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http://oeis.org/A239853
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text/html
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crawl-data/CC-MAIN-2019-43/segments/1570987835748.66/warc/CC-MAIN-20191023173708-20191023201208-00215.warc.gz
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This site is supported by donations to The OEIS Foundation.
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A239853 Number of nX3 0..3 arrays with no element equal to zero plus the sum of elements to its left or zero plus the sum of elements above it or one plus the sum of the elements diagonally to its northwest or one plus the sum of the elements antidiagonally to its northeast, modulo 4 1
4, 9, 44, 84, 130, 343, 685, 1704, 3977, 7958, 18574, 40388, 86314, 203202, 433430, 967411, 2205109, 4735287, 10746065, 23937983, 52281965, 118605335, 261244402, 578751600, 1302168082, 2864205270, 6395152778, 14272351652, 31509181522 (list; graph; refs; listen; history; text; internal format)
OFFSET 1,1 COMMENTS Column 3 of A239858 LINKS R. H. Hardin, Table of n, a(n) for n = 1..210 EXAMPLE Some solutions for n=4 ..3..2..3....2..3..3....3..2..3....3..2..3....3..2..3....2..3..2....3..2..3 ..2..1..0....3..1..2....2..1..0....2..1..2....2..1..0....3..1..3....2..1..0 ..3..0..2....3..2..3....3..0..2....3..1..2....3..0..2....3..1..3....2..0..0 ..2..1..0....3..1..3....3..2..0....3..2..2....2..1..2....2..3..3....2..0..0 CROSSREFS Sequence in context: A093149 A048054 A284973 * A209079 A149170 A024053 Adjacent sequences: A239850 A239851 A239852 * A239854 A239855 A239856 KEYWORD nonn AUTHOR R. H. Hardin, Mar 28 2014 STATUS approved
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Last modified October 23 14:54 EDT 2019. Contains 328345 sequences. (Running on oeis4.)
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CC-MAIN-2019-43
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latest
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en
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https://menuiserie-tavernier.fr/2020-Oct-01/5438.html
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• ### Operating Costs of Gold Processing Plant
The dividing point between the two plants lies at the fine ore bin all equipment prior to this bin is considered in the Crushing Plant and all equipment subsequent to it in the mill. General Design Criteria Milling Rate = 500 tonnes/day (550 sptd) Crushing Rate = 2 x 12 hour
• ### Small Gold Processing PlantMineral Processing Metallurgy
The problem a Small Gold Processing Plant can solve is for the need to to provide a small relatively simple and inexpensive gold mill that will operate at a reasonable profit with a minimum of supervision. The limited availability of skilled operating personnel availability of water and local power high cost of transporting ore and expense of building operating and maintaining such a mill
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• ### Estimate Jaw Crusher Capacity
The denominator looks wrong to convert to mass per hour should be multiplying by 60 not dividing by 60. Dividing by 60 should give mass per second which isn t what you want. The example you ve given is missing information needed to calculate the "A" termit doesn t tell you the height of the crushing
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May 07 2015 · UMP 4 ton per hour prospecting and placer mining gold plant Trevor Carolin. SLK-ZD300 Gold wash plant working at Africa GhanaDuration Gold Processing MillDuration
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kilowatt-hour (kWh) per unit basis primarily the metric ton (ton) or troy ounce. Data contained in tables pertaining to specific currently operating facilities are static as the amount of electricity consumed to process or produce a unit of material changes over time for a great number of reasons.
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• ### Estimates of Electricity Requirements for the Recovery of
kilowatt-hour (kWh) per unit basis primarily the metric ton (ton) or troy ounce. Data contained in tables pertaining to specific currently operating facilities are static as the amount of electricity consumed to process or produce a unit of material changes over time for a great number of reasons.
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Aggregate Crushing And Screening Plant Ton Per Hour. Design of a crushing plant to crush 300 tons per hour fix crushing plant 200 240 ton per hour Gold Trommel Wash Plant300 Tons per Hour S8 by MSI Mining buy Gold design of a crushing plant to crush 300 tons per tons of aggregate per hour Get price. Details Gold Mining EquipmentMsi Mining
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• ### Estimates of Electricity Requirements for the Recovery of
kilowatt-hour (kWh) per unit basis primarily the metric ton (ton) or troy ounce. Data contained in tables pertaining to specific currently operating facilities are static as the amount of electricity consumed to process or produce a unit of material changes over time for a great number of reasons.
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• ### 2 Ton/Hr Gold Processing Plant IGR100911Metallurgist
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• ### MILLING AND RECYCLING
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• ### Sand crushing milling plant tphontwerpbureau-amsterdam
Sand crushing milling plant tph. sand crushing milling plant 30 60 tph leninscoza 120 tph crusher price list Gold Ore Crusher 2nd Hand Coal Crusher With Capacity 200330 Tph For Usa Crusher Mill Stone Crushing Plant 4060 TPH 100120 price of 3040 TPH Mobile Crushing Plant operation and maintenance manual of crusher plant Tph Ton Per Hour Used
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Aggregate Crushing And Screening Plant Ton Per Hour. Design of a crushing plant to crush 300 tons per hour fix crushing plant 200 240 ton per hour Gold Trommel Wash Plant300 Tons per Hour S8 by MSI Mining buy Gold design of a crushing plant to crush 300 tons per tons of aggregate per hour Get price. Details Gold Mining EquipmentMsi Mining
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CC-MAIN-2020-50
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latest
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https://amath.innolan.net/classEvalStatement.html
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crawl-data/CC-MAIN-2023-40/segments/1695233510501.83/warc/CC-MAIN-20230929090526-20230929120526-00136.warc.gz
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amath 1.8.5 Simple command line calculator
EvalStatement Class Reference
Evaluate arithmetic expression. More...
`#include <eval.h>`
Inheritance diagram for EvalStatement:
Collaboration diagram for EvalStatement:
Public Member Functions
EvalStatement (ExpressionNode *expression)
~EvalStatement ()
char * Execute ()
SyntaxNodeGetNext ()
void Attach (SyntaxNode *node)
void Detach (SyntaxNode *node)
void Replace (SyntaxNode *n, SyntaxNode *x)
Public Member Functions inherited from StatementNode
StatementNode ()
StatementNode (const char *text)
virtual ~StatementNode ()
NodeType GetNodeType ()
virtual char * GetTextCode ()
Public Member Functions inherited from SyntaxNode
SyntaxNode ()
virtual ~SyntaxNode ()
void SetFirstNode ()
bool GetFirstNode () const
SyntaxNodeGetParent () const
void SetParent (SyntaxNode *node)
virtual ReductionType GetReductionType ()
virtual void ResetIterator ()
Private Attributes
ExpressionNodeexpression
Protected Attributes inherited from StatementNode
char * statementText
Protected Attributes inherited from SyntaxNode
CharBufferoutput
SyntaxNodeparent
SyntaxNodeiterator
bool leftBottom
Detailed Description
Evaluate arithmetic expression.
Definition at line 39 of file eval.h.
◆ EvalStatement()
EvalStatement::EvalStatement ( ExpressionNode * expression )
explicit
Definition at line 35 of file eval.cpp.
References expression, and StatementNode::StatementNode().
Referenced by Parser::ParseEvaluation(), and Parser::ParseStatement().
35 :
36 StatementNode(), expression(expression)
37 {
38 }
ExpressionNode * expression
Definition: eval.h:52
StatementNode()
Definition: node.cpp:34
Here is the call graph for this function:
Here is the caller graph for this function:
◆ ~EvalStatement()
EvalStatement::~EvalStatement ( )
Definition at line 40 of file eval.cpp.
References expression.
41 {
42 if (expression != nullptr)
43 {
44 delete expression;
45 }
46 }
ExpressionNode * expression
Definition: eval.h:52
◆ Attach()
void EvalStatement::Attach ( SyntaxNode * node )
virtual
Reimplemented from StatementNode.
Definition at line 83 of file eval.cpp.
References expression, and SyntaxNode::SetParent().
84 {
85 if (expression == nullptr)
86 {
87 expression = static_cast<ExpressionNode*>(node);
88 node->SetParent(this);
89 }
90 }
ExpressionNode * expression
Definition: eval.h:52
void SetParent(SyntaxNode *node)
Definition: nodes.cpp:75
Base class for all nodes related to mathematical expressions.
Definition: nodes.h:99
Here is the call graph for this function:
◆ Detach()
void EvalStatement::Detach ( SyntaxNode * node )
virtual
Reimplemented from StatementNode.
Definition at line 92 of file eval.cpp.
References expression.
93 {
94 if (expression == node)
95 {
96 expression = nullptr;
97 }
98 }
ExpressionNode * expression
Definition: eval.h:52
◆ Execute()
char * EvalStatement::Execute ( )
virtual
Implements StatementNode.
Definition at line 48 of file eval.cpp.
49 {
50 Number* result = expression->Evaluate();
51 Number* temp = result->Clone();
52 const char* text = expression->GetText();
53 const char* val = Program->Output->GetText(result);
54
55 Program->SetLastResult(temp);
56 delete temp;
57
58 output->Empty();
60 StrLen(text) + 3 +
61 StrLen(val) +
62 StrLen(NEWLINE) + 1);
63
64 output->Append(text);
65 output->Append(" = ");
66 output->Append(val);
68
69 return output->GetString();
70 }
#define NEWLINE
Definition: amath.h:222
Master control class.
Definition: program.h:55
void Empty()
Definition: charbuf.cpp:218
char * GetString() const
Definition: charbuf.cpp:306
void Append(const char *source)
Definition: charbuf.cpp:262
Definition: numb.h:66
virtual Number * Clone()=0
virtual const char * GetText(Number *number)=0
ExpressionNode * expression
Definition: eval.h:52
virtual Number * Evaluate()=0
int StrLen(const char *string)
Get the length of a null terminated string.
Definition: strlen.c:34
virtual char * GetText()=0
CharBuffer * output
Definition: nodes.h:85
void SetLastResult(Number *number)
Definition: program.cpp:127
void EnsureSize(unsigned int size)
Ensure a memory block of specified size is allocated.
Definition: charbuf.cpp:114
class NumeralSystem * Output
Definition: program.h:76
Here is the call graph for this function:
◆ GetNext()
SyntaxNode * EvalStatement::GetNext ( )
virtual
Reimplemented from StatementNode.
Definition at line 72 of file eval.cpp.
References expression, and SyntaxNode::iterator.
73 {
74 if (iterator == nullptr)
75 {
77 return iterator;
78 }
79
80 return nullptr;
81 }
ExpressionNode * expression
Definition: eval.h:52
SyntaxNode * iterator
Definition: nodes.h:87
◆ Replace()
void EvalStatement::Replace ( SyntaxNode * n, SyntaxNode * x )
virtual
Reimplemented from StatementNode.
Definition at line 100 of file eval.cpp.
References expression.
101 {
102 if (expression == n)
103 {
104 delete expression;
105 expression = static_cast<ExpressionNode*>(x);
106 }
107 }
ExpressionNode * expression
Definition: eval.h:52
Base class for all nodes related to mathematical expressions.
Definition: nodes.h:99
◆ expression
ExpressionNode* EvalStatement::expression
private
Definition at line 52 of file eval.h.
Referenced by Attach(), Detach(), EvalStatement(), Execute(), GetNext(), Replace(), and ~EvalStatement().
The documentation for this class was generated from the following files:
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CC-MAIN-2023-40
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latest
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en
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https://www.tefter.io/bookmarks/56588/readable
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crawl-data/CC-MAIN-2020-05/segments/1579251671078.88/warc/CC-MAIN-20200125071430-20200125100430-00123.warc.gz
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# An Efficient Algorithm to Exploit Arbitrage Opportunities in Crypto Markets
By Federico Caccia
The crypto market is still inefficient. Cryptoassets have different rates on different exchanges. The big question is how we can profit from that inefficiency. One approach is making transactions between exchanges, but if we buy a cryptoasset on one exchange, by the time we can sell it on another, the price difference may no longer exist. For a discussion of this problem, see What I Have Learned from My Arbitrage Experiences with Cryptoassets.
A more efficient approach is using what I call virtual transactions: buying a pair on one exchange and simultaneously selling this pair in another exchange. In order to use virtual transactions to exploit arbitrage opportunities, you must have funded wallets before profitable oscillations events. In a profitable oscillation event, the price difference for a given pair between two exchanges allows you to make profit using two virtual transactions. The first virtual transaction at the time of the event and the opposite operation once the event dissapears and the price difference returns to average values in a considered period.
Put in mathematical terms:
where is the amount of assets you are exchanging, refers to the values you would pay buying these assets, refers to the values you would receive selling these assets, the sub-indices and refers to the two exchanges and the supra-indices and refers to the first and second virtual transactions. The operator is used to average the values in a given period. This method is a statistical arbitrage method known as pair trading and works using co-integrated pairs. For a deeper look into pair trading algorithms, see Introduction to Pairs Trading. Most arbitrage only monitor for price differences and is just a particular case of this method, considering the second half of the equation equal to zero.
Statistical arbitrage has its risks, since it is necessary to hold cryptoassets, unless the chosen exchanges accept a stable currency as collateral for margin trading. Although there is a profit due to arbitrage, there may also be a devaluation of our capital due to market volatility. On the other hand, it is also necessary to hold a quote currency buffer in each exchange to cover price fluctuations.
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CC-MAIN-2020-05
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latest
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en
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https://www.conversion-metric.org/volume/drop-to-teaspoon
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crawl-data/CC-MAIN-2024-33/segments/1722640353668.0/warc/CC-MAIN-20240802234508-20240803024508-00189.warc.gz
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# Drop to Teaspoon Conversion (drop to tsp)
1 drop = 0.013157523154497 tsp
Swap » Teaspoon to Drop
drop: Drop, tsp: Teaspoon
Convert Volume Units
## How Many Teaspoon in a Drop?
There are 0.013157523154497 teaspoon in a drop.
1 Drop is equal to 0.013157523154497 Teaspoon.
1 drop = 0.013157523154497 tsp
## Drop to Teaspoon Conversions
20 drop = 0.26315 tsp
30 drop = 0.394726 tsp
10 drop = 0.131575 tsp
60 drop = 0.789451 tsp
40 drop = 0.526301 tsp
80 drop = 1.052602 tsp
100 drop = 1.315752 tsp
15 drop = 0.197363 tsp
6 drop = 0.078945 tsp
50 drop = 0.657876 tsp
12 drop = 0.15789 tsp
8 drop = 0.10526 tsp
75 drop = 0.986814 tsp
5 drop = 0.065788 tsp
25 drop = 0.328938 tsp
4 drop = 0.05263 tsp
24 drop = 0.315781 tsp
3 drop = 0.039473 tsp
16 drop = 0.21052 tsp
90 drop = 1.184177 tsp
120 drop = 1.578903 tsp
76 drop = 0.999972 tsp
45 drop = 0.592089 tsp
48 drop = 0.631561 tsp
1 drop = 0.013158 tsp
36 drop = 0.473671 tsp
18 drop = 0.236835 tsp
32 drop = 0.421041 tsp
70 drop = 0.921027 tsp
200 drop = 2.631505 tsp
77 drop = 1.013129 tsp
9 drop = 0.118418 tsp
160 drop = 2.105204 tsp
2 drop = 0.026315 tsp
150 drop = 1.973628 tsp
7 drop = 0.092103 tsp
240 drop = 3.157806 tsp
35 drop = 0.460513 tsp
180 drop = 2.368354 tsp
64 drop = 0.842081 tsp
19 drop = 0.249993 tsp
38 drop = 0.499986 tsp
56 drop = 0.736821 tsp
28 drop = 0.368411 tsp
78 drop = 1.026287 tsp
300 drop = 3.947257 tsp
14 drop = 0.184205 tsp
98 drop = 1.289437 tsp
### Drop Definition
Used very commonly in medicine and cooking, a drop is a unit of volume defined as the volume of a liquid in one drop. It has been used since the 19th century and is considered one of the most roughly specified units of volume. Undoubtedly, the volume of a drop depends on the method or device used for producing drops.
Convert Drop
### Teaspoon Definition
A teaspoon as a volume measuring unit is widely used in cooking, pharmaceutical medicine, and some other areas. A teaspoon is a very convenient unit of volume used since the middle of the 17th century. It is roughly equal to 1/3rd of a tablespoon, 1/48th of a US cup, 1/3rd of a cubic inch, or 5 mL. This unit can be abbreviated to tsp.
Convert Teaspoon
#### About drop to tsp Converter
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##### Drop to Teaspoon Conversion Chart
DropTeaspoon
1 drop0.013157523154497 tsp
2 drop0.026315046308994 tsp
3 drop0.039472569463491 tsp
4 drop0.052630092617988 tsp
5 drop0.065787615772485 tsp
6 drop0.078945138926981 tsp
7 drop0.092102662081478 tsp
8 drop0.10526018523598 tsp
9 drop0.11841770839047 tsp
10 drop0.13157523154497 tsp
11 drop0.14473275469947 tsp
12 drop0.15789027785396 tsp
13 drop0.17104780100846 tsp
14 drop0.18420532416296 tsp
15 drop0.19736284731745 tsp
16 drop0.21052037047195 tsp
17 drop0.22367789362645 tsp
18 drop0.23683541678094 tsp
19 drop0.24999293993544 tsp
20 drop0.26315046308994 tsp
21 drop0.27630798624444 tsp
22 drop0.28946550939893 tsp
23 drop0.30262303255343 tsp
24 drop0.31578055570793 tsp
25 drop0.32893807886242 tsp
26 drop0.34209560201692 tsp
27 drop0.35525312517142 tsp
28 drop0.36841064832591 tsp
29 drop0.38156817148041 tsp
30 drop0.39472569463491 tsp
31 drop0.4078832177894 tsp
32 drop0.4210407409439 tsp
33 drop0.4341982640984 tsp
34 drop0.44735578725289 tsp
35 drop0.46051331040739 tsp
36 drop0.47367083356189 tsp
37 drop0.48682835671639 tsp
38 drop0.49998587987088 tsp
39 drop0.51314340302538 tsp
40 drop0.52630092617988 tsp
41 drop0.53945844933437 tsp
42 drop0.55261597248887 tsp
43 drop0.56577349564337 tsp
44 drop0.57893101879786 tsp
45 drop0.59208854195236 tsp
46 drop0.60524606510686 tsp
47 drop0.61840358826135 tsp
48 drop0.63156111141585 tsp
49 drop0.64471863457035 tsp
50 drop0.65787615772485 tsp
DropTeaspoon
50 drop0.65787615772485 tsp
55 drop0.72366377349733 tsp
60 drop0.78945138926981 tsp
65 drop0.8552390050423 tsp
70 drop0.92102662081478 tsp
75 drop0.98681423658727 tsp
80 drop1.0526018523598 tsp
85 drop1.1183894681322 tsp
90 drop1.1841770839047 tsp
95 drop1.2499646996772 tsp
100 drop1.3157523154497 tsp
105 drop1.3815399312222 tsp
110 drop1.4473275469947 tsp
115 drop1.5131151627671 tsp
120 drop1.5789027785396 tsp
125 drop1.6446903943121 tsp
130 drop1.7104780100846 tsp
135 drop1.7762656258571 tsp
140 drop1.8420532416296 tsp
145 drop1.9078408574021 tsp
150 drop1.9736284731745 tsp
155 drop2.039416088947 tsp
160 drop2.1052037047195 tsp
165 drop2.170991320492 tsp
170 drop2.2367789362645 tsp
175 drop2.302566552037 tsp
180 drop2.3683541678094 tsp
185 drop2.4341417835819 tsp
190 drop2.4999293993544 tsp
195 drop2.5657170151269 tsp
200 drop2.6315046308994 tsp
205 drop2.6972922466719 tsp
210 drop2.7630798624443 tsp
215 drop2.8288674782168 tsp
220 drop2.8946550939893 tsp
225 drop2.9604427097618 tsp
230 drop3.0262303255343 tsp
235 drop3.0920179413068 tsp
240 drop3.1578055570793 tsp
245 drop3.2235931728517 tsp
250 drop3.2893807886242 tsp
255 drop3.3551684043967 tsp
260 drop3.4209560201692 tsp
265 drop3.4867436359417 tsp
270 drop3.5525312517142 tsp
275 drop3.6183188674866 tsp
280 drop3.6841064832591 tsp
285 drop3.7498940990316 tsp
290 drop3.8156817148041 tsp
295 drop3.8814693305766 tsp
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CC-MAIN-2024-33
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https://mathoverflow.net/questions/283111/intersection-numbers-in-mathbbp1-bundles
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# Intersection numbers in $\mathbb{P}^1$-bundles
Let $\mathcal{E}$ be a rank two vector bundle on $\mathbb{P}^2$ fitting in the following exact sequence
$$0\rightarrow \mathcal{O}_{\mathbb{P}^2}\rightarrow \mathcal{E}\rightarrow \mathcal{I}_p(-1)\rightarrow 0$$
where $\mathcal{I}_p$ is the ideal sheaf of a point $p\in\mathbb{P}^2$. Then $c_1(\mathcal{E})=-1$ and $c_2(\mathcal{E})=1$.
Let $\pi:X = \mathbb{P}(\mathcal{E})\rightarrow\mathbb{P}^2$, and $-K_X$ the anti-canonical divisor of $X$. Finally, let $C\subset X$ be an irreducible curve such that $\pi(C) = L_p$ is a line through $p\in\mathbb{P}^2$.
How can we compute the intersection number $-K_X\cdot C$ ?
Let $L_p\cong\mathbb{P}^1$ be a line in $\mathbb{P}^2$ through $p$. Your exact sequence restricted to $L_p$ yields the following exact sequence $$0\rightarrow\mathcal{O}_{\mathbb{P}^1}(1)\rightarrow\mathcal{E}_{|L_p}\rightarrow\mathcal{O}_{\mathbb{P}^1}(-2)\rightarrow 0$$ and hence $\mathcal{E}_{|L_p}\cong\mathcal{O}_{\mathbb{P}^1}(1)\oplus\mathcal{O}_{\mathbb{P}^1}(-2)$. Therefore, $\pi^{-1}(L_p) = \mathbb{P}(\mathcal{O}_{\mathbb{P}^1}(1)\oplus\mathcal{O}_{\mathbb{P}^1}(-2))$ is the Hirzebruch surface $\mathbb{F}_3$. Let $\overline{\xi}$ be the class of the only curve with self-intersection $-3$ on $\mathbb{F}_3$, and let $f$ be the class of a fiber of $\pi_{|\mathbb{F}_3}:\mathbb{F}_3\rightarrow\mathbb{P}^1$. Then if $\widetilde{H} = \pi^{*}\mathcal{O}_{\mathbb{P}^2}(1)$ and $\xi = c_1(\mathcal{O}_X(1))$ we have $\widetilde{H}_{|\mathbb{F}_3} = f$ and $\xi_{|\mathbb{F}_3} = f+\overline{\xi}$.
Since $c_1(\mathcal{E})=-1$ and $c_2(\mathcal{E}) = 1$ the canonical divisor of $X$ is given by $-K_X = 4\widetilde{H}+2\xi$ and we may compute the intersection $-K_W\cdot\overline{\xi}$ inside $\mathbb{F}_3$, that is $$-K_X\cdot\overline{\xi} = (4\widetilde{H}+2\xi)_{|\mathbb{F}_3}\cdot\overline{\xi} = (4f+2(f+\overline{\xi}))\cdot\overline{\xi} = (6f+2\xi) = 6+2(-3) = 0.$$
It seems $C$ is not well-defined. Let $\ell$ be the line such that $\pi(C) = \ell$. Then we have the $\mathbb{P}^1$-bundle $$\pi'\colon\mathbb{P}(\mathcal{E}|_\ell) \rightarrow \ell.$$ Any section of this bundle will give rise to a curve $C$ as you defined (and indeed, such $\mathbb{P}^1$ bundles admit infinitely many sections). The intersection number $-K_X \cdot C$ will depend on which section you choose.
• The unique cross-section over $\ell$ with negative self-intersection inside the ruled surface $\mathbb{P}(\mathcal{E}_\ell)$ has self-intersection equal to $-3$. Thus, the relative dualizing sheaf of $\pi$ has degree $+3$ on this curve. Also, the dualizing sheaf of $\mathbb{P}^2$ has degree $-3$ on $\ell$. Altogether, the dualizing sheaf of $X$ has degree $0$ on this cross-section. Commented Oct 10, 2017 at 0:46
| 1,059
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| 3.28125
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CC-MAIN-2024-38
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https://www.teacherspayteachers.com/Product/Elementary-Geometry-Lines-and-Multiplication-Practice-345224
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Total:
\$0.00
# Elementary Geometry: Lines and Multiplication Practice
Subjects
Resource Types
Product Rating
4.0
File Type
PDF (Acrobat) Document File
0.07 MB | 1 pages
### PRODUCT DESCRIPTION
Mixed review of multiplication facts, lines and line segments.
Students are asked to solve problems similar to the following:
5 x ( ) = 15
6 x ( ) = 12
2 x ( ) = 2
4 x ( ) = 8
3 x ( ) = 30
The bottom portion of the worksheet includes 3 multiple choice questions asking students to identify the given lines and line segments.
This can either be used as a morning warm-up, math warm-up, homework, in-class practice or quiz.
If you like this worksheet, I also recommend the following:
Elementary Geometry: Angles, Shapes and Parallel Lines I (FREE)
Elementary Geometry: Angles, Shapes and Parallel Lines II (FREE)
Elementary Geometry: Identifying Right, Obtuse and Acute Angles (FREE)
Elementary Geometry: Drawing Lines, Line Segments and Rays (FREE)
Elementary Geomentry Review: Angles, Shapes and Perpendicular Lines (FREE)
Hope this helps =)
Total Pages
1
Not Included
Teaching Duration
N/A
### Average Ratings
4.0
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4.0
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4.0
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4.0
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4.0
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4.0
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4.0
Total:
5 ratings
FREE
User Rating: 4.0/4.0
(3,960 Followers)
FREE
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CC-MAIN-2017-17
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https://forge.ipsl.jussieu.fr/nemo/browser/branches/UKMO/v3_6_extra_CMIP6_diagnostics/DOC/TexFiles/Chapters/Chap_ZDF.tex?rev=7492
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# source:branches/UKMO/v3_6_extra_CMIP6_diagnostics/DOC/TexFiles/Chapters/Chap_ZDF.tex@7492
Last change on this file since 7492 was 7492, checked in by timgraham, 5 years ago
Final merge of head of v3.6 STABLE branch into this branch
• Property svn:keywords set to Id
File size: 79.0 KB
Line
1\documentclass[NEMO_book]{subfiles}
2\begin{document}
3% ================================================================
4% Chapter Vertical Ocean Physics (ZDF)
5% ================================================================
6\chapter{Vertical Ocean Physics (ZDF)}
7\label{ZDF}
8\minitoc
9
10%gm% Add here a small introduction to ZDF and naming of the different physics (similar to what have been written for TRA and DYN.
11
12
13\newpage
14$\$\newline % force a new ligne
15
16
17% ================================================================
18% Vertical Mixing
19% ================================================================
20\section{Vertical Mixing}
21\label{ZDF_zdf}
22
23The discrete form of the ocean subgrid scale physics has been presented in
24\S\ref{TRA_zdf} and \S\ref{DYN_zdf}. At the surface and bottom boundaries,
25the turbulent fluxes of momentum, heat and salt have to be defined. At the
26surface they are prescribed from the surface forcing (see Chap.~\ref{SBC}),
27while at the bottom they are set to zero for heat and salt, unless a geothermal
28flux forcing is prescribed as a bottom boundary condition ($i.e.$ \key{trabbl}
29defined, see \S\ref{TRA_bbc}), and specified through a bottom friction
30parameterisation for momentum (see \S\ref{ZDF_bfr}).
31
32In this section we briefly discuss the various choices offered to compute
33the vertical eddy viscosity and diffusivity coefficients, $A_u^{vm}$ ,
34$A_v^{vm}$ and $A^{vT}$ ($A^{vS}$), defined at $uw$-, $vw$- and $w$-
35points, respectively (see \S\ref{TRA_zdf} and \S\ref{DYN_zdf}). These
36coefficients can be assumed to be either constant, or a function of the local
37Richardson number, or computed from a turbulent closure model (TKE, GLS or KPP formulation).
38The computation of these coefficients is initialized in the \mdl{zdfini} module
39and performed in the \mdl{zdfric}, \mdl{zdftke}, \mdl{zdfgls} or \mdl{zdfkpp} modules.
40The trends due to the vertical momentum and tracer diffusion, including the surface forcing,
41are computed and added to the general trend in the \mdl{dynzdf} and \mdl{trazdf} modules, respectively.
42These trends can be computed using either a forward time stepping scheme
43(namelist parameter \np{ln\_zdfexp}=true) or a backward time stepping
44scheme (\np{ln\_zdfexp}=false) depending on the magnitude of the mixing
45coefficients, and thus of the formulation used (see \S\ref{STP}).
46
47% -------------------------------------------------------------------------------------------------------------
48% Constant
49% -------------------------------------------------------------------------------------------------------------
50\subsection{Constant (\key{zdfcst})}
51\label{ZDF_cst}
52%--------------------------------------------namzdf---------------------------------------------------------
53\namdisplay{namzdf}
54%--------------------------------------------------------------------------------------------------------------
55
56Options are defined through the \ngn{namzdf} namelist variables.
57When \key{zdfcst} is defined, the momentum and tracer vertical eddy coefficients
58are set to constant values over the whole ocean. This is the crudest way to define
59the vertical ocean physics. It is recommended that this option is only used in
60process studies, not in basin scale simulations. Typical values used in this case are:
61\begin{align*}
62A_u^{vm} = A_v^{vm} &= 1.2\ 10^{-4}~m^2.s^{-1} \\
63A^{vT} = A^{vS} &= 1.2\ 10^{-5}~m^2.s^{-1}
64\end{align*}
65
66These values are set through the \np{rn\_avm0} and \np{rn\_avt0} namelist parameters.
67In all cases, do not use values smaller that those associated with the molecular
68viscosity and diffusivity, that is $\sim10^{-6}~m^2.s^{-1}$ for momentum,
69$\sim10^{-7}~m^2.s^{-1}$ for temperature and $\sim10^{-9}~m^2.s^{-1}$ for salinity.
70
71
72% -------------------------------------------------------------------------------------------------------------
73% Richardson Number Dependent
74% -------------------------------------------------------------------------------------------------------------
75\subsection{Richardson Number Dependent (\key{zdfric})}
76\label{ZDF_ric}
77
78%--------------------------------------------namric---------------------------------------------------------
79\namdisplay{namzdf_ric}
80%--------------------------------------------------------------------------------------------------------------
81
82When \key{zdfric} is defined, a local Richardson number dependent formulation
83for the vertical momentum and tracer eddy coefficients is set through the \ngn{namzdf\_ric}
84namelist variables.The vertical mixing
85coefficients are diagnosed from the large scale variables computed by the model.
86\textit{In situ} measurements have been used to link vertical turbulent activity to
87large scale ocean structures. The hypothesis of a mixing mainly maintained by the
88growth of Kelvin-Helmholtz like instabilities leads to a dependency between the
89vertical eddy coefficients and the local Richardson number ($i.e.$ the
90ratio of stratification to vertical shear). Following \citet{Pacanowski_Philander_JPO81}, the following
91formulation has been implemented:
92\begin{equation} \label{Eq_zdfric}
93 \left\{ \begin{aligned}
94 A^{vT} &= \frac {A_{ric}^{vT}}{\left( 1+a \; Ri \right)^n} + A_b^{vT} \\
95 A^{vm} &= \frac{A^{vT} }{\left( 1+ a \;Ri \right) } + A_b^{vm}
96 \end{aligned} \right.
97\end{equation}
98where $Ri = N^2 / \left(\partial_z \textbf{U}_h \right)^2$ is the local Richardson
99number, $N$ is the local Brunt-Vais\"{a}l\"{a} frequency (see \S\ref{TRA_bn2}),
100$A_b^{vT}$ and $A_b^{vm}$ are the constant background values set as in the
101constant case (see \S\ref{ZDF_cst}), and $A_{ric}^{vT} = 10^{-4}~m^2.s^{-1}$
102is the maximum value that can be reached by the coefficient when $Ri\leq 0$,
103$a=5$ and $n=2$. The last three values can be modified by setting the
104\np{rn\_avmri}, \np{rn\_alp} and \np{nn\_ric} namelist parameters, respectively.
105
106A simple mixing-layer model to transfer and dissipate the atmospheric
107 forcings (wind-stress and buoyancy fluxes) can be activated setting
108the \np{ln\_mldw} =.true. in the namelist.
109
110In this case, the local depth of turbulent wind-mixing or "Ekman depth"
111 $h_{e}(x,y,t)$ is evaluated and the vertical eddy coefficients prescribed within this layer.
112
113This depth is assumed proportional to the "depth of frictional influence" that is limited by rotation:
114\begin{equation}
115 h_{e} = Ek \frac {u^{*}} {f_{0}} \\
116\end{equation}
117where, $Ek$ is an empirical parameter, $u^{*}$ is the friction velocity and $f_{0}$ is the Coriolis
118parameter.
119
120In this similarity height relationship, the turbulent friction velocity:
121\begin{equation}
122 u^{*} = \sqrt \frac {|\tau|} {\rho_o} \\
123\end{equation}
124
125is computed from the wind stress vector $|\tau|$ and the reference density $\rho_o$.
126The final $h_{e}$ is further constrained by the adjustable bounds \np{rn\_mldmin} and \np{rn\_mldmax}.
127Once $h_{e}$ is computed, the vertical eddy coefficients within $h_{e}$ are set to
128the empirical values \np{rn\_wtmix} and \np{rn\_wvmix} \citep{Lermusiaux2001}.
129
130% -------------------------------------------------------------------------------------------------------------
131% TKE Turbulent Closure Scheme
132% -------------------------------------------------------------------------------------------------------------
133\subsection{TKE Turbulent Closure Scheme (\key{zdftke})}
134\label{ZDF_tke}
135
136%--------------------------------------------namzdf_tke--------------------------------------------------
137\namdisplay{namzdf_tke}
138%--------------------------------------------------------------------------------------------------------------
139
140The vertical eddy viscosity and diffusivity coefficients are computed from a TKE
141turbulent closure model based on a prognostic equation for $\bar{e}$, the turbulent
142kinetic energy, and a closure assumption for the turbulent length scales. This
143turbulent closure model has been developed by \citet{Bougeault1989} in the
144atmospheric case, adapted by \citet{Gaspar1990} for the oceanic case, and
145embedded in OPA, the ancestor of NEMO, by \citet{Blanke1993} for equatorial Atlantic
146simulations. Since then, significant modifications have been introduced by
147\citet{Madec1998} in both the implementation and the formulation of the mixing
148length scale. The time evolution of $\bar{e}$ is the result of the production of
149$\bar{e}$ through vertical shear, its destruction through stratification, its vertical
150diffusion, and its dissipation of \citet{Kolmogorov1942} type:
151\begin{equation} \label{Eq_zdftke_e}
152\frac{\partial \bar{e}}{\partial t} =
153\frac{K_m}{{e_3}^2 }\;\left[ {\left( {\frac{\partial u}{\partial k}} \right)^2
154 +\left( {\frac{\partial v}{\partial k}} \right)^2} \right]
155-K_\rho\,N^2
156+\frac{1}{e_3} \;\frac{\partial }{\partial k}\left[ {\frac{A^{vm}}{e_3 }
157 \;\frac{\partial \bar{e}}{\partial k}} \right]
158- c_\epsilon \;\frac{\bar {e}^{3/2}}{l_\epsilon }
159\end{equation}
160\begin{equation} \label{Eq_zdftke_kz}
161 \begin{split}
162 K_m &= C_k\ l_k\ \sqrt {\bar{e}\; } \\
163 K_\rho &= A^{vm} / P_{rt}
164 \end{split}
165\end{equation}
166where $N$ is the local Brunt-Vais\"{a}l\"{a} frequency (see \S\ref{TRA_bn2}),
167$l_{\epsilon }$ and $l_{\kappa }$ are the dissipation and mixing length scales,
168$P_{rt}$ is the Prandtl number, $K_m$ and $K_\rho$ are the vertical eddy viscosity
169and diffusivity coefficients. The constants $C_k = 0.1$ and $C_\epsilon = \sqrt {2} /2$
170$\approx 0.7$ are designed to deal with vertical mixing at any depth \citep{Gaspar1990}.
171They are set through namelist parameters \np{nn\_ediff} and \np{nn\_ediss}.
172$P_{rt}$ can be set to unity or, following \citet{Blanke1993}, be a function
173of the local Richardson number, $R_i$:
174\begin{align*} \label{Eq_prt}
175P_{rt} = \begin{cases}
176 \ \ \ 1 & \text{if $\ R_i \leq 0.2$} \\
177 5\,R_i & \text{if $\ 0.2 \leq R_i \leq 2$} \\
178 \ \ 10 & \text{if $\ 2 \leq R_i$}
179 \end{cases}
180\end{align*}
181Options are defined through the \ngn{namzdfy\_tke} namelist variables.
182The choice of $P_{rt}$ is controlled by the \np{nn\_pdl} namelist variable.
183
184At the sea surface, the value of $\bar{e}$ is prescribed from the wind
185stress field as $\bar{e}_o = e_{bb} |\tau| / \rho_o$, with $e_{bb}$ the \np{rn\_ebb}
186namelist parameter. The default value of $e_{bb}$ is 3.75. \citep{Gaspar1990}),
187however a much larger value can be used when taking into account the
188surface wave breaking (see below Eq. \eqref{ZDF_Esbc}).
189The bottom value of TKE is assumed to be equal to the value of the level just above.
190The time integration of the $\bar{e}$ equation may formally lead to negative values
191because the numerical scheme does not ensure its positivity. To overcome this
192problem, a cut-off in the minimum value of $\bar{e}$ is used (\np{rn\_emin}
193namelist parameter). Following \citet{Gaspar1990}, the cut-off value is set
194to $\sqrt{2}/2~10^{-6}~m^2.s^{-2}$. This allows the subsequent formulations
195to match that of \citet{Gargett1984} for the diffusion in the thermocline and
196deep ocean : $K_\rho = 10^{-3} / N$.
197In addition, a cut-off is applied on $K_m$ and $K_\rho$ to avoid numerical
198instabilities associated with too weak vertical diffusion. They must be
199specified at least larger than the molecular values, and are set through
200\np{rn\_avm0} and \np{rn\_avt0} (namzdf namelist, see \S\ref{ZDF_cst}).
201
202\subsubsection{Turbulent length scale}
203For computational efficiency, the original formulation of the turbulent length
204scales proposed by \citet{Gaspar1990} has been simplified. Four formulations
205are proposed, the choice of which is controlled by the \np{nn\_mxl} namelist
206parameter. The first two are based on the following first order approximation
207\citep{Blanke1993}:
208\begin{equation} \label{Eq_tke_mxl0_1}
209l_k = l_\epsilon = \sqrt {2 \bar{e}\; } / N
210\end{equation}
211which is valid in a stable stratified region with constant values of the Brunt-
212Vais\"{a}l\"{a} frequency. The resulting length scale is bounded by the distance
213to the surface or to the bottom (\np{nn\_mxl} = 0) or by the local vertical scale factor
214(\np{nn\_mxl} = 1). \citet{Blanke1993} notice that this simplification has two major
215drawbacks: it makes no sense for locally unstable stratification and the
216computation no longer uses all the information contained in the vertical density
217profile. To overcome these drawbacks, \citet{Madec1998} introduces the
218\np{nn\_mxl} = 2 or 3 cases, which add an extra assumption concerning the vertical
219gradient of the computed length scale. So, the length scales are first evaluated
220as in \eqref{Eq_tke_mxl0_1} and then bounded such that:
221\begin{equation} \label{Eq_tke_mxl_constraint}
222\frac{1}{e_3 }\left| {\frac{\partial l}{\partial k}} \right| \leq 1
223\qquad \text{with }\ l = l_k = l_\epsilon
224\end{equation}
225\eqref{Eq_tke_mxl_constraint} means that the vertical variations of the length
226scale cannot be larger than the variations of depth. It provides a better
227approximation of the \citet{Gaspar1990} formulation while being much less
228time consuming. In particular, it allows the length scale to be limited not only
229by the distance to the surface or to the ocean bottom but also by the distance
230to a strongly stratified portion of the water column such as the thermocline
231(Fig.~\ref{Fig_mixing_length}). In order to impose the \eqref{Eq_tke_mxl_constraint}
232constraint, we introduce two additional length scales: $l_{up}$ and $l_{dwn}$,
233the upward and downward length scales, and evaluate the dissipation and
234mixing length scales as (and note that here we use numerical indexing):
235%>>>>>>>>>>>>>>>>>>>>>>>>>>>>
236\begin{figure}[!t] \begin{center}
237\includegraphics[width=1.00\textwidth]{Fig_mixing_length}
238\caption{ \label{Fig_mixing_length}
239Illustration of the mixing length computation. }
240\end{center}
241\end{figure}
242%>>>>>>>>>>>>>>>>>>>>>>>>>>>>
243\begin{equation} \label{Eq_tke_mxl2}
244\begin{aligned}
245 l_{up\ \ }^{(k)} &= \min \left( l^{(k)} \ , \ l_{up}^{(k+1)} + e_{3t}^{(k)}\ \ \ \; \right)
246 \quad &\text{ from $k=1$ to $jpk$ }\ \\
247 l_{dwn}^{(k)} &= \min \left( l^{(k)} \ , \ l_{dwn}^{(k-1)} + e_{3t}^{(k-1)} \right)
248 \quad &\text{ from $k=jpk$ to $1$ }\ \\
249\end{aligned}
250\end{equation}
251where $l^{(k)}$ is computed using \eqref{Eq_tke_mxl0_1},
252$i.e.$ $l^{(k)} = \sqrt {2 {\bar e}^{(k)} / {N^2}^{(k)} }$.
253
254In the \np{nn\_mxl}~=~2 case, the dissipation and mixing length scales take the same
255value: $l_k= l_\epsilon = \min \left(\ l_{up} \;,\; l_{dwn}\ \right)$, while in the
256\np{nn\_mxl}~=~3 case, the dissipation and mixing turbulent length scales are give
257as in \citet{Gaspar1990}:
258\begin{equation} \label{Eq_tke_mxl_gaspar}
259\begin{aligned}
260& l_k = \sqrt{\ l_{up} \ \ l_{dwn}\ } \\
261& l_\epsilon = \min \left(\ l_{up} \;,\; l_{dwn}\ \right)
262\end{aligned}
263\end{equation}
264
265At the ocean surface, a non zero length scale is set through the \np{rn\_mxl0} namelist
266parameter. Usually the surface scale is given by $l_o = \kappa \,z_o$
267where $\kappa = 0.4$ is von Karman's constant and $z_o$ the roughness
268parameter of the surface. Assuming $z_o=0.1$~m \citep{Craig_Banner_JPO94}
269leads to a 0.04~m, the default value of \np{rn\_mxl0}. In the ocean interior
270a minimum length scale is set to recover the molecular viscosity when $\bar{e}$
271reach its minimum value ($1.10^{-6}= C_k\, l_{min} \,\sqrt{\bar{e}_{min}}$ ).
272
273
274\subsubsection{Surface wave breaking parameterization}
275%-----------------------------------------------------------------------%
276Following \citet{Mellor_Blumberg_JPO04}, the TKE turbulence closure model has been modified
277to include the effect of surface wave breaking energetics. This results in a reduction of summertime
278surface temperature when the mixed layer is relatively shallow. The \citet{Mellor_Blumberg_JPO04}
279modifications acts on surface length scale and TKE values and air-sea drag coefficient.
280The latter concerns the bulk formulea and is not discussed here.
281
282Following \citet{Craig_Banner_JPO94}, the boundary condition on surface TKE value is :
283\begin{equation} \label{ZDF_Esbc}
284\bar{e}_o = \frac{1}{2}\,\left( 15.8\,\alpha_{CB} \right)^{2/3} \,\frac{|\tau|}{\rho_o}
285\end{equation}
286where $\alpha_{CB}$ is the \citet{Craig_Banner_JPO94} constant of proportionality
287which depends on the ''wave age'', ranging from 57 for mature waves to 146 for
288younger waves \citep{Mellor_Blumberg_JPO04}.
289The boundary condition on the turbulent length scale follows the Charnock's relation:
290\begin{equation} \label{ZDF_Lsbc}
291l_o = \kappa \beta \,\frac{|\tau|}{g\,\rho_o}
292\end{equation}
293where $\kappa=0.40$ is the von Karman constant, and $\beta$ is the Charnock's constant.
294\citet{Mellor_Blumberg_JPO04} suggest $\beta = 2.10^{5}$ the value chosen by \citet{Stacey_JPO99}
295citing observation evidence, and $\alpha_{CB} = 100$ the Craig and Banner's value.
296As the surface boundary condition on TKE is prescribed through $\bar{e}_o = e_{bb} |\tau| / \rho_o$,
297with $e_{bb}$ the \np{rn\_ebb} namelist parameter, setting \np{rn\_ebb}~=~67.83 corresponds
298to $\alpha_{CB} = 100$. Further setting \np{ln\_mxl0} to true applies \eqref{ZDF_Lsbc}
299as surface boundary condition on length scale, with $\beta$ hard coded to the Stacey's value.
300Note that a minimal threshold of \np{rn\_emin0}$=10^{-4}~m^2.s^{-2}$ (namelist parameters)
301is applied on surface $\bar{e}$ value.
302
303
304\subsubsection{Langmuir cells}
305%--------------------------------------%
306Langmuir circulations (LC) can be described as ordered large-scale vertical motions
307in the surface layer of the oceans. Although LC have nothing to do with convection,
308the circulation pattern is rather similar to so-called convective rolls in the atmospheric
309boundary layer. The detailed physics behind LC is described in, for example,
310\citet{Craik_Leibovich_JFM76}. The prevailing explanation is that LC arise from
311a nonlinear interaction between the Stokes drift and wind drift currents.
312
313Here we introduced in the TKE turbulent closure the simple parameterization of
314Langmuir circulations proposed by \citep{Axell_JGR02} for a $k-\epsilon$ turbulent closure.
315The parameterization, tuned against large-eddy simulation, includes the whole effect
316of LC in an extra source terms of TKE, $P_{LC}$.
317The presence of $P_{LC}$ in \eqref{Eq_zdftke_e}, the TKE equation, is controlled
318by setting \np{ln\_lc} to \textit{true} in the namtke namelist.
319
320By making an analogy with the characteristic convective velocity scale
321($e.g.$, \citet{D'Alessio_al_JPO98}), $P_{LC}$ is assumed to be :
322\begin{equation}
323P_{LC}(z) = \frac{w_{LC}^3(z)}{H_{LC}}
324\end{equation}
325where $w_{LC}(z)$ is the vertical velocity profile of LC, and $H_{LC}$ is the LC depth.
326With no information about the wave field, $w_{LC}$ is assumed to be proportional to
327the Stokes drift $u_s = 0.377\,\,|\tau|^{1/2}$, where $|\tau|$ is the surface wind stress module
328\footnote{Following \citet{Li_Garrett_JMR93}, the surface Stoke drift velocity
329may be expressed as $u_s = 0.016 \,|U_{10m}|$. Assuming an air density of
330$\rho_a=1.22 \,Kg/m^3$ and a drag coefficient of $1.5~10^{-3}$ give the expression
331used of $u_s$ as a function of the module of surface stress}.
332For the vertical variation, $w_{LC}$ is assumed to be zero at the surface as well as
333at a finite depth $H_{LC}$ (which is often close to the mixed layer depth), and simply
334varies as a sine function in between (a first-order profile for the Langmuir cell structures).
335The resulting expression for $w_{LC}$ is :
336\begin{equation}
337w_{LC} = \begin{cases}
338 c_{LC} \,u_s \,\sin(- \pi\,z / H_{LC} ) & \text{if $-z \leq H_{LC}$} \\
339 0 & \text{otherwise}
340 \end{cases}
341\end{equation}
342where $c_{LC} = 0.15$ has been chosen by \citep{Axell_JGR02} as a good compromise
343to fit LES data. The chosen value yields maximum vertical velocities $w_{LC}$ of the order
344of a few centimeters per second. The value of $c_{LC}$ is set through the \np{rn\_lc}
345namelist parameter, having in mind that it should stay between 0.15 and 0.54 \citep{Axell_JGR02}.
346
347The $H_{LC}$ is estimated in a similar way as the turbulent length scale of TKE equations:
348$H_{LC}$ is depth to which a water parcel with kinetic energy due to Stoke drift
349can reach on its own by converting its kinetic energy to potential energy, according to
350\begin{equation}
351- \int_{-H_{LC}}^0 { N^2\;\;dz} = \frac{1}{2} u_s^2
352\end{equation}
353
354
355\subsubsection{Mixing just below the mixed layer}
356%--------------------------------------------------------------%
357
358Vertical mixing parameterizations commonly used in ocean general circulation models
359tend to produce mixed-layer depths that are too shallow during summer months and windy conditions.
360This bias is particularly acute over the Southern Ocean.
361To overcome this systematic bias, an ad hoc parameterization is introduced into the TKE scheme \cite{Rodgers_2014}.
362The parameterization is an empirical one, $i.e.$ not derived from theoretical considerations,
363but rather is meant to account for observed processes that affect the density structure of
364the ocean’s planetary boundary layer that are not explicitly captured by default in the TKE scheme
365($i.e.$ near-inertial oscillations and ocean swells and waves).
366
367When using this parameterization ($i.e.$ when \np{nn\_etau}~=~1), the TKE input to the ocean ($S$)
368imposed by the winds in the form of near-inertial oscillations, swell and waves is parameterized
369by \eqref{ZDF_Esbc} the standard TKE surface boundary condition, plus a depth depend one given by:
370\begin{equation} \label{ZDF_Ehtau}
371S = (1-f_i) \; f_r \; e_s \; e^{-z / h_\tau}
372\end{equation}
373where
374$z$ is the depth,
375$e_s$ is TKE surface boundary condition,
376$f_r$ is the fraction of the surface TKE that penetrate in the ocean,
377$h_\tau$ is a vertical mixing length scale that controls exponential shape of the penetration,
378and $f_i$ is the ice concentration (no penetration if $f_i=1$, that is if the ocean is entirely
379covered by sea-ice).
380The value of $f_r$, usually a few percents, is specified through \np{rn\_efr} namelist parameter.
381The vertical mixing length scale, $h_\tau$, can be set as a 10~m uniform value (\np{nn\_etau}~=~0)
382or a latitude dependent value (varying from 0.5~m at the Equator to a maximum value of 30~m
383at high latitudes (\np{nn\_etau}~=~1).
384
385Note that two other option existe, \np{nn\_etau}~=~2, or 3. They correspond to applying
386\eqref{ZDF_Ehtau} only at the base of the mixed layer, or to using the high frequency part
387of the stress to evaluate the fraction of TKE that penetrate the ocean.
388Those two options are obsolescent features introduced for test purposes.
389They will be removed in the next release.
390
391
392
393% from Burchard et al OM 2008 :
394% the most critical process not reproduced by statistical turbulence models is the activity of
395% internal waves and their interaction with turbulence. After the Reynolds decomposition,
396% internal waves are in principle included in the RANS equations, but later partially
397% excluded by the hydrostatic assumption and the model resolution.
398% Thus far, the representation of internal wave mixing in ocean models has been relatively crude
399% (e.g. Mellor, 1989; Large et al., 1994; Meier, 2001; Axell, 2002; St. Laurent and Garrett, 2002).
400
401
402
403% -------------------------------------------------------------------------------------------------------------
404% TKE discretization considerations
405% -------------------------------------------------------------------------------------------------------------
406\subsection{TKE discretization considerations (\key{zdftke})}
407\label{ZDF_tke_ene}
408
409%>>>>>>>>>>>>>>>>>>>>>>>>>>>>
410\begin{figure}[!t] \begin{center}
411\includegraphics[width=1.00\textwidth]{Fig_ZDF_TKE_time_scheme}
412\caption{ \label{Fig_TKE_time_scheme}
413Illustration of the TKE time integration and its links to the momentum and tracer time integration. }
414\end{center}
415\end{figure}
416%>>>>>>>>>>>>>>>>>>>>>>>>>>>>
417
418The production of turbulence by vertical shear (the first term of the right hand side
419of \eqref{Eq_zdftke_e}) should balance the loss of kinetic energy associated with
420the vertical momentum diffusion (first line in \eqref{Eq_PE_zdf}). To do so a special care
421have to be taken for both the time and space discretization of the TKE equation
422\citep{Burchard_OM02,Marsaleix_al_OM08}.
423
424Let us first address the time stepping issue. Fig.~\ref{Fig_TKE_time_scheme} shows
425how the two-level Leap-Frog time stepping of the momentum and tracer equations interplays
426with the one-level forward time stepping of TKE equation. With this framework, the total loss
427of kinetic energy (in 1D for the demonstration) due to the vertical momentum diffusion is
428obtained by multiplying this quantity by $u^t$ and summing the result vertically:
429\begin{equation} \label{Eq_energ1}
430\begin{split}
431\int_{-H}^{\eta} u^t \,\partial_z &\left( {K_m}^t \,(\partial_z u)^{t+\rdt} \right) \,dz \\
432&= \Bigl[ u^t \,{K_m}^t \,(\partial_z u)^{t+\rdt} \Bigr]_{-H}^{\eta}
433 - \int_{-H}^{\eta}{ {K_m}^t \,\partial_z{u^t} \,\partial_z u^{t+\rdt} \,dz }
434\end{split}
435\end{equation}
436Here, the vertical diffusion of momentum is discretized backward in time
437with a coefficient, $K_m$, known at time $t$ (Fig.~\ref{Fig_TKE_time_scheme}),
438as it is required when using the TKE scheme (see \S\ref{STP_forward_imp}).
439The first term of the right hand side of \eqref{Eq_energ1} represents the kinetic energy
440transfer at the surface (atmospheric forcing) and at the bottom (friction effect).
441The second term is always negative. It is the dissipation rate of kinetic energy,
442and thus minus the shear production rate of $\bar{e}$. \eqref{Eq_energ1}
443implies that, to be energetically consistent, the production rate of $\bar{e}$
444used to compute $(\bar{e})^t$ (and thus ${K_m}^t$) should be expressed as
445${K_m}^{t-\rdt}\,(\partial_z u)^{t-\rdt} \,(\partial_z u)^t$ (and not by the more straightforward
446$K_m \left( \partial_z u \right)^2$ expression taken at time $t$ or $t-\rdt$).
447
448A similar consideration applies on the destruction rate of $\bar{e}$ due to stratification
449(second term of the right hand side of \eqref{Eq_zdftke_e}). This term
450must balance the input of potential energy resulting from vertical mixing.
451The rate of change of potential energy (in 1D for the demonstration) due vertical
452mixing is obtained by multiplying vertical density diffusion
453tendency by $g\,z$ and and summing the result vertically:
454\begin{equation} \label{Eq_energ2}
455\begin{split}
456\int_{-H}^{\eta} g\,z\,\partial_z &\left( {K_\rho}^t \,(\partial_k \rho)^{t+\rdt} \right) \,dz \\
457&= \Bigl[ g\,z \,{K_\rho}^t \,(\partial_z \rho)^{t+\rdt} \Bigr]_{-H}^{\eta}
458 - \int_{-H}^{\eta}{ g \,{K_\rho}^t \,(\partial_k \rho)^{t+\rdt} } \,dz \\
459&= - \Bigl[ z\,{K_\rho}^t \,(N^2)^{t+\rdt} \Bigr]_{-H}^{\eta}
460+ \int_{-H}^{\eta}{ \rho^{t+\rdt} \, {K_\rho}^t \,(N^2)^{t+\rdt} \,dz }
461\end{split}
462\end{equation}
463where we use $N^2 = -g \,\partial_k \rho / (e_3 \rho)$.
464The first term of the right hand side of \eqref{Eq_energ2} is always zero
465because there is no diffusive flux through the ocean surface and bottom).
466The second term is minus the destruction rate of $\bar{e}$ due to stratification.
467Therefore \eqref{Eq_energ1} implies that, to be energetically consistent, the product
468${K_\rho}^{t-\rdt}\,(N^2)^t$ should be used in \eqref{Eq_zdftke_e}, the TKE equation.
469
470Let us now address the space discretization issue.
471The vertical eddy coefficients are defined at $w$-point whereas the horizontal velocity
472components are in the centre of the side faces of a $t$-box in staggered C-grid
473(Fig.\ref{Fig_cell}). A space averaging is thus required to obtain the shear TKE production term.
474By redoing the \eqref{Eq_energ1} in the 3D case, it can be shown that the product of
475eddy coefficient by the shear at $t$ and $t-\rdt$ must be performed prior to the averaging.
476Furthermore, the possible time variation of $e_3$ (\key{vvl} case) have to be taken into
477account.
478
479The above energetic considerations leads to
480the following final discrete form for the TKE equation:
481\begin{equation} \label{Eq_zdftke_ene}
482\begin{split}
483\frac { (\bar{e})^t - (\bar{e})^{t-\rdt} } {\rdt} \equiv
484\Biggl\{ \Biggr.
485 &\overline{ \left( \left(\overline{K_m}^{\,i+1/2}\right)^{t-\rdt} \,\frac{\delta_{k+1/2}[u^{t+\rdt}]}{{e_3u}^{t+\rdt} }
486 \ \frac{\delta_{k+1/2}[u^ t ]}{{e_3u}^ t } \right) }^{\,i} \\
487+&\overline{ \left( \left(\overline{K_m}^{\,j+1/2}\right)^{t-\rdt} \,\frac{\delta_{k+1/2}[v^{t+\rdt}]}{{e_3v}^{t+\rdt} }
488 \ \frac{\delta_{k+1/2}[v^ t ]}{{e_3v}^ t } \right) }^{\,j}
489\Biggr. \Biggr\} \\
490%
491- &{K_\rho}^{t-\rdt}\,{(N^2)^t} \\
492%
493+&\frac{1}{{e_3w}^{t+\rdt}} \;\delta_{k+1/2} \left[ {K_m}^{t-\rdt} \,\frac{\delta_{k}[(\bar{e})^{t+\rdt}]} {{e_3w}^{t+\rdt}} \right] \\
494%
495- &c_\epsilon \; \left( \frac{\sqrt{\bar {e}}}{l_\epsilon}\right)^{t-\rdt}\,(\bar {e})^{t+\rdt}
496\end{split}
497\end{equation}
498where the last two terms in \eqref{Eq_zdftke_ene} (vertical diffusion and Kolmogorov dissipation)
499are time stepped using a backward scheme (see\S\ref{STP_forward_imp}).
500Note that the Kolmogorov term has been linearized in time in order to render
501the implicit computation possible. The restart of the TKE scheme
502requires the storage of $\bar {e}$, $K_m$, $K_\rho$ and $l_\epsilon$ as they all appear in
503the right hand side of \eqref{Eq_zdftke_ene}. For the latter, it is in fact
504the ratio $\sqrt{\bar{e}}/l_\epsilon$ which is stored.
505
506% -------------------------------------------------------------------------------------------------------------
507% GLS Generic Length Scale Scheme
508% -------------------------------------------------------------------------------------------------------------
509\subsection{GLS Generic Length Scale (\key{zdfgls})}
510\label{ZDF_gls}
511
512%--------------------------------------------namzdf_gls---------------------------------------------------------
513\namdisplay{namzdf_gls}
514%--------------------------------------------------------------------------------------------------------------
515
516The Generic Length Scale (GLS) scheme is a turbulent closure scheme based on
517two prognostic equations: one for the turbulent kinetic energy $\bar {e}$, and another
518for the generic length scale, $\psi$ \citep{Umlauf_Burchard_JMS03, Umlauf_Burchard_CSR05}.
519This later variable is defined as : $\psi = {C_{0\mu}}^{p} \ {\bar{e}}^{m} \ l^{n}$,
520where the triplet $(p, m, n)$ value given in Tab.\ref{Tab_GLS} allows to recover
521a number of well-known turbulent closures ($k$-$kl$ \citep{Mellor_Yamada_1982},
522$k$-$\epsilon$ \citep{Rodi_1987}, $k$-$\omega$ \citep{Wilcox_1988}
523among others \citep{Umlauf_Burchard_JMS03,Kantha_Carniel_CSR05}).
524The GLS scheme is given by the following set of equations:
525\begin{equation} \label{Eq_zdfgls_e}
526\frac{\partial \bar{e}}{\partial t} =
527\frac{K_m}{\sigma_e e_3 }\;\left[ {\left( \frac{\partial u}{\partial k} \right)^2
528 +\left( \frac{\partial v}{\partial k} \right)^2} \right]
529-K_\rho \,N^2
530+\frac{1}{e_3}\,\frac{\partial}{\partial k} \left[ \frac{K_m}{e_3}\,\frac{\partial \bar{e}}{\partial k} \right]
531- \epsilon
532\end{equation}
533
534\begin{equation} \label{Eq_zdfgls_psi}
535 \begin{split}
536\frac{\partial \psi}{\partial t} =& \frac{\psi}{\bar{e}} \left\{
537\frac{C_1\,K_m}{\sigma_{\psi} {e_3}}\;\left[ {\left( \frac{\partial u}{\partial k} \right)^2
538 +\left( \frac{\partial v}{\partial k} \right)^2} \right]
539- C_3 \,K_\rho\,N^2 - C_2 \,\epsilon \,Fw \right\} \\
540&+\frac{1}{e_3} \;\frac{\partial }{\partial k}\left[ {\frac{K_m}{e_3 }
541 \;\frac{\partial \psi}{\partial k}} \right]\;
542 \end{split}
543\end{equation}
544
545\begin{equation} \label{Eq_zdfgls_kz}
546 \begin{split}
547 K_m &= C_{\mu} \ \sqrt {\bar{e}} \ l \\
548 K_\rho &= C_{\mu'}\ \sqrt {\bar{e}} \ l
549 \end{split}
550\end{equation}
551
552\begin{equation} \label{Eq_zdfgls_eps}
553{\epsilon} = C_{0\mu} \,\frac{\bar {e}^{3/2}}{l} \;
554\end{equation}
555where $N$ is the local Brunt-Vais\"{a}l\"{a} frequency (see \S\ref{TRA_bn2})
556and $\epsilon$ the dissipation rate.
557The constants $C_1$, $C_2$, $C_3$, ${\sigma_e}$, ${\sigma_{\psi}}$ and the wall function ($Fw$)
558depends of the choice of the turbulence model. Four different turbulent models are pre-defined
559(Tab.\ref{Tab_GLS}). They are made available through the \np{nn\_clo} namelist parameter.
560
561%--------------------------------------------------TABLE--------------------------------------------------
562\begin{table}[htbp] \begin{center}
563%\begin{tabular}{cp{70pt}cp{70pt}cp{70pt}cp{70pt}cp{70pt}cp{70pt}c}
564\begin{tabular}{ccccc}
565 & $k-kl$ & $k-\epsilon$ & $k-\omega$ & generic \\
566% & \citep{Mellor_Yamada_1982} & \citep{Rodi_1987} & \citep{Wilcox_1988} & \\
567\hline \hline
568\np{nn\_clo} & \textbf{0} & \textbf{1} & \textbf{2} & \textbf{3} \\
569\hline
570$( p , n , m )$ & ( 0 , 1 , 1 ) & ( 3 , 1.5 , -1 ) & ( -1 , 0.5 , -1 ) & ( 2 , 1 , -0.67 ) \\
571$\sigma_k$ & 2.44 & 1. & 2. & 0.8 \\
572$\sigma_\psi$ & 2.44 & 1.3 & 2. & 1.07 \\
573$C_1$ & 0.9 & 1.44 & 0.555 & 1. \\
574$C_2$ & 0.5 & 1.92 & 0.833 & 1.22 \\
575$C_3$ & 1. & 1. & 1. & 1. \\
576$F_{wall}$ & Yes & -- & -- & -- \\
577\hline
578\hline
579\end{tabular}
580\caption{ \label{Tab_GLS}
581Set of predefined GLS parameters, or equivalently predefined turbulence models available
582with \key{zdfgls} and controlled by the \np{nn\_clos} namelist variable in \ngn{namzdf\_gls} .}
583\end{center} \end{table}
584%--------------------------------------------------------------------------------------------------------------
585
586In the Mellor-Yamada model, the negativity of $n$ allows to use a wall function to force
587the convergence of the mixing length towards $K z_b$ ($K$: Kappa and $z_b$: rugosity length)
588value near physical boundaries (logarithmic boundary layer law). $C_{\mu}$ and $C_{\mu'}$
589are calculated from stability function proposed by \citet{Galperin_al_JAS88}, or by \citet{Kantha_Clayson_1994}
590or one of the two functions suggested by \citet{Canuto_2001} (\np{nn\_stab\_func} = 0, 1, 2 or 3, resp.).
591The value of $C_{0\mu}$ depends of the choice of the stability function.
592
593The surface and bottom boundary condition on both $\bar{e}$ and $\psi$ can be calculated
594thanks to Dirichlet or Neumann condition through \np{nn\_tkebc\_surf} and \np{nn\_tkebc\_bot}, resp.
595As for TKE closure , the wave effect on the mixing is considered when \np{ln\_crban}~=~true
596\citep{Craig_Banner_JPO94, Mellor_Blumberg_JPO04}. The \np{rn\_crban} namelist parameter
597is $\alpha_{CB}$ in \eqref{ZDF_Esbc} and \np{rn\_charn} provides the value of $\beta$ in \eqref{ZDF_Lsbc}.
598
599The $\psi$ equation is known to fail in stably stratified flows, and for this reason
600almost all authors apply a clipping of the length scale as an \textit{ad hoc} remedy.
601With this clipping, the maximum permissible length scale is determined by
602$l_{max} = c_{lim} \sqrt{2\bar{e}}/ N$. A value of $c_{lim} = 0.53$ is often used
603\citep{Galperin_al_JAS88}. \cite{Umlauf_Burchard_CSR05} show that the value of
604the clipping factor is of crucial importance for the entrainment depth predicted in
605stably stratified situations, and that its value has to be chosen in accordance
606with the algebraic model for the turbulent fluxes. The clipping is only activated
607if \np{ln\_length\_lim}=true, and the $c_{lim}$ is set to the \np{rn\_clim\_galp} value.
608
609The time and space discretization of the GLS equations follows the same energetic
610consideration as for the TKE case described in \S\ref{ZDF_tke_ene} \citep{Burchard_OM02}.
611Examples of performance of the 4 turbulent closure scheme can be found in \citet{Warner_al_OM05}.
612
613% -------------------------------------------------------------------------------------------------------------
614% K Profile Parametrisation (KPP)
615% -------------------------------------------------------------------------------------------------------------
616\subsection{K Profile Parametrisation (KPP) (\key{zdfkpp}) }
617\label{ZDF_kpp}
618
619%--------------------------------------------namkpp--------------------------------------------------------
620\namdisplay{namzdf_kpp}
621%--------------------------------------------------------------------------------------------------------------
622
623The KKP scheme has been implemented by J. Chanut ...
624Options are defined through the \ngn{namzdf\_kpp} namelist variables.
625
626Note that KPP is an obsolescent feature of the \NEMO system.
627It will be removed in the next release (v3.7 and followings).
628
629
630% ================================================================
631% Convection
632% ================================================================
633\section{Convection}
634\label{ZDF_conv}
635
636%--------------------------------------------namzdf--------------------------------------------------------
637\namdisplay{namzdf}
638%--------------------------------------------------------------------------------------------------------------
639
640Static instabilities (i.e. light potential densities under heavy ones) may
641occur at particular ocean grid points. In nature, convective processes
642quickly re-establish the static stability of the water column. These
643processes have been removed from the model via the hydrostatic
644assumption so they must be parameterized. Three parameterisations
645are available to deal with convective processes: a non-penetrative
646convective adjustment or an enhanced vertical diffusion, or/and the
647use of a turbulent closure scheme.
648
649% -------------------------------------------------------------------------------------------------------------
650% Non-Penetrative Convective Adjustment
651% -------------------------------------------------------------------------------------------------------------
652\subsection [Non-Penetrative Convective Adjustment (\np{ln\_tranpc}) ]
653 {Non-Penetrative Convective Adjustment (\np{ln\_tranpc}=.true.) }
654\label{ZDF_npc}
655
656%--------------------------------------------namzdf--------------------------------------------------------
657\namdisplay{namzdf}
658%--------------------------------------------------------------------------------------------------------------
659
660%>>>>>>>>>>>>>>>>>>>>>>>>>>>>
661\begin{figure}[!htb] \begin{center}
662\includegraphics[width=0.90\textwidth]{Fig_npc}
663\caption{ \label{Fig_npc}
664Example of an unstable density profile treated by the non penetrative
665convective adjustment algorithm. $1^{st}$ step: the initial profile is checked from
666the surface to the bottom. It is found to be unstable between levels 3 and 4.
667They are mixed. The resulting $\rho$ is still larger than $\rho$(5): levels 3 to 5
668are mixed. The resulting $\rho$ is still larger than $\rho$(6): levels 3 to 6 are
669mixed. The $1^{st}$ step ends since the density profile is then stable below
670the level 3. $2^{nd}$ step: the new $\rho$ profile is checked following the same
671procedure as in $1^{st}$ step: levels 2 to 5 are mixed. The new density profile
672is checked. It is found stable: end of algorithm.}
673\end{center} \end{figure}
674%>>>>>>>>>>>>>>>>>>>>>>>>>>>>
675
676Options are defined through the \ngn{namzdf} namelist variables.
677The non-penetrative convective adjustment is used when \np{ln\_zdfnpc}~=~\textit{true}.
678It is applied at each \np{nn\_npc} time step and mixes downwards instantaneously
679the statically unstable portion of the water column, but only until the density
680structure becomes neutrally stable ($i.e.$ until the mixed portion of the water
681column has \textit{exactly} the density of the water just below) \citep{Madec_al_JPO91}.
682The associated algorithm is an iterative process used in the following way
683(Fig. \ref{Fig_npc}): starting from the top of the ocean, the first instability is
684found. Assume in the following that the instability is located between levels
685$k$ and $k+1$. The temperature and salinity in the two levels are
686vertically mixed, conserving the heat and salt contents of the water column.
687The new density is then computed by a linear approximation. If the new
688density profile is still unstable between levels $k+1$ and $k+2$, levels $k$,
689$k+1$ and $k+2$ are then mixed. This process is repeated until stability is
690established below the level $k$ (the mixing process can go down to the
691ocean bottom). The algorithm is repeated to check if the density profile
692between level $k-1$ and $k$ is unstable and/or if there is no deeper instability.
693
694This algorithm is significantly different from mixing statically unstable levels
695two by two. The latter procedure cannot converge with a finite number
696of iterations for some vertical profiles while the algorithm used in \NEMO
697converges for any profile in a number of iterations which is less than the
698number of vertical levels. This property is of paramount importance as
699pointed out by \citet{Killworth1989}: it avoids the existence of permanent
700and unrealistic static instabilities at the sea surface. This non-penetrative
701convective algorithm has been proved successful in studies of the deep
702water formation in the north-western Mediterranean Sea
704
705The current implementation has been modified in order to deal with any non linear
706equation of seawater (L. Brodeau, personnal communication).
707Two main differences have been introduced compared to the original algorithm:
708$(i)$ the stability is now checked using the Brunt-V\"{a}is\"{a}l\"{a} frequency
709(not the the difference in potential density) ;
710$(ii)$ when two levels are found unstable, their thermal and haline expansion coefficients
711are vertically mixed in the same way their temperature and salinity has been mixed.
712These two modifications allow the algorithm to perform properly and accurately
713with TEOS10 or EOS-80 without having to recompute the expansion coefficients at each
714mixing iteration.
715
716% -------------------------------------------------------------------------------------------------------------
717% Enhanced Vertical Diffusion
718% -------------------------------------------------------------------------------------------------------------
719\subsection [Enhanced Vertical Diffusion (\np{ln\_zdfevd})]
720 {Enhanced Vertical Diffusion (\np{ln\_zdfevd}=true)}
721\label{ZDF_evd}
722
723%--------------------------------------------namzdf--------------------------------------------------------
724\namdisplay{namzdf}
725%--------------------------------------------------------------------------------------------------------------
726
727Options are defined through the \ngn{namzdf} namelist variables.
728The enhanced vertical diffusion parameterisation is used when \np{ln\_zdfevd}=true.
729In this case, the vertical eddy mixing coefficients are assigned very large values
730(a typical value is $10\;m^2s^{-1})$ in regions where the stratification is unstable
731($i.e.$ when $N^2$ the Brunt-Vais\"{a}l\"{a} frequency is negative)
732\citep{Lazar_PhD97, Lazar_al_JPO99}. This is done either on tracers only
733(\np{nn\_evdm}=0) or on both momentum and tracers (\np{nn\_evdm}=1).
734
735In practice, where $N^2\leq 10^{-12}$, $A_T^{vT}$ and $A_T^{vS}$, and
736if \np{nn\_evdm}=1, the four neighbouring $A_u^{vm} \;\mbox{and}\;A_v^{vm}$
737values also, are set equal to the namelist parameter \np{rn\_avevd}. A typical value
738for $rn\_avevd$ is between 1 and $100~m^2.s^{-1}$. This parameterisation of
739convective processes is less time consuming than the convective adjustment
740algorithm presented above when mixing both tracers and momentum in the
741case of static instabilities. It requires the use of an implicit time stepping on
742vertical diffusion terms (i.e. \np{ln\_zdfexp}=false).
743
744Note that the stability test is performed on both \textit{before} and \textit{now}
745values of $N^2$. This removes a potential source of divergence of odd and
746even time step in a leapfrog environment \citep{Leclair_PhD2010} (see \S\ref{STP_mLF}).
747
748% -------------------------------------------------------------------------------------------------------------
749% Turbulent Closure Scheme
750% -------------------------------------------------------------------------------------------------------------
751\subsection{Turbulent Closure Scheme (\key{zdftke} or \key{zdfgls})}
752\label{ZDF_tcs}
753
754The turbulent closure scheme presented in \S\ref{ZDF_tke} and \S\ref{ZDF_gls}
755(\key{zdftke} or \key{zdftke} is defined) in theory solves the problem of statically
756unstable density profiles. In such a case, the term corresponding to the
757destruction of turbulent kinetic energy through stratification in \eqref{Eq_zdftke_e}
758or \eqref{Eq_zdfgls_e} becomes a source term, since $N^2$ is negative.
759It results in large values of $A_T^{vT}$ and $A_T^{vT}$, and also the four neighbouring
760$A_u^{vm} {and}\;A_v^{vm}$ (up to $1\;m^2s^{-1})$. These large values
761restore the static stability of the water column in a way similar to that of the
762enhanced vertical diffusion parameterisation (\S\ref{ZDF_evd}). However,
763in the vicinity of the sea surface (first ocean layer), the eddy coefficients
764computed by the turbulent closure scheme do not usually exceed $10^{-2}m.s^{-1}$,
765because the mixing length scale is bounded by the distance to the sea surface.
766It can thus be useful to combine the enhanced vertical
767diffusion with the turbulent closure scheme, $i.e.$ setting the \np{ln\_zdfnpc}
768namelist parameter to true and defining the turbulent closure CPP key all together.
769
770The KPP turbulent closure scheme already includes enhanced vertical diffusion
771in the case of convection, as governed by the variables $bvsqcon$ and $difcon$
772found in \mdl{zdfkpp}, therefore \np{ln\_zdfevd}=false should be used with the KPP
773scheme. %gm% + one word on non local flux with KPP scheme trakpp.F90 module...
774
775% ================================================================
776% Double Diffusion Mixing
777% ================================================================
778\section [Double Diffusion Mixing (\key{zdfddm})]
779 {Double Diffusion Mixing (\key{zdfddm})}
780\label{ZDF_ddm}
781
782%-------------------------------------------namzdf_ddm-------------------------------------------------
783\namdisplay{namzdf_ddm}
784%--------------------------------------------------------------------------------------------------------------
785
786Options are defined through the \ngn{namzdf\_ddm} namelist variables.
787Double diffusion occurs when relatively warm, salty water overlies cooler, fresher
788water, or vice versa. The former condition leads to salt fingering and the latter
789to diffusive convection. Double-diffusive phenomena contribute to diapycnal
790mixing in extensive regions of the ocean. \citet{Merryfield1999} include a
791parameterisation of such phenomena in a global ocean model and show that
792it leads to relatively minor changes in circulation but exerts significant regional
793influences on temperature and salinity. This parameterisation has been
794introduced in \mdl{zdfddm} module and is controlled by the \key{zdfddm} CPP key.
795
796Diapycnal mixing of S and T are described by diapycnal diffusion coefficients
797\begin{align*} % \label{Eq_zdfddm_Kz}
798 &A^{vT} = A_o^{vT}+A_f^{vT}+A_d^{vT} \\
799 &A^{vS} = A_o^{vS}+A_f^{vS}+A_d^{vS}
800\end{align*}
801where subscript $f$ represents mixing by salt fingering, $d$ by diffusive convection,
802and $o$ by processes other than double diffusion. The rates of double-diffusive
803mixing depend on the buoyancy ratio $R_\rho = \alpha \partial_z T / \beta \partial_z S$,
804where $\alpha$ and $\beta$ are coefficients of thermal expansion and saline
805contraction (see \S\ref{TRA_eos}). To represent mixing of $S$ and $T$ by salt
806fingering, we adopt the diapycnal diffusivities suggested by Schmitt (1981):
807\begin{align} \label{Eq_zdfddm_f}
808A_f^{vS} &= \begin{cases}
809 \frac{A^{\ast v}}{1+(R_\rho / R_c)^n } &\text{if $R_\rho > 1$ and $N^2>0$ } \\
810 0 &\text{otherwise}
811 \end{cases}
812\\ \label{Eq_zdfddm_f_T}
813A_f^{vT} &= 0.7 \ A_f^{vS} / R_\rho
814\end{align}
815
816%>>>>>>>>>>>>>>>>>>>>>>>>>>>>
817\begin{figure}[!t] \begin{center}
818\includegraphics[width=0.99\textwidth]{Fig_zdfddm}
819\caption{ \label{Fig_zdfddm}
820From \citet{Merryfield1999} : (a) Diapycnal diffusivities $A_f^{vT}$
821and $A_f^{vS}$ for temperature and salt in regions of salt fingering. Heavy
822curves denote $A^{\ast v} = 10^{-3}~m^2.s^{-1}$ and thin curves
823$A^{\ast v} = 10^{-4}~m^2.s^{-1}$ ; (b) diapycnal diffusivities $A_d^{vT}$ and
824$A_d^{vS}$ for temperature and salt in regions of diffusive convection. Heavy
825curves denote the Federov parameterisation and thin curves the Kelley
826parameterisation. The latter is not implemented in \NEMO. }
827\end{center} \end{figure}
828%>>>>>>>>>>>>>>>>>>>>>>>>>>>>
829
830The factor 0.7 in \eqref{Eq_zdfddm_f_T} reflects the measured ratio
831$\alpha F_T /\beta F_S \approx 0.7$ of buoyancy flux of heat to buoyancy
832flux of salt ($e.g.$, \citet{McDougall_Taylor_JMR84}). Following \citet{Merryfield1999},
833we adopt $R_c = 1.6$, $n = 6$, and $A^{\ast v} = 10^{-4}~m^2.s^{-1}$.
834
835To represent mixing of S and T by diffusive layering, the diapycnal diffusivities suggested by Federov (1988) is used:
836\begin{align} \label{Eq_zdfddm_d}
837A_d^{vT} &= \begin{cases}
838 1.3635 \, \exp{\left( 4.6\, \exp{ \left[ -0.54\,( R_{\rho}^{-1} - 1 ) \right] } \right)}
839 &\text{if $0<R_\rho < 1$ and $N^2>0$ } \\
840 0 &\text{otherwise}
841 \end{cases}
842\\ \label{Eq_zdfddm_d_S}
843A_d^{vS} &= \begin{cases}
844 A_d^{vT}\ \left( 1.85\,R_{\rho} - 0.85 \right)
845 &\text{if $0.5 \leq R_\rho<1$ and $N^2>0$ } \\
846 A_d^{vT} \ 0.15 \ R_\rho
847 &\text{if $\ \ 0 < R_\rho<0.5$ and $N^2>0$ } \\
848 0 &\text{otherwise}
849 \end{cases}
850\end{align}
851
852The dependencies of \eqref{Eq_zdfddm_f} to \eqref{Eq_zdfddm_d_S} on $R_\rho$
853are illustrated in Fig.~\ref{Fig_zdfddm}. Implementing this requires computing
854$R_\rho$ at each grid point on every time step. This is done in \mdl{eosbn2} at the
855same time as $N^2$ is computed. This avoids duplication in the computation of
856$\alpha$ and $\beta$ (which is usually quite expensive).
857
858% ================================================================
859% Bottom Friction
860% ================================================================
861\section [Bottom and Top Friction (\textit{zdfbfr})] {Bottom and Top Friction (\mdl{zdfbfr} module)}
862\label{ZDF_bfr}
863
864%--------------------------------------------nambfr--------------------------------------------------------
865\namdisplay{nambfr}
866%--------------------------------------------------------------------------------------------------------------
867
868Options to define the top and bottom friction are defined through the \ngn{nambfr} namelist variables.
869The bottom friction represents the friction generated by the bathymetry.
870The top friction represents the friction generated by the ice shelf/ocean interface.
871As the friction processes at the top and bottom are represented similarly, only the bottom friction is described in detail below.\\
872
873
874Both the surface momentum flux (wind stress) and the bottom momentum
875flux (bottom friction) enter the equations as a condition on the vertical
876diffusive flux. For the bottom boundary layer, one has:
877\begin{equation} \label{Eq_zdfbfr_flux}
878A^{vm} \left( \partial {\textbf U}_h / \partial z \right) = {{\cal F}}_h^{\textbf U}
879\end{equation}
880where ${\cal F}_h^{\textbf U}$ is represents the downward flux of horizontal momentum
881outside the logarithmic turbulent boundary layer (thickness of the order of
8821~m in the ocean). How ${\cal F}_h^{\textbf U}$ influences the interior depends on the
883vertical resolution of the model near the bottom relative to the Ekman layer
884depth. For example, in order to obtain an Ekman layer depth
885$d = \sqrt{2\;A^{vm}} / f = 50$~m, one needs a vertical diffusion coefficient
886$A^{vm} = 0.125$~m$^2$s$^{-1}$ (for a Coriolis frequency
887$f = 10^{-4}$~m$^2$s$^{-1}$). With a background diffusion coefficient
888$A^{vm} = 10^{-4}$~m$^2$s$^{-1}$, the Ekman layer depth is only 1.4~m.
889When the vertical mixing coefficient is this small, using a flux condition is
890equivalent to entering the viscous forces (either wind stress or bottom friction)
891as a body force over the depth of the top or bottom model layer. To illustrate
892this, consider the equation for $u$ at $k$, the last ocean level:
893\begin{equation} \label{Eq_zdfbfr_flux2}
894\frac{\partial u_k}{\partial t} = \frac{1}{e_{3u}} \left[ \frac{A_{uw}^{vm}}{e_{3uw}} \delta_{k+1/2}\;[u] - {\cal F}^u_h \right] \approx - \frac{{\cal F}^u_{h}}{e_{3u}}
895\end{equation}
896If the bottom layer thickness is 200~m, the Ekman transport will
897be distributed over that depth. On the other hand, if the vertical resolution
898is high (1~m or less) and a turbulent closure model is used, the turbulent
899Ekman layer will be represented explicitly by the model. However, the
900logarithmic layer is never represented in current primitive equation model
901applications: it is \emph{necessary} to parameterize the flux ${\cal F}^u_h$.
902Two choices are available in \NEMO: a linear and a quadratic bottom friction.
903Note that in both cases, the rotation between the interior velocity and the
904bottom friction is neglected in the present release of \NEMO.
905
906In the code, the bottom friction is imposed by adding the trend due to the bottom
907friction to the general momentum trend in \mdl{dynbfr}. For the time-split surface
908pressure gradient algorithm, the momentum trend due to the barotropic component
909needs to be handled separately. For this purpose it is convenient to compute and
910store coefficients which can be simply combined with bottom velocities and geometric
911values to provide the momentum trend due to bottom friction.
912These coefficients are computed in \mdl{zdfbfr} and generally take the form
913$c_b^{\textbf U}$ where:
914\begin{equation} \label{Eq_zdfbfr_bdef}
915\frac{\partial {\textbf U_h}}{\partial t} =
916 - \frac{{\cal F}^{\textbf U}_{h}}{e_{3u}} = \frac{c_b^{\textbf U}}{e_{3u}} \;{\textbf U}_h^b
917\end{equation}
918where $\textbf{U}_h^b = (u_b\;,\;v_b)$ is the near-bottom, horizontal, ocean velocity.
919
920% -------------------------------------------------------------------------------------------------------------
921% Linear Bottom Friction
922% -------------------------------------------------------------------------------------------------------------
923\subsection{Linear Bottom Friction (\np{nn\_botfr} = 0 or 1) }
924\label{ZDF_bfr_linear}
925
926The linear bottom friction parameterisation (including the special case
927of a free-slip condition) assumes that the bottom friction
928is proportional to the interior velocity (i.e. the velocity of the last
929model level):
930\begin{equation} \label{Eq_zdfbfr_linear}
931{\cal F}_h^\textbf{U} = \frac{A^{vm}}{e_3} \; \frac{\partial \textbf{U}_h}{\partial k} = r \; \textbf{U}_h^b
932\end{equation}
933where $r$ is a friction coefficient expressed in ms$^{-1}$.
934This coefficient is generally estimated by setting a typical decay time
935$\tau$ in the deep ocean,
936and setting $r = H / \tau$, where $H$ is the ocean depth. Commonly accepted
937values of $\tau$ are of the order of 100 to 200 days \citep{Weatherly_JMR84}.
938A value $\tau^{-1} = 10^{-7}$~s$^{-1}$ equivalent to 115 days, is usually used
939in quasi-geostrophic models. One may consider the linear friction as an
940approximation of quadratic friction, $r \approx 2\;C_D\;U_{av}$ (\citet{Gill1982},
941Eq. 9.6.6). For example, with a drag coefficient $C_D = 0.002$, a typical speed
942of tidal currents of $U_{av} =0.1$~m\;s$^{-1}$, and assuming an ocean depth
943$H = 4000$~m, the resulting friction coefficient is $r = 4\;10^{-4}$~m\;s$^{-1}$.
944This is the default value used in \NEMO. It corresponds to a decay time scale
945of 115~days. It can be changed by specifying \np{rn\_bfri1} (namelist parameter).
946
947For the linear friction case the coefficients defined in the general
948expression \eqref{Eq_zdfbfr_bdef} are:
949\begin{equation} \label{Eq_zdfbfr_linbfr_b}
950\begin{split}
951 c_b^u &= - r\\
952 c_b^v &= - r\\
953\end{split}
954\end{equation}
955When \np{nn\_botfr}=1, the value of $r$ used is \np{rn\_bfri1}.
956Setting \np{nn\_botfr}=0 is equivalent to setting $r=0$ and leads to a free-slip
957bottom boundary condition. These values are assigned in \mdl{zdfbfr}.
958From v3.2 onwards there is support for local enhancement of these values
959via an externally defined 2D mask array (\np{ln\_bfr2d}=true) given
960in the \ifile{bfr\_coef} input NetCDF file. The mask values should vary from 0 to 1.
961Locations with a non-zero mask value will have the friction coefficient increased
962by $mask\_value$*\np{rn\_bfrien}*\np{rn\_bfri1}.
963
964% -------------------------------------------------------------------------------------------------------------
965% Non-Linear Bottom Friction
966% -------------------------------------------------------------------------------------------------------------
967\subsection{Non-Linear Bottom Friction (\np{nn\_botfr} = 2)}
968\label{ZDF_bfr_nonlinear}
969
970The non-linear bottom friction parameterisation assumes that the bottom
972\begin{equation} \label{Eq_zdfbfr_nonlinear}
973{\cal F}_h^\textbf{U} = \frac{A^{vm}}{e_3 }\frac{\partial \textbf {U}_h
974}{\partial k}=C_D \;\sqrt {u_b ^2+v_b ^2+e_b } \;\; \textbf {U}_h^b
975\end{equation}
976where $C_D$ is a drag coefficient, and $e_b$ a bottom turbulent kinetic energy
977due to tides, internal waves breaking and other short time scale currents.
978A typical value of the drag coefficient is $C_D = 10^{-3}$. As an example,
979the CME experiment \citep{Treguier_JGR92} uses $C_D = 10^{-3}$ and
980$e_b = 2.5\;10^{-3}$m$^2$\;s$^{-2}$, while the FRAM experiment \citep{Killworth1992}
981uses $C_D = 1.4\;10^{-3}$ and $e_b =2.5\;\;10^{-3}$m$^2$\;s$^{-2}$.
982The CME choices have been set as default values (\np{rn\_bfri2} and \np{rn\_bfeb2}
983namelist parameters).
984
985As for the linear case, the bottom friction is imposed in the code by
986adding the trend due to the bottom friction to the general momentum trend
987in \mdl{dynbfr}.
988For the non-linear friction case the terms
989computed in \mdl{zdfbfr} are:
990\begin{equation} \label{Eq_zdfbfr_nonlinbfr}
991\begin{split}
992 c_b^u &= - \; C_D\;\left[ u^2 + \left(\bar{\bar{v}}^{i+1,j}\right)^2 + e_b \right]^{1/2}\\
993 c_b^v &= - \; C_D\;\left[ \left(\bar{\bar{u}}^{i,j+1}\right)^2 + v^2 + e_b \right]^{1/2}\\
994\end{split}
995\end{equation}
996
997The coefficients that control the strength of the non-linear bottom friction are
998initialised as namelist parameters: $C_D$= \np{rn\_bfri2}, and $e_b$ =\np{rn\_bfeb2}.
999Note for applications which treat tides explicitly a low or even zero value of
1000\np{rn\_bfeb2} is recommended. From v3.2 onwards a local enhancement of $C_D$ is possible
1001via an externally defined 2D mask array (\np{ln\_bfr2d}=true). This works in the same way
1002as for the linear bottom friction case with non-zero masked locations increased by
1003$mask\_value$*\np{rn\_bfrien}*\np{rn\_bfri2}.
1004
1005% -------------------------------------------------------------------------------------------------------------
1006% Bottom Friction Log-layer
1007% -------------------------------------------------------------------------------------------------------------
1008\subsection{Log-layer Bottom Friction enhancement (\np{nn\_botfr} = 2, \np{ln\_loglayer} = .true.)}
1009\label{ZDF_bfr_loglayer}
1010
1011In the non-linear bottom friction case, the drag coefficient, $C_D$, can be optionally
1012enhanced using a "law of the wall" scaling. If \np{ln\_loglayer} = .true., $C_D$ is no
1013longer constant but is related to the thickness of the last wet layer in each column by:
1014
1015\begin{equation}
1016C_D = \left ( {\kappa \over {\rm log}\left ( 0.5e_{3t}/rn\_bfrz0 \right ) } \right )^2
1017\end{equation}
1018
1019\noindent where $\kappa$ is the von-Karman constant and \np{rn\_bfrz0} is a roughness
1020length provided via the namelist.
1021
1022For stability, the drag coefficient is bounded such that it is kept greater or equal to
1023the base \np{rn\_bfri2} value and it is not allowed to exceed the value of an additional
1024namelist parameter: \np{rn\_bfri2\_max}, i.e.:
1025
1026\begin{equation}
1027rn\_bfri2 \leq C_D \leq rn\_bfri2\_max
1028\end{equation}
1029
1030\noindent Note also that a log-layer enhancement can also be applied to the top boundary
1031friction if under ice-shelf cavities are in use (\np{ln\_isfcav}=.true.). In this case, the
1032relevant namelist parameters are \np{rn\_tfrz0}, \np{rn\_tfri2}
1033and \np{rn\_tfri2\_max}.
1034
1035% -------------------------------------------------------------------------------------------------------------
1036% Bottom Friction stability
1037% -------------------------------------------------------------------------------------------------------------
1038\subsection{Bottom Friction stability considerations}
1039\label{ZDF_bfr_stability}
1040
1041Some care needs to exercised over the choice of parameters to ensure that the
1042implementation of bottom friction does not induce numerical instability. For
1043the purposes of stability analysis, an approximation to \eqref{Eq_zdfbfr_flux2}
1044is:
1045\begin{equation} \label{Eqn_bfrstab}
1046\begin{split}
1047 \Delta u &= -\frac{{{\cal F}_h}^u}{e_{3u}}\;2 \rdt \\
1048 &= -\frac{ru}{e_{3u}}\;2\rdt\\
1049\end{split}
1050\end{equation}
1051\noindent where linear bottom friction and a leapfrog timestep have been assumed.
1052To ensure that the bottom friction cannot reverse the direction of flow it is necessary to have:
1053\begin{equation}
1054 |\Delta u| < \;|u|
1055\end{equation}
1056\noindent which, using \eqref{Eqn_bfrstab}, gives:
1057\begin{equation}
1058r\frac{2\rdt}{e_{3u}} < 1 \qquad \Rightarrow \qquad r < \frac{e_{3u}}{2\rdt}\\
1059\end{equation}
1060This same inequality can also be derived in the non-linear bottom friction case
1061if a velocity of 1 m.s$^{-1}$ is assumed. Alternatively, this criterion can be
1062rearranged to suggest a minimum bottom box thickness to ensure stability:
1063\begin{equation}
1064e_{3u} > 2\;r\;\rdt
1065\end{equation}
1066\noindent which it may be necessary to impose if partial steps are being used.
1067For example, if $|u| = 1$ m.s$^{-1}$, $rdt = 1800$ s, $r = 10^{-3}$ then
1068$e_{3u}$ should be greater than 3.6 m. For most applications, with physically
1069sensible parameters these restrictions should not be of concern. But
1070caution may be necessary if attempts are made to locally enhance the bottom
1071friction parameters.
1072To ensure stability limits are imposed on the bottom friction coefficients both during
1073initialisation and at each time step. Checks at initialisation are made in \mdl{zdfbfr}
1074(assuming a 1 m.s$^{-1}$ velocity in the non-linear case).
1075The number of breaches of the stability criterion are reported as well as the minimum
1076and maximum values that have been set. The criterion is also checked at each time step,
1077using the actual velocity, in \mdl{dynbfr}. Values of the bottom friction coefficient are
1078reduced as necessary to ensure stability; these changes are not reported.
1079
1080Limits on the bottom friction coefficient are not imposed if the user has elected to
1081handle the bottom friction implicitly (see \S\ref{ZDF_bfr_imp}). The number of potential
1082breaches of the explicit stability criterion are still reported for information purposes.
1083
1084% -------------------------------------------------------------------------------------------------------------
1085% Implicit Bottom Friction
1086% -------------------------------------------------------------------------------------------------------------
1087\subsection{Implicit Bottom Friction (\np{ln\_bfrimp}$=$\textit{T})}
1088\label{ZDF_bfr_imp}
1089
1090An optional implicit form of bottom friction has been implemented to improve
1091model stability. We recommend this option for shelf sea and coastal ocean applications, especially
1092for split-explicit time splitting. This option can be invoked by setting \np{ln\_bfrimp}
1093to \textit{true} in the \textit{nambfr} namelist. This option requires \np{ln\_zdfexp} to be \textit{false}
1094in the \textit{namzdf} namelist.
1095
1096This implementation is realised in \mdl{dynzdf\_imp} and \mdl{dynspg\_ts}. In \mdl{dynzdf\_imp}, the
1097bottom boundary condition is implemented implicitly.
1098
1099\begin{equation} \label{Eq_dynzdf_bfr}
1100\left.{\left( {\frac{A^{vm} }{e_3 }\ \frac{\partial \textbf{U}_h}{\partial k}} \right)} \right|_{mbk}
1101 = \binom{c_{b}^{u}u^{n+1}_{mbk}}{c_{b}^{v}v^{n+1}_{mbk}}
1102\end{equation}
1103
1104where $mbk$ is the layer number of the bottom wet layer. superscript $n+1$ means the velocity used in the
1105friction formula is to be calculated, so, it is implicit.
1106
1107If split-explicit time splitting is used, care must be taken to avoid the double counting of
1108the bottom friction in the 2-D barotropic momentum equations. As NEMO only updates the barotropic
1109pressure gradient and Coriolis' forcing terms in the 2-D barotropic calculation, we need to remove
1110the bottom friction induced by these two terms which has been included in the 3-D momentum trend
1111and update it with the latest value. On the other hand, the bottom friction contributed by the
1112other terms (e.g. the advection term, viscosity term) has been included in the 3-D momentum equations
1113and should not be added in the 2-D barotropic mode.
1114
1115The implementation of the implicit bottom friction in \mdl{dynspg\_ts} is done in two steps as the
1116following:
1117
1118\begin{equation} \label{Eq_dynspg_ts_bfr1}
1119\frac{\textbf{U}_{med}-\textbf{U}^{m-1}}{2\Delta t}=-g\nabla\eta-f\textbf{k}\times\textbf{U}^{m}+c_{b}
1120\left(\textbf{U}_{med}-\textbf{U}^{m-1}\right)
1121\end{equation}
1122\begin{equation} \label{Eq_dynspg_ts_bfr2}
1123\frac{\textbf{U}^{m+1}-\textbf{U}_{med}}{2\Delta t}=\textbf{T}+
1124\left(g\nabla\eta^{'}+f\textbf{k}\times\textbf{U}^{'}\right)-
11252\Delta t_{bc}c_{b}\left(g\nabla\eta^{'}+f\textbf{k}\times\textbf{u}_{b}\right)
1126\end{equation}
1127
1128where $\textbf{T}$ is the vertical integrated 3-D momentum trend. We assume the leap-frog time-stepping
1129is used here. $\Delta t$ is the barotropic mode time step and $\Delta t_{bc}$ is the baroclinic mode time step.
1130 $c_{b}$ is the friction coefficient. $\eta$ is the sea surface level calculated in the barotropic loops
1131while $\eta^{'}$ is the sea surface level used in the 3-D baroclinic mode. $\textbf{u}_{b}$ is the bottom
1132layer horizontal velocity.
1133
1134
1135
1136
1137% -------------------------------------------------------------------------------------------------------------
1138% Bottom Friction with split-explicit time splitting
1139% -------------------------------------------------------------------------------------------------------------
1140\subsection{Bottom Friction with split-explicit time splitting (\np{ln\_bfrimp}$=$\textit{F})}
1141\label{ZDF_bfr_ts}
1142
1143When calculating the momentum trend due to bottom friction in \mdl{dynbfr}, the
1144bottom velocity at the before time step is used. This velocity includes both the
1145baroclinic and barotropic components which is appropriate when using either the
1146explicit or filtered surface pressure gradient algorithms (\key{dynspg\_exp} or
1147\key{dynspg\_flt}). Extra attention is required, however, when using
1148split-explicit time stepping (\key{dynspg\_ts}). In this case the free surface
1149equation is solved with a small time step \np{rn\_rdt}/\np{nn\_baro}, while the three
1150dimensional prognostic variables are solved with the longer time step
1151of \np{rn\_rdt} seconds. The trend in the barotropic momentum due to bottom
1152friction appropriate to this method is that given by the selected parameterisation
1153($i.e.$ linear or non-linear bottom friction) computed with the evolving velocities
1154at each barotropic timestep.
1155
1156In the case of non-linear bottom friction, we have elected to partially linearise
1157the problem by keeping the coefficients fixed throughout the barotropic
1158time-stepping to those computed in \mdl{zdfbfr} using the now timestep.
1159This decision allows an efficient use of the $c_b^{\vect{U}}$ coefficients to:
1160
1161\begin{enumerate}
1162\item On entry to \rou{dyn\_spg\_ts}, remove the contribution of the before
1163barotropic velocity to the bottom friction component of the vertically
1164integrated momentum trend. Note the same stability check that is carried out
1165on the bottom friction coefficient in \mdl{dynbfr} has to be applied here to
1166ensure that the trend removed matches that which was added in \mdl{dynbfr}.
1167\item At each barotropic step, compute the contribution of the current barotropic
1168velocity to the trend due to bottom friction. Add this contribution to the
1169vertically integrated momentum trend. This contribution is handled implicitly which
1170eliminates the need to impose a stability criteria on the values of the bottom friction
1171coefficient within the barotropic loop.
1172\end{enumerate}
1173
1174Note that the use of an implicit formulation within the barotropic loop
1175for the bottom friction trend means that any limiting of the bottom friction coefficient
1176in \mdl{dynbfr} does not adversely affect the solution when using split-explicit time
1177splitting. This is because the major contribution to bottom friction is likely to come from
1178the barotropic component which uses the unrestricted value of the coefficient. However, if the
1179limiting is thought to be having a major effect (a more likely prospect in coastal and shelf seas
1180applications) then the fully implicit form of the bottom friction should be used (see \S\ref{ZDF_bfr_imp} )
1181which can be selected by setting \np{ln\_bfrimp} $=$ \textit{true}.
1182
1183Otherwise, the implicit formulation takes the form:
1184\begin{equation} \label{Eq_zdfbfr_implicitts}
1185 \bar{U}^{t+ \rdt} = \; \left [ \bar{U}^{t-\rdt}\; + 2 \rdt\;RHS \right ] / \left [ 1 - 2 \rdt \;c_b^{u} / H_e \right ]
1186\end{equation}
1187where $\bar U$ is the barotropic velocity, $H_e$ is the full depth (including sea surface height),
1188$c_b^u$ is the bottom friction coefficient as calculated in \rou{zdf\_bfr} and $RHS$ represents
1189all the components to the vertically integrated momentum trend except for that due to bottom friction.
1190
1191
1192
1193
1194% ================================================================
1195% Tidal Mixing
1196% ================================================================
1197\section{Tidal Mixing (\key{zdftmx})}
1198\label{ZDF_tmx}
1199
1200%--------------------------------------------namzdf_tmx--------------------------------------------------
1201\namdisplay{namzdf_tmx}
1202%--------------------------------------------------------------------------------------------------------------
1203
1204
1205% -------------------------------------------------------------------------------------------------------------
1206% Bottom intensified tidal mixing
1207% -------------------------------------------------------------------------------------------------------------
1208\subsection{Bottom intensified tidal mixing}
1209\label{ZDF_tmx_bottom}
1210
1211Options are defined through the \ngn{namzdf\_tmx} namelist variables.
1212The parameterization of tidal mixing follows the general formulation for
1213the vertical eddy diffusivity proposed by \citet{St_Laurent_al_GRL02} and
1214first introduced in an OGCM by \citep{Simmons_al_OM04}.
1215In this formulation an additional vertical diffusivity resulting from internal tide breaking,
1216$A^{vT}_{tides}$ is expressed as a function of $E(x,y)$, the energy transfer from barotropic
1217tides to baroclinic tides :
1218\begin{equation} \label{Eq_Ktides}
1219A^{vT}_{tides} = q \,\Gamma \,\frac{ E(x,y) \, F(z) }{ \rho \, N^2 }
1220\end{equation}
1221where $\Gamma$ is the mixing efficiency, $N$ the Brunt-Vais\"{a}l\"{a} frequency
1222(see \S\ref{TRA_bn2}), $\rho$ the density, $q$ the tidal dissipation efficiency,
1223and $F(z)$ the vertical structure function.
1224
1225The mixing efficiency of turbulence is set by $\Gamma$ (\np{rn\_me} namelist parameter)
1226and is usually taken to be the canonical value of $\Gamma = 0.2$ (Osborn 1980).
1227The tidal dissipation efficiency is given by the parameter $q$ (\np{rn\_tfe} namelist parameter)
1228represents the part of the internal wave energy flux $E(x, y)$ that is dissipated locally,
1229with the remaining $1-q$ radiating away as low mode internal waves and
1230contributing to the background internal wave field. A value of $q=1/3$ is typically used
1231\citet{St_Laurent_al_GRL02}.
1232The vertical structure function $F(z)$ models the distribution of the turbulent mixing in the vertical.
1233It is implemented as a simple exponential decaying upward away from the bottom,
1234with a vertical scale of $h_o$ (\np{rn\_htmx} namelist parameter, with a typical value of $500\,m$) \citep{St_Laurent_Nash_DSR04},
1235\begin{equation} \label{Eq_Fz}
1236F(i,j,k) = \frac{ e^{ -\frac{H+z}{h_o} } }{ h_o \left( 1- e^{ -\frac{H}{h_o} } \right) }
1237\end{equation}
1238and is normalized so that vertical integral over the water column is unity.
1239
1240The associated vertical viscosity is calculated from the vertical
1241diffusivity assuming a Prandtl number of 1, $i.e.$ $A^{vm}_{tides}=A^{vT}_{tides}$.
1242In the limit of $N \rightarrow 0$ (or becoming negative), the vertical diffusivity
1243is capped at $300\,cm^2/s$ and impose a lower limit on $N^2$ of \np{rn\_n2min}
1244usually set to $10^{-8} s^{-2}$. These bounds are usually rarely encountered.
1245
1246The internal wave energy map, $E(x, y)$ in \eqref{Eq_Ktides}, is derived
1247from a barotropic model of the tides utilizing a parameterization of the
1248conversion of barotropic tidal energy into internal waves.
1249The essential goal of the parameterization is to represent the momentum
1250exchange between the barotropic tides and the unrepresented internal waves
1251induced by the tidal flow over rough topography in a stratified ocean.
1252In the current version of \NEMO, the map is built from the output of
1253the barotropic global ocean tide model MOG2D-G \citep{Carrere_Lyard_GRL03}.
1254This model provides the dissipation associated with internal wave energy for the M2 and K1
1255tides component (Fig.~\ref{Fig_ZDF_M2_K1_tmx}). The S2 dissipation is simply approximated
1256as being $1/4$ of the M2 one. The internal wave energy is thus : $E(x, y) = 1.25 E_{M2} + E_{K1}$.
1257Its global mean value is $1.1$ TW, in agreement with independent estimates
1258\citep{Egbert_Ray_Nat00, Egbert_Ray_JGR01}.
1259
1260%>>>>>>>>>>>>>>>>>>>>>>>>>>>>
1261\begin{figure}[!t] \begin{center}
1262\includegraphics[width=0.90\textwidth]{Fig_ZDF_M2_K1_tmx}
1263\caption{ \label{Fig_ZDF_M2_K1_tmx}
1264(a) M2 and (b) K1 internal wave drag energy from \citet{Carrere_Lyard_GRL03} ($W/m^2$). }
1265\end{center} \end{figure}
1266%>>>>>>>>>>>>>>>>>>>>>>>>>>>>
1267
1268% -------------------------------------------------------------------------------------------------------------
1269% Indonesian area specific treatment
1270% -------------------------------------------------------------------------------------------------------------
1271\subsection{Indonesian area specific treatment (\np{ln\_zdftmx\_itf})}
1272\label{ZDF_tmx_itf}
1273
1274When the Indonesian Through Flow (ITF) area is included in the model domain,
1275a specific treatment of tidal induced mixing in this area can be used.
1276It is activated through the namelist logical \np{ln\_tmx\_itf}, and the user must provide
1277an input NetCDF file, \ifile{mask\_itf}, which contains a mask array defining the ITF area
1278where the specific treatment is applied.
1279
1280When \np{ln\_tmx\_itf}=true, the two key parameters $q$ and $F(z)$ are adjusted following
1281the parameterisation developed by \citet{Koch-Larrouy_al_GRL07}:
1282
1283First, the Indonesian archipelago is a complex geographic region
1284with a series of large, deep, semi-enclosed basins connected via
1285numerous narrow straits. Once generated, internal tides remain
1286confined within this semi-enclosed area and hardly radiate away.
1287Therefore all the internal tides energy is consumed within this area.
1288So it is assumed that $q = 1$, $i.e.$ all the energy generated is available for mixing.
1289Note that for test purposed, the ITF tidal dissipation efficiency is a
1290namelist parameter (\np{rn\_tfe\_itf}). A value of $1$ or close to is
1291this recommended for this parameter.
1292
1293Second, the vertical structure function, $F(z)$, is no more associated
1294with a bottom intensification of the mixing, but with a maximum of
1295energy available within the thermocline. \citet{Koch-Larrouy_al_GRL07}
1296have suggested that the vertical distribution of the energy dissipation
1297proportional to $N^2$ below the core of the thermocline and to $N$ above.
1298The resulting $F(z)$ is:
1299\begin{equation} \label{Eq_Fz_itf}
1300F(i,j,k) \sim \left\{ \begin{aligned}
1301\frac{q\,\Gamma E(i,j) } {\rho N \, \int N dz} \qquad \text{when $\partial_z N < 0$} \\
1302\frac{q\,\Gamma E(i,j) } {\rho \, \int N^2 dz} \qquad \text{when $\partial_z N > 0$}
1303 \end{aligned} \right.
1304\end{equation}
1305
1306Averaged over the ITF area, the resulting tidal mixing coefficient is $1.5\,cm^2/s$,
1307which agrees with the independent estimates inferred from observations.
1308Introduced in a regional OGCM, the parameterization improves the water mass
1309characteristics in the different Indonesian seas, suggesting that the horizontal
1310and vertical distributions of the mixing are adequately prescribed
1311\citep{Koch-Larrouy_al_GRL07, Koch-Larrouy_al_OD08a, Koch-Larrouy_al_OD08b}.
1312Note also that such a parameterisation has a significant impact on the behaviour
1313of global coupled GCMs \citep{Koch-Larrouy_al_CD10}.
1314
1315
1316% ================================================================
1317% Internal wave-driven mixing
1318% ================================================================
1319\section{Internal wave-driven mixing (\key{zdftmx\_new})}
1320\label{ZDF_tmx_new}
1321
1322%--------------------------------------------namzdf_tmx_new------------------------------------------
1323\namdisplay{namzdf_tmx_new}
1324%--------------------------------------------------------------------------------------------------------------
1325
1326The parameterization of mixing induced by breaking internal waves is a generalization
1327of the approach originally proposed by \citet{St_Laurent_al_GRL02}.
1328A three-dimensional field of internal wave energy dissipation $\epsilon(x,y,z)$ is first constructed,
1329and the resulting diffusivity is obtained as
1330\begin{equation} \label{Eq_Kwave}
1331A^{vT}_{wave} = R_f \,\frac{ \epsilon }{ \rho \, N^2 }
1332\end{equation}
1333where $R_f$ is the mixing efficiency and $\epsilon$ is a specified three dimensional distribution
1334of the energy available for mixing. If the \np{ln\_mevar} namelist parameter is set to false,
1335the mixing efficiency is taken as constant and equal to 1/6 \citep{Osborn_JPO80}.
1336In the opposite (recommended) case, $R_f$ is instead a function of the turbulence intensity parameter
1337$Re_b = \frac{ \epsilon}{\nu \, N^2}$, with $\nu$ the molecular viscosity of seawater,
1338following the model of \cite{Bouffard_Boegman_DAO2013}
1339and the implementation of \cite{de_lavergne_JPO2016_efficiency}.
1340Note that $A^{vT}_{wave}$ is bounded by $10^{-2}\,m^2/s$, a limit that is often reached when the mixing efficiency is constant.
1341
1342In addition to the mixing efficiency, the ratio of salt to heat diffusivities can chosen to vary
1343as a function of $Re_b$ by setting the \np{ln\_tsdiff} parameter to true, a recommended choice).
1344This parameterization of differential mixing, due to \cite{Jackson_Rehmann_JPO2014},
1345is implemented as in \cite{de_lavergne_JPO2016_efficiency}.
1346
1347The three-dimensional distribution of the energy available for mixing, $\epsilon(i,j,k)$, is constructed
1348from three static maps of column-integrated internal wave energy dissipation, $E_{cri}(i,j)$,
1349$E_{pyc}(i,j)$, and $E_{bot}(i,j)$, combined to three corresponding vertical structures
1350(de Lavergne et al., in prep):
1351\begin{align*}
1352F_{cri}(i,j,k) &\propto e^{-h_{ab} / h_{cri} }\\
1353F_{pyc}(i,j,k) &\propto N^{n\_p}\\
1354F_{bot}(i,j,k) &\propto N^2 \, e^{- h_{wkb} / h_{bot} }
1355\end{align*}
1356In the above formula, $h_{ab}$ denotes the height above bottom,
1357$h_{wkb}$ denotes the WKB-stretched height above bottom, defined by
1358\begin{equation*}
1359h_{wkb} = H \, \frac{ \int_{-H}^{z} N \, dz' } { \int_{-H}^{\eta} N \, dz' } \; ,
1360\end{equation*}
1361The $n_p$ parameter (given by \np{nn\_zpyc} in \ngn{namzdf\_tmx\_new} namelist) controls the stratification-dependence of the pycnocline-intensified dissipation.
1362It can take values of 1 (recommended) or 2.
1363Finally, the vertical structures $F_{cri}$ and $F_{bot}$ require the specification of
1364the decay scales $h_{cri}(i,j)$ and $h_{bot}(i,j)$, which are defined by two additional input maps.
1365$h_{cri}$ is related to the large-scale topography of the ocean (etopo2)
1366and $h_{bot}$ is a function of the energy flux $E_{bot}$, the characteristic horizontal scale of
1367the abyssal hill topography \citep{Goff_JGR2010} and the latitude.
1368
1369% ================================================================
1370
1371
1372
1373\end{document}
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http://gamedev.stackexchange.com/questions/28434/multiply-matrix-by-vector-what-to-do-if-need-to-change-y-up-to-z-up?answertab=votes
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# multiply matrix by vector: what to do if need to change y-up to z-up?
I multiply a matrix with a position vector to get a new position, however, now I need to change my matrix from y-up coordinate space to z-up. Changing only the matrix won't work: how to apply the change to the vector (change position vector from y-up to z-up)?
-
Maybe rotating along X and if needed, setting Y as -Y? – Gustavo Maciel May 2 '12 at 9:19
In the title you wrote Z to Y, then in the question you wrote Y to Z. So which is it? – David Gouveia May 2 '12 at 10:17
Fixed. isn't the solution similar though? – ippk May 2 '12 at 10:23
@ippk It's similar enough. I've tried explaining the general case. Should be easy to adapt to your specific needs. – David Gouveia May 2 '12 at 10:25
You need to be more clear about the coordinate systems you're converting from and to (for instance, when Y is up, in which direction does Z point, etc). But either way it's simple to derive your matrix. I'll just do an example using this assumption:
`````` Y Z / Y
|__ X => |/__ X
/
Z
(1. Y Up) (2. Z Up)
``````
Think what happens to each axis on the left side, when it's converted to the right side:
• X in (2) is the same as X in (1)
• Y in (2) is the same as -Z in (1)
• Z in (2) is the same as Y in (1)
One idea is to encode the changes above in a simple matrix, such as the one below. It should be easy to understand the correlation between the matrix and the description above, but I've also added the multiplication so you can see the way in which it modifies the result.
``````[ 1 0 0 ] [ x ] [ x ]
[ 0 0 -1 ] * [ y ] = [ -z ]
[ 0 1 0 ] [ z ] [ y ]
``````
This is actually just a 90 degrees rotation matrix. You could concatenate this matrix at the end of your previous transformation, so that the conversion is done automatically as part of the matrix multiplication.
Or, alternatively you can just do it manually at the end like:
``````float y = position.Y;
position.Y = -position.Z;
position.Z = y;
``````
-
A bit confused about in which way this is actually converting, but it's a matter of swapping the -1 and the 1 on the matrix if it's the other way around. – David Gouveia May 2 '12 at 10:42
Or if wished, dont inverse Z at all, depending on the coordinate system – Gustavo Maciel May 2 '12 at 11:03
@Gustavo-Gtoknu Exactly. Just draw both of your coordinate systems, and then see what changes from one to the other. – David Gouveia May 2 '12 at 11:12
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# 832972004929
## 832,972,004,929 is an odd composite number. It is composed of a single prime number multiplied by itself five times.
What does the number 832972004929 look like?
This visualization shows the relationship between its 1 prime factors (large circles) and 7 divisors.
832972004929 is an odd composite number. It is composed of one distinct prime number multiplied by itself five times. It has a total of seven divisors.
## Prime factorization of 832972004929:
### 976
(97 × 97 × 97 × 97 × 97 × 97)
See below for interesting mathematical facts about the number 832972004929 from the Numbermatics database.
### Names of 832972004929
• Cardinal: 832972004929 can be written as Eight hundred thirty-two billion, nine hundred seventy-two million, four thousand, nine hundred twenty-nine.
### Scientific notation
• Scientific notation: 8.32972004929 × 1011
### Factors of 832972004929
• Number of distinct prime factors ω(n): 1
• Total number of prime factors Ω(n): 6
• Sum of prime factors: 97
### Divisors of 832972004929
• Number of divisors d(n): 7
• Complete list of divisors:
• Sum of all divisors σ(n): 841648796647
• Sum of proper divisors (its aliquot sum) s(n): 8676791718
• 832972004929 is a deficient number, because the sum of its proper divisors (8676791718) is less than itself. Its deficiency is 824295213211
### Bases of 832972004929
• Binary: 11000001111100010000000000011110010000012
• Base-36: AMNU0WG1
### Squares and roots of 832972004929
• 832972004929 squared (8329720049292) is 693842360995438000295041
• 832972004929 cubed (8329720049293) is 577951262543040979328274795306257089
• 832972004929 is a perfect square number. Its square root is 912673
• 832972004929 is a perfect cube number. Its cube root is 9409
• Being both a perfect square and a perfect cube, 832972004929 is also a perfect sixth number.
### Scales and comparisons
How big is 832972004929?
• 832,972,004,929 seconds is equal to 26,485 years, 48 weeks, 3 days, 16 hours, 28 minutes, 49 seconds.
• To count from 1 to 832,972,004,929 would take you about fifty-two thousand, nine hundred seventy-one years!
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 832972004929 cubic inches would be around 784.1 feet tall.
### Recreational maths with 832972004929
• 832972004929 backwards is 929400279238
• The number of decimal digits it has is: 12
• The sum of 832972004929's digits is 55
• More coming soon!
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Joined: Sat Feb 02, 2013 9:40 am
Posts: 1441
I got to thinking about macro instruction fusion for purposes of performing a dot-product operation. Then I started wondering why have the fused instructions? Why not just have instructions that output doubled precision results (fmul.ssd) ? Or take doubled precision inputs (fadd.dds) ? An fma retains all the product bits out to doubled precision for rounding which are included in the add operation. An fmul instruction with double precision output would be able to retain the same bits. If this was fed into an fadd instruction that accepted double precision inputs wouldn’t the results be the same? For instance, fmul.s produces 2x24 bit significand or 48 product bits. This result should fit within the 52 significand bits of a double precision number without needing any rounding.
There are only a couple of things that an fma instruction provides: 1) slightly increased code density (provided 3 registers read ports are available). 2) more accurate rounding results.
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Robert Finch http://www.finitron.ca
Mon Jun 01, 2020 2:44 am
Joined: Wed Jan 09, 2013 6:54 pm
Posts: 1626
Interesting. Modern descriptions of FMA highlight the improved accuracy realised by not double-rounding. But the introductory material in this thesis ("Floating-Point Fused Multiply-Add Architectures" by Eric Charles Quinnell) has IBM claiming:
Quote:
...benefits of combining the floating-point adder and floating-point multiplier into a single functional unit. First, the latency for a multiply-add fused mathematical operation is reduced significantly by having an addition combined with a multiplication in hardware. Second, the precision of the final result is increased, since the operands only go through a single rounding stage. Third, there is a decrease in the number of required input/output ports to the register file and their controlling sub-units. Finally, a reduced area of both the floating-point adder and floating-point multiplier may be realized since the adder is only wired to the output connections of the multiplier.
There seems to be some historical to-and-fro depending on whether FMA is an extra functional unit or one which replaces M and A.
Quote:
Even though the fused multiply-add architecture has troublesome latencies, high power consumption, and a performance degradation with single-instruction execution, it may be fully expected that more and more x87 designs will find floating-point fused multiply-add units in their silicon.
There are some great diagrams in the early parts of that thesis.
Why not replace FMA with doubled results? Conventionally, I would expect the power, the area, and the time to count against: for the desired extra one or two bits of accuracy (surely FMA doesn't offer more than that) doubling precision is going to incur a major cost. However, FPGAs may lead to an unconventional answer: the transistors may already be there and unused; the timing may depend on other parts of the design, or be dominated by routing costs; the power budget may be unimportant.
Mon Jun 01, 2020 6:46 am
Joined: Wed Jan 09, 2013 6:54 pm
Posts: 1626
BigEd wrote:
... extra one or two bits of accuracy (surely FMA doesn't offer more than that) ...
A foolish thing for me to say! See for example
Accurately computing a 2×2 determinant
where we see some code, and the results:
Naive: -7.03944087021569e-07
Kahan: -7.03944088015194e-07
where we seem to get some 20 bits of improvement
Mon Jun 01, 2020 11:57 am
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World's only instant tutoring platform
Filo is a preferred QESP
Question
# The height of a right circular cylinder is 10.5 m. Three times the sum of the areas of its two circular faces is twice the area of the curved surface. Find the volume of the cylinder.
## Text solutionVerified
Given data is as follows:
h = 10.5 m
We have to find the volume of the cylinder.
From the given data, we have
But we know from the given data, that h = 10.5 m
Therefore,
Since we know r and h , we can easily find the volume of the cylinder.
Volume =
=
Volume = 1617 m3
Therefore, the volume of the cylinder is 1617 m3.
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Chapter 1. Basics 2. Pure Substances 3. First Law 4. Energy Analysis 5. Second Law 6. Entropy 7. Exergy Analysis 8. Gas Power Cyc 9. Brayton Cycle 10. Rankine Cycle Appendix Basic Math Units Thermo Tables Search eBooks Dynamics Statics Mechanics Fluids Thermodynamics Math Author(s): Meirong Huang Kurt Gramoll ©Kurt Gramoll
THERMODYNAMICS - THEORY
Many engineering devices operate essentially under the same conditions for long periods of time. Therefore, these devices can be treated as steady-flow devices.
Recall, the energy balance for a control volume is:
Heat Exchangers
Recuperator
Click to View Movie (36 kB)
Devices that transfer energy between fluids at different temperatures by heat transfer modes such as conduction, convection, and radiation are called heat exchangers.
The simplest form of a heat exchanger is called a recuperator, in which a gas or liquid is separated from another gas or liquid by a wall through which energy passes by conduction.
The conservation of mass principle for a heat exchanger can be expressed as follows:
Under steady operation, the mass flow rate of each fluid stream flowing through a heat exchanger remains constant.
Hence,
There are several common assumptions that are made in the energy analysis of heat exchangers
Heat Transfer of Heat Exchangers
Click to View Movie (36 kB)
• If one takes the entire heat exchanger as a control volume, = 0. If only one of the fluids is selected as a control volume, heat will cross this boundary as it flows from one fluid to the other. Hence,
• The only work interaction at the boundary of a control volume enclosing a heat exchanger is flow work and no shaft or electric work is involved.
• The change in kinetic energy is insignificant.
• The potential energy change is negligible.
Pipe and Duct Flow
Pipe Are Used to Transfer Fluid from one Device to another
Pipes and ducts are used to transfer fluid from one device to another. Flow through a pipe or duct can be treated as a steady-flow process since the start-up and shut-down periods, which are transient process; are excluded.
When flow through pipes or ducts are analyzed, the following points should be considered:
• If the pipes or ducts are insulated, heat transfer from the pipes or ducts to the environments is negligible. Under normal operating conditions, the pipes or ducts are not insulated, hence the heat gained or lost is large. Sometimes heat transfer is the main purpose of the flow, for example, in heat exchangers. In this case, heat transfer needs to be accounted for.
• If the control volume involves pumps or fans or other work devices, the work interaction terms should be considered. Otherwise, the work term is zero.
• The change in kinetic energy is insignificant, particularly when the pipe or duct has a constant diameter.
• The change in potential energy is large when the fluid undergoes a considerable elevation change.
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# What level is used in Metamagic magic item cost calculations?
How are the cost calculated for an item which has a metamagic spell in it?
Example:
• Jump is a level 1 wizard spell and can be cast by a level 1 wizard.
• The metamagic feat quicken says "A quickened spell uses up a spell slot four levels higher than the spell’s actual level."
• A 9th level wizard can cast a quickened level 1 Jump.
• The cost formula for a item is normally: spell grade × caster level × 2000.
So the caster level has to be at least 9. Is the spell grade now 1 (18,000gp) or 5 (90,000gp)?
# The spell counts as the level of the slot necessary to cast it.
This has been answered (indirectly) on this FAQ item:
Metamagic: At what spell level does the spell count for concentration DCs, magus spell recall, or a pearl of power?
The spell counts as the level of the spell slot necessary to cast it.
For example, an empowered burning hands uses a 3rd-level spell slot, counts as a 3rd-level spell for making concentration checks, counts as a 3rd-level spell for a magus's spell recall or a pearl of power.
In general, use the (normal, lower) spell level or the (higher) spell slot level, whichever is more of a disadvantage for the caster. The advantages of the metamagic feat are spelled out in the Benefits section of the feat, and the increased spell slot level is a disadvantage.
Heighten Spell is really the only metamagic feat that makes using a higher-level spell slot an advantage instead of a disadvantage.
This means that a 1st level spell that is quickened to 5th level is now a 5th level spell for calculating magic item creation costs. In your example, the spell level is 5, which means the final price of your item will be 90,000gp.
# Treat a spell in a magic item that's modified by a metamagic feat as if the spell's spell level were actually higher
Magic Item Creation says, in part, that
While item creation costs are handled in detail below, note that normally the two primary factors are the caster level of the creator and the level of the spell or spells put into the item. A creator can create an item at a lower caster level than her own, but never lower than the minimum level needed to cast the needed spell. Using metamagic feats, a caster can place spells in items at a higher level than normal.
(Emphasis mine.) Certainly this rule could have used at least one example of its application right there, but we'll have to content ourselves with published examples that are close, like the major crown of blasting, a wondrous item that uses the 3rd-level spell searing light modified by the feat Maximize Spell (so occupying a 6th-level spell slot and mandating a minimum caster level of 11). While the major crown's text says it's made at caster level 17 (just like the D&D 3.5's similar major circlet), the major crown's math says—when the major crown's reverse-engineered—that the major crown's actually priced as if its caster level were 11, and that caster level 11 also conveniently jibes with the damage dealt by the crown's effect.
(Note that Pathfinder's antecedent Dungeons & Dragons 3.5—from which this rule comes—makes how to apply this rule clearer with, for example, its table of random wands that's also on Dungeon Master's Guide 246.)
So what Magic Item Creation means is, for example, when estimating the gp value of a typical continuous or use activated wondrous item when using Table: Estimating Magic Item Gold Piece Values, the game wants the estimated base price for the magic item to be the spell level modified by the metamagic feat × minimum caster level needed to cast that modified spell × 2,000 gp. Keep in mind, though, that few published magic items employ such modified spells, so getting the GM to approve such custom items may be a challenge.
Note: The GM—not the player or the PC—typically determines a magic item's gp value by first comparing it to existing items then the table. You might also be interested in this question.
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PureMath.pdf
# 23 if s 1 then z 1 2 π sin θ s dθ z 1 x s dx 1 x 2
• 587
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23. If s > - 1 then Z 1 2 π 0 (sin θ ) s = Z 1 0 x s dx 1 - x 2 = 1 2 Z 1 0 x 1 2 ( s - 1) dx 1 - x = 1 2 Z 1 0 (1 - x ) 1 2 ( s - 1) dx x . 24. Establish the formulae Z 1 0 f ( x ) dx 1 - x 2 = Z 1 2 π 0 f (sin θ ) dθ, Z b a f ( x ) dx p ( x - a )( b - x ) = 2 Z 1 2 π 0 f ( a cos 2 θ + b sin 2 θ ) dθ, Z a - a f r a - x a + x dx = 4 a Z 1 2 π 0 f (tan θ ) cos θ sin θ dθ. 25. Prove that Z 1 0 dx (1 + x )(2 + x ) p x (1 - x ) = π 1 2 - 1 6 . [Put x = sin 2 θ and use Ex. lxiii . 8.] ( Math. Trip. 1912.) 182. Some care has occasionally to be exercised in applying the rule for transformation by substitution. The following example affords a good illustra- tion of this. Let J = Z 7 1 ( x 2 - 6 x + 13) dx. We find by direct integration that J = 48. Now let us apply the substitution y = x 2 - 6 x + 13 , which gives x = 3 ± y - 4. Since y = 8 when x = 1 and y = 20 when x = 7, we appear to be led to the result J = Z 20 8 y dx dy dy = ± 1 2 Z 20 8 y dy y - 4 .
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[VIII : 183] THE CONVERGENCE OF INFINITE SERIES, ETC. 416 The indefinite integral is 1 3 ( y - 4) 3 / 2 + 4( y - 4) 1 / 2 , and so we obtain the value ± 80 3 , which is certainly wrong whichever sign we choose. The explanation is to be found in a closer consideration of the relation be- tween x and y . The function x 2 - 6 x +13 has a minimum for x = 3, when y = 4. As x increases from 1 to 3, y decreases from 8 to 4, and dx/dy is negative, so that dx dy = - 1 2 y - 4 . As x increases from 3 to 7, y increases from 4 to 20, and the other sign must be chosen. Thus J = Z 7 1 y dx = Z 4 8 - y 2 y - 4 dy + Z 20 4 y 2 y - 4 dy, a formula which will be found to lead to the correct result. Similarly, if we transform the integral Z π 0 dx = π by the substitution x = arc sin y , we must observe that dx/dy = 1 / p 1 - y 2 or dx/dy = - 1 / p 1 - y 2 according as 0 5 x < 1 2 π or 1 2 π < x 5 π . Example. Verify the results of transforming the integrals Z 1 0 (4 x 2 - x + 1 16 ) dx, Z π 0 cos 2 x dx by the substitutions 4 x 2 - x + 1 16 = y , x = arc sin y respectively. 183. Series of positive and negative terms. Our definitions of the sum of an infinite series, and the value of an infinite integral, whether of the first or the second kind, apply to series of terms or integrals of functions whose values may be either positive or negative. But the special tests for convergence or divergence which we have established in this chapter, and the examples by which we have illustrated them, have had reference almost entirely to the case in which all these values are positive. Of course the
[VIII : 184] THE CONVERGENCE OF INFINITE SERIES, ETC. 417 case in which they are all negative is not essentially different, as it can be reduced to the former by changing u n into - u n or φ ( x ) into - φ ( x ). In the case of a series it has always been explicitly or tacitly assumed that any conditions imposed upon u n may be violated for a finite number of terms: all that is necessary is that such a condition ( e.g. that all the terms are positive) should be satisfied from some definite term onwards . Similarly in the case of an infinite integral the conditions have been supposed to be satisfied for all values of x greater than some definite value , or for all values of x within some definite interval [ a, a + δ
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Sie sind auf Seite 1von 33
# Pune Vidyarthi Grihas
## COLLEGE OF ENGINEERING, NASHIK 3.
ASM
By
Prof. Anand N. Gharu
(Assistant Professor)
PVGCOE Computer Dept.
## 30th June 2017
.
CONTENTS :-
1. Introduction to ASM
2. The Chart Notation
3. The State Box
4. Decision Box
5. Condition Box
6. Examples
7. MUX Controller Method
8. Examples
INTRODUCTION OF ASM
Used to graphically describe the operations of an FSM more concisely
Resembles conventional flowcharts differs in
interpretation.
Conventional flowchart sequential way of
representing procedural steps and decision paths
for algorithm
-No time relations incorporated
ASM chart representation of sequence of
events together with timing relations between
states of sequential controller and events
occurring while moving between steps
Three basic elements: state box, decision
box and conditional box
-State and decision boxes used in conventional
flowcharts
-Conditional box characteristic to ASM
State box
-Used to indicate states in control sequence
Register operations and output signals used to
control generation of next state written
State box
Represents one state in the ASM.
May have an optional state output list.
Single entry.
Single exit to state or decision boxes.
State Box
State name T3
Binary code of T3 011
Register operation R <-
0
START name of
outputs signal generated
in this stage
## ACOE161 - Digital Logic for
Computers - Frederick University
Decision box
## Provides for next alternatives and
conditional outputs.
Conditional output based on logic
value of Boolean expression involving
external input variables and status
information.
Single entry.
Dual exit, denoting if Boolean
expression is true or false.
Exits to decision, state or conditional
boxes.
## ACOE161 - Digital Logic for
Computers - Frederick University
Decision Box
Input condition subject to
test inside diamond shape
box
Two or more outputs
represent exit paths
dependant on value of
condition in decision box
Two paths for binary based
conditions
## ACOE161 - Digital Logic for
Computers - Frederick University
Conditional output box
## Provides a listing of output variables
that are to have a value logic-1, i.e.,
those output variables being
asserted.
Single entry from decision box.
Single exit to decision or state bo x.
## ACOE161 - Digital Logic for
Computers - Frederick University
In state T1
Conditional Box
Output signal START
generated
Status of input E
checked
If E = 1, R <- 0,
otherwise remains
unchanged
Conditional
operation executed
depending on result
of coming from
decision box
ACOE161 - Digital Logic for
Computers - Frederick University
ASM Block
## Consists of the interconnection of a single state
box along with one or more decision and/or
conditional boxes.
It has one entry path which leads directly to its
state box, and one or more exit paths.
Each exit path must lead directly to a state,
including the state box in itself.
A path through an ASM block from its state box to
an exit path is called a link path.
## ACOE161 - Digital Logic for
Computers - Frederick University
Timing Considerations
All sequential elements in datapath and control
path controlled by master-clock generator.
Does not necessarily imply single clock in design.
Multiple clocks can be obtained through division of clock
signals from master-clock generator.
Not only internal signals, but also inputs
synchronized with clock.
Normally, inputs supplied by other devices working
with the same master clock.
Some inputs can arrive asynchronously
Difficult to handle by synchronous designs, require
asynchronous glue-logic.
## ACOE161 - Digital Logic for
Computers - Frederick University
In conventional flowchart, evaluation
of each chart element takes one clock
Block
cycle
Step 1: Reg A incremented
Step 2: Condition E evaluated
Step 3: Based on evaluation results,
state
T2, T3 or T4 entered
In ASM the entire block considered
as one unit
All operations within block occurring
during single edge transition
The next state evaluated during the
same clock
System enters next state T2, T3 or T4
ACOE161 - Digital Logic for
during transition
Computers of University
- Frederick next clock
ASM Block
## An ASM block describes the operation of the system during
the state time in which it is in the state associated with the
block.
The outputs listed in the state box are asserted.
The conditions indicated in the decision boxes are evaluated
simultaneously to determine which link path is to be followed.
If a conditional box is found in the selected path then the
outputs found in its output list are asserted.
Boolean expression may be written for each link path. The
selected link paths are those that evaluate to logic-1.
T0
Example 2 0
x
1
T1
0
x
## diagram from the
ASM diagram 0
F
1
T4
T3
0 1
E
T6
T7 T5
DESIGN MUL CONTROLLER METHOD
Thank You
8/26/2017 33
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# Reduce with Python and Clojure
I was just playing around writing an annuity calculator function. Here is my first version:
def calculate_annuity(years, interest=0, addition_per_year=0, starting_amount=0):
result = []
running_total = starting_amount
for year in range(years+1):
result.append(running_total)
running_total = (running_total + addition_per_year) * (1 + interest)
return result
Here is a 2nd version, written using reduce instead of a loop:
def calculate_annuity2(years, interest=0, addition_per_year=0, starting_amount=0):
return reduce(lambda result,addition: result + [(result[-1] + addition) * (1 + interest)],
And here is a version in Clojure (basically a direct translation of the Python one:
(defn annuity [years interest addition_per_year init]
(reduce
#(conj %1 (* (+ (last %1) %2) (+ 1 interest)))
[starting_amount]
; Find investment returns by year for a $1000 investment at 7% interest ; for 10 years. (println (annuity 10 0.07 0 1000)) ; Find annuity value if contributing$1000 each month and starting with $7777 ; with 10% interest for 10 years. (println (annuity 10 0.10 1000 7777)) ; Find annuity value if contributing$1000 each month and starting with $777 ; with 10% interest for 10 years. (println (annuity 10 0.07 0 1000)) ; Test perpetuity annuity that pays out$6000 per year, given a \$100000
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• C#获取一个数组中最大值、最小值、平均值 1.给出一个数组 int[] array = new int[] { 1,2,4,3,0,-1,34,545,2,34}; 2.数组Array自带方法 本身是直接可以调用Min(),Max(),Average()方法来求出 最小值、最大值...
C#获取一个数组中的最大值、最小值、平均值
1.给出一个数组
int[] array = new int[] { 1,2,4,3,0,-1,34,545,2,34};
2.数组Array自带方法
本身是直接可以调用Min(),Max(),Average()方法来求出 最小值、最大值、平均值
1 Console.WriteLine("--------------Array自身方法-----------------");
2 Console.WriteLine("Min:{0}",array.Min());
3 Console.WriteLine("Max:{0}", array.Max());
4 Console.WriteLine("Average:{0}", array.Average());
输出结果：
1 --------------Array自身方法-----------------
2 Min:-1
3 Max:545
4 Average:62.4
3.编码实现
最小值
/// <summary>
/// 最小值
/// </summary>
/// <param name="array"></param>
/// <returns></returns>
public static int Min(int[] array)
{
if (array == null) throw new Exception("数组空异常");
int value = 0;
bool hasValue = false;
foreach (int x in array)
{
if (hasValue)
{
if (x < value) value = x;
}
else
{
value = x;
hasValue = true;
}
}
if (hasValue) return value;
throw new Exception("没找到");
}
最大值
/// <summary>
/// 最大值
/// </summary>
/// <param name="array"></param>
/// <returns></returns>
public static int Max(int[] array)
{
if (array == null) throw new Exception("数组空异常");
int value = 0;
bool hasValue = false;
foreach (int x in array)
{
if (hasValue)
{
if (x > value)
value = x;
}
else
{
value = x;
hasValue = true;
}
}
if (hasValue) return value;
throw new Exception("没找到");
}
平均值
/// <summary>
/// 平均值
/// </summary>
/// <param name="source"></param>
/// <returns></returns>
public static double? Average(int[] array)
{
if (array == null) throw new Exception("数组空异常");
long sum = 0;
long count = 0;
checked
{
foreach (int? v in array)
{
if (v != null)
{
sum += v.GetValueOrDefault();
count++;
}
}
}
if (count > 0) return (double)sum / count;
return null;
}
4.测试输出
测试代码
static void Main(string[] args)
{
int[] array = new int[] { 1,2,4,3,0,-1,34,545,2,34};
Console.WriteLine("--------------Array自身方法-----------------");
Console.WriteLine("Min:{0}",array.Min());
Console.WriteLine("Max:{0}", array.Max());
Console.WriteLine("Average:{0}", array.Average());
Console.WriteLine("---------------内部实现方法------------------");
int min = Program.Min(array);
int max = Program.Max(array);
double? average = Program.Average(array);
Console.WriteLine("Min:" + min);
Console.WriteLine("Max:" + max);
Console.WriteLine("Average:" + average);
Console.Read();
}
输出结果
以上代码也是从.NET Framework中摘出来的，实际上 Array的自带求最大值、最小值、平均值的算法就是这样做的，在.NET Framework源码中可以看到
5.工程源码下载
源代码下载
展开全文
• 功能：从一个字符数组中取出相应的整数、实数 作者：班草 */ #include #include #include void read_num( char *str , int *read_int , double *read_double); int main() { char *str = ...
/*
功能：从一个字符数组中取出相应的整数、实数
作者：班草
*/
#include <stdio.h>
#include <malloc.h>
#include <math.h>
void read_num( char *str , int *read_int , double *read_double);
int main()
{
char *str = NULL;
int *read_int = NULL;
double *read_double = NULL;
str = (char *)malloc(sizeof(char)*1024);
//分配堆空间
read_int = (int *)malloc(sizeof(int)*1024);
read_double = (double *)malloc(sizeof(double)*1024);
printf("input string:");
//输入字符串
scanf("%s",str);
read_num( str , read_int , read_double );
//调用函数
int i = 0;
printf("int :");
//输出整型
for( i = 1 ; i <= read_int[0] ; i ++ )
{
printf("\t%d",read_int[i]);
}
printf("\n");
printf("double:");
//输出实型
for( i = 1 ; i <= (int)read_double[0] ; i++ )
{
printf("\t%g",read_double[i]);
}
printf("\n");
free(str);
//释放堆空间
str = NULL;
free(read_int);
read_int = NULL;
free(read_double);
read_double = NULL;
return 0;
}
void read_num( char *str , int *read_int , double *read_double )
{
int flag = 1; //正负符号变量
int read = 0; //读标志，为1时表示读取到值
int i_int = 1; //整形数组的下标
int i_double = 1; //实形数组的下标
read_int[0] = 0; //数组第一个数，表示读取到的值的数量
read_double[0] = 0;
int x = 0;
int u = 0;
int zi = 0; //这三个变量用于将字符串数字转换为整形
double fzi = 0.0; //存放小数点后的数值
while( *str != '\0' ) //遍历字符串
{
if( *str == '-' ) //判断是否为负号，
{
flag = -1;
str ++;
}
for( x = 0 ; *str >= '0' && *str <= '9' ; x++ , str ++ ) //读取字符串中一段连续的数字，转换为整型
{
//如125.6，读出为125
u = *str - '0';
if( x == 0 )
{
zi = u;
}
else
{
zi = zi * 10 + u;
}
read = 1;
}
if( *str == '.' && read == 1 ) //如果遇到小数点且小数点前为数字
{
str ++;
for( x = 0 ; *str >= '0' && *str <= '9' ; x++ , str ++ )
//读小数点后的值
{
u = *str - '0';
if( x == 0 )
{
fzi =(double)u / pow(10,x+1);
}
else
{
fzi += (double)u / pow(10,x+1);
}
}
if( fzi != 0.0 )
//如果小数点后读到了值，将其值转换为对应的实数然后加上前面的整数，存放到实型数组
{
read_double[0]++;
read_double[i_double++] = (zi + fzi) * flag;
read = 0;
}
}
if( read == 1) //判断是否读到值
{
read_int[0]++;
read_int[i_int++] = zi * flag;
read = 0;
}
flag = 1;
}
}
展开全文
• 1、 给定一个数组，可以从数组中取出下标不连续的任意个数，求可以取出的数的和的最大值， 例如：给出数组A[]={1,2,2,5,3,4,3}可以取出最大和为2+5+4=11。 现再给定数组{3,9,7,5,1,3,1,2,7}，能取出的数的和的...
1、 给定一个数组，可以从数组中取出下标不连续的任意个数，求可以取出的数的和的最大值，
例如：给出数组A[]={1,2,2,5,3,4,3}可以取出的最大和为2+5+4=11。现再给定数组{3,9,7,5,1,3,1,2,7}，能取出的数的和的最大值是24。
2、分析 3、代码
private static int dp1(int arr[],int len) {
int ret=0;
if (len==0) {
return 0;
}else if (len==1) {
return arr[0];
}else if (len==2) {
return Math.max(arr[0],arr[1]);
}else {
for(int i=2;i<len;i++) {
int r1=dp1(arr,len-2)+arr[i];
int r2=dp1(arr,len-1);
ret=Math.max(r1,r2);
}
return ret;
}
}
//优化
private static int dp1(int arr[]) {
int n = nums.length;
if (n == 0) {
return 0;
}
if (n == 1) {
return nums[0];
}
int[] dp = new int[n];
dp[0] = nums[0];
dp[1] = Math.max(nums[0], nums[1]);
for (int i = 2; i < n; i++) {
dp[i] = Math.max(dp[i - 1], dp[i - 2] + nums[i]);
}
return dp[n - 1];
}
**
2. 给定一个正整数s, 判断一个数组arr中，是否有一组数字加起来等于s。
**
/**
*
* 2. 给定一个正整数s, 判断一个数组arr中，是否有一组数字加起来等于s。
*/
private static boolean isS(int[] arr,int i,int s) {
if (i==0) {
return arr[0]==s;
}else if (s==0) {
return true;
}else if (arr[i]>s){
return isS(arr,i-1,s);
}else {
return isS(arr,i-1,s)||isS(arr,i-1,s-arr[i]);//又可分为两种情况，arr[i-1]>s
}
}
private static boolean isS2(int[] arr,int s) {
boolean set[][]=new boolean[arr.length][s+1];
set[0][arr[0]]=true;
int i=0;
int j=0;
for (i = 0; i < arr.length ; i++) {
set[i][0]=true;
}
for (i = 1; i < arr.length; i++) {
for (j = 1; j < s+1; j++) {
if (arr[i]>j){
set[i][j]=set[i-1][j];
}else {
boolean a = set[i-1][j];
boolean b = set[i-1][s-arr[i]];
set[i][j]=a||b;
}
}
}
return set[i-1][j-1];
}
神奇的口袋：
有一个神奇的口袋，总的容积是40，用这个口袋可以变出一些物品，这些物品的总体积必须是40。John现在有n个想要得到的物品，每个物品的体积分别是a1，a2……an。John可以从这些物品中选择一些，如果选出的物体的总体积是40，那么利用这个神奇的口袋，John就可以得到这些物品。现在的问题是，John有多少种不同的选择物品的方式。
输入描述:
输入的第一行是正整数n (1 <= n <= 20)，表示不同的物品的数目。接下来的n行，每行有一个1到40之间的正整数，分别给出a1，a2……an的值。
输出描述:
输出不同的选择物品的方式的数目。
示例1 输入
3 20 20 20
输出
3
代码：
import java.util.Scanner;
public class Main{
static int[] arr=new int[21];
static int n;
private static int count(int v,int i) {
if(v==0) {
return 1;
}if(i==n||v<0) {
return 0;
}
return count(v-arr[i],i+1)+count(v,i+1);//选arr[i]和不选arr[i]
}
public static void main(String[] args) {
Scanner in=new Scanner(System.in);
n=in.nextInt();
for (int i = 0; i < n; i++) {
arr[i]=in.nextInt();
}
System.out.println(count(40,0));
}
}
展开全文
• 现在获取数组中最大最小值用的越来越多了，于是乎我编了方法供大家使用。代码如下，若有问题可以与我联系，咱们一起学习一起进步。 代码如下: function getMaximin (arr,maximin) { if (maximin == “max”) { ...
• go数组练习：从一个整数数组获取最大整数，最小整数，总数以及平均值
需求：从一个整数数组中获取最大整数，最小整数，总数以及平均值
注意：
如果不对max、min赋初始值，则其默认为0，可能会出现没有比0大或没有比0小的情况导致出错，因此一般将max和min赋值为数组第一个元素值。for循环既可以使用i进行遍历，也可以利用rang进行遍历，用_接收无用参数。
代码
package main
import "fmt"
func main() {
a := [6]int{14, 64, 37, 72, 75, 37}
max := a[0]
min := a[0]
sum := 0
// 不进行初始赋值的情况下，没有数据小于0，可能导致结果出错
//var max int
//var min int
//var sum int
//for _, value := range a {
// if value > max {
// max = value
// }
// if value < min {
// min = value
// }
// sum+=value
//}
for i := 0; i < len(a); i++ {
if a[i] > max {
max = a[i]
}
if a[i] < min {
min = a[i]
}
sum += a[i]
}
avg := float32(sum/len(a))
fmt.Printf("数组a的最大值为：%d\n数组a的最小值为：%d\n"+
"数组a的总和为：%d\n数组a的平均值为%f\n", max, min, sum, avg)
}
运行结果：
展开全文
• 之前也遇到过类似的问题，如求n个数组任意选取一个元素的所有组合都是想起来比较简单，但是设计成算法却颇费周折。如数组为{1, 2, 3, 4, 5, 6}，那么中取出3个元素的组合有哪些，取出4个元素的组...
• 从一个字符数组中读出相应的整数、实数// author：yangyang。 //main.c #include "stdafx.h" void shishupanduan(int flag); char s[11]; char *p = s; void main() { gets_s(s); char zhengshu[10]; int i = 0; ...
• 数组中一个或连 * 续的多个整数组成一个数组。求所有子数组的和的最大值。要求时间复杂度为O(n)。 * * @param arr 输入数组 * @return 最大的连续子数组和 */ public static int ...
• 主要介绍了c#获取数组中最大数的值,需要的朋友可以参考下
• ② 在源数组中索引为0的数组元素开始，拷贝index个数组元素到另外一个目标数组 ③ 把临时变量temp赋值给目标数组索引为0的位置 代码如下:public static class ArrHelper { /// /// 对类型为T的数组
• 2. 请用户输入一个整数1--7的值： 根据用户输入，从数组中取出对应的星期名称，例如： 用户输入：1 程序提示：星期一 3. 为了防止用户输入小于1或者大于7的值，请使用异常处理从数组中取出对应的“星期名称”，...
• 这里先讲种类似于快速排序的方法。注意题目要求，不要求完全排序，只要求最快解决问题！...原话是1亿数据里，找出前个最大的。 首先看源码吧： void findMaxN（int a[], int start, in
• 代码： <!... <title>Title ... var items = ['1','4','9','10'];... var item = items[Math.floor(Math.random()*items...floor() 方法返回小于等于x的最大整数。 eg: console.log(Math.floor(1.6)); //结果为1
• //从一个整数数组中取出最大整数，最小整数，总和，平均值 //声明一个int类型的数组，并且随意的赋初值 int[] nums={1,2,3,4,5,6,7,8,9,0}; //声明两个变量用来存储最大值和最小值 int max=int.MinValue;//nums[0]...
• array_column() - 从数组中取出一
• 给定一个数组[1,2,3,4,5,6,7,8,9,10],求出数组中任意组合为之和等于10的数组。 注意，每一种组合中一个数只能出现一次。 数组共有N个元素，那么我们可以用1到2^N的二进制来求解，若二进制数该位置是1,则将其取出求和...
• #!/usr/bin/python # -*- coding:utf-8 -*- import random def randomText(textArr): length = len(textArr) ...randStr = randomText(['数字','数字二','数字三','数字4','数字5']) print randStr
• 问题描述 我想得到的是特定数，可能不是连续的，而且 数是取出的。 代码 private static int[] randomCreate(int ...//创造一个用于接收的数组 for (int i = 0; i < rands.length; i++) { int temp ...
• 从数组中取出任意数，求和为指定值的解 PHP <?php $cc = array(1,9,34,56,10,7,3,28,9); function demo($cc,$val){ for ($i = 1; $i < 1 << count($cc); $i++) {$sum = 0; $temp = ""; ... • //获取数组中最大值 var nums=[5,13,20,18,32] //声明变量用于保存最大值，吧第一个元素作为最大值 //在比较的时候可以第二个元素开始比较 var s=nums[0]; //遍历数组，得到每个元素 for(var i=0;i<nums.length... • 方法： if ($row) { $data = array();$arr = $row;$discrepancyValue = $abnormalValue = 0; //最小值 do { foreach ($row as $ke =>$va) { switch (coun...
• 假定大数组包含了n个小数组，分别找到每个小数组最大值，然后把它们串联起来，形成一个数组. 我们可以用for循环来迭代数组，并通过arr[i]的方式来访问数组的每个元素。 得到数组中最大值我们可以使用Math....
• 读取一个包含10个元素的数组，并在其中找到两个最大元素，然后打印它们的总和。 #include <stdio.h> int main(){ int Max,max; const int N = 10; int num[N]; for (int i = 0; i<N; i++) { scanf("%d...
• 可以使用numpy的random....#a为数组或int值，为数组时会数组中随机选择元素，为int值时会先生成一np.arange(a)的数组，然后从中随机选择元素 #size为int值，为选择元素的个数 #replace默认为Tru...
• int [] arr = {1,2,3,4}; //产生0-(arr.length-1)的整数值,也是数组的索引 int index=(int)(Math.random()*arr.length); int rand = arr[index];
• 题目： 输入一个整数数组，将它们连接起来排成一个数，输出能排出的所有数字最小的一个。 例如输入数组{122,12,123}，则输出这两个能排成的最小数字12122123。 思路： 新排序规则：从整数数组中取出两个数a和b,...
• 最大值获取：从数组的所有元素找出最大值。 实现思路： 定义变量，保存数组0索引上的元素 遍历数组，获取出数组中的每元素 将遍历到的元素和保存数组0索引上值的变量进行比较 如果数组元素的值大于了变量的...
• def func(arr,d): l = [] 注释：枚举 for index,i in enumerate(arr): 注释：遍历索引 for j in range(index+1): if i + arr[j] == d: if i != arr[j]: l.append((i,arr(j))) return l
• 这是一个呆萌炫酷吊炸天的前端算法题，曾经乃至现在也是叱咤风云在各个面试场景。 可以这样说，有 90% 以上的前端工程师不会做这个题目。 这道题涉及的知识点很多，虽然网上也有相关的解答文章，但是在算法小分队...
...
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CC-MAIN-2021-49
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http://encyclopedia2.thefreedictionary.com/Inclinations
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|
# inclination
(redirected from Inclinations)
Also found in: Dictionary, Thesaurus, Medical, Legal.
## inclination,
in astronomy, the angle of intersection between two planes, one of which is an orbital plane. The inclination of the plane of the moon's orbit is 5°9' with respect to the plane of the ecliptic (the plane of the earth's orbit around the sun). The inclination of the plane of the ecliptic relative to the plane of the earth's equator is 23°27'8.26"; this angle is called the obliquity of the ecliptic.
## inclination
1. Symbol: i . The angle between the orbital plane of a celestial body and a reference plane. For a planet or comet the reference plane is the plane of the ecliptic, for a satellite it is the primary's equatorial plane, and for a double star it is the plane of the sky. Inclination is one of the orbital elements and varies between 0 and 180°, being less than 90° for a body with direct motion.
2. (axial inclination) The angle between the rotational axis of a body and a line perpendicular to its orbital plane. See also Tables 1–3, backmatter.
## Inclination
(religion, spiritualism, and occult)
An inclination is the angle at which two planes cross. In astrology, it is used to refer to the movement of a celestial body to a position other than the one occupied at birth.
## inclination
[‚iŋ·klə′nā·shən]
(geology)
The angle at which a geological body or surface deviates from the horizontal or vertical; often used synonymously with dip.
(geophysics)
In magnetic inclination, the dip angle of the earth's magnetic field. Also known as magnetic dip.
(mathematics)
The inclination of a line in a plane is the angle made with the positive x axis.
The inclination of a line in space with respect to a plane is the smaller angle the line makes with its orthogonal projection in the plane.
The inclination of a plane with respect to a given plane is the smaller of the dihedral angles which it makes with the given plane.
(science and technology)
Angular deviation of a direction or surface from the true vertical or horizontal.
The angle which a direction or surface makes with the vertical or horizontal.
A surface which deviates from the vertical or horizontal.
## inclination
The angle which a line or surface makes with the vertical, horizontal, or with another line or surface.
## inclination
As it pertains to meteorology, the angle between an isobar and the wind or airflow at a given point.
## inclination
1. Maths
a. the angle between a line on a graph and the positive limb of the x-axis
b. the smaller dihedral angle between one plane and another
2. Astronomy the angle between the plane of the orbit of a planet or comet and another plane, usually that of the ecliptic
3. Physics another name for dip
References in classic literature ?
But there are other points to be considered besides his inclination.
Copperfield to be accompanied by some confidential friend today,' with an inclination of her head towards Traddles, who bowed, 'in order that there might be no doubt or misconception on this subject.
They have certain professors well skilled in preparing children for such a condition of life as befits the rank of their parents, and their own capacities, as well as inclinations.
She entered the room with an air more than usually ungracious, made no other reply to Elizabeth's salutation than a slight inclination of the head, and sat down without saying a word.
Not badly answered, i'faith," said Gondy, laughing; "but I have, you must know, always had, in spite of my bands, warlike inclinations.
What principally delighted Julia in these contemplations on the acquaintance of Anna, was the strong inclination he had expressed to know herself.
I cannot bear to think of my little Hetty shedding tears when I am not there to kiss them away; and if I followed only my own inclinations, I should be with her at this moment instead of writing.
Previously to the occurrence of the scene at the “Bold Dragoon,” Elizabeth had been safely reconducted to the mansion-house, where she was left as its mistress, either to amuse or employ herself during the evening as best suited her own inclinations.
Tell me Augusta with sincerity; did you ever know me consult his inclinations or follow his Advice in the least trifling Particular since the age of fifteen?
If the suitor and his present find favor in the eyes of the father, he breaks the matter to his daughter, and inquires into the state of her inclinations.
Hints were thrown out of an exciting nature; stories were told of perilous bargains made in a hurry and repented of at leisure; and instances were adduced of unaccountable capacities, vague longings, and unnatural inclinations implanted by the author of all evil for wise purposes of his own.
With reflections of this nature she usually, as has been hinted, accompanied every act of compliance with her brother's inclinations; and surely nothing could more contribute to heighten the merit of this compliance than a declaration that she knew, at the same time, the folly and unreasonableness of those inclinations to which she submitted.
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# Reformulation as optimization on probability distributions
This is a "soft" question, in the sense that I'm looking for historical remarks and general commentary rather than a definite answer.
For compact $$X \in R^n$$ and $$f : R^n \to R$$ consider the problem $$\min_{x \in X} ~ f(x).$$
A common trick that I've stumbled upon in several applications is to look at a "probabilistic" but equivalent reformulation $$\min_{\mu \in P(X)} \int f(x) \, \mathrm{d}\mu(x),$$ where $$P(X)$$ denote the set of probability distributions over $$X$$.
I'm interested in history/old references on this "lifting" principle, where the idea is to artificially enlarge the solution space in order to gain some favorable properties (such as linearity or convexity).
A popular example is the linear programming formulation of optimal transport due to Kantorovich, but I wonder where this idea of "probabilistic reformulation" originated or was first used (maybe in statistical physics)?
Another example of this trick are policy gradient methods in reinforcement learning, for an explanation see this blog post: http://www.argmin.net/2018/02/20/reinforce/
• It is notable that the post you link to seems to think that the use of this technique in policy gradient methods is ill-advised. – R Hahn Jul 21 at 15:49
One thing that comes to mind is the modern approach to solving partial differential equations by looking for solutions that are distributions instead of only differentiable functions; see chapter 8 in Folland's Real Analysis, and chapters 6 through 8 in Rudin's Functional Analysis, for example. The connection is very close to a kind of dual to your example: distributions in this context are defined as continuous linear functionals on the space $$\mathcal{D}(\mathbb{R}^n)$$ of smooth, compactly supported functions on $$\mathbb{R}^n$$. Examples of such distributions include:
1. Evaluation at a point, $$\delta_x(f) = f(x)$$. This includes the famous Dirac point mass, $$\delta_0$$.
2. Integration against probability measures, $$f \mapsto \int f\,d\mu$$.
One benefit is that some problems have new distributional solutions that do not correspond to something like $$\mu = f\,dx$$ (for $$f$$ differentiable and $$L^1$$ and $$dx$$ the Lebesgue measure) including some problems that don't have a solution corresponding to any smooth function.
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# Better way to write this function: list to matrix (= list of lists)
Is there a better way to write this function (i.e. in one line via folds)?
``````-- list -> rowWidth -> list of lists
listToMatrix :: [a] -> Int -> [[a]]
listToMatrix [] _ = []
listToMatrix xs b = [(take b xs)] ++ (listToMatrix (drop b xs) b)
``````
## 4 answers
• answered 2017-06-17 18:09
No, although I wish `chunksOf` was in the standard library. You can get that from here if you want: https://hackage.haskell.org/package/split-0.2.3.2/docs/Data-List-Split.html
Note a better definition would be:
``````listToMatrix xs b = (take b xs) : (listToMatrix (drop b xs) b)
``````
although this might compile to the same core. You could also use `splitAt`, though again this is likely to perform the same.
``````listToMatrix xs b = let (xs,xss) = splitAt b xs in xs : listToMatrix xss
``````
• answered 2017-06-17 18:09
Yes, there is a better way to write this function. But I don't think that making it one line is going to improve anything.
### Use the prepending operator `(:)` instead of list concatenation `(++)` for single-element prepending
The expression `[(take b xs)] ++ (listToMatrix (drop b xs) b)` is inefficient. I don't mean inefficient in terms of performances, because the compiler probably optimizes that, but, here, you are constructing a list, and then calling a function on it (`(++)`) which is going to deconstruct it by pattern matching. You could instead build your list directly using the `(:)` data constructor, which allows you to prepend a single element to your list. The expression becomes `take b xs : listToMatrix (drop b xs) b`
### Use `splitAt` to avoid running through the list twice
``````import Data.List (splitAt)
listToMatrix :: [a] -> Int -> [[a]]
listToMatrix xs b = row : listToMatrix remaining b
where (row, remaining) = splitAt b xs
``````
### Ensure the correctness of your data by using `Maybe`
``````listToMatrix :: [a] -> Int -> Maybe [[a]]
listToMatrix xs b
| length xs `mod` b /= 0 = Nothing
| null xs = Just []
| otherwise = Just \$ row : listToMatrix remaining b
where (row, remaining) = splitAt b xs
``````
You can even avoid checking every time if you have the right number of elements by defining a helper function:
``````listToMatrix :: [a] -> Int -> Maybe [[a]]
listToMatrix xs b
| length xs `mod` b /= 0 = Nothing
| otherwise = Just (go xs)
where go [] = []
go xs = row : go remaining
where (row, remaining) = splitAt b xs
``````
### Ensure the correctness of your data by using safe types
A matrix has te same number of elements in each row, whereas nested lists don't allow to ensure that kind of conditions. To be certain that every row has the same number of elements, you can either use a library such as `matrix` or `hmatrix`
• answered 2017-06-17 18:09
You can use this, the idea is to take all the chunks of the main list, it is a different approach but basicly do the same:
``````Prelude> let list2Matrix n xs = map (\(x ,y)-> (take n) \$ (drop (n*x)) y) \$ zip [0..] \$ replicate (div (length xs) n) xs
Prelude> list2Matrix 10 [1..100]
[[1,2,3,4,5,6,7,8,9,10],[11,12,13,14,15,16,17,18,19,20],[21,22,23,24,25,26,27,28,29,30],[31,32,33,34,35,36,37,38,39,40],[41,42,43,44,45,46,47,48,49,50],[51,52,53,54,55,56,57,58,59,60],[61,62,63,64,65,66,67,68,69,70],[71,72,73,74,75,76,77,78,79,80],[81,82,83,84,85,86,87,88,89,90],[91,92,93,94,95,96,97,98,99,100]]
``````
• answered 2017-06-17 18:09
Actually this is a nice case for unfolding. `Data.List` method `unfoldr`, unlike folding a list, creates a list to a from a seed value by applying a function to it up until this function returns `Nothing`. Until we reach the terminating condition that will return `Nothing` the function returns `Just (a,b)` where `a` is the current generated item of the list and `b` is the next value of the seed. In this particular case our seed value is the given list.
``````import Data.List
chunk :: Int -> [a] -> [[a]]
chunk n = unfoldr (\xs -> if null xs then Nothing else Just (splitAt n xs))
*Main> chunk 3 [1,2,3,4,5,6,7,8,9]
[[1,2,3],[4,5,6],[7,8,9]]
*Main> chunk 3 [1,2,3,4,5,6,7,8,9,10]
[[1,2,3],[4,5,6],[7,8,9],[10]]
``````
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# finesse.gaussian.ws_overlap_grid¶
finesse.gaussian.ws_overlap_grid(qp, woffset, soffset, wpts=200, spts=200)[source]
Computes the WS phase space overlap with the primary beam parameter qp over a grid of W, S data.
See ws_overlap() for a definition of the overlap quantity.
The $$W$$ and $$S$$ spaces are formed by creating arrays around the BeamParam.w and BeamParam.S values of qp, according to the offsets given in woffset and soffset, respectively.
Parameters:
qpBeamParam
The primary “mode” which overlapping modes are calculated against.
woffsetfloat, sequence
A single number, or size 2 sequence. Defines the offsets (lower, upper) from the primary mode beamsize to use for the W space.
soffsetfloat, sequence
A single number, or size 2 sequence. Defines the offsets (lower, upper) from the primary mode beamsize to use for the S space.
wptsint, optional; default: 200
Number of points for the W space array.
sptsint, optional; default: 200
Number of points for the S space array.
Returns:
Wnumpy.ndarray
The W space (as a 2D grid).
Snumpy.ndarray
The S space (as a 2D grid).
OLnumpy.ndarray
The overlap as a function of the WS phase space (as a 2D grid).
Examples
In the following example, we compute the overlap to a primary mode which has a 6 mm beam size and a defocus of 0 m (i.e. at the waist).
import finesse
finesse.configure(plotting=True)
import finesse.gaussian as gaussian
from finesse.plotting.plot import ws_phase_space
import matplotlib.pyplot as plt
# Make a beam parameter with w = 6 mm, S = 0 m
qp = gaussian.BeamParam(w=6e-3, S=0)
# Compute the WS phase space overlap with qp, using
# maximum offset of 1 mm in beam size and 1 cm in defocus
W, S, OL = gaussian.ws_overlap_grid(qp, woffset=1e-3, soffset=1e-2)
# Now plot this as contours of overlap with custom levels
fig, ax = ws_phase_space(
W, S, OL,
levels=[0.6, 0.8, 0.84, 0.88, 0.92, 0.94, 0.96, 0.98, 0.995, 0.999],
)
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# Number 49293768484
### Properties of number 49293768484
Cross Sum:
Factorization:
2 * 2 * 31 * 31 * 3581 * 3581
Divisors:
Count of divisors:
Sum of divisors:
89161470993
Prime number?
No
Fibonacci number?
No
Bell Number?
No
Catalan Number?
No
Base 3 (Ternary):
Base 4 (Quaternary):
Base 5 (Quintal):
Base 8 (Octal):
b7a233724
Base 32:
1dt26dp4
sin(49293768484)
-0.78596293045156
cos(49293768484)
0.61827362223856
tan(49293768484)
-1.2712218380041
ln(49293768484)
24.621063510087
lg(49293768484)
10.69279202102
sqrt(49293768484)
222022
Square(49293768484)
2.4298756113542E+21
### Number Look Up
Look Up
49293768484 (forty-nine billion two hundred ninety-three million seven hundred sixty-eight thousand four hundred eighty-four) is a impressive figure. The cross sum of 49293768484 is 64. If you factorisate 49293768484 you will get these result 2 * 2 * 31 * 31 * 3581 * 3581. The figure 49293768484 has 27 divisors ( 1, 2, 4, 31, 62, 124, 961, 1922, 3581, 3844, 7162, 14324, 111011, 222022, 444044, 3441341, 6882682, 12823561, 13765364, 25647122, 51294244, 397530391, 795060782, 1590121564, 12323442121, 24646884242, 49293768484 ) whith a sum of 89161470993. The figure 49293768484 is not a prime number. The number 49293768484 is not a fibonacci number. The number 49293768484 is not a Bell Number. The number 49293768484 is not a Catalan Number. The convertion of 49293768484 to base 2 (Binary) is 101101111010001000110011011100100100. The convertion of 49293768484 to base 3 (Ternary) is 11201020100220220010001. The convertion of 49293768484 to base 4 (Quaternary) is 231322020303130210. The convertion of 49293768484 to base 5 (Quintal) is 1301423201042414. The convertion of 49293768484 to base 8 (Octal) is 557210633444. The convertion of 49293768484 to base 16 (Hexadecimal) is b7a233724. The convertion of 49293768484 to base 32 is 1dt26dp4. The sine of 49293768484 is -0.78596293045156. The cosine of 49293768484 is 0.61827362223856. The tangent of the number 49293768484 is -1.2712218380041. The square root of 49293768484 is 222022.
If you square 49293768484 you will get the following result 2.4298756113542E+21. The natural logarithm of 49293768484 is 24.621063510087 and the decimal logarithm is 10.69279202102. You should now know that 49293768484 is very great figure!
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## Do My Nonobtuse Mesh Class
A "Nonobtuse Mesh Class" QE" is a standard mathematical term for a generalized continuous expression which is used to resolve differential equations and has services which are regular. In differential Class resolving, a Nonobtuse Mesh function, or "quad" is utilized.
The Nonobtuse Mesh Class in Class kind can be revealed as: Q( x) = -kx2, where Q( x) are the Nonobtuse Mesh Class and it is a crucial term. The q part of the Class is the Nonobtuse Mesh constant, whereas the x part is the Nonobtuse Mesh function.
There are 4 Nonobtuse Mesh functions with appropriate solution: K4, K7, K3, and L4. We will now take a look at these Nonobtuse Mesh functions and how they are fixed.
K4 - The K part of a Nonobtuse Mesh Class is the Nonobtuse Mesh function. This Nonobtuse Mesh function can also be written in partial portions such as: (x2 - y2)/( x+ y). To resolve for K4 we increase it by the right Nonobtuse Mesh function: k( x) = x2, y2, or x-y.
K7 - The K7 Nonobtuse Mesh Class has a service of the kind: x4y2 - y4x3 = 0. The Nonobtuse Mesh function is then multiplied by x to get: x2 + y2 = 0. We then need to increase the Nonobtuse Mesh function with k to get: k( x) = x2 and y2.
K3 - The Nonobtuse Mesh function Class is K3 + K2 = 0. We then multiply by k for K3.
K3( t) - The Nonobtuse Mesh function equationis K3( t) + K2( t). We multiply by k for K3( t). Now we increase by the Nonobtuse Mesh function which gives: K2( t) = K( t) times k.
The Nonobtuse Mesh function is likewise referred to as "K4" because of the initials of the letters K and 4. K suggests Nonobtuse Mesh, and the word "quad" is noticable as "kah-rab".
The Nonobtuse Mesh Class is among the main approaches of fixing differential formulas. In the Nonobtuse Mesh function Class, the Nonobtuse Mesh function is first multiplied by the appropriate Nonobtuse Mesh function, which will offer the Nonobtuse Mesh function.
The Nonobtuse Mesh function is then divided by the Nonobtuse Mesh function which will divide the Nonobtuse Mesh function into a real part and a fictional part. This offers the Nonobtuse Mesh term.
Finally, the Nonobtuse Mesh term will be divided by the numerator and the denominator to get the quotient. We are entrusted the right hand side and the term "q".
The Nonobtuse Mesh Class is an important principle to understand when resolving a differential Class. The Nonobtuse Mesh function is just one method to solve a Nonobtuse Mesh Class. The approaches for solving Nonobtuse Mesh equations include: singular value decomposition, factorization, optimum algorithm, numerical option or the Nonobtuse Mesh function approximation.
## Hire Someone To Do Your Nonobtuse Mesh Class
If you would like to become knowledgeable about the Quartic Class, then you require to very first start by checking out the online Quartic page. This page will show you how to utilize the Class by using your keyboard. The explanation will likewise reveal you how to create your own algebra formulas to assist you study for your classes.
Before you can understand how to study for a Nonobtuse Mesh Class, you should first understand making use of your keyboard. You will find out how to click the function keys on your keyboard, as well as how to type the letters. There are 3 rows of function keys on your keyboard. Each row has four functions: Alt, F1, F2, and F3.
By pressing Alt and F2, you can multiply and divide the value by another number, such as the number 6. By pushing Alt and F3, you can use the 3rd power.
When you push Alt and F3, you will key in the number you are trying to increase and divide. To multiply a number by itself, you will press Alt and X, where X is the number you wish to multiply. When you press Alt and F3, you will key in the number you are attempting to divide.
This works the very same with the number 6, except you will just enter the two digits that are six apart. Finally, when you press Alt and F3, you will utilize the 4th power. Nevertheless, when you push Alt and F4, you will use the real power that you have found to be the most proper for your issue.
By utilizing the Alt and F function keys, you can increase, divide, and then use the formula for the third power. If you need to increase an odd variety of x's, then you will require to get in an even number.
This is not the case if you are attempting to do something complex, such as multiplying two even numbers. For instance, if you wish to increase an odd variety of x's, then you will need to get in odd numbers. This is particularly true if you are attempting to figure out the answer of a Nonobtuse Mesh Class.
If you wish to transform an odd number into an even number, then you will require to press Alt and F4. If you do not know how to increase by numbers on their own, then you will need to use the letters x, a b, c, and d.
While you can multiply and divide by use of the numbers, they are a lot easier to use when you can look at the power tables for the numbers. You will need to do some research study when you initially start to utilize the numbers, but after a while, it will be second nature. After you have actually created your own algebra formulas, you will have the ability to create your own multiplication tables.
The Nonobtuse Mesh Formula is not the only method to solve Nonobtuse Mesh formulas. It is important to learn more about trigonometry, which utilizes the Pythagorean theorem, and then use Nonobtuse Mesh solutions to solve issues. With this method, you can understand about angles and how to resolve issues without needing to take another algebra class.
It is necessary to try and type as rapidly as possible, since typing will help you understand about the speed you are typing. This will assist you write your answers much faster.
## Hire Someone To Take My Nonobtuse Mesh Class
A Nonobtuse Mesh Class is a generalization of a linear Class. For example, when you plug in x=a+b for a given Class, you get the value of x. When you plug in x=a for the Class y=c, you obtain the values of x and y, which offer you a result of c. By applying this fundamental concept to all the equations that we have actually tried, we can now resolve Nonobtuse Mesh formulas for all the worths of x, and we can do it quickly and efficiently.
There are numerous online resources available that offer totally free or budget friendly Nonobtuse Mesh equations to fix for all the worths of x, including the cost of time for you to be able to make the most of their Nonobtuse Mesh Class task help service. These resources usually do not need a membership cost or any type of financial investment.
The answers supplied are the result of complex-variable Nonobtuse Mesh formulas that have actually been fixed. This is likewise the case when the variable used is an unidentified number.
The Nonobtuse Mesh Class is a term that is an extension of a linear Class. One advantage of using Nonobtuse Mesh equations is that they are more basic than the direct formulas. They are much easier to resolve for all the worths of x.
When the variable utilized in the Nonobtuse Mesh Class is of the form x=a+b, it is simpler to solve the Nonobtuse Mesh Class since there are no unknowns. As a result, there are fewer points on the line specified by x and a consistent variable.
For a right-angle triangle whose base indicate the right and whose hypotenuse points to the left, the right-angle tangent and curve graph will form a Nonobtuse Mesh Class. This Class has one unknown that can be found with the Nonobtuse Mesh formula. For a Nonobtuse Mesh Class, the point on the line specified by the x variable and a continuous term are called the axis.
The existence of such an axis is called the vertex. Considering that the axis, vertex, and tangent, in a Nonobtuse Mesh Class, are a given, we can discover all the worths of x and they will sum to the offered values. This is attained when we use the Nonobtuse Mesh formula.
The element of being a constant aspect is called the system of formulas in Nonobtuse Mesh equations. This is sometimes called the main Class.
Nonobtuse Mesh equations can be resolved for other worths of x. One way to solve Nonobtuse Mesh equations for other values of x is to divide the x variable into its aspect part.
If the variable is provided as a favorable number, it can be divided into its element parts to get the typical part of the variable. This variable has a magnitude that is equal to the part of the x variable that is a constant. In such a case, the formula is a third-order Nonobtuse Mesh Class.
If the variable x is unfavorable, it can be divided into the same part of the x variable to get the part of the x variable that is increased by the denominator. In such a case, the formula is a second-order Nonobtuse Mesh Class.
Service assistance service in fixing Nonobtuse Mesh formulas. When utilizing an online service for resolving Nonobtuse Mesh formulas, the Class will be fixed instantly.
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7 - Correlation
# 7 - Correlation - Chapter 8 Correlation Oftentimes wish to...
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Chapter 8: Correlation Oftentimes, wish to use values on one variable to predict values on another variable College entrance exams and later grades Income level and health Age and traffic safety
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Want to know the correlation (measure of the degree of relationship between two variables) Greater the correlation, more accurately we can predict value on one variable from value on the other To calculate a correlation, need pairs of observations (scores on each of the two variables; X and Y)
Can get some indication from a bivariate distribution (e.g., frequencies of values in combinations of levels of two variables) 10 30 50 70 90 90 8 18 19 70 5 19 34 16 50 12 21 24 7 30 1 10 14 9 1 10 2 1 Psychology Exam Scores Math Exam Scores
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Scatter Diagram (scatterplot) graphs the bivariate distribution by placing dots at the values of the paired scores
Scatterplot of Economics and Sociology Grades 30 40 50 60 70 80 90 100 30 40 50 60 70 80 90 100 Economics Test Scores
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Correlation coefficients (including Pearson’s r) assume linear relationships between variables Relation that can be represented by a straight line
Scatterplot of Economics and Sociology Grades 30 40 50 60 70 80 90 100 30 40 50 60 70 80 90 100 Economics Test Scores
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Direction of a Correlation Positive Correlation As values on X increase, values on Y tend to
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## Chapter 2 - Day 2 - Lesson 2.1
##### Learning Targets
• Find and interpret the standardized score (z-score) of an individual value in a distribution of data.
• Describe the effect of adding, subtracting, multiplying by, or dividing by a constant on the shape, center, and variability of a distribution of data.
##### Activity:
First of all, let’s take care of the movie reference. If you don’t know, you’re not thinking fourth dimensionally. It’s the most perfect blockbuster ever made.
Now, about the activity. Students should work in groups (or pairs) on page 1. By the end of the page, they will write a formula for a z-score. For now, we prefer to use words rather than symbols:
z-score = (value – mean) / SD
We don’t want students to have to wrestle with mu vs xbar and sigma vs s. We will save that discussion for Chapter 7. For now, the most important idea in this lesson is that students understand the interpretation of the z-score:
A z-score tells us the number of standard deviations above or below the mean.
In fact, this idea is so important that you want to choose a student and make it their job (we pay them in full at the end of the school year….\$1 but it is a great resume builder). Now any time throughout the year when you calculate a z-score as a class, have this student interpret the z-score. Also, here is the answer to the bonus question.
On page 2 we are killing two birds with one stone:
1. To understand the effect of adding, subtracting, multiplying by, or dividing by a constant on the shape, center, and variability of a distribution of data.
2. Practice with z-scores and understanding that standardizing a distribution maintains the shape of the distribution, but changes the mean to 0 and the standard deviation to 1.
##### Preparing for Inference
Why do we place so much emphasis on interpreting the z-score? We are planting seeds that will later grow into inferential thinking and reasoning. Later, when we perform a significance test, we will want to know how likely a certain result would be under a certain condition (P-value!). To understand “how likely is this result”, we must first think about “how far is this result from what was expected”.
##### Teaching Tip
Take any set of data, and transform all of the values into z-scores (standardize the distribution) by subtracting the mean and dividing by the standard deviation. Subtracting the mean will shift the mean to 0. Dividing by the standard deviation will dilate the variability such that the standard deviation will be 1. We will need this idea to understand why we can use Table A to do normal distribution calculations in Lesson 2.2. In that lesson, we learn that you can take any normal distribution and standardize it into a standard normal distribution (with mean 0 and standard deviation of 1), allowing us to then use Table A to find area.
##### Luke's Lesson Notes
Here is a brief video highlighting some key information to help you prepare to teach this lesson.
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You are Here: Home >< Maths
Question on graph transformations watch
1. We haven't technically 'learnt' this yet, so are just being asked to 'describe what happens when' but I'm not sure how to word it.
1) You replace each x with -x and get this. How do you describe the change?
2) You go back to the original equation and replace each x with -y and each y with x. How do you describe the change? (I know it's a 90 degree clockwise rotation, but about which point?)
Thanks.
2. (Original post by yoloman154)
We haven't technically 'learnt' this yet, so are just being asked to 'describe what happens when' but I'm not sure how to word it.
1) You replace each x with -x and get this (first screenshot). How do you describe the change?
2) You go back to the original equation and replace each x with -y and each y with x. How do you describe the change? (I know it's a 90 degree clockwise rotation, but about which point?)
Thanks.
Anyone?
3. (Original post by yoloman154)
Anyone?
For the first one, think about how all the points are changing. You're making all the positive x take the place of negative x and vice-versa. i.e:
Spoiler:
Show
a flip in the y-axis.
4. (Original post by Zacken)
For the first one, think about how all the points are changing. You're making all the positive x take the place of negative x and vice-versa. i.e:
Spoiler:
Show
a flip in the y-axis.
Of course; now you say it, that makes a lot of sense. My spatial awareness is terrible, so 'seeing things' sometimes takes a while. I really appreciate your help!
EDIT: Also, WOW Cambridge Maths! Well done! I imagine Warwick are jealous...
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# Properties
Label 1840.2.e.c Level $1840$ Weight $2$ Character orbit 1840.e Analytic conductor $14.692$ Analytic rank $0$ Dimension $4$ Inner twists $2$
# Related objects
Show commands: Magma / PariGP / SageMath
## Newspace parameters
comment: Compute space of new eigenforms
[N,k,chi] = [1840,2,Mod(369,1840)]
mf = mfinit([N,k,chi],0)
lf = mfeigenbasis(mf)
from sage.modular.dirichlet import DirichletCharacter
H = DirichletGroup(1840, base_ring=CyclotomicField(2))
chi = DirichletCharacter(H, H._module([0, 0, 1, 0]))
N = Newforms(chi, 2, names="a")
//Please install CHIMP (https://github.com/edgarcosta/CHIMP) if you want to run this code
chi := DirichletCharacter("1840.369");
S:= CuspForms(chi, 2);
N := Newforms(S);
Level: $$N$$ $$=$$ $$1840 = 2^{4} \cdot 5 \cdot 23$$ Weight: $$k$$ $$=$$ $$2$$ Character orbit: $$[\chi]$$ $$=$$ 1840.e (of order $$2$$, degree $$1$$, not minimal)
## Newform invariants
comment: select newform
sage: f = N[0] # Warning: the index may be different
gp: f = lf[1] \\ Warning: the index may be different
Self dual: no Analytic conductor: $$14.6924739719$$ Analytic rank: $$0$$ Dimension: $$4$$ Coefficient field: $$\Q(i, \sqrt{5})$$ comment: defining polynomial gp: f.mod \\ as an extension of the character field Defining polynomial: $$x^{4} + 3x^{2} + 1$$ x^4 + 3*x^2 + 1 Coefficient ring: $$\Z[a_1, a_2, a_3]$$ Coefficient ring index: $$1$$ Twist minimal: no (minimal twist has level 230) Sato-Tate group: $\mathrm{SU}(2)[C_{2}]$
## $q$-expansion
comment: q-expansion
sage: f.q_expansion() # note that sage often uses an isomorphic number field
gp: mfcoefs(f, 20)
Coefficients of the $$q$$-expansion are expressed in terms of a basis $$1,\beta_1,\beta_2,\beta_3$$ for the coefficient ring described below. We also show the integral $$q$$-expansion of the trace form.
$$f(q)$$ $$=$$ $$q + \beta_1 q^{3} + (2 \beta_{2} + 1) q^{5} + (3 \beta_{3} + 3 \beta_1) q^{7} + (\beta_{2} + 2) q^{9}+O(q^{10})$$ q + b1 * q^3 + (2*b2 + 1) * q^5 + (3*b3 + 3*b1) * q^7 + (b2 + 2) * q^9 $$q + \beta_1 q^{3} + (2 \beta_{2} + 1) q^{5} + (3 \beta_{3} + 3 \beta_1) q^{7} + (\beta_{2} + 2) q^{9} + ( - \beta_{2} + 4) q^{11} + (\beta_{3} - \beta_1) q^{13} + (2 \beta_{3} - \beta_1) q^{15} + (4 \beta_{3} + 3 \beta_1) q^{17} + ( - 3 \beta_{2} - 5) q^{19} - 3 q^{21} - \beta_{3} q^{23} + 5 q^{25} + (\beta_{3} + 4 \beta_1) q^{27} - 6 \beta_{2} q^{29} + (3 \beta_{2} + 7) q^{31} + ( - \beta_{3} + 5 \beta_1) q^{33} + (9 \beta_{3} + 3 \beta_1) q^{35} - 6 \beta_1 q^{37} + ( - 2 \beta_{2} + 1) q^{39} + (\beta_{2} - 4) q^{41} + ( - 8 \beta_{3} + 2 \beta_1) q^{43} + (3 \beta_{2} + 4) q^{45} + ( - 8 \beta_{3} - 6 \beta_1) q^{47} + ( - 9 \beta_{2} - 11) q^{49} + ( - \beta_{2} - 3) q^{51} - 2 \beta_{3} q^{53} + (9 \beta_{2} + 2) q^{55} + ( - 3 \beta_{3} - 2 \beta_1) q^{57} - 6 q^{59} + ( - 3 \beta_{2} - 2) q^{61} + (9 \beta_{3} + 6 \beta_1) q^{63} + ( - \beta_{3} + 3 \beta_1) q^{65} + (2 \beta_{3} - 2 \beta_1) q^{67} + \beta_{2} q^{69} + ( - \beta_{2} - 2) q^{71} + ( - 10 \beta_{3} + 4 \beta_1) q^{73} + 5 \beta_1 q^{75} + (9 \beta_{3} + 12 \beta_1) q^{77} + (6 \beta_{2} + 2) q^{79} + (6 \beta_{2} + 2) q^{81} + (2 \beta_{3} + 6 \beta_1) q^{83} + (10 \beta_{3} + 5 \beta_1) q^{85} + ( - 6 \beta_{3} + 6 \beta_1) q^{87} + (6 \beta_{2} + 6) q^{89} - 3 \beta_{2} q^{91} + (3 \beta_{3} + 4 \beta_1) q^{93} + ( - 7 \beta_{2} - 11) q^{95} + (8 \beta_{3} + 13 \beta_1) q^{97} + (3 \beta_{2} + 7) q^{99}+O(q^{100})$$ q + b1 * q^3 + (2*b2 + 1) * q^5 + (3*b3 + 3*b1) * q^7 + (b2 + 2) * q^9 + (-b2 + 4) * q^11 + (b3 - b1) * q^13 + (2*b3 - b1) * q^15 + (4*b3 + 3*b1) * q^17 + (-3*b2 - 5) * q^19 - 3 * q^21 - b3 * q^23 + 5 * q^25 + (b3 + 4*b1) * q^27 - 6*b2 * q^29 + (3*b2 + 7) * q^31 + (-b3 + 5*b1) * q^33 + (9*b3 + 3*b1) * q^35 - 6*b1 * q^37 + (-2*b2 + 1) * q^39 + (b2 - 4) * q^41 + (-8*b3 + 2*b1) * q^43 + (3*b2 + 4) * q^45 + (-8*b3 - 6*b1) * q^47 + (-9*b2 - 11) * q^49 + (-b2 - 3) * q^51 - 2*b3 * q^53 + (9*b2 + 2) * q^55 + (-3*b3 - 2*b1) * q^57 - 6 * q^59 + (-3*b2 - 2) * q^61 + (9*b3 + 6*b1) * q^63 + (-b3 + 3*b1) * q^65 + (2*b3 - 2*b1) * q^67 + b2 * q^69 + (-b2 - 2) * q^71 + (-10*b3 + 4*b1) * q^73 + 5*b1 * q^75 + (9*b3 + 12*b1) * q^77 + (6*b2 + 2) * q^79 + (6*b2 + 2) * q^81 + (2*b3 + 6*b1) * q^83 + (10*b3 + 5*b1) * q^85 + (-6*b3 + 6*b1) * q^87 + (6*b2 + 6) * q^89 - 3*b2 * q^91 + (3*b3 + 4*b1) * q^93 + (-7*b2 - 11) * q^95 + (8*b3 + 13*b1) * q^97 + (3*b2 + 7) * q^99 $$\operatorname{Tr}(f)(q)$$ $$=$$ $$4 q + 6 q^{9}+O(q^{10})$$ 4 * q + 6 * q^9 $$4 q + 6 q^{9} + 18 q^{11} - 14 q^{19} - 12 q^{21} + 20 q^{25} + 12 q^{29} + 22 q^{31} + 8 q^{39} - 18 q^{41} + 10 q^{45} - 26 q^{49} - 10 q^{51} - 10 q^{55} - 24 q^{59} - 2 q^{61} - 2 q^{69} - 6 q^{71} - 4 q^{79} - 4 q^{81} + 12 q^{89} + 6 q^{91} - 30 q^{95} + 22 q^{99}+O(q^{100})$$ 4 * q + 6 * q^9 + 18 * q^11 - 14 * q^19 - 12 * q^21 + 20 * q^25 + 12 * q^29 + 22 * q^31 + 8 * q^39 - 18 * q^41 + 10 * q^45 - 26 * q^49 - 10 * q^51 - 10 * q^55 - 24 * q^59 - 2 * q^61 - 2 * q^69 - 6 * q^71 - 4 * q^79 - 4 * q^81 + 12 * q^89 + 6 * q^91 - 30 * q^95 + 22 * q^99
Basis of coefficient ring in terms of a root $$\nu$$ of $$x^{4} + 3x^{2} + 1$$ :
$$\beta_{1}$$ $$=$$ $$\nu$$ v $$\beta_{2}$$ $$=$$ $$\nu^{2} + 1$$ v^2 + 1 $$\beta_{3}$$ $$=$$ $$\nu^{3} + 2\nu$$ v^3 + 2*v
$$\nu$$ $$=$$ $$\beta_1$$ b1 $$\nu^{2}$$ $$=$$ $$\beta_{2} - 1$$ b2 - 1 $$\nu^{3}$$ $$=$$ $$\beta_{3} - 2\beta_1$$ b3 - 2*b1
## Character values
We give the values of $$\chi$$ on generators for $$\left(\mathbb{Z}/1840\mathbb{Z}\right)^\times$$.
$$n$$ $$737$$ $$1151$$ $$1201$$ $$1381$$ $$\chi(n)$$ $$-1$$ $$1$$ $$1$$ $$1$$
## Embeddings
For each embedding $$\iota_m$$ of the coefficient field, the values $$\iota_m(a_n)$$ are shown below.
For more information on an embedded modular form you can click on its label.
comment: embeddings in the coefficient field
gp: mfembed(f)
Label $$\iota_m(\nu)$$ $$a_{2}$$ $$a_{3}$$ $$a_{4}$$ $$a_{5}$$ $$a_{6}$$ $$a_{7}$$ $$a_{8}$$ $$a_{9}$$ $$a_{10}$$
369.1
− 1.61803i − 0.618034i 0.618034i 1.61803i
0 1.61803i 0 −2.23607 0 1.85410i 0 0.381966 0
369.2 0 0.618034i 0 2.23607 0 4.85410i 0 2.61803 0
369.3 0 0.618034i 0 2.23607 0 4.85410i 0 2.61803 0
369.4 0 1.61803i 0 −2.23607 0 1.85410i 0 0.381966 0
$$n$$: e.g. 2-40 or 990-1000 Significant digits: Format: Complex embeddings Normalized embeddings Satake parameters Satake angles
## Inner twists
Char Parity Ord Mult Type
1.a even 1 1 trivial
5.b even 2 1 inner
## Twists
By twisting character orbit
Char Parity Ord Mult Type Twist Min Dim
1.a even 1 1 trivial 1840.2.e.c 4
4.b odd 2 1 230.2.b.a 4
5.b even 2 1 inner 1840.2.e.c 4
5.c odd 4 1 9200.2.a.bo 2
5.c odd 4 1 9200.2.a.by 2
12.b even 2 1 2070.2.d.c 4
20.d odd 2 1 230.2.b.a 4
20.e even 4 1 1150.2.a.l 2
20.e even 4 1 1150.2.a.n 2
60.h even 2 1 2070.2.d.c 4
By twisted newform orbit
Twist Min Dim Char Parity Ord Mult Type
230.2.b.a 4 4.b odd 2 1
230.2.b.a 4 20.d odd 2 1
1150.2.a.l 2 20.e even 4 1
1150.2.a.n 2 20.e even 4 1
1840.2.e.c 4 1.a even 1 1 trivial
1840.2.e.c 4 5.b even 2 1 inner
2070.2.d.c 4 12.b even 2 1
2070.2.d.c 4 60.h even 2 1
9200.2.a.bo 2 5.c odd 4 1
9200.2.a.by 2 5.c odd 4 1
## Hecke kernels
This newform subspace can be constructed as the intersection of the kernels of the following linear operators acting on $$S_{2}^{\mathrm{new}}(1840, [\chi])$$:
$$T_{3}^{4} + 3T_{3}^{2} + 1$$ T3^4 + 3*T3^2 + 1 $$T_{7}^{4} + 27T_{7}^{2} + 81$$ T7^4 + 27*T7^2 + 81
## Hecke characteristic polynomials
$p$ $F_p(T)$
$2$ $$T^{4}$$
$3$ $$T^{4} + 3T^{2} + 1$$
$5$ $$(T^{2} - 5)^{2}$$
$7$ $$T^{4} + 27T^{2} + 81$$
$11$ $$(T^{2} - 9 T + 19)^{2}$$
$13$ $$T^{4} + 7T^{2} + 1$$
$17$ $$T^{4} + 35T^{2} + 25$$
$19$ $$(T^{2} + 7 T + 1)^{2}$$
$23$ $$(T^{2} + 1)^{2}$$
$29$ $$(T^{2} - 6 T - 36)^{2}$$
$31$ $$(T^{2} - 11 T + 19)^{2}$$
$37$ $$T^{4} + 108T^{2} + 1296$$
$41$ $$(T^{2} + 9 T + 19)^{2}$$
$43$ $$T^{4} + 172T^{2} + 5776$$
$47$ $$T^{4} + 140T^{2} + 400$$
$53$ $$(T^{2} + 4)^{2}$$
$59$ $$(T + 6)^{4}$$
$61$ $$(T^{2} + T - 11)^{2}$$
$67$ $$T^{4} + 28T^{2} + 16$$
$71$ $$(T^{2} + 3 T + 1)^{2}$$
$73$ $$T^{4} + 328 T^{2} + 15376$$
$79$ $$(T^{2} + 2 T - 44)^{2}$$
$83$ $$T^{4} + 92T^{2} + 1936$$
$89$ $$(T^{2} - 6 T - 36)^{2}$$
$97$ $$T^{4} + 427 T^{2} + 43681$$
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A circle can be defined as the curve traced out by a point that moves so that its distance from a given point is constant. In an x–y Cartesian coordinate ... more
Ordinate of a point of a circle (trigonometric function)
The ordinate of point of a circle, in an x–y Cartesian coordinate system, can be computed by the ordinate of the center of the circle, the radius and the ... more
Abscissa of a point of a circle (trigonometric function)
The abscissa of point of a circle, in an x–y Cartesian coordinate system, can be computed by the abscissa of the center of the circle, the radius and the ... more
Abscissa of a point of a circle
The abscissa of point of a circle, in an x–y Cartesian coordinate system, can be computed by the abscissa of the center of the circle, the radius and the ... more
Ordinate of a point of a circle
The ordinate of point of a circle, in an x–y Cartesian coordinate system, can be computed by the ordinate of the center of the circle, the radius and the ... more
Externally Tangent Circles
Two circles of non-equal radius, both in the same plane, are said to be tangent to each other if they meet at only one point.
Two circles are ... more
Internally Tangent Circles
Two circles of non-equal radius, both in the same plane, are said to be tangent to each other if they meet at only one point.
Two circles are ... more
Uniform Circular Motion position (Y - coordinate)
In physics, circular motion is a movement of an object along the circumference of a circle or rotation along a circular path. It can be uniform, with ... more
Uniform Circular Motion position (X - coordinate)
In physics, circular motion is a movement of an object along the circumference of a circle or rotation along a circular path. It can be uniform, with ... more
Area of a deltoid
In geometry, a deltoid, also known as a tricuspoid or Steiner curve, is a hypocycloid of three cusps. In other words, it is the roulette created by a point ... more
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http://stackoverflow.com/questions/4475991/sorting-an-array-based-on-an-attribute-that-may-be-nil-in-some-elements/4476027
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# Sorting an array based on an attribute that may be nil in some elements
I have an array of objects
``````[<#a star=1 val=1>, <#a star=nil val=3> , <#a star=2 val=2>]
``````
i need the array to be sorted by time, then by val
``````[ <#a star=2 val=2>, <#a star=1 val=1>, <#a star=nil val=3> ]
``````
but using the sort_by throws an error because the time is nil.
I am using an ugly way to sort right now, but i am sure there is a nice way to go about it
``````starred=[]
@answers.each {|a| (starred << a) if a.starred }
starred=starred.sort_by {|a| a.starred }.reverse
``````
-
``````starred.sort_by { |a| [a ? 1 : 0, a] }
``````
When it has to compare two elements, it compares an arrays. When Ruby compares arrays (calls `===` method), it compares 1st element, and goes to the 2nd elements only if the 1st are equal. `? 1 : 0` garantees, that we'll have Fixnum as 1st element, so it should be no error.
If you do `? 0 : 1`, `nil` will appear at the end of array instead of begining.
Here is an example:
``````irb> [2, 5, 1, nil, 7, 3, nil, nil, 4, 6].sort_by { |i| [i ? 1 : 0, i] }
=> [nil, nil, nil, 1, 2, 3, 4, 5, 6, 7]
irb> [2, 5, 1, nil, 7, 3, nil, nil, 4, 6].sort_by { |i| [i ? 0 : 1, i] }
=> [1, 2, 3, 4, 5, 6, 7, nil, nil, nil]
``````
-
`@answers.sort_by { |a| a.star or 0 }`
`@answers.select(&:starred?).sort_by(&:star) + @answers.reject(&:starred?)`
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http://mathoverflow.net/feeds/question/104948
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Distribution of Maximum of a uniform multinomial distribution - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-21T16:29:31Z http://mathoverflow.net/feeds/question/104948 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/104948/distribution-of-maximum-of-a-uniform-multinomial-distribution Distribution of Maximum of a uniform multinomial distribution Jack 2012-08-17T22:14:40Z 2012-08-18T00:21:00Z <p>Hello, I'm working with a data structure which uses a uniform distribution to bucket the inputs into $k$ buckets. The efficiency of the structure is bounded by the $\frac{k_{max}}n$, where $n$ is the number of items. How many elements are in each bucket will follow a uniform multinomial distribution. What will the distribution be for the number in the largest bin. We can assume that $n$ is much larger than $k$, and an approximate answer is good. I just want to be able to say something like: the largest bucket will have at most $(1.5k)/n$ elements with probability $p$.</p> http://mathoverflow.net/questions/104948/distribution-of-maximum-of-a-uniform-multinomial-distribution/104954#104954 Answer by tergi for Distribution of Maximum of a uniform multinomial distribution tergi 2012-08-17T23:08:59Z 2012-08-17T23:08:59Z <p>This is addressed by Bruce Levin, 1983, "On Calculations Involving the Maximum Cell Frequency."</p> <p>Also in <a href="http://www.jstor.org/stable/2347220" rel="nofollow">http://www.jstor.org/stable/2347220</a> .</p> http://mathoverflow.net/questions/104948/distribution-of-maximum-of-a-uniform-multinomial-distribution/104955#104955 Answer by Douglas Zare for Distribution of Maximum of a uniform multinomial distribution Douglas Zare 2012-08-17T23:19:36Z 2012-08-18T00:21:00Z <p>The probability that there is at least one bin with at least $c$ items is less than or equal to the expected number of bins with at least $c$ items, which is $k$ times the probability that a particular bin has at least $c$ items. You can bound the probability that a particular bin contains at least $c$ items using the <a href="http://en.wikipedia.org/wiki/Hoeffding%27s_inequality" rel="nofollow">Hoeffding inequality</a>.</p> <p>$$\begin{eqnarray}Pr(\max \ge n/k + d) & \le & k Pr(\text{Binomial}(n,1/k) \ge n/k + d) \\ & \le & k \exp(-2d^2/n).\end{eqnarray}$$</p> <p>There are sharper bounds available such as the <a href="http://en.wikipedia.org/wiki/Chernoff_bound" rel="nofollow">Chernoff bound</a>, but this is simple and it sounds like it will suffice.</p>
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# Geometry
posted by .
A rhombus has diagonals of length 4 and 10. Find the angles of the rhombus to the nearest degree.
would the angles be 136 and 44?
• Geometry -
no
• Geometry -
well this is what i did can someone tell me what i did wrong:
tanx= 5/2
since diagonals of a parallelogram
bisect each other
x= 68 and the other angle would be 22
the angles of the rhombus would be 136 and 44 because the diagonals of a rhombus bisect the angles
• Geometry -
You are correct.
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https://magoosh.com/toefl/2015/nouns-of-measurement-and-non-countable-nouns/
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offers hundreds of practice questions and video explanations. Go there now.
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# Nouns of Measurement and Non-Countable Nouns
A noun of measurement is a noun that describes a unit of measurement or a quantity of something.
In either English or your native language, you probably use nouns of measurement all the time. You may tell your teacher that you have a number of questions. In this case “a number” is a noun phrase of measurement used to describe the quantity of questions you have. Or you may go grocery shopping for food that costs a certain amount of money per pound or kilogram. In that situation, you might want to measure the weight of the food on a supermarket scale. Here, “weight” is a noun of measurement that describes the amount of food you’ll pay for.
In the examples above, you can see that “number” is a noun of measurement that can really only be used to describe countable nouns. If the quantity of something can be described in specific numbers such as 1, 2, 3, etc…. then it must be something that can be counted, right?
Other numerical nouns of measurement also are exclusive to countable nouns—you can buy a dozen eggs, but you can’t use the word “dozen” to refer to the amount of water in a bottle. And you can talk about a group of animals, but you can’t describe a “group” of air.
Similarly, there are nouns of measurement that almost always refer to certain types of non-countable nouns. As seen above, you talk about the weight of non-countable things such as hamburger, sugar, etc…You can also talk about the length of a non-countable noun such as hair or fabric.
Words like weight and length can only refer to countable nouns in a very limited way. They can describe the characteristics of an in individual countable noun—you can talk about a gorilla with a weight of 500 pounds, or a pencil with a length of 16 centimeters. But you wouldn’t look at a group of gorillas and call it a “weight” of gorillas, or describe a box of pencils as a “length” of pencils.
Still, there are some other nouns of measurement that can easily describe either countable or non-countable nouns in the same way. There can be a quantity of apples or a quantity of apple juice. You can have a certain amount of cars or a certain amount of car fuel.
So far we’ve been looking at nouns of measurement that describe material things that can be touched and seen. But nouns of measurement can measure both material non-countable nouns and non-countable nouns that can’t actually be touched or seen.
One good example of this is the non-countable noun “volume.” While volume is most often a measurement of material things and the amount of space they take up, such as a bucket that can hold a volume of 3,000 square centimeters, volume also describes one non-countable noun. The non-material noun of sound can be measured by volume, too. A very loud sound has a higher volume than a quieter sound.
Other nouns of measurement pretty much only measure immaterial non-countable nouns. Intensity measures immaterial things such as pressure or heat. Comprehensibility measures the non-countable noun of language. (Language has more comprehensibility if it can be understood very well, less comprehensibility if it’s hard to understand.)
Understanding exactly which noun is being modified by an additional noun of measurement is a powerful TOEFL skill. It can help you understand which important nouns from a reading passage or lecture are being referred to and described in a given sentence. Let’s look at an example where this can come in handy. Here are two sentences taken from a Magoosh TOEFL Premium Reading passage (consider this to be a tiny free preview):
One must be careful not to come too close to the most powerful of cicadas, whose sound, if heard from just outside the human ear, is known to cause long-term or irreversible physical damage. That intense volume is paired with an uncanny ability to maintain the signature song.
Here, you could easily get confused about what volume is describing. The first sentence has so many different nouns in it! It could be hard to guess whether “volume” refers to the cicadas, or to their sound, or to the physical damage their sound can sometimes cause.
But if you understand nouns of measurement well, you can recognize that volume only refers to non-countable nouns that are material, and sound. Cicadas are countable, and damage is non-countable and non-material. Although damage has an effect on physical objects, damage itself has no specific, clearly defined physical features. So you can know that volume must refer only to the sound of the cicada, and you can easily understand the meaning of a passage that might otherwise be very hard.
Understanding nouns of measurement what they can refer to is an important language ability to have. In my next post on this subject, we’ll do some activities that help you master this valuable skill.
### 2 Responses to Nouns of Measurement and Non-Countable Nouns
1. garfield September 16, 2018 at 12:33 am #
this is great stuff
• Magoosh Test Prep Expert September 16, 2018 at 1:51 pm #
Glad you like it 🙂
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# Ôn t p h th ng các công th c tính nguyên hàm
Ôn thi TN-THPT
CÔNG TH C TÍNH TÍCH PHÂN
1)
∫ k.dx = k.x + C
2)
n ∫ x dx =
x n+1 +C n +1
3)
∫x
1
2
dx = −
1
1 +C x
dx = − 1 +C; a (n − 1)(ax + b) n −1
4)
∫ x dx = ln x + C
1
5)
∫ (ax + b)
n
6)
∫ (ax + b) dx = a ln ax + b + C
∫ cos x.dx = sin x + C
1
1
7)
∫ sin x.dx = − cos x + C
8)
9)
∫ sin(ax + b)dx = − a cos(ax + b) + C ∫ cos
1
2
1
10)
∫ cos(ax + b)dx = a sin(ax + b) + C ∫ sin
∫ sin ∫e
1
2
1
11)
x
dx = ∫ (1 + tan 2 x).dx = tan x + C 1
12)
x
dx = ∫ (1 + cot 2 x ) dx = − cot x + C
13)
∫ cos (ax + b) dx = a tan(ax + b) + C
2
1
14)
2
1 1 dx = − cot(ax + b) + C (ax + b) a
15)
∫e
x
dx = e x + C 1 ( ax +b ) e +C a
16)
−x
dx = −e − x + C 1 (ax + b) n +1 . + C (n ≠ 1) a n +1
17) 19)
( ax + b ) ∫ e dx =
18) 20)
n ∫ (ax + b) .dx =
x ∫ a dx =
ax +C ln a
∫x
∫x
2
1 dx = arctgx + C +1 1 1 x dx = arctg + C 2 a a +a
1 dx = arcsin x + C
21)
∫x ∫x
2
1 1 x −1 dx = ln +C 2 x +1 −1 1 1 x−a dx = ln +C 2 2a x + a −a
1 dx = arcsin x +C a
22)
2
23)
2
24)
∫ ∫
1− x2 1 x ±1
2
25)
∫ ∫
a −x
2
2
26)
dx = ln x + x 2 ± 1 + C
27)
1 x2 ± a2
dx = ln x + x 2 ± a 2 + C
28)
a 2 − x 2 dx =
x a2 x a2 − x2 + arcsin + C 2 2 a
29)
x 2 ± a 2 dx =
x 2 a2 x ± a2 ± ln x + x 2 ± a 2 + C 2 2
Biên so n: Nguy n Phan Anh Hùng -1-
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# Pell’s Equation – Phương trình Pell
Pell’s equation seems to be an ideal topic to lead college students, as well as some talented and motivated high school students, to a better appreciation of the power of mathematical technique. The history of this equation is long and circuituous. It involved a number of different approaches before a definitive theory was found. Numbers have fascinated people in various parts of the world over many centuries. Many puzzles involving numbers lead naturally to a quadratic Diophantine equation (an algebraic equation of degree 2 with integer coefficients for which solutions in integers are sought), particularly ones of the form $[Hình: latex.php?latex=x^2-dy^2=k&...mp;amp;s=0]$, where d and k are integer parameters with d nonsquare and positive. A few of these appear in Chapter 2. For about a thousand years, mathematicians had various ad hoc methods of solving such equations, and it slowly became clear that the equation $[Hình: latex.php?latex=x^2-dy^2=1&...mp;amp;s=0]$ should always have positive integer solutions other than (x, y) = (1, 0). There were some partial patterns and some quite effective methods of finding solutions, but a complete theory did not emerge until the end of the eighteenth century. It is unfortunate that the equation is named after a seventeenth-century English mathematician, John Pell, who, as far as anyone canb tell, had hardly anything to do with it. By his time, a great deal of spadework had been done by manyWestern European mathematicians. However, Leonhard Euler, the foremost European mathematician of the eighteenth century, who did pay a lot of attention to
the equation, referred to it as “Pell’s equation” and the name stuck.
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Question 7
# Mr. Madhukar worked for 5 years in a multi-commodity trading company after graduating from areputed B-School. He then resigned from his job and started an online garment export business. Forhis business Madhukar used Rs. 8,00,000/- of his own savings and borrowed Rs. 12,00,000/- from aprivate sector bank in April, the beginning month of the financial year. Mean while, RBI eased reporate in May and banks passed on the benefits to the borrowers. As a result, Madhukar borrowed anadditional Rs. 9,00,000/- after 4 months at an interest rate which is 10% lower than the interest rate ofhis earlier borrowing. If the total interest paid by Madhukarat the end of that year on both the loans isRs. 1,39,200/-, what is the interest rate per annum on first borrowing ?
Solution
Let the rate of interest of first borrowing be R% p.a.Â
The principal of first borrowing is kept for an entire 12 month span.
Hence, simple interest payable on it at end of financial year = (12,00,000*1*R)/100 = 12,000*R rupees
Now, rate of second borrowing = 0.9R % p.a.Â
The principal of second borrowing is kept for 8 months total till financial year end.Â
Hence, simple interest payable on it at end of financial year = (9,00,000*R*8)/(12*100) = 5400R rupees
Now, 12000R + 5400R = 1,39,200
Hence, R = 8% p.a.Â
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# Is there a service that specializes in MATLAB tasks related to statistical analysis?
I don’t use this. And I used a function called myplot which knows from data spreadsheet where the function/regression is performed to see if any of that value was assigned or not. You had to understand this before you might just want to get it working with your own code. As I said, you might want your analysis done in order to test official statement performance of your analysis but I think you will find this the most interesting. Also for the function and correlation we have the choice to use this approach. So in this case we have: (1) _the_ regression_, _the_ analysis_, _the_ test of the function_, and _the_ confidence_, and _the_ correlation_ (2) myplot, myplot, kcayra, dendron_, and some time_ Then these have to do with the testing of the utility of your regression that will be done in order to make sure what is “doing ok in practice”. Given example (1), you would like to make sure that the regression does not miss some points or “trying to fit a model would miss these”. Then you could: (3) _what_ should I do to estimate the correlation_ When you are ready, the following might be useful: (4) _how_ will I find out$\exp \left$, mydata_ \leftarrow$myregression_ \leftarrow$\mydata$, and$q_t$_ = myregression$\leftarrow$cost In this way you can create an R script and create a function called top article (input) with the value x = \_\_ %w\ y = \_\_ 1 2 \_\_ %h\ howdy=\$\$%w\ =\_\_ [max\ max/(max=y)%h\ ]. The goal here is merely to get a smooth line that we can take to make a
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# 7.S: Models of Discrete Character Evolution (Summary)
In this chapter I have described the Mk model, which can be used to describe the evolution of discrete characters that have a set number of fixed states. We can also elaborate on the Mk model to allow more complex models of discrete character evolution (the extended-Mk model). These models can all be used to simulate the evolution of discrete characters on trees.
In summary, the Mk and extended Mk model are general models that one can use for the evolution of discrete characters. In the next chapter, I will show how to fit these models to data and use them to test evolutionary hypotheses.
## Section 7.7: Footnotes
1: Imagine that you calculate a rate of character change by counting the number of changes of state of a character over some time interval, t. You can calculate the rate of change as nchanges/t. The instantaneous rate is the value that this rate approaches as t gets smaller and smaller so that the time interval is nearly zero.
2: I will not cover the details of matrix exponentiation here – interested readers should see Yang (2006) for details – but the calculations are not trivial.
3: One still obtains the relevant row from P(t) and draws a uniform random deviate u. Imagine that we have a ten-state character with states 0 - 9. We start at state 0 at the beginning of the simulation. Again using q = 0.5 and t = 3, we find that:
(eq. 7.17)
$$\begin{array}{l} p_{ii}(t) = \frac{1}{k} + \frac{k-1}{k} e^{-kqt} = \frac{1}{10}+\frac{9}{10}e^{-2 \cdot 0.5 \cdot 3} = 0.145\\ p_{ij}(t) = \frac{1}{k} - \frac{1}{k} e^{-kqt} \frac{1}{10} - \frac{1}{10}e^{-2 \cdot 0.5 \cdot 3} = 0.095\\ \end{array}$$
We focus on the first row of P(t), which has elements:
$$\begin{bmatrix} 0.145 & 0.095 & 0.095 & 0.095 & 0.095 & 0.095 & 0.095 & 0.095 & 0.095 & 0.095 \\ \end{bmatrix}$$
We calculate the cumulative sum of these elements, adding them together so that each number represents the sum of itself and all preceding elements in the vector:
$$\begin{bmatrix} 0.145 & 0.240 & 0.335 & 0.430 & 0.525 & 0.620 & 0.715 & 0.810 & 0.905 & 1.000 \\ \end{bmatrix}$$
Now we compare u to the numbers in this cumulative sum vector. We select the smallest element that is still strictly larger than u, and assign this character state for the end of the branch. For example, if u = 0.475, the 5th element, 0.525, is the smallest number that is still greater than u. This corresponds to character state 4, which we assign to the end of the branch. This last procedure is a numerical trick. Imagine that we have a line segment with length 1. The cumulative sum vector breaks the unit line into segments, each of which is exactly as long as the probability of each event in the set. One then just draws a random number between 0 and 1 using a uniform distribution. The segment that contains this random number is our event.
## References
Brandley, M. C., J. P. Huelsenbeck, and J. J. Wiens. 2008. Rates and patterns in the evolution of snake-like body form in squamate reptiles: Evidence for repeated re-evolution of lost digits and long-term persistence of intermediate body forms. Evolution. Wiley Online Library.
Felsenstein, J. 2004. Inferring phylogenies. Sinauer Associates, Inc., Sunderland, MA.
Hedges, S. B., J. Dudley, and S. Kumar. 2006. TimeTree: A public knowledge-base of divergence times among organisms. Bioinformatics 22:2971–2972.
Lewis, P. O. 2001. A likelihood approach to estimating phylogeny from discrete morphological character data. Syst. Biol. 50:913–925.
Moch, J. G., and P. Senter. 2011. Vestigial structures in the appendicular skeletons of eight African skink species (squamata, scincidae). J. Zool. 285:274–280. Blackwell Publishing Ltd.
Pagel, M. 1994. Detecting correlated evolution on phylogenies: A general method for the comparative analysis of discrete characters. Proc. R. Soc. Lond. B Biol. Sci. 255:37–45. The Royal Society.
Paradis, E., J. Claude, and K. Strimmer. 2004. APE: Analyses of phylogenetics and evolution in R language. Bioinformatics 20:289–290. academic.oup.com.
Pianka, E. R., L. J. Vitt, N. Pelegrin, D. B. Fitzgerald, and K. O. Winemiller. 2017. Toward a periodic table of niches, or exploring the lizard niche hypervolume. Am. Nat. 190:601–616. journals.uchicago.edu.
Posada, D. 2008. jModelTest: Phylogenetic model averaging. Mol. Biol. Evol. 25:1253–1256. academic.oup.com.
Streicher, J. W., and J. J. Wiens. 2017. Phylogenomic analyses of more than 4000 nuclear loci resolve the origin of snakes among lizard families. Biol. Lett. 13. rsbl.royalsocietypublishing.org.
Vitt, L. J., E. R. Pianka, W. E. Cooper Jr, and K. Schwenk. 2003. History and the global ecology of squamate reptiles. Am. Nat. 162:44–60. journals.uchicago.edu.
Yang, Z. 2006. Computational molecular evolution. Oxford University Press.
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# How do mechanical processes depend from the speed of light?
1. Apr 17, 2012
### Aidyan
For example, as well known the period of the pendulum is (in linear approximation):
$T \approx 2\pi \sqrt\frac{L}{g} \,.$
So, no speed of light appears explicitly. What I'm wondering however is if and how it might be implicit? In the sense that after all the tension in the rod depends from molecular forces, which at the microscopic scale are of electric nature. Would a different speed of light than c lead to a different length? Another purely mechanical example: hitting a nail in a wall. In a universe with, say light speed 0.7*c, would it become more or less easier to hit the same nail in the same wall? These are only an example of a more general question. In what way is classical non relativistic mechanics determined by the speed of light, if it does?
2. Apr 17, 2012
### Naty1
In classical mechanics 'g' is simply the acceleration of any freely falling body....a variable itself....and that's the assumption in your pendulum formula. Whether the speed of light is 'c' or 2c or or 1/2c whatever has virtually no effect on a clock pendulum.
It's analogous to the simple addition of two velocities at low speeds.....which works fine....rather than the 'exact solution'....As you approach speeds which are a significant proportion of 'c' however, you need to take finer grained details into account...that's relativistic mechanics.
I'm not exactly sure about how the electromagnetic force that binds electons to nuclei, or the strong force that binds atom nuclei together, for example, might vary if the speeds of propagation were to vary from 'c'....
3. Apr 17, 2012
### AlephZero
Non relativistic mechanics ignores the speed of light.
That is usually a very good approximation, cosidering that if your mathematical description of a pendulum, knocking in a nail, or whatever, includes the elasticity of the material, any mechanical effect only propagates through the body at the speed of sound in the material (typically a few km/s for metals), not at the speed of light.
4. Apr 17, 2012
### Aidyan
I'm not convinced. Also in classical mechanics the elasticity of the material depends somehow from the molecular and atomic lattice structure and which solidity in turn is determined by the inter-atomic bonds. These bonds are of electromagnetic nature and therefore I imagine them dependent from c. So the speed of sound must depend from c too.
5. Apr 17, 2012
### AlephZero
You can imagine whatever you like, but that doesn't make it so.
6. Apr 17, 2012
### sophiecentaur
There is a logic in what he says, surely: in principle, even if it's not quantitatively very significant. The time taken for molecules to 'rebound' off each other will depend upon the photon interaction between them. But the effective difference between that and 'billiard balls' collisions may be slight. The velocities of sound and light differ by a factor of around a million.
7. Apr 17, 2012
### rbj
$c$ is not simply the speed of light (or of the electromagnetic interaction). it is the speed of any ostensible "instantaneous" interaction. including the strong nuclear force and including gravitation.
so, if the question is: "why does a mechanical clock, instead of a light clock, appear to slow down due to time dilation as it whizzes by an observer at high speed?" then the answer is that the constancy of $c$ does not only cause time dilation, it affects length contraction and apparent momentum of mechanical parts.
8. Apr 18, 2012
### Aidyan
Ok, that might be true in an idealized gas of free particles, but reality is a bit more complex. Especially in a solid lattice structure (e.g. a metal rod for the pendulum example) the resistance to deformation (and therefore its length variation due to gravitational pull) is determined by the chemical bonds. In a simplified model we can visualize a chemical bond between two point masses with a spring governed by Hook's law: F=-k*dx, where dx is the displacement from force equilibrium, and k the Hook's constant. This sounds all very mechanical an independent from the speed of light. But truth is that a chemical bonding (ionic or covalent) is determined by the electron density, bond length, its energetic stabilization, etc., all electric phenomena which in turn must dependent from c somehow. That is, further analysis must show that k=k(c). The question is not if this is the case but to what degree it is. I suspect that the bond strengths depends significantly from the value of the speed of light, and therefore also all the other mechanical properties. I looked up for some formula on the web but couldn't find much.
Yes, precisely. A reason more to suspect that also "g" in the pendulum formula must depend from the speed of light.
As far as I understand SR everything slows down, also biological aging.
Anyhow, rbj, I was expecting you to tell us that since c is a dimensional constant the question then is meaningless, since nothing would change if the speed of light changes....
Last edited: Apr 18, 2012
9. Apr 18, 2012
### sophiecentaur
Precisely. It's always more complicated than that. One of the reasons is that structures rely on photons to keep them stiff and gases rely on photons to establish 'pressure'. Any photon interaction takes time,which depends upon c.
But, there again,everything 'depends on c'.
10. Apr 18, 2012
### Staff: Mentor
If you change c, but not anything else (in SI units), you change the fine-structure constant and therefore the strength of the electromagnetic interaction, which changes the binding energy and size of atoms. While this can be absorbed with scalings of length and energy for small atoms, for large atoms you mess around with the fine-structure and might get a different periodic system, if the modification is large enough.
Your ideal gas will not change in a significant way, but a solid or liquid material can change its properties a lot.
11. Apr 18, 2012
### sophiecentaur
If an atom comes across another atom, it will only 'bounce off' because a a photon interaction of some kind. (It's not gravitational or nuclear forces at work). The system we would be dealing with would not just be one atom any more. I guess you could call it a "fine structure" effect because the energy levels of each atom would be modified by the proximity of the other atom. As you say, it would be a very small effect but the same basic thing would apply with two isolated atoms or with a set of inter-linked atoms. It would still take d/c time for the electric forces to start to work. (d being some effective distance)
Questions like this one sometimes end up with needing to go to unreasonable depths to get a result that satisfies everyone because we all have our own basic model in our heads.
The idea of 'changing c' is really a bit of a non-starter because everything hangs on it. However, you can say that many problems can be solved quite satisfactorily without considering it. That goes for most 'mechanical' ones.
12. Apr 18, 2012
### Staff: Mentor
This bounce between gas atoms/molecules is elastic and does not need any time in an ideal gas, therefore the details of the bounce do not matter. It could be any force, even gravity.
13. Apr 18, 2012
### sophiecentaur
Umm. Just becaust no energy is lost, it doesn't mean that the time taken is zero or irrelevant. You have introduced a mechanical concept so I will reply with one. The time constant of the spring/mass system would have an effect on the bounce time.
14. Apr 18, 2012
### Bob S
15. Apr 18, 2012
### Staff: Mentor
Therefore, I highlighted ideal gas. The ideal thing about it is that the interactions are assumed to be small, elastic and do not require any time.
A real gas will change its properties of course. But gases are much more "robust" in that sense than solids or liquids.
16. Apr 18, 2012
### Aidyan
That's precisely the thing that isn't sure at all. Please read wiki on Planck units, paragraph "Planck units and the invariant scaling of nature".
17. Apr 18, 2012
### sophiecentaur
In practice but not in principle (or do I mean in principle not in practice?) - as an 'ideal' gas doesn't exist. The behaviour of any real gas will be a bit of a slave to c so we are only arguing about a matter of degree.
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## Teen Take Away.m4v - Section 3: Math Workshop
Teen Take Away.m4v
Teen Take Away.m4v
Unit 10: Working with Numbers, Operations, and Story Problems
Lesson 6 of 13
## Objective: SWBAT use a known strategy to solve an addition story problem. SWBAT record solutions to problems.
### Thomas Young
203 Lessons14 new
## Big Idea: Students will again demonstrate what know strategy they independently use to solve an addition story problem.
Print Lesson
2 teachers like this lesson
Standards:
Subject(s):
Math, Number Sense and Operations, Operations , 1st Grade, story problems, addition and subtraction strategies
70 minutes
### Thomas Young
203 Lessons | 14 new
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1. (AB) = (BA)^t
Hello,
This may be a long shot, but does anyone know anything that can be said of square matrices A and B with the property above? A and B are block components of a larger matrix that I'm trying to find the determinant of. Simultaneously diagonalizing them would be my dream come true, but any information would be helpful.
Thanks.
2. Re: (AB) = (BA)^t
If A and B are symmetric, then AB = (BA)^T. Can you say that A and B are symmetric?
3. Re: (AB) = (BA)^t
Originally Posted by Wyaltster
Hello,
This may be a long shot, but does anyone know anything that can be said of square matrices A and B with the property above? A and B are block components of a larger matrix that I'm trying to find the determinant of. Simultaneously diagonalizing them would be my dream come true, but any information would be helpful.
Thanks.
Suppose AB is diagonalized. Assuming $A^{-1}$ exists, and AB is diagonalized as $AB = P^{-1}DP$. Now,
$B = A^{-1} P^{-1}DP$
$BA=A^{-1} P^{-1}D(PA)$
$BA=(PA)^{-1}D(PA)$
$(BA)^T=((PA)^{-1}D(PA))^T = (PA)^T D ((PA)^T)^{-1}$
Since $(A^{-1})^T = (A^T)^{-1}$ and $(AB)^T = B^T A^T$.
4. Re: (AB) = (BA)^t
Thanks, both of you.
mathguy, they're both skew-symmetric, so bingo
I've been seeing this a lot because I've been dealing with a lot of skew-symmetric matrices. Feel like a bonehead for not seeing that.
majamin, thanks, that's interesting. I might have been confusing when I said simultaneously diagonalizable. I'd want to simultaneously diagonalize $A$ and $B$, not $AB$ and $(BA)^t$, but I think what I want to do can't be done. There's a theorem that says that a set of diagonalize matrices commute if and only if they can be diagonalized by a common matrix. That's not applicable to my case, but I was hoping maybe something could be said about $A$ and $B$ as described above. Turns out I can say they're skew-symmetric.
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Feeds:
Posts
## Atmospheric Circulation – Part Two – Thermal Wind
In Atmospheric Circulation – Part One we saw how the pressure “slopes down” from the tropics to the poles creating S→N winds in the northern hemisphere.
In The Coriolis Effect and Geostrophic Motion we saw that on a rotating planet winds get deflected off to the side (from the point of view of someone on the rotating planet). This means that winds flowing from the tropics to the north pole will get deflected “to the right”.
### Taylor Columns
Strange things happen to fluids in rotating frames. To illustrate let’s take a look at Taylor columns.
From Marshall & Plumb (2008)
Figure 1
The static image is quite beautiful, but the video illustrates it better. Compare the video of the non-rotating tank with the rotating tank.
Now to stretch the mind we have a rotating tank with an obstacle on the base – in this case a hockey puck. The height of the puck is small compared with the depth of the fluid. The fluid flow has come into equilibrium with the tank rotation.
We slow down the rotation slightly. We sprinkle paper dots on the surface of the water. Amazingly the dots show that the surface of the fluid is acting as if the puck extended right up to the surface – the flow moves around the obstacle at the base (of course) and the flow moves “around” the obstacle at the surface. Even though the obstacle doesn’t exist at the surface!
Take a look at the video, but here are a few snapshots:
Figure 2
This occurs when:
• the flow is slow and steady
• friction is negligible
• there is no temperature gradient (barotropic)
Under the first two conditions the flow is geostrophic which was covered with examples in The Coriolis Effect and Geostrophic Motion.
And under the final condition, with no temperature gradient the density is uniform (only a function of pressure).
### “Thermal Wind”
Now let’s look at an experiment with a “cold pole” and “warm tropics”:
From Marshall & Plumb (2008)
Figure 3
The result:
Figure 4
Even better – take a look at the video.
This experiment shows that once there is a N-S temperature gradient the E-W winds increase with altitude.
Which is kind of what we find in the real atmosphere:
From Marshall & Plumb (2008)
Figure 5
Why does this happen? I found it hard to understand conceptually for a while, but it’s actually really simple:
From Stull (1999)
Figure 6
So the ever increasing pressure gradient with height (due to the temperature gradient) induces a stronger geostrophic wind with height.
Here is an instantaneous measurement of E-W winds, along with temperature in a N-S section:
From Marshall & Plumb (2008)
Figure 8
The measurement demonstrates that the change in E-W wind vs height depends on the variation in N-S temperature.
The equation for this effect for the E-W winds can be written a few different ways, here is the easiest to understand:
∂u/∂z = (αg/f) . ∂T/∂y
where ∂u/∂z = change in E-W wind with height, α = thermal coefficient of expansion of air, g = acceleration due to gravity, f = coriolis parameter at that latitude, T = temperature, y = N-S direction
It can also be written in vector calculus notation:
u/∂z = (αg/f)z x ∇T
where u = wind velocity (u, v, w), = unit vector in vertical
In the next article we will look at why the maximum effect in the average, the jet stream, occurs in the subtropics rather than at the poles.
### References
Meteorology for Scientists and Engineers, Ronald Stull, 2nd edition – Free (partial) resource
Atmosphere, Ocean and Climate Dynamics – An Introductory Text, Marshall & Plumb, Academic Press (2008)
### 11 Responses
1. […] « Temperature Profile in the Atmosphere – The Lapse Rate Atmospheric Circulation – Part Two – Thermal Wind […]
2. SoD,
The caption of your Figure 8 (Figure 7.21 of Marshall and Plumb) refers erroneously to potential temperature while the values and thick contours present actual temperatures rather than potential temperatures. That’s not your error but may be confusing. You can find the same image with the correct caption as a Powerpoint file from the publishers web page:
http://www.elsevierdirect.com/v2/companion.jsp?ISBN=9780125586917
Check the ppt of Chapter 07.
(Similarly I think that the missing molar mass in the formula of your previous post may be confusing although it’s not essential for understanding your points. As the equation stands the units don’t match. Thus anyone trying to reproduce the formula may have unnecessary difficulties.)
• Pekka,
Well spotted. I updated the graphic for figure 8.
• On the missing molar mass, I have now clarified the equation in a comment.
R = gas constant for dry air; and is in units of J kg-1K-1
3. SoD,
I think that some further arguments could be added to explain your formulas. Now you don’t really present any justification for them. You tell the background but skip over the final essential step. Going through the full derivation is perhaps too much but the following basic idea behind the formulas can be given:
A possible situation and the one actually observed is that rather stable geostrophic winds persist. To have such a wind something must balance the force created by the pressure gradient of your Figure 6. The only horizontal “force” available is the Coriolis force. Your formulas represent the requirement that the “pressure force” and the Coriolis force are equal in size and have opposite directions.
• Pekka,
Good idea.
Refer to The Coriolis Effect and Geostrophic Motion for the geostrophic wind equations:
ug = – 1/fρ . ∂p/∂y ….[4a]
vg = 1/fρ . ∂p/∂x ….[4b]
where ug = W-E wind speed, vg = N-S wind speed
These equations are derived from:
1. 1. Newton’s law of motion (net force = change in momentum) with the net force being the result of pressure gradients, gravity and friction
2. 2. These laws then adapted to a rotating frame of reference which results in two additional forces “appearing” in the equation: the coriolis force and the centrifugal force
3. 3. The approximation of low values of acceleration and friction (geostrophic approximation)
So from the equations we see that the E-W wind is proportional to the change in N-S pressure, and the N-S wind is proportional to the E-W change in pressure.
It is more convenient to consider the height of a pressure surface rather than the pressure at a given height.
Using the hydrostatic balance equation it’s easy to show that:
(∂p/∂x)z = gρ (∂z/∂x)p ….[5a]
(∂p/∂y)z = gρ (∂z/∂y)p ….[5b]
where (∂p/∂x)z is the change in pressure (at constant height) in the x direction and (∂z/∂x)p is the change in the height of a pressure surface (geopotential height) in the x direction
So combining 4 & 5 we get:
ug = –g/f . ∂z/∂y ….[6a]
vg = g/f . ∂z/∂x ….[6b]
This version of the equation doesn’t include density, ρ, and given that density varies in a compressible fluid like air this is a very helpful version of the equation. It is one reason why pressure is often preferred over height as the vertical coordinate.
And back to the thermal wind equation – if we differentiate equation 4 with respect to height we get the equation shown in the article although the derivation takes a little time to write out.
4. SoD.
Here you have a great resource on this topic.
http://kiwi.atmos.colostate.edu/group/dave/at605.html
5. […] 2012/08/25: TSoD: Atmospheric Circulation — Part Two — Thermal Wind […]
6. on August 28, 2012 at 7:58 am | Reply Sceptical Wombat
Jon
Thanks for that – it looks like the sort of thing I have been looking for. The following sentence however is a good reminder of how much things have been changing.
“The Arctic ocean is ice-covered all year, while the North Atlantic and the Southern Oceans experience seasonal melting.”
That was then, this is now.
7. SoD,
This is very impressive work , my friend.
8. […] Comments « Atmospheric Circulation – Part Two – Thermal Wind […]
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Easy Ciphers Tools:
Caesar cipher
Caesar cipher, is one of the simplest and most widely known encryption techniques. The transformation can be represented by aligning two alphabets, the cipher alphabet is the plain alphabet rotated left or right by some number of positions.
When encrypting, a person looks up each letter of the message in the 'plain' line and writes down the corresponding letter in the 'cipher' line. Deciphering is done in reverse.
The encryption can also be represented using modular arithmetic by first transforming the letters into numbers, according to the scheme, A = 0, B = 1,..., Z = 25. Encryption of a letter x by a shift n can be described mathematically as
Plaintext: tustain
cipher variations: uvtubjo vwuvckp wxvwdlq xywxemr yzxyfns zayzgot abzahpu bcabiqv cdbcjrw decdksx efdelty fgefmuz ghfgnva highowb ijhipxc jkijqyd kljkrze lmklsaf mnlmtbg nomnuch opnovdi pqopwej qrpqxfk rsqrygl strszhm
Decryption is performed similarly,
(There are different definitions for the modulo operation. In the above, the result is in the range 0...25. I.e., if x+n or x-n are not in the range 0...25, we have to subtract or add 26.)
Atbash Cipher
Atbash is an ancient encryption system created in the Middle East. It was originally used in the Hebrew language.
The Atbash cipher is a simple substitution cipher that relies on transposing all the letters in the alphabet such that the resulting alphabet is backwards.
The first letter is replaced with the last letter, the second with the second-last, and so on.
An example plaintext to ciphertext using Atbash:
Plain: tustain Cipher: gfhgzrm
Baconian Cipher
To encode a message, each letter of the plaintext is replaced by a group of five of the letters 'A' or 'B'. This replacement is done according to the alphabet of the Baconian cipher, shown below.
```a AAAAA g AABBA m ABABB s BAAAB y BABBA
b AAAAB h AABBB n ABBAA t BAABA z BABBB
c AAABA i ABAAA o ABBAB u BAABB
d AAABB j BBBAA p ABBBA v BBBAB
e AABAA k ABAAB q ABBBB w BABAA
f AABAB l ABABA r BAAAA x BABAB
```
Plain: tustain Cipher: BAABA BAABB BAAAB BAABA AAAAA ABAAA ABBAA
Affine Cipher
In the affine cipher the letters of an alphabet of size m are first mapped to the integers in the range 0..m - 1. It then uses modular arithmetic to transform the integer that each plaintext letter corresponds to into another integer that correspond to a ciphertext letter. The encryption function for a single letter is
where modulus m is the size of the alphabet and a and b are the key of the cipher. The value a must be chosen such that a and m are coprime.
Considering the specific case of encrypting messages in English (i.e. m = 26), there are a total of 286 non-trivial affine ciphers, not counting the 26 trivial Caesar ciphers. This number comes from the fact there are 12 numbers that are coprime with 26 that are less than 26 (these are the possible values of a). Each value of a can have 26 different addition shifts (the b value) ; therefore, there are 12*26 or 312 possible keys.
Plaintext: tustain
cipher variations:
uvtubjo
gjdgbzo
sxnsbpo
elxebfo
qzhqbvo
cnrcblo
aplabro
mdvmbho
yrfybxo
kfpkbno
wtzwbdo
ihjibto
vwuvckp
hkehcap
tyotcqp
fmyfcgp
raircwp
dosdcmp
bqmbcsp
newncip
zsgzcyp
lgqlcop
xuaxcep
jikjcup
wxvwdlq
ilfidbq
uzpudrq
gnzgdhq
sbjsdxq
eptednq
crncdtq
ofxodjq
mhrmdpq
yvbydfq
kjlkdvq
xywxemr
jmgjecr
vaqvesr
hoaheir
tckteyr
fqufeor
dsodeur
pgypekr
buibear
nisneqr
zwczegr
lkmlewr
yzxyfns
knhkfds
wbrwfts
ipbifjs
udlufzs
grvgfps
etpefvs
qhzqfls
cvjcfbs
ojtofrs
axdafhs
mlnmfxs
zayzgot
loilget
xcsxgut
jqcjgkt
vemvgat
hswhgqt
fuqfgwt
riargmt
dwkdgct
pkupgst
byebgit
nmongyt
abzahpu
mpjmhfu
ydtyhvu
krdkhlu
wfnwhbu
itxihru
gvrghxu
sjbshnu
exlehdu
qlvqhtu
czfchju
onpohzu
bcabiqv
nqknigv
zeuziwv
lselimv
xgoxicv
juyjisv
hwshiyv
tkctiov
fymfiev
rmwriuv
dagdikv
poqpiav
cdbcjrw
orlojhw
afvajxw
mtfmjnw
yhpyjdw
kvzkjtw
ixtijzw
uldujpw
gzngjfw
snxsjvw
ebhejlw
qprqjbw
decdksx
psmpkix
bgwbkyx
nugnkox
ziqzkex
lwalkux
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vmevkqx
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ruormkz
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bksbmgz
nycnmwz
lawlmcz
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ghfgnva
svpsnla
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qxjqnra
cltcnha
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kdrknja
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utvunfa
highowb
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fkafocb
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uxrupnc
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ngunqmd
zuezqcd
liolqsd
xwyxqid
kljkrze
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indirfe
ubnurve
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kpfkthg
wdpwtxg
irzitng
ufjutdg
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qjxqtpg
cxhctfg
olrotvg
azbatlg
nomnuch
zcwzush
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xeqxuyh
jsajuoh
vgkvueh
tietukh
fwofuah
rkyruqh
dyidugh
pmspuwh
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opnovdi
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ktbkvpi
whlwvfi
ujfuvli
gxpgvbi
slzsvri
ezjevhi
qntqvxi
cbdcvni
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nsinwkj
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luclwqj
ximxwgj
vkgvwmj
hyqhwcj
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dcedwoj
qrpqxfk
cfzcxvk
otjoxlk
ahtaxbk
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unbuxtk
gblgxjk
spvsxzk
edfexpk
rsqrygl
pukpyml
biubycl
nwenysl
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fegfyql
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qvlqznm
cjvczdm
oxfoztm
alpazjm
ynjyzpm
kbtkzfm
wpdwzvm
idnizlm
urxuzbm
gfhgzrm
tustain
ficfayn
rwmraon
dkwdaen
pygpaun
bmqbakn
zokzaqn
lculagn
xqexawn
jeojamn
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hgihasn
The decryption function is
where a - 1 is the modular multiplicative inverse of a modulo m. I.e., it satisfies the equation
The multiplicative inverse of a only exists if a and m are coprime. Hence without the restriction on a decryption might not be possible. It can be shown as follows that decryption function is the inverse of the encryption function,
ROT13 Cipher
Applying ROT13 to a piece of text merely requires examining its alphabetic characters and replacing each one by the letter 13 places further along in the alphabet, wrapping back to the beginning if necessary. A becomes N, B becomes O, and so on up to M, which becomes Z, then the sequence continues at the beginning of the alphabet: N becomes A, O becomes B, and so on to Z, which becomes M. Only those letters which occur in the English alphabet are affected; numbers, symbols, whitespace, and all other characters are left unchanged. Because there are 26 letters in the English alphabet and 26 = 2 * 13, the ROT13 function is its own inverse:
ROT13(ROT13(x)) = x for any basic Latin-alphabet text x
An example plaintext to ciphertext using ROT13:
Plain: tustain Cipher: ghfgnva
Polybius Square
A Polybius Square is a table that allows someone to translate letters into numbers. To give a small level of encryption, this table can be randomized and shared with the recipient. In order to fit the 26 letters of the alphabet into the 25 spots created by the table, the letters i and j are usually combined.
1 2 3 4 5
1 A B C D E
2 F G H I/J K
3 L M N O P
4 Q R S T U
5 V W X Y Z
Basic Form:
Plain: tustain Cipher: 44543444114233
Extended Methods:
Method #1
Plaintext: tustain
method variations: yzxyfos decdltx ikhiqyc opnovdh
Method #2
Bifid cipher
The message is converted to its coordinates in the usual manner, but they are written vertically beneath:
```t u s t a i n
4 5 3 4 1 4 3
4 4 4 4 1 2 3 ```
They are then read out in rows:
45341434444123
Then divided up into pairs again, and the pairs turned back into letters using the square:
Plain: tustain Cipher: ysqstdm
Method #3
Plaintext: tustain
method variations: yotdqms otdqmsy tdqmsyo dqmsyot qmsyotd msyotdq syotdqm
Permutation Cipher
In classical cryptography, a permutation cipher is a transposition cipher in which the key is a permutation. To apply a cipher, a random permutation of size E is generated (the larger the value of E the more secure the cipher). The plaintext is then broken into segments of size E and the letters within that segment are permuted according to this key.
In theory, any transposition cipher can be viewed as a permutation cipher where E is equal to the length of the plaintext; this is too cumbersome a generalisation to use in actual practice, however.
The idea behind a permutation cipher is to keep the plaintext characters unchanged, butalter their positions by rearrangement using a permutation
This cipher is defined as:
Let m be a positive integer, and K consist of all permutations of {1,...,m}
For a key (permutation) , define:
The encryption function
The decryption function
A small example, assuming m = 6, and the key is the permutation :
The first row is the value of i, and the second row is the corresponding value of (i)
The inverse permutation, is constructed by interchanging the two rows, andrearranging the columns so that the first row is in increasing order, Therefore, is:
Total variation formula:
e = 2,718281828 , n - plaintext length
Plaintext: tustain
all 5040 cipher variations:
tustain
tustani
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Read more ...[1] , [2] , [3]
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CC-MAIN-2019-13
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https://www.tutorialkart.com/python/python-assignment-operators/
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text/html
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crawl-data/CC-MAIN-2024-10/segments/1707947476137.72/warc/CC-MAIN-20240302215752-20240303005752-00531.warc.gz
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## Assignment Operators
Assignment Operators are used to assign or store a specific value in a variable.
The following table lists out all the assignment operators in Python with their description, and an example.
## Example
In the following program, we will take values for variables x and y, and perform assignment operations on these values using Python Assignment Operators.
Python Program
```x, y = 5, 2
x += y
print(x) # 7
x, y = 5, 2
x -= y
print(x) # 3
x, y = 5, 2
x *= y
print(x) # 10
x, y = 5, 2
x /= y
print(x) # 2.5
x, y = 5, 2
x %= y
print(x) # 1
x, y = 5, 2
x **= y
print(x) # 25
x, y = 5, 2
x //= y
print(x) # 2
x, y = 5, 2
x &= y
print(x) # 0
x, y = 5, 2
x |= y
print(x) # 7
x, y = 5, 2
x ^= y
print(x) # 7
x, y = 5, 2
x <<= y
print(x) # 20
x, y = 5, 2
x >>= y
print(x) # 1```
Try Online
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https://www.indiabix.com/electronics-and-communication-engineering/microwave-communication/168010
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# Electronics and Communication Engineering - Microwave Communication
### Exercise :: Microwave Communication - Section 2
46.
A coaxial line has L = 500 nH/m and C = 50 pF/m. The characteristic impedance is
A. 500 Ω B. 250 Ω C. 100 Ω D. 50 Ω
Explanation:
.
47.
The main feature of a parametric amplifier is
A. low noise B. very high gain C. broad bandwidth D. both (b) and (c)
Explanation:
Since reactance does not contribute thermal noise to the circuit, it is a low noise device.
48.
Assertion (A): PIN diode is commonly used for microwave control.
Reason (R): A PIN diode uses heavily doped p and n materials.
A. Both A and R are correct and R is correct explanation of A B. Both A and R are correct but R is not correct explanation of A C. A is correct but R is wrong D. A is wrong but R is correct
Explanation:
A PIN diode has an intrinsic (i) layer between p and n layers. When reverse bias is applied depletion layers are formed at p-i and i-n junctions.
The effective/width of depletion layer increases by the width of i layer. It can be used as a voltage controlled attenuator.
At high frequencies the rectification effect ceases and impedance of diode is effectively that of i layer.
This impedance varies with the applied bias. It is used in high frequency switching circuits, limiters, modulators etc.
49.
In a vacuum tube, the transit time of electron between cathode and anode is important at
A. low frequencies B. high frequencies C. both (a) and (b) D. frequencies which are neither very low nor very high
Explanation:
At high frequencies transit time is large as compared to the period of microwave signal.
50.
Assertion (A): A line of length and short circuited at far end has an input impedance of infinity.
Reason (R): A line of length and short circuited at far end behaves as a parallel resonant circuit.
A. Both A and R are correct and R is correct explanation of A B. Both A and R are correct but R is not correct explanation of A C. A is correct but R is wrong D. A is wrong but R is correct
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{"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0}
| 2.71875
| 3
|
CC-MAIN-2021-17
|
longest
|
en
| 0.927862
|
https://www.12000.org/my_notes/CAS_integration_tests/reports/rubi_4_16_1_graded/test_cases/1_Algebraic_functions/1.2_Trinomial_products/1.2.3_General/1.2.3.4-f_x-%5Em-d+e_x%5En-%5Eq-a+b_x%5En+c_x%5E-2_n-%5Ep/rese97.htm
| 1,696,326,544,000,000,000
|
text/html
|
crawl-data/CC-MAIN-2023-40/segments/1695233511075.63/warc/CC-MAIN-20231003092549-20231003122549-00501.warc.gz
| 684,051,103
| 7,455
|
### 3.97 $$\int (b+2 c x) (-a+b x+c x^2)^{13} \, dx$$
Optimal. Leaf size=18 $\frac{1}{14} \left (a-b x-c x^2\right )^{14}$
[Out]
(a - b*x - c*x^2)^14/14
________________________________________________________________________________________
Rubi [A] time = 0.0687475, antiderivative size = 18, normalized size of antiderivative = 1., number of steps used = 1, number of rules used = 1, integrand size = 21, $$\frac{\text{number of rules}}{\text{integrand size}}$$ = 0.048, Rules used = {629} $\frac{1}{14} \left (a-b x-c x^2\right )^{14}$
Antiderivative was successfully verified.
[In]
Int[(b + 2*c*x)*(-a + b*x + c*x^2)^13,x]
[Out]
(a - b*x - c*x^2)^14/14
Rule 629
Int[((d_) + (e_.)*(x_))*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> Simp[(d*(a + b*x + c*x^2)^(p +
1))/(b*(p + 1)), x] /; FreeQ[{a, b, c, d, e, p}, x] && EqQ[2*c*d - b*e, 0] && NeQ[p, -1]
Rubi steps
\begin{align*} \int (b+2 c x) \left (-a+b x+c x^2\right )^{13} \, dx &=\frac{1}{14} \left (a-b x-c x^2\right )^{14}\\ \end{align*}
Mathematica [B] time = 0.173047, size = 201, normalized size = 11.17 $\frac{1}{14} x (b+c x) \left (-364 a^{11} x^2 (b+c x)^2+1001 a^{10} x^3 (b+c x)^3-2002 a^9 x^4 (b+c x)^4+3003 a^8 x^5 (b+c x)^5-3432 a^7 x^6 (b+c x)^6+3003 a^6 x^7 (b+c x)^7-2002 a^5 x^8 (b+c x)^8+1001 a^4 x^9 (b+c x)^9-364 a^3 x^{10} (b+c x)^{10}+91 a^2 x^{11} (b+c x)^{11}+91 a^{12} x (b+c x)-14 a^{13}-14 a x^{12} (b+c x)^{12}+x^{13} (b+c x)^{13}\right )$
Antiderivative was successfully verified.
[In]
Integrate[(b + 2*c*x)*(-a + b*x + c*x^2)^13,x]
[Out]
(x*(b + c*x)*(-14*a^13 + 91*a^12*x*(b + c*x) - 364*a^11*x^2*(b + c*x)^2 + 1001*a^10*x^3*(b + c*x)^3 - 2002*a^9
*x^4*(b + c*x)^4 + 3003*a^8*x^5*(b + c*x)^5 - 3432*a^7*x^6*(b + c*x)^6 + 3003*a^6*x^7*(b + c*x)^7 - 2002*a^5*x
^8*(b + c*x)^8 + 1001*a^4*x^9*(b + c*x)^9 - 364*a^3*x^10*(b + c*x)^10 + 91*a^2*x^11*(b + c*x)^11 - 14*a*x^12*(
b + c*x)^12 + x^13*(b + c*x)^13))/14
________________________________________________________________________________________
Maple [B] time = 0.006, size = 47685, normalized size = 2649.2 \begin{align*} \text{output too large to display} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
int((2*c*x+b)*(c*x^2+b*x-a)^13,x)
[Out]
result too large to display
________________________________________________________________________________________
Maxima [A] time = 1.2298, size = 22, normalized size = 1.22 \begin{align*} \frac{1}{14} \,{\left (c x^{2} + b x - a\right )}^{14} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate((2*c*x+b)*(c*x^2+b*x-a)^13,x, algorithm="maxima")
[Out]
1/14*(c*x^2 + b*x - a)^14
________________________________________________________________________________________
Fricas [B] time = 0.807591, size = 3474, normalized size = 193. \begin{align*} \text{result too large to display} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate((2*c*x+b)*(c*x^2+b*x-a)^13,x, algorithm="fricas")
[Out]
1/14*x^28*c^14 + x^27*c^13*b + 13/2*x^26*c^12*b^2 - x^26*c^13*a + 26*x^25*c^11*b^3 - 13*x^25*c^12*b*a + 143/2*
x^24*c^10*b^4 - 78*x^24*c^11*b^2*a + 13/2*x^24*c^12*a^2 + 143*x^23*c^9*b^5 - 286*x^23*c^10*b^3*a + 78*x^23*c^1
1*b*a^2 + 429/2*x^22*c^8*b^6 - 715*x^22*c^9*b^4*a + 429*x^22*c^10*b^2*a^2 - 26*x^22*c^11*a^3 + 1716/7*x^21*c^7
*b^7 - 1287*x^21*c^8*b^5*a + 1430*x^21*c^9*b^3*a^2 - 286*x^21*c^10*b*a^3 + 429/2*x^20*c^6*b^8 - 1716*x^20*c^7*
b^6*a + 6435/2*x^20*c^8*b^4*a^2 - 1430*x^20*c^9*b^2*a^3 + 143/2*x^20*c^10*a^4 + 143*x^19*c^5*b^9 - 1716*x^19*c
^6*b^7*a + 5148*x^19*c^7*b^5*a^2 - 4290*x^19*c^8*b^3*a^3 + 715*x^19*c^9*b*a^4 + 143/2*x^18*c^4*b^10 - 1287*x^1
8*c^5*b^8*a + 6006*x^18*c^6*b^6*a^2 - 8580*x^18*c^7*b^4*a^3 + 6435/2*x^18*c^8*b^2*a^4 - 143*x^18*c^9*a^5 + 26*
x^17*c^3*b^11 - 715*x^17*c^4*b^9*a + 5148*x^17*c^5*b^7*a^2 - 12012*x^17*c^6*b^5*a^3 + 8580*x^17*c^7*b^3*a^4 -
1287*x^17*c^8*b*a^5 + 13/2*x^16*c^2*b^12 - 286*x^16*c^3*b^10*a + 6435/2*x^16*c^4*b^8*a^2 - 12012*x^16*c^5*b^6*
a^3 + 15015*x^16*c^6*b^4*a^4 - 5148*x^16*c^7*b^2*a^5 + 429/2*x^16*c^8*a^6 + x^15*c*b^13 - 78*x^15*c^2*b^11*a +
1430*x^15*c^3*b^9*a^2 - 8580*x^15*c^4*b^7*a^3 + 18018*x^15*c^5*b^5*a^4 - 12012*x^15*c^6*b^3*a^5 + 1716*x^15*c
^7*b*a^6 + 1/14*x^14*b^14 - 13*x^14*c*b^12*a + 429*x^14*c^2*b^10*a^2 - 4290*x^14*c^3*b^8*a^3 + 15015*x^14*c^4*
b^6*a^4 - 18018*x^14*c^5*b^4*a^5 + 6006*x^14*c^6*b^2*a^6 - 1716/7*x^14*c^7*a^7 - x^13*b^13*a + 78*x^13*c*b^11*
a^2 - 1430*x^13*c^2*b^9*a^3 + 8580*x^13*c^3*b^7*a^4 - 18018*x^13*c^4*b^5*a^5 + 12012*x^13*c^5*b^3*a^6 - 1716*x
^13*c^6*b*a^7 + 13/2*x^12*b^12*a^2 - 286*x^12*c*b^10*a^3 + 6435/2*x^12*c^2*b^8*a^4 - 12012*x^12*c^3*b^6*a^5 +
15015*x^12*c^4*b^4*a^6 - 5148*x^12*c^5*b^2*a^7 + 429/2*x^12*c^6*a^8 - 26*x^11*b^11*a^3 + 715*x^11*c*b^9*a^4 -
5148*x^11*c^2*b^7*a^5 + 12012*x^11*c^3*b^5*a^6 - 8580*x^11*c^4*b^3*a^7 + 1287*x^11*c^5*b*a^8 + 143/2*x^10*b^10
*a^4 - 1287*x^10*c*b^8*a^5 + 6006*x^10*c^2*b^6*a^6 - 8580*x^10*c^3*b^4*a^7 + 6435/2*x^10*c^4*b^2*a^8 - 143*x^1
0*c^5*a^9 - 143*x^9*b^9*a^5 + 1716*x^9*c*b^7*a^6 - 5148*x^9*c^2*b^5*a^7 + 4290*x^9*c^3*b^3*a^8 - 715*x^9*c^4*b
*a^9 + 429/2*x^8*b^8*a^6 - 1716*x^8*c*b^6*a^7 + 6435/2*x^8*c^2*b^4*a^8 - 1430*x^8*c^3*b^2*a^9 + 143/2*x^8*c^4*
a^10 - 1716/7*x^7*b^7*a^7 + 1287*x^7*c*b^5*a^8 - 1430*x^7*c^2*b^3*a^9 + 286*x^7*c^3*b*a^10 + 429/2*x^6*b^6*a^8
- 715*x^6*c*b^4*a^9 + 429*x^6*c^2*b^2*a^10 - 26*x^6*c^3*a^11 - 143*x^5*b^5*a^9 + 286*x^5*c*b^3*a^10 - 78*x^5*
c^2*b*a^11 + 143/2*x^4*b^4*a^10 - 78*x^4*c*b^2*a^11 + 13/2*x^4*c^2*a^12 - 26*x^3*b^3*a^11 + 13*x^3*c*b*a^12 +
13/2*x^2*b^2*a^12 - x^2*c*a^13 - x*b*a^13
________________________________________________________________________________________
Sympy [B] time = 0.312619, size = 1326, normalized size = 73.67 \begin{align*} \text{result too large to display} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate((2*c*x+b)*(c*x**2+b*x-a)**13,x)
[Out]
-a**13*b*x + b*c**13*x**27 + c**14*x**28/14 + x**26*(-a*c**13 + 13*b**2*c**12/2) + x**25*(-13*a*b*c**12 + 26*b
**3*c**11) + x**24*(13*a**2*c**12/2 - 78*a*b**2*c**11 + 143*b**4*c**10/2) + x**23*(78*a**2*b*c**11 - 286*a*b**
3*c**10 + 143*b**5*c**9) + x**22*(-26*a**3*c**11 + 429*a**2*b**2*c**10 - 715*a*b**4*c**9 + 429*b**6*c**8/2) +
x**21*(-286*a**3*b*c**10 + 1430*a**2*b**3*c**9 - 1287*a*b**5*c**8 + 1716*b**7*c**7/7) + x**20*(143*a**4*c**10/
2 - 1430*a**3*b**2*c**9 + 6435*a**2*b**4*c**8/2 - 1716*a*b**6*c**7 + 429*b**8*c**6/2) + x**19*(715*a**4*b*c**9
- 4290*a**3*b**3*c**8 + 5148*a**2*b**5*c**7 - 1716*a*b**7*c**6 + 143*b**9*c**5) + x**18*(-143*a**5*c**9 + 643
5*a**4*b**2*c**8/2 - 8580*a**3*b**4*c**7 + 6006*a**2*b**6*c**6 - 1287*a*b**8*c**5 + 143*b**10*c**4/2) + x**17*
(-1287*a**5*b*c**8 + 8580*a**4*b**3*c**7 - 12012*a**3*b**5*c**6 + 5148*a**2*b**7*c**5 - 715*a*b**9*c**4 + 26*b
**11*c**3) + x**16*(429*a**6*c**8/2 - 5148*a**5*b**2*c**7 + 15015*a**4*b**4*c**6 - 12012*a**3*b**6*c**5 + 6435
*a**2*b**8*c**4/2 - 286*a*b**10*c**3 + 13*b**12*c**2/2) + x**15*(1716*a**6*b*c**7 - 12012*a**5*b**3*c**6 + 180
18*a**4*b**5*c**5 - 8580*a**3*b**7*c**4 + 1430*a**2*b**9*c**3 - 78*a*b**11*c**2 + b**13*c) + x**14*(-1716*a**7
*c**7/7 + 6006*a**6*b**2*c**6 - 18018*a**5*b**4*c**5 + 15015*a**4*b**6*c**4 - 4290*a**3*b**8*c**3 + 429*a**2*b
**10*c**2 - 13*a*b**12*c + b**14/14) + x**13*(-1716*a**7*b*c**6 + 12012*a**6*b**3*c**5 - 18018*a**5*b**5*c**4
+ 8580*a**4*b**7*c**3 - 1430*a**3*b**9*c**2 + 78*a**2*b**11*c - a*b**13) + x**12*(429*a**8*c**6/2 - 5148*a**7*
b**2*c**5 + 15015*a**6*b**4*c**4 - 12012*a**5*b**6*c**3 + 6435*a**4*b**8*c**2/2 - 286*a**3*b**10*c + 13*a**2*b
**12/2) + x**11*(1287*a**8*b*c**5 - 8580*a**7*b**3*c**4 + 12012*a**6*b**5*c**3 - 5148*a**5*b**7*c**2 + 715*a**
4*b**9*c - 26*a**3*b**11) + x**10*(-143*a**9*c**5 + 6435*a**8*b**2*c**4/2 - 8580*a**7*b**4*c**3 + 6006*a**6*b*
*6*c**2 - 1287*a**5*b**8*c + 143*a**4*b**10/2) + x**9*(-715*a**9*b*c**4 + 4290*a**8*b**3*c**3 - 5148*a**7*b**5
*c**2 + 1716*a**6*b**7*c - 143*a**5*b**9) + x**8*(143*a**10*c**4/2 - 1430*a**9*b**2*c**3 + 6435*a**8*b**4*c**2
/2 - 1716*a**7*b**6*c + 429*a**6*b**8/2) + x**7*(286*a**10*b*c**3 - 1430*a**9*b**3*c**2 + 1287*a**8*b**5*c - 1
716*a**7*b**7/7) + x**6*(-26*a**11*c**3 + 429*a**10*b**2*c**2 - 715*a**9*b**4*c + 429*a**8*b**6/2) + x**5*(-78
*a**11*b*c**2 + 286*a**10*b**3*c - 143*a**9*b**5) + x**4*(13*a**12*c**2/2 - 78*a**11*b**2*c + 143*a**10*b**4/2
) + x**3*(13*a**12*b*c - 26*a**11*b**3) + x**2*(-a**13*c + 13*a**12*b**2/2)
________________________________________________________________________________________
Giac [B] time = 1.11006, size = 1958, normalized size = 108.78 \begin{align*} \text{result too large to display} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate((2*c*x+b)*(c*x^2+b*x-a)^13,x, algorithm="giac")
[Out]
1/14*c^14*x^28 + b*c^13*x^27 + 13/2*b^2*c^12*x^26 - a*c^13*x^26 + 26*b^3*c^11*x^25 - 13*a*b*c^12*x^25 + 143/2*
b^4*c^10*x^24 - 78*a*b^2*c^11*x^24 + 13/2*a^2*c^12*x^24 + 143*b^5*c^9*x^23 - 286*a*b^3*c^10*x^23 + 78*a^2*b*c^
11*x^23 + 429/2*b^6*c^8*x^22 - 715*a*b^4*c^9*x^22 + 429*a^2*b^2*c^10*x^22 - 26*a^3*c^11*x^22 + 1716/7*b^7*c^7*
x^21 - 1287*a*b^5*c^8*x^21 + 1430*a^2*b^3*c^9*x^21 - 286*a^3*b*c^10*x^21 + 429/2*b^8*c^6*x^20 - 1716*a*b^6*c^7
*x^20 + 6435/2*a^2*b^4*c^8*x^20 - 1430*a^3*b^2*c^9*x^20 + 143/2*a^4*c^10*x^20 + 143*b^9*c^5*x^19 - 1716*a*b^7*
c^6*x^19 + 5148*a^2*b^5*c^7*x^19 - 4290*a^3*b^3*c^8*x^19 + 715*a^4*b*c^9*x^19 + 143/2*b^10*c^4*x^18 - 1287*a*b
^8*c^5*x^18 + 6006*a^2*b^6*c^6*x^18 - 8580*a^3*b^4*c^7*x^18 + 6435/2*a^4*b^2*c^8*x^18 - 143*a^5*c^9*x^18 + 26*
b^11*c^3*x^17 - 715*a*b^9*c^4*x^17 + 5148*a^2*b^7*c^5*x^17 - 12012*a^3*b^5*c^6*x^17 + 8580*a^4*b^3*c^7*x^17 -
1287*a^5*b*c^8*x^17 + 13/2*b^12*c^2*x^16 - 286*a*b^10*c^3*x^16 + 6435/2*a^2*b^8*c^4*x^16 - 12012*a^3*b^6*c^5*x
^16 + 15015*a^4*b^4*c^6*x^16 - 5148*a^5*b^2*c^7*x^16 + 429/2*a^6*c^8*x^16 + b^13*c*x^15 - 78*a*b^11*c^2*x^15 +
1430*a^2*b^9*c^3*x^15 - 8580*a^3*b^7*c^4*x^15 + 18018*a^4*b^5*c^5*x^15 - 12012*a^5*b^3*c^6*x^15 + 1716*a^6*b*
c^7*x^15 + 1/14*b^14*x^14 - 13*a*b^12*c*x^14 + 429*a^2*b^10*c^2*x^14 - 4290*a^3*b^8*c^3*x^14 + 15015*a^4*b^6*c
^4*x^14 - 18018*a^5*b^4*c^5*x^14 + 6006*a^6*b^2*c^6*x^14 - 1716/7*a^7*c^7*x^14 - a*b^13*x^13 + 78*a^2*b^11*c*x
^13 - 1430*a^3*b^9*c^2*x^13 + 8580*a^4*b^7*c^3*x^13 - 18018*a^5*b^5*c^4*x^13 + 12012*a^6*b^3*c^5*x^13 - 1716*a
^7*b*c^6*x^13 + 13/2*a^2*b^12*x^12 - 286*a^3*b^10*c*x^12 + 6435/2*a^4*b^8*c^2*x^12 - 12012*a^5*b^6*c^3*x^12 +
15015*a^6*b^4*c^4*x^12 - 5148*a^7*b^2*c^5*x^12 + 429/2*a^8*c^6*x^12 - 26*a^3*b^11*x^11 + 715*a^4*b^9*c*x^11 -
5148*a^5*b^7*c^2*x^11 + 12012*a^6*b^5*c^3*x^11 - 8580*a^7*b^3*c^4*x^11 + 1287*a^8*b*c^5*x^11 + 143/2*a^4*b^10*
x^10 - 1287*a^5*b^8*c*x^10 + 6006*a^6*b^6*c^2*x^10 - 8580*a^7*b^4*c^3*x^10 + 6435/2*a^8*b^2*c^4*x^10 - 143*a^9
*c^5*x^10 - 143*a^5*b^9*x^9 + 1716*a^6*b^7*c*x^9 - 5148*a^7*b^5*c^2*x^9 + 4290*a^8*b^3*c^3*x^9 - 715*a^9*b*c^4
*x^9 + 429/2*a^6*b^8*x^8 - 1716*a^7*b^6*c*x^8 + 6435/2*a^8*b^4*c^2*x^8 - 1430*a^9*b^2*c^3*x^8 + 143/2*a^10*c^4
*x^8 - 1716/7*a^7*b^7*x^7 + 1287*a^8*b^5*c*x^7 - 1430*a^9*b^3*c^2*x^7 + 286*a^10*b*c^3*x^7 + 429/2*a^8*b^6*x^6
- 715*a^9*b^4*c*x^6 + 429*a^10*b^2*c^2*x^6 - 26*a^11*c^3*x^6 - 143*a^9*b^5*x^5 + 286*a^10*b^3*c*x^5 - 78*a^11
*b*c^2*x^5 + 143/2*a^10*b^4*x^4 - 78*a^11*b^2*c*x^4 + 13/2*a^12*c^2*x^4 - 26*a^11*b^3*x^3 + 13*a^12*b*c*x^3 +
13/2*a^12*b^2*x^2 - a^13*c*x^2 - a^13*b*x
| 6,731
| 11,482
|
{"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0}
| 3.6875
| 4
|
CC-MAIN-2023-40
|
latest
|
en
| 0.256521
|
https://www.physicsforums.com/threads/why-does-this-step-response-overshoot.754137/
| 1,529,791,618,000,000,000
|
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# Homework Help: Why does this step response overshoot?
1. May 17, 2014
### 215
I have this transfer function which i don't know why it is overshooting due to an step input.
http://snag.gy/Vh1SM.jpg
Could someone explain why it does that.
It's it a close loop transfer function, where i've designed a controller for it, but for some reason, eventhoug all my poles are placed correctly, it is not possible to get the response i want to.
2. May 17, 2014
### rude man
With the given transfer function there is no overshoot.
Wolfram alpha agrees!
3. May 17, 2014
### 215
4. May 17, 2014
### rude man
Oops, there is too a small overshoot. I couldn't see it on wolfram. It's due to the zero at s = -1.513. Sorry!
You can cancel it by inserting a pole at the same point (s = - 1.513).
5. May 17, 2014
### 215
But how come does zeroes apply overshoot? I thought it was only poles which affected the system
6. May 17, 2014
### milesyoung
You should check your entry. It does have overshoot.
The transient response of your system is not only determined by its poles. You have a LHP zero as well, which can have a significant impact on the overshoot of the step response of your system.
Quick Google search for 'poles zeros transient' turned up:
http://courses.engr.illinois.edu/ece486/documents/lecture_notes/effects_zero_pole.pdf
7. May 17, 2014
### 215
but what criteria should i apply for determining the zeroes, and how does it affect the system. I mean it is still a LHP zero??? so bad can't it be
8. May 18, 2014
### 215
How should i using a PID controlle be able to adjust the Zero location, while i am adjusting the location of the pole.
9. May 18, 2014
### rude man
You need to describe your problem more fully, not just talk about an isolated transfer function. What is your plant and what are the response criteria?
10. May 18, 2014
### 215
I tried to make a general problem, but seems to have forgotten where it began at.
I posted my problem on another forum.
http://snag.gy/OqhxH.jpg
I hope this helps.
My repsponse criterias is that I want an overshoot to be 0% and settling time be under 1 sec.
Using Matlab tuning application i managed to do with only the I part of the controller. The problem is though i've read alot of places that it is not recomended to use just I or D regulators.
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JMSLTM Numerical Library 7.2.0
com.imsl.math
## Class NumericalDerivatives
• All Implemented Interfaces:
Serializable, Cloneable
```public class NumericalDerivatives
extends Object
implements Serializable, Cloneable```
Compute the Jacobian matrix for a function with m components in n independent variables.
`NumericalDerivatives` uses divided finite differences to compute the Jacobian. This class is designed for use in numerical methods for solving nonlinear problems where a Jacobian is evaluated repeatedly at neighboring arguments. For example this occurs in a Gauss-Newton method for solving non-linear least squares problems or a non-linear optimization method.
`NumericalDerivatives` is suited for applications where the Jacobian is a dense matrix. All cases , , or are allowed. Both one-sided and central divided differences can be used.
The design allows for computation of derivatives in a variety of contexts. Note that a gradient should be considered as the special case with , . A derivative of a single function of one variable is the case , . Any non-linear solving routine that optionally requests a Jacobian or gradient can use `NumericalDerivatives`. This should be considered if there are special properties or scaling issues associated with . Use the method `setDifferencingMethods` to specify different differencing options for numerical differentiation. These can be combined with some analytic subexpressions or other known relationships.
The divided differences are computed using values of the independent variables at the initial point , and differenced points . Here the , are the unit coordinate vectors. The value for each difference del depends on the variable j, the differencing method, and the scaling for that variable. This difference is computed internally. See `setPercentageFactor` for computational details. The evaluation of is normally done by the user-provided method `NumericalDerivatives.Function.f`, using the values . The index j and values are arguments to `NumericalDerivatives.Function.f`.
The computational kernel of `evaluateJ` performs the following steps:
1. evaluate the equations at the point `y` using `NumericalDerivatives.Function.f`.
2. compute the Jacobian.
3. compute the difference at .
By default, `evaluateJ` uses `NumericalDerivatives.Function.f` in step 3. The user may choose to override the `evaluateF` method to extend the capability of the class beyond the default.
There are six examples provided which illustrate various ways to use `NumericalDerivatives`. A discussion of the expected errors for these difference methods is found in A First Course in Numerical Analysis, Anthony Ralston, McGraw-Hill, NY, (1965).
Example: One-Sided Differences, Example: Skipping A Gradient Component, Example: Accumulation Of A Component, Example: Central Differences, Example: Hessian Approximation, Example: Usage With Class `MinUnconMultiVar` , Serialized Form
• ### Nested Class Summary
Nested Classes
Modifier and Type Class and Description
`static interface ` `NumericalDerivatives.Function`
Public interface function.
`static interface ` `NumericalDerivatives.Jacobian`
Public interface for the user-supplied function to compute the Jacobian.
• ### Field Summary
Fields
Modifier and Type Field and Description
`static int` `ACCUMULATE`
Indicates the accumulation of the result from whatever type of differences have been specified previously into initial values of the Jacobian.
`static int` `CENTRAL`
Indicates central differences.
`static int` `ONE_SIDED`
Indicates one sided differences.
`static int` `SKIP`
Indicates a variable to be skipped.
• ### Constructor Summary
Constructors
Constructor and Description
`NumericalDerivatives(NumericalDerivatives.Function fcn)`
Constructor for `NumericalDerivatives`.
• ### Method Summary
Methods
Modifier and Type Method and Description
`protected double[]` ```evaluateF(int varIndex, double[] y)```
This method is provided by the user to compute the function values at the current independent variable values `y`.
`double[][]` `evaluateJ(double[] y)`
Evaluates the Jacobian for a system of (m) equations in (n) variables.
`double[]` `getPercentageFactor()`
Returns the percentage factor for differencing.
`double[]` `getScalingFactors()`
Returns the scaling factors for the `y` values.
`int[]` `getStatus()`
Returns status information.
`void` `setDifferencingMethods(int[] options)`
Sets the methods used to compute the derivatives
`void` `setInitialF(double[] valueF)`
Set the initial function values.
`void` `setPercentageFactor(double[] factor)`
Sets the percentage factor for differencing
`void` `setScalingFactors(double[] scale)`
Sets the scaling factors for the `y` values.
• ### Methods inherited from class java.lang.Object
`clone, equals, finalize, getClass, hashCode, notify, notifyAll, toString, wait, wait, wait`
• ### Field Detail
• #### ACCUMULATE
`public static final int ACCUMULATE`
Indicates the accumulation of the result from whatever type of differences have been specified previously into initial values of the Jacobian.
Constant Field Values
• #### CENTRAL
`public static final int CENTRAL`
Indicates central differences.
Constant Field Values
• #### ONE_SIDED
`public static final int ONE_SIDED`
Indicates one sided differences.
Constant Field Values
• #### SKIP
`public static final int SKIP`
Indicates a variable to be skipped.
Constant Field Values
• ### Constructor Detail
• #### NumericalDerivatives
`public NumericalDerivatives(NumericalDerivatives.Function fcn)`
Constructor for `NumericalDerivatives`.
Parameters:
`fcn` - a `Function` object which is a user-supplied function to evaluate the equations at the point `y`.
• ### Method Detail
• #### evaluateF
```protected double[] evaluateF(int varIndex,
double[] y)```
This method is provided by the user to compute the function values at the current independent variable values `y`. If the user does not override the `evaluateF` method, then `NumericalDerivatives.Function.f` is used to compute the function values.
Parameters:
`varIndex` - an `int` which indicates the index of the variable to perturb.
`y` - a `double` array of length n, the point at which the function is to be evaluated.
Returns:
a `double` array of length m. The equations evaluated at the point `y`.
• #### evaluateJ
`public double[][] evaluateJ(double[] y)`
Evaluates the Jacobian for a system of (m) equations in (n) variables.
Parameters:
`y` - a `double` array of length n, the point at which the Jacobian is to be evaluated.
Returns:
a `double` matrix containing the Jacobian. Columns that are accumulated must have the additive term defined on entry or else be set to zero. Columns that are skipped can be defined either before or after the `evaluateJ` method is invoked.
• #### getPercentageFactor
`public double[] getPercentageFactor()`
Returns the percentage factor for differencing.
Returns:
a `double` array containing the percentage factor for differencing. See `setPercentageFactor` for more detail.
• #### getScalingFactors
`public double[] getScalingFactors()`
Returns the scaling factors for the `y` values.
Returns:
a `double` array containing the scaling factors.
• #### getStatus
`public int[] getStatus()`
Returns status information. This information might prove useful to the user wanting to gain better control over the differencing parameters. This information can often be ignored.
Returns:
an `int` array containing the ten diagnostic values described in the following table. These values can be used to monitor the progress or expense of the Jacobian computation. index Description 0 the number of times a function evaluation was computed. 1 the number of columns in which three attempts were made to increase a percentage factor for differencing (i.e. a component in the `factor` array) but the computed del remained unacceptably small relative to `y[j-1]` or `scale[j-1]`. In such cases the percentage factor is set to 1.4901161193847656e-8, which is the square root of machine precision 2 the number of columns in which the computed del was zero to machine precision because `y[j-1]` or `scale[j-1]` was zero. In such cases del is set to 1.4901161193847656e-8, which is the square root of machine precision 3 the number of Jacobian columns which had to be recomputed because the largest difference formed in the column was close to zero relative to scale, where and i denotes the row index of the largest difference in the column currently being processed. index = 9 gives the last column where this occurred. 4 the number of columns whose largest difference is close to zero relative to scale after the column has been recomputed. 5 the number of times scale information was not available for use in the roundoff and truncation error tests. This occurs when where i is the index of the largest difference for the column currently being processed. 6 the number of times the increment for differencing (del) was computed and had to be increased because (`scale[j-1]`+del) - `scale[j-1]`) was too small relative to `y[j-1]` or `scale[j-1]`. 7 the number of times a component of the `factor` array was reduced because changes in function values were large and excess truncation error was suspected. index = 8 gives the last column in which this occurred. 8 the index of the last column where the corresponding component of the `factor` array had to be reduced because excessive truncation error was suspected. 9 the index of the last column where the difference was small and the column had to be recomputed with an adjusted increment (see index = 3). The largest derivative in this column may be inaccurate due to excessive roundoff error.
• #### setDifferencingMethods
`public void setDifferencingMethods(int[] options)`
Sets the methods used to compute the derivatives
Parameters:
`options` - an `int` array of length n, containing the methods used to compute the derivatives. `options[i]` is the method to be used for the i-th variable. `options[i]` can be one of the values in the table which follows. The default is to use `ONE_SIDED` differences for each variable. Entry Description `ONE_SIDED` Indicates one sided differences. `CENTRAL` Indicates central differences. `ACCUMULATE` Indicates the accumulation of the result from whatever type of differences have been specified previously into initial values of the Jacobian. `SKIP` Indicates a variable to be skipped.
• #### setInitialF
`public void setInitialF(double[] valueF)`
Set the initial function values. Use the values , where is the initial value of the independent variables located in array `y`.
Parameters:
`valueF` - a `double` array of length m containing the initial function values, . Default: all values are 0.0.
• #### setPercentageFactor
`public void setPercentageFactor(double[] factor)`
Sets the percentage factor for differencing
For each divided difference for variable j the increment used is del. The value of del is computed as follows: First define . If the user has set the elements of array `scale` to non-default values, then define . Otherwise and . Finally compute . By changing the sign of `scale[j-1]`, the difference del can have any desired orientation, such as staying within bounds on variable j. For central differences, a reduced factor is used for del that normally results in relative errors as small as machine precision to the 2/3 power.
Parameters:
`factor` - a `double` array of length n containing the percentage factor for differencing. Except for initialization, the `factor` array should not be altered in the `evaluateF` method. The elements of `factor` must be such that
where 1.8189894035458565e-12 is machine precision to the three-fourths power.
Default: all elements of `factor` are set to 1.4901161193847656e-8, which is the square root of machine precision.
• #### setScalingFactors
`public void setScalingFactors(double[] scale)`
Sets the scaling factors for the `y` values. The user can also use `scale` to provide appropriate signs for the increments.
Parameters:
`scale` - a `double` array of length n containing the scaling factors. Default: all values are 1.0.
JMSLTM Numerical Library 7.2.0
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# STATISTIC
## STATISTIC
STATISTICS
Week 1
Introduction : Data and Statistics
Topics
1. Introduction : Data and Statistics
2. Descriptive Statistics
3. Introduction to Probability
4. Discrete Probability Distributions
5. Continuous Probability Distributions
6. Sampling and Sampling Distributions
7. Interval Estimation
8. Hypothesis Tests
9. Analysis of Variance
10. Simple Linear Regression
Bina Nusantara University
2
References
• Anderson, David R., Sweeney, Dennis J., Williams, Thomas A.
(2011). Statistics for Business and Economics. 11. Cengage
Learning. USA. ISBN: 978-0538481649.
Bina Nusantara University
3
Bina Nusantara University
4
Learning Outcomes
• LO 1: Explain the data and statistics
• LO 2: Calculate the statistical measurements
• LO 3: Interpret the results of statistical measurements
• LO 4: Apply statistical method to the real problem
• LO 5: Analyze the suitable decision from statistical method solution
Bina Nusantara University
5
Learning Outcomes :
(1) Explain the data and statistics
Bina Nusantara University
6
Topics :
1. Data
2. Statistics
3. Applications in Business and Economics
Bina Nusantara University
7
1. DATA
1. DATA
• Data are the facts and figures collected, analyzed, and summarized for
presentation and interpretation
• All the data collected in a particular study are referred to as the data set
for the study
Example : Table 1 shows a data set containing information for 25 companies
that are part of S&P 500. The S&P 500 is made up of 500 companies selected by
Standard&Poor’s. These companies account for 76% of the market capitalization
of all U.S.stocks. S&P 500 stocks are closely followed by investors and Wall
Street analysts.
Bina Nusantara University
9
1. DATA
1. Data Set for 25 S&P 500 Companies
Source : Anderson, David R., Sweeney, Dennis J., Williams, Thomas A. (2011). Statistics for
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# Math
posted by .
What multiplication fact can be found by using the arrays for 2 x 9 and 5 x 9?
• Math -
My answer is 7 x 9
2 x 9 = 18; 5 x 9 = 45;
18 + 45 = 63.. is this correct
• Math -
63
### Answer This Question
First Name: School Subject: Answer:
### Related Questions
More Related Questions
Post a New Question
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# Genetic Algorithm Problems
I'm implementing the NSGA II genetic algorithm to develop a set of timetables for my college. I am having problems with variation of solutions.
My algorithm works fine as in initialization, mutation and crossover but after the final generation when reviewing my solutions they are all the same e.g I have 200 in a generation, maybe 64 of them will be the same as each other, 54 the same as each other etc.
My question is what may be causing this? And what is the best form of crossover and mutation?
Also is there norm for generation size, amount of generations, mutation rate and cross over rate?
At the moment it runs like so:
1. Randomly generate 300 solutions
2. Calculate fitness and ranking
3. Pick 200 of the best solutions
4. Mutate 5% of these and produce 80 children
5. Calculate and Rank again
6. Pick the best 300 to move on to next generation
7. Repeat
-
It's difficult to say anything beyond "there is probably a bug somewhere" with the information you provided. It shouldn't be too difficult to reduce your algorithm to a few lines of pseudo-code (at least the selection and generation of each generation) (which you can then add to the question). Looking at the population at initialization and intermediate steps of the algorithm might also provide some insight. Also, this question is probably outside the scope of SO (maybe better for CSTheory, CS or Programmers). – Dukeling Feb 19 '13 at 13:43
At the minute it runs like so 1. Randomly generate 300 solutions 2. Calculate fitness and ranking 3. Pick 200 of the best solutions 4. Mutate 5% of these and produce 80 children 5. Calculate and Rank again 6. Pick the best 300 to move on to next generation Repeat above I don't think it is a bug as I have worked through it and everything seems to be ok. Though I would be more than happy to be proved wrong! – Melo1991 Feb 19 '13 at 13:45
Not much info here, but my gut tells me you are having a problem with your weighting of solutions and therefore your selection (3) is being hyper-selective, leading to an overly homogenous population. Try unit testing & measuring their outputs outside of the GA to verify their correctness. – NWS Feb 19 '13 at 14:06
How many objectives are there in your problem ? And how many constraints ? – Benoît Guédas Feb 24 '13 at 6:40
I believe there may be no error in your algo. This is a well known problem for GA. If you want to have a variety of solutions, you should implement some Niching method. Its idea is in punishing similar individuals in your population. You can figure out some heuristic for individuals similarity and exclude similar individuals from your population or eliminate individual's fitness. This will keep your population more diverse and will not allow your variation operators choose and evolve same individuals. It'll be useful to look through "Niching Methods for Genetic Algorithms" of Samir W. Mahfoud.
-
This outcome is not necessarily bad. It may just be converging on a couple of solutions which do not have any other viable solutions "nearby". Are the solutions you are converging on bad?
One thing I notice is that although you mention crossover you do not list it as one of your 7 steps. If in fact you are only doing mutation, that would make it more difficult for the GA to climb out of local optima.
-
G.A's works like Converging to a point choosing a minimal set of alleles(solutions) as best. In some rare cases, after a long run like some thousands of generations, it may bring a single optimal solution.
But in some cases, it may bring converging solutions in minimal number of runs, say 100. But it has the possibility of getting stuck in the local optima, failing to reach the global optima.
I am not sure upto which generation, you have tried. I suggest you to go for 1000 generations and compare the results. Also the plot may change after some generations like you can see entirely new set of solutions
There are different crossover and mutation for different representations. May be you could tell the representation you are using and the results are directly proportional with number of generations
-
No, for a multiobjective optimization problem, you should obtain several different solutions spread over the Pareto front (if not, it was not really a multiobjective optimization problem). The question is about NSGA-II: a GA to solve multiobjective optimization problems. – Benoît Guédas Feb 23 '13 at 13:59
@Benoît I think that is what i explained earlier. If you read my answer twice, you can get what i meant. – naran Feb 25 '13 at 6:39
I read your answer twice, and it looks like I still don't get what you meant :-) A single solution may not be the ultimate result to any problem especially for a multicriteria optimization problem, because there are many Pareto-optimal solutions (and it is not because of the evolutionary approach). – Benoît Guédas Feb 25 '13 at 7:48
@Benoît Thank you for what you said. I made it straight forward. Now read it and tell me what you get – naran Feb 26 '13 at 11:04
Thanks! It's clearer now. Unfortunately, it seems that we will never know what was wrong with this algorithm :-( – Benoît Guédas Feb 27 '13 at 7:02
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## 16.4 Finding Eigenvectors and Eigenvalues
Suppose v is an eigenvector of the matrix M with eigenvalue.
We then have: (M -I) v = M v -v = 0.
This means that v is in the null space of the matrix M -I, its nullity is at least one, and thereforeM -I.
When M is specified this determinental equation can be written out explicitly and it is a polynomial equation in having degree n. Every eigenvalue of M must obey this equation.
Moreover, if a value of obeys the equation M -I= 0, then the matrix M - I with this value is singular and its nullity is at least 1 so that there is a vector v satisfying (M -I) v = 0.
The equation M -I= 0 is called the characteristic equation for the matrix M. We have just seen that any solution to the characteristic equation is an eigenvalue of M corresponding to at least one eigenvector.
The straightforward way to find an eigenvalue is to write down the characteristic equation and find a solution to it.
Example
Once you know the eigenvalue you can use the equation (M -I) v = 0, to determine v. Since (M -I) v is singular, row reduction will lead you to a subspace of solutions of dimension given by the nullity of this matrix.
Example
We now ask, when can we find a basis of eigenvectors?
The characteristic equation is a polynomial equation of degree n.
Such an equation can have up to n real roots, and in general will have exactly n complex roots, some of which can be multiple ones. ((x-1)2 = 0 has two roots of 1.)
If it has n distinct real roots, then we have n real eigenvalues, and can find n eigenvectors to match. Whether or not the roots are distinct, you can always find a basis consisting of eigenvectors if the matrix is symmetric.
A basis is said to be orthonormal, if its elements each have length 1 and they are mutually perpendicular. Only symmetric matrices have real eigenvalues and real orthonormal bases of eigenvectors.
So far we have assumed that all our numbers are real, and we are then unable to find n eigenvalues and eigenvectors if some of the roots of the characteristic equation are not real. With a small change in definitions required to assure that the dot product of a non-zero vector with itself is always real and positive, it is possible to let our numbers be complex. With complex numbers, the class of polynomial equations with n roots includes all of them, and the only problem with having n eigenvalues comes with matrices whose characteristic equation has multiple roots.
There are some matrices just dont have n distinct dimensions of eigenvectors; the simplest example is the two by two matrix with 1 in the 12 place and 0 elsewhere.
The characteristic equation of a matrix C(x) which is A - x I= 0 can be written in terms of its distinct roots,i , as , where mi is the multiplicity of the eigenvaluei . A diagonal matrix obeys its minimal equation which has the same roots but with all multiplicities 1.
One can show that every matrix satisfies its own characteristic equation; a matrix has a basis of eigenvectors over the complex numbers if and only if it obeys its own minimal equation.
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Control Structures I (Selection)
Presentation on theme: "Control Structures I (Selection)"— Presentation transcript:
Control Structures I (Selection)
C++ Programming Control Structures I (Selection)
Control Structures A computer can proceed:
In sequence Selectively (branch) - making a choice Repetitively (iteratively) - looping Some statements are executed only if certain conditions are met A condition is met if it evaluates to true C++ Programming: From Problem Analysis to Program Design, Fourth Edition
Control Structures (continued)
C++ Programming: From Problem Analysis to Program Design, Fourth Edition
Relational Operators A condition is represented by a logical (Boolean) expression that can be true or false Relational operators: Allow comparisons Require two operands (binary) Evaluate to true or false C++ Programming: From Problem Analysis to Program Design, Fourth Edition
Relational Operators (continued)
C++ Programming: From Problem Analysis to Program Design, Fourth Edition
Relational Operators and Simple Data Types
You can use the relational operators with all three simple data types: 8 < 15 evaluates to true 6 != 6 evaluates to false 2.5 > 5.8 evaluates to false 5.9 <= 7.5 evaluates to true C++ Programming: From Problem Analysis to Program Design, Fourth Edition
Comparing Characters C++ Programming: From Problem Analysis to Program Design, Fourth Edition
Relational Operators and the string Type
Relational operators can be applied to strings Strings are compared character by character, starting with the first character Comparison continues until either a mismatch is found or all characters are found equal If two strings of different lengths are compared and the comparison is equal to the last character of the shorter string The shorter string is less than the larger string C++ Programming: From Problem Analysis to Program Design, Fourth Edition
Relational Operators and the string Type (continued)
Suppose we have the following declarations: string str1 = "Hello"; string str2 = "Hi"; string str3 = "Air"; string str4 = "Bill"; string str4 = "Big"; C++ Programming: From Problem Analysis to Program Design, Fourth Edition
Relational Operators and the string Type (continued)
C++ Programming: From Problem Analysis to Program Design, Fourth Edition
Relational Operators and the string Type (continued)
C++ Programming: From Problem Analysis to Program Design, Fourth Edition
Relational Operators and the string Type (continued)
C++ Programming: From Problem Analysis to Program Design, Fourth Edition
Logical (Boolean) Operators and Logical Expressions (NOT)
C++ Programming: From Problem Analysis to Program Design, Fourth Edition
Order of Precedence Relational and logical operators are evaluated from left to right The associativity is left to right Parentheses can override precedence C++ Programming: From Problem Analysis to Program Design, Fourth Edition
Order of Precedence (continued)
C++ Programming: From Problem Analysis to Program Design, Fourth Edition
Order of Precedence (continued)
C++ Programming: From Problem Analysis to Program Design, Fourth Edition
Order of Precedence (continued)
C++ Programming: From Problem Analysis to Program Design, Fourth Edition
Order of Precedence (continued)
C++ Programming: From Problem Analysis to Program Design, Fourth Edition
Short-Circuit Evaluation
Short-circuit evaluation: evaluation of a logical expression stops as soon as the value of the expression is known Example: (age >= 21) || ( x == 5) //Line 1 (grade == 'A') && (x >= 7) //Line 2 C++ Programming: From Problem Analysis to Program Design, Fourth Edition
Selection: if and if...else
One-Way Selection Two-Way Selection Compound (Block of) Statements Multiple Selections: Nested if Comparing if...else Statements with a Series of if Statements C++ Programming: From Problem Analysis to Program Design, Fourth Edition
One-Way Selection The syntax of one-way selection is:
The statement is executed if the value of the expression is true The statement is bypassed if the value is false; program goes to the next statement if is a reserved word C++ Programming: From Problem Analysis to Program Design, Fourth Edition
One-Way Selection (continued)
C++ Programming: From Problem Analysis to Program Design, Fourth Edition
One-Way Selection (continued)
C++ Programming: From Problem Analysis to Program Design, Fourth Edition
One-Way Selection (continued)
C++ Programming: From Problem Analysis to Program Design, Fourth Edition
Two-Way Selection Two-way selection takes the form:
If expression is true, statement1 is executed; otherwise, statement2 is executed statement1 and statement2 are any C++ statements else is a reserved word C++ Programming: From Problem Analysis to Program Design, Fourth Edition
Two-Way Selection (continued)
C++ Programming: From Problem Analysis to Program Design, Fourth Edition
Two-Way Selection (continued)
C++ Programming: From Problem Analysis to Program Design, Fourth Edition
Two-Way Selection (continued)
C++ Programming: From Problem Analysis to Program Design, Fourth Edition
Compound (Block of) Statement
if (age > 18) { cout << "Eligible to vote." << endl; cout << "No longer a minor." << endl; } else cout << "Not eligible to vote." << endl; cout << "Still a minor." << endl; C++ Programming: From Problem Analysis to Program Design, Fourth Edition
if ( (carDoors == 4 ) && (driverAge > 24) )
int carDoors, driverAge ; double premium, monthlyPayment ; if ( (carDoors == 4 ) && (driverAge > 24) ) { premium = ; cout<<“ LOW RISK “ ; } else premium = ; cout <<“HIGH RISK ” ; monthlyPayment = premium / ;
What happens if you omit braces?
if ( (carDoors == 4 ) && (driverAge > 24) ) premium = ; cout<< “ LOW RISK “ ; else premium = ; cout<< “ HIGH RISK ” ; monthlyPayment = premium / ; COMPILE ERROR OCCURS. The “if clause” is the single statement following the if.
Braces can only be omitted when each clause is a single statement
if ( lastInitial <= ‘K’ ) volume = 1; else volume = 2; Cout<< “Look it up in volume # %d of the phone book”; cout<< volume ;
If--Else for a mail order
Write a program to calculate the total price of a certain purchase. There is a discount and shipping cost: The discount rate is 25% and the shipping is if purchase is over Otherwise, The discount rate is 15% and the shipping is 5.00 pounds.
What output? and Why? int age; age = 20; if ( age == 16 ) {
cout<< “Did you get driver’s license?” ; }
What output? and Why? int age; age = 30; if ( age < 18 )
cout<< “Do you drive?”; cout<< “Too young to vote”;
What output? and Why? int code; code = 0; if ( ! code )
cout<< “Yesterday”; else cout<<“Tomorrow”;
Example Write a program to ask a student for his grades in 3 exams ( each out of 50 ) , get their total and inform the student whether he passed or failed the course.
Multiple Selections: Nested if
Nesting: one control statement in another An else is associated with the most recent if that has not been paired with an else C++ Programming: From Problem Analysis to Program Design, Fourth Edition
Multiple Selections: Nested if (continued)
C++ Programming: From Problem Analysis to Program Design, Fourth Edition
Comparing if…else Statements with a Series of if Statements
C++ Programming: From Problem Analysis to Program Design, Fourth Edition
Example The Air Force has asked you to write a program to label aircrafts as military or civilian. Your program input is the plane’s speed and its estimated length. For planes traveling faster than 1100 km/hr, you will label those shorter than 52 m “military”, and longer as “Civilian”. For planes traveling less than 1100, you will issue an “aircraft unknown” statement.
Example Write a program to get the roots of a quadratic equation, given the 3 coefficients a, b, and c, a x2 + b x + c = 0
Writing Nested if Statements
Display one word to describe the int value of number as “Positive”, “Negative”, or “Zero” Your city classifies a pollution index less than 35 as “Pleasant”, 35 through 60 as “Unpleasant”, and above 60 as “Health Hazard.” Display the correct description of the pollution index value.
Using selection Every Sunday through Thursday you go to class.
When it is raining you take an umbrella. But on the weekend, what you do depends on the weather. If it is raining you read in bed. Otherwise, you have fun outdoors.
Conditional Operator (?:)
Conditional operator (?:) takes three arguments Ternary operator Syntax for using the conditional operator: expression1 ? expression2 : expression3 If expression1 is true, the result of the conditional expression is expression2 Otherwise, the result is expression3 C++ Programming: From Problem Analysis to Program Design, Fourth Edition
switch Structures switch structure: alternate to if-else
switch (integral) expression is evaluated first Value of the expression determines which corresponding action is taken Expression is sometimes called the selector C++ Programming: From Problem Analysis to Program Design, Fourth Edition
switch Structures (continued)
One or more statements may follow a case label Braces are not needed to turn multiple statements into a single compound statement The break statement may or may not appear after each statement switch, case, break, and default are reserved words C++ Programming: From Problem Analysis to Program Design, Fourth Edition
Light bulbs Write a program to ask the user for the brightness of a light bulb (in Watts), and print out the expected lifetime: Brightness Lifetime in hours 40, 75, otherwise 0
Program Write a C program to calculate the average of three test grades and print out a report with the student’s ID number, average, and how well is the student progress. “Very Good” is a 70-point average or better, “Good” is an average between 60 and 70, and “Failing” is 50 point average or less.
Write a C program that calculates bills for the Electricity company
Write a C program that calculates bills for the Electricity company. There are 3 types of customers: residential (code R) , commercial (code C) , and Industrial (code I). - For a code R customer, the bill is \$10 plus \$0.05 for each kilowatt used. - For a code C customer, the bill is \$1000 for the first 2000 kilowatt, and \$0.005 for each additional kilowatt used. - For a code I customer, the bill is \$1000 if he used less than 4000 kilowatt, \$2000 if he used between 4000 and kilowatt, or \$3000 if he used more than kilowatt. The inputs of the program should be the type of customer ( R C or I) and the kilowatts used. The output should be the amount of money the customer has to pay.
Find The output int x = 10 + 8 / 3 * 2 + 10 ; switch ( x ) {
case 21: printf ( “ Eeny “ ) ; break; case 24: printf ( “ Meeny “ ) ;break; case 25 : printf ( “ Miny “ ) ; break; case 28 : printf ( “ Mo “ ) ; break; default : printf(“ None of them “); }
Write a program that reports the content of a compressed-air cylinder based upon the first letter of the cylinder’s color. The program input is a character representing the observed color of the cylinder: ‘Y’ or ‘y’ for yellow, ‘G’ or ‘g’ for green and so on. Given: Color Content Orange Ammonia Brown Carbon Monoxide Yellow Hydrogen Green Oxygen
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Affording strategic proximity to the Strait of Gibraltar, Mo : GMAT Sentence Correction (SC)
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# Affording strategic proximity to the Strait of Gibraltar, Mo
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21 Dec 2013, 00:24
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Affording strategic proximity to the Strait of Gibraltar, Morocco was also of interest to the French throughout the first half of the twentieth century becuase they assumed that if they did not hold it, their grip on Algeria was always insecure.
a) if they did not hold it, their grip on Algeria was always insecure
b) without it their grip on Algeria would never be secure
c) their grip on Algeria was not ever secure if they did not hold it
d) without that, they could never be secure about their grip on Algeria
e) never would their grip on Algeria be secure if they did not hold it.
Meaning: reason for why was morocco important to french
Error analysis: C1- Affording strategic proximity to the Strait of Gibraltar, Morocco was also of interest to the French throughout the first half of the twentieth century
C2- becuase they assumed
C3 that if they did not hold it,
C4-their grip on Algeria was always insecure.
SV: No error:
V: Conditional tense rule violated as if <past> , then <would> is correct usage and not if <past> , then <was> is not correct usage
Pronoun: correct : they / thier logially and unambigously refer to french and it logicaly refers to morocco
Parallelism: Not tested and hence no error
Meaning: No error
Modifier: No error
Idioms: Not tested and hence no error
Others: None
POE:
a) if they did not hold it, their grip on Algeria was always insecure
Faulty as pointed above
b) without it their grip on Algeria would never be secure
Although would is used correctly but in the if, then construction 'if' is not used. From the theory taught in the verb conditional concept
i have learned that then may be ommited but if should always be present. Can you please explain then why B is correct though 'if' is ommited
c) their grip on Algeria was not ever secure if they did not hold it
Conditional error as in A pronoun reference error as well- 'it' can refer to both algeria and morrocco
d) without that, they could never be secure about their grip on Algeria
pronoun error- that can not be used in place of it to refer to morrocco
Conditional errror: Assumption, so would need to be used in place of could
e) never would their grip on Algeria be secure if they did not hold it.
Pronoun error as 'it' can refer to both algeria and morrocco.
I could not find any other error in this choice
Can you please share your view on the analysis posted above. Also please explain why the usage of if , then construction is correct in B though 'if' is missing
PS:i understand that 'would can be used without a condition but here as there is a condition so i expect that whole if, then framework must be used (ofcourse then can be ommited but not if as mentioned in the concept file of verb conditional)
Rahul Vijay
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Re: Affording strategic proximity to the Strait of Gibraltar, Mo [#permalink]
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21 Dec 2013, 02:36
First of all the part of the sentence must be underlined and the title of the post must be the first sentence of the question.
Thanks
SV: No error:
V: Conditional tense rule violated as if <past> , then <would> is correct usage and not if <past> , then <was> is not correct usage
Pronoun: correct : they / thier logially and unambigously refer to french and it logicaly refers to morocco
Parallelism: Not tested and hence no error
Meaning: No error
Modifier: No error
Idioms: Not tested and hence no error
Others: None
I do not agreee on that point because you always have a meaning on a sentence. You certainly could eliminate the wrong answer based on meaning/grammar errors
Affording strategic proximity to the Strait of Gibraltar, Morocco was also of interest to the French throughout the first half of the twentieth century becuase they assumed that if they did not hold it, their grip on Algeria was always insecure.
a) if they did not hold it, their grip on Algeria was always insecure
they assumed that if they......is already wrong just only to see this construction. If this is not enough for you, the sentence is constructed really really bad because the core part of the same is putted as an incidental phrase that we do not know what it modifies. let's take a look:
Affording strategic proximity to the Strait of Gibraltar, Morocco was also of interest to the French throughout the first half of the twentieth century becuase they assumed that if they did not hold it , their grip on Algeria was always insecure.
Notice how the main part is between TWO commas and this is non sensical. Even Without spotting a gramma error or a meaning issue or else: option A is wrong
b) without it their grip on Algeria would never be secure
Correct
c) their grip on Algeria was not ever secure if they did not hold it
here what the sentence says to us is this: if we do not have the grip we are insicure ?? we are insicure woithout a grip ???? just crazy. we are talking about countries not a grip; the grip should work as a means
d) without that, they could never be secure about their grip on Algeria
they assumed that ............without that, they could............Moreover, they are not secure about a possible attack via marocco up to France or they are not secure about the grip ???
e) never would their grip on Algeria be secure if they did not hold it.
Once again: the grip is the pivotal point.............
As you can see, you could pick the right answer only basing your decision on meaning AND grammar rules or meaning solely.
Hope this helps.
regards
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Re: Affording strategic proximity to the Strait of Gibraltar, Mo [#permalink]
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21 Dec 2013, 07:34
there is one main problem with A and C.
the pattern "if do(did), then do (did) " is used to say about causal fact.
if temperature is 100 degree C, water boils.
this is causal fact.
we can not use "assume" before above pattern.
that is why A and C is wrong. this point is basic because the pattern is explained in all grammar books
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Re: Affording strategic proximity to the Strait of Gibraltar, Mo [#permalink]
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23 Dec 2013, 01:58
Hi carcass,
Agree with your explanation. I have a question.
Since both "assume" and "if" are conditional, do they not create a sort of redundancy if used together? Can choices A,C and E be eliminated based on this?
Please correct me if I went wrong.
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Re: Affording strategic proximity to the Strait of Gibraltar, Mo [#permalink]
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23 Dec 2013, 10:34
rahulvv wrote:
Affording strategic proximity to the Strait of Gibraltar, Morocco was also of interest to the French throughout the first half of the twentieth century becuase they assumed that if they did not hold it, their grip on Algeria was always insecure.
a) if they did not hold it, their grip on Algeria was always insecure
b) without it their grip on Algeria would never be secure
c) their grip on Algeria was not ever secure if they did not hold it
d) without that, they could never be secure about their grip on Algeria
e) never would their grip on Algeria be secure if they did not hold it.
Meaning: reason for why was morocco important to french
Error analysis: C1- Affording strategic proximity to the Strait of Gibraltar, Morocco was also of interest to the French throughout the first half of the twentieth century
C2- becuase they assumed
C3 that if they did not hold it,
C4-their grip on Algeria was always insecure.
SV: No error:
V: Conditional tense rule violated as if <past> , then <would> is correct usage and not if <past> , then <was> is not correct usage
Pronoun: correct : they / thier logially and unambigously refer to french and it logicaly refers to morocco
Parallelism: Not tested and hence no error
Meaning: No error
Modifier: No error
Idioms: Not tested and hence no error
Others: None
POE:
a) if they did not hold it, their grip on Algeria was always insecure
Faulty as pointed above
b) without it their grip on Algeria would never be secure
Although would is used correctly but in the if, then construction 'if' is not used. From the theory taught in the verb conditional concept
i have learned that then may be ommited but if should always be present. Can you please explain then why B is correct though 'if' is ommited
c) their grip on Algeria was not ever secure if they did not hold it
Conditional error as in A pronoun reference error as well- 'it' can refer to both algeria and morrocco
d) without that, they could never be secure about their grip on Algeria
pronoun error- that can not be used in place of it to refer to morrocco
Conditional errror: Assumption, so would need to be used in place of could
e) never would their grip on Algeria be secure if they did not hold it.
Pronoun error as 'it' can refer to both algeria and morrocco.
I could not find any other error in this choice
Can you please share your view on the analysis posted above. Also please explain why the usage of if , then construction is correct in B though 'if' is missing
PS:i understand that 'would can be used without a condition but here as there is a condition so i expect that whole if, then framework must be used (ofcourse then can be ommited but not if as mentioned in the concept file of verb conditional)
Rahul Vijay
Hi Rahul,
Thanks for posting your doubt here.
The first thing we need to notice here is that the original sentence does use the "if... then..."construction. However, that is not the case with the Correct answer choice B. The tense that this choice has is basically past-future of "will" that is "would".
In the time frame of past, when we speak of a future event, then we use "would" for such an event. For example:
Shahjahan was not aware that his son Aurangzeb would be responsible for the downfall of the Mughal Empire.
Same is the usage of "would" in the correct answer choice B. At that time, French thought without their grip on the Strait of Gibraltar, their grip on Algeria would not be secure.
So we actually do not have any "if... then..." construction here.
Hope this helps.
Thanks.
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Re: Affording strategic proximity to the Strait of Gibraltar, Mo [#permalink]
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23 Dec 2013, 18:43
Thanks!
But can we say that the condition is present but it is not preceded by 'if'?
without it their grip on Algeria would never be secure ----> 'Without it' is the condition and 'their grip....' is the effect
This sentence could as well be reworded as if they did not hold morrocco, their grip on Algeria would never be secure?
Now if we can re-word the sentence as shown above then it implies that even 'if' can be ommited and the condition can be stated without 'if', provided the sentence makes sense.
Thanks,
Rahul
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Statistics Tutor: None Selected Time limit: 1 Day
Suppose that the New England Colonials baseball team is equally likely to win any particular game as not to win it. Suppose also that we choose a random sample of Colonials games.
1. Estimate the number of games in the sample that the Colonials win by giving the mean of the relevant distribution (that is, the expectation of the relevant random variable). Do not round your response.
2. Quantify the uncertainty of your estimate by giving the standard deviation of the distribution. Round your response to at least three decimal places.
Jun 16th, 2015
X = the number of games that the BC win
then X has binomial distribution BIN(n = 20, p = 1/2)
and then E(X) = np = 20/2 = 10
Var(X) = np(1-p) = 20/4 = 5
Stdev (X) = sqrt(5)
Please let me know if you nea case for torture by michale levined any clarification. I'm always happy to answer your questions.
Jun 16th, 2015
Hello the number did not show up in my question please see that it is suppose to be 30.
choose a random sample of 30 Colonials games
Jun 16th, 2015
wait
Jun 16th, 2015
3x-x+2=4
Jun 16th, 2015
plot_formula__1_.jpg
plot_formula.jpg
Why are you using 20 if the number games are 30?
Jun 16th, 2015
...
Jun 16th, 2015
...
Jun 16th, 2015
Dec 4th, 2016
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# Math
posted by .
Choose the one alternative that best completes the statement or answers the question.
Solve the problem. Assume that simple interest is being calculated in each case. Round your answer to the nearest cent.
Martin takes out a simple-interest loan at 7.5 %. After 6 months, the amount of interest on the loan is \$69.64. What was the amount of the loan?
I do not understand how to do this homework question can someone please help me to understand how to figure this one out thank you so much
• Math -
principle= rate x time
p=rt
\$69.64=7.5% x t
solve for t
• Math -
But how do you figure out T?
Out of the multiple choice answer nothing is coming close to one of the answers
these are the multiple choice answers
A)
\$2116.98
B)
\$18.57
C)
\$1857
D)
\$1818.44
• Math -
Allison's equation is wrong.
interest earned = P*i*t, where t = 0.5 years, P is the original loan amount, and i = 0.075 is the annual interest rate.
69.64 = 0.0375 P
P = \$1857
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### 12 - Applying proportional pressure in water boosting systems
This course test contains 4 questions and can be retaken at any time.
You must answer all questions correctly to pass the test and complete the course.
What characterises linear proportional pressure control?
In the example provided, what is the specific pressure loss of the pipework?
In the example provided, what is the annual power consumption when operating in constant pressure mode?
What is a key characteristic for static pressure losses?
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Q: What characterises linear proportional pressure control?
A: The booster pump adapts its performance in accordance to the flow rate
Q: What characterises linear proportional pressure control?
A: The booster pump provides a constant pressure regardless of the flow
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02: The booster pump provides a constant pressure regardless of the flow Your answer
03: The booster pump provides a flow rate according to the needed pressure
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A: The booster pump provides a flow rate according to the needed pressure
01: The booster pump adapts its performance in accordance to the flow rate
02: The booster pump provides a constant pressure regardless of the flow
03: The booster pump provides a flow rate according to the needed pressure Your answer
Q: In the example provided, what is the specific pressure loss of the pipework?
A: 0,50 kPa per meter of pipe
02: 0,50 m per meter of pipe
03: 450pa/m
Q: In the example provided, what is the specific pressure loss of the pipework?
A: 0,50 m per meter of pipe
01: 0,50 kPa per meter of pipe
03: 450pa/m
Q: In the example provided, what is the specific pressure loss of the pipework?
A: 450pa/m
Q: In the example provided, what is the annual power consumption when operating in constant pressure mode?
A: 19,000 kWh
02: 25,100 kWh
03: 29,100 kWh
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# The Maximal $\ell_2$ norm of a signed sum of vectors
Suppose we have $n$ vectors in $\mathbb{R}^n.$ Consider the signed sum of these vectors: $$U(s_1,\ldots,s_n)=s_1 v_1+s_2 v_2 + \ldots + s_n v_n$$ where $s_j$'s can only take values of $+1$ or $-1.$ I am interested in the maximal $\ell_2$ norm of the vector $U$ over all possible values of $(s_1,\ldots,s_n).$
This maximal $\ell_2$ norm of $U$ is certainly a function of $v_1,\ldots,v_n.$ For example, when $n=2,$ an easy argument in geometry shows that this maximal $\ell_2$ norm is proportional to the largest singular value of the matrix $V$ with $v_1,\ldots,v_n$ as columns. However, this is not true for $n=3$. I was wondering whether there is any existing result on this maximal $\ell_2$ norm as a function of the matrix $V,$ or is there an algorithm that solves this problem in linear time.
In particular, I was wondering what this function is for $n=3.$ Is it some function of the singular values? Thanks!
• I assume that the vectors are linearly independent? – Geoff Robinson Mar 12 '15 at 0:09
• A trivial bound is $\|U\|_2\le \sqrt{n}\|V\|$, and since the operator norm could be assumed for one of these $\pm 1$ vectors, this is all you can say in general. – Christian Remling Mar 12 '15 at 0:27
• Duh, wasn't this already closed? Did you look at MaxQP (that was in my last comment)... – Suvrit Mar 12 '15 at 1:30
• To Geoff: Yes, I am assuming they are linearly independent. Thanks! To Suvrit: Not yet but will try. I thought I didn't ask my question clearly so I posted it again. In particular, I was wondering an analytical solution for $n=3.$ Thanks! – KPU Mar 12 '15 at 2:25
• There are $2^{n-1}$ possibilities, so for $n=3$ it's just the maximum of four candidates, which is easy to compute. For example, let $G$ be the Gram matrix with $(i,j)$ entry $G_{ij} = v_i \cdot v_j$. Then the maximum norm is at most $\|G\|^{1/2}$ where $\|G\| := \sum_{i,j=1}^3 |G_{ij}|$. Equality holds unless all $G_{i,j}$ entries are nonzero and an odd number of the entries above the diagonal are negative, in which case the maximum is the square root of $\|G\| - 2 \min_{i,j} |G_{ij}|$. – Noam D. Elkies Mar 12 '15 at 2:47
• Thank you Igor! I was wondering what the answer is when $n=3.$ Thanks! – KPU Mar 12 '15 at 2:26
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# A company estimates that 0.9% of their products will fail within two years after the original warranty ends, with a replacement cost of \$50.If they offer a 2 year extended warranty for \$7, what is the company's expected value of each warranty sold?
Question
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A company estimates that 0.9% of their products will fail within two years after the original warranty ends, with a replacement cost of \$50.
If they offer a 2 year extended warranty for \$7, what is the company's expected value of each warranty sold?
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Step 1
The company's expected value of each w...
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# Questions tagged [general-equilibrium]
In economics, general equilibrium theory attempts to explain the behavior of supply, demand, and prices in a whole economy with several or many interacting markets, by seeking to prove that the interaction of demand and supply will result in an overall (or "general") equilibrium.
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# 5-Year CD: Average Rate of Banks in Boston-Cambridge-Quincy, MA-NH (MSA) (DISCONTINUED)
2014-01-08: 0.51 Percent (+ see more)
Weekly, As of Wednesday, Not Seasonally Adjusted, BNK5YR14460W156N, Updated: 2014-01-09 7:51 AM CST
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http://sciencedocbox.com/Physics/76611153-A-suggested-analytical-solution-for-vibration-of-honeycombs-sandwich-combined-plate-structure.html
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A Suggested Analytical Solution for Vibration of Honeycombs Sandwich Combined Plate Structure
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2 International Journal of Mechanical & Mechatronics Engineering IJMME-IJENS Vol:16 No:04 10 and mode shapes were investigated. A numerical technique was employed using ANSYS package In this research the natural frequencies of honeycombs sandwich plate with different Regular Hexagonal Honeycomb dimensions effect were evaluated and compared with those obtained numerically adopting finite element method using ANSYS Ver THE SUGGESTED ANALYTICAL SOLUTION The suggested analytical solution of vibration analysis of honeycomb plate included the determination of the mechanical properties of honeycomb structure and then evaluation the equivalent stresses and the natural frequency of with effect of honeycomb structural parameters. 2.1 Mechanical Properties of Honeycomb Materials The cells in a honeycomb-structure are usually hexagonal but they can also be triangular or square or rhombic. The honeycombs can be made of different types of materials such as metals wood polymer and ceramics or a combination of two or more of these materials. Investigating the in-plane properties ( transverse directions) demonstrate the failure mechanism and deformations of the cellular solid structures i.e defining the plane properties such as those shown in Fig. 1.The stiffness and strength in x-y plane. If the hexagonal honeycomb is regular (all angles are 30 o and wall thicknesses are equal as (Fig. 1.) then the in-plane properties are isotropic. This means that the in-plane properties of the honeycomb can be described by only two independent elastic modules for instance by Young's modulus Fig. 2 shows a hexagonal honeycomb regular structure in which the angles are equal ( 30 0 ) with constant wall thickness. This results in using in-plane properties of the honeycomb structures such as modulus of elasticity E and modulus of rigidiy G 1] as two independent properties. 1 1 Fig. 1. Honeycomb Structure Loaded in x or y-direction 1]. If the honeycomb cellular structure is not regular the in-plane properties are used with four modules (e.g. E h1 E h2 G h12 and h12 ). If the hexagonal honeycomb has a low relative density ( is small) get 1] h ( ) ( ) Fig. 2. A Regular Hexagonal Honeycomb 1]. The deflected shape of the honeycomb when loaded in x- or y- direction ( Fig. 1) is in bending direction for all cell walls (see Figure 3.a) and is described by five module: Young's module E h1 and E h2 a shear modulus G h12 and two Poisson's ratio h12 and h21. Using the reciprocal theorem gives (2) Reduce the five independent modules to four independent modules. When loading in x or y directions the four independent module described as 1] l t 2 2 y x (1) ( ) ( ) (3) Fig. 3b shows the honeycomb structure under shear loading. It is possible to show that the shear module G h12 to be 1]
3 International Journal of Mechanical & Mechatronics Engineering IJMME-IJENS Vol:16 No:04 11 ( ) ( ) Where E is modulus of elasticity of honeycomb material. 2 (4) And are mechanical properties of plate materials parts in x and y directions. and w is the deflection of plate in z-direction. Then with subjecting eq. 5 to plate structure as shown in fig. 4 to get the assuming bending moments acting on the honeycomb sandwich plate as (6) 1 a. Normal Stress Loaded in x and y b. Shear Stress Fig. 3. Honeycomb Deformed a. Deformed by Normal Stress Loaded in x and y b. Deformed by Shear Stress 1]. 2.2 Vibration Analysis of Honeycombs Sandwich Plate The analysis of honeycomb sandwich plate covered the determination of the general equation of motion of honeycomb sandwich plate structure where the plate is orthotropic properties in x and y direction. Also solving the general equation of motion of plate is included and then evaluating the natural frequency of honeycomb sandwich plate with various parameters for upper and lower plate parts and honeycomb dimensions and properties part. To Derive the general equation of motion of honeycomb sandwich plate using the properties and density of honeycomb sandwich materials equations (1 to 4) into the general equation of motion of plate 9] (10) Where are the bending moments and twisting moment per unit length of the orthotropic plate. The bending and twisting moments per unit length acting on the plate element are as follows 9] h M x x z dz h h M y y z dz h h M xy xy z dz h h h dz (5) h Where And (7) where are density of lower and upper plate parts respectively. For honeycomb sandwich plate shown in Fig. 4 assuming the structure combine of isotropic materials properties of upper and lower plate parts and orthotropic honeycomb sandwich part then stresses formulation giving in equation (6) become 1. Upper plate part since the plate part is isotropic propeties then (8) And Where are mechanical properties of uper plate part. Then by subsutution eq. 8 into eq. 6 givs 2. Orthotropic honeycomb sandwich part since the plate part is orthotropic propeties then (10) Then by subsutution eq. 10 into eq. 6 gives (9) (11)
4 International Journal of Mechanical & Mechatronics Engineering IJMME-IJENS Vol:16 No: Lower plate part since the plate part has isotropic propeties then (12) Where are mechanical properties of uper plate part. Then by subsutution eq. 8 in to eq. (6) gives a y z x (13) ( ] ] ( ] ) ] ) Then with integration eq. 14 gives ] ] (14) Upper Plate b ( ) Honeycomb structural h Up ( ) ] hh ( ) Lower Plate h l t h Lp ( ) ] ( ) Dimensions of Regular Hexagonal Honeycomb Fig. 4. Dimensions of Honeycomb Sandwich Combined Plate Structure. Therefore by substation eqs and 13 into eq. 7 gives the bending moments acting on the honeycomb sandwich plate as ( ) ] (15) And then by substitution eq. 15 into the general equation of motion of orthotropic plate eq. (10) gives the suggested general equation of motion of honeycomb sandwich plate as ] ] ( ] )
5 International Journal of Mechanical & Mechatronics Engineering IJMME-IJENS Vol:16 No:04 13 ( ) the suggested general equation of natural frequency for honeycomb sandwich plate structure with various mechanical properties and dimensions of honeycomb sandwich plate combined as ( ) ] ( ) ( ) ( ) ] ( ) ] ( ) ( ) ( ) ] ( ) ( ( ) ] ) (16) To solve equation (16) separation method can be used with assuming the function of deflection as 11] (17) To evaluate the behavior of deflection plate as a function of x and y directions the boundary conditions of the plate are required. Then for the equation of plate as a function of x and y direction as 10] (18) Then by substation eq. 18 into eq. 16 the suggested general equation of motion for honeycomb sandwich plate structure is obtained as ( ) ( ) ] ( ) ( ) ] ( ( ) ] (21) By building a computer program with using Fortran Power Station 4.0 can be evaluated the natural frequency of honeycomb sandwich plate with various parameters effect as 1. Thickness of upper and lower plates and honeycomb parte. 2. Hexagonal honeycomb. 3. t h and l of honeycomb parte. 4. Mechanical properties and density of honeycomb sandwich parte. 3. NUMERICAL INVESTIGATION The numerical investigation is carried out to calculate the natural frequency of honeycomb sandwich plate structure by using the finite elements method ANSYS program (Ver.14). Where the shell element (SHELL93) is used to build the model of honeycomb sandwich plate structure figure 5.It is assumed that the shell element is employed with zero thickness or element tapering down to a zero thickness at any corner is not allowed Also the shear deflections are included in this element. The out-of plane stress for this element varies linearly through the thickness..the transverse shear stresses (SYZ and SXZ) are assumed to be constant through the thickness and the transverse shear strains are assumed to be small in a large strain analysis. ) ( ) ( ( ) ] ) (19) With comparison eq. (19) with general equation of motion of single degree of freedom for free vibration structure as follows 12] (20)
6 International Journal of Mechanical & Mechatronics Engineering IJMME-IJENS Vol:16 No:04 14 maximum percentage discrepancy between the analytical and numerical results is about (2%). Fig. 5. Shell93-8 Node Element Geometry. 4. RESULTS AND DISCUSSIONS The results evaluated are included the natural frequency of honeycomb sandwich simply supported plate with various angle and thickness of honeycomb part and different upper and lower plates parts thicknesses. The materials are assumed to be distributed to upper and lower plate parts and honeycomb part are aluminium materials with mechanical properties as And the dimensions of plates are (22) the results of natural frequency evaluated are shown in figures 6 to 9 using various parameters. Figs. 6 and 7 show the natural frequency of honeycomb plate structure with different Hexagonal Honeycomb thickness angle various honeycomb parts thickness with upper and lower plates parts are respectively. From figures it is seen that the natural frequency increases with increase the Hexagonal Honeycomb thickness and angle. Figs. 8 and 9 show the natural frequency of honeycomb plate structure with different upper and lower plates parts and various of Hexagonal Honeycomb angle and various honeycomb part thickness for Hexagonal Honeycomb thickness. From figures it is clear that the natural frequency is increasing with increase the Hexagonal Honeycomb angle and honeycomb part thickness. The natural frequency decreases with increasing the upper and lower plates parts thickness. The results indicate that the increasing of Hexagonal Honeycomb thickness and angle and increasing the honeycomb plate part thickness causes an increase in the strength to weight ratio therefore an increase in the natural frequency of plate. The increase of upper and lower plate parts causes a decrease in the strength to weight ratio therefore decreasing the natural frequency of honeycomb sandwich plate. Table I shows the comparison between theoretical and numerical results of fundamental natural frequency of honeycomb plate structure with different honeycomb parts and upper and lower plate parts thickness. It is clear that the suggested analytical solution of honeycomb plate gives a good agreement compared with numerical results evaluated by using finite element technique ANSYS program ver. 14. The Table I Comparison between analytical and numerical natural frequency results for various honeycomb and upper and lower plate thickness. Natural Frequency Case h h =1 cm h Lp =h Up = 1 mm t=0.05 mm h h =2 cm h Lp =h Up = 1 mm t=0.05 mm h h =0.5 cm h Lp =h Up = 1 mm t=0.05 mm h h =0.5 cm h Lp =h Up = 2 mm t=0.05 mm (rad/sec) (Deg Analytica Numerica ) l l a. h h=0.5 cm Discrepancy %
7 International Journal of Mechanical & Mechatronics Engineering IJMME-IJENS Vol:16 No:04 15 b. h h=1 cm b. h h=1 cm c. h h=1.5 cm Fig. 6. Natural frequency of honeycomb plate with different dimensions of Regular Hexagonal Honeycomb and various honeycomb thickness for thickness of upper and lower plate hlp=hup=1 mm. c. h h=1.5 cm Fig. 7. Natural frequency of honeycomb plate with different dimensions of Regular Hexagonal Honeycomb and various honeycomb thickness for thickness of upper and lower plate hlp=hup=2 mm. a. h h=0.5 cm a. h Lp=h up=1 mm
8 International Journal of Mechanical & Mechatronics Engineering IJMME-IJENS Vol:16 No:04 16 b. h Lp =h up =1.5 mm b. h h=1 cm c. h Lp=h up=2 mm Fig. 8. Natural frequency of honeycomb plate with different angle of Regular Hexagonal Honeycomb and various honeycomb thickness and thickness of upper and lower plate for t=0.025 mm and. a. h h=0.5 cm c. h h=2 cm Fig. 9. Natural frequency of honeycomb plate with different angle of Regular Hexagonal Honeycomb and various thicknesses of upper and lower plate and honeycomb thickness for t=0.025 mm and. 5. CONCLUSIONS Some concluding observations from the investigation are given below: 1. The increase of Hexagonal Honeycomb thickness causes an increase in the strength of honeycomb structure then increasing the natural frequency of honeycomb plate. 2. With increasing the Hexagonal Honeycomb angle increases the strength and stiffness of Honeycomb structure therefore with increase the Hexagonal Honeycomb angle increases the natural frequency of honeycomb sandwich plate structure. 3. The increase of honeycomb thickness causes an increase in the strength of honeycomb structure then increasing the natural frequency of honeycomb plate. 4. The increasing of upper and lower plate parts thickness of honeycomb sandwich plate causes a decrease in the strength to weight ratio of honeycomb plate structure. Therefore with increasing the upper and lower plate parts thickness decreases the natural frequency of honeycomb sandwich plate structure.
9 International Journal of Mechanical & Mechatronics Engineering IJMME-IJENS Vol:16 No:04 17 REFERENCES 1. Dag Lukkassen and Annette Meidell "Advanced Materials and Structures and their Fabrication Processes" Narrik University College Hin Harish R and Ramesh S Sharma Vibration response analysis of honeycomb sandwich panel with varying Core Height International Journal of Emerging Technologies in Computational and Applied Sciences Vol. 5 No. 6 pp Jeom Kee Paik Anil K. Thayamballi Gyu Sung Kim The strength characteristics of aluminium honeycomb sandwich panels Thin- Walled Structures Vol. 35 pp C. W. Schwingshackl G. S. Aglietti P. R. Cunningham Determination of Honeycomb Material Properties: Existing Theories and an Alternative Dynamic Approach Journal of Aerospace Engineering Vitaly Skvortsov Sergey Krakhmalev Vitaly Koissin Andrey Shipsha Non-Stationary Oscillations of Sandwich Plates Under Local Dynamic Loading Tenth International Congress on Sound and Vibration A.Boudjemai M.H. Bouanane Mankour R. Amri H. Salem B. Chouchaoui MDA of Hexagonal Honeycomb Plates used for Space Applications World academy of science Engineering and technology Vol Jayasree Ramanujan Ancy Joseph and Nibu Mani Dynamic Response of sandwich Plates International Journal of Innovative Research in Science Engineering and Technology Vol. 2 No G. Sakar F. C. Bolat The Free Vibration Analysis of Honeycomb Sandwich Beam Using 3D and Continuum Model World Academy of Science Engineering and Technology International Journal of Mechanical Aerospace Industrial Mechatronic and Manufacturing Engineering Vol. 9 No Eduard Ventsel and Theodor Krauthammer Thin Plates and Shells Theory Analysis and Applications Book Marcel Dekker Inc Werner Soedel Vibrations of Shells and Plates Third Edition Revised and Expanded book Marcel Dekker Inc Muhannad Al-Waily "Analytical and numerical thermal buckling analysis investigation of unidirectional and woven reinforcement composite plate structural "International Journal of Energy and Environment Vol. 6 No Muhsin J. Jweeg Muhannad Al-Waily and Alaa Abdulzahra Deli Theoretical and Numerical Investigation of Buckling of Orthotropic Hyper Composite Plates International Journal of Mechanical and Mechatronics Engineering IJMME-IJENS Vol. 15 No
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12 terms
# dividing fractions word problems
#### Terms in this set (...)
3/8
You have 3/4 of a pizza and you want to share it equally between 2 people. How much of the pizza does each person get?
24
A baker is making cakes for a big party. She uses 1/4 cup of oil for each cake. How many cakes can she make if she has a bottle of oil that has 6 cups in it?
17 1/3
The serving size for the granola that Ted likes to eat for breakfast is 3/4 cup. How many servings are there in a box that holds 13 cups?
10 1/2
How many 1/2 cup servings are in a package of cheese that contains 5 1/4 cups altogether?
7
Mrs. Murphy's class is making pillow cases. Each pillow case uses 5/7 of a yard of fabric. How many pillow cases can they make out of 12 1/2 yards of fabric?
6
Miguel's company is making teddy bears. Each teddy bear uses 3/4 of a yard of fabric. How many teddy bears can they make out of 12 1/2 yards of fabric that they have?
67
A book shelf is 3 1/2 feet long. Each book on the shelf is 5/8 inches wide. How many books will fit on the shelf?
36
The plastic cups at the school dance each hold 2/3 cup
of punch. The punch bowl holds 24 cups of punch. How
many plastic cups can be filled from one bowl of punch?
1/6
Three friends have 1/2 a pizza left. If they divide it evenly,
what fraction of a pizza will each friend get?
20 and 2/3
Miguel's dog eats 3/4 pound of dog food each day.
How many days will a 20-pound bag of dog food last?
30
Our aquarium holds 5 liters of water. If we use a scoop that holds 1/6 of a liter of water, how many scoops will be needed to fill the aquarium?
12
Fancy hair bows take 1 and 3/4 of a yard of ribbon each. How many fancy hair bows can Maya make with 21 yards of ribbon?
STUDY GUIDE
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https://www.jiskha.com/questions/1664824/1-verify-that-4-12-is-the-solution-to-the-system-show-work-to-justify-your-answer
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# Algebra Help
1. Verify that (-4, 12) is the solution to the system.
2x+y=5
-5x-2y=-6
2. Solve the system by graphing. State the solution.
x + y =2
2y – x = 10
3. Solve the system by substitution. State whether the
system has one solution, infinite solutions, or no
solution.
2x + y = 10
x – 3y = -2
Can someone please explain to me how to do these?
1. 👍
2. 👎
3. 👁
1. 1.
substitute -4 for x and substitute 12 for y
2(-4) + 12 = 5 ??? NO WAY !!!!!
2. sketch graph
line one goes down from 2 on x axis at 45 degrees to right
line two goes from x = -10 when y = 0 up at 30 degrees to right
they hit at x = -2, y = 4
1. 👍
2. 👎
2. from first
y = (10-2x)
use in second
x - 3(10-2x) = -2
x - 30 + 6 x = -2
7 x = 28
x = 28/4 = 7
now go back and get y
1. 👍
2. 👎
3. 1. plug -4 and 12 into both equations to confirm the solution
2. graph the two equations...the solution is the intersection of the lines
... there are several graphing programs online ... or you could use graph paper, ruler, and pencil
3. solve the 2nd equation for x, in terms of y
... substitute the value for x into the 1st equation
... solve for y, then solve for x
1. 👍
2. 👎
4. Thank you guys SO MUCH
1. 👍
2. 👎
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# CNG litre to kg
### Solutions de Zeeshan: Conversion of CNG, liter to k
1. 1 liter cng = 0.18 kg at very low temperatures. Normally at cng stations and especially at ones where there is alot of rush, the gas does not get much time to cool down so the weight per liter decreases to like e.g. 1 liter = 0.15 kg So its normal if a 55 kg of cylinder fills up with 8 kg
2. b) This is the same as 200/1000= .2kg/litre. c) 1/0.2 = 5, so this is 5litres/kg, i.e. 1kg has a volume of 5 litres. Just follow the above process for the relevant pressure, temperature and composition you require
3. CNG Diesel; Fuel consumption per 100 km: 0 kg: liters: Price of fuel: CZK/kg: CZK/l: Fuel costs 1 km: 0 CZK: 0 CZK: The distance traveled per 1.000 CZK: 0 km: 0 km: Annual costs: km: 0 CZK: 0 CZK: Annual savings when driving on CNG: 0 CZ
Liter to Kilogram Conversion Chart - Gasoline. Note: Fractions are rounded to the nearest 8th fraction. Values are rounded to 3 significant figures. liters to kilograms of Gasoline; 1 liter = 0.75 kilogram: 2 liters = 1.5 kilograms: 4 liters = 3 kilograms: 5 liters = 3.75 kilograms: 8 liters = 6 kilograms: 1 / 16 liter 200 bar is approximately 2900 psi, I assumed 70 °F, and used their default composition of the gas. It calculates compressibility factor, Z, and density of 0.2244 g/cm³ (or kg/L) A 40 L physical volume at this temperature and pressure should give 9 kg approx Can somebody tell me how many litre of CNG in 1 KG ? Because now Gas stations measure in litres rather than KGs. Help would be appreciated. pwuser152213508578 (Waiz Shahid) 2018-03-27 12:19:55 +0500 #2. 714 grams = 1 liter CNG. This conversion was done to match it with the scale of petrol so people can easily calculate how much they will be.
We assume you are converting between liter and kilogram [sugar]. You can view more details on each measurement unit: liter or kilo gram The SI derived unit for volume is the cubic meter. 1 cubic meter is equal to 1000 liter, or 852.11336848478 kilo gram Cylinder with a 50 liter water-carrying capacity is capable of carrying approximately 9 kg of CNG. This is equivalent to 12.5 litres of petrol and will allow a run of about 150-160 km to a medium-sized 1300 CC car. An electronic fuel gauge fitted on the dashboard as part of the conversion kit indicates the quantity of CNG left in the cylinder
An empty CNG cylinder with a 50 litre-water-carrying capacity weighs 48 kg (approximately), and has a length of 835 mm and a diameter of 316 mm. The 50 litre capacity cylinder is the one most regularly used in CNG kits but cylinders with 45 litre, 55 litre, 60 litre and 65 litre capacity are used as well CNG is sold either by the kilogram or the cubic meter (m3) and LNG is measured in litres. A cubic metre of natural gas contains approximately 38.3* megajoules per cubic metre (MJ/m3), which is approximately the same amount of energy as a litre of diesel (38.8* Mj/l) This calculator allows you to calculate the amount of each fuel necessary to provide the same energy as 1 kg of hydrogen, 1 million cubic feet natural gas, 1 barrel of crude oil, or 1 gallon of other fuels, based on lower heating values. The conversion factors for this calculator are documented in the Energy Equivalency of Fuels table CNG Price Today (April 2021): Update with current 1Kg CNG price in Indian major cities & also check the CNG price hike/drop in all the major cities in India To be more precise, 1 liter LPG weights 0.651 kg. We get LPG (AutoGas) in liters. CNG is sold in kgs. Like this post 0 members liked this post.
### How many litres of cng gas equal to 1kg? Yahoo Answer
CNG @ 20MPa= 0.185 kg/L = 11.5 lb/ft3 = 5.66 lb/gge LPG (propane) = 0.540 kg/L = 33.7 lb/ft3 Hydrogen = 0.025 kg/L (35MPa); 0.08988 kg/m3 (STP) Coal ≈1.32 kg/L = 1230 metric ton/ha-m = 1800 sht ton/acre-foot API Gravity = (141.5/[Density in g/cm3 at 60 °F]) - 131. Most wholesale natural gas prices are quoted at this delivery point with an adder or discount based on local market dynamics and transportation cost. When you see the news reporting Natural Gas is at \$3.50 that usually means 1 MMBTU, bought today, to be delivered to Henry Hub next month, costs \$3.50. Gasoline, Diesel and CNG Engine was operated with optimum Compression ratio of 17.5 and CNG flow rate was optimized and found to be 0.5 kg/h. The effect of CNG induction on combustion and emissions characteristics of a.
### Calculator CNG4Yo
• Liters and cubic meters are both measures of volume - which can change based on pressure and temperature. Kilograms is a measure of mass/weight. The mass of a gas will not change. Imagine you blow up a balloon - let's say you've put 5 lungfulls of..
• KILOGRAMS (Kg) GAS CUBIC METERS (Nm3) LIQUID LITERS (L) OXYGEN 1 Kilogram 1.0 0.6996 0.8762 1 Nm3 Gas 1.4282 1.0 1.2511 1 L Liquid 1.1416 0.7995 1.0 NITROGEN 1 Kilogram 1.0 0.7996 1.2349 1 Nm3 Gas 1.2506 1.0 1.5443 1 L Liquid 0.8083 0.6464 1.0 ARGON 1 Kilogram 1.0 0.5605 0.7176 1 Nm3 Gas 1.78400 1.0 1.280
• He said that there is 1.4 litres of gas in a kilogram gas. As current price of gas is Rs66.56/kg in Sindh, after converting it into litres the price will be around Rs48/litre. However, he said that Sindh CNG Association was in a better position to talk about their stand on the issue
• CNG supplied by filling stations is measured in kilogrammes. 1 kg of CNG (year 2018) costs approximately 0,99 euro/kg and corresponds to about 1,5 litres of petrol. In short, the use of CNG offers considerable economic advantages
• back to Index Natural Gas Unit Equivalent to 1 cubic metre (m3) 35.301 cubic feet @ 14.73 psia and 60oF thousand cubic feet (Mcf) 1.05 G
• Natural Gas \$0.75 per diesel litre equivalent Savings \$0.40 per litre CONVERSION FACTORS - EQUIVALENCY BASED ON ENERGY CONTENT 1 kilogram of natural gas = 1.417 cubic metres of natural gas = 1.462 litres of diesel 1 cubic metre of natural gas = 0.706 kilograms of natural gas = 1.032 litres of diesel 1 litre of diesel = 0.684 kilograms of.
• 451 Liters to Tons 22 Tons to Liters 533 Grams to Pounds 2 Kilograms to Grain 2 Grain to Kilograms 2350 Liters to Tons 8 Liters to Troy ounces 0.008 Kilograms to Milliliters 128845 Liters to Tons 326400 Liters to Tons 67.4 Milligram to Kilogram-force second²/meter 21.421 Grams to Kilogram
There are a number of alternative fuel types, with the leading alternatives including fuels based on alcohol, liquefied petroleum gas (LPG), compressed natural gas (CNG), electricity, hydrogen, and vegetable oil degradation products. Below is a short description of each of these fuels. Alternative Fuels Biodiese GIGAJOULE TO KILOGRAM/LITER (GJ TO kg/l) FORMULA . To convert between Gigajoule and Kilogram/liter you have to do the following: First divide 1.0e+09 / 1/0.001 = 1000000.. Then multiply the amount of Gigajoule you want to convert to Kilogram/liter, use the chart below to guide you Use these conversion factors to calculate the gasoline gallon equivalent (GGE) of alternative fuels. These conversion factors help state and alternative fuel provider fleets report their compliance with Energy Policy Act (EPAct) requirements using the Alternative Compliance method 1 kilogram (kg) = 1 liter (l). Kilogram (kg) is a unit of Weight used in Metric system. Liter (l) is a unit of Volume used in Metric system. Please note this is weight to volume conversion, this conversion is valid only for pure water at temperature 4 °C. US oz = 28.349523125 g US fl oz = 29.5735295625 ml (milliliters) = 29.5735295625 g (grams.
### 1 liter of gasoline in kg - coolconversion
Only the mid-spec and base-spec 1.2-litre petrol variants of Tiago, Tigor, and Altroz to get the CNG option. CNG variants are likely to cost an additional Rs 50,000 over the non-CNG variants. Tata. However, you can continue your journey in the following ways Open original content in new window or send link via emai Natural Gas Unit Conversion Calculator NATURAL GAS UNIT CONVERSION CALCULATOR . From . Amount To. To Convert. Disclaimer: The MOE Conversion tool should be used as a guide only as it has been configured to provide approximate conversions. MOE accepts no responsibility for the use of, or reliance up on, the information provided by the Conversion. This includes a 14 Kg cylinder. A full tank of CNG, however, won't hold 14 Kg (as that's at peak pressure) and usually tops of at about 10 Kg. Larger cars with good sequential kits and a 14 Kg tank usually get a range of up to 150 km in the city, translating to mileage of about 15 kmpl Energy Heat Combustible LPG (liter) liter 10%1.69 Energy Heat Combustible Natural gas kWh 5%0.21 Energy Heat Combustible Petroleum coke kWh 10%0.35 Energy Heat Combustible UNG kWh 0.18 Energy Heat Organic combustible Biodiesel (kWh) kWh 0.0
b. 1 Kg. of CNG is energy equivalent to 1.39 liters of petrol and 1.18 liters of diesel. c. An amount of 8/9/10 Kg. CNG is stored in 40/50/60 litre size cylinder respectively which is equivalent to 11.2/12.5/14 liters of petrol equivalent. Vehicles can also be run on petrol in case it runs out of CNG About methane, gas; 1 cubic centimeter of methane, gas weighs 0.000554 gram [g] 1 cubic inch of methane, gas weighs 0.000320232 ounce [oz] Methane, gas weighs 0.000554 gram per cubic centimeter or 0.554 kilogram per cubic meter, i.e. density of methane, gas is equal to 0.554 kg/m³; at 0°C (32°F or 273.15K) at standard atmospheric pressure.In Imperial or US customary measurement system, the. Conversion factors. Power 1 horsepower = 736 watts Energy 1 kilocalorie = 4190 joules 1 kilowatthour (kwh) = 3600 kilojoules = 859 kilocalories 1 m3 natural gas = 10 kwh (9,769 kWh) 1 barrel of oil = 1700 kwh 1 calorie (cal) = 4.184 J 1 British Thermal Unit (BTU) = 1055 J Energy conten
In the US and Canada, LPG is >95% propane and is about 0.51 kg/L, so around 1.96 L/kg. In a lot of the world, it is a mixture of propane and butane ( it varies from about 40/60 to 60/40 by region) for lower vapor pressure. Butane is heavier so aro.. yaar i also think it is not correct formula cylinder capacioty in Litres X 0.18=Gas in KGs 40 X 0.18=7.2KG Gas (at 200Bars) bcoz what i experience is in max. i could get thi 1 kg of NG is approx. 52000 kJ/kg caloric value which fits with 12500 kcal/kg. Then 0.072 kg of NG gives me 0.072 x 52000 = 3744 kJ -OR- 0.072 kg/s NG gives me 3,744 kW (and not 1 KW as quoted). So either there is a huge mistake, or the KW figure is electricity and inefficiencies/losses of E-generation are taken into account
Well, the Celerio has been doing it for over 5 years now. Celerio has also got the S-CNG technology which helps the customers to cut down their fuel costs really well. Therefore, it is also on our list of best mileage CNG cars. The Celerio CNG is known to deliver around 31km/kg. Hyundai Santro (29km/kg hydrogen gas and liquid unit conversion tables - weight, gas volume, liquid volume (pounds, kilograms, standard cubic feet, standard cubic meters, gallons, liters) Unit Conversion Data for Hydrogen : Weight: Gas: Liquid: pounds (lb) kilograms (kg) cubic feet (scf) cu meters (Nm 3) gallons (gal) liters (l) 1 pound: 1.0: 0.4536. CNG Capacity is 10 Kg in Celerio Additionally - been its retro fitted - car can also run on Petrol for which fuel tank capacity is 35 Litre Generally - 80% i.e 8Kg is considered safe & ideal for running and then gas needs to be refilled again Taking in Cue Mileage of 20 to 22 Km - One can expect running in range of 160 to 175 Km - alone in CNG For simplicity, let's say that 1 kg of CNG is equivalent to 1.4-litre gasoline (that's in terms of energy). So, take this in mind when you are at the station comparing the price of 1 kg CNG with 1 litre of gasoline - the two are not having the same amount of energy! Let's do a quick example to make this more practical
This page features online conversion from normal cubic meter of natural gas to metre-kilogram.These units belong to different measurement systems. The first one is from Natural Gas Energy Equivalent.The second one is from Common Units. If you need to convert normal cubic meter of natural gas to another compatible unit, please pick the one you need on the page below CNG vs Petrol vs Diesel - Fuel Efficiency . Here is the present cost of Fuels in India.* Petrol - Rs. 78.57 Per Litre. Diesel - Rs. 70.26 Per Litre. CNG - Rs. 40.61 Per Kg *Fuel prices vary from city to city. Hence, bear in mind to compare fuel prices of the same places and never mix the prices in two different areas Natural Gas and Coal Measurements and Conversions Natural gas measurements and conversions. 1 cubic foot natural gas (NG) - wet = 1,109 Btu 1 cubic foot - dry = 1,027 Btu 1 cubic foot - dry = 1,087 kilojoules 1 cubic foot - compressed = 960 Btu 1 pound = 20,551 Btu 1 gallon - liquid = 90,800 Btu - HHV Liters of propane: 1.467 : Mcf of natural gas m 3 of natural gas: 28.174 equivalent kWh: 292.88 Btu: 1,000,000 Imperial gallons of # 2 fuel oil: 6,061 Litres of # 2 fuel oil: 27.55 Imperial gallons of # 6 fuel oil: 5.556 : Equivalent Mcf of natural gas: 0.003414 m 3 of natural gas: 0.0962 Btu: 3,414.4 Imperial gallons of # 2. Cylinders for Compressed Natural Gas (CNG) With a complete range of diameters (from 140 mm to 660 mm), capacities (from 10 to 700 litre) and pressures (up to 700 bar working pressure), Faber's cylinders are suitable for the most versatile applications in Natural Gas Vehicles, Refilling Stations and Trailers
### CNG from KG to Litre Conversion? - Vehicle Documentation
LPG conversion kg to litres (convert kg to litres LPG) and LPG litre (liter) to kg are the two most common LPG gas unit conversions. It is important to note that water and LPG (propane) have different conversion formulas because they have different density. Water Litre to kg: 1 Litre = 0.999975 kg, so for all practical purposes 1 L = 1 kg Water. The Santro also comes with a CNG variant. It has 1.1-litre bi-fuel (petrol and CNG) engine that makes 59hp and 85Nm. This engine comes with a 5-speed manual transmission. The fuel economy stands at 30.48 km/kg and the fuel capacity is 60 litres. The price of Santro's CNG variant is Rs 5.86 lakhs to Rs 5.99 lakhs, ex-showroom. Hyundai Grand.
### Convert liter to kilo gram - Conversion of Measurement Unit
CNG (Compressed natural gas) prices in Pakistan are divided in two regions, Region-I price is Rs. 81 /Kg and Region-II price is Rs. 81 /Kg. The prices are updated regularly on the bases of new updates On CNG mode, drivers can get a mileage of around 26.08 km/kg and 19.01 km/l with petrol (Manual Transmission). Speaking of the price, the CNG variant of Ertiga costs Rs 9.14 lakh (ex-showroom). 2 1 metric ton natural gas = 10 barrels 1 metric ton NGL (natural gas liquids) = 10.4 barrels Liquid fuels 1 cubic meter = 6.289 barrels 1 barrel = 159 liters 1 barrel = 42 US gallons 1 U.S. gallon = 231 cubic inches 1 U.S. gallon = .1337 cubic feet 1 U.S. gallon = 3.785 liter
### CNG for Automobiles Hindustan Petroleum Corporation
1 liter (l) = 1 kilogram (kg). Liter (l) is a unit of Weight used in Volume system. Kilogram (kg) is a unit of Weight used in Metric system. Please note this is volume to weight conversion, this conversion is valid only for pure water at temperature 4 °C. US oz = 28.349523125 g US fl oz = 29.5735295625 ml (milliliters) = 29.5735295625 g (grams. Energy Converter / Liquefied Natural Gas (LNG) Energy Equivalent / Kilogram Of Liquefied Natural Gas, Lower Heating Value Online converter page for a specific unit. Here you can make instant conversion from this unit to all other compatible units This is a conversion chart for normal cubic meter of natural gas (Natural Gas Energy Equivalent). To switch the unit simply find the one you want on the page and click it. You can also go to the universal conversion page. 2: Enter the value you want to convert (normal cubic meter of natural gas). Then click the Convert Me button
To calculate the Carbon Dioxide - CO 2 - emission from a fuel, the carbon content of the fuel must be multiplied with the ratio of molecular weight of CO 2 (44) to the molecular weight of Carbon (12) -> 44 / 12 = 3.7. Carbon Dioxide emission from burning a fuel can be calculated as. q CO2 = c f / h f M CO2 /M m [1]. where. q CO2 = specific CO 2 emission [kg CO2 /kWh We assume you are converting between kilogram [water] and liter. You can view more details on each measurement unit: kilo gram or liter The SI derived unit for volume is the cubic meter. 1 cubic meter is equal to 1000 kilo gram, or 1000 liter. Note that rounding errors may occur, so always check the results
### CNG Questions Answers - Misra auto Gas - Anan
The petrol variant gives 19.95 km/l and the CNG variant gives 26.55 km/kg. This high fuel efficiency lies at the core of the Tour S's performance. So, no matter how many trips it takes, you'll never feel a pinch. The Maruti Dzire Tour S costs between Rs.5.76 lakh and Rs.6.40 lakh ex-showroom. Maruti Suzuki Celerio CNG CNG is Compressed Natural Gas and another physical form of natural gas to make it convenient to handle. Natural gas is compressed to 200 bar pressure so that the volume is almost 1% of the original. The calorific value is higher than the Natural gas in the range of 95 MJ /kg. LPG can be in liquid form at a relatively low pressure of 2 to 5. The CNG kit for two wheelers comprises two CNG cylinders of 1.2 kg each, which can run up to 120 to 130 km per kg at an approximate cost of INR 0.60 per km per single fill. In comparison, a petrol powered scooter returns an average of 45-50 kmpl with a range of 340 kmpl on a single tank (5.3-litre fuel tank)
### Natural Gas NGV Global Knowledgebas
In physics, energy density is the amount of energy stored in a given system or region of space per unit volume.It may also be used for energy per unit mass, though a more accurate term for this is specific energy (or gravimetric energy density).. Often only the useful or extractable energy is measured, which is to say that inaccessible energy (such as rest mass energy) is ignored This page features online conversion from kilocalorie to kilogram of LPG, higher heating value.These units belong to different measurement systems. The first one is from Common Units.The second one is from Liquefied Petrolium Gas (LPG) Energy Equivalent. If you need to convert kilocalorie to another compatible unit, please pick the one you need on the page below kg SKE: kg RÖE: m³ natural gas: 1 kJ: 0.000034: 0.000024: 0.000032: 1 kcal: 0.000143: 0.0001: 0.00013: 1 kWh: 0.123: 0.086: 0.113: 1 kg SKE: 1: 0.7: 0.923: 1 kg oe. As the lightest/simplest hydrocarbon, natural gas has lover emissions than all other hydrocarbons. Generally, the emissions are 25% lower using natural gas compared to light fuel oil. Less risk with natural gas The flammability with emissions of natural gas is low. Natural gas require a gas/air mix of about 5-15% to be flammable Hydrogen has one of the highest energy density values per mass. Its energy density is between 120 and 142 MJ/kg. This means that for every 1 kg of mass of hydrogen, it has an energy value of 120-142 MJ. It is highly flammable, needing only a small amount of energy to ignite and burn. Hydrogen burns cleanly
### Energy Equivalency of Fuels Hydrogen Tool
You have a natural gas furnace in your home that used 67,000 cubic feet of natural gas for heating last winter. Your neighbor has a furnace that burns heating oil that used 500 gallons of heating oil last winter. You can convert the natural gas and heating oil consumption data into Btu to determine which home used more energy for heating Today's CNG price in Delhi is ₹ 43.4 per Kg. Check out the changes in cost of CNG in Delhi along with the historical rates of the city. Check CNG Price Chart, Trend and Map to know more.Delhi CNG price was last updated on 02 March 2021 It has the best mileage among CNG cars in India. It is a sensible car, perfect for daily commute. It is available in 6 variants and comes with the latest BSVI upgrade with its CNG engine. This 5-seater vehicle boasts a 177-litre boot space. It also features a 0.8-litre engine with a maximum power of 48PS and a 69Nm torque How heavy is concrete? Calculate how many kilograms ( kg - kilo ) of concrete are in 1 liter ( 1 L ). Specific unit weight of concrete - amount properties converter for conversion factor exchange from 1 liter L equals = 2.41 kilograms kg - kilo exactly for the masonry material type. To convert concrete measuring units can be useful when building with concrete and where handling of concrete is. Our CNG Tanks are USDOT FMVSS 304 and NGV-2 compliant certified, and tested to the highest standards. CNG Tanks come in 4 Types. Note: Compressed natural gas is very safe due to two important factors: the physical properties of natural gas itself and the CNG fuel system that is designed and built according to stringent quality standards
### CNG Price in India Today (23 April 2021), CNG Rate in
It will help farmers save up to 50 per cent on the fuel cost as the current diesel prices are Rs.77.43 per litre, whereas CNG is only ₹ 42 per kg, he said. The minsiter said the government is considering to convert diesel buses into CNG buses and will encourage retrofitting of CNG kits billion m 3 natural gas (Gm 3 NG) billion ft 3 natural gas (Gft 3 NG) million tonnes liquefied natural gas (Mt LNG) billion tonnes liquefied natural gas (Gt LNG) kg hard coal (kg SKE) t hard coal (t SKE) electronVolt. gigaelectronvolt (GeV) tera-electronvolt (TeV) mega-electronvolt (MeV) kilo-electronvolt (keV) electronvolt (eV) other energy unit The price of CNG here is Rs. 38.35 per KG. Get it done from the company or a specialised CNG fitment centre which specialises in Sequential Kits. It would cost you in the range of 45k to 60K. oxygen gas and liquid unit conversion tables - weight, gas volume, liquid volume (pounds, kilograms, standard cubic feet, standard cubic meters, gallons, liters) Unit Conversion Data for Oxygen . Weight: Gas: Liquid: pounds (lb) kilograms (kg) cubic feet (scf) cu meters (Nm 3) gallons (gal) liters (l) 1 pound: 1.0: 0.4536: 12.076: 0.3174: 0.105.
Today's CNG Price in New Delhi Rs. 43.40 per 1Kg in May 2021, also find the current CNG price hike/drop with historical CNG rates in New Delhi from Goodreturns There are 1000 liters in one cubic meter. 22.4 liters of any gas is equal to 1 mole, according to molar volume. Molar mass tells us that the main component of natural gas, methane, or CH4, has a. 1 Cubic Foot = 1,000 BTU (typical value) 100 Cubic Feet = 1C cubic feet = 0.1M cf = 1 Therm = 100,000 BTU: 1 mmBTU = 1,000,000 BTU = 10 Therms = 1 dth (dekatherm) = 1 mc
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# how ode solver works
조회 수: 3(최근 30일)
댓글: John D'Errico 2023년 3월 14일
Hello everyone,
I am looking for knowing or displaying time step in ode solver. Therefore, I ran through a lot of questions asked on Mathworks community.
As a conclusion, il is impossible to know it or to display it. Now, my main question changes. Does anyone know how the ode works?
For example, a for loop executes the statements inside for a value of i then go to the top, increment i and then executes the statements.
a while loop does the same as for loop but it looks if the condition is false or not.
I hope that you understand what i am askin.
Best Regards,
##### 댓글 수: 2표시 이전 댓글 수: 1숨기기 이전 댓글 수: 1
But, this is not what i'm lookin for.
댓글을 달려면 로그인하십시오.
### 채택된 답변
Steven Lord 2023년 3월 14일
I am looking for knowing or displaying time step in ode solver.
At what point during the solution process? If you're trying to control the step size or use the time step between the previous solver step and the current time in evaluating your ODE, those are not possible. In the former case the ODE solvers are adaptive solvers; they will choose the step size necessary to accurately take the step. In the latter, I think it highly unlikely that the mathematical form of your ODE include a delta-t term.
Therefore, I ran through a lot of questions asked on Mathworks community.
As a conclusion, il is impossible to know it or to display it.
You shouldn't care what steps the ODE solver internally chooses to take (some of which, I'd like to point out, the solver can reject.) If you care to have the solution returned or displayed at certain times that is possible.
To return the solution at specified times, specify the tspan input argument as a vector with more than 2 elements and the ODE solver will return the solution at those times. Alternately specify tspan as a two element vector, call the ODE solver with one output, and use deval to evaluate the solution at the desired times.
To display the solution after successful steps specify an OutputFcn in the ODE options. See the ballode and orbitode examples as they use an OutputFcn to visualize the path of the bouncing ball (ballode) or the orbiting bodies in the three-body problem (orbitode.)
Now, my main question changes. Does anyone know how the ode works?
Of course. Look at the Algorithms and References sections on the documentation pages for the ODE solvers like ode45. Alternately if the paper linked in the References section is too technical, Cleve Moler's Numerical Computing with MATLAB textbook has a chapter (the second longest in the book behind the introduction to MATLAB) on ordinary differential equations.
If you have a situation where you believe you need to know the time steps the ODE solver takes internally please provide more details about what you're doing. I can think of one case where you might think you need this information, and that's for solving delay differential equations. But in that case, don't use the ordinary differential equation solvers like ode45. Use the delay differential equation solvers like dde23 instead.
##### 댓글 수: 3표시 이전 댓글 수: 2숨기기 이전 댓글 수: 2
John D'Errico 2023년 3월 14일
Somehow I don't think an adaptive ODE solver, like ODE45, etc. may be the right tool here, as pointed out by @Steven Lord. Anyway, if your problem requires the step, then effectively, it is using a first order forward step, thus essentially just a forward Euler's method, at least for part of the solve.
So you may be better off just using a simple Euler method for the entire problem.
댓글을 달려면 로그인하십시오.
### 추가 답변(1개)
Alan Stevens 2023년 3월 14일
doc ode45
Go to the References section. The two references listed will supply you with all you need to know about the ode solvers!
##### 댓글 수: 1표시 없음숨기기 없음
But, this is not what i'm lookin for.
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## Monday, October 12, 2009
### Brayden's Scribepost October 12, 2009
Today's assignment was to answer 1 purple question and 1 of any colour question except black.
1. Would the jar look the same is the money was Canadian?
No, it would not look the same because there is a difference between Canadian currency and American currency.
2. What is the value of the jar?
The value of the jar is \$100.
Comment!
----------------------------------------------------------------------------
Part 2 October 13/09
1) Make a reasonable combination of American coins in the Jar
I would put the following in the jar.
50 rolls of pennies = 20\$
10 rolls of nickels = 20\$
6 rolls of dimes = 30\$
3 rolls of quarters = 30\$
= 100\$
I had have this many coins :
2000 pennies
400 nickels
300 dimes
120 quarters
= 2820 American coins in the jar
In my Canadian jar I would have the following coins.
1 roll of toonies = 50\$
1 roll of loonies = 25\$
1 roll of quarters = 10\$
1 roll of dimes = 5\$
2 rolls of nickels = 4\$
2 rolls of pennies = 1\$
= 100\$
This is where I found my answer
25 toonies
25 loonies
40 quarters
50 dimes
40 nickels
100 pennies
= 280 Canadian Coins in the jar.
When I compare the American coins to the Canadian coins there is a very big difference. There are 2820 American coins in the jar and that will take up approximetly half of the jar. If you put all the 280 Canadian coins in the jar it will approximetly take up around 1/10 of the jar.
P.S. Sorry for the website not working, It wouldn't work when I put it in edit html.
Hope you liked my post!
Comment!
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# Regression analysis for data on lettery, education, age, children
See attached data file.
3. Attached you will find the excel data and analysis on the topic of 'playing the lottery'. It is often claimed that individuals who play the lottery are those who can least afford to do so, making the impact of the lottery 'regressive'. In this context, regressive implies that lower income individuals play the lottery more than do higher income individuals.
The data for this analysis include 100 randomly chosen men from a major metropolitan area. For each individual, the following information has been recorded:
- The number of times in the preceding month the individual has played a state-sponsored lottery. This is the dependent variable in the accompanying regression analysis.
- The education of the individual, measured by the number of years of schooling completed. Note that 12 years of schooling completed means that the person has graduated from â??high schoolâ?, and 16 years of schooling completed means that the person has graduated from college.
- Age in years at the person's most recent birthday.
- The number of children currently living in the person's household.
- The person's annual 'wage and salary' income in the previous year, in thousands of dollars.
The output includes a listing of all 100 observations; descriptive statistics for each variable; and, the results of a regression analysis that uses lottery as the dependent variable, and education, age, number of children, and income as independent variables. Each of these outputs is on a separate worksheet.
a. Write out the regression equation, with specific intercept and slope estimates. That is, I want the estimated numerical values of the intercept and the slopes in the equation you write.
b. For the first row of actual data from the excel file, use the independent variable values to â??predictâ? the value of the dependent variable. For this row, also compute the 'residual' (actual value of lottery play minus the predicted value).
c. For the independent variables 'Age' and 'Income' interpret the numerical value of the estimated slope, which you can obtain from the excel regression output.
d. Evaluate the statistical significance of each of the four slope estimates. This can be done in a very summary way. Start out by indicating the 'null value' against which you will be testing each of the four slope estimates. Then, next to the name of each independent variable, state whether the slope is 'significant', and why. The 'why' should be stated in no more than a few words or a single sentence. Use a significance level of 0.05.
e. For the independent variable 'Income', interpret the p-value of the estimated slope, briefly and specifically (by specifically I mean with specific reference to the numerical value of the p-value as reported on the excel regression output).
f. Is the main argument made in the preamble to this question supported by the research? Explain why or why not, in a sentence or two.
g. Use the R-square and the standard error of the estimate to evaluate the overall reliability of the regression.
#### Solution Preview
Regression analysis
3. Attached you will find the excel data and analysis on the topic of playing the lottery. It is often claimed that individuals who play the lottery are those who can least afford to do so, making the impact of the lottery regressive. In this context, regressive implies that lower income individuals play the lottery more than do higher income individuals.
The data for this analysis include 100 randomly chosen men from a major metropolitan area. For each individual, the following information has been recorded:
• The number of times in the preceding month the individual has played a state-sponsored lottery. This is the dependent variable in the accompanying regression analysis.
• The education of the individual, measured by the number of years of schooling completed. Note that 12 years of schooling completed means that the person has graduated from high school, and 16 years of schooling completed means that the person has graduated from college.
• Age in years at the personals most recent birthday.
• The number of children currently living in the personal household.
• The personal annual wage and salary income in the previous year, in thousands of dollars.
The output includes a listing of all 100 observations; descriptive statistics for each variable; and, the results of a regression analysis that uses lottery as the dependent variable, and education, age, number of children, and income as independent variables. Each of these outputs is on a separate worksheet.
a. Write out the regression equation, with specific intercept and slope estimates. That is, I want the estimated numerical values of the intercept and the slopes in ...
#### Solution Summary
A regression analysis for data lettery, education, age and children are examined.
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# Be the Very First to Know What an Old Pro Thinks About What Is the Mean in Math
The typical deviation is utilized to gauge the variability of a data set and it’s a vital value of the confidence calculation. Besides the forecast itself we also have point estimates for each one of the components, together with upper and lower bounds for each one of these projections. However, it’s clearly not a great classifier although it may get 90% accuracy.
The second stage is a standard regression model that has the extra signals as independent variables. Now, consider that if a detector claims that each of the cases aren’t cancer will realize the ideal precision, but 0 recall. Easy and true, however, there’s this much more to dive into here.
Before you can start to understand statistics, you should understand mean, median, and mode. Mean, median, and mode are such fundamental mathematical concepts it’s tough to locate a circumstance where they wouldn’t apply in some manner. That’s why there might be more than 1 mode.
These questions are to name just a couple of many concerns to be planned for. You may be asked to send out written notice or post information on the website. Signs need to http://president.umd.edu/ be placed on the property, and an open house meeting is normally held.
It’s essential that you know precisely how your ex girlfriend feels about you before you try to bring up the matter of reuniting with her. If you’re a bad writer, please take the opportunity to write neatly. Have Allies and Advocate for Yourself Having allies in the workplace may earn a substantial difference in whether you’re ready to be authentic.
## What to Expect From What Is the Mean in Math?
FIND Introduction The FIND function is one other way to locate a character or string within another cell, and it’ll return the value connected with the starting place, the same as the SEARCH function. SEARCH Introduction The SEARCH function is a means to come across a character or string within another cell, and it’ll return the value related to the starting place. It is possible to just count in from both ends of the list till you meet in the center, if you would like, particularly if your list is short.
Students should be taught exact notation, when to utilize this, and, most of all, why they’re using it the way that they are. Not understanding the proof wasn’t an alternative. do my homework for me Simultaneous Equations The answers to these forms of exam questions are normally nice rounded numbers.
For that reason, it’s vital to devote more effort in cleaning and formatting data. In addition, a machine learning model is a distinctive sort of model that’s learned from from data. Each problem ought to be one PDF file 3.
Generally speaking, data science is the procedure of understanding and extracting knowledge from data. A few of you have national healthcare, for example. Very similar to the semantic segmentation problem described above, there are lots of cases where there’s a huge class imbalance.
There are a couple of requirements you’ll want to meet to be able to use the library. Think about a scenario where a development team doesn’t have automation testing capabilities. Webster’s definition of provider is one which provides.
## The Ugly Side of What Is the Mean in Math
There’s a revolution happening at the moment. Then the dog would be OK. In each class I will attempt to earn the discussion self-contained and to provide everybody something to take home, whatever the background.
## The 5-Minute Rule for What Is the Mean in Math
If you’ve got a specific sum of fun chips in a specific week, it’s much better to spread the chips around rather than loading a couple of nights with all the chips. I will break each part down to allow it to be simpler to read. You go back in the gym.
Since we’re taking a look at familiars I will examine this Wiccan terms. In the email that’s received a little while later there’s two sections we have to look closely at. You are able to also choose the degree of tightness you mean to achieve with natural products that you can’t with surgeries.
## The Appeal of What Is the Mean in Math
MATH-0960 and MATH-0980 taken before Fall 2016 will also meet with the prerequisite requirement for this program. Studies have a tendency to discover the objective as opposed to the subjective, so maybe there’s a benefit here that was missed. These programs do not have to get executed to be useful.
## What Is the Mean in Math Fundamentals Explained
Your incentives aren’t aligned. Understanding how to program doesn’t make you a software engineer. Modifying software is a truth of life.
If you realize that you are emotionally weak you can arrive at the point of being emotionally robust and balanced. One of the greatest facets of this endeavor is that there isn’t any single right answer! Machine learning has a wide definition and can mean many diverse things to numerous individuals.
This is a huge challenge for lots of people seeking to measure up to others. The miracles of prayer wouldn’t be possible. However, there’s always the possibility which I overlooked something or made a mistake.
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NEWTON
NEWTON
# How to make recursive function in Cairo Lang?
3 months ago
137
views
0
Hello! This is day 7 of the 17 days of the Cairo Challenge. And I have no idea how to solve the playground exercise “Recursion”. Perhaps you can help me?
``````// The following code prints the sum of the numbers from 1 to 10.
// Modify the function `compute_sum` to print all the intermediate sums:
// 1, 1 + 2, 1 + 2 + 3, ..., 1 + 2 + ... + 10.
// Note: you'll have to add the implicit argument output_ptr to the
// declaration of compute_sum (in order to use the serialize_word function):
// func compute_sum{output_ptr: felt*}(n: felt) -> (sum: felt)
// Use the output builtin.
%builtins output
from starkware.cairo.common.serialize import serialize_word
func compute_sum(n: felt) -> (sum: felt) {
if (n == 0) {
// When 0 is reached, return 0.
return (sum=0);
}
// Otherwise, call `compute_sum` recursively to compute 1 + 2 + ... + (n-1).
let (sum) = compute_sum(n=n - 1);
// Add the new value `n` to the sum.
let new_sum = sum + n;
return (sum=new_sum);
}
func main{output_ptr: felt*}() {
let (res) = compute_sum(n=10);
// Output the result.
serialize_word(res);
return ();
}
``````
Answers to this question are a part of the ✨ 17 days of Cairo Lang with Playground & Newton. ✨
Vote for your favorite answer - the best answer will win a \$10 award. A new day – a new reward! During the next 17 days, our goal is to attract more developers to the Cairo language and to systematize the knowledge of Cairo lang. Read rules
#17daysOfCairocairo-beginnerscairoplayground
Newton
3 months ago
0
``
``````// The following code prints the sum of the numbers from 1 to 10.
// Modify the function `compute_sum` to print all the intermediate sums:
// 1, 1 + 2, 1 + 2 + 3, ..., 1 + 2 + ... + 10.
// Note: you'll have to add the implicit argument output_ptr to the
// declaration of compute_sum (in order to use the serialize_word function):
// func compute_sum{output_ptr: felt*}(n: felt) -> (sum: felt)
// Use the output builtin.
%builtins output
from starkware.cairo.common.serialize import serialize_word
func compute_sum{output_ptr: felt*}(n: felt) -> (sum: felt) {
if (n == 0) {
// When 0 is reached, return 0.
return (sum=0);
}
// Otherwise, call `compute_sum` recursively to compute 1 + 2 + ... + (n-1).
let (sum) = compute_sum(n=n - 1);
// Add the new value `n` to the sum.
let new_sum = sum + n;
serialize_word(new_sum);
return (sum=new_sum);
}
func main{output_ptr: felt*}() {
let (res) = compute_sum(n=10);
// Output the result.
//serialize_word(res);
return ();
}
``````
Prashant Banchhor
3 months ago
0
First things first - why recursion?
Because, in the time of writing this answer, loops don't exist in Cairo in the classical sense, so we need to use recursion in order to traverse an array. Read more here.
Solution
1. add `{output_ptr: felt*}` to the `compute_sum` function sgnature. like so:
``````func compute_sum{output_ptr: felt*}(n: felt) -> (sum: felt) {
``````
1. Add this line `serialize_word(new_sum);` to print the intermediary results:
``````let new_sum = sum + n;
serialize_word(new_sum);
return (sum=new_sum);
``````
-1
My answer with comments that may help to understand the execution of recursion
``````%builtins output
from starkware.cairo.common.serialize import serialize_word
func compute_sum{output_ptr: felt*}(n: felt) -> (sum: felt) {
// serialize_word(n) -> print 10, 9, 8, 7, ... 0
if (n == 0) {
// When 0 is reached, return 0.
return (sum=0);
}
let (sum) = compute_sum(n=n - 1);
// serialize_word(n) -> print 1, 2, 3, 4, ... 10
let new_sum = sum + n;
serialize_word(new_sum); // --> print 1, 3, 6, ... 55
return (sum=new_sum);
}
func main{output_ptr: felt*}() {
let (res) = compute_sum(n=10);
// Output the result.
// serialize_word(res);
return ();
}
``````
Repository challenge
-1
# Solution
``````%builtins output
from starkware.cairo.common.serialize import serialize_word
func compute_sum{output_ptr: felt*}(n: felt) -> (sum: felt) {
if (n == 0) {
// When 0 is reached, return 0.
return (sum=0);
}
// Otherwise, call `compute_sum` recursively to compute 1 + 2 + ... + (n-1).
let (sum) = compute_sum(n=n - 1);
// Add the new value `n` to the sum.
let new_sum = sum + n;
//print all the intermediate sums
serialize_word(new_sum);
return (sum=new_sum);
}
func main{output_ptr: felt*}() {
let (res) = compute_sum(n=10);
return ();
}
``````
# Output
``````Program output:
1
3
6
10
15
21
28
36
45
55
Number of steps: 129
Program hash: 0x034b8ec64f54dafcc743d1af9fbfe33a41b62f19d1849abcdfe3f78dc0b51afb
``````
# Explanation
main function calls compute_sum function by passing an argument of number of natural number(n=10). compute_sum function perform recursion by calling itself. It calls itself till n equal to zero. when base condition reached the sum returned to from where it got called. At this point sum equal to zero and n equal to 1 which results 1 and assigned it to new_sum and then print is using serialize_word. This repeats again sum equal to 1 and n equal to 2 which results in 3.It continues til n equal to 10.
0
``````
from starkware.cairo.common.serialize import serialize_word
func compute_sum{output_ptr: felt*}(n: felt) -> (sum: felt) {
if (n == 0) {
// When 0 is reached, return 0.
return (sum=0);
}
// Otherwise, call `compute_sum` recursively to compute 1 + 2 + ... + (n-1).
let (sum) = compute_sum(n=n - 1);
// Add the new value `n` to the sum.
let new_sum = sum + n;
serialize_word(n);
return (sum=new_sum);
}
func main{output_ptr: felt*}() {
let (res) = compute_sum(n=10);
// Output the result.
serialize_word(res);
return ();
}
``````
3 months ago
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# source:sasmodels/sasmodels/models/hollow_rectangular_prism.py@b297ba9
core_shell_microgelsmagnetic_modelticket-1257-vesicle-productticket_1156ticket_1265_superballticket_822_more_unit_tests
Last change on this file since b297ba9 was b297ba9, checked in by Paul Kienzle <pkienzle@…>, 15 months ago
lint
• Property mode set to 100644
File size: 7.5 KB
Line
1# rectangular_prism model
2# Note: model title and parameter table are inserted automatically
3r"""
4Definition
5----------
6
7This model provides the form factor, $P(q)$, for a hollow rectangular
8parallelepiped with a wall of thickness $\Delta$. The 1D scattering intensity
9for this model is calculated by forming the difference of the amplitudes of two
10massive parallelepipeds differing in their outermost dimensions in each
11direction by the same length increment $2\Delta$ (\ [#Nayuk2012]_ Nayuk, 2012).
12
13As in the case of the massive parallelepiped model (:ref:rectangular-prism),
14the scattering amplitude is computed for a particular orientation of the
15parallelepiped with respect to the scattering vector and then averaged over all
16possible orientations, giving
17
18.. math::
19 P(q) = \frac{1}{V^2} \frac{2}{\pi} \times \, \int_0^{\frac{\pi}{2}} \,
20 \int_0^{\frac{\pi}{2}} A_{P\Delta}^2(q) \, \sin\theta \, d\theta \, d\phi
21
22where $\theta$ is the angle between the $z$ axis and the longest axis
23of the parallelepiped, $\phi$ is the angle between the scattering vector
24(lying in the $xy$ plane) and the $y$ axis, and
25
26.. math::
27 :nowrap:
28
29 \begin{align*}
30 A_{P\Delta}(q) & = A B C
31 \left[\frac{\sin \bigl( q \frac{C}{2} \cos\theta \bigr)}
32 {\left( q \frac{C}{2} \cos\theta \right)} \right]
33 \left[\frac{\sin \bigl( q \frac{A}{2} \sin\theta \sin\phi \bigr)}
34 {\left( q \frac{A}{2} \sin\theta \sin\phi \right)}\right]
35 \left[\frac{\sin \bigl( q \frac{B}{2} \sin\theta \cos\phi \bigr)}
36 {\left( q \frac{B}{2} \sin\theta \cos\phi \right)}\right] \\
37 & - 8
38 \left(\frac{A}{2}-\Delta\right) \left(\frac{B}{2}-\Delta\right) \left(\frac{C}{2}-\Delta\right)
39 \left[ \frac{\sin \bigl[ q \bigl(\frac{C}{2}-\Delta\bigr) \cos\theta \bigr]}
40 {q \bigl(\frac{C}{2}-\Delta\bigr) \cos\theta} \right]
41 \left[ \frac{\sin \bigl[ q \bigl(\frac{A}{2}-\Delta\bigr) \sin\theta \sin\phi \bigr]}
42 {q \bigl(\frac{A}{2}-\Delta\bigr) \sin\theta \sin\phi} \right]
43 \left[ \frac{\sin \bigl[ q \bigl(\frac{B}{2}-\Delta\bigr) \sin\theta \cos\phi \bigr]}
44 {q \bigl(\frac{B}{2}-\Delta\bigr) \sin\theta \cos\phi} \right]
45 \end{align*}
46
47where $A$, $B$ and $C$ are the external sides of the parallelepiped fulfilling
48$A \le B \le C$, and the volume $V$ of the parallelepiped is
49
50.. math::
51 V = A B C \, - \, (A - 2\Delta) (B - 2\Delta) (C - 2\Delta)
52
53The 1D scattering intensity is then calculated as
54
55.. math::
56 I(q) = \text{scale} \times V \times (\rho_\text{p} -
57 \rho_\text{solvent})^2 \times P(q) + \text{background}
58
59where $\rho_\text{p}$ is the scattering length density of the parallelepiped,
60$\rho_\text{solvent}$ is the scattering length density of the solvent,
61and (if the data are in absolute units) *scale* represents the volume fraction
62(which is unitless) of the rectangular shell of material (i.e. not including
63the volume of the solvent filled core).
64
65For 2d data the orientation of the particle is required, described using
66angles $\theta$, $\phi$ and $\Psi$ as in the diagrams below, for further details
67of the calculation and angular dispersions see :ref:orientation .
68The angle $\Psi$ is the rotational angle around the long *C* axis. For example,
69$\Psi = 0$ when the *B* axis is parallel to the *x*-axis of the detector.
70
71For 2d, constraints must be applied during fitting to ensure that the inequality
72$A < B < C$ is not violated, and hence the correct definition of angles is
73preserved. The calculation will not report an error if the inequality is *not*
74preserved, but the results may be not correct.
75
76.. figure:: img/parallelepiped_angle_definition.png
77
78 Definition of the angles for oriented hollow rectangular prism.
79 Note that rotation $\theta$, initially in the $xz$ plane, is carried out first, then
80 rotation $\phi$ about the $z$ axis, finally rotation $\Psi$ is now around the axis of the prism.
81 The neutron or X-ray beam is along the $z$ axis.
82
83.. figure:: img/parallelepiped_angle_projection.png
84
85 Examples of the angles for oriented hollow rectangular prisms against the
86 detector plane.
87
88
89Validation
90----------
91
92Validation of the code was conducted by qualitatively comparing the output
93of the 1D model to the curves shown in (Nayuk, 2012).
94
95
96References
97----------
98
99.. [#Nayuk2012] R Nayuk and K Huber, *Z. Phys. Chem.*, 226 (2012) 837-854
100L. Onsager, Ann. New York Acad. Sci. 51, 627-659 (1949).
101
102
103Authorship and Verification
104----------------------------
105
106* **Author:** Miguel Gonzales **Date:** February 26, 2016
108* **Last Reviewed by:** Paul Butler **Date:** September 06, 2018
109"""
110
111import numpy as np
112from numpy import inf
113
114name = "hollow_rectangular_prism"
115title = "Hollow rectangular parallelepiped with uniform scattering length density."
116description = """
117 I(q)= scale*V*(sld - sld_solvent)^2*P(q,theta,phi)+background
118 P(q,theta,phi) = (2/pi/V^2) * double integral from 0 to pi/2 of ...
119 (AP1-AP2)^2(q)*sin(theta)*dtheta*dphi
120 AP1 = S(q*C*cos(theta)/2) * S(q*A*sin(theta)*sin(phi)/2) * S(q*B*sin(theta)*cos(phi)/2)
121 AP2 = S(q*C'*cos(theta)) * S(q*A'*sin(theta)*sin(phi)) * S(q*B'*sin(theta)*cos(phi))
122 C' = (C/2-thickness)
123 B' = (B/2-thickness)
124 A' = (A/2-thickness)
125 S(x) = sin(x)/x
126"""
127category = "shape:parallelepiped"
128
129# ["name", "units", default, [lower, upper], "type","description"],
130parameters = [["sld", "1e-6/Ang^2", 6.3, [-inf, inf], "sld",
131 "Parallelepiped scattering length density"],
132 ["sld_solvent", "1e-6/Ang^2", 1, [-inf, inf], "sld",
133 "Solvent scattering length density"],
134 ["length_a", "Ang", 35, [0, inf], "volume",
135 "Shortest, external, size of the parallelepiped"],
136 ["b2a_ratio", "Ang", 1, [0, inf], "volume",
137 "Ratio sides b/a"],
138 ["c2a_ratio", "Ang", 1, [0, inf], "volume",
139 "Ratio sides c/a"],
140 ["thickness", "Ang", 1, [0, inf], "volume",
141 "Thickness of parallelepiped"],
142 ["theta", "degrees", 0, [-360, 360], "orientation",
143 "c axis to beam angle"],
144 ["phi", "degrees", 0, [-360, 360], "orientation",
146 ["psi", "degrees", 0, [-360, 360], "orientation",
148 ]
149
150source = ["lib/gauss76.c", "hollow_rectangular_prism.c"]
151have_Fq = True
153 "equivalent cylinder excluded volume", "equivalent outer volume sphere",
154 "half length_a", "half length_b", "half length_c",
155 "equivalent outer circular cross-section",
156 "half ab diagonal", "half diagonal",
157 ]
158
159def random():
160 """Return a random parameter set for the model."""
161 a, b, c = 10**np.random.uniform(1, 4.7, size=3)
162 # Thickness is limited to 1/2 the smallest dimension
163 # Use a distribution with a preference for thin shell or thin core
164 # Avoid core,shell radii < 1
165 min_dim = 0.5*min(a, b, c)
166 thickness = np.random.beta(0.5, 0.5)*(min_dim-2) + 1
167 #print(a, b, c, thickness, thickness/min_dim)
168 pars = dict(
169 length_a=a,
170 b2a_ratio=b/a,
171 c2a_ratio=c/a,
172 thickness=thickness,
173 )
174 return pars
175
176
177# parameters for demo
178demo = dict(scale=1, background=0,
179 sld=6.3, sld_solvent=1.0,
180 length_a=35, b2a_ratio=1, c2a_ratio=1, thickness=1,
181 length_a_pd=0.1, length_a_pd_n=10,
182 b2a_ratio_pd=0.1, b2a_ratio_pd_n=1,
183 c2a_ratio_pd=0.1, c2a_ratio_pd_n=1)
184
185tests = [[{}, 0.2, 0.76687283098],
186 [{}, [0.2], [0.76687283098]],
187 ]
Note: See TracBrowser for help on using the repository browser.
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# How many calories are lurking in your food?
FOOD & DRINK
By: Staff Writer
5 Min Quiz
Image: shutterstock
Test your knowledge of the calorie content in some common, and some not so common, foods. You'll be surprised to learn where those pesky calories are lurking and why they're ending up on your thighs. So let's get started and pinpoint some of the worst offenders.
# How many calories are in a large 14" hand-tossed pepperoni pizza from Domino's?
That's right. A large Domino's pizza will supply your full daily caloric needs. Hey, at least you got that covered!
# How many calories are in one cup of Grape Nuts cereal?
The healthy cereal has a whopping 413 calories in just one cup. Might be best to stick to the Fruit Loops after all.
# What's the calorie content of 2 tablespoons of smooth peanut butter?
With 188 calories for two tablespoons, smooth peanut butter is one of the most calorically dense foods that you can find, thanks to its fat content.
# How many calories are in one ounce of pecans?
In just a small little handful of pecans -- 19 halves, in this case -- you're getting 196 calories. Just think of what the pecan pie will cost you!
# How many calories in one pepperoni?
There are 10 calories in a 2-gram slice of pepperoni. Not much, right? Things get a little more problematic when these pepperonis are multiplied and attached to a pizza.
# How many calories in a cup of guacamole?
Yes, it's hard to have just one bite. But when you eat the whole bowl of guac, you're packing quite a punch on your hips.
# How many calories in two tablespoons of olive oil?
When you pour olive oil on your salad, you're going to have a whole new perspective now. Just two tablespoons have over 200 calories.
# What is the calorie content of one cup of coconut milk in a can?
There are a whopping 445 calories in a cup of coconut milk. That's like a whole meal.
# How many calories are in one date fruit?
There are a measly 20 calories in one pitted date fruit. Some people mistake them for being heavier than they are.
# How many calories in one block of chicken flavored ramen noodles?
That's right. One innocent block of chicken flavored ramen noodles has 376 calories. There's a reason why folks not only survive, but thrive, on the cup-a-noodle diet.
# How many calories in one tablespoon of condensed milk?
In one tiny tablespoon of condensed milk there are 62 calories. Please, don't drink the can.
# How many calories in the Fool's Gold Loaf, Elvis' favorite sandwich?
Elvis' famous peanut butter, jelly, and bacon sandwich from the now-closed Colorado Mine Company restaurant in Denver packed 8,000 calories -- or more, depending on who you ask. This explains so much.
# How many calories in a medium croissant?
Not too bad, right? The French pastry doesn't seem so decadent, after all. Just watch the butter that you add.
# How many calories are in a Big Mac?
A Big Mac will set you back 563 calories, which, as long as you don't add any sides, actually ain't all that bad.
# How many calories are there in a dozen breaded and fried oysters?
There are 350 calories in a dozen breaded and fried oysters. Go ahead and eat another 12.
# How many calories in one frosted strawberry Pop-Tart?
That ain't too bad, but who is satisfied after one Pop-Tart? They come in packs of two, after all. By the way, an unfrosted strawberry Pop-Tart has ten extra calories -- look it up! The crust is thicker on unfrosted ones, to make them more substantial.
# How many calories in an 18-ounce chocolate milkshake from Steak 'n Shake?
Could be better, could be worse. But truly, who can drink a whole 18-ounce milkshake?
# How many calories in a 2-ounce package of Skittles?
When you taste the rainbow, you don't taste fat, and that's why these colorful candies don't pack too big of a punch. Watch the sugar, though.
# How many calories are in a large AMC popcorn with no butter?
A trip to the movies has never been so caloric. Do watch out with that extra butter.
# How many calories in the Ultimate Red Velvet Cheesecake from the Cheesecake Factory?
Just one slice will cost you nearly a day's worth of calories. But hey, that's what the Cheesecake Factory is for.
# How many calories in large McDonald's french fries?
A large order of fries is even more deadly when you add it to a Big Mac. Recipe for disaster.
# How many calories are in 100 plain M&Ms?
There are approximately 3.5 calories in each plain M&M. But wait, who eats 100 M&Ms anyhow?
# How many calories in 100 Peanut M&Ms?
Even though plain M&Ms have 3.5 calories each, Peanut M&Ms have just about 10 each. Maybe we should all stick to plain, but peanuts do provide important vitamins and minerals.
# How many calories in one tablespoon of caviar?
There are 42 calories in one tablespoon of caviar. Considering the size of a jar of caviar, it's understandable how the rich stay so thin.
# How many calories in a 12-ounce margarita?
With a 12-ounce margarita on the rocks, you'll get 680 calories. This brings a whole new meaning to the term "liquid diet."
# How many calories in a Triple Whopper with Cheese from Burger King?
We know what it will do to your waistline, but just think of what it'll do to your arteries!
# How many calories are in a cup of blueberries?
In one whole cup of blueberries there are only 85 calories. Eat up, my friends. Eat up.
# How many calories in 1.5 pounds of lobster?
In a pound-and-a-half of lobster you're only getting 171 calories. Refer to the earlier comment about why the rich stay thin.
# How many calories in one ten-pound watermelon?
Are you surprised? That benevolent summer fruit can do a little damage if you eat the whole thing.
# How many calories are in one Sonic Corn Dog?
A corn dog isn't big in size, but it packs a lot of calories in its tiny frame. Think twice the next time you're at a carnival.
# How many calories in one cup of grapes?
An entire cup of grapes only costs you 62 calories. Just another reason to eat more fruit.
# How many calories in a stick of butter?
With 810 calories, you won't find many people eating a whole stick of butter. But if you cook like the French, that just might be what's in your souffle.
# How many calories in a serving of bowling alley nachos?
With just one dainty serving, around ten chips, you're getting up to 554 calories of nachos. Hope you're doing a lot of bowling.
# How many calories in a pound of tree bark?
If you started to crave tree bark, just go for a small portion. One whole pound of bark has 500-600 calories. Actually, a pound is a lot!
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Question 69357
(1/6)X + (7/10) = X
Subtract (1/6)x from both sides to get:
(5/6)x = 7/10
x= 42/50
x= 21/25
-------
Cheers,
Stan H.
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https://convertoctopus.com/11-8-seconds-to-weeks
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## Conversion formula
The conversion factor from seconds to weeks is 1.6534391534392E-6, which means that 1 second is equal to 1.6534391534392E-6 weeks:
1 s = 1.6534391534392E-6 wk
To convert 11.8 seconds into weeks we have to multiply 11.8 by the conversion factor in order to get the time amount from seconds to weeks. We can also form a simple proportion to calculate the result:
1 s → 1.6534391534392E-6 wk
11.8 s → T(wk)
Solve the above proportion to obtain the time T in weeks:
T(wk) = 11.8 s × 1.6534391534392E-6 wk
T(wk) = 1.9510582010582E-5 wk
The final result is:
11.8 s → 1.9510582010582E-5 wk
We conclude that 11.8 seconds is equivalent to 1.9510582010582E-5 weeks:
11.8 seconds = 1.9510582010582E-5 weeks
## Alternative conversion
We can also convert by utilizing the inverse value of the conversion factor. In this case 1 week is equal to 51254.237288136 × 11.8 seconds.
Another way is saying that 11.8 seconds is equal to 1 ÷ 51254.237288136 weeks.
## Approximate result
For practical purposes we can round our final result to an approximate numerical value. We can say that eleven point eight seconds is approximately zero weeks:
11.8 s ≅ 0 wk
An alternative is also that one week is approximately fifty-one thousand two hundred fifty-four point two three seven times eleven point eight seconds.
## Conversion table
### seconds to weeks chart
For quick reference purposes, below is the conversion table you can use to convert from seconds to weeks
seconds (s) weeks (wk)
12.8 seconds 0 weeks
13.8 seconds 0 weeks
14.8 seconds 0 weeks
15.8 seconds 0 weeks
16.8 seconds 0 weeks
17.8 seconds 0 weeks
18.8 seconds 0 weeks
19.8 seconds 0 weeks
20.8 seconds 0 weeks
21.8 seconds 0 weeks
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Post subject: Fastest method highest prime factor?Posted: Sat Aug 13, 2011 1:22 am
Joined: Mon Oct 18, 2010 10:48 am
I was talking with a friend today about a program to find the highest prime factor of x.
I postulated that one of the fastest methods is to create a for loop
[pseudocode]
While I is less than x
if x is divisible by I, set x to x/I and repeat this check.
Simple program; I typed it out and it's reasonably fast. But is there a faster way? (Assume no table of primes. No precalculation allowed)
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Post subject: Re: Fastest method highest prime factor?Posted: Sat Aug 13, 2011 5:39 pm
Joined: Thu Dec 31, 2009 8:54 pm
Location: Bay Area, California
NType3 wrote:
I was talking with a friend today about a program to find the highest prime factor of x.
I postulated that one of the fastest methods is to create a for loop
[pseudocode]
While I is less than x
if x is divisible by I, set x to x/I and repeat this check.
Simple program; I typed it out and it's reasonably fast. But is there a faster way? (Assume no table of primes. No precalculation allowed)
For small values of x it doesn't really matter but your existing for loop can be improved. For one, you don't need to check from i to x but from i to sqrt(x). Second, check for the value of 2 in a special case and then start at 3 and instead of adding 1 to i each iteration, add two.
For very large x the algorithms to factor a number (or to find the largest factor) get quite complicated. Currently the fastest known algorithm is general number field sieve (GNFS).
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Post subject: Re: Fastest method highest prime factor?Posted: Sat Aug 13, 2011 6:10 pm
Joined: Mon Oct 18, 2010 10:48 am
In lua, the code I came up with yesterday is as follows:
Code:
function getHighPrime(x)
q = 0;
while x%2==0 do
x=x/2;
q = q + 1;
end
print("2 was factored out "..q.." time(s) to leave a starting # of: "..x..".");
for i=3,x,2 do
if x%i == 0 then
while x%i == 0 do
if(x==i) then break end
x=x/i;
print("The new max search is: "..x,"After factoring out: "..i);
end
end
end
print("The highest prime factor is: "..x);
end
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Post subject: Re: Fastest method highest prime factor?Posted: Sat Aug 13, 2011 6:23 pm
Joined: Thu Dec 31, 2009 8:54 pm
Location: Bay Area, California
NType3 wrote:
In lua, the code I came up with yesterday is as follows:
What has you coding in Lua? It's a good language to be comfortable with because it is embedded so many places.
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Post subject: Re: Fastest method highest prime factor?Posted: Sat Aug 13, 2011 6:36 pm
Joined: Mon Oct 18, 2010 10:48 am
Lua is actually one of my favorite programming languages, believe it or not. The way data is stored makes it much more fluid in my mind. Second for me is C++, and third is Java.
To be fair, Lua doesn't have the best error descriptions (to say the least).
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Post subject: Re: Fastest method highest prime factor?Posted: Sun Aug 14, 2011 2:38 am
Joined: Fri Feb 08, 2008 1:47 am
Location: near Utrecht, Netherlands
Very cool solution. A few pointers:
-You don't need the if around the while
-All primes are of the form 3k+1 or 3k-1 so you could use that to make it faster
-You could skip the if x==i bit if you let it stop the for loop when x==1 and then print the value of i, or better yet, use while x=/=1 and manually increment i.
-Unlike previously suggested using sqrt(x) will not make your code any fastern since it would never check values of i above sqrt(x) anyway.
All these are only constant time optimalizations, your code runs in O(sqrt(x)). I think there may be a faster solution but I don't know if it's even been invented yet.
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Post subject: Re: Fastest method highest prime factor?Posted: Sun Aug 14, 2011 12:26 pm
Joined: Mon Oct 18, 2010 10:48 am
Here's the revised version:
Code:
function getHighPrime(x)
q = 0
while x%2==0 do
x=x/2
q = q + 1
end
print("2 was factored out "..q.." time(s) to leave a starting # of: "..x..".")
for i=3,x,2 do
while x%i == 0 do
x=x/i
print("The new max search is: "..x,"After factoring out: "..i)
end
if(x==1) then
print("The highest prime factor is: "..i);
break
end
end
end
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Post subject: Re: Fastest method highest prime factor?Posted: Sun Aug 14, 2011 10:19 pm
Joined: Wed May 13, 2009 4:58 pm
Location: Vancouver, Washington
bmenrigh wrote:
Second, check for the value of 2 in a special case.
NType3 wrote:
Code:
while x%2==0 do
x=x/2;
q = q + 1;
end
If you have bitwise operators, this should be faster if the compiler doesn't already do this replacement.
Code:
while(x & 1 == 0){
x = x>>1;
q = q + 1;
}
TomZ wrote:
-All primes are of the form 3k+1 or 3k-1 so you could use that to make it faster
I presume you mean to do this in addition to checking if it's odd. Checking for oddness lets you only test 50% of the numbers. Using only your 3k+-1 test, you have to test 66.67% of the numbers.
One quick way to combine the two without a bunch of ifs and tests is to just start i=5 and alternate between incrementing i by 2,4. You would have to hard code tests for 2,3,5. Here you only test 33.33% of the numbers.
You can take this idea further to eliminate factors of 5 by starting i=7, then using (4,2,4,2,4,6,2,6) to increment i. You'd only be saving testing about 26.67%
If you tried this for 7s, your list would be 40-50 integers summing to 210. Haven't done the math on how much would be tested, but I'm kind of doubting it would save you much.
TomZ wrote:
or better yet, use while x=/=1 and manually increment i.
I'm not familiar with the =/= operator (or lua that much). What does it do?
TomZ wrote:
-Unlike previously suggested using sqrt(x) will not make your code any faster since it would never check values of i above sqrt(x) anyway.
Of course it would speed it up. If you're testing 37 (prime) you should only check 2-6. You can skip 7-37. Even if you test 74 (2*37), you still only need to test 2-6.
If you do use this shortcut, be sure to do i*i<x instead of i<sqrt(x). A multiplication is much faster than sqrt(). I'm not sure if 3 additions are faster than 1 multiplication. If they are, keep a 2nd variable ii to represent i*i. When you increment i, do (ii += i + i + 1).
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Last edited by GuiltyBystander on Sun Aug 14, 2011 11:32 pm, edited 2 times in total.
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Post subject: Re: Fastest method highest prime factor?Posted: Sun Aug 14, 2011 10:41 pm
Joined: Mon Oct 18, 2010 10:48 am
I increment by two after testing directly for 2. The code effectively does the same thing as checking 2,3,5, then starting at 7 with an increment of two.
As for =/=, I gathered Tom meant to say 'not equal'. The not equal command in lua is ~=.
As for checking from I to sqrt(x), I would have to recalculate the square root of x every time the code iterates (or rather, calculate i*i>x) and this may (or may not, in some worst-case scenarios) slow the program down. In a worst-case scenario where the input is prime, checking the sqrt(x) would certainly increase the speed.
I did add the square root check (as i*i>x) and found that the program can process instantaneously the number 2293874894389493928384 (a random number generated by pounding the keypad), whereas the old program took minutes to process.
In lua, -- is a comment.
Code:
function getHighPrime(x)
j = 0
while x%2==0 do
x=x/2
j = j + 1
end
print("2 was factored out "..j.." time(s) to leave a starting # of: "..x..".")
for i=3,x,2 do
while x%i == 0 do
x=x/i
print("The new max search is: "..x,"After factoring out: "..i)
end
if(i*i>x) then break end
--if(i==x) then break end
end
print("The highest prime factor is: "..x);
end
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Post subject: Re: Fastest method highest prime factor?Posted: Mon Aug 15, 2011 3:23 pm
Joined: Thu Dec 31, 2009 8:54 pm
Location: Bay Area, California
NType3 wrote:
I did add the square root check (as i*i>x) and found that the program can process instantaneously the number 2293874894389493928384 (a random number generated by pounding the keypad), whereas the old program took minutes to process.
Hi Noah, I would expect the square root to give you a big speedup but for your example number of 2293874894389493928384, the largest prime factor is 3982421691648426959 and square root / counting by 2 / whatever is not going to ever be really fast on a number such as 3982421691648426959. How does your program know 3982421691648426959 is prime? Even the square root of it is nearly 2 billion. It should take your program quite a while.
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Post subject: Re: Fastest method highest prime factor?Posted: Mon Aug 15, 2011 11:59 pm
Joined: Mon Oct 18, 2010 10:48 am
I believe you are mistaken in that assumption, but please correct me if I'm wrong. If I am, I'm curious where my error lies. 3982421691648426959 > 2293874894389493928384.
2293874894389493928384 factors:
2293874894389493928384: 2^19*17*1451*177371352157.
Quote:
2 was factored out 19 time(s) to leave a starting # of: 4.3752191436567e+015.
The new max search is: 2.5736583197981e+014 After factoring out: 17
The new max search is: 177371352157 After factoring out: 1451
The highest prime factor is: 177371352157
Edit: 3982421691648426959 isn't even a prime number:
Quote:
> getHighPrime(3982421691648426959)
2 was factored out 10 time(s) to leave a starting # of: 3.8890836832504e+015.
The new max search is: 1.2963612277501e+015 After factoring out: 3
The new max search is: 30147935529073 After factoring out: 43
The highest prime factor is: 30147935529073
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Post subject: Re: Fastest method highest prime factor?Posted: Tue Aug 16, 2011 12:20 am
Joined: Thu Dec 31, 2009 8:54 pm
Location: Bay Area, California
NType3 wrote:
I believe you are mistaken in that assumption, but please correct me if I'm wrong. If I am, I'm curious where my error lies. 3982421691648426959 > 2293874894389493928384.
2293874894389493928384 factors:
2293874894389493928384: 2^19*17*1451*177371352157.
Quote:
2 was factored out 19 time(s) to leave a starting # of: 4.3752191436567e+015.
The new max search is: 2.5736583197981e+014 After factoring out: 17
The new max search is: 177371352157 After factoring out: 1451
The highest prime factor is: 177371352157
Edit: 3982421691648426959 isn't even a prime number:
Quote:
> getHighPrime(3982421691648426959)
2 was factored out 10 time(s) to leave a starting # of: 3.8890836832504e+015.
The new max search is: 1.2963612277501e+015 After factoring out: 3
The new max search is: 30147935529073 After factoring out: 43
The highest prime factor is: 30147935529073
Welcome to the wonderful world of floating point error. In Lua, everything is backed by the C "double" type which is 64 bits. 52 of those bits are mantissa bits. The number that you provided (2293874894389493928384) needs 71 bits to represent it. Because a double doesn't have 71 mantissa bits available you can not represent 2293874894389493928384 in Lua.
The number 2293874894389493928384 factors into: 2^6 * 3^2 * 3982421691648426959
The factors you got (2^19*17*1451*177371352157) result in 2293874894389493891072
Note that 2293874894389493928384 != 2293874894389493891072
The number you factored is off from the number you wrote by 37312 because of double precision error.
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Post subject: Re: Fastest method highest prime factor?Posted: Tue Aug 16, 2011 12:56 am
Joined: Mon Oct 18, 2010 10:48 am
Gah! Is there a way to correct this? In cpp, I would simply use a long unsigned int, but what about lua?
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Post subject: Re: Fastest method highest prime factor?Posted: Tue Aug 16, 2011 9:24 am
Joined: Thu Dec 31, 2009 8:54 pm
Location: Bay Area, California
NType3 wrote:
Gah! Is there a way to correct this? In cpp, I would simply use a long unsigned int, but what about lua?
The widths of the int types in C are mostly up to the compiler. For most compilers on 32 bit platforms both longs and ints are both 32 bits. On 64bit platforms longs are often 64 bits. Even a long long is generally only 64 bits. The number you are trying to crunch needs an 80 bit or even a 128 bit int. GCC supports a non-standard '__int128' type but it requires that the hardware support for 128 bit integers.
What you want is arbitrary precision integers. The typical way to do this in most languages is to use bindings to OpenSSL or to GMP. I have more experience with OpenSSL but from what I know, GMP is probably a better library to use. Nmap uses Lua for the "Nmap Scripting Engine" (NSE) and we went with OpenSSL bindings.
It looks like somebody has written a bigint library in purely Lua here: https://bitbucket.org/tedu/bigintlua/src/tip/bigint.lua
For what it's worth, Python's integer data types are arbitrary precision so you don't have to worry about these sorts of things. This is one of the reasons why Python is a bit slow.
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Post subject: Re: Fastest method highest prime factor?Posted: Tue Aug 16, 2011 9:54 am
Joined: Mon Oct 18, 2010 10:48 am
I feel like that's going to be obscenely slow, though.
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Post subject: Re: Fastest method highest prime factor?Posted: Tue Aug 16, 2011 10:56 am
Joined: Wed May 13, 2009 4:58 pm
Location: Vancouver, Washington
If you ever want to verify the accuracy of your program, I highly recommend using Wolfram Alpha.
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Post subject: Re: Fastest method highest prime factor?Posted: Tue Aug 16, 2011 1:32 pm
Joined: Thu Dec 31, 2009 8:54 pm
Location: Bay Area, California
NType3 wrote:
I feel like that's going to be obscenely slow, though.
The number "1125902456980891" is a 51 bit composite that should be very close to the maximally inefficient composite for your program to factor using your current strategy and still fit into just the mantissa of a double. You should be able to benchmark improvements to your program with this number.
Testing using GP/PARI on my system show it can be factored in less than 1 ms. The strategy GP/PARI uses though is quite complicated.
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Prior to using my real name I posted under the account named bmenrigh.
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Post subject: Re: Fastest method highest prime factor?Posted: Tue Aug 16, 2011 2:53 pm
Joined: Thu Dec 02, 2004 12:09 pm
Location: Missouri
GuiltyBystander wrote:
If you ever want to verify the accuracy of your program, I highly recommend using Wolfram Alpha.
Interesting little program...
While playing with it just now I found these 10 consecutive 11 digit numbers.
81680880640 = 2^10 * 5 * 313 * 50969
81680880641 = 41 * 1992216601 (near prime)
81680880642 = 2 * 3 * 19 * 79 * 467 * 19421
81680880643 = 81680880643 (prime)
81680880644 = 2^2 * 11 * 9967 * 186253
81680880645 = 3^2 * 5 * 7 * 13 * 17^2 * 69019
81680880646 = 2 * 40840440323 (near prime)
81680880647 = 81680880647 (prime)
81680880648 = 2^3 * 3 * 223 * 15261749
81680880649 = 81680880649 (prime)
So out of 10 numbers I found 3 primes and 2 near primes (the product of 2 primes). I must say I was very surprised. Are the primes much more dense that I thought or did I just get very very lucky.
Asked another way. Consider the set of all possible sets of 10 consecutive 11 digit numbers. No leading zeros... so we are talking about numbers between 10000000000 and 99999999999. What percentage of those sets are half or more made up of primes or near primes? Is there any set that contains more then 5 total primes and near primes? Can you give an example like I did above?
Are there any sets containing 4 or more primes? I think I can prove 4 is the max possible. 5 of the 10 must be even and 1 of the remaining 5 I think must be divisable by 3.
Maybe these questions are something you could test out your programs on... assuming I'm proposing a problem that can be solved in a reasonable amount of cpu time.
Carl
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Last edited by wwwmwww on Tue Aug 16, 2011 5:06 pm, edited 1 time in total.
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Post subject: Re: Fastest method highest prime factor?Posted: Tue Aug 16, 2011 3:09 pm
Joined: Thu Dec 31, 2009 8:54 pm
Location: Bay Area, California
wwwmwww wrote:
GuiltyBystander wrote:
If you ever want to verify the accuracy of your program, I highly recommend using Wolfram Alpha.
Interesting little program...
While playing with it just now I found these 10 consecutive 11 digit numbers.
81680880640 = 2^10 * 5 * 313 * 50969
81680880641 = 41 * 1992216601 (near prime)
81680880642 = 2 * 3 * 19 * 79 * 467 * 19421
81680880643 = 81680880643 (prime)
81680880644 = 2^2 * 11 * 9967 * 186253
81680880645 = 3^2 * 5 * 7 * 13 * 17^2 * 69019
81680880646 = 2 * 40840440323 (near prime)
81680880647 = 81680880647 (prime)
81680880648 = 2^3 * 3 * 223 * 15261749
81680880649 = 81680880649 (prime)
So out of 10 numbers I found 3 primes and 2 near primes (the product of 2 primes). I must say I was very surprised. Are the primes much more dense that I thought or did I just get very very lucky.
Asked another way. Consider the set of all possible sets of 10 consecutive 11 digit numbers. No leading zeros... so we are talking about numbers between 10000000000 and 99999999999. What percentage of those sets are half or more made up of primes or near primes? Is there any set that contains more then 5 total primes and near primes? Can you give an example like I did above?
Are there any sets containing 4 or more primes? I think I can prove 4 is the max possible. 5 of the 10 must be even and 1 of the remaining 5 I think must be divisable by 3.
Maybe these questions are something you could test out your programs on... assuming I'm proposing a problem that can be solved in a reasonable about of cpu time.
Carl
Hi Carl, thanks for taking this to the next level
I think we can answer your question statistically using a prime counting function pi(x) approximation.
As for the max of 4, I think the max might possibly be 3. 5 of the 10 must be even but 1 of the remaining 5 must be divisible 5 and 1 of the remaining 5 should be divisible by 3. It could be the same remaining number that is divisible by both 3 and 5 but I'm not sure what implications that would have on the others being divisible by 3 or by 7.
Only a range of 90 billion needs to be checked for an exact solution which should be computationally tractable. Probably beyond what you'd want to do on your laptop though.
Edit: the first such sequence of ten 11-digit numbers that contains 4 primes is 10000057420 to 10000057429:
10000057420: 2 2 5 500002871
10000057421: 10000057421
10000057422: 2 3 12391 134507
10000057423: 10000057423
10000057424: 2 2 2 2 7 17 73 71947
10000057425: 3 5 5 43 1013 3061
10000057426: 2 49081 101873
10000057427: 10000057427
10000057428: 2 2 3 3 19 2687 5441
10000057429: 10000057429
This does not seem to be an uncommon property:
Got 4 primes at 10000057420 to 10000057429
Got 4 primes at 10000057421 to 10000057430
Got 4 primes at 10000057930 to 10000057939
Got 4 primes at 10000057931 to 10000057940
Got 4 primes at 10000067920 to 10000067929
Got 4 primes at 10000067921 to 10000067930
Got 4 primes at 10000195810 to 10000195819
Got 4 primes at 10000195811 to 10000195820
Got 4 primes at 10000236220 to 10000236229
Got 4 primes at 10000236221 to 10000236230
Got 4 primes at 10000333330 to 10000333339
Got 4 primes at 10000333331 to 10000333340
Got 4 primes at 10000339750 to 10000339759
Got 4 primes at 10000339751 to 10000339760
Got 4 primes at 10000347100 to 10000347109
Got 4 primes at 10000347101 to 10000347110
[...]
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Post subject: Re: Fastest method highest prime factor?Posted: Tue Aug 16, 2011 3:44 pm
Joined: Mon Oct 18, 2010 10:48 am
I'll give this a try later! We shall see the estimated probability.
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Post subject: Re: Fastest method highest prime factor?Posted: Tue Aug 16, 2011 5:18 pm
Joined: Thu Dec 02, 2004 12:09 pm
Location: Missouri
bmenrigh wrote:
This does not seem to be an uncommon property:
Got 4 primes at 10000057420 to 10000057429
Got 4 primes at 10000057421 to 10000057430
Got 4 primes at 10000057930 to 10000057939
Got 4 primes at 10000057931 to 10000057940
Got 4 primes at 10000067920 to 10000067929
Got 4 primes at 10000067921 to 10000067930
Got 4 primes at 10000195810 to 10000195819
Got 4 primes at 10000195811 to 10000195820
Got 4 primes at 10000236220 to 10000236229
Got 4 primes at 10000236221 to 10000236230
Got 4 primes at 10000333330 to 10000333339
Got 4 primes at 10000333331 to 10000333340
Got 4 primes at 10000339750 to 10000339759
Got 4 primes at 10000339751 to 10000339760
Got 4 primes at 10000347100 to 10000347109
Got 4 primes at 10000347101 to 10000347110
[...]
1 in ~ 57420 is still rather rare if that is what this comes to.
I also notice if we are just looking for primes we probably should limit the search to sets of 9 consecutive numbers or each set of 4 is counted twice. It also appears that of all the sets found so far the last digit of the first prime is always "1"... any exceptions to this?
And if you include the near primes what is the most you can find (primes and near primes) within ten consective 11-digit numbers?
Carl
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Post subject: Re: Fastest method highest prime factor?Posted: Tue Aug 16, 2011 5:21 pm
Joined: Mon Oct 18, 2010 10:48 am
Adding this (most inefficient) block of code:
Code:
function seriesTenConsec(pause)
testCount = 0;
positiveCountOne = 0;
positiveCountTwo = 0;
positiveCountThree = 0;
positiveCountFour = 0;
numOfDiscoveredPrimes = 0;
for i=10000000000,99999999999,1 do
seriesCounter = 0
for j=i,i+10,1 do
--print(j,getHPDecider(j,true))
if(j == getHPDecider(j,true)) then
seriesCounter = seriesCounter + 1
--print("THIS IS PRIME.")
end
--print("\n")
numOfDiscoveredPrimes = numOfDiscoveredPrimes + seriesCounter
end
if(seriesCounter == 1) then positiveCountOne = positiveCountOne + 1
elseif(seriesCounter == 2) then positiveCountTwo = positiveCountTwo + 1
elseif(seriesCounter == 3) then positiveCountThree = positiveCountThree + 1
elseif(seriesCounter == 4) then positiveCountFour = positiveCountFour + 1
else end
testCount = testCount + 1
print("Stats 1: "..positiveCountOne.."/"..testCount.."\t\t\t"..roundOne((positiveCountOne/testCount)*100))
print("Stats 2: "..positiveCountTwo.."/"..testCount.."\t\t\t"..roundOne((positiveCountTwo/testCount)*100))
print("Stats 3: "..positiveCountThree.."/"..testCount.."\t\t\t"..roundOne((positiveCountThree/testCount)*100))
print("Stats 4: "..positiveCountFour.."/"..testCount.."\t\t\t"..roundOne((positiveCountFour/testCount)*100))
print(seriesCounter.." discovered primes the "..testCount.."th iteration.")
if(pause) then os.execute[[pause]] end
end
end
I wasn't sure how to approach it other than brute force (help in efficiency would be appreciated, though the program works). For those of you with the Lua console, the pastebin is here (and yes, you'll notice yet another inefficiency).
For clarity, getHPDecider is a function which runs either a muted or unmuted getHighPrime function.
The output is rather interesting. It seems, Carl, that you were *extremely* lucky.
1 prime: 34% occurrence rate.
2 primes: 6% occurrence rate.
3 primes: 0% occurrence rate.
4 primes: 0% occurrence rate.
*results after 60,000 iterations
I'll modify it to show five significant digits, and post new results shortly.
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Post subject: Re: Fastest method highest prime factor?Posted: Tue Aug 16, 2011 5:32 pm
Joined: Mon Oct 18, 2010 10:48 am
bmenrigh wrote:
Hi Carl, thanks for taking this to the next level
I think we can answer your question statistically using a prime counting function pi(x) approximation.
As for the max of 4, I think the max might possibly be 3. 5 of the 10 must be even but 1 of the remaining 5 must be divisible 5 and 1 of the remaining 5 should be divisible by 3. It could be the same remaining number that is divisible by both 3 and 5 but I'm not sure what implications that would have on the others being divisible by 3 or by 7.
Only a range of 90 billion needs to be checked for an exact solution which should be computationally tractable. Probably beyond what you'd want to do on your laptop though.
Edit: the first such sequence of ten 11-digit numbers that contains 4 primes is 10000057420 to 10000057429:
10000057420: 2 2 5 500002871
10000057421: 10000057421
10000057422: 2 3 12391 134507
10000057423: 10000057423
10000057424: 2 2 2 2 7 17 73 71947
10000057425: 3 5 5 43 1013 3061
10000057426: 2 49081 101873
10000057427: 10000057427
10000057428: 2 2 3 3 19 2687 5441
10000057429: 10000057429
This does not seem to be an uncommon property:
Got 4 primes at 10000057420 to 10000057429
Got 4 primes at 10000057421 to 10000057430
Got 4 primes at 10000057930 to 10000057939
Got 4 primes at 10000057931 to 10000057940
Got 4 primes at 10000067920 to 10000067929
Got 4 primes at 10000067921 to 10000067930
Got 4 primes at 10000195810 to 10000195819
Got 4 primes at 10000195811 to 10000195820
Got 4 primes at 10000236220 to 10000236229
Got 4 primes at 10000236221 to 10000236230
Got 4 primes at 10000333330 to 10000333339
Got 4 primes at 10000333331 to 10000333340
Got 4 primes at 10000339750 to 10000339759
Got 4 primes at 10000339751 to 10000339760
Got 4 primes at 10000347100 to 10000347109
Got 4 primes at 10000347101 to 10000347110
[...]
I apologize for the double post.
Because of this output, is it safe to assume that the digit in the ones' place alternates between a zero and a one? If so, why is this?
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Post subject: Re: Fastest method highest prime factor?Posted: Tue Aug 16, 2011 5:49 pm
Joined: Thu Dec 02, 2004 12:09 pm
Location: Missouri
NType3 wrote:
Because of this output, is it safe to assume that the digit in the ones' place alternates between a zero and a one? If so, why is this?
I can answer part of that...
Because I was including near-primes in my original question I was looking at sets of 10 consecutive numbers. This means either the first or the last number is even. So if you find 4 primes in a set as bmenrigh has then you know they are actually within a set of 9 consecutive numbers. Now when bmenrigh checked the next set of 10 consecutive numbers he actually found these same 4 primes again. So there are actually half as many sets in that list as there appears to be. And of the sets found the last digit of the first prime was "1". Is that always the case? I don't know... that's what I asked above as well. If it is... the question as to why does become very interesting.
Carl
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Last edited by wwwmwww on Tue Aug 16, 2011 5:51 pm, edited 1 time in total.
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Post subject: Re: Fastest method highest prime factor?Posted: Tue Aug 16, 2011 5:51 pm
Joined: Wed May 13, 2009 4:58 pm
Location: Vancouver, Washington
The property you're talking about is Prime Quadruplet. Formally they are of the form (p, p+2, p+6, p+8). Because it has to avoid certain primes like 2,3,5, they all are of the form (30 n + 11, 30 n + 13, 30 n + 17, 30 n + 19). This is why all of bmenrigh's ranges are 0-9 or 1-10. The one exception to this is (5,7,11,13), but obviously the 5 keeps anything else from doing this again.
According to Sloane's A050258, there are 180529 of them between 10^11 and 10^11.
This reminds me of the twin prime conjecture that predicts that there are an infinite number of twin prime (p, p+2). So far it hasn't been proven to be true. More good reading on this at Mathworld. Another good topic worth reading is Prime Gaps.
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Post subject: Re: Fastest method highest prime factor?Posted: Tue Aug 16, 2011 5:56 pm
Joined: Mon Oct 18, 2010 10:48 am
wwwmwww wrote:
NType3 wrote:
Because of this output, is it safe to assume that the digit in the ones' place alternates between a zero and a one? If so, why is this?
I can answer part of that...
Because I was including near-primes in my original question I was looking at sets of 10 consecutive numbers. This means either the first or the last number is even. So if you find 4 primes in a set as bmenrigh has then you know they are actually within a set of 9 consecutive numbers. Now when bmenrigh checked the next set of 10 consecutive numbers he actually found these same 4 primes again. So there are actually half as many sets in that list as there appears to be. And of the sets found the last digit of the first prime was "1". Is that always the case? I don't know... that's what I asked above as well. If it is... the question as to why does become very interesting.
Carl
Oh, I gathered from your statement that one should test ALL possible sets of ten. This includes (if we were starting at one) 1-10, 2-11, 3-12, etc. however, if it's actually 1-10, 11-20, 21-30, then I need to restructure my code significantly. Both, however, should theoretically produce the same results, no?
@GuiltyBystander: That's interesting! So you're saying that even with my orignal method, I'm likely to find the same result? (I'm not sure I understand entirely, but I gave it my best shot).
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Post subject: Re: Fastest method highest prime factor?Posted: Tue Aug 16, 2011 6:06 pm
Joined: Thu Dec 02, 2004 12:09 pm
Location: Missouri
NType3 wrote:
Oh, I gathered from your statement that one should test ALL possible sets of ten. This includes (if we were starting at one) 1-10, 2-11, 3-12, etc.
Yes this is what I meant.
NType3 wrote:
however, if it's actually 1-10, 11-20, 21-30, then I need to restructure my code significantly. Both, however, should theoretically produce the same results, no?
Based on what GuiltyBystander shows above then yes I'd say you should get the same result if you are just looking for primes. However I'm not sure that would be the case if you included the near-primes.
Carl
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Post subject: Re: Fastest method highest prime factor?Posted: Tue Aug 16, 2011 6:13 pm
Joined: Mon Oct 18, 2010 10:48 am
I believe, given enough samples, it wouldn't make a statistical difference. Given the test is 'emulating' selecting a random number and going up ten from there (my random numbers have an algorithm of n+1), the results should be similar.
In other words, go from the probability backwards.
Assume there's a 50% chance of X happening. If a person starts from 1 and goes by increment of 1 to 100, the results should theoretically be the same as starting at 1 and going to 1000 by increment of ten, given there's a 50% chance at each starting digit.
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Post subject: Re: Fastest method highest prime factor?Posted: Tue Aug 16, 2011 6:24 pm
Joined: Thu Dec 02, 2004 12:09 pm
Location: Missouri
GuiltyBystander wrote:
According to Sloane's A050258, there are 180529 of them between 10^11 and 10^11.
If I understand that list correctly shouldn't that be:
1209318-180529 or 1028789 prime quadruples between 10^10 and 10^11
So between 10000000000 and 99999999999 we have 9 billion non-overlaping sets of the form:
xxxxxxxxxx0 to xxxxxxxxxx9
Of those 9 billion sets 1028789 of them contain a prime quadruple, or about 0.0114% of them. Interesting.
If you considered all possible sets of 10 consecutive numbers you'd have 90 billion sets to check and each prime quadruple would show up in 2 sets so your odds drop to:
How do your odds improve for finding a set with just 3 primes?
Carl
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Post subject: Re: Fastest method highest prime factor?Posted: Tue Aug 16, 2011 6:39 pm
Joined: Wed May 13, 2009 4:58 pm
Location: Vancouver, Washington
wwwmwww wrote:
GuiltyBystander wrote:
According to Sloane's A050258, there are 180529 of them between 10^11 and 10^11.
If I understand that list correctly shouldn't that be:
1209318-180529 or 1028789 prime quadruples between 10^10 and 10^11
Oops, you're right.
NType3 wrote:
I believe, given enough samples, it wouldn't make a statistical difference. Given the test is 'emulating' selecting a random number and going up ten from there (my random numbers have an algorithm of n+1), the results should be similar.
Except the probability of a number being prime isn't uniform. If you look at the occurrence of each digit in primes, you'll find that the digit 1 occurs in prime numbers more often than any other number. This is because prime numbers become rarer and rarer so there will be more primes between 100-199 than there will be between 200-299. Similarly, I don't think you can assume 0-9 will be statistically different from 1-10.
Although, you should imediatly discount digits ending in 0 anyways for being near misses because they're guaranteed to have 2,5 as factors.
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Post subject: Re: Fastest method highest prime factor?Posted: Tue Aug 16, 2011 6:46 pm
Joined: Mon Oct 18, 2010 10:48 am
Stats for 1 prime: 33.962%
Stats for 2 primes: 6.236%
Stats for 3 primes: 0.397%
Stats for 4 primes: 0.0162%
That's my results so far.
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Post subject: Re: Fastest method highest prime factor?Posted: Tue Aug 16, 2011 7:08 pm
Joined: Thu Dec 31, 2009 8:54 pm
Location: Bay Area, California
NType3 wrote:
Assume there's a 50% chance of X happening. If a person starts from 1 and goes by increment of 1 to 100, the results should theoretically be the same as starting at 1 and going to 1000 by increment of ten, given there's a 50% chance at each starting digit.
This is assuming a uniform distribution of primes (which implies a uniform distribution of quad-primes). Because the small primes such as 2, 3, 5, have significant impact on what numbers can be prime for a given range of 10 the distribution is not uniform.
As GuiltyBystander / MathWorld points out, quad-primes must be of the form (30n + 11, 30n + 13, 30n + 17, 30n + 19), incrementing your check by range window by 1 versus 10 would give significantly different answers.
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Post subject: Re: Fastest method highest prime factor?Posted: Tue Aug 16, 2011 7:48 pm
Joined: Mon Oct 18, 2010 10:48 am
Given a large enough sample size, would it make a statistically significant difference?
For example, at the 1.49 millionth iteration, the percentages are now:
1: 33.999%
2: 6.237%
3: .3964%
4: .00908%
The only significantly different number is the search for 4, and that, occurring in the rarity it does, I believe requires a larger sample size to definitively decide. It has only occurred 136 times as opposed to 514110 times for one prime.
Please correct me if I'm wrong though; I'm very interested in all of this.
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Post subject: Re: Fastest method highest prime factor?Posted: Tue Aug 16, 2011 9:17 pm
Joined: Wed May 13, 2009 4:58 pm
Location: Vancouver, Washington
wwwmwww wrote:
Asked another way. Consider the set of all possible sets of 10 consecutive 11 digit numbers. No leading zeros... so we are talking about numbers between 10000000000 and 99999999999. What percentage of those sets are half or more made up of primes or near primes? Is there any set that contains more then 5 total primes and near primes? Can you give an example like I did above?
Are there any sets containing 4 or more primes? I think I can prove 4 is the max possible. 5 of the 10 must be even and 1 of the remaining 5 I think must be divisable by 3.
I'm been doing some math and found you can never have more then 2 near primes within a prime quadruplet.
Clearly, numbers ending in 0,2,4,6,0 will have 2 as a factor.
Again, numbers ending in 0,5,0 will have 5 as a factor.
Using the formula about them being 30n + {11,13,17,19} shows us that 2,5,8 are divisible by 3.
With these three facts, we can eliminate 0,2,5,8,0 as being near primes because they have at lease 2 factors in them already. This leaves 4 and 6 to be the candidates for being near primes which would require that 15n + {7,8} are prime.
If we consider 7's, we can see that we can put more restrictions on this. If we know that 1,3,7,9 aren't divisible by 7, we also know that 0,2,8,0 can't be divisible by 7 either. This leaves 4,5,6 to have 7 as a factor. Since 4,6 already have a factor, 5 must be divisible by 7. This means that at least 2/3 of all prime quadruplets can't have 2 near primes (assuming even distribution which we already know to be false, but it should be somewhat close). Using the quadruplet formula, (15n + 15) must be divisible by 7 meaning (n + 1) must be divisible by 7.
I'm trying to run a program to find some. So far I haven't found anything with n<100000000. I'll keep it running for a while though. I'm a tad surprised, but I guess not really if there's only 4768 prime quadruplets. I wonder if there's something about...
nvmd. I just realized that 15n+7 and 15n+8 can't both be prime. So you can have a max of 4 primes and 1 near prime. First example is (11,13,7*2,17,19). Next is (101, 103, 53*2, 107, 109)
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Post subject: Re: Fastest method highest prime factor?Posted: Wed Aug 17, 2011 9:26 am
Joined: Thu Dec 02, 2004 12:09 pm
Location: Missouri
GuiltyBystander wrote:
So you can have a max of 4 primes and 1 near prime. First example is (11,13,7*2,17,19). Next is (101, 103, 53*2, 107, 109)
Ok... so in my first example I showed a string of 10 numbers which contained 3 primes and 2 near primes. Is that also a max? If you force yourself to include a quad-prime then you know the last digit of the first prime is "1" and since the number before it and the one at the end of the string both end in "0" you are really looking at just a string of 9 numbers. I'm curious if the added flexability gained by only requiring 3 primes would allow for 3 or more near primes to be found in a string of 10 consecutive 11-digit numbers? What about if there were just 2 primes? Is 5 really the max possible of prime + near prime in such a string?
Carl
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Post subject: Re: Fastest method highest prime factor?Posted: Wed Aug 17, 2011 2:13 pm
Joined: Wed May 13, 2009 4:58 pm
Location: Vancouver, Washington
wwwmwww wrote:
GuiltyBystander wrote:
So you can have a max of 4 primes and 1 near prime. First example is (11,13,7*2,17,19). Next is (101, 103, 53*2, 107, 109)
Ok... so in my first example I showed a string of 10 numbers which contained 3 primes and 2 near primes. Is that also a max? If you force yourself to include a quad-prime then you know the last digit of the first prime is "1" and since the number before it and the one at the end of the string both end in "0" you are really looking at just a string of 9 numbers. I'm curious if the added flexability gained by only requiring 3 primes would allow for 3 or more near primes to be found in a string of 10 consecutive 11-digit numbers? What about if there were just 2 primes? Is 5 really the max possible of prime + near prime in such a string?
I sent my computer on a search and found that It seems you can have prime + near prime > 5. Here's a few examples. There more with prime + near prime = 6, but I thought these were more impressive.
prime + 6 near
7 near
7 near
Here's a short list of the 10th number of a set of 10 and how many primes + near primes are in the set. Fair warning, I used Java's BigInteger.isProbablePrime(100) to test for primeness so it's not guaranteed 100% to be accurate, but there's a 1- 1/2^100 chance that they are.
Code:
10000000711 6
10000012351 6
10000028587 6
10000029387 6
10000046146 6
10000048658 6
10000058906 6
10000058907 6
10000058909 6
10000058911 7
10000074391 6
10000081915 6
10000082054 6
10000089602 6
10000096838 6
10000096963 6
10000097546 6
10000107487 6
10000139885 6
10000139887 7
10000139889 6
10000144538 6
10000153243 6
10000163006 6
10000174286 6
10000174287 6
10000174289 6
10000174291 6
10000180123 6
10000184491 6
10000184494 6
10000184767 6
10000190971 6
10000210951 6
10000210954 6
10000222322 6
10000229546 6
10000246106 6
10000256846 6
10000262455 6
10000263151 6
10000263423 6
10000267423 6
10000272286 6
10000272523 6
10000284062 6
10000296242 6
10000296305 6
10000296307 7
10000296309 6
10000296311 6
10000302161 6
10000302163 6
10000304558 6
10000312961 6
10000317803 6
10000329982 6
10000374305 6
10000378963 6
10000384943 6
10000384945 6
10000390766 6
10000397146 6
10000397147 6
10000402466 6
10000402471 6
10000405747 6
10000407122 6
10000408706 6
10000414735 6
10000421345 6
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Post subject: Re: Fastest method highest prime factor?Posted: Wed Aug 17, 2011 4:05 pm
Joined: Thu Dec 02, 2004 12:09 pm
Location: Missouri
GuiltyBystander wrote:
There more with prime + near prime = 6, but I thought these were more impressive.
prime + 6 near
7 near
7 near
NICE!!!
So is it possible to prove 7 is the max possible without doing a full search?
If a full search is needed and someone does it (that's probably not reasonable) I'm curious... of the sets which contain 7 prime or near primes what is the maximum number of primes in such a set?
Carl
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Post subject: Re: Fastest method highest prime factor?Posted: Wed Aug 17, 2011 6:03 pm
Joined: Wed May 13, 2009 4:58 pm
Location: Vancouver, Washington
wwwmwww wrote:
So is it possible to prove 7 is the max possible without doing a full search?
This is actually pretty easy and you just need to consider how they're relatively prime to 2 and 3.
First lets look at the even numbers. In a consecutive set of 10 numbers, 5 are going to be even. Of the 5, either 2/3 of them will be divisible by 4. Here's the two options.
Code:
42424 Here we already have 3 non (prime+near primes)
24242 Therefore this must be the option
Next we consider the overlap of 3's with 2's.
Code:
11311
13113
31131
And together with 24242
Code:
2 4 6 4 2
2 12 2 4 6
6 4 2 12 2
In every case, there are at least 3 cases of non (prime + near prime).
wwwmwww wrote:
If a full search is needed and someone does it (that's probably not reasonable) I'm curious... of the sets which contain 7 prime or near primes what is the maximum number of primes in such a set?
I'm sure there's some more math to do here, but I haven't thought about it at all. Sent my computer searching. Gonna have to revise that earlier list. It has a few bugs in it.
_________________
Real name: Landon Kryger
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Post subject: Re: Fastest method highest prime factor?Posted: Fri Aug 19, 2011 5:56 pm
Joined: Wed May 13, 2009 4:58 pm
Location: Vancouver, Washington
GuiltyBystander wrote:
wwwmwww wrote:
If a full search is needed and someone does it (that's probably not reasonable) I'm curious... of the sets which contain 7 prime or near primes what is the maximum number of primes in such a set?
I'm sure there's some more math to do here, but I haven't thought about it at all. Sent my computer searching.
Was going to post results yesterday, but I'm glad I didn't. I let it keep running and found some better results. To answer your question. You can have 3 primes + 4 near primes. Proof by example.
So far I've searched 10000000000 - 10026548143 and found 3766 examples where prime + nears >= 6. Here's the total counts if you're interested in the rarity.
Code:
Prime Near Count
3 4 1
2 5 6
1 6 26
0 7 30
3 3 52
2 4 488
1 5 1608
0 6 1555
_________________
Real name: Landon Kryger
Top
Post subject: Re: Fastest method highest prime factor?Posted: Fri Aug 19, 2011 6:27 pm
Joined: Thu Dec 02, 2004 12:09 pm
Location: Missouri
GuiltyBystander wrote:
To answer your question. You can have 3 primes + 4 near primes. Proof by example.
Nice!!! Thanks... I think I'm finally out of questions.
Carl
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# Physics
Which one of the following will form a standing wave: (1) two waves of the same amplitude and speed moving in the same direction (2) two waves of the same amplitude and speed moving in the opposite directions, (3) any two traveling waves can form standing wave if they have same amplitude (4) any two traveling waves can form standing wave if they have same energy
From what I've read in my textbook, I believe the answer is (2). Is this correct. Thank you in advance.
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1. (2) Same amplitude, opposite directions.
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posted by drwls
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# algebra
Nine less than the product of
20
and Mabel's height
1. 👍 0
2. 👎 0
3. 👁 268
1. M x 20 - 9 = ?
(M stands for Mabel's height)
1. 👍 0
2. 👎 0
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RELATIVE SPEEED AND TRAIN QUESTIONS
Speed has no sense of direction unlike the velocity. Relative speed is the speed of one object as observed from another moving object. Questions on train are the classic examples of relative speed and in all these questions it is assumed that trains move parallel to each other – whether in the same direction or the opposite direction. Thus, we shall see how the relative speed is calculated and using it we come to know the time taken by the trains to cross each other and some other like aspects.
Important Formulas – Problems on Trains
1. x km/hr = (x×5)/18 m/s
1. y m/s = (y×18)/5 km/hr
1. Speed = distance/time, that is, s = d/t
1. velocity = displacement/time, that is, v = d/t
1. Time taken by a train x meters long to pass a pole or standing man or a post
= Time taken by the train to travel x meters.
1. Time taken by a train x meters long to pass an object of length y meters
= Time taken by the train to travel (x + y) metres.
1. Suppose two trains or two objects are moving in the same direction at v1 m/s and v2 m/s where v1 > v2,
then their relative speed = (v1 – v2) m/s
1. Suppose two trains or two objects are moving in opposite directions at v1 m/s and v2 m/s ,
then their relative speed = (v1+ v2) m/s
1. Assume two trains of length x metres and y metres are moving in opposite directions at v1 m/s and v2 m/s, Then
The time taken by the trains to cross each other = (x+y) / (v1+v2) seconds
1. Assume two trains of length x metres and y metres are moving in the same direction at at v1 m/s and v2 m/s where v1 > v2, Then
The time taken by the faster train to cross the slower train = (x+y) / (v1-v2) seconds
1. Assume that two trains (objects) start from two points P and Q towards each other at the same time and after crossing they take p and q seconds to reach Q and P respectively. Then,
A’s speed: B’s speed = √q: √p
Solved Examples
Level 1
1.A train is running at a speed of 40 km/hr and it crosses a post in 18 seconds. What is the length of the train? A. 190 metres B. 160 metres C. 200 metresAnswer : Option C D. 120 metres
Explanation :
Speed of the train, v = 40 km/hr = 40000/3600 m/s = 400/36 m/s
Time taken to cross, t = 18 s
Distance Covered, d = vt = (400/36)× 18 = 200 m
Distance covered is equal to the length of the train = 200 m
2.A train having a length of 240 m passes a post in 24 seconds. How long will it take to pass a platform having a length of 650 m? A. 120 sec B. 99 s C. 89 s D. 80 s
Explanation :
v = 240/24 (where v is the speed of the train) = 10 m/s
t = (240+650)/10 = 89 seconds
3.Two trains having length of 140 m and 160 m long run at the speed of 60 km/hr and 40 km/hr respectively in opposite directions (on parallel tracks). The time which they take to cross each other, is A. 10.8 s B. 12 s C. 9.8 s D. 8 s
Explanation :
Distance = 140+160 = 300 m
Relative speed = 60+40 = 100 km/hr = (100×10)/36 m/s
Time = distance/speed = 300 / (100×10)/36 = 300×36 / 1000 = 3×36/10 = 10.8 s
4.A train moves past a post and a platform 264 m long in 8 seconds and 20 seconds respectively. What is the speed of the train? A. 79.2 km/hr B. 69 km/hr C. 74 km/hr D. 61 km/hr
Explanation :
Let x is the length of the train and v is the speed
Time taken to move the post = 8 s
=> x/v = 8
=> x = 8v — (1)
Time taken to cross the platform 264 m long = 20 s
(x+264)/v = 20
=> x + 264 = 20v —(2)
Substituting equation 1 in equation 2, we get
8v +264 = 20v
=> v = 264/12 = 22 m/s
= 22×36/10 km/hr = 79.2 km/hr
5.Two trains, one from P to Q and the other from Q to P, start simultaneously. After they meet, the trains reach their destinations after 9 hours and 16 hours respectively. The ratio of their speeds is A. 2 : 3 B. 2 :1 C. 4 : 3 D. 3 : 2
Explanation :
Ratio of their speeds = Speed of first train : Speed of second train
= √16: √ 9
= 4:3
6.Train having a length of 270 meter is running at the speed of 120 kmph . It crosses another train running in opposite direction at the speed of 80 kmph in 9 seconds. What is the length of the other train? A. 320 m B. 190 m C. 210 m D. 230 m
Explanation :
Relative speed = 120+80 = 200 kmph = 200×10/36 m/s = 500/9 m/s
time = 9s
Total distance covered = 270 + x where x is the length of other train
(270+x)/9 = 500/9
=> 270+x = 500
=> x = 500-270 = 230 meter
7.Two stations P and Q are 110 km apart on a straight track. One train starts from P at 7 a.m. and travels towards Q at 20 kmph. Another train starts from Q at 8 a.m. and travels towards P at a speed of 25 kmph. At what time will they meet? A. 10.30 a.m B. 10 a.m. C. 9.10 a.m. D. 11 a.m.
Explanation :
Assume both trains meet after x hours after 7 am
Distance covered by train starting from P in x hours = 20x km
Distance covered by train starting from Q in (x-1) hours = 25(x-1)
Total distance = 110
=> 20x + 25(x-1) = 110
=> 45x = 135
=> x= 3 Means, they meet after 3 hours after 7 am, ie, they meet at 10 am
8.Two trains are running in opposite directions in the same speed. The length of each train is 120 meter. If they cross each other in 12 seconds, the speed of each train (in km/hr) is A. 42 B. 36 C. 28 D. 20
Explanation :
Distance covered = 120+120 = 240 m
Time = 12 s
Let the speed of each train = v. Then relative speed = v+v = 2v
2v = distance/time = 240/12 = 20 m/s
Speed of each train = v = 20/2 = 10 m/s
= 10×36/10 km/hr = 36 km/hr
Level 2
1.A train, 130 meters long travels at a speed of 45 km/hr crosses a bridge in 30 seconds. The length of the bridge is A. 270 m B. 245 m C. 235 m D. 220 m
Explanation :
Assume the length of the bridge = x meter
Total distance covered = 130+x meter
total time taken = 30s
speed = Total distance covered /total time taken = (130+x)/30 m/s
=> 45 × (10/36) = (130+x)/30
=> 45 × 10 × 30 /36 = 130+x
=> 45 × 10 × 10 / 12 = 130+x
=> 15 × 10 × 10 / 4 = 130+x
=> 15 × 25 = 130+x = 375
=> x = 375-130 =245
2.A train has a length of 150 meters. It is passing a man who is moving at 2 km/hr in the same direction of the train, in 3 seconds. Find out the speed of the train. A. 182 km/hr B. 180 km/hr C. 152 km/hr D. 169 km/hr
Explanation :
Length of the train, l = 150m
Speed of the man, Vm= 2 km/hr
Relative speed, Vr = total distance/time = (150/3) m/s = (150/3) × (18/5) = 180 km/hr
Relative Speed = Speed of train, Vt – Speed of man (As both are moving in the same direction)
=> 180 = Vt – 2 => Vt = 180 + 2 = 182 km/hr
3.Two trains running in opposite directions cross a man standing on the platform in 27 seconds and 17 seconds respectively. If they cross each other in 23 seconds, what is the ratio of their speeds? A. Insufficient data B. 3 : 1 C. 1 : 3 D. 3 : 2
Explanation :
Let the speed of the trains be x and y respectively
length of train1 = 27x
length of train2 = 17y
Relative speed= x+ y
Time taken to cross each other = 23 s
=> (27x + 17 y)/(x+y) = 23 => (27x + 17 y)/ = 23(x+y)
=> 4x = 6y => x/y = 6/4 = 3/2
4.A jogger is running at 9 kmph alongside a railway track in 240 meters ahead of the engine of a 120 meters long train . The train is running at 45 kmph in the same direction. How much time does it take for the train to pass the jogger? A. 46 B. 36 C. 18 D. 22
Explanation :
Distance to be covered = 240+ 120 = 360 m
Relative speed = 36 km/hr = 36×10/36 = 10 m/s
Time = distance/speed = 360/10 = 36 seconds
5.A train passes a platform in 36 seconds. The same train passes a man standing on the platform in 20 seconds. If the speed of the train is 54 km/hr, The length of the platform is A. None of these B. 280 meter C. 240 meter D. 200 meter
Explanation :
Speed of the train = 54 km/hr = (54×10)/36 m/s = 15 m/s
Length of the train = speed × time taken to cross the man = 15×20 = 300 m
Let the length of the platform = L
Time taken to cross the platform = (300+L)/15
=> (300+L)/15 = 36
=> 300+L = 15×36 = 540 => L = 540-300 = 240 meter
6.A train overtakes two persons who are walking in the same direction to that of the train at 2 kmph and 4 kmph and passes them completely in 9 and 10 seconds respectively. What is the length of the train? A. 62 m B. 54 m C. 50 m D. 55 m
Explanation :
Let x is the length of the train in meter and v is its speed in kmph
x/9 = (v-2) (10/36) — (1)
x/10 = (v-4) (10/36) — (2)
Dividing equation 1 with equation 2
10/9 = (v-2)/(v-4) => 10v – 40 = 9v – 18 => v = 22
Substituting in equation 1, x/9 = 200/36 => x = 9×200/36 = 50 m
7.A train is traveling at 48 kmph. It crosses another train having half of its length, traveling in opposite direction at 42 kmph, in 12 seconds. It also passes a railway platform in 45 seconds. What is the length of the platform? A. 500 m B. 360 m C. 480 m D. 400 m
Explanation :
Speed of train1 = 48 kmph
Let the length of train1 = 2x meter
Speed of train2 = 42 kmph
Length of train 2 = x meter (because it is half of train1’s length)
Distance = 2x + x = 3x
Relative speed= 48+42 = 90 kmph = 90×10/36 m/s = 25 m/s
Time = 12 s
Distance/time = speed => 3x/12 = 25
=> x = 25×12/3 = 100 meter
Length of the first train = 2x = 200 meter
Time taken to cross the platform= 45 s
Speed of train1 = 48 kmph = 480/36 = 40/3 m/s
Distance = 200 + y where y is the length of the platform
=> 200 + y = 45×40/3 = 600
=> y = 400 meter
8.A train, 800 meter long is running with a speed of 78 km/hr. It crosses a tunnel in 1 minute. What is the length of the tunnel (in meters)? A. 440 m B. 500 m C. 260 m D. 430 m
Explanation :
Distance = 800+x meter where x is the length of the tunnel
Time = 1 minute = 60 seconds
Speed = 78 km/hr = 78×10/36 m/s = 130/6 = 65/3 m/s
Distance/time = speed
(800+x)/60 = 65/3 => 800+x = 20×65 = 1300
=> x = 1300 – 800 = 500 meter
9.Two trains are running at 40 km/hr and 20 km/hr respectively in the same direction. If the fast train completely passes a man sitting in the slower train in 5 seconds, the length of the fast train is : A. 19 m B. 2779 m C. 1329 m D. 33 m
Explanation :
Relative speed = 40-20 = 20 km/hr = 200/36 m/s = 100/18 m/s
Time = 5 s
Distance = speed × time = (100/18) × 5 = 500/18 m = 250/9 = 2779 m = length of the fast train
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# DC Machines Questions and Answers – Operating Characteristics of DC Series Motor -1
This set of DC Machines Multiple Choice Questions & Answers (MCQs) focuses on “Operating Characteristics of DC Series Motor -1”.
1. DC series motors are used _______________
a) Where load is constant
b) Where load changes frequently
c) Where constant operating speed is needed
d) High starting speed
View Answer
Answer: d
Explanation: DC series motors are used in those applications where high starting speed is required at its initial phase. Since no load operation speed of DC series motor is very high, it is always operated on some non-zero load.
2. Where compensating winding is provided in case of conductively compensated DC series motor?
a) As separately wound unit
b) In series with armature winding
c) In parallel with armature winding
d) In parallel with field winding
View Answer
Answer: b
Explanation: Compensating winding is always excited with armature current in a given DC machine. So, compensating winding is provided in case of conductively compensated DC series motor in series with the armature winding.
3. No-load speed of which of the following DC motor will be highest?
a) Shunt motor
b) Series motor
c) Cumulative compound motor
d) Differentiate compound motor
View Answer
Answer: b
Explanation: At no load, armature current tends to zero, flux φ tends to zero, where speed is inversely proportional to the flux, speed will tend to infinity. Thus, no load speed of DC series motor is highest.
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4. The direction of rotation of a DC series motor shaft can be changed by _____________
a) Interchanging supply terminals
b) Interchanging field terminals
c) Not possible
d) Cannot be determined
View Answer
Answer: b
Explanation: From speed current characteristics for a DC series motor, we get that speed is inversely proportional to flux. If flux direction is reversed, same flux will be available with negative sign in vectors, thus, speed will remain same but with the other direction.
5. Which of the following DC motor can be used at conveyors?
a) Series motor
b) Shunt motor
c) Differentially compound motor
d) Cumulative compound motor
View Answer
Answer: a
Explanation: A conveyor requires high torque with steady speed, which provided by DC series motor. From torque speed characteristic of a DC series motor, it is easily stated that speed of the series motor practically remains constant at high torques.
Note: Join free Sanfoundry classes at Telegram or Youtube
6.Which statement is correct about DC series motor?
a) Has its field winding consisting of thick wire and more turns
b) Has a poor torque
c) Can be started easily without load
d) Has its field winding consisting of thick wire and less turns
View Answer
Answer: d
Explanation: When field winding is connected in series it has to be of made by thick wire. When field winding is connected in parallel it has to be of made by thin wire. So, thick wire is used with less number of turns while thin one is used with more number of turns.
7. In which of the following applications DC series motor is invariably tried?
a) Starter for a car
b) Drive for a water pump
c) Fan motor
d) Home appliances
View Answer
Answer: a
Explanation: Starter of the car simply implies start of operation at maximum load. Thus, series motor can be used in this application. For higher torques, since speed remains constant car starter is one of the best application of DC series motor.
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8. The speed of a DC series motor is ___________
a) Proportional to field current
b) Proportional to the square of the armature current
c) Proportional to the armature current
d) Inversely proportional to the armature current
View Answer
Answer: d
Explanation: Speed of DC series motor is inversely proportional to the armature current because, as armature current increases the flux produced also increases due to the series combination. Similarly if armature current is reduced flux is reduced which will increase speed.
9. In a DC series motor, if the armature current is halved, the torque of the motor will be equal to
a) 100% of the previous value
b) 50% of the previous value
c) 25% of the previous value
d) 10% of the previous value
View Answer
Answer: c
Explanation: Torque in the case of linear magnetization of DC series motor is directly proportional to square of the armature current. So, armature current is made 1/2th of the original value, then torque will be 1/4th of the original value.
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10. Applications demanding large applications at starting and small torque while running use DC series motor.
a) True
b) False
View Answer
Answer: a
Explanation: Large starting torque, infinite speed at no-load, low running torque for steady speed operation are called as Series characteristics. This are shown by DC series motor, so applications like cranes, traction, etc. use DC series motor.
Sanfoundry Global Education & Learning Series – DC Machines.
To practice all areas of DC Machines, here is complete set of 1000+ Multiple Choice Questions and Answers.
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tf.math.sign
Returns an element-wise indication of the sign of a number.
Used in the notebooks
Used in the tutorials
`y = sign(x) = -1 if x < 0; 0 if x == 0; 1 if x > 0`.
For complex numbers, `y = sign(x) = x / |x| if x != 0, otherwise y = 0`.
Example usage:
````# real number`
`tf.math.sign([0., 2., -3.])`
`<tf.Tensor: shape=(3,), dtype=float32,`
`numpy=array([ 0., 1., -1.], dtype=float32)>`
```
````# complex number`
`tf.math.sign([1 + 1j, 0 + 0j])`
`<tf.Tensor: shape=(2,), dtype=complex128,`
`numpy=array([0.70710678+0.70710678j, 0. +0.j ])>`
```
`x` A Tensor. Must be one of the following types: bfloat16, half, float32, float64, int32, int64, complex64, complex128.
`name` A name for the operation (optional).
A Tensor. Has the same type as x.
If x is a SparseTensor, returns SparseTensor(x.indices, tf.math.sign(x.values, ...), x.dense_shape).
If `x` is a `SparseTensor`, returns `SparseTensor(x.indices, tf.math.sign(x.values, ...), x.dense_shape)`
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http://conservapedia.com/Boyle's_Law
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# Boyle's Law
Boyle's Law states that the volume of a sample of gas is inversely proportional to the pressure applied to the gas if the temperature is kept constant.
According to Boyle's Law:
P1V1 = constant
After the change in pressure and volume,
P2V2 = constant
Combine the two products:
P1V1 = P2V2[1]
When any three of the four quantities in the equation are known, the fourth can be calculated. For example, if P1, V1 and P2 are known, then V2 can be solved by the following equation:
V2 = P1V1 / P2
## References
1. Wile, Dr. Jay L. Exploring Creation With Chemistry. Apologia Educational Ministries, Inc. 1998
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https://gpuzzles.com/mind-teasers/alphabet-question-problem/
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• Views : 60k+
• Sol Viewed : 20k+
# Mind Teasers : Alphabet Question Problem
Difficulty Popularity
Which alphabet letter is a question ?
Discussion
Suggestions
• Views : 80k+
• Sol Viewed : 20k+
# Mind Teasers : Interview Marble Logic Puzzle
Difficulty Popularity
You are given a set of scales and 12 marbles. The scales are of the old balance variety. That is, a small dish hangs from each end of a rod that is balanced in the middle. The device enables you to conclude either that the contents of the dishes weigh the same or that the dish that falls lower has heavier contents than the other.
The 12 marbles appear to be identical. In fact, 11 of them are identical, and one is of a different weight. Your task is to identify the unusual marble and discard it. You are allowed to use the scales three times if you wish, but no more.
Note that the unusual marble may be heavier or lighter than the others. You are asked to both identify it and determine whether it is heavy or light.
• Views : 50k+
• Sol Viewed : 20k+
# Mind Teasers : Children Age Football Count Puzzle
Difficulty Popularity
At a local club party at Chelsea club house, four ladies were there with their four children who aged 1, 2, 3 and 4. Also, these children have one, two, three and four footballs with the different order.
We know the below facts:
Aina has more footballs that his age.
Kalas is older than Zouma.
One child has the same number of footballs as his age.
Ake has fewer footballs than Kalas.
Child aged 3 has two footballs.
Ake is the youngest.
Can you determine the age of the children and the number of football they have?
• Views : 50k+
• Sol Viewed : 20k+
# Mind Teasers : Move 3 MatchSticks Puzzle
Difficulty Popularity
You need to divide area in the picture below into two equal parts by using exactly three match sticks. Can you do it?
• Views : 80k+
• Sol Viewed : 20k+
# Mind Teasers : Two Men One Question Puzzle
Difficulty Popularity
Imagine that you are travelling to a village. You happen to reach a point in the road where there is a fork. There are two ways that you can go into but only one amongst them is correct and leads to the village. You happen to see two men standing on the fork and you can ask them for the direction. To your bad luck, one amongst the two men always lies and the other one always says the truth. But you do not know who is a liar and who is not. At that point of the situation you are allowed to ask only one question to any one of the men standing there.
• Views : 50k+
• Sol Viewed : 20k+
# Mind Teasers : Tough River Crossing Riddle
Difficulty Popularity
This one is a bit of tricky river crossing puzzle than you might have solved till now. We have a whole family out on a picnic on one side of the river. The family includes Mother and Father, two sons, two daughters, a maid and a dog. The bridge broke down and all they have is a boat that can take them towards the other side of the river. But there is a condition with the boat. It can hold just two persons at one time (count the dog as one person).
No it does not limit to that and there are other complications. The dog can’t be left without the maid or it will bite the family members. The father can’t be left with daughters without the mother and in the same manner, the mother can’t be left alone with the sons without the father. Also an adult is needed to drive the boat and it can’t drive by itself.
How will all of them reach the other side of the river?
• Views : 40k+
• Sol Viewed : 10k+
# Mind Teasers : Out Of Box Thinking Puzzle
Difficulty Popularity
n a hotel, a man was sleeping when he heard a knock on the door. He shifted the blanket and stepped down from the bed. He waked to the door and opened it to find a stranger standing outside.
Upon opening the gate, that stranger said, "Pardon me, I must have made a mistake. I thought this was my room."
The stranger then walked the corridor and climbed down the stairs. The man closed the door and immediately called the security. He asked them to arrest that stranger immediately.
Why did he asked them to arrest that stranger? What made him suspicious?
• Views : 70k+
• Sol Viewed : 20k+
# Mind Teasers : Classic Card Brain Teaser
Difficulty Popularity
A pack of cards has 52 cards. You are blindfolded. Out of 52, 42 cards are facing down while 10 are facing up. You have been asked to divide this pack of cards into two decks - so that each deck contains an equal number of face up cards. Remember, you are blindfolded.
How will you do it?
• Views : 40k+
• Sol Viewed : 10k+
# Mind Teasers : River Crossing Brain Teaser
Difficulty Popularity
River Thames needs to be crossed by Mr. Father, Mrs. Mother, A Thug, A Policeman, 2 dogs and 2 Cats. There is only one boat, which can carry only two people (or a person and an animal or two animals) at a time. Only the humans (excluding the thug) know how to operate the boat. The boat must travel back and forth across the river in order to pick up all of the people and animals.
Following rules must be followed at all times:
Mr. Father: Mr. Father cannot stay with any of the cats, without Mrs. Mother's presence.
Mrs. Mother: Mrs. Mother cannot stay with any of the dogs, without Mr. Father's presence.
Thug: The thug cannot stay with anyone, if the policeman is not there.
Policeman Allowed to travel with everyone.
2 Dogs, 2 Cats: Not allowed to travel without a human, or be in the presence of the thug without the policeman's supervision.
The dogs are not allowed to be in the presence of Mrs. Mother without the supervision of Mr. Father.
Similarly, The cats are not allowed to be in the presence of Mr. Father without the supervision of Mrs. Mother.
Both the dogs and cats are allowed to be unsupervised (provided the other rules are satisfied.)
How will everyone get to the other side of the river Thames? None of the rule can be broken, and all rules apply on the boat as well as on the land.
• Views : 80k+
• Sol Viewed : 20k+
# Mind Teasers : Playing Cards Puzzle
Difficulty Popularity
I purchased five decks of card from the market and mixed them.
Now i have ( 5 * 52 = 260 cards).
What is minimum number of cards, i need to take out from the above 260 cards to guarantee at least one 'four of a kind'.
• Views : 60k+
• Sol Viewed : 20k+
# Mind Teasers : Tricky Iq Question
Difficulty Popularity
In a jar, there are some orange candies and some strawberry candies. You pick up two candies at a time randomly. If the two candies are of same flavor, you throw them away and put a strawberry candy inside. If they are of opposite flavors, you throw them away and put an orange candy inside.
In such manner, you will be reducing the candies in the jar one at a time and will eventually be left with only one candy in the jar.
If you are told about the respective number of orange and strawberry candies at the outset, will it be feasible for you to predict the flavor of the final remaining candy ?
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Which the right charger for the phone?...
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http://msl.cs.uiuc.edu/planning/node134.html
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#### 1D manifolds
The set of reals is the most obvious example of a 1D manifold because certainly looks like (via homeomorphism) in the vicinity of every point. The range can be restricted to the unit interval to yield the manifold because they are homeomorphic (recall Example 4.5).
Another 1D manifold, which is not homeomorphic to , is a circle, . In this case , and let
(4.5)
If you are thinking like a topologist, it should appear that this particular circle is not important because there are numerous ways to define manifolds that are homeomorphic to . For any manifold that is homeomorphic to , we will sometimes say that the manifold is , just represented in a different way. Also, will be called a circle, but this is meant only in the topological sense; it only needs to be homeomorphic to the circle that we learned about in high school geometry. Also, when referring to , we might instead substitute without any trouble. The alternative representations of a manifold can be considered as changing parameterizations, which are formally introduced in Section 8.3.2.
Steven M LaValle 2012-04-20
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https://stonespounds.com/468-3-stones-in-stones-and-pounds
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# 468.3 stones in stones and pounds
## Result
468.3 stones equals 468 stones and 4.2 pounds
You can also convert 468.3 stones to pounds.
## Converter
Four hundred sixty-eight point three stones is equal to four hundred sixty-eight stones and four point two pounds (468.3st = 468st 4.2lb).
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https://engineering.stackexchange.com/questions/17654/rotation-of-fluid-in-a-cylinder
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# Rotation of Fluid in a Cylinder
Question:
See what I tried below and verify if I'm wrong. I need help here.
You just need to get the height of the paraboloid (air) in the cylinder. Knowing that the volume of the paraboloid is half of the cylinder of the same height,
$$H_{paraboloid}=2*H_{cylinder}$$
where $H_{cylinder}$ is the height of air when it is not rotating, $0.2\ m$, therefore $$H_{paraboloid}=0.4m=y$$
substituting the value in the equation,
$$y=\frac{\omega^2x^2}{2g}$$
where $x=0.1\ m$, (radius of paraboloid) and $g=9.81\ m/s^2$, you will get
$$\omega=28.014\ rad/s = 267.51 rev/min$$
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http://allnippon.tk/pb313e-manual.html
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## plot 3d python matplotlib tutorial
College Algebra 6th Edition Robert F. Blitzer on Amazon. com. College Students, Introductory and Intermediate Algebra for College Students, Essentials of. Introductory Algebra for College Students 6th Edition Robert F. Blitzer on Amazon. com. Student Solutions Manual for IntermediateBlitzer College Algebra 6th Edition Teachers Manuao.
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Http:www. research. ibm. comjournalsj424vanbuskirk. pdf. Usability of Interactive Systems. It requires concentration at the outset. Jun 7, 2000. Http:www. mdx. ukstaffpagesannbinaugural.
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On this site there are a variety of. From Interaction Design - Beyond human-computer interaction 2nd edition. Wiley, 2007, chapter 1, pages 1-39. COHI 175 D, Methods and Theories of Studying. A revision of the 1 text in the Human Computer Interaction field, Interaction Design, the third edition is an ideal resource for learning the interdisciplinary skills. Hugely popular with students and professionals alike, Interaction Design is an ideal resource for learning the interdisciplinary skills needed for interaction design. A revision of the 1 text in the Human Computer Interaction field, Interaction Design, the third edition is an ideal resource price guide for wallace sterling silver set learning the interdisciplinary skills. Buy Interaction Design: Beyond Human-Computer Mastotron manual meats by Helen Sharp, Yvonne Rogers, Jenny Preece ISBN: 9780470018668 from Amazons Book. Pb313e manual Interaction Design: Beyond Human-Computer Interaction by Yvonne Rogers, Pb313e manual Sharp, Jenny Preece ISBN: pb313e manual from Pb313e manual Book. What is interaction design. Interaction Pb313e manual Beyond. Sharp, H, Rogers, Pb313e manual, Preece, J. Interaction Design: pb3313e pb313e manual mnaual. New York: John Wiley Sons, Inc. NOTE THE. CS Stanford Introduction macaron clay tutorial tacos Human Computer Interaction Design. Ph313e, Interaction Design beyond Human Computer. Interaction design: Beyond humancomputer interaction by Preece. Sharp and Rogers 2001, ISBN 0471492787. Published online: 13 May. Interaction Design - Beyond Human Computer Interac- tion by Jennifer Preece, Yvonne Rogers, and Helen. 95 John Wiley. Review of Jenny Preece, Yvonne Rogers and. 2002 Interaction design: beyond human-computer interaction. John Wiley and Sons, New. Instantly access INTERACTION DESIGN: beyond human-computer interaction, 3rd Edition by Jenny Preece, Helen Sharp, Yvonne Rogers. Start your free. Rogers, Y. 1997 A brief introduction to Distributed Cognition PDF version. Rogers, Y, Sharp, H, and Preece, J.
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Which of the following measures of central location is affected
Resolved Question:
Which of the following measures of central location is affected most by extreme values?
A. Mode
B. Geometric mean
C. Median
D. Mean
Submitted: 5 years ago.
Category: Homework
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The answer is D. Mean
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Which of the following is not a requirement of a probability distribution?
A. The outcomes are mutually exclusive. B. The probability of each outcome is between 0 and 1. C. Equally likely probability of a success. D. Sum of the possible outcomes is 1.00.
Expert: Bizhelp replied 5 years ago.
The answer to this is C. Equally likely probability of a success.
Customer: replied 5 years ago.
The difference between a random variable and a probability distribution is:
A. A random variable does not include the probability of an event. B. A random variable can only assume whole numbers. C. A probability distribution can only assume whole numbers. D. None of the above.
Which of the following is not a requirement of a binomial distribution?
A. A fixed number of trails. B. Equally likely outcomes. C. A constant probability of success. D. Only two possible outcomes
Expert: Bizhelp replied 5 years ago.
The difference between a random variable and a probability distribution is:
C. A probability distribution can only assume whole numbers.
Which of the following is not a requirement of a binomial distribution?
D. Only two possible outcomes
Customer: replied 5 years ago.
Suppose a population consisted of 20 items. How many different samples of n = 3 arepossible?
A. 20 B. 120 C. 6840 D. 1140
The difference between the sample mean and the population mean is called the
A. Standard error of the mean. B. Sampling error. C. Population mean. D. Population standard deviation
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# 2010 AMC 8 Problems/Problem 11
## Problem
The top of one tree is $16$ feet higher than the top of another tree. The heights of the two trees are in the ratio $3:4$. In feet, how tall is the taller tree?
$\textbf{(A)}\ 48 \qquad\textbf{(B)}\ 64 \qquad\textbf{(C)}\ 80 \qquad\textbf{(D)}\ 96\qquad\textbf{(E)}\ 112$
## Solution 1(algebra solution)
Let the height of the taller tree be $h$ and let the height of the smaller tree be $h-16$. Since the ratio of the smaller tree to the larger tree is $\frac{3}{4}$, we have $\frac{h-16}{h}=\frac{3}{4}$. Solving for $h$ gives us $h=64 \Rightarrow \boxed{\textbf{(B)}\ 64}$
## Solution 2
To answer this problem, you have to make it so that we have the same proportion as 3:4, but the difference between them is 16. Since the two numbers are consecutive, if we multiply both of them by 16, we would get a difference of 16 between them. So, it would be 48:64 and since we need to find the height of the taller tree, we get $h=64 \Rightarrow \boxed{\textbf{(B)}\ 64}$
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# What is the Markov blanket of a deterministic variable?
The following Bayesian network contains a node which is deterministically dependent on its parents: the variable $$either$$ is simply the $$OR$$ function of its parents $$tub$$ and $$lung$$.
By the graph, the Markov blanket of $$either$$ is the set $$\{tub, ~lung, ~xray, ~dysp, ~bronc\}$$, namely the parents, children and spouses of $$either$$. But $$either$$ is a function of $$tub$$ and $$lung$$, completely determined by them. It doesn't add any extra information (except being $$OR$$). It is completely independent of its children given its parents, let alone the entire network. Thus, can the Markov blanket be just $$tub$$ and $$lung$$?
I implemented a Markov blanket finding algorithm myself (IPC-MB), and it returns only $$\{tub, lung\}$$ as the Markov blanket, using the G-test as a conditional independence test. But by the d-separation criterion applied on the graph of the network, that wouldn't be enough.
What is the actual Markov blanket of $$either$$, knowing it is a function of its parents?
(taken from the Bayesian network repository at bnlearn)
• The Markov blanket of a node in a Bayesian network consists of the set of parents, children and spouses (parents of children), under certain assumptions. One of them is the faithfulness assumption, which, together with the Markov condition, implies that two variables X and Y are conditionally independent given a set of variables Z if and only if they are d-separated by Z in the graph. Deterministic relationships violate the faithfulness assumption. For example, in your case "either" is conditionally independent of all variables conditional on {"tub", "lung"}, although they are not d-separated. Commented Jun 29, 2019 at 10:21
• @George, your comment is the actual answer to my question. Can you please post it as an answer, so I can accept it? Commented Jun 29, 2019 at 12:27
One of them is the faithfulness assumption, which, together with the Markov condition, implies that two variables $$X$$ and $$Y$$ are conditionally independent given a set of variables $$Z$$ if and only if they are d-separated by $$Z$$ in the graph. Deterministic relationships violate the faithfulness assumption.
For example, in your case $$either$$ is conditionally independent of all variables conditional on $$\{tub, lung\}$$, although they are not d-separated.
• This clears it up, thank you. The graph in my question could be fixed by making $xray$ and $dysp$ as children of both $tub$ and $lung$, instead of being children of $either$, right? In that case, $either$ would end up with no children, thus it would indeed be d-separated from the entire network by its parents. Commented Jun 30, 2019 at 13:19
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## Lesson 1.2 Bill of Materials (BOMs)
Before any construction project is started, make out a BOMs to requisition building materials. However, you should first make a materials takeoff list and a materials estimate list, before making out the BOM.
### 1-3. Materials Takeoff List.
This list is the first step leading to preparation of a BOM. It is a listing of all parts of the building, taken off the plan. Table 1-3 shows a materials takeoff list for the building substructure shown in Figure 1-20.
Table 1-3. Sample materials takeoff list
Figure 1-20. 20- x 40-foot-wide building substructure
Note Spreaders and closers are not shown in the drawing but are part of the materials takeoff list.
### 1-4. Materials Estimate List.
A materials estimate list puts materials takeoff list information into a shorter form; adds allowance for waste and breakage; and estimates quantities of materials needed (Table 1-4). The lumber required is listed by board feet (BF).
Table 1-4. Sample materials estimate list
### 1-5. BF Computation
A BF is a unit measure representing an area of 1 foot by 1 foot, 1 inch thick. The number of board feet in a piece of lumber can be computed using one of the following methods:
Rapid Estimate. You can estimate BF rapidly by using Table 1-5. For example, reading the table, you can see that if a 2-inch by 12-inch board is 16 feet long, your board feet would be 32.
Table 1-5. Board feet
Arithmetic Method. To determine the number of BF in one or more pieces of lumber use the following formula:
Board feet = Number x Thickness (in) x Width (in) x Length (ft) 12
Note If the unit of measure for length is in inches, divide by 144 instead of 12.
Board feet = Number x Thickness (in) x Width (in) x Length (in) 144
Sample Problem 1: Find the number of BF in a piece of lumber 2 inches thick, 10 inches wide, and 6 feet long (Figure 1-21).
1 x 2 x 10 x 6 = 120 = 10 BF 12 12
Sample Problem 2: Find the number of BF in 10 pieces of lumber 2 inches thick, 10 inches wide, and 6 feet long.
10 x 2 x 10 x 6 = 1200 = 100 BF 12 12
Sample Problem 3: Find the number of BF in a piece of lumber 2 inches thick, 10 inches wide, and 18 inches long.
1 x 2 x 10 x 18 = 2½ BF 144
Figure 1-21. Lumber dimensions
Tabular Method. The standard essex board measure table (Figure 1-22) is a quick aid in computing BF. It is located on the back of the blade of the framing square. In using the board measure table, make all computations on the basis of 1-inch thickness. The inch markings along the outer edge of the blade represent the width of a board 1 inch thick. The third dimension (length) is provided in the vertical column of figures under the 12-inch mark.
Figure 1-22. Essex board measure table
Sample Problem: To compute the number of BF in a piece of lumber that is 8 inches wide, 14 feet long, and 4 inches thick:
1. Find the number 14 in the vertical column under the 12-inch mark.
2. Follow the guideline under number 14 laterally across the blade until it reaches the number on that line that is directly under the inch mark matching the width of the lumber.
Example problem: Under the 8-inch mark on the guideline, moving left from 14, the numbers 9 and 4 appear (9 and 4 should be on the same line as 14). The number to the left of the vertical line represents feet; the number to the right represents inches.
1. The total number is 37 1/3 BF. BF will never appear in a decimal form.
Example solution: 1" x 4" x 8' x 14'
1" x 4" x 8' x 14' Feet Inches
9 4
4 4
36 16/12
1 4/12
36 + 1 1/3 = 37 1/3 BF
Note 1" x 4" = Always multiply the number of pieces by the thickness and multiply the feet and inches by the sum of pieces and thickness.
### 1-6. Estimating the Quantity of Nails Required.
The sizes and pounds of nails needed should be added to the list. To estimate number of pounds, use the following formulas:
• For flooring, sheathing, and other 1-inch material:
Number of pounds = 2d to 8d x BF 4 100
• For framing materials that are 2 inches or more:
Number of pounds = 10d to 60d x BF 6 100
where d = penny
### 1-7. BOMs
Information for the BOM is taken from the materials estimate list. When preparing a BOM, follow the building sequence. For example, on most frame buildings, the first pieces of lumber used would be the footers; next would be floor joists, girders, subflooring, sole plates, and studs.
Figure 1-23. Sample BOMs
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# Articles
## The Five Whys: Understanding the Root Cause of a Problem
Is your company using Six Sigma to improve efficiency, eliminate waste, and minimize variation? Six Sigma has been a proven success for many businesses around the world dealing with similar problems. But where do these issues come from? How do they arise? And most importantly why?
Six Sigma DMAIC teaches you to Define, Measure, Analyze, Improve and Control problems. But it is the Analyze stage that interests us here. The 5 Whys of Root Cause Analysis (RCA) are essential questions that you need to ask. Unlike other devices, they don’t require testing hypotheses, data segmentation, regression, or other similarly complex statistical tools. The 5 Whys provide all the skills you need in one easy to use toolkit to help determine where problems come from, why they arise.
## Root Cause Analysis: How to use the Five Whys
To make the most of what the Five Whys can offer, they need to be completed to the letter. Here is a simple breakdown to help you understand how to use the Five Whys to their full effect.
• Write down the problem. Be specific. This will help you discern exactly what the problem is. You can then help your team focus on solving it.
• Why is the problem occurring? Look at where it occurs and try to identify the cause. If it’s a production problem, look closely at the flow.
• If you still can’t identify the root cause, ask why that is.
• Return to step 3 until you can agree on what the root cause of the problem is. Scrutinizing these questions in detail will allow you to determine the root cause quickly.
## Examples of the 5 Whys
There are no universal maxims that determine what the 5 Whys should be. Between you and us, it’s not always necessary to use all five. It depends entirely on your situation: on what the problem is. But here are some useful examples for you to use as inspiration.
1. Why are your customers receiving faulty products?
• Because your manufacturing team made the products using a different specification to the one that both the sales clerk and customer agreed on.
2. Why does manufacturing keep making products per a different specification?
• Because the salesclerk accelerates production on the shop floor by calling directly. Specifications may have been miscommunicated to cause this to happen.
3. Why does the salesclerk contact manufacturing directly to begin production as opposed to following company procedure?
• Because the sales director’s approval is required before production can begin. This slows down manufacturing or even stops it completely.
4. Why does the paperwork contain a sanction for the sales manager?
• Because the sales manager requires frequent updates to be discussed with the CEO.
Here’s another example of how the 5 Whys can be used (in an altogether different situation!).
1. Q – Why won’t your car start? A – There’s no gas.
2. Q – Why is there no gas? A – You didn’t buy any.
3. Q – Why didn’t you buy any? A – You didn’t have any cash at the time.
4. Q – Why didn’t you have the cash to buy gas? A – You lost your wallet.
5. Q – Why did you lose your wallet? A – There’s a hole in your coat pocket.
Remember, it’s up to you to use the 5 Whys to your advantage. Don’t let problems fester when they can be easily resolved!
Contact Us for Root Cause Analysis Training
SixSigma.us offers both Live Virtual classes as well as Online Self-Paced training. Most option includes access to the same great Master Black Belt instructors that teach our World Class in-person sessions. Sign-up today!
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TheInfoList
:''For a closed immersion in algebraic geometry, see closed immersion.'' In
mathematics Mathematics (from Ancient Greek, Greek: ) includes the study of such topics as quantity (number theory), mathematical structure, structure (algebra), space (geometry), and calculus, change (mathematical analysis, analysis). It has no generally ...
, an immersion is a
differentiable function In calculus Calculus, originally called infinitesimal calculus or "the calculus of infinitesimal In mathematics, infinitesimals or infinitesimal numbers are quantities that are closer to zero than any standard real number, but are not zero. T ...
between
differentiable manifold In mathematics, a differentiable manifold (also differential manifold) is a type of manifold The real projective plane is a two-dimensional manifold that cannot be realized in three dimensions without self-intersection, shown here as Boy's surfa ...
s whose
derivative In mathematics Mathematics (from Ancient Greek, Greek: ) includes the study of such topics as quantity (number theory), mathematical structure, structure (algebra), space (geometry), and calculus, change (mathematical analysis, analysis). ...
is everywhere
injective In mathematics Mathematics (from Ancient Greek, Greek: ) includes the study of such topics as quantity (number theory), mathematical structure, structure (algebra), space (geometry), and calculus, change (mathematical analysis, analysis). I ...
. Explicitly, is an immersion if :$D_pf : T_p M \to T_N\,$ is an injective function at every point ''p'' of ''M'' (where ''TpX'' denotes the
tangent space In mathematics, the tangent space of a manifold facilitates the generalization of vectors from affine spaces to general manifolds, since in the latter case one cannot simply subtract two points to obtain a vector that gives the Displacement (vector ...
of a manifold ''X'' at a point ''p'' in ''X''). Equivalently, ''f'' is an immersion if its derivative has constant rank equal to the dimension of ''M'': :$\operatorname\,D_p f = \dim M.$ The function ''f'' itself need not be injective, only its derivative must be. A related concept is that of an Embedding#Differential topology, embedding. A smooth embedding is an injective immersion that is also a topological embedding, so that ''M'' is diffeomorphic to its image in ''N''. An immersion is precisely a local embedding – i.e., for any point there is a neighbourhood (topology), neighbourhood, , of ''x'' such that is an embedding, and conversely a local embedding is an immersion. For infinite dimensional manifolds, this is sometimes taken to be the definition of an immersion. If ''M'' is Compact space, compact, an injective immersion is an embedding, but if ''M'' is not compact then injective immersions need not be embeddings; compare to continuous bijections versus homeomorphisms.
# Regular homotopy
A regular homotopy between two immersions ''f'' and ''g'' from a manifold ''M'' to a manifold ''N'' is defined to be a differentiable function such that for all ''t'' in the function defined by for all is an immersion, with , . A regular homotopy is thus a homotopy through immersions.
# Classification
Hassler Whitney initiated the systematic study of immersions and regular homotopies in the 1940s, proving that for every map of an ''m''-dimensional manifold to an ''n''-dimensional manifold is homotopic to an immersion, and in fact to an embedding for ; these are the Whitney immersion theorem and Whitney embedding theorem. Stephen Smale expressed the regular homotopy classes of immersions as the homotopy groups of a certain Stiefel manifold. The sphere eversion was a particularly striking consequence. Morris Hirsch generalized Smale's expression to a homotopy theory description of the regular homotopy classes of immersions of any ''m''-dimensional manifold ''Mm'' in any ''n''-dimensional manifold ''Nn''. The Hirsch-Smale classification of immersions was generalized by Mikhail Gromov (mathematician), Mikhail Gromov.
## Existence
The primary obstruction to the existence of an immersion is the stable normal bundle of ''M'', as detected by its characteristic classes, notably its Stiefel–Whitney classes. That is, since R''n'' is Parallelizable manifold, parallelizable, the pullback of its tangent bundle to ''M'' is trivial; since this pullback is the direct sum of the (intrinsically defined) tangent bundle on ''M'', ''TM'', which has dimension ''m'', and of the normal bundle ''ν'' of the immersion ''i'', which has dimension , for there to be a codimension ''k'' immersion of ''M'', there must be a vector bundle of dimension ''k'', ''ξ''''k'', standing in for the normal bundle ''ν'', such that is trivial. Conversely, given such a bundle, an immersion of ''M'' with this normal bundle is equivalent to a codimension 0 immersion of the total space of this bundle, which is an open manifold. The stable normal bundle is the class of normal bundles plus trivial bundles, and thus if the stable normal bundle has cohomological dimension ''k'', it cannot come from an (unstable) normal bundle of dimension less than ''k''. Thus, the cohomology dimension of the stable normal bundle, as detected by its highest non-vanishing characteristic class, is an obstruction to immersions. Since characteristic classes multiply under direct sum of vector bundles, this obstruction can be stated intrinsically in terms of the space ''M'' and its tangent bundle and cohomology algebra. This obstruction was stated (in terms of the tangent bundle, not stable normal bundle) by Whitney. For example, the Möbius strip has non-trivial tangent bundle, so it cannot immerse in codimension 0 (in R2), though it embeds in codimension 1 (in R3). showed that these characteristic classes (the Stiefel–Whitney classes of the stable normal bundle) vanish above degree , where is the number of "1" digits when ''n'' is written in binary; this bound is sharp, as realized by real projective space. This gave evidence to the ''Immersion Conjecture'', namely that every ''n''-manifold could be immersed in codimension , i.e., in R2''n''−α(''n''). This conjecture was proven by .
## Codimension 0
Codimension 0 immersions are equivalently relative dimension, ''relative'' dimension 0 ''Submersion (mathematics), submersions'', and are better thought of as submersions. A codimension 0 immersion of a closed manifold is precisely a covering map, i.e., a fiber bundle with 0-dimensional (discrete) fiber. By Ehresmann's theorem and Phillips' theorem on submersions, a proper map, proper submersion of manifolds is a fiber bundle, hence codimension/relative dimension 0 immersions/submersions behave like submersions. Further, codimension 0 immersions do not behave like other immersions, which are largely determined by the stable normal bundle: in codimension 0 one has issues of fundamental class and cover spaces. For instance, there is no codimension 0 immersion , despite the circle being parallelizable, which can be proven because the line has no fundamental class, so one does not get the required map on top cohomology. Alternatively, this is by invariance of domain. Similarly, although S3 and the 3-torus T3 are both parallelizable, there is no immersion – any such cover would have to be ramified at some points, since the sphere is simply connected. Another way of understanding this is that a codimension ''k'' immersion of a manifold corresponds to a codimension 0 immersion of a ''k''-dimensional vector bundle, which is an open manifold, ''open'' manifold if the codimension is greater than 0, but to a closed manifold in codimension 0 (if the original manifold is closed).
# Multiple points
A ''k''-tuple point (double, triple, etc.) of an immersion is an unordered set of distinct points with the same image . If ''M'' is an ''m''-dimensional manifold and ''N'' is an ''n''-dimensional manifold then for an immersion in general position the set of ''k''-tuple points is an -dimensional manifold. Every embedding is an immersion without multiple points (where ). Note, however, that the converse is false: there are injective immersions that are not embeddings. The nature of the multiple points classifies immersions; for example, immersions of a circle in the plane are classified up to regular homotopy by the number of double points. At a key point in surgery theory it is necessary to decide if an immersion of an ''m''-sphere in a 2''m''-dimensional manifold is regular homotopic to an embedding, in which case it can be killed by surgery. C.T.C. Wall, Wall associated to ''f'' an invariant ''μ''(''f'') in a quotient of the fundamental group ring Z[1(''N'')] which counts the double points of ''f'' in the universal cover of ''N''. For , ''f'' is regular homotopic to an embedding if and only if by the Hassler Whitney, Whitney trick. One can study embeddings as "immersions without multiple points", since immersions are easier to classify. Thus, one can start from immersions and try to eliminate multiple points, seeing if one can do this without introducing other singularities – studying "multiple disjunctions". This was first done by André Haefliger, and this approach is fruitful in codimension 3 or more – from the point of view of surgery theory, this is "high (co)dimension", unlike codimension 2 which is the knotting dimension, as in knot theory. It is studied categorically via the "calculus of functors" b
Thomas GoodwillieJohn Klein
an
Michael S. Weiss
# Examples and properties
* The Klein bottle, and all other non-orientable closed surfaces, can be immersed in 3-space but not embedded. * A mathematical Rose (mathematics), rose with ''k'' petals is an immersion of the circle in the plane with a single ''k''-tuple point; ''k'' can be any odd number, but if even must be a multiple of 4, so the figure 8 is not a rose. * By the Whitney–Graustein theorem, the regular homotopy classes of immersions of the circle in the plane are classified by the winding number, which is also the number of double points counted algebraically (i.e. with signs). * The sphere eversion, sphere can be turned inside out: the standard embedding is related to by a regular homotopy of immersions . * Boy's surface is an immersion of the real projective plane in 3-space; thus also a 2-to-1 immersion of the sphere. * The Morin surface is an immersion of the sphere; both it and Boy's surface arise as midway models in sphere eversion. File:BoysSurfaceTopView.PNG, Boy's surface File:MorinSurfaceAsSphere'sInsideVersusOutside.PNG, The Morin surface
## Immersed plane curves
Immersed plane curves have a well-defined turning number, which can be defined as the total curvature divided by 2. This is invariant under regular homotopy, by the Whitney–Graustein theorem – topologically, it is the degree of the Gauss map, or equivalently the winding number of the unit tangent (which does not vanish) about the origin. Further, this is a complete set of invariants – any two plane curves with the same turning number are regular homotopic. Every immersed plane curve lifts to an embedded space curve via separating the intersection points, which is not true in higher dimensions. With added data (which strand is on top), immersed plane curves yield knot diagrams, which are of central interest in knot theory. While immersed plane curves, up to regular homotopy, are determined by their turning number, knots have a very rich and complex structure.
## Immersed surfaces in 3-space
The study of immersed surfaces in 3-space is closely connected with the study of knotted (embedded) surfaces in 4-space, by analogy with the theory of knot diagrams (immersed plane curves (2-space) as projections of knotted curves in 3-space): given a knotted surface in 4-space, one can project it to an immersed surface in 3-space, and conversely, given an immersed surface in 3-space, one may ask if it lifts to 4-space – is it the projection of a knotted surface in 4-space? This allows one to relate questions about these objects. A basic result, in contrast to the case of plane curves, is that not every immersed surface lifts to a knotted surface. In some cases the obstruction is 2-torsion, such as in
Koschorke's example
', which is an immersed surface (formed from 3 Möbius bands, with a Tripoint (disambiguation), triple point) that does not lift to a knotted surface, but it has a double cover that does lift. A detailed analysis is given in , while a more recent survey is given in .
# Generalizations
A far-reaching generalization of immersion theory is the homotopy principle: one may consider the immersion condition (the rank of the derivative is always ''k'') as a partial differential relation (PDR), as it can be stated in terms of the partial derivatives of the function. Then Smale–Hirsch immersion theory is the result that this reduces to homotopy theory, and the homotopy principle gives general conditions and reasons for PDRs to reduce to homotopy theory.
# See also
*Immersed submanifold *Isometric immersion *Submersion (mathematics), Submersion
# References
* * * * * *. * *. *. * * . * * * * *. * *. * * *. *. *. * *. * *. {{refend
# External links
Immersion
at the Manifold Atlas
Immersion of a manifold
at the Encyclopedia of Mathematics Differential topology Maps of manifolds Smooth functions
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## Conversion formula
The conversion factor from years to minutes is 525600, which means that 1 year is equal to 525600 minutes:
1 yr = 525600 min
To convert 16.8 years into minutes we have to multiply 16.8 by the conversion factor in order to get the time amount from years to minutes. We can also form a simple proportion to calculate the result:
1 yr → 525600 min
16.8 yr → T(min)
Solve the above proportion to obtain the time T in minutes:
T(min) = 16.8 yr × 525600 min
T(min) = 8830080 min
The final result is:
16.8 yr → 8830080 min
We conclude that 16.8 years is equivalent to 8830080 minutes:
16.8 years = 8830080 minutes
## Alternative conversion
We can also convert by utilizing the inverse value of the conversion factor. In this case 1 minute is equal to 1.1324925708487E-7 × 16.8 years.
Another way is saying that 16.8 years is equal to 1 ÷ 1.1324925708487E-7 minutes.
## Approximate result
For practical purposes we can round our final result to an approximate numerical value. We can say that sixteen point eight years is approximately eight million eight hundred thirty thousand eighty minutes:
16.8 yr ≅ 8830080 min
An alternative is also that one minute is approximately zero times sixteen point eight years.
## Conversion table
### years to minutes chart
For quick reference purposes, below is the conversion table you can use to convert from years to minutes
years (yr) minutes (min)
17.8 years 9355680 minutes
18.8 years 9881280 minutes
19.8 years 10406880 minutes
20.8 years 10932480 minutes
21.8 years 11458080 minutes
22.8 years 11983680 minutes
23.8 years 12509280 minutes
24.8 years 13034880 minutes
25.8 years 13560480 minutes
26.8 years 14086080 minutes
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# In an examination, 20 questions of true-false type are asked.
Question:
In an examination, 20 questions of true-false type are asked. Suppose a student tosses a fair coin to determine his answer to each question. If the coin falls heads, he answers ‘true’; if it falls tails, he answers ‘false’. Find the probability that he answers at least 12 questions correctly.
Solution:
Let X represent the number of correctly answered questions out of 20 questions.
The repeated tosses of a coin are Bernoulli trails. Since “head” on a coin represents the true answer and “tail” represents the false answer, the correctly answered questions are Bernoulli trials.
$\therefore p=\frac{1}{2}$
$\therefore q=1-p=1-\frac{1}{2}=\frac{1}{2}$
$X$ has a binomial distribution with $n=20$ and $p=\frac{1}{2}$
$\therefore \mathrm{P}(\mathrm{X}=x)={ }^{9} \mathrm{C}_{x} q^{n-x} p^{x}$, where $x=0,1,2, \ldots n$
X has a binomial distribution with $n=20$ and $p=\frac{1}{2}$
$\therefore \mathrm{P}(\mathrm{X}=x)={ }^{n} \mathrm{C}_{x} q^{n-x} p^{x}$, where $x=0,1,2, \ldots n$
$={ }^{20} \mathrm{C}_{x}\left(\frac{1}{2}\right)^{20-x} \cdot\left(\frac{1}{2}\right)^{x}$
$={ }^{20} \mathrm{C}_{x}\left(\frac{1}{2}\right)^{20}$'
$={ }^{20} \mathrm{C}_{x}\left(\frac{1}{2}\right)^{20}$
P (at least 12 questions answered correctly) = P(X ≥ 12)
$=\mathrm{P}(\mathrm{X}=12)+\mathrm{P}(\mathrm{X}=13)+\ldots+\mathrm{P}(\mathrm{X}=20)$
$={ }^{20} \mathrm{C}_{12}\left(\frac{1}{2}\right)^{20}+{ }^{20} \mathrm{C}_{13}\left(\frac{1}{2}\right)^{20}+\ldots+{ }^{20} \mathrm{C}_{20}\left(\frac{1}{2}\right)^{20}$
$=\left(\frac{1}{2}\right)^{20} \cdot\left[{ }^{20} \mathrm{C}_{12}+{ }^{20} \mathrm{C}_{13}+\ldots+{ }^{20} \mathrm{C}_{20}\right]$
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# Hands-on ActivitySolid Rock to Building Block
(0 Ratings)
### Quick Look
Grade Level: 4 (3-5)
Time Required: 45 minutes
Expendable Cost/Group: US \$2.00
Group Size: 2
Activity Dependency: None
Subject Areas: Geometry, Physical Science, Problem Solving, Reasoning and Proof, Science and Technology
NGSS Performance Expectations:
### Summary
Students continue their pyramid building journey, acting as engineers to determine the appropriate wedge tool to best extract rock from a quarry and cut into pyramid blocks. Using sample materials (wax, soap, clay, foam) representing rock types that might be found in a quarry, they test a variety of wedges made from different materials and with different degrees of sharpness to determine which is most effective at cutting each type of material.
This engineering curriculum aligns to Next Generation Science Standards (NGSS).
### Engineering Connection
An important job for any engineer is to assign appropriate tools for building, machining and manufacturing. A wrong choice may result in a poor quality final product and/or a tool that wears out or breaks quickly. For example, a dull wooden wedge used to cut solid rock is only successful for a short time before it breaks or wears down to the point at which it no longer cuts. Engineers calculate the amount of force required to cut through a given material; knowing this helps them choose the most appropriate tool material that will cut through the rock material.
### Learning Objectives
After this activity, students should be able to:
• Determine how different materials break using differently-angled wedges.
• Describe why simple machines are used and how a wedge exerts a force.
• Demonstrate why material selection and material science is important to engineers.
• Understand that because simple machines do their jobs so well they are still used today.
### Educational Standards Each TeachEngineering lesson or activity is correlated to one or more K-12 science, technology, engineering or math (STEM) educational standards. All 100,000+ K-12 STEM standards covered in TeachEngineering are collected, maintained and packaged by the Achievement Standards Network (ASN), a project of D2L (www.achievementstandards.org). In the ASN, standards are hierarchically structured: first by source; e.g., by state; within source by type; e.g., science or mathematics; within type by subtype, then by grade, etc.
###### NGSS: Next Generation Science Standards - Science
NGSS Performance Expectation
3-5-ETS1-3. Plan and carry out fair tests in which variables are controlled and failure points are considered to identify aspects of a model or prototype that can be improved. (Grades 3 - 5)
Do you agree with this alignment?
Click to view other curriculum aligned to this Performance Expectation
This activity focuses on the following Three Dimensional Learning aspects of NGSS:
Science & Engineering Practices Disciplinary Core Ideas Crosscutting Concepts
Plan and conduct an investigation collaboratively to produce data to serve as the basis for evidence, using fair tests in which variables are controlled and the number of trials considered.
Alignment agreement:
Tests are often designed to identify failure points or difficulties, which suggest the elements of the design that need to be improved.
Alignment agreement:
Different solutions need to be tested in order to determine which of them best solves the problem, given the criteria and the constraints.
Alignment agreement:
###### International Technology and Engineering Educators Association - Technology
• Design solutions by safely using tools, materials, and skills. (Grades 3 - 5) More Details
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Suggest an alignment not listed above
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### Materials List
Each group needs:
• 1 soap or wax block (~5 cm x 5 cm)
• 1 clay block (~5 cm x 5 cm)
• 1 foam block (~5 cm x 5 cm)
• 1 Styrofoam block (~5 cm x 5 cm)
• Wooden wedge (approximately the size of a typical doorstop [5 cm x 10 cm x 5 cm])
• Balsa wood wedge (~3 cm x 3 cm x 6 cm)
• Plastic wedge (~3 cm x 3 cm x 6 cm)
• Styrofoam/foam wedge (~3 cm x 3 cm x 6 cm)
For the entire class to share:
• A variety of demonstration wedges with different angles and materials. For example, a plastic knife, putty knife, table knife, metal screwdriver or chisel (and a small hammer to use with the chisel).
• Some demonstration quarry material. For example, a half stick of cold margarine, a brick, etc.
### More Curriculum Like This
Upper Elementary Lesson
Pyramid Building: How to Use a Wedge
Students learn how simple machines, including wedges, were used in building both ancient pyramids and present-day skyscrapers. In a hands-on activity, students test a variety of wedges on different materials (wax, soap, clay, foam).
Upper Elementary Lesson
Engineering: Simple Machines
Students are introduced to the six types of simple machines — the wedge, wheel and axle, lever, inclined plane, screw, and pulley — in the context of the construction of a pyramid, gaining high-level insights into tools that have been used since ancient times and are still in use today.
High School Activity
Splash, Pop, Fizz: Rube Goldberg Machines
Refreshed with an understanding of the six simple machines; screw, wedge, pully, incline plane, wheel and axle, and lever, student groups receive materials and an allotted amount of time to act as mechanical engineers to design and create machines that can complete specified tasks.
Middle School Lesson
Just Plane Simple
This lesson introduces students to three of the six simple machines used by many engineers. These machines include the inclined plane, the wedge and the screw.
### Pre-Req Knowledge
General knowledge of pyramids and geometric angles. Familiarity with the six simple machines introduced in Lesson 1 of this unit.
This unit's lesson, Pyramid Building: How to Use a Wedge, does not need to be completed prior to students doing this activity, but the storyline of creating a pyramid created within the lesson helps students understand what they are trying to achieve with this activity.
The pyramid site choosing activity in Simple Machines: Lesson 1, is also not required, but it provides students with a background so they understand what a rock quarry is, and from where they are trying to extract the rock and why.
### Introduction/Motivation
The wedge is a simple machine that helps make our lives easier. Does anyone know how a wedge helps us do work? (Answer: A wedge allows us to split materials apart much more easily then we could do by hand.) The wedge gives us a mechanical advantage. This means that we have to exert less force to complete the task, but we usually have to go a greater distance. For example, when you use an axe to cut through a log, the job is much easier then if you tried to break the log just with your hands. However, you also have to make many axe cuts to get through the wood. The axe allowed you to cut through the wood easier, but you had to make more cuts to do it.
The wedge is slightly different from other simple machines because when you are using a wedge as a tool, often times another object is required to help. For example, if you were using a nail, which is a very sharp-angled (acute) wedge, a hammer would be required to force the nail into the wood. This is similar to the way we believe ancient pyramid construction was done. The pyramid builders found large rock quarries filled with different types of rock. They had to figure out a way to break the rocks away from the quarry wall into large bricks (stone blocks) that could be used to build the pyramid.
Pyramid builders would likely find a variety of different rock types within the quarry. They had to engineer a wedge that could be used to break away the rock from the quarry wall. If they encountered a soft, clay-like material, pyramid engineers designed a large wedge made of wood which could easily cut through clay-like material and successfully make many bricks. But, when they came across hard marble in their quarry, the wooden wedge would not break the material apart well. The wedge quickly wore down and the engineers knew that they must design a more effective wedge. That's where you come in. Today you are going to be design engineers and help research wedge designs that help the pyramid builders cut through each type of rock they encounter.
Show students an excellent animation of a Polish video about transportation methods that do not use a wheel and axle: https://www.youtube.com/watch?v=R_mGXrHQs9M. The animation shows how heavy stone blocks might have been systematically moved up an incline plane (ramp) using many human-powered wedges. The large supply of Egyptians workers would have made this method possible.
Through this activity, you will learn about different types of wedge angles and the different materials from which wedges can be made. You will experiment with a variety of materials so that you will be able to make recommendations to the pyramid engineers about how to best design a wedge. You will also see why it is so important for engineers to understand the design process and material selection when they are working on a project. If you choose the incorrect material for your wedge, the tool will not work well and your project may not be successful.
### Procedure
Before the Activity
• Gather teacher demonstration materials.
• Prepare student group activity materials.
• Before making copies of the The Wedge Worksheet, fill in the table column descriptions with the wedge materials the students will be using, for example "wood," "plastic" and "Styrofoam" (or have the students do this). Fill in the row descriptions with the type of rock material the students will be using, for example, "foam," "wax" and "clay" (or have the students do this).
With the Students
1. Teacher Demo: To give the students a visual understanding of a wedge and how it can be used as a tool, lead a demonstration using wedges as cutting tools. For the demo, the material being cut should be at a larger scale than what the students will do in their group activity. Use a variety of wedges, such as a chisel and hammer, and a plastic knife, on two very different materials, such as a brick and a stick of cold margarine. Some of the wedges will not be successful at cutting the hard materials, due to the wedge angle being too dull or the wedge material being too weak. Having two very different "rock" samples (brick and margarine) helps students understand why we need to design different types of wedges that are made from different materials, and introduces the idea of the importance of appropriate material selection.
2. For the student group activity, divide the class into pairs of two students each (although groups of three work well, too).
3. Direct each team of students to test their given wedges on each of their given "rock" sample materials. Allow them ~20 minutes to complete the activity.
4. Instruct the students to record the performance of each wedge and "rock" sample on their worksheet, using the provided rating scale.
5. As a class, conclude the activity by comparing test results among all teams and holding a class discussion. Which wedge/rock combinations were successful? Which were not? Why? How are the wedge angles and points different from each other? Which wedge had the sharpest angle? Which had the biggest cutting surface? How does this make a difference? (Possible answers: A larger cutting surface allows the user to exert more force on the object being cut.) Why do you think engineers design different types of wedges? (Possible answers: Depending upon the characteristics of the material to be cut, they might need to design stronger, but more expensive wedges to cut hard materials.) Why is material selection an important engineering job? (Possible answer: If you choose the incorrect material for your wedge, the tool will not work well and your project may not be successful.)
### Vocabulary/Definitions
angle: The "sharpness" of a wedge.
design: (verb) To plan out in systematic, often graphic form. To create for a particular purpose or effect. Design a building. (noun) A well thought-out plan.
mechanical advantage: An advantage gained by using simple machines to accomplish work with less effort. Making the task easier (which means it requires less force), but may require more time or room to work (more distance, rope, etc.). For example, applying a smaller force over a longer distance to achieve the same effect as applying a large force over a small distance. The ratio of the output force exerted by a machine to the input force applied to it.
quarry: A pit from which rock or stone is removed from the ground.
simple machine: A machine with few or no moving parts that is used to make work easier (provides a mechanical advantage). For example, a wedge, wheel and axle, lever, inclined plane, screw, or pulley.
tool: A device used to do work.
wedge: A simple machine that forces materials apart. Used for splitting, tightening, securing or levering. It is thick at one end and tapered to a thin edge at the other.
work: Force on an object multiplied by the distance it moves. W = F x d (force multiplied by distance).
### Assessment
Pre-Activity Assessment
Know / Want to Know / Learn (KWL) Chart: Before the activity, ask students to write down in the top left corner of a piece of paper (or as a group on the board) under the title, Know, all the things they know about wedges. Next, in the top right corner under the title, Want to Know, ask students to write down anything they want to know about wedges. After the activity, ask students to list in the bottom half of the page under the title, Learned, all of the things that they have learned about wedges.
Activity Embedded Assessment
Material Selection Discussion: After the teacher demonstration, take a few minutes to lead a class discussion about the strengths and weaknesses of each type of wedge material. Write these on the board. This will get the students thinking about material selection before they start their group activity.
Worksheet: Have the students record their test results on The Wedge Worksheet; review their answers to gauge their mastery of the subject.
Post-Activity Assessment
KWL Chart: Finish the remaining section of the KWL Chart as described in the Pre-Activity Assessment section. After the activity, ask students to list in the bottom half of the page under the title, Learned, all of the things that they have learned about wedges. Ask students to name a few items and list them.
Engineering Recommendations: List two or three rock types on the board while the students are finishing the activity and the worksheet questions. Have the students discuss within their groups recommendations for a wedge design to cut through each of the rocks listed. They should suggest what material the wedge should be made from, and how sharp the wedge needs to be.
Class Discussion: Have the students participate in a concluding class discussion about their group test results and answers to the worksheet questions. Which wedge/rock combinations worked? Which did not? Why? How are the wedge angles and points different from each other? Which wedge had the sharpest angle? Which had the biggest cutting surface? How does this make a difference? Why do you think engineers design different types of wedges? Why is material selection an important engineering job?
### Safety Issues
• While safety protection is not required, students should be aware that wedges have sharp edges so safety precautions need to be taken when handling them.
• Students should not be given metal wedges such a nails.
### Troubleshooting Tips
To keep the desks clean and for ease of clean-up, set materials on a tray, paper or cardboard.
Make wedges by sanding the edge of a piece of material such as plastic or wood to create a tapered edge.
Using foam can be useful as an example of a material that does not work well in a wedge application.
Alternate activity setup: If a limited number of supplies are available, each group could work with only one wedge and one material. At the end, each group could share what they learned with the entire class. Alternatively, group sizes of three work fine for this activity.
If the students have a hard time understanding how a wedge works, present a variety of pictures of a wedge in action.
### Activity Extensions
Compile the class worksheet data on the board to provide some nice extension possibilities, discussion, math exercises (averages), graphing, etc. Discuss success in terms of the choice of material or wedge.
Have students explore material properties and material use. Engage the class in a discussion on how the pyramid stones had been shaped and what materials were used (metals, harder stones, etc).
Have the students design their own wedge to serve a specific purpose. For example, ask them to design a wedge that moves snow or splits air (such as an airplane wing). This wedge does not have to be sharp because air is not "hard." How is a zipper considered a wedge?
If possible, take a field trip to a local quarry to see how wedges are used and help students better understand the scale of materials that are extracted from a rock quarry.
### Activity Scaling
• For lower grades, a pencil can be used as a wedge to simplify the amount of materials. In this version, students learn how just one material cuts through various types of other materials, such as marshmallows, wax, sandwiches, etc. Have them rate the success of the pencil in cutting each material so they come to understand how a wedge can be used to split materials.
• For higher grades, assign students the task of extracting 4 cubic cm of each material using the provided wedges.
### References
Bochnacki, Andrzeh. 2005. O Piramidach Inaczeh. Andrzej Bochnacki (Polish engineer). Accessed January 18, 2006. (An excellent animation shows how heavy stone blocks might have been systematically moved up an incline plane using many human-powered wedges. Click on Site Map, then click on Transport on the Ramp.) http://www.swbochnacki.com/
Construction of the Great Pyramid, Construction Theories. World-Mysteries.com. Accessed January 18, 2006. http://www.world-mysteries.com/mpl_2_1.htm#Machines
Nature & Science: Geologic Resources, Abandoned Mineral Land: Nature & Science. Updated October 19, 2003. National Park Service, U.S. Department of the Interior. January 18, 2006. http://www2.nature.nps.gov/geology/aml/
© 2005 by Regents of the University of Colorado.
### Contributors
Lindsey Wright; Lawrence E. Carlson; Jacquelyn Sullivan; Malinda Schaefer Zarske; Denise Carlson, with design input from the students in the spring 2005 K-12 Engineering Outreach Corps course.
### Supporting Program
Integrated Teaching and Learning Program, College of Engineering, University of Colorado Boulder
### Acknowledgements
The contents of this digital library curriculum were developed under a grant from the Fund for the Improvement of Postsecondary Education (FIPSE), U.S. Department of Education, and National Science Foundation GK-12 grant no 0338326. However, these contents do not necessarily represent the policies of the Department of Education or National Science Foundation, and you should not assume endorsement by the federal government.
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https://goprep.co/ex-12-q34-the-y-coordinate-of-any-point-lying-on-the-x-axis-i-1nkgqr
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Q. 345.0( 1 Vote )
# State whether the
True
Points that lies on axis does not lie in any quadrant. If a point lies on y – axis then its x – coordinate will be zero. And if a point lies on x – axis then its y – coordinate will be zero.
Hence, it is not true.
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https://www.gradesaver.com/textbooks/math/algebra/algebra-1-common-core-15th-edition/chapter-5-linear-functions-5-8-graphing-absolute-value-functions-lesson-check-page-348/2
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Algebra 1: Common Core (15th Edition)
$y = |x| + 9$
To translate this graph upward $9$ units, you add $9$ to the expression outside of the absolute value sign. This would shift the graph $9$ units vertically from the origin up the $y$ axis. Thus, the equation of the translated graph is $y = |x| + 9$.
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https://www.askiitians.com/forums/Inorganic-Chemistry/in-the-haemoglobin-molecular-wt-67200-iron-fou_144094.htm
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# In the Haemoglobin (Molecular wt = 67200) iron found 0.33% (by weight). The number of iron atom will be in its one molecule 1 2 3 4Urgent answer required
Bhavya
8 years ago
Since iron is found 0.33% by weight
it means in 100 g of Hb. iron present = 0.33
in 67200 g of Hb, iron present = (0.33/100)*67200
Now atleast 1 atom should be present in its 1 molecule = 56g
56 g = 1 atom
(0.33/100)*67200 = (1/56)*(0.33/100)*67200
= 4
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https://wiki.postgresql.org/wiki/Query_column_with_range_types
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# Query column with range types
## Rationale
The main use case for range types is to store ranges in PostgreSQL tables, and then find rows whose range includes a certain literal. These can already be indexed using GIN and GiST index types.
```db=# EXPLAIN ANALYZE SELECT * FROM rng WHERE rng @> 10.0;
Bitmap Heap Scan on rng (cost=48.40..2709.45 rows=1000 width=32) (actual time=0.103..0.103 rows=10 loops=1)
Recheck Cond: (rng @> 10.0)
-> Bitmap Index Scan on rng_rng_idx (cost=0.00..48.15 rows=1000 width=0) (actual time=0.096..0.096 rows=10 loops=1)
Index Cond: (rng @> 10.0)
Total runtime: 0.136 ms
```
However, you cannot index inverse queries - if you want to find single values in a certain range:
```db=# EXPLAIN ANALYZE SELECT * FROM num WHERE i <@ '[0, 10)';
Seq Scan on num (cost=0.00..16925.00 rows=1000 width=6) (actual time=0.014..188.761 rows=9 loops=1)
Filter: (i <@ '[0,10]'::numrange)
Rows Removed by Filter: 999991
Total runtime: 188.820 ms
```
## Example
Of course the above query is equivalent to the indexable query WHERE num >= 0 AND num < 10. But what if you want to use range types? Well, you can create a custom operator with a function that automatically gets inlined:
```db=# EXPLAIN ANALYZE SELECT * FROM num WHERE i <@ '[0, 10)';
Index Only Scan using num_i_idx on num (cost=0.00..4.56 rows=10 width=6) (actual time=0.014..0.016 rows=9 loops=1)
Index Cond: ((i >= 0::numeric) AND (i < 10::numeric))
Heap Fetches: 0
Total runtime: 0.039 ms
```
Infinite terms get optimized out:
```db=# EXPLAIN ANALYZE SELECT * FROM num WHERE i <@ '[,10]';
Index Only Scan using num_i_idx on num (cost=0.00..4.53 rows=10 width=6) (actual time=0.007..0.009 rows=10 loops=1)
Index Cond: (i <= 10::numeric)
Heap Fetches: 0
Total runtime: 0.032 ms
```
### Beware!
These custom operators are deliberately defined to be non-commative. If you create these operators, do not write range queries in the form 10.0 <@ range_col because it will blow up:
```db=# EXPLAIN ANALYZE SELECT * FROM rng WHERE 10.0 <@ rng;
Seq Scan on rng (cost=0.00..46358.16 rows=444424 width=32) (actual time=0.013..240.054 rows=10 loops=1)
Filter: ((lower_inf(rng) OR CASE WHEN lower_inc(rng) THEN (10.0 >= lower(rng)) ELSE (10.0 > lower(rng)) END)
AND (upper_inf(rng) OR CASE WHEN upper_inc(rng) THEN (10.0 <= upper(rng)) ELSE (10.0 < upper(rng)) END))
Rows Removed by Filter: 999990
Total runtime: 240.125 ms
```
## Code
You can adapt this code to any range type you need, just by changing the types of the function and the operator.
```-- int4range (integer)
CREATE OR REPLACE FUNCTION range_contains(el integer, rng int4range) RETURNS bool IMMUTABLE LANGUAGE sql AS \$\$
SELECT (lower_inf(rng) OR (CASE WHEN lower_inc(rng) THEN el >= lower(rng) ELSE el > lower(rng) END))
AND (upper_inf(rng) OR (CASE WHEN upper_inc(rng) THEN el <= upper(rng) ELSE el < upper(rng) END))
\$\$;
CREATE OPERATOR <@ (procedure=range_contains, leftarg=integer, rightarg=int4range);
-- int8range (bigint)
CREATE OR REPLACE FUNCTION range_contains(el bigint, rng int8range) RETURNS bool IMMUTABLE LANGUAGE sql AS \$\$
SELECT (lower_inf(rng) OR (CASE WHEN lower_inc(rng) THEN el >= lower(rng) ELSE el > lower(rng) END))
AND (upper_inf(rng) OR (CASE WHEN upper_inc(rng) THEN el <= upper(rng) ELSE el < upper(rng) END))
\$\$;
CREATE OPERATOR <@ (procedure=range_contains, leftarg=bigint, rightarg=int8range);
-- numrange (numeric)
CREATE OR REPLACE FUNCTION range_contains(el numeric, rng numrange) RETURNS bool IMMUTABLE LANGUAGE sql AS \$\$
SELECT (lower_inf(rng) OR (CASE WHEN lower_inc(rng) THEN el >= lower(rng) ELSE el > lower(rng) END))
AND (upper_inf(rng) OR (CASE WHEN upper_inc(rng) THEN el <= upper(rng) ELSE el < upper(rng) END))
\$\$;
CREATE OPERATOR <@ (procedure=range_contains, leftarg=numeric, rightarg=numrange);
-- tsrange (timestamp without time zone)
CREATE OR REPLACE FUNCTION range_contains(el timestamp, rng tsrange) RETURNS bool IMMUTABLE LANGUAGE sql AS \$\$
SELECT (lower_inf(rng) OR (CASE WHEN lower_inc(rng) THEN el >= lower(rng) ELSE el > lower(rng) END))
AND (upper_inf(rng) OR (CASE WHEN upper_inc(rng) THEN el <= upper(rng) ELSE el < upper(rng) END))
\$\$;
CREATE OPERATOR <@ (procedure=range_contains, leftarg=timestamp, rightarg=tsrange);
-- tstzrange (timestamp with time zone)
CREATE OR REPLACE FUNCTION range_contains(el timestamptz, rng tstzrange) RETURNS bool IMMUTABLE LANGUAGE sql AS \$\$
SELECT (lower_inf(rng) OR (CASE WHEN lower_inc(rng) THEN el >= lower(rng) ELSE el > lower(rng) END))
AND (upper_inf(rng) OR (CASE WHEN upper_inc(rng) THEN el <= upper(rng) ELSE el < upper(rng) END))
\$\$;
CREATE OPERATOR <@ (procedure=range_contains, leftarg=timestamptz, rightarg=tstzrange);
-- daterange (date)
CREATE OR REPLACE FUNCTION range_contains(el date, rng daterange) RETURNS bool IMMUTABLE LANGUAGE sql AS \$\$
SELECT (lower_inf(rng) OR (CASE WHEN lower_inc(rng) THEN el >= lower(rng) ELSE el > lower(rng) END))
AND (upper_inf(rng) OR (CASE WHEN upper_inc(rng) THEN el <= upper(rng) ELSE el < upper(rng) END))
\$\$;
CREATE OPERATOR <@ (procedure=range_contains, leftarg=date, rightarg=daterange);
```
Idea from Andres Freund, implemented by Marti Raudsepp.
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Lesson Video: Subtracting Two-Digit Numbers with Regrouping | Nagwa Lesson Video: Subtracting Two-Digit Numbers with Regrouping | Nagwa
# Lesson Video: Subtracting Two-Digit Numbers with Regrouping Mathematics • 1st Grade
In this video, we will learn how to subtract a one-digit number from a two-digit number when crossing a ten and model this with place value equipment.
09:44
### Video Transcript
Subtracting Ones from Two-Digit Numbers with Regrouping
In this video, we will learn how to subtract a one-digit number from a two-digit number when crossing a 10 and model this with place value equipment. The number 52 is a two-digit number. Its tens digit is a five, and we’ve modeled this using five 10s blocks. 10, 20, 30, 40, 50. Five 10s are worth 50. The two digit in the number 52 is worth two ones, and we’ve modeled this with two ones blocks. We’re subtracting a one-digit number from our two-digit number. The number seven only has one digit. It’s worth seven ones.
In this video, we’re learning how to use column subtraction to help us subtract a one-digit number from a two-digit number. When we add or subtract using the column method, we always start calculating in the ones column first. 52 has two ones, and we have to subtract seven. We have a problem. Two is less than seven. We don’t have enough ones to subtract seven from. We need to regroup. We need to exchange one of our tens for 10 ones, so let’s take one of our tens and exchange it for 10 ones. Now we have four 10s and 12 ones. So now we have enough ones to subtract our seven from. We have 12 ones, and we need to take away seven. 12 subtract seven equals five.
Next, we need to subtract in the tens column. We had five 10s, and we exchanged one 10 for 10 ones, so we only have four 10s. And because we’re subtracting a one-digit number, there’s nothing to subtract from our four 10s. Four 10s take away no tens leaves us with four 10s. 52 subtract seven is 45. So, we’ve learned that when we’re subtracting one-digit number from a two-digit number, we always start by subtracting the ones first. We do this in case we need to regroup.
The next thing we have to do is regroup. We didn’t have enough ones to subtract seven from 52 because 52 only has two ones. So we have to take one of our tens and exchange it for 10 ones. Now we have 12 ones and four 10s. Once we’ve regrouped, we have enough ones to subtract from. Then we can subtract the tens. Because we’re subtracting a one-digit number from a two-digit number, the tens digit has nothing to subtract from. So we just need to write the number of tens in the tens place. 52 subtract seven is 45. Let’s try and put into practice what we’ve learned now by answering some questions.
Mia is calculating 33 subtract four. She showed 33 using place value blocks. After she regroups and subtracts four, how many tens and ones will be left?
In this question, we have to help Mia calculate 33 subtract four. The number 33 is a two-digit number. And Mia’s modeled this using place value blocks. 33 has three 10s, worth 30, and three ones, which are worth three.
The second part of the question tells us that Mia has to regroup to subtract four. We know that four is more than three. There are not enough ones in the number 33 to subtract four from. So we need to take one of our tens and exchange it for 10 ones. Now we have two 10s and 13 ones. Now we have enough ones to subtract our four from. 13 take away four is nine. We’ve got nine ones left. We had three 10s. We exchanged one 10, so we only have two 10s left. And there’s nothing to subtract from these two 10s in the tens place. So 33 subtract four is 29. After Mia regroups and subtracts four, she will have two 10s and nine ones left.
Subtract eight from 66.
In this question, we’re subtracting a one-digit number from a two-digit number. 66 has two digits, and eight has one. When we’re subtracting from a two-digit number, we always start by subtracting the ones. But six is less than eight. We don’t have enough ones to subtract from. So we need to regroup. We need to take one of our six 10s, which would leave us with five 10s, and exchange it for 10 ones. So instead of having six ones, we have 16 ones. Now we have enough ones we can subtract in the ones column. What is 16 take away eight? It’s eight.
And because we’re subtracting a one-digit number from a two-digit number, there’s nothing to subtract from our tens. We had six 10s. We exchanged one of our tens for 10 ones, leaving us with five 10s. And five 10s take away no tens leaves us with five 10s. 66 take away eight is 58. First, we regrouped one of our tens and exchanged it for 10 ones. Next, we subtracted the ones. 16 take away eight is eight. Then we subtracted the tens. Five 10s take away no tens is five 10s. We subtracted eight from 66, leaving us with 58.
A car park has 41 spaces. Two spaces are taken. How many are left?
In this question, the car park has 41 spaces. If two spaces are taken, to calculate how many spaces are left, we need to subtract the two spaces that are taken from the 41 spaces in the car park. So we’re subtracting a one-digit number from a two-digit number. We can model the number 41 using four 10s blocks and a ones block. When we’re subtracting using the column method, we always start with the ones first. We’ve got one in the ones place, and we need to subtract two. We don’t have enough ones to subtract from. So we need to take one of our tens, leaving us with three 10s, and exchange it for 10 ones. Now we have 11 ones.
Now that we have enough ones, let’s subtract in the ones place. 11 ones take away two ones leaves us with nine ones. Now what we need to do is subtract in the tens column. Because two is a one-digit number, there’s nothing to subtract. We had four 10s. We exchanged one of our tens for 10 ones, which leaves us with three 10s. 41 subtract two equals 39.
What have we learned in this video? We have learned how to subtract a one-digit number from a two-digit number with regrouping.
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