ( x ) t k 1 2 x ( t - 1 t ) k -1 BesselJ k x t k 1 2 x t -1 1 t -1 [/itex]
Rule Form
Cell[BoxData[RowBox[List[RowBox[List["HoldPattern", "[", RowBox[List[UnderoverscriptBox["\[Sum]", RowBox[List["k_", "=", RowBox[List["-", "\[Infinity]"]]]], "\[Infinity]"], RowBox[List[RowBox[List["BesselJ", "[", RowBox[List["k_", ",", "x_"]], "]"]], " ", SuperscriptBox["t_", "k_"]]]]], "]"]], "\[RuleDelayed]", SuperscriptBox["\[ExponentialE]", RowBox[List[FractionBox["1", "2"], " ", "x", " ", RowBox[List["(", RowBox[List["t", "-", FractionBox["1", "t"]]], ")"]]]]]]]]]
Date Added to functions.wolfram.com (modification date)
2001-10-29
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U.S. Department of Transportation
1200 New Jersey Avenue, SE
Washington, DC 20590
202-366-4000
Federal Highway Administration Research and Technology
Coordinating, Developing, and Delivering Highway Transportation Innovations
This report is an archived publication and may contain dated technical, contact, and link information
Federal Highway Administration > Publications > Research Publications > Safety > 98096 > Modeling Intersection Crash Counts and Traffic Volume - Final Report
Publication Number: FHWA-RD-98-096 Date: September 1997
Modeling Intersection Crash Counts and Traffic Volume - Final Report
7. CONCLUSIONS ON MODELING INTERSECTION CRASHES IN RELATION TO TRAFFIC VOLUMES AS EXPOSURE MEASURES
7.1 Relations between crash counts and traffic volumes at four–leg signalized intersections
7.2 What can currently be done?
7.3 Substantive research needed
7.4 Methodological research needs
7.1 Relations between crash counts and traffic volumes at four–leg signalized intersections
The currently readily available exposure measures at intersections are the average traffic volumes on the intersecting roads. Intuitively, one expects crashes in intersections to result from the interaction of the two traffic streams. The simplest mathematical function expressing an interaction is z=a*x*y. This expression is too rigid because it has only one parameter, a. A simple generalization is z=a*xbyc. It is often used, usually in logarithmic from as a log–linear model.
A critical question is, is this form adequate to represent the actual relation between crashes and traffic volumes, or are there better relationships? The simplest way to obtain an approximation to the "true" relationship between crashes and traffic volumes is to smooth crash counts over the two traffic volumes. We did this for three data sets: four–leg signalized and stop–controlled intersections in Washtenaw County, Michigan; four–leg signalized intersections in California; and four–leg signalized intersections in Minnesota.
The Washtenaw County data suffered from a selection bias because only intersections with more than a certain number of crashes were included. Such data are rarely selected in practice. For this data set, crash counts increased with increasing major volume as well as with increasing minor volume. The conventional log–linear model appeared to be an acceptable qualitative representation. Quantitatively, however, there were complex systematic deviations between the data and the model. There was practically no relationship between crash counts and traffic volumes for crashes at stop–controlled intersections in this data set.
The large number of intersections in the California data set allowed detailed analysis. A fairly simple visual, but analytically complex, relationship was apparent. The most obvious feature of the surface was a "ridge" at minor road volumes of 20,000 vpd. Up to that value, crash counts increased nearly linearly with the minor volume. Beyond that volume, they initially dropped rapidly, and then leveled off. The relationship with the major volume was not that pronounced but showed a fairly strong increase at low volumes, no or moderate increase at middle volumes, and an irregular decrease at high volumes.
Crashes within the intersection itself showed a similar but much less pronounced pattern with a weaker variation with the volumes. Crashes on the major approach showed a pattern very similar to that of all crashes, while those on the minor approach showed a definitely different pattern. There was, however, a "ridge" beyond which again crashes declined, a very strong increase with minor volume, and relatively little variation with major volume. It was obvious that a log–linear model could not even roughly approximate the actual surface.
One possible reason for a deviation from the expected pattern is that intersections that otherwise would have very high crash counts have been "improved" so as to reduce the crash risk. However, none of the intersection characteristics given in the data file that might reduce the crash risk appeared more frequently in the areas of the diagrams where the unexpected decline of the crash counts occurred. Thus, there must be either other intersection features not available in our data file, or the relationship between crash counts and traffic volumes must be far from log–linear.
The analysis of Minnesota intersections was limited by their low number. They showed a complex pattern. The relationship of major volume to crash counts was nearly a step function, which was approximately constant for low volumes, even more so for high volumes, with a "ramp" connecting the two levels. Strong smoothing that came close to fitting a plane to the data points resulted in a surface that increased with both volumes.
Crashes within the intersection itself showed a slightly different pattern. The pattern for the major volume again had two levels connected by a ramp, but there was a fairly strong increase with minor volume. Crashes on the approaches showed no clear pattern. Strong smoothing revealed only a weak increase with major volume and a stronger increase with minor volume.
A log–linear function was qualitatively similar to the more strongly smoothed surface, but deviated quantitatively. It could not represent the less strongly smoothed surface showing two levels and a connecting ramp.
If the crash risk in specific intersection maneuvers, such as turning left, turning right, going straight, etc., were the same across and independent of the volumes, but not across maneuvers, the frequencies of crashes reflecting such maneuvers would be proportional to the frequencies of the maneuvers. If an intersection had many high–risk maneuvers, one would expect more crashes than at intersections with comparable volumes but with fewer such maneuvers. Therefore, we also explored possible relationships between the frequencies of crash types, and of total crashes, in the Minnesota and California data sets. We found none.
Our three data sets showed very different relationships between crash counts at four–leg signalized intersections and the traffic volumes on the intersecting roads. None of them could be adequately represented by the conventional log–linear model. Either other intersection characteristics that were not readily available in our data sets had a strong influence on crash counts, or average annual daily traffic is not an adequate exposure measure.
Top
7.2 What can currently be done?
Considering our negative conclusions about the usefulness of using conventional mathematical models to represent relationships between crash counts at signalized intersections and traffic volumes, what can be done? Smoothing is a promising alternative because it allows the fitting of even complicated surfaces by a simple process and avoids arbitrary assumptions. Using a function of two volumes, as done in this study, is a relatively simple matter. If more than two volumes, or other variables, especially categorical variables, are desired, the procedures have to be extended and refined, as discussed in subsequent sections.
We cannot rule out the possibility that someone may find manageable and not too ad–hoc mathematical expression for the relationship between traffic volumes and crash counts. By ad hoc we mean mathematical expressions selected specifically to fit the data sets studied, without consideration whether they could plausibly be extended to other data sets. However, such mathematical expressions have to be validated by more detailed criteria than correlative coefficients, likelihood ratios, or similar aggregate measures to be acceptable substitutes for smoothed surfaces.
What can be done in practice with such smoothed (or validated analytical) relationships? They can be used to compare the experience of an individual intersection with that expected from the relationship. If the difference is sufficiently large then the crash experience of that intersection should be studied in detail. Criteria for what is considered sufficiently large still have to be selected. Possibly, an explanation that suggests either which countermeasure should be applied to that intersection or which features of that intersection might have the beneficial effect of a crash countermeasure could be found.
This approach has its limitations. It can work only for intersections in an "area," defined by combinations of the two volumes, with enough data points in the "area" so that the individual peculiarities of the intersections will average out. It will not work for relatively isolated intersections near the boundary of the area covered by intersections. There, the smoothed surface will be "pulled" toward the value of each individual intersection. Even if the actual crash count for an intersection may be much higher than to be expected from the "true" relationship between crashes and volumes, this deviation may not be recognizable.
Top
7.3 Substantive research needed
Before one can realistically think about modeling intersection crash counts, one needs to develop a more realistic logical and functional structure for such models. A first step is to re–think the concept of exposure. As already discussed, traffic volumes on the intersecting roads are conceptually unsatisfactory. Only in the simplest case of uncontrolled intersections can one expect crash counts to be log–linear or similar functions of the volumes. An exposure measure should count the opportunities for collisions. These depend heavily on the type and characteristics of traffic control provided. Promising steps to develop more meaningful exposure measures have been taken. However, much more work on the problem using different perspectives is needed.
Another aspect is that intersection crashes, and even more so, intersection–related crashes, are very inhomogeneous. It cannot be expected that a single model will describe their frequency in a manner reflecting crash causation. Therefore, a closer examination of intersection crash types should be made and classes for meaningful modeling must be identified. This might require performing nearly a "clinical" analysis of individual crashes.
In reality, many intersections have certain features exactly because they had much crash experience. This creates relationships that make the standard statistical models uninterpretable. Either much more sophisticated models have to be developed, or different techniques used.
One alternative to the conventional approach, that of using a large set of intersections and including many variables in a complicated model, is to select intersections that are matched in many respects as closely as practical, and differ only in one or very few characteristics to be studied. This is much more likely to isolate any effect of such characteristics. If this is done, in turn, with many different subsets of intersections, a realistic model may be built in a stepwise fashion.
Top
7.4 Methodological research needs
Before smoothing can be used routinely to model relationships between crash counts and exposure measures and other intersection characteristics, additional research needs to be done.
A realistic model will contain one or several exposure measures that are continuous variables (or counts that can be treated as continuous variables), intersection characteristics that will usually be described by 0/1 categorical variables, and possibly other continuous variables, such as travel speeds. In principle, one can smooth over all continuous variables simultaneously, but one cannot smooth over the categorical variables. They have to be accommodated by either additive or multiplicative terms, or the entire data set may have to be split according to a categorical variable, or combinations of several categorical variables, and each part modeled separately. Criteria have to be developed to decide when each of these treatments is appropriate.
Though it is possible to smooth data sets with a large number of continuous independent variables, it is not very useful. The data can be stored in a computer, or in hardcopy tables, and the smoothed value can be calculated for any combination of the independent variables. However, if the number of variables is greater than two, or at most three, the smoothed surface cannot be visualized or intuitively assessed for overall shape and smoothness. As a practical matter, one wants to separate the model into additive or multiplicative components, each of which can be studied and assessed separately. Indeed, this is the same approach used in analytical modeling, where one uses additive or multiplicative terms. If interactions have to be considered, they are introduced as additional additive or multiplicative terms. Since one can easily visualize a surface smoothed over two variables, one only needs to determine how to separate a model into components representing main effects, or interactions of any two variables. Research is needed to learn how to do this best and how to assess the adequacy of such additive or multiplicative models.
If one deals with experimental data where by design the data points can cover the range of the variables of interest more or less uniformly, smoothing by standard methods can give a good representation of the relationship, and the deviations of the individual points from the surface can give a good idea of the random variability of the data points.
In the case of intersections, and probably also other highway locations, the situation is different. Most observations are concentrated in only part of the entire area covered with observations. Toward the edges of this area, observations become more sparse and may be isolated. This poses a dilemma for smoothing. In the areas densely covered with observations, a narrow smoothing window may be appropriate, because it can well represent a complex relationship and still provide adequate smoothing. Where the points are more isolated, such a narrow window is no longer appropriate, because in extreme cases it may result in a perfect or at least very good fit to any single point or to a combination of only a few points. This can result in erratic behavior of the smoothed surface toward its boundaries. To avoid this, one might enlarge the smoothing window. While this has the effect of giving a smoothed surface near the boundaries of the covered area, it can result in smoothing out important details in the area well covered with points. Techniques should be developed that avoid this, for instance, by using a window with adoptive size, or by identifying parts of the smoothed surface that depend on only a few data points.
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FHWA-RD-98-096
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# Cellular Phone Hacking and Phreaking
### Cellular Phone Authentication FAQ
> Very interesting! I haven't heard anything about ORYX...
> Was it already implemented in any cellular network?
> For which cellular system was it originally intended (IS-54, IS-136,
> CDMA, GSM ?)
So far as I know, it hasn't been implemented anywhere yet. I don't
know what it was intended for -- it was described in the Common Cryptographic
Algorithms specification (which describes crypto-primitives which are
common to many or all of the North American cellular standards).
It definitely wasn't intended for GSM -- this was by the North American
standards body (TIA/CTIA/etc.), which has nothing to do with GSM.
> I mean how many values of A-key exist for any given RANDU/AUTHU pair?
Well, the A-key is 64 bits, so there are 2^64 possible A-key values.
(The A-key checksum is only for use when typing in the A-key by hand
Those additional checksum digits are then securely erased after entry
of the A-key. Typically the A-key should be entered before the customer
even sees the cellphone, or entered once by the customer when they start
service. The A-key checksum is merely intended to detect typing errors,
from what I can tell.)
Therefore, I don't think the A-key checksum will help in determining
the A-key.
Now if you have a RAND/AUTHU pair, I believe the AUTHU value is 18 bits
long, so in theory this means that we can eliminate all but 1 out of
every 2^18 possibilities for the A-key. In other words, 2^{64-18} = 2^46
possibilities will remain. With two RAND/AUTHU pairs, 2^28 possibilities
will remain, etc. With four RAND/AUTHU pairs, the A-key will (theoretically)
be determined uniquely.
However, even though the A-key will be determined uniquely, actually
finding that unique value is very difficult -- it requires much computation.
The most straightforward approach, brute force exhaustive key search,
would require 2^64 computations, which is probably too much to be feasible
today.
> Perhaps if we know both RANDU and AUTHU as a shortcut, and
> we could use the A-key checksum as another shortcut, it would
> take less than 100 hours of Pentium 200 to calculate all the possible
> (not 2^64, probably 2^8 or less) values of A-key. That's what I want
> to know. However, I'm not an expert in crypto, unfortunately :)
Well, that's the \$64,000 question. I'm not aware of any shortcuts which
let you recover the A-key quickly given a few known RANDU/AUTHU values.
There are a number of cryptographers who have looked for such a shortcut;
it's not known whether one exists. This is the subject of active research.
I can say that CAVE appears to be cryptographically more secure than
any of the other cryptographic primitives used in the North American
cellphones. (That shouldn't be too surprising, since the industry
has a lot of money riding on the security of CAVE.)
> I understand that the CAVE is factory built-in in all newer AMPS/DAMPS
> cellular phones. Upon reception of RANDU (24 bit argument) it calculates
> AUTHU (18 bit function) according to one-way hash CAVE algorithm using
> additional arguments MIN1 (24 bit), MIN2 (8 bit), ESN (32 bit) and SSD-A
> (64 bit).
Right.
I do know this. Suppose we could get the full output (all 16 bytes) of
CAVE after hashing MIN1,MIN2,ESN,SSD-A, instead of just 18 bits (AUTHU)
selected from the output. Then there is a partial shortcut which lets
you reduce the amount of work required from 2^64 to 2^56. That would
mean that one could recover an SSD-A-key in about the same amount of time
that it took to break DES recently.
> I'm interested in the reverse process: knowing RANDU, AUTHU, MIN1, MIN2
> and ESN can anybody write a PC program to calculate all the values of
> A-key (how many such values do exist and how long will it take to
> calculate all of them with Pentium 200?). What if we know SSD-A as well?
SSD-A is generated (if I remember correctly) as the output of CAVE
applied to some different inputs, including the A-key. So it seems
that if you could find a shortcut to quickly recover SSD-A knowing
just RANDU,AUTHU,MIN1,MIN2,ESN... then the same shortcut might very
well also let you recover the A-key konwing SSD-A,MIN1,MIN2,ESN.
New!
A-Key Calculator (Win32) program for cellular industry
Description: this is an indispensable tool for cellular industry staff including technicians and security managers, as well as cellular shops and dealers. Now generating a valid A-Key with 6-digit checksum can be done on the spot, provided you know the phone's ESN number. This is sometimes required for testing the phone's A-Key entry function as well as for the phone's activation. It used to be possible to obtain a valid A-Key with cheksum only from the cellular operator itself, now everyone can easily generate the secret checksum value, enter those magic numbers into the phone just using the phone's own keypad and see how the phone is accepting the new A-Key by displaying "VALID". The A-Key Calculator program is not free, a single computer license costs US\$49.00. The evaluation period is 30 days.
New servicing software for CDMA phones
(reading and writing ESN, MIN, SCM, A-Key, SSD, reading the SPC code, removing the operator's binding, installing the new PRL list, reprogramming the American operator subsidized phones to work in any CDMA network, calculating the ESN checksum, etc.)
AMPS/DAMPS IS-54b Cellular Protocol
(the new TIA/EIA-627 version)
The Source of Standards and Technical Papers for Cellular Industry
(IS-54b, IS-136, IS-95, IS-88, IS-41, TSB-51 and many more)
Obtaining a Copy of CAVE Algorithm is not Easy
Call Processing Software
(CDMA, IS-136)
INfusion IS-41C Authentication Center
Cellular Authentication as seen by NACN
Cracking CDMA and TDMA Algorithms
GSM Technical Specifications List
GSM Security and Encryption
GSM Security Questions
Overview Of GSM
GSM Encryption Algorithm A5
The Source Code of A5 Algorithm
GSM Encryption Algorithms A3 & A8
Very good overview of smartcards
The mobile phone meets the Internet
ESN Manufacturer's Code Assignment - AMPS/DAMPS
Intercepting GSM Phone Traffic by Professionals
Cellular Phone Fraud Countermeasures (Corsair)
This was written 13 years ago! - AMPS/ETACS/NMT
The Case of Bowitz
Cellular Phreaking
British H.P.A.
Cellular Resources
Realm by SorcereR
Hacking the Ericsson 688 (GSM)
GSM Phone Secret Codes
Fraud & Con Schemes
CTIA Against Fraudulent Cloning
Cellular Telephone FraudVirus in Cellular Phone - Is It Possible?
Yet Another Hacker!
General HPA Archive by Stoned Spirit
AMPS/DAMPS/IS-54b/IS-136 Roaming in Russia/CIS
Radio Communication Products and Services - Buy & Sell Forum (mainly in Russian)
Cellular Phone Secrets Forum (with Search Function!
)
CDMA Software Discussion Board (recommended)
BEPCTAK (Russian hacking site)
### Inquiry Form
Please describe here what you are looking for (software, hardware, information)
Не забудьте переключить кодировку на KOI-8!
| 1,705
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