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# BCPA/PlotBCPA
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(Created page with "<pre> PlotBCPA <- function (t, x, ws, pp, threshold=10) { plot(t, x, type = "n", main = "", ylim = c(min(x), max(x) * 1.3)) x.breaks <- ws\$Break x.model <- ...")
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## Revision as of 09:57, 19 July 2011
```PlotBCPA <- function (t, x, ws, pp, threshold=10)
{
plot(t, x, type = "n", main = "", ylim = c(min(x), max(x) *
1.3))
x.breaks <- ws\$Break
x.model <- ws\$Model
x.breaks <- x.breaks[x.model > 0]
x.model <- x.model[x.model > 0]
mids <- hist(x.breaks, breaks = t, plot = F)\$mid
freq <- hist(x.breaks, breaks = t, plot = F)\$count
goodbreaks <- mids[freq > threshold]
freq <- freq[freq > threshold]
break.cols <- heat.colors(max(freq))
break.cols <- break.cols[length(break.cols):1]
abline(v = goodbreaks, lwd = freq, col = "orange")
lines(t, x, col = "darkgrey")
lines(t, pp\$mu.hat, lwd = 2)
lines(t, pp\$mu.hat + pp\$s.hat, col = 2, lwd = 1.5)
lines(t, pp\$mu.hat - pp\$s.hat, col = 2, lwd = 1.5)
rho.hat <- pp\$rho.hat
rho.int <- round((rho.hat - min(rho.hat, na.rm = 1)) * 1000) +
1
rho.col <- topo.colors(max(rho.int, na.rm=TRUE))[rho.int]
rho.cex <- (rho.hat - min(rho.hat, na.rm=TRUE))/(diff(range(rho.hat))) *
1.5 + 0.5
points(t, x, col = rho.col, pch = 19)
rho.index <- quantile(0:max(rho.int))
rho.index[1] <- 1
legend.cols <- topo.colors(max(rho.int, na.rm=TRUE))[round(rho.index)]
legend.rhos <- round(quantile((min(rho.hat) * 1000):(max(rho.hat) *
1000))/1000, 2)
legend("bottomright", bg = "white", legend = c(expression(hat(rho)),
legend.rhos), pch = c(0, rep(19, 5)), ncol = 2, col = c(0,
legend.cols), xjust = 0.5, yjust = 0.5, cex = 1.2)
legend("topright", bg = "white", legend = c(expression(hat(mu)),
expression(hat(mu) %+-% hat(sigma))), lty = 1, lwd = 2:1,
col = 1:2, xjust = 0.5, yjust = 0.5, cex = 1.2)
}
```
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Tools
EBM tools for the five stages of Evidence-Based Medicine
Asking focused questions is directly relevant to patients’ problems and phrased to direct your search to relevant and precise answers.
Finding the Evidence
Finding the evidence is the second step in evidence-based practice, after formulating a clearly focused clinical question.
Critical Appraisal tools
Critical appraisal worksheets to help you appraise the reliability, importance and applicability of clinical evidence.
Making a Decision
The integration of relevant evidence with clinical experience forms the cornerstone of decision making in evidence-based practice.
CATMaker and EBM Calculators
Put the “like” into your likelihood ratios with the CEBM’s CATmaker and EBM calculators.
Finding the Evidence 1 - Using PICO to formulate a search question
A mini tutorial on finding the evidence provided by Neal Thurley and Owen Coxall who regularly tutor on CEBM courses. It is recommended that you watch them in order.
Finding the evidence 2 - Turning search terms into a search strategy
This mini tutorial shows you how to convert your search question into a search strategy that will work on any bibliographic database.
Finding the evidence 3 - Turning your search strategy into results: searching PubMed
A video demonstration of how to get the best out of PubMed.
Glossary
A glossary of EBM terms
PaT Plot tool for randomised trials
This simple program will enable you to depict any randomised trial graphically however many components the interventions have.
Study Designs
This short article gives a brief guide to the different study types and a comparison of the advantages and disadvantages of the different types of study.
Number Needed to Treat (NNT)
The Number Needed to Treat (NNT) is the number of patients you need to treat to prevent one additional bad outcome (death, stroke, etc.).
SpPin and SnNout
To understand what is meant by the terms SpPin and SnNout, we need to understand the notions of sensitivity and specificity.
Pretest Probability
Pretest Probability is defined as the probability of a patient having the target disorder before a diagnostic test result is known.
Likelihood Ratios
The Likelihood Ratio (LR) is the likelihood that a given test result would be expected in a patient with the target disorder compared to the likelihood that that same result would be expected in a patient without the target disorder.
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https://math.stackexchange.com/questions/1343497/what-is-the-definition-of-numbers?noredirect=1
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# what is the definition of numbers? [duplicate]
Well the question may seem obvious but I can't really find a proper answer to this. Mathematics all seem kind of difficult to understand so please help. I believe it is a quantity. Thanks in advance.i am mainly confused about using numbers and using numbers with units.the second one makes sense but i cant understand the first one
## marked as duplicate by Thomas Andrews, mweiss, Simon S, JMoravitz, user147263 Jun 29 '15 at 19:44
• The question in the following link is a little different, but I think the answer is spectacular, and also answers your question. math.stackexchange.com/questions/199676/… – Alex S Jun 29 '15 at 16:53
• In general, the term "number" is used fairly loosely in mathematics, unfortunately, so there isn't a general rule. – Thomas Andrews Jun 29 '15 at 16:57
• In my opinion, "numbers" can be added, subtracted, multiplied and divided; roughly speaking, numbers is elements of a some field. – Michael Galuza Jun 29 '15 at 17:12
Perhaps rather strangely, there is not a simple standard definition of "number".
In the third century BC, Euclid took "number" to mean one of $2,3,4,5,\ldots$, a sequence starting with $2$ and closed under the operation of adding a unit, i.e. if $n$ is a number, then so is $n+1$, so the sequence continues infinitely. Euclid did not consider the "unit" to be a number. That is consistent at least with the meaning of the expression "A number of students asked this question." One wouldn't say that if just one student had done so.
Euclid proved a number of propositions about numbers, including things about divisibility. He introduced what is now the oldest algorithm still in standard use, his algorithm for finding the largest divisor that two numbers share in common. He proved that the list of prime numbers keeps going no matter how far it is extended. (His proof was that if $S$ is any set of prime numbers then $1+\prod S$ has at least one prime factor other than the prime numbers that are in $S$. In fact every prime factor of that number fails to be in $S$.)
Today "numbers" are taken to include $0$ and $1$ and negative integers $-1,-2,-3,\ldots$ and rational numbers, e.g. $10/7$, that result from dividing one integer by another, and "real" numbers that are not rational. "Real" numbers are identified with points on what you may have heard called the "number line". (In a sense Euclid dealt with irrational numbers and in another sense he didn't. He showed that no pair of what he called "numbers" had the same ratio as the ratio of the length of the diagonal of a square to the length of the side. But he didn't consider such ratios to be "numbers".
("$\alpha\rho\iota\theta\mu\omicron\sigma$" is the Greek word used by Euclid, usually (or always?) translated as "number".))
Conventionally "numbers" are taken to include "imaginary" numbers and "complex" numbers. A complex number is a sum of a real number and an imaginary number.
The concept of "number" is sometimes taken to include cardinal numbers of infinite sets, and sometimes taken to include "nonstandard" real numbers, some of which are infinitely large or infinitely small. Occasionally, someone considers quaternions to be "numbers", and that's really only a question of how words should be conventionally used.
Formally, we would describe numbers as "mathematical objects used to count, measure or label". However, this definition seems to apply more to real numbers, that is numbers we use in everyday life (2, 5, $\pi$ ...). There exist certain quantities, which we call numbers, that cannot be assigned to a measure or label in real life. Therefore, depending on how exact you want to be, different definitions will be applicable. Here are some I have found, and the link corresponding to them.
"Mathematical objects used to count, measure or label" - wikipedia
"an arithmetical value, expressed by a word, symbol, or figure, representing a particular quantity and used in counting and making calculations and for showing order in a series or for identification." - New Oxford American Dictionary
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# Chi-squared Test of Independence
It is a non-parametric test to determine if there is a significant relationship between two categorical variables. The frequency of one variable is compared with frequency of second variable .
Contact at TJT@TechnicalJockey.com , if you are looking for an Instructor Based Online Training or Corporate Training in R !
The assumptions of chi-square test are :
1. The data in the cells should be frequencies or counts of cases.
2. The levels of the variables are mutually exclusive .
3. Each subject may contribute data to one and only one cell .
4. The groups must be independent . There is no interdependency between groups while comparing the groups.
5. The variables should be categorical or we can change data in categorical form .
6. The sample data are displayed in a contingency table , the expected frequency count for each cell of the table is at least 5.
Expected frequencies :
The expected frequency is calculated for each cell in a contingency table. The expected frequency is calculated as :
E = nr X nc /n
Where
E - represents the cell expected value,
nr - represents the total number of sample observations for row for level r
nc - represents the total number of sample observations for column for level c
n - represents the total sample size
Contact at TJT@TechnicalJockey.com , if you are looking for an Instructor Based Online Training or Corporate Training in R !
Test statistic :
The test statistic of the chi-square test is -
χ2 = ∑ (O-E)2/ E
Where
O - Observed value
E - Expected value
χ2 - the chi-square value
∑ - Calculate summation of all values in cell
Null hypothesis : Assumes that there is no association between the two variables .
Alternative hypothesis : Assumes that there is an association between the two variables.
If p-value > 0.05 , then null hypothesis is true. If p-value is less than 0.05 then alternative hypothesis is true.
Degrees of freedom :
The number of degrees of freedom can be defined as the minimum number of independent coordinates that can specify the position of the system completely.
The degrees of freedom , df = (Number of rows -1) X (Number of columns - 1)
We will use housetasks data set from STHDA .
We import dataset using online link.
We import dataset by using read.delim() function.
We are installing "gplots" library for visualization.
install.packages("gplots")
We load "gplots" library using following code:
library("gplots")
We want to create a table format to store the dataset . To convert dataset into a table , we used as.matrix() function to convert in matrix form and then convert matrix into a table format by using as.table() function on it.
We transform dt table to represent rows values corresponds to values in table .
t(dt)
We are using baloonplot to plot data in a dot form. In this plot , dot is bigger if the value of the variable is larger. We used label = FALSE to not show the values of the elements on the plot. We used show.margins = FALSE to not print the total sum of rows and columns in the plot.
Contact at TJT@TechnicalJockey.com , if you are looking for an Instructor Based Online Training or Corporate Training in R !
balloonplot(t(dt), main ="housetasks", xlab ="", ylab="", label = FALSE, show.margins = FALSE)
We are installing "graphics" library for advanced visualization. We load "graphics" library as:
install.packages("graphics")
library("graphics")
We are using mosaicplot to plot the work associated with Husband and Wife .
The argument shade is used to color the graph
The argument las=2 produces vertical labels.
From this plot , we can see that housetasks Laundry, Main_meal , Dinner and breakfast(blue color) are mainly done by the wife .
The chi-square test can be done as :
chisq
Output :
Here , X-squared = 1944.5 means chi-square value is 1944.5 and the degrees of freedom is 36. The p-value is less than 2.2e-16
We can see observed frequency by using following code :
chisq\$observed
We can see expected frequency by using following code :
round(chisq\$expected,2)
Pearson residual
The Pearson residuals can be used to check the model fit at each observation for generalized linear models. The Pearson residual for a cell in a two-way table is :
r = O - E / √ E
We can calculate residuals by following code :
round(chisq\$residuals, 3)
The chi-square statistic is the sum of the contributions from each of the individual cells.
If an individual contribution is high, it is either because the expected value is low or the difference between the observed and the expected is reasonably high. If the independent variable has more than two values, you might like to consider whether the distinction between a specific value and all the others would be significant.
We can see chi-square vale as :
We can also find contribution of each combination of pairs in chi-square test.
It is the ratio of squared residual value and chi-square value.
contrib <- 100*chisq\$residuals^2/chisq\$statistic
round(contrib, 3)
Contact at TJT@TechnicalJockey.com , if you are looking for an Instructor Based Online Training or Corporate Training in R !
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# How do you convert time to longitude?
To find the longitude, simply compare your local time with the time at a known location and multiply by 15 °/hr. For example, if your local solar time differs from another location by 4 hours, then you are 4 x 15 = 60° longitude away from that location.
## How do you convert time into longitude?
Finding Longitude. Because one day is 24 hours long one can easily use time to calculate longitude. One hour of time difference corresponds to 15° of longitude (360°/24 hours = 15°/hour). Suppose an observer sets his accurate watch to 12:00 at noon in Greenwich, England and then travels a great distance.
## How does time determine longitude?
In fact, if you know the difference in time between Greenwich noon (what your clock says) and local noon (when the sun is at the highest point), you can find your longitude. This time difference is essentially your longitude.
## How do you convert longitude to time zone?
Longitude is usually measured in degrees, minutes and seconds. Here's how longitude converts into your personal time zone: 15 degrees of longitude = 1 hour difference; 1 degree longitude = 4 minutes difference. 15 minutes of longitude = 1 minute difference; 1 minute of longitude = 4 seconds difference.
## What is the formula for longitude?
Latitude = 90 - θ, where θ is the angle measured from the equator (0°) to the point of interest (90° at the North Pole, -90° at the South Pole). Longitude = φ, where φ is the angle measured from the Prime Meridian (0°) to the point of interest, either to the east (0° to 180°) or to the west (0° to -180°).
## What is 1 longitude equal to?
One degree of latitude equals approximately 364,000 feet (69 miles), one minute equals 6,068 feet (1.15 miles), and one-second equals 101 feet. One-degree of longitude equals 288,200 feet (54.6 miles), one minute equals 4,800 feet (0.91 mile), and one second equals 80 feet.
## How do you calculate longitude from local time to GMT?
Note that the world's time zones change every 15 degrees of longtitude. To calculate, take the difference between local time and the GMT. For example, if it's 7 am local and 12 Noon GMT, the location is somwhere around 75 degrees W longitude, that is 5 x 15.
## How do you find latitude and longitude from time?
Time the difference between local noon, measured from the sun, and the 12 o'clock noon pips on the radio. If you reach local noon before the radio signal, you're east of the Prime Meridian (0º longitude). If the radio signals noon while the sun is still climbing, you're on a western longitude.
## How much longitude equals one hour of time?
In 24hrs, the Earth completes one rotation of 360° along its axis. 1 degree of longitude is equal to 4 minutes. 15-degree longitude is 1 hour.
## What is the format for longitude minutes?
Each degree on a map is divided into 60 minutes, or 1/60 of a degree. To calculate the number of minutes represented by the decimal parts of your latitude and longitude, multiply the decimal parts by 60. The minutes are written with an apostrophe. For example, 37.44 minutes would be written as 37.44'.
## Why is longitude so hard to calculate?
Latitude relates to something physical (the equator) and can be determined from the position of heavenly bodies such as the Sun or pole star, but longitude is more difficult because there are no natural references from which to measure.
## Does time equate to longitude?
A navigator would compare the time at local noon (when the sun is at its highest point in the sky) to an onboard clock that was set to Greenwich Mean Time (the time at the prime meridian). Each hour of difference between local noon and the time in Greenwich equals 15 degrees of longitude.
## Can you determine longitude without a clock?
In principle, you could tell your longitude by observing the angle between the Moon and a particular star then consulting an almanac, which catalogued the time at Greenwich based on the position of a range of celestial objects.
## How many minutes are in 1 degree longitude?
Latitude and longitude are measuring lines used for locating places on the surface of the Earth. They are angular measurements, expressed as degrees of a circle. A full circle contains 360°. Each degree can be divided into 60 minutes, and each minute is divided into 60 seconds.
## How is latitude calculated?
1 The latitude of a point on the Earth's surface is determined by the angle (ø) between the point and the equator, passing through Earth's center (Peter Mercator [Public domain], via Wikimedia Commons). One degree of latitude is divided into 60 minutes (').
## How much time is between 2 longitudes?
The correct answer is 4 minutes. The time difference between any two longitudes is 4 minutes. The places located in the East of the Prime Meridian are ahead in time of the places located in the Western Hemisphere. So if it is 12 noon at Greenwich (0 degrees), it would be 12:04 pm at 1-degree meridian, and so on.
## Is 1 degree longitude equal to 4 minutes or 60 minutes?
One degree of longitude is divided into 60 minutes.
## Why is longitude in minutes?
Sixty was used because it's convenient, being evenly divisible by many numbers. Navigation techniques developed while minutes and seconds were still common and we've kept them, much as we have for hours of time, because changing is difficult and there's little reason to do so.
## What longitude does time start?
The international date line, established in 1884, passes through the mid-Pacific Ocean and roughly follows a 180 degrees longitude north-south line on the Earth. It is located halfway around the world from the prime meridian — the 0 degrees longitude line in Greenwich, England.
## How did sailors calculate longitude?
Before the invention of GPS, sailors used a method called celestial navigation to calculate longitude. This involved using celestial bodies such as the sun, moon, and stars to determine their position at sea.
## How did sailors calculate latitude?
Astrolabe - The astrolabe is a circular device, often made of brass or wood, used to measure latitude. This Arabic device was created in the 2nd century and perfected over time. To measure latitude, sailors would point the astrolabe towards the sun during the day and the North Star at night.
## Is GMT based on longitude?
Greenwich Mean Time (GMT), the name for mean solar time of the longitude (0°) of the Royal Greenwich Observatory in England. The meridian at this longitude is called the prime meridian or Greenwich meridian.
## Does longitude affect time?
Explanation: In theory every 15 degrees change in longitude from line zero in England represents a new time zone or a 1 hour difference. 36024 = 15 degrees. In practice political factors influence where the time zones actually fall.
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Chapter 6 and 7 Review
Chapter
Chapter 7
Section
Chapter 6 and 7 Review
Solutions 38 Videos
Write the first three terms of the sequence. Describe the pattern in words.
t_n = 3n -2
Q1a
Write the first three terms of the sequence. Describe the pattern in words.
\displaystyle f(n) = 4^n + 1
Q1b
Write the first three terms of the sequence. Describe the pattern in words.
\displaystyle t_n = 5n^2 -16
Q1c
Write the first three terms of the sequence. Describe the pattern in words.
\displaystyle f(n) = \frac{n^3}{2}+ 2
Q1d
Write the first three terms of the sequence. Describe the pattern in words.
\displaystyle t_1 =1, t_n = 2t_{n-1} + 5
Q1e
Write the first three terms of the sequence. Describe the pattern in words.
\displaystyle t_1 = -2, t_n = (t_{n-1})^2 -8
Q1f
Graph the first eight terms of each sequence in question 1.
Q2
The value of a new car bought for $45 000 depreciates at a rate of 15% in the first year and 5% every year after that. a) Determine the value of the car at the end of the first year, the second year, and the third year. Write these values as a sequence. b) Determine an explicit formula for the value of the car at the end of year 11. c) What is the value of the car at the end of year 20? Is this realistic? Explain your thinking. Buy to View Q3 Given the graph, write the sequence of terms and determine a recursion formula using function notation. Buy to View Coming Soon Q4 Write a recursion formula and an explicit formula for each sequence. 5, 7, 9, 11, .... Buy to View Q5a Write a recursion formula and an explicit formula for each sequence. 2, 4, 16, 256, ... Buy to View Q5b Investigate the prime—numbered rows of Pascal’s triangle. Describe the property that is common to these rows but not common to the rows that are not prime-numbered. Buy to View Q6 Expand and simplify each binomial. \displaystyle (2x + 5)^7 Buy to View Q7a Expand and simplify each binomial. \displaystyle (a^2 -3b)^5 Buy to View Q7b Expand and simplify each binomial. \displaystyle (\frac{2}{x} + x^2)^6 Buy to View Q7c Expand and simplify each binomial. \displaystyle (5 - \frac{3}{\sqrt{n}})^4 Buy to View Q7d State whether the sequence is arithmetic, geometric, or neither. Determine a defining equation for those that are arithmetic or geometric, and find the 12th term. 6, 11, 17, 29, ... Buy to View Q8a State whether the sequence is arithmetic, geometric, or neither. Determine a defining equation for those that are arithmetic or geometric, and find the 12th term. -3, 1, 5, 9, .... Buy to View Q8b State whether the sequence is arithmetic, geometric, or neither. Determine a defining equation for those that are arithmetic or geometric, and find the 12th term. 3, 12, 48, 192, .... Buy to View Q8c State whether the sequence is arithmetic, geometric, or neither. Determine a defining equation for those that are arithmetic or geometric, and find the 12th term. 2 657 205, -885 735, 295 245, -98415, .... Buy to View Q8d The fourth term of an arithmetic sequence is 6 and the seventh term is 27. Determine the first and second terms. Buy to View Q9 How many terms are in the geometric sequence 7, 21, 63, 3 720 087? Buy to View Q10 A new golf course has 480 lifetime members 3 weeks after it opens and 1005 lifetime members 6 weeks after it opens. Assume the membership m‘crease is arithmetic. a) Determine the general term that represents the sequence for this situation. b) How many members were there at the end of the fifth week? c) When will there be over 2000 members? Buy to View Q11 Determine the sum of the first 10 terms of each series. 2 + 9 + 16 + 23 + .... Buy to View Q12a Determine the sum of the first 10 terms of each series. 5 -25 + 125 -625 + .... Buy to View Q12b Determine the sum of the first 10 terms of each series. 256 + 128 + 64 + 32 + .... Buy to View Q12c Determine the sum of the first 10 terms of each series. - \frac{1}{3} - \frac{5}{6} - \frac{4}{3} - \frac{11}{6} - ... Buy to View Q12d A ball is dropped from a height of 160 cm. Each time it drops, it rebounds to 80% of its previous height. a) What is the height of the rebound after the 8th bounce? b) What is the total distance travelled at the time of the 15th bounce? Buy to View Q13 Define each term. a) principal b) amount c) simple interest d) compound interest e) annuity f) present value g) compounding period Buy to View Coming Soon Q14 Determine the interest earned for each investment. a)$1000 is invested for 8 months at 5% per year simple interest.
b) $800 is placed into an account that pays 2.5% annual simple interest for 40 weeks. c) A 90-day$10 000 treasury bill earns simple interest at a rate of 4.8% per year.
Q15
Alex deposited $500 into an account at 3% simple interest. a) Write an equation to relate the amount in the account to time. b) Sketch a graph of this relation for 1 year. c) How long does it take to earn$10 in interest?
Q16
Sarah invests $750 in a term deposit, at 4.5% per annum, compounded semi-annually, for 5 years. How much interest will Sarah earn? Buy to View Q17 Abdul decides to invest some money so that he can have$10 000 for a down payment on a new car in 5 years. He is considering two investment options: Account A pays 3.5% per annum, compounded semi-annually. Account B pays 3.2% per annum, compounded monthly.
a) Compare the present values of the two options.
b) Which account is the better choice for Abdul? Explain your reasoning.
To save for her university education, Shally will deposit $50 into an account at the end of each month for the next 3 years. She expects the interest rate to be 1.5% per year, compounded monthly, over that time. How much will she have saved after 3 years? Buy to View Q19 Wayne is 16 years old. To become a millionaire by the time he is 50 years old, how much does Wayne need to invest, at the end of every 6 months, at 4% per year, compounded semi-annually? Buy to View Q20 What annual interest rate, compounded monthly, is needed for Eva to accumulate$20 000 by depositing $300 at the end of each month for 5 years? Buy to View Q21 A rental contract calls for a down payment of$1000, and $500 to be paid at the end of each month for 3 years. If interest is at 4.5% per year, compounded monthly, what is the present value of this rental contract? Buy to View Q22 A retirement account contains$100 000. Barb would like to withdraw equal amounts of money at the end of every 3 months for 15 years. If interest is 5% per year, compounded quarterly, what will the size of Barb’s withdrawals be?
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10sE2FirstFewProblems
# 10sE2FirstFewProblems - Prof Girardi Math 142 Typical First...
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Prof. Girardi Math 142 Typical First Few Problems on Sequences & Series Exam If you do not make at least a 50% on this exam’s first few problems , then your total exam score will be only as many points was you managed to get on the first few problems . Here are some typical first few problems . 1. Fill-in-the blanks/boxes. All series are understood to be X n =1 . Hint: You should NOT write the words absolute nor conditional on Problem 1! 1a. Sequences (Afterall, this is needed for Geometric Series!) Let -∞ < r < . (Fill-in-the blanks with exists or does not exist, i.e. DNE ) If | r | < 1, then lim n →∞ r n If | r | > 1, then lim n →∞ r n If r = 1, then lim n →∞ r n If r = - 1, then lim n →∞ r n 1b. Geometric Series where -∞ < r < . The series r n converges if and only if | r | diverges if and only if | r | 1c. p -series where 0 < p < . The series 1 n p converges if and only if p diverges if and only if p 1d. Integral Test for a positive-termed series a n where a n 0. Let f : [1 , ) R be so that a n = f ( ) for each n N f is a function f is a function f is a function .
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# Help with calculus 2!!! its difficult..10 points?
A force of 3 pounds is required to hold a spring stretched 0.5 feet beyond its natural length. How much work (in foot-pounds) is done in stretching the spring from its natural length to 1 feet beyond its natural length?
can anyone solve that???
Relevancia
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# math
posted by .
A certain computer can perform 105 calculations per second. How many calculations can it perform in 10 seconds?
• math -
10 seconds is 10 times as big as 1 second, so you will get 10 times as many calculations.
PS DON'T buy this tar baby of a computer!
• math -
If you are from connections academy you are not to use the provided computer to cheat or look up answers. The student who posted this will be getting a phone call and immediately removed from the online program and sent back to public school. We do not tolerate this cheating behavior.
• math -
Connections police, you are fake. Sorry but you are.
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Home > Documents > HW1 6130 Solution
# HW1 6130 Solution
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Homework I (due January 21, 2010)
1. Between what two circle numbers on the Porton graticule will the image of a 1-m particle fall when viewed with a 43 objective and a 15 eyepiece? Circle
number 10 has an image diameter of 9.9 mm. (10%)Sol>
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2. (a) A straight fiber has an aspect ratio of 5:1. If the fiber is viewedperpendicular to its axis, what is the maximum ratio of Ferets diameter toprojected area diameter? (b) For particles having a square projected crosssection, what is the ratio of Ferets diameter (averaged over all orientations) to
the projected area diameter? (20%)Sol>
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3. A particle has a smooth, circular cross section of diameter dp. The fractaldimension of the particle's perimeter is to be estimated by fitting a four-sided
and a six-sided regular inscribed polygon to the circle. Based only on thisinformation, what fractal dimension is estimated for this step size range. [Hint:
The length b of the side of a regular inscribed polygon with n sides is b = dp
sin(180o/n).] (15%)
Sol>
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4. Fumed silica produced by aerosol process is an agglomerate following a powerlaw (fractal-like). Professor Ulrich just produced some fumed silica following
an algorithm that generates the structure with Df= 2. The radius of gyration of
the agglomerate, Rg, equals 1 m; the primary particle size, dpo, equals 5 nm;
and the prefactor, A, equals 1.23. (a) What is the total number of primaryparticles in the agglomerate? (b) The agglomerate is heated and coalesces to
form a spherical particle. What is the radius of the particle? (c) Ifdp0 is doubled,what is the new value ofRg? Assume the same growth algorithm for large and
small primary particles; the number of particles remains the same. (d) An
agglomerate with Rg= 0.5 m, composed of the same primary particles (dpo = 5
nm) with the same Dfcollides and sticks to the original Rg= 1 m agglomerate.
What is the new value ofRg, assuming that Dfis preserved? (20%)
Sol>
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5. Measurement from a utility boiler shows the aerosol has a lognormal size
distribution with a geometric mean particle diameter of 0.3 m and g of 1.5. If
the number concentration is 106 #/cc, what is the mass concentration? Particles
are spherical with a density of 4.5 g/cm
3
. (15%)Sol>
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6. You are given the following data, obtained by sequential sieving of a sample ofgranite dust:
Sieve opening (m) Mass captured on Sieve (g)
200 4.0100 21.650 38.4
40 8.0Final Pan 8.0
Total 80.0
Determine the mass median diameter and geometric standard deviation of thisdistribution using log-probability graph paper. Use the appropriate conversion
equation to determine the count median diameter and mass mean diameter.(15%)
Sol>
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Ref> Log-Probability graph
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7. (a) An aerosol is mixed with radon gas, resulting in a surface coating ofradioactive radon decay products on the particles. The aerosol is then divided
into eight aerodynamic size groups, and the radioactivity of each size group ismeasured. How can this information be used to calculate the count median
diameter if the distribution is lognormal? All particles have the same densityand are geometrically similar. Assume the log-probability graph paper is
available. (b) For particles less than 0.05 m, light extinction efficiency is
proportional to dp4. If an aerosol is lognormally distributed with a CMD of 0.01
m and a GSD of 1.8, what is the diameter of average extinction efficiency?
(20%)
Sol>
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8. A powder company generates their powder using an aerosol technique. Themeasurement of the generated powder indicates that 16% of the total particles is
contributed by particles less than 0.3 m and 2.5% by particles larger than 2.4
m. Assuming the powder is lognormally distributed, calculate the dpg and g.
(10%)Sol>
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# What is Vitamix C and G-Series?
• C-Series Standard (variable-speed settings) • C-Series Standard Program (variable-speed settings and presets) • C-series 5300 (A C-series model with a more powerful motor) • G-Series Standard (variable-speed settings.
Contents
## Which Vitamix is the C-Series?
Vitamix High Performance Blender C Series 6500.
## Is Gustavo Fring his real name?
Giancarlo Esposito, who plays fast food mogul and crystal meth mastermind Gustavo “Gus” Fring on AMC’s “Breaking Bad,” told producers who kept inviting him back for guest appearances that he wasn’t interested: He wanted a regular role on the series, one that would let him explore all of his character’s many secrets.
## How do you solve a proportion question?
12:5816:48Ratio and Proportion (10 Solved Problems.
## How do you find first proportion?
The proportion formula is used to depict if two ratios or fractions are equal. We can find the missing value by dividing the given values. The proportion formula can be given as a: b::c : d = a/b = c/d where a and d are the extreme terms and b and c are the mean terms.
## Was Howard Donald on The Masked Singer?
“I’m so made up that I made it to the final, I got a bronze medal.” Before Howard was unmasked, former EastEnders actress Tamzin Outhwaite was unveiled as Scarecrow. Once Howard was revealed, he was followed by runner up Bonnie Langford as Squirrel and then Louis Smith was crowned the winner as Carwash.
## Which is formula of proportion?
The proportion formula is used to depict if two ratios or fractions are equal. We can find the missing value by dividing the given values. The proportion formula can be given as a: b::c : d = a/b = c/d where a and d are the extreme terms and b and c are the mean terms.
## Who was Scarecrow masked dancer?
actress Tamsin Outhwait.
## Who is the Blue Sonic?
Sonic the Hedgehog is the main character of the series and debuted on the game of the same name. He’s a blue hedgehog and is generally regarded as one of the most popular video game characters of all time. He’s best known for his speed, which overtime seems to have increased.
## Is Sonic inappropriate?
Sonic the Hedgehog is rated PG for action, some violence, rude humor and brief mild language. The Sonic movie is generally safe for kids with a few things to watch for like fart jokes and a couple of bad words, but nothing major.
## Can you mix cloves and cinnamon?
While clove has a nice hot and pungent flavor, cinnamon has mildly sweet tones. Together they make a combination that spells blockbuster. The spices also happen to be a treasure trove of antioxidants.
## Which brand of cake mixer is the best?
Our top picksBest Overall Stand Mixer. Artisan Design Series 5-Qt. Mixer KitchenAid. Best Value Stand Mixer. Electric Stand Mixer Hamilton Beach. Best Multi-Purpose Stand Mixer. SM-50TQ Stand Mixer Cuisinart. Best Compact Stand Mixer. Artisan Mini Stand Mixer. Best Large Stand Mixer. High Performance Stand Mixer Wolf Gourmet.8 Apr 202.
Frog hip-hopped to The Masked Singer season 3 finale, in which he was unmasked as rapper and TV personality Bow Wow. And although he finished in third place, Bow Wow beat out 15 other celebrities along the way and rediscovered his love for performing.
## Can a 4 year old watch Sonic 2?
With the full PG rating breakdown above, I would not recommend Sonic the Hedgehog 2 movie to kids younger than the age of 5-6 years old.
## Can you freeze Christmas mince pies?
According to Good Housekeeping, uncooked mince pies can be wrapped in clingfilm and frozen in their tins for up to three months, before being baked in the oven for around half an hour to get them piping hot once more.
## What is ratio and proportion for Class 6?
If two ratios are equal we say that they are in proportion and use the symbol to equate the two ratios. and say that 2, 4, 60 and 120 are in proportion. We say that 2, 5, 60 and 15 are not in proportion. So, if two ratios are not equal, then we say that they are not in proportion.
## Is Moxie based on a true story?
While the film Moxie leads many people to believe that the story is based on the true story of the director or Amy Poehler. However, this isn’t true. The film has been adapted from a 2017 novel of the same name by popular Jennifer Mathieu.
## How Long Will homemade mince pies keep in a tin?
According to Good Housekeeping, uncooked mince pies can be wrapped in clingfilm and frozen in their tins for up to three months, before being baked in the oven for around half an hour to get them piping hot once more.
## How accurate is Moxie?
Moxie Review The movie has received mostly positive reviews and currently stands at a 6.7 rating on IMDb. Fans have reviewed the film much better than the film critics. The film accurately portrays issues that girls and young women face in their lives as they go through high school and college.
## Is the girl band in Moxie real?
Writing a concert into the story was a great move for a movie with zines and leather jackets, too. And featuring The Linda Lindas instead of a fake band or a bigger band with no connection to riot grrrl or punk rock was kind of brilliant.
## Is KitchenAid a planetary mixer?
NSF Certified: KitchenAid Commercial series planetary stand mixers are Underwriters Laboratories (UL) listed and National Sanitation Foundation (NSF) certified, meaning mixers from this series are safe in commercial foodservice applications.
## What is defined as processed food?
By definition, a processed food is a food item that has had a series of mechanical or chemical operations performed on it to change or preserve it. Processed foods are those that typically come in a box or bag and contain more than one item on the list of ingredients.
## Are The Linda Lindas in Moxie?
The Linda Lindas – Performance video for MOXIE premiere.
## Who invented apple cobbler?
Mary Hinman Abel. It was part of a series of menus to feed a family on thirteen cents a day.
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# Intersection of two planes in R4
1. Dec 20, 2012
### Gauss M.D.
1. The problem statement, all variables and given/known data
I have two planes in R4, namely {[2, 0, 0, 1], [1, 1, 2, 0]} and {[-2, 0, 0, 1], [0, 1, -1, 0]}.
2. Relevant equations
3. The attempt at a solution
Tried to row eliminate, didn't work. Tried figuring out a normal equation, but clearly that won't work in R4. Don't know what to do really.
2. Dec 20, 2012
### HallsofIvy
Staff Emeritus
Two vectors do not define a plane if R4. I suspect you mean the subspaces that are spanned by the two vectors, planes that include the origin. It looks to me like the only point of intersection is the origin.
Last edited: Dec 20, 2012
3. Dec 20, 2012
### Staff: Mentor
What does your notation mean? All points a*(2, 0, 0, 1)+b*(1, 1, 2, 0), for example?
You can set this equal to a corresponding expression for the second plane, and find all points which can satisfy both conditions at the same time (there are not many points in the intersection...)
4. Dec 20, 2012
### Gauss M.D.
"What does your notation mean? All points a*(2, 0, 0, 1)+b*(1, 1, 2, 0), for example?
You can set this equal to a corresponding expression for the second plane, and find all points which can satisfy both conditions at the same time (there are not many points in the intersection...)"
Yes, my notation meant the span of the vectors in brackets.
Your method is what I would do with two planes in R3. But in R3, you have normal equations for planes. In R4, you don't. So how do you set the corresponding plane equations equal to each other?
5. Dec 20, 2012
### Staff: Mentor
You don't need normal equations for those planes. But if you like them, you can use them in R^4, too - you get two equations per plane, and the intersection has to satisfy them all.
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# carpentry 1- basic math concepts-part-B
Home > Preview
The flashcards below were created by user heeres on FreezingBlue Flashcards.
1. fractions get___ when multiplied.
smaller
2. multiply a fraction by a whole number. 2 converts to?
2/1
3. when multiplying fractions reduce equation to simplest terms before
solving
4. the reciprocal of 1/2 is?
2/1
5. when dividing fractions you have to change ÷to× and?
invert the last fraction
6. convert 3/8 to a decimal?
3÷8=0.375
7. convert 0.25 to a common fraction?
25/100 = 5/20 = 1/4
8. to convert mm to inches divide by?
25.4
9. to convert inches to mm multiply by?
25.4
10. to convert sq m to sq ft multiply by?
10.764
11. to convert sq ft to sq m divide by?
10.764
12. 3-4-5 rule is an example of?
pythagorean theorem
13. find the hypotenuse?
c²=a²+b²
14. find the height?
a²=c²-b²
15. find the base?
b²=c²-a²
16. pythagorean theorom equations only work on what type of triangle?
right angle, 90°
17. always remember to use √ when____equation
completing
18. 2.54 cm =
1 inch
### Card Set Information
Author: heeres ID: 265839 Filename: carpentry 1- basic math concepts-part-B Updated: 2014-03-10 22:17:12 Tags: math Folders: carpentry math Description: carpentry 1- basic math concepts-part-B Show Answers:
Home > Flashcards > Print Preview
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Fun & Easy Science for Kids
# All About Bridges: Importance and Types
Remember the story of the Three Billy Goats Gruff? They needed to cross a river to get to the green grass on the other side. Of course, the mean troll played a big part in the story, but what about the bridge? If there hadn’t been a bridge, there would have been no story. And, those poor little goats would have gone hungry.
A beam bridge is made by building two piers or towers. Boards are placed over the piers to make a bridge. Keep reading all about bridges below.
Bridges have always been important because they allow people to travel over canyons, rivers and streams. There are over 500,000 bridges in the United States. Some bridges are small, while other bridges span 7,000 feet or more. Bridges seem like a simple invention, but they’re actually quite complex. Engineers who build bridges must think about the distance to be covered, as well as the type of traffic that will cross the bridge. A bridge for cars and trains must be stronger than a simple bridge for foot travel.
A truss bridge is stronger than a beam bridge because the trusses help distribute some of the weight. The trusses run horizontally between each beam.
## Fun Facts about Bridges for Kids
• Have you ever built a bridge? Stretching a log over a stream is a simple type of bridge. Dropping stones in a creek to pass is also a type of bridge.
• You might have seen beam bridges over small rivers and creeks. A beam bridge is made by building two piers or towers. Boards are placed over the piers to make a bridge. The boards press down on the piers, so these types of bridges aren’t very strong. They are usually pretty short.
• What would happen if you reinforced a beam bridge with metal poles pieced together in a pattern of squares or triangles? You’d have a truss bridge. A truss bridge is stronger than a beam bridge because the trusses help distribute some of the weight. The trusses run horizontally between each beam.
• Arch bridges are very strong. Ancient Romans made arch bridges out of stone. These bridges were made of columns joined by arches.
• Suspension bridges like the San Francisco Bay Bridge use a completely different approach. Tall towers are built. Then long metal cables attached to the towers suspend the road in place. Piers and trusses are usually built underneath for additional support. The Chinese built the first suspension bridges in 300 A.D. They used twisted bamboo fibers instead of steel cables.
Suspension and Arch bridges are very strong type of bridges.
### Bridge Vocabulary
1. Reinforce: strengthen
2. Complex: complicated
Suspension bridges like the San Francisco Bay Bridge use a completely different approach.
Watch beautiful bridges in this video:
A video all about bridges around the world.
### Bridge Q&A
Question: How safe are bridges?
Answer: Bridges are very safe as long as they are properly constructed and maintained. Some bridges can even withstand sever earthquakes and other natural disasters, but not all can. Some bridges might be ancient and not as strong as more modern ones, should disaster strike.
You may cut-and-paste the below MLA and APA citation examples:
### MLA Style Citation
Declan, Tobin. " Facts About Bridges for Kids ." Easy Science for Kids, Jun 2018. Web. 19 Jun 2018. < http://easyscienceforkids.com/all-about-bridges/ >.
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## 10 to the power of 14,205
What is 10 to the power of 14205 (10^14205)? The answer is a Quadrilliquattuortrigintaseptingentillion. This is based on the Conway-Wechsler system for naming numbers.
## 10 to the power of 14,205 written out
10^14205 is written out as:
1,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,00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| 3
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CC-MAIN-2020-24
|
latest
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en
| 0.60477
|
https://www.turtlediary.com/quiz/area-of-squares-rectangles-word-problems.html?app=1?topicname=beg.htm.html?topicname=beginner?topicname=beg.html?topicname=beginner
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text/html
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crawl-data/CC-MAIN-2024-18/segments/1712296817382.50/warc/CC-MAIN-20240419074959-20240419104959-00483.warc.gz
| 936,787,063
| 32,343
|
# Area of Squares and Rectangles: Word Problems
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https://www.turtlediary.com/quiz/area-of-squares-rectangles-word-problems.html?app=1?topicname=beg.htm.html?topicname=beginner?topicname=beg.html?topicname=beginner
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The area is the amount of surface enclosed by a closed figure.
Area of a rectangle = length × width
Area of a square = side × side
For example, consider this situation.
The area of a rectangular pool is 500 square feet. The width of the pool is 20 feet.
Let's find the length of the pool.
We know,
Length × width = area Length × 20 = 500 Length = 500 ÷ 20 = 25
So, the length of the pool is 25 feet.
ds
A B C D E F G H I J K L M N O P Q R S T U V W X Y Z
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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}
| 3.84375
| 4
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CC-MAIN-2024-18
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latest
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en
| 0.870137
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https://istopdeath.com/find-the-slope-and-y-intercept-1-5-12-5-0-8/
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text/html
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crawl-data/CC-MAIN-2022-49/segments/1669446710218.49/warc/CC-MAIN-20221127073607-20221127103607-00208.warc.gz
| 364,095,585
| 16,282
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# Find the Slope and Y-Intercept (-1.5,-12.5) , (0,-8)
,
Find the value of the slope.
Slope is equal to the change in over the change in , or rise over run.
The change in is equal to the difference in x-coordinates (also called run), and the change in is equal to the difference in y-coordinates (also called rise).
Substitute in the values of and into the equation to find the slope.
Simplify.
Simplify the numerator.
Multiply by .
Simplify the denominator.
Multiply by .
Divide by .
Find the value of the y-intercept.
Substitute the value of into the slope-intercept form of the equation, .
Substitute the value of into the slope-intercept form of the equation, .
Substitute the value of into the slope-intercept form of the equation, .
Rewrite the equation as .
Multiply by .
Move all terms not containing to the right side of the equation.
Add to both sides of the equation.
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# Petrov classification
The space-time of general relativity theory consists of a connected Hausdorff manifold of dimension , called the space-time, which admits a metric tensor of Lorentz signature . A tensor of particular importance from both the physical and mathematical viewpoint is the Weyl tensor . This tensor has the property that if the metric is replaced by a metric which is conformally related to it (so that for a nowhere-zero real-valued function on ), then it is unchanged. If is zero on , then is locally conformally related to a flat metric on . The tensor is given in components in terms of the curvature components , the Ricci tensor components , and the Ricci scalar (and with square brackets denoting the usual skew-symmetrization of indices) by
(a1)
In the late 1940s and early 1950s, A.Z. Petrov developed some elegant algebraic techniques which led to the classification of the Weyl tensor and which now bears his name [a1]. Essentially, one takes the Weyl tensor with components and, making use of the symmetries of this tensor, writes them in "block index" form where each capital index can take a value and represents a pair of skew indices according to the scheme:
represents ,
represents ,
represents ,
represents ,
represents ,
represents . Thus, becomes , etc. Then one converts the resulting -matrix to a complex -matrix which, in a well-defined sense, is equivalent to it and which, by the symmetries of the Weyl tensor, is symmetric and trace-free. One then proceeds to classify the Weyl tensor by the Jordan form of this matrix which, from the trace-free condition, must either be (in Segre's notation; see Segre classification) , , , , , or the zero matrix at a point of (the type being dependent on the point!). The trace-free condition shows that in the last three types all eigenvalues are zero. These possibilities are the Petrov types and they are conventionally labelled I, D, II, N, III, and 0, respectively. If a space-time has the same Petrov type at each point it is said to be of that type. A space-time of type 0 is conformally flat. For a vacuum space-time (one for which the Ricci tensor is zero on ) the Riemann tensor and the Weyl tensor are equal and so the Petrov classification applies to the Riemann tensor. Although it is more convenient to "go complex" to achieve these results, it is not necessary. One can work entirely in the field of real numbers, but the loss of algebraic closure means that one must deal with the case of complex eigenvalues as well as the (real) Jordan forms.
In the 1960s much further work was done on the Petrov classification, see [a2], [a3], [a4], [a5]. Associated with each Petrov type at at point in are "principal" null directions (some possibly coincident). These directions are instrumental in the physical interpretation of the Petrov type and there are four distinct ones associated with type I, two with type D (both repeated), three with type II (one repeated), two with type III (one triply repeated), and one with type N (quadruply repeated). These directions have a rather elegant relationship with the Weyl tensor and were explored in detail by L. Bel and R.K. Sachs. R. Penrose was able to give a simpler approach to them using spinors instead of tensors [a4].
One of the early uses of the Petrov classification was in the finding of exact solutions of the Einstein equations [a6]. The idea was that one looked for space-times of a particular Petrov type (together with other restrictions) to simplify the situation. The Petrov classification also fits elegantly into those powerful calculatory techniques usually referred to as the Newman–Penrose formalism [a7].
#### References
[a1] A.Z. Petrov, "Einstein spaces" , Pergamon (1969) [a2] L. Bel, Cah. de Phys. , 16 (1962) pp. 59 [a3] R.K. Sachs, "Gravitational waves in general relativity VI: The outgoing radiation condition" Proc. R. Soc. London , A264 (1961) pp. 309 [a4] R. Penrose, "A spinor approach to general relativity" Ann. Phys. , 10 (1960) pp. 171–201 [a5] J. Géhéniau, "Une classification des espaces einsteiniens" C.R. Acad. Sci. Paris , 244 (1957) pp. 723 [a6] D. Kramer, H. Stephani, M.A.H. MacCallum, E. Herlt., "Exact solutions of Einstein's field equations" , Cambridge Univ. Press (1980) [a7] E.T. Newman, R. Penrose, "An approach to gravitational radiation by a method of spin coefficients" J. Math. Phys. , 3 (1962) pp. 566
How to Cite This Entry:
Petrov classification. G.S. Hall (originator), Encyclopedia of Mathematics. URL: http://www.encyclopediaofmath.org/index.php?title=Petrov_classification&oldid=15004
This text originally appeared in Encyclopedia of Mathematics - ISBN 1402006098
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» » » Given that $$\alpha$$ and $$\beta$$ are the roots of the equation $$3x^2 - 5x - 7 = 0$$, find the roots of $$\alpha^2 + \beta^2 - \alpha \beta$$....
# Given that $$\alpha$$ and $$\beta$$ are the roots of the equation $$3x^2 - 5x - 7 = 0$$, find the roots of $$\alpha^2 + \beta^2 - \alpha \beta$$....
### Question
Given that $$\alpha$$ and $$\beta$$ are the roots of the equation $$3x^2 - 5x - 7 = 0$$, find the roots of $$\alpha^2 + \beta^2 - \alpha \beta$$.
### Options
A)
$$-\frac{36}{9}$$
B)
$$\frac{4}{9}$$
C)
$$\frac{46}{9}$$
D)
$$\frac{88}{9}$$
## Discussion (4)
• The answer to the question. Hope it's understood
• Anthony
3x$$a^{b}2$$ - 5x - 7 = 0
a = 3, b = -5, c = -7
$$\alpha^{2}+ \beta^{2} - \alpha\beta Since (\alpha + \beta)^{2} - 2\alpha\beta is the same as \alpha^{2} + \beta^{2} Then ( \alpha + \beta )^{2} - 2\alpha\beta - \alpha\beta ( \alpha + \beta )^{2} - 3\alpha\beta \alpha + \beta = \cfrac{-b}{a}, \cfrac{5}{3} \alpha\beta = \cfrac{c}{a}, \cfrac{-7}{3} \therefore (\cfrac{5}{3})^{2} - (3 \times -\cfrac{7}{3}) \cfrac{25}{9} + 7 \frac{25 + 63}{9} = \cfrac{88}{9}$$
1. Please can you solve it and snap it... I do not understand it like this
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# Computational convergence
G
#### Guest
I'm trying to solve a nonlinear matrix system in excel. I have a rigged
method for implementing the Newton Raphson method, but it takes time because
it has to compute N gradient vectors, invert the NxN matrix of gradients,
compute X_n+1, calculate, reset initial values, and repeat. Any thoughts on
an easy (and fast) way to do this in excel?
Ad
H
#### Harlan Grove
bangecon said:
I'm trying to solve a nonlinear matrix system in excel. I have a rigged
method for implementing the Newton Raphson method, but it takes time
because it has to compute N gradient vectors, invert the NxN matrix of
gradients, compute X_n+1, calculate, reset initial values, and repeat.
Any thoughts on an easy (and fast) way to do this in excel?
Why aren't you using Solver? It includes both Newton and Conjugate methods
for multiple variable problems, and it's implemented in binary rather than
interpretted Excel formulas, so it should be much faster (and likely also
better able to maintain precision).
More to the point, what you're doing is likely ill-suited to Excel (or any
other spreadsheet). Handling large, nonlinear systems is better done in true
math software like MatLab. If you don't want to spend money, there's also
SciLab and Octave.
G
#### Guest
As nearly as i can tell from Solver help, this is designed for scalar
targets. I'm looking to make an N-vector converge. I would much rather do
it in a program for matrix algebra, e.g. "R", which is freeware, but there
are others working with the same project that need to be able to "see" how it
is converging.
J
Ad
H
#### Harlan Grove
bangecon said:
As nearly as i can tell from Solver help, this is designed for scalar
targets. I'm looking to make an N-vector converge. I would much rather
do it in a program for matrix algebra, e.g. "R", which is freeware, but
there are others working with the same project that need to be able to
"see" how it is converging.
....
Depending on what sort of convergence you want to achieve, you could use a
scalar statistic based on your N objective functions. So if you want
(f(x,y,z),g(x,y,z),h(x,y,z))
to converge to (f0,g0,h0), you could try minimizing the scalar statistic
SUM((f(x,y,z)-f0)^2,(g(x,y,z)-g0)^2,(h(x,y,z)-h0)^2)
Difficult to be more precise without any details about what you're trying to
do.
## Ask a Question
Want to reply to this thread or ask your own question?
You'll need to choose a username for the site, which only take a couple of moments. After that, you can post your question and our members will help you out.
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EENS 1110 Physical Geology Tulane University Prof. Stephen A. Nelson Mass Movements
Mass movements (also called mass-wasting) is the down-slope movement of Regolith (loose uncemented mixture of soil and rock particles that covers the Earth's surface) by the force of gravity without the aid of a transporting medium such as water, ice, or wind. Still, as we shall see, water plays a key role. Mass movements are part of a continuum of erosional processes between weathering and stream transport. Mass movement causes regolith and rock to move down-slope where sooner or later the loose particles will be picked up by another transporting agent and eventually moved to a site of deposition such as an ocean basin or lake bed. Mass movement processes are occurring continuously on all slopes; some act very slowly, others occur very suddenly, often with disastrous results. In this discussion, we hope to answer the following questions: What determines whether or not a slope is stable or unstable? How are the different mass movement processes classified? How are mass movement events triggered? What observations might suggest that the area is unstable and may start to move? How can we mitigate against mass movement hazards? We start with a discussion of the forces acting at the surface that cause mass movements. Gravity Gravity is the main force responsible for mass movements.
Gravity is a force that acts everywhere on the Earth's surface, pulling everything in a direction toward the center of the Earth. On a flat surface, parallel to the Earth's surface, the force of gravity acts downward. So long as the material remains on the flat surface it will not move under the force of gravity. Of course if the material forming the flat surface becomes weak or fails, then the unsupported support mass will move downward.
On a slope, the force of gravity can be resolved into two components: a component acting perpendicular to the slope, and a component acting parallel to the slope.
The perpendicular component of gravity, gp, helps to hold the object in place on the slope. The component of gravity acting parallel to the slope, gs, causes a shear stress parallel to the slope and helps to move the object in the down-slope direction. On a steeper slope, the shear stress component of gravity, gs, increases, and the perpendicular component of gravity, gp, decreases. Another force resisting movement down the slope is grouped under the term shear strength and includes frictional resistance and cohesion among the particles that make up the object. When the sheer stress becomes greater than the combination of forces holding the object on the slope, the object will move down-slope. Thus, down-slope movement is favored by steeper slope angles (increasing the shear stress) and anything that reduces the shear strength (such as lowering the cohesion among the particles or lowering the frictional resistance. For unconsolidated material, the angle that forms a stable slope is called the angle of repose.
The Role of Water Although water is not always directly involved as the transporting medium in mass movement processes, it does play an important role. Addition of water from rainfall or snow melt adds weight to the slope. Water can seep into the soil or rock and replace the air in the pore space or fractures. Since water is heavier than air, this increases the weight of the soil. If the material becomes saturated with water, vibrations could cause liquifaction to occur, just like often happens during earthquakes. Water can reduce the friction along a sliding surface. Water has the ability to change the angle of repose (the slope angle which is the stable angle for the slope). Think about building a sand castle on the beach. If the sand is totally dry, it is impossible to build a pile of sand with a steep face like a castle wall. If the sand is somewhat wet, however, one can build a vertical wall. If the sand is too wet, then it flows like a fluid and cannot remain in position as a wall.
Dry unconsolidated grains will form a pile with a slope angle determined by the angle of repose. The angle of repose is the steepest angle at which a pile of unconsolidated grains remains stable, and is controlled by the frictional contact between the grains. In general, for dry materials the angle of repose increases with increasing grain size, but usually lies between about 30 and 45 o. Coarser grained and angular particles have a steeper angle of repose than fine grained and rounded particles.
Slightly wet unconsolidated materials exhibit a very high angle of repose because surface tension between the water and the grains tends to hold the grains in place.
When the material becomes saturated with water, the angle of repose is reduced to very small values and the material tends to flow like a fluid. This is because the water gets between the grains and eliminates grain to grain frictional contact.
Troublesome Earth Materials Expansive and Hydrocompacting Soils - These are soils that contain a high proportion of a type of clay mineral called smectites or montmorillinites. Such clay minerals expand when they become wet as water enters the crystal structure and increases the volume of the mineral. When such clays dry out, the loss of water causes the volume to decrease and the clays to shrink or compact (This process is referred to as hydrocompaction).
Another material that shows similar swelling and compaction as a result of addition or removal of water is peat. Peat is organic-rich material accumulated in the bottoms of swamps as decaying vegetable matter.
Sensitive Soils - In some soils the clay minerals are arranged in random fashion, with much pore space between the individual grains. This is often referred to as a "house of cards" structure. Often the grains are held in this position by salts (such as gypsum, calcite, or halite) precipitated in the pore space that "glue" the particles together.As water infiltrates into the pore spaces, as discussed above, it can both be absorbed onto the clay minerals, and can dissolve away the salts holding the "house of cards" together. Compaction of the soil or shaking of the soil can thus cause a rapid change in the structure of the material. The clay minerals will then line up with one another and the open space will be reduced. But this may cause a loss in shear strength of the soil and result in slippage down slope or liquefaction. Some water saturated clays are stable so long as they aren’t disturbed, but when shaking occurs, just like sands, they can turn into a runny fluid. These are referred to as quick clays.
Weak Materials and Structures Rocks often contain planar structures that become slippage surfaces if weight is added or support is removed. Bedding Planes - These are basically planar layers of rocks upon which original deposition occurred. Since they are planar and since they may have a dip down-slope, they can form surfaces upon which sliding occurs, particularly if water can enter along the bedding plane to reduce cohesion. In the diagram below, note how the slope above the road on the left is inherently less stable than the slope above the road on the right. Weak Layers - Some rocks are stronger than others. In particular, clay minerals generally tend to have a low shear strength. If a weak rock or soil occurs between stronger rocks or soils, the weak layer will be the most likely place for failure to occur, especially if the layer dips in a down-slope direction as in the illustration above. Similarly, loose unconsolidated sand has no cohesive strength. A layer of such sand then becomes a weak layer in the slope. Joints & Fractures - If the joints are parallel to the slope they may become a sliding surface. Combined with joints running perpendicular to the slope (as seen in the sandstone body in the illustration above), the joint pattern results in fractures along which blocks can become loosened to slide down-slope. Foliation Planes - Foliation in metamorphic rocks is mainly caused by the alignment of sheet silicate minerals. Because the sheet silicates can break easily parallel to their sheet structure, the foliation or schistosity may become a slip surface, particularly if it it dips in the down-slope direction.
Mass Movement Processes The down-slope movement of material, whether it be bedrock, regolith, or a mixture of these, is commonly referred to as a landslide. All of these processes generally grade into one another, so classification of mass movement processes is somewhat difficult. We will use a common classification of mass movements, which divides the processes into two broad categories and further subdivides these categories. Slope Failures - a sudden failure of the slope resulting in transport of debris down hill by sliding, rolling, falling, or slumping. Sediment Flows - material flows down hill mixed with water or air.
Slope Failures Slumps - types of slides wherein downward rotation of rock or regolith occurs along a curved surface. The upper surface of each slump block remains relatively undisturbed, as do the individual blocks. Slumps leave arcuate scars or depressions on the hill slope. Heavy rains or earthquakes usually trigger slumps.
Rock Falls and Debris Falls - Rock falls occur when a piece of rock on a steep slope becomes dislodged and falls down the slope. Debris falls are similar, except they involve a mixture of soil, regolith, and rocks. A rock fall may be a single rock, or a mass of rocks, and the falling rocks can dislodge other rocks as they collide with the cliff. At the base of most cliffs is an accumulation of fallen material termed talus. The slope of the talus is controlled by the angle of repose for the size of the material. Since talus results from the accumulation of large rocks or masses of debris the angle of repose is usually greater than it would be for sand.
Rock Slides and Debris Slides - Rock slides and debris slides result when rocks or debris slide down a pre-existing surface, such as a bedding plane or joint surface. Piles of talus are common at the base of a rock slide or debris slide.
Sediment Flows Sediment flows occur when sufficient force is applied to rocks and regolith that they begin to flow down slope. A sediment flow is a mixture of rock, regolith with some water. They can be broken into two types depending on the amount of water present. Slurry Flows- are sediment flows that contain between about 20 and 40% water. As the water content increases above about 40% slurry flows grade into streams. Granular Flows - are sediment flows that contain between 20 and 0% water. Note that granular flows are possible with little or no water. Fluid-like behavior is given these flows by mixing with air. Each of these classes of sediment flows can be further subdivided on the basis of the velocity at which flowage occurs.
Slurry Flows (high amounts of water) Solifluction - flowage at rates measured on the order of centimeters per year of regolith containing water. Solifluction produces distinctive lobes on hill slopes (see figure 16.2d) in your text). These occur in areas where the soil remains frozen and is then is thawed for a short time to become saturated with water . Debris Flows- these occur at higher velocities than solifluction, and often result from heavy rains causing saturation of the soil and regolith with water. They sometimes start with slumps and then flow down hill forming lobes with an irregular surface consisting of ridges and furrows. Mudflows- a highly fluid, high velocity mixture of sediment and water that has a consistency of wet concrete. These usually result from heavy rains in areas where there is an abundance of unconsolidated sediment that can be picked up by streams. Thus, after a heavy rain streams can turn into mudflows as they pick up more and more loose sediment. Mudflows can travel for long distances over gently sloping stream beds. Because of their high velocity and long distance of travel they are potentially very dangerous. Mudflows on volcanoes are called lahars.
Granular Flows (low amounts of water) Creep- the very slow, usually continuous movement of regolith down slope. Creep occurs on almost all slopes, but the rates vary. Evidence for creep is often seen in bent trees, offsets in roads and fences, and inclined utility poles (see figure 16.2c in your text). Earthflows - are usually associated with heavy rains and move at velocities between several cm/yr and 100s of m/day. They usually remain active for long periods of time. They generally tend to be narrow tongue-like features that begin at a scarp or small cliff Grain Flows - usually form in relatively dry material, such as a sand dune, on a steep slope. A small disturbance sends the dry unconsolidated grains moving rapidly down slope. Debris Avalanches - These are very high velocity flows of large volume mixtures of rock and regolith that result from complete collapse of a mountainous slope. They move down slope and then can travel for considerable distances along relatively gentle slopes. They are often triggered by earthquakes and volcanic eruptions. Snow avalanches are similar, but usually involve only snow.
Mass-Movements in Cold Climates Mass movements in cold climates is governed by the fact that water is frozen as ice during long periods of the year. Ice, although it is solid, does have the ability to flow, and freezing and thawing cycles can also contribute to movement. Rock Glaciers - a lobe of ice-cemented rock debris (mostly rocks with ice between the blocks) that slowly moves downhill (see figure 16.2e in your text). Subaqueous Mass movement Mass movements also occur on slopes in the ocean basins. Most slope failure can occur due to over-accumulation of sediment on slope or in a submarine canyon, or could occur as a result of a shock like an earthquake. 3 types can occur - (1) Submarine slumps – Coherent blocks break and slip, similar to slumps on land. (2) Submarine debris flows – Moving material breaks apart and flows, similar to debris flows on land. (3) Sediment moves as a turbulent cloud, called a turbidity current. (See figure 16.10 in your text). Gigantic submarine slope failures are widespread on the ocean floor, particularly around islands like Hawaii and off the east and gulf coasts of North America. They are much larger than land-based slope failures and are an important process sculpting adjacent land. When they occur, they create catastrophic tsunamis. (See figures 16.11 in your text).
Triggering Events A mass movement can occur any time a slope becomes unstable. Sometimes, as in the case of creep or solifluction, the slope is unstable all of the time and the process is continuous. But other times, triggering events can occur that cause a sudden instability to occur. Here we discuss major triggering events, but it should be noted that it if a slope is very close to instability, only a minor event may be necessary to cause a failure and disaster. This may be something as simple as an ant removing the single grain of sand that holds the slope in place. Shocks and vibrations - A sudden shock, such as an earthquake may trigger slope instability. Minor shocks like heavy trucks rambling down the road, trees blowing in the wind, or human made explosions can also trigger mass movement events.
Slope Modification - Modification of a slope either by humans or by natural causes can result in changing the slope angle so that it is no longer at the angle of repose. A mass movement can then restore the slope to its angle of repose. Undercutting - streams eroding their banks or surf action along a coast can undercut a slope making it unstable.
Changes in Hydrologic Characteristics - heavy rains can saturate regolith reducing grain to grain contact and reducing the angle of repose, thus triggering a mass movement. Heavy rains can also saturate rock and increase its weight. Changes in the groundwater system can increase or decrease fluid pressure in rock and also trigger mass movements. Changes in slope strength - Weathering creates weaker material, and thus leads to slope failure. Vegetation holds soil in place and slows the influx of water. Trees put down roots that hold the ground together and strengthen the slope. Removal of tress and vegetation either by humans or by a forest fire, often results in slope failures in the next rainy season.
Volcanic Eruptions - produce shocks like explosions and earthquakes. They can also cause snow to melt or discharges from crater lakes, rapidly releasing large amounts of water that can be mixed with regolith to reduce grain to grain contact and result in debris flows, mudflows, and landslides.
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# Calculating Wattage on resistors
Status
Not open for further replies.
#### McGuinn
##### New Member
Could someone verify my maths in this one...
I have an air-cooled resistor which lies in series with a 55W lamp, and it's used to dip the lamp on a car when driving during the day. The car runs 12V DC.
It's resistance is 1.7 ohms, and the coil had burned out due to dampness and such.
My plan is to replace the wire wound unit with a standard high wattage resistor of a similar resistance.
My maths goes as follows:
The resistance of the 55W lamp is:
E^2/W
144/55 = 2.61818 ohms.
The combined resistance of the two components is 2.61818 + 1.7 = 4.3818 ohms.
This will dissapate the following watts in total:
E^2/R = W
144/4.3818 = 32.863 W.
The wattage per component is the ratio of the resistance:
(W/R)*R1 = R1W
(32.863/4.3818)*1.7 = 12.465W for the 1.7 ohm resistor.
(W/R)*R2 = R2W
(32.863/4.3818)*2.61818 = 19.1979W for the 2.61818 ohm light.
So that means I need a minimum wattage of a ~16W resistor.
Is my calculation of the wattage correct or am I making a mistake?
#### niq_ro
##### New Member
in a car E=14V not 12V and resistance of bulb light is a non-linear resistance...
I think a true mode to put a exactly aditional resistor is to make more experiment... live... to determine rezistance bulb light with voltage
#### Attachments
• light.jpg
22.8 KB · Views: 610
#### McGuinn
##### New Member
I'll measure the voltage on the car (I think 13.4 is the accepted voltage when the engine is running) and I'll check the voltage drop across the two parts.
#### Nigel Goodwin
##### Super Moderator
McGuinn said:
I'll measure the voltage on the car (I think 13.4 is the accepted voltage when the engine is running) and I'll check the voltage drop across the two parts.
13.8V is the nominal voltage for a car battery.
Have you considered getting a replacement resistor off of a scrap car in a scrapyard? - probably easier, and cheaper, than getting a suitably large wirewound - and it would plug directly in.
I don't know what sort of car you have, but my 1990 Vauxhall Astra had 'dim dip' headlamps, using a similar sounding resistor. Presumably pretty well any car with this system would use a similar value?.
#### ljcox
##### Well-Known Member
Your mistake is in assuming that the lamp resistance is constant. In fact it is non linear (positive temperature coefficient) as niq_ro pointed out.
So the lamp has minimum resistance when it is cold, thus the current will be much higher initially. So the resistor has to withstand a higher power dissipation initially than when the lamp resistance has stabilised.
Len
#### McGuinn
##### New Member
Well hang on, I didn't measure the lamp resistance, I used it's wattage (55W) to calculate it. As the lamp is not produced by a single vendor, I am trying to compensate for small changes in resistance.
The reason for looking for a resistor based solution is because the cars (Alfa 145) are becoming rare, and while I could get one from a scrapper, I want to see if I can make a viable replacement to what Alfa fitted.
#### Nigel Goodwin
##### Super Moderator
I think we've lost the plot here!.
You already know the value of the resistor - 1.7 ohm.
Where are you going to find a high wattage 1.7 ohm resistor?, except as a specific spare part.
Assuming you can find them, I would suggest you get a 25W one, as long as you mount it on a decent size heatsink. If you're not planning using a heatsink then go for a 50W or 75W one.
#### McGuinn
##### New Member
That's the thing... I can't find a 25W 1.7 ohm resistor, so I need to use a pair of 3.4 ohm ones in parallel. I need to know whether two 10W ones will be suitable....
#### Nigel Goodwin
##### Super Moderator
McGuinn said:
That's the thing... I can't find a 25W 1.7 ohm resistor, so I need to use a pair of 3.4 ohm ones in parallel. I need to know whether two 10W ones will be suitable....
Try them and see! - but I would have thought they will probably run too hot. Assuming it's dissipating something like 16W, two 10W resistors will be running too near their maximum ratings and you probably won't be able to touch them , and they will probably melt insulation off wires.
Does your car use one for each light?, or one for the pair?. If it's one for each light then measure the voltage across the existing resistor when it's turned on
#### john1
##### Active Member
Those old meter shunts with the brass ends were sometimes around that value.
You can adjust their value a small amount by filing them narrower.
#### panic mode
##### Well-Known Member
Does it have to be resistor?
I'd rather put small PWM circuit that would:
- run much cooler than series resistor
- have adjustment so you can set the brightness to whatever you choose.
There are so many very simple circuits and here is something just
to get you interested:
http://www.solorb.com/elect/solarcirc/pwm2/index.html
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Math and Arithmetic
# What is the standard form to 693?
###### Wiki User
Writing a number in standard form simply means to express the number in its 'normal' form. Therefore, the way you wrote the number is the standard form for your example.
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## Related Questions
693 693 / 3 = 231 693 / 7 = 99 693 / 9 = 77 693 / 11 = 63
3 divided by 693 (3 ÷ 693) is 693 divided into 3 which in long division is:........ 0.00432...... ------------693 | 3.00000........ 2 772........ -------........... 2280........... 2079........... ------............. 2010............. 1386............. ------............... 624and so on3 ÷ 693 ≈ 0.0043
In standard form it is 4.86*104. Standard form and scientific form are the same.
The standard form is: 34,729,348,729,432,794,823,947,239,472,394
###### Math and ArithmeticPercentages, Fractions, and Decimal ValuesNumbers
Copyright ยฉ 2021 Multiply Media, LLC. All Rights Reserved. The material on this site can not be reproduced, distributed, transmitted, cached or otherwise used, except with prior written permission of Multiply.
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# Contributed Numbers
### Psychometric function slopes (Burr)
From: David Burr <dave@neuro.in.pi.cnr.it>
What an interesting idea. I'd like to see the list when you've got it.
One obvious number is that the slope of the psychometric function for
almost anything (on log co-ords) is 3. (Weibull function) or a sigma
of about 4 dBs for a cummulative Gaussian. I guess Pelli showed this for
contrast, but it seems to work for orientation, signal-to-noise measures,
vernier or whatever, provided you taker logs, although I don't know if it's
been formally documented.
### Eccentricity (Anderson)
From: cha@shifter.wustl.edu (Charles H. Anderson)
>I am interested in collecting up some experimentally obtained numbers
>concerning human visual performance that people have found useful in
>designing experiments or in thinking about models of human vision.
Brian,
Here are two factors that play a major role in my thinking about human
visual performance. The first deals with how visual acuity changes
with eccentricity, the second with the spatial extent of the window of
visual attention.
The resolution, dE, at the retinal ganglion cell level, primarily Parvo,
changes with eccentricity according to the equation, with all units
in degrees.
dE ~ a(E+E0) ~ 0.01(E+1.3)
These numbers are for the macaque monkey. My guess for human values
is that a~0.006-0.008, and E0 is somewhere between 1.0 and 1.5 for
both monkey and humans. Visual acuity drops by a factor of 2 at
and eccentricity of E0~1.0 degrees. The linear fall off in resolution
reflects the underlying scale invariance of the system.
The anisotropy with angle around the fovea is small except along
the horizontal meridian toward the blind spot, where the resolution
could be a factor of 2 smaller in size, i.e. higher acuity.
As I recall this is quite good out to 40 to 50 degrees.
One has to be careful not to confuse this equation with grating
threshold sensitivity since the Magno system, more specifically the
non-linear subpopulation of the Magno system, can respond to higher
spatial frequencies, especially in the periphery. This response
however is only to the power, all phase information is lost. I have
estimated that information carried in the optic nerve as "dynamic
texture", i.e. the power of high temporal-spatial frequencies,
exceeds that of color. This information is used in manner similar
as color is for both searching for targets as well as aiding in
recognition. It also bypasses the cognitive system altogether and
is used to initiate protective reflexes against in coming objects.
Another interesting numberical factor is the sampling spacing of the
Magno system is about a factor of 3 larger than that of the Parvo
system at all eccentrities. There is some data, which Van Essen
believes, that suggests this rises to about 5 in the center of the
fovea.
Reference:
Van Essen, D.C. and Anderson, C.H., Information processing
strategies and pathways in the primate retina and visual cortex'',
in Introduction to Neural and Electronic Networks ,
eds. S.F. Zornetzer, J.L. Davis, and C. Lau,
Academic Press, Orlando, Florida, 1990.
An updated revised version will be available shortly.
The second factor is that the relative spatial extent of the window of
visual attention spans a diameter of about 30 nodes, where the
absolute spatial size of the nodes depends on the scale at which
covert attention is set. I first settled on this factor using the
observation that grating sensitivity increases with the number of
cycles displayed up to about 10-15 cycles. To first order this is
independent of spatial frequency. Using the Nyquist limit, this means
the number of samples involved in the analysis is about 20-30. This is
much smaller than the number of cycles you can actually perceive. The
equation above would suggest that the number of resolvable points
would be = 2*E/dE ~ 200!
Since then I have found many examples that show there is
something special about a spatial extent of 30 nodes. George
Sperling has shown that sign language can be perceived with
better than 90% accuracy using a display 28 by 20 pixels in size.
Note that this holds irrespective of the distance between the
viewer and the screen. We have summarized some of the supporting
experiments in
Van Essen, D.C., Olshausen, B., Anderson, C.H. and Gallant, J.L.,
Pattern recognition, attention, and
information bottlenecks in the primate visual system.''
In: Proc. SPIE Conf. on Visual Information Processing:
From Neurons to Chips, vol. 1473, 17-28, 1991.
The machine vision project I initiated at the David Sarnoff Labs
back in 1983 was inspired by these observations. The neurobiological
circuitry for processing visual information using a window of
attention that can be translated and zoomed in scale have been
reported in.
Olshausen, B., Anderson, C. and Van Essen D., A
neurobiological model of visual attention and invariant pattern
recognition based on dynamic routing of information'',
J. Neuroscience, {\bf 13},pp. 4700-4719, 1993.
A second more detailed paper has been accepted for publication in
the Journal of Computational Neuroscience.
"A Multiscale Dynamic Routing Circuit for Forming Size- and
Position-Invariant Object Representations"
Bruno A. Olshausen, Charles H. Anderson, and David C. Van Essen
I believe there will also be a paper appearing in Science shortly by
Ed Conner, Van Essen, and Gallant on experiments showing cells in area
V4 encode information in a local coordinate frame that moves with
covert attention.
### Attention (Biederman)
From: ib@rana.usc.edu (Irving Biederman)
Dear Brian:
A number that I have found useful--and an important constraint in
trying to understand models of visual recognition--is 100 msec. It is the
exposure duration, followed by a mask, required to recognize an object or
scene. This value is not noticeablly increased (if at all) if RSVP
presentations are employed rather than single trial presentations. Also,
in RSVP a negative detection task ("Press the key if you see a picture that
is NOT a mode of transportation") is not much more difficult than
positively specified basic level classes (e.g., "Hit the key if you see a
chair").
References:
Single trial:
Biederman, I., Rabinowitz J., Glass, A. L., \& Stacy, E. W., Jr. (1974).
On the information extracted from a glance at a scene. Journal of
Experimental Psychology, 103, 597-600.
RSVP (Good review):
Intraub, H. (1981). Identification and naming of briefly glimpsed visual
scenes. In D. F. Fisher, R. A. Monty, and J. W. Senders (Eds.).
Eyemovements: Cognition and Visual Perception (Pp. 181-190). Hillsdale,
N. J.: Erlbaum.
While we are in the temporal domain, another result that I find
useful is that cells in the anterior reaches of IT show tuned responding
(e.g., if a face cell to a picture of a face) in under 100 msec from the
presentation of the picture. Reference:
Baylis, G. C., Rolls, E. T., and Leonard, C. M. (1987). Functional
subdivisions of the temporal lobe neurocortex. Journal of Neuroscience, 7,
330-342.
Though these cells will continue to fire for about 400 msec, much
of their information (as estimated from a population code of face cells) is
in the first 50 msec. Reference:
Tovee, M. J., \& Rolls, E. T. (1995?) Information encoding in short firing
rate epochs by single neurons in the primate temporal visual cortex.
Visual Cognition, in press (?)
As for a non-temporal constant:
Only two or three simple parts (in their appropriate relation)
rather than the whole object are almost always sufficient for basic level
classification. Ref:
Biederman, I. (1987). Recognition-by-Components: A Theory of Human Image
Understanding. Psychological Review, 94, 115-147.
I liked the two numbers you mentioned. (By the way, is the value
for the number of cones per degree of visual angle only at a .5 deg radius
around fixation, rather than throughout the cone regions of the eye?)
If it would not be too much trouble, I would enjoy seeing your
final collection.
Hope all is going well with you. My best wishes for the New Year,
Irv
### Visual angles (Teller)
From: Davida Teller <dteller@u.washington.edu>
Brian -- I don't know if this qualifies, but here is my favorite "Rule of
Thumb": your thumbnail at arm's length is a bout a degree of visual
angle! Also, the sun and the moon each subtend about 1/2 degree.
In another area, by behavioral testing, an infant's acuity in
cycles/degree is roughly numerically equal to its age in months (5
months, 5 c/d, etc). dt
### Visual angles (Brown)
From: dsbrown@hkpcc.hkp.hk (Brian Brown)
Brian
Undergraduates (and postgraduates) often have trouble estimating angular
subtenses.... the 'Rule of thumb'.. is that the thumbnail at arms length
covers about 1.25 degrees (or at least mine does), and can be used then to
estimate angular sizes and distances in the real world for any object that
can be aligned with the thumb.
And of course 1cm at 57 cm is 1 degree.
Your human lens (60 Diopters) is wrong... that's the total power of the
eye, cornea, lens and effectivity of the various optical components.
Here are a few more:
Corneal radius: about 7.8 mm
Corneal power: about 42D (as you see, leaving only 18D for the lens and
effectivity of the corneal power and lens power)
Corneal diameter: 12 mm
Axial length: about 25 mm
Change in refractive power: about 3D/mm change in axial length
Time taken for a saccadic eye movement in ms:
20 + twice the amplitude in degrees,
so that a 10 deg saccade takes about 40 ms.
Robinson DA. The mechanics of human saccadic eye movement. J Physiol 180,
569-590, 1964.
Handbook of Ophthalmic Optics, published by the
Carl Zeiss Co
7082 Oberkochen
West Germany
### Visual angles (Knoblauch)
From: Ken Knoblauch <Ken.Knoblauch@cismibm.univ-lyon1.fr>
The constant that I run into most frequently lately is 57.3 cm, which
is the distance at which 1 cm subtends 1 deg, but everyone knows that.
### Units (Mulligan)
From: "Jeffrey B. Mulligan" <jbm@vision.arc.nasa.gov>
candelas per m^2 x area of pupil in mm^2 = photopic trolands
approximate relative luminous efficiencies of typical rgb monitor:
R:G:B = 3:6:1
### Cortex and motion (Britten)
From: khbritten@ucdavis.edu (Ken Britten)
10^6 fibers/optic nerve (OK, I don't use it in my work, but it's such a nice
round number)
30 cortical areas for vision
approx 1 degree spatial scale for "short-range" motion, centrally. (Braddick)
V1 RF size = .1 * eccentricity
MT RF size = .8 * eccentricity (Maunsell \& Van Essen)
OK, that comes from the top of my head, maybe more will bubble up later...
### Chromatic aberration (Howarth)
From: Peter A Howarth <P.Howarth@lut.ac.uk>
2 Diopters ....... the approximate amount of longitudinal
chromatic aberration of the human eye over the visible spectrum
The reference I like is Howarth and Bradley, Vision Research 26, 361-366, 1986
but probably a better one is Wald and Griffin 1947, JOSA, 37, 321-336
### Cortex (Shadlen)
From: mike@monkeybiz.stanford.edu (Mike Shadlen)
Brian,
Here are some #s for you -- I suspect you already know these well.
120,000 neurons per mm^3 in monkey striate cortex
or 200,000 under 1 mm^2 of cortical surface.
(O'Kusky and Colonnier, J Comp Neurol 1982, 210:278)
(see also, Peters' chapter in Cerebral cortex, Vol 6, Joes and Peters
eds. 1987, New York, Plenum)
This one's sort of obscure, but I think about it a lot:
150 neurons in a microcolumn of cortex defined by a cylinder the
diameter of a layer 5 pyramidal cell's dendritic tree.
The real numbers are 143 for monkey and 203 for cat.
(Peters and Yilmaz, Cerebral Cortex 1993, 3:49-68)
(Peters and Sethares, J Comp Neurol 1991, 306:1-23)
### Search (Wolfe)
From: wolfe@search.bwh.harvard.edu (Jeremy Wolfe)
Hi Brian
My favorite number these days is 50 msec/item as an
estimate of the rate at which serial attention moves from
item to item in visual search. There are various theoretical
complications surrounding this but it does useful work for me.
Can I get a copy of the list that you compile? What is it
for, anyway?
jeremy
### Cortex (Kersten)
From: Daniel Kersten <kersten@mach.psych.umn.edu>
Hi Brian,
I don't know if this really satisfies the "usefulness" criteria as
well as #cones/deg. You may be familiar with Cherniak's article below
that was stimulated by the discrepancies he noted in the literature
citations regarding how big the human cortex is (area estimates differed by a
factor of 10, I believe...again this discrepancy probably has more
to do with "who really cares", than any inherent difficult in making
the estimates more precise.)
If you plan on publishing any of these, they should probably be
double-checked. I just pulled them off of an old spreadsheet I filled
out when I first read Cherniak's article.
If one assumes 40% of the connections carry visual information (a
visual scientist estimate, linguists probably put it closer to 1%),
one can calculate that there are 1,990 miles
of visual connections--a number neither very useful or very
believable--but may be fun to quote.
------------------------
Cherniak, J. of Cog. Neurosc., 1990, vol 2., pp 58-68
Area of cortex 160000 mm^2
Thickness of cortex 2 mm
Volume of cortex = 320000 mm^3 0.32 liters
Cortex synapse density 4000 synapse/neuron
Cortex connectivity 200000000 synapses/mm^3
connectivity/neuron 5 mm
connection length/mm^3 25000 mm
neuron density in cortex 50000 neurons/mm^3
Total brain volume 1.4 liters
Total neurons in cortex 1.60E+10
### Optics (Adams)
From: aadams@garnet.berkeley.edu (Anthony Adams)
Brian,
I'll give it some thought
Incidently 60D is about the power of the human EYE- not the human lens
(which is only about 9-11D IN the eye.
Tony
optic nerve head is 5x7degrees angle (Vertical=7)
20/200 letter )'the big E") is just under one degree angular subtense (50
min exactly)
### Optics (Levick)
From: William.Levick@anu.edu.au (William Levick)
Brian: You might consider including the following formula, although it is
not strictly based on experimentally measured quantities:
B = dE (approximate, for relatively small E)
where: B (radians)is the angular diameter of out-of-focus blur circle
d (meters) is the diameter of the entrance pupil
E (diopters) is the out-of focus error of the eye.
It is probably well known but I was not aware of a citation when I presented
it in:
Levick, W.R. (1972). Receptive fields of retinal ganglion cells. Pp.
531-566 in: Handbook of Sensory Physiology, Vol VII/2. Physiology of
Photoreceptor Organs, ed. by M.G.F. Fuortes. Springer Verlag: Berlin.
A simplified form, together with other data is given on pp. 538-539.
Good luck with your collection!
Bill Levick
### Visual angles (O’Shea)
From: "Robert P. O'Shea" <R-OSHEA@rivendell.otago.ac.nz>
The visual angle of the first joint of the thumb held
at arm's length is approximately 2 deg.
O'Shea, R. P. (1991). Thumb's rule tested: Visual angle of thumb's
width is about 2 deg. Perception, 20, 415-418.
(By the way, if my memory serves, Helmholtz is the source of the
number: 120 cones per degree of visual angle.
Helmholtz, H. (1873). The recent progress of the theory of vision. In
Popular lectures on scientific subjects (pp. 197-316). London:
Longmans, Green \& Co.)
Cheers,
Robert.
### Retina (Schein)
From: Stan Schein <stan@psych.ucla.edu>
Dear Brian, 1/5/95
A number that is related to 120 cones/deg of visual angle is foveal
retinal magnification, which is about 290 um/deg in human and 210
um/deg in macaque monkey. (I specify foveal, because the distance
from the posterior nodal point to the retina declines with
eccentricity, so retinal magnification does proportionately.
### Stereo, visual sensitivity, many others (Backus)
From: ben@john.Berkeley.EDU (Ben Backus)
Dear Brian,
What a good idea. I hope you plan to make your collection available soon.
Useful vision numbers:
The stereoscopic threshold for a step in depth on a surface is 3 sec of
arc. The threshold for detecting that points are not coplanar is 30 sec.
[Stevenson, S. B., Cormack, L. K., \& Schor, C. M. (1989). Hyperacuity,
superresolution and gap resolution in human stereopsis. Vision Res, 29(11),
1597-605.]
The eyes are 6 cm apart
Best contrast sensitivity is at 3 cycles/deg. [Van Nes, F.L., \& Bouman,
M.A. (1967). Spatial modulation tranfer in the human eye. Journal of the
Optical Society of America, 57(3), 401-406.]
Max spatial resolution is 60 cycles/deg [Campbell \& Green, 1965?]
Visible spectrum runs 400-700 nm
Other useful numbers:
A 1 cm wide object at 57 cm distance subtends 1 degree of visual angle;
the width of one's thumb in cm is its angular subtense in degrees at arm's
length.
Luminance in cd/m^2 of starlight is 10^-3, moonlight is 10^-1, indoor
lighting is 10^2, and sunlight is 10^5. [Hood \& Finkelstein, Ch. 5 in some
book from which I have an unlabeled reprint. Maybe it's: Handbook of
perception and human performance / editors, Kenneth R. Boff, Lloyd Kaufman,
James P. Thomas. New York : Wiley, c1986.]
Useful numbers I don't remember but hope will be in your collection:
The field of view for both eyes together
The binocular region's field of view
Axial length of eyeball
Rod and cone integration times
Easy way to remember how much light a troland is
Wavelength of peak sensitivities of L, M, and S cones
Fastest simple RT for flash detection; Typical V1 neuron's RT to a
flash of light.
Size of a V1 foveal hypercolumn in deg visual angle
Number of different V1 hypercolumns used to tile retinotopic visual space
Hope I've helped.
--Ben
### Time, cones, solid angles (Nilsson)
From: Thomy Nilsson <NILSSON@upei.ca>
Dear Brian,
Here is a less common "fact". The minimum interval at which two brief
(1 ms), small (30') pulses of light can be discriminated from a single
equal energy pulse is about 15- 20 ms and at photpic luminance seems not
to vary with luminance. (Vision Res '79, ARVO '92).
Your data on the number of cones per degree visual angle is not clear.
Do you mean the diameters of 120 cones across a distance on the retina
subtended by 1 degree or an area on the retinae with a diameter equal to
1 degree?
When it come to solid angles, here is what I've found to be a
good approximation for that elusive unite the steradian: Hold your arm
straight out infront of you; bend the elbo 90 degrees; now rotate the bent
portion as much as possible about the axis of the straight portion. The
resulting area that would be swept by a 180 sweep with a center at the mid
point of the bent forearm equal about 1 steradian. Surprizingly
big. Smaller or larger people's portions of arm lengths seem to maintain
this principle.
Would you send me list of the "facts' when it is completed? Thanks,
Thomy
### Numbers (Cormack)
From: CORMACK@PSYVAX.PSY.UTEXAS.EDU (Lawrence K. Cormack)
Brian,
I constantly find myself using:
360/2=PI or approx. 57.3
I believe many of us use a multiple of this as the viewing
distance in our experiments for obvious reasons. I
don't really know a ref., but I suspect it would be Euclid.
-Lar
----------------
Lawrence K. Cormack
University of Texas
ph. (512) 471-1649
fax: (512) 471-5935
### Optics (Moreland)
From: "J.D. Moreland" <coa09@cc.keele.ac.uk>
Dear Brian,
The use of age-matched controls is good practice in many clinical studies
but is not appropriate for diabetes since lens absorbance changes are
accelerated. Ideally, lens absorbance should be matched individually but,
where the required facilities for this are not available, simple formulae
may be utilised to define lens-equivalent age controls.
Thus:
E = A + 2.54T - 3.8 for 20 <= E <= 60
E = 0.30A + 0.76T + 40.9 for 60 < E < 80
where E is the age of a normal lens having the same absorbance spectrum as
that of a diabetic patient of age A and with a diabetes duration T > 1.5.
For T up to 1.5, E = A.
E, A and T are all in years.
See: J D Moreland. "Lens-equivalent age controls for diabetics".
Invest Ophthalmol Vis Sci 1993, 34, 281-282. (Letter to the Editor)
With best wishes,
Jack Moreland.
### Visual angles (Bach)
From: bach@sun1.ruf.uni-freiburg.de (Michael Bach)
I like the simple equation: at 57cm distance (e.g. to your stimulus) one
degree visual angle covers 1 centimeter.
Good idea, greetings, Michael.
Dr. Michael Bach, Dept. of Ophthalmology, University of Freiburg,
D-79106 Freiburg, Germany. ++049(761)270-4060. bach@sun1.ruf.uni-freiburg.de
### Eyes (Ahumada)
From: Al Ahumada <al@vision.arc.nasa.gov>
Model number of eyes per subject: 2
### Visual angles (Tootell)
From: roger@nmr-m.mgh.harvard.edu (Roger Tootell)
Brian:
A conversion factor that I use alot, and which does not seem to be as
widely known as it should be, is:
1 Diopter = 0.57 degrees. Another handy factoid known to most
physiologists is that at 57 cm, one degree is one centimeter wide; at 57
inches one degree is one inch wide, etc. If this collection is for your
book, it sounds like a good idea.
1 diopter refracts light by 0.57 degrees. Useful when calculating the
effect of various lenses and prisms.
Roger
### Binocular vision (Pietsch)
From: reading@ucs.indiana.edu
My favorite one involves the enhancement of visual performance gained by
going binocular. It applies to absolute light detection thresholds,
brightness discrimination, critical flicker frequency, and visual acuity.
The gain in sensitivity is always [more or less] equal to the square root
of 2! You can arrive at this result through sampling theory,probability
summation, or areal summation [an adaptation of Piper's law to binocular
vision]. One of the implications of this approach is that a three eyed
individual should enjoy an enhancement equal to the square root of three!
[as suggested by the late Fergus Campbell}
### Optics (Spillmann)
From: Lothar Spillmann <spillman@ruf.uni-freiburg.de>
Gullstrand's schematic eye puts the refractive power of the
human cornea at 43.05 dptrs and that of the lens at 20.28 (unaccommodated)
and 30.13 dptrs (fully accommodated), respectively. The result of accommodation
is to change the EQUIVALENT power of the eye as a whole from 59.60 dptrs
to 68.22 dptrs, a difference of 8.62 dptrs.
Source: Davson, H. (Ed.): The Eye, vol. 4, page 105 (1962), Academic
Press, New York-London.
Lothar Spillmann, Phone (49-761) 270-9572, Fax (49-761) 270-9500
Arbeitsgruppe Hirnforschung, Institut fuer Biophysik
Hansastrasse 9, D-79104 Freiburg, Germany
### Optics (Enoch)
From: jmenoch@garnet.berkeley.edu (Jay Enoch)
In studies where the Stiles-Crawford effect is of importance, or
in ocular ray tracing, a simple relationship relating displacement
in the entrance pupil of the eye to change in the angle of incidence
at the retina is as follows:
1.0 mm in the entrance pupil of the eye = 2.5 degrees change in the
angle of incidence
I first encountered this derivation (based on the emmetropized
Gullstrand eye) in a monograph produced for the Air Force by
Brian O'Brien and Norma Miller (post 2nd World War). I have
separately derived it an reported it in many papers (including
my book on Vertebrate Photoreceptor Optics). Interestingly,
I pursued the issue using the Pommerantzef Schematic Eye
derived for wide angle fundus photography. This same relationship
holds for the peripheral retina as well for at least 30 degrees
eccentricity (i.e., it may vary slightly, but only by a few tenths
of a degree per mm displacement).
Hope this helps.
Jay M. Enoch jmenoch@garnet.berkeley.edu
### Optics (Mershon)
From: mershon@unity.ncsu.edu (Don Mershon)
I saw your request (for "useful" numbers) which was distributed via
CVNet. I am interested in whether you plan to circulate the results of
your survey (perhaps to the same audience).
As a suggestion for your endevour: it would probably be most helpful if
there were more specific identification of the application of such
values. For example, you mentioned the number of cones per degree ---
should one assume that the number refers to the fovea? Would it perhaps
be a good idea to indicate the limits (e.g., +/- X deg.)? Similarly, you
mention that the human "lens" provides about 60 diopters of optical
power. Actually, the cornea provides about 40 diopters, while the lens
itself provides a variable amount (from 20-32 D), assuming that I am
remembering correctly. And, of course, again, one should specify that
this would only be correct for a young eye.
Good luck with your project,
Don Mershon (N.C. State University)
### Squid’s eyes, the brain, subject contours, many (Cavanagh)
From: "Patrick Cavanagh" <patrick@burrhus.harvard.edu>
Brian,
I share some of your numerophiliac tendancies and will give just a stream of
consciousness recall of the useless numbers which I can recall, some of which
you should verify before using.
Size of giant squid's eye: 19 feet
Percent of all receptors in body which are on the retinae: 70%
Number of moons which will tile the sky: 100,000
Hyperacuity equivalent to: seeing an eye movement at a distance of one mile
Number of neurons in the brain in 1974: 10 billion
Number of neurons in the brain in 1994: 100 billion
Reason mirrors reverse left-right, not up-down: left-right symmetry of body
Earliest subjective contour: 4000 BC
Number of seizures induced by TMS: 2
Number of seizures induced by research fMRI: 0
Duration of Hess positive rod afterimage: 1 minute
Percent of submitted articles which never get published: 0%
Movie image rate: 24 per second
Movie frame rate: 72 per second (each shown 3 times)
Earliest motion model: Al-Haytham, 1024AD
Earliest pop-out model: Al-Haytham, 1024AD
And finally, my question of the day:
Why can lobsters survive out of water for several days? How do they breathe?
Number of watts required to power brain: 10 watts
Percent of total body energy required to run brain: 10%
Percent total input energy from breakfast: 10%
Conclusion: don't forget to eat breakfast
Will computers replace brains: No
Why: brains fit in a shoe box, run on 10 watts, and are easily reproduced with
cheap labor
Patrick
### Units, levels, optics (Burns)
From: sburns@vision.eri.harvard.edu (Steve Burns)
Brian:
Went over your list (thanks) and have some additions.
1. A troland (td) is produced on the retina when the eye is looking at a
surface of 1 cd/m^2 through a pupil of area 1 mm^2. This makes it easy to
compute the td value of a surface or TV monitor if you know the approximate
pupil size.
2. For a standard eye 1 td produces 0.0035 lumens/m^2 on the retina.
3. From experience with a lot of aged individuals, the steady state diameter
of the pupil, even at very high light levels (bleaching), never is less than
2 mm. Transient constriction very occaisionally will make the pupil smaller
than this.
### Yuks (Groh)
From: groh@monkeybiz.stanford.edu (Jennifer M. Groh)
Hi Brian,
Don't forget my favorite "facts":
1. The brain contains 10^12 neurons, 10^14 of which are in the
cerebellum.
Reference: somebody famous.
2. We only use 10% of our brains.
Reference: the popular press.
### More on Psychometric function slopes (Eskew)
From: ESKEW@neu.edu (Rhea Eskew)
Dear Brian and David:
I had intended to send you one or two useful vision 'constants', but I
didn't get around to it. Sorry. I'm motivated by your last message,
however, to argue against the submission made by David Burr, who said
the slope (beta) of (almost any) Weibull psychometric function is 3.
I (and Stromeyer) consistently find lower betas for the detection of
spots. Part of the difference is that neither Charles nor I use the
best procedures for measuring beta (we pool data across sessions, for
instance, which is likely to lower the estimate), but I'm fairly
confident that even if we did not do that we'd consistently get betas
lower than 3.
More important than the absolute magnitude of beta, however, is the fact
that we (and others) have demonstrated that beta varies across spectral
conditions and tasks. I'll mention the three cases I've been most
involved in: (a) chromatic betas are lower than luminance betas for
detecting foveal gaussian blobs (1deg s.d.) (about 1.7 vs 2.4 in my lab
-- not a large difference but a very consistent one). We haven't
published this work yet but some was reported at ARVO last year. (b)
With a 1deg foveal spot, a luminance pedestal approximately halves the
chromatic detection slope (from about 2.2 to about 1.3 --ie, near
ideal). Eskew et al JOSA 1991 vol 8 p 394-403. In that paper we show
that this change in beta isn't simply uncertainly reduction. (c) The
slopes for detection of chromatic and luminance spots increases in the
periphery (with no pedestals). See the 91 JOSA paper, Fig. 3, and
especially Stromeyer et al Vision Research 1992 vol 32, pp 1865-1873,
Fig. 7.
I wish the psychometric slope truly was constant -- life would be
simpler. But I don't believe it.
By the way, one of the constants I should have sent you before is that
red-green chromatic sensitivity is roughly 10X greater than luminance
sensitivity when sensitivity is defined as (1/sqrt[L^2 + M^2 + S^2]),
and L, M, \& S are cone contrasts -- this is for 200 ms flashes. For
flashes shorter than about 45 msec (for our bright adapting conditions),
the ratio drops to about about 3X and should stay that way for all
shorter flashes, assuming Bloch's law (Eskew et al, Vision Res 1994 vol
34, pp 3217-3137). So half a log unit is due to temporal integration,
some is due to spatial integration, and some must be due to other
factors.
Best,
Rhea
### Optics, retina, cortex, more (Wooding)
From: David Wooding <WDRDSW@CARDIFF.AC.UK>
Dr.Wandell,
Thanks for the summary of numbers. A great idea!
I have been a bit busy of late, but was intending to submit
something. Perhaps you would like to include the following, some of
which people suggested should be included. I can expand on
references if you like.
Thanks,
David.
------------------------------------------------------------
THE OPTICAL PATH
----------------
Amount of light lost through scattering in the eye: 3-4%
Lens absorbance peaks at: 365-368nm (near UV)
Said and Weale, 1959; Cooper and Robson, 1969
THE RETINA
----------
Thickness of retina: 350 microns
%age of light incident on photoreceptors which is actually absorbed
by them: 10% at 500nm
Maximum rod density occurs: 10-15 degrees off axis
Osterberg, 1935
Position of cones: central 1-2 degrees of fovea
Number of rods in human eye: 110-125 million
Osterberg, 1935
Number of cones in human eye: 6.3-6.8 million
Dimensions of rod: 30 microns long by 1 micron diam
Dimensions of cone: 50-80 microns long by 5 microns diam
Relative abundance of long and middle wavelength to short cones: 100
Amplification of energy provided by rod: 10e4 - 10e5
Short-wavelength limit of visual sensitivity (human lens): 380nm
Peak absorption of Rhodopsin (rod pigment): 500nm
Extent of macular pigment: up to 8 degrees from fovea
THE CORTEX
----------
approx %age of human cortex occupied by visual system: 60%
approx no.neurons in human visual system: 10e8
V1 surface area devoted to central 10 degs of visual field: 65%
VISUAL FIELDS
-------------
MONOCULAR:
Horizontal dimensions from central fixation: 60 degs nasally,
100 degs temporally
Vertical dimensions from central fixation: 60 degs above,
75 degs below
BINOCULAR:
Horizontal: 200 degrees
Vertical: 135 degrees
Width of area of steroscopic vision: 120 degrees
Height of area of stereoscopic vision: 135 degrees
SENSITIVITY
-----------
Dark adaptation (cones): 10 minutes
Dark adaptation (rods): 40 minutes
COLOUR VISION
-------------
Incidence of anomalous trichromacy in human popn:
1 in 100 (male), 1 in 10000 (female)
Incidence of protanopia and deuteranopia (dichromacy):
1 in 100 (male), 1 in 10000 (female)
Incidence of tritanopia (dichromacy):
1 in 10000
Incidence of rod monochromacy:
1 in 10000
Incidence of cone monochromacy:
1 in 100000
EYE MOVEMENTS
-------------
Slow drifts: velocity 6mins/s, amplitude 0.8-6.0 mins
Microsaccades: duration 10-20ms, amp < 50mins
(elicited by positional error of > 15mins)
Tremors: frequency 30Hz+, amp 20mins
Smooth pursuit latency: 120ms (typical)
Smooth pursuit tracking limit: 30 degs/s
Vergence: velocity 15degs/s, latency 160ms, duration 800ms
Maximum size for natural saccades: approx 15ms
Bahill et al., 1975
### Visual sensitivity (Pelli)
From: "Denis Pelli" <denis-pelli@isr.syr.edu>
Subject:numbers 2:39 AM
1/13/1995
hi brian
great list. Here's my contribution:
The contrast threshold (delta L over L) for a static edge at photopic
luminances is 1%. This predicts the visibility of a wide variety of large
sharp edged objects, including letters.
This is an old result, as John explains:
Robson, J. G. (1993) Contrast sensitivity: One hundred years of clinical
measurement. In R. Shapley and D. Man-Kit Lam (Eds.), Contrast Sensitivity
(pp. 253-266), Cambridge, MA: MIT Press.
regards
denis
### Visual sensitivity (Knoblauch)
From: Ken Knoblauch <Ken.Knoblauch@cismibm.univ-lyon1.fr>
Brian,
Notably missing from your list:
Cut-off frequency for chromatic gratings: ~10-12 c/deg (e.g., Mullen, 1985,
J. Physiol).
Minimum bandwidth for optimal reading: 2 cycles/character
(relatively independent of size) Legge et al., 1985, Vis. Res.
(this article and its companions contain a number of fundamental
constants related to reading text).
etc.
Other readings (Baker)
Sender: mrbaker@vax.ox.ac.uk (Mark Baker)
Dear Brian,
The list of useful numbers seems a great idea! Have you looked in
Wyszecki \& Styles "Color Science."? It is full of gems of that sort
and might make excellent additional source material. I've only got the
reference for the 1967 edition but I think that there is a fatter even
more comprehensive 1988(?) edition too . . .
Incidentally - the Eye-Movement network keeps a number of useful
files on the Mailbase fileserver - members can have them e-mailed to
them by sending the fileserver a message - would you like to have
a copy of your file put there too? I'll put details of joining on the
end of this message.
Keep up the good work!
Regards
Mark
Dr Mark R Baker (MRBAKER@UK.AC.OX.VAX)
### Optics (Anderton)
From: (Phil Anderton) <P.Anderton@unsw.EDU.AU>
Hi,
I just read your list on CVNet, and one useful number which is not
there is an approximate rule-of-thumb for deriving angular magnification by
a "spectacle magnifier". This is a lens of high positive dioptric
power held at the normal spectacle plane. The object of interest placed
close to the anterior focal plane of the lens, and the eye views an
enlarged virtual image of the object. Assuming a normal working distance of
250mm (without the lens), the angular magnification induced by the lens is
about 1/4 the dioptric power. For example, a 16D lens gives a retinal image
about 4 times larger than 'normal'. Beware though, this is very 'rough'.
Philip Anderton.
### Visual sensitivity, optics, sampling (Teller)
From: Davida Teller <dteller@u.washington.edu>
Brian -- and of course there's that wonderful convergence of numbers,
that grating acuity, the cut-off frequency of the optics, and the Nyquist
limit of the foveal photoreceptors are all about 60 c/d. dt
Return-Receipt-To: "M. Angeles Losada Binue" <iodma15@cc.csic.es>
Hi Brian
One of the main lines here has been modeling the eye and measuring the
ocular MTF using a double-pass method. I think you may be interested
in these two papers:
1. Accommodation-dependent model of the human eye with
aspherics. R. Navarro, J. Santamaria \& J. Bescos JOSA A 2, 8 pp1273 (85).
2. Monochromatic MTF of the human eye for different pupil
diameters: an analytical expression. JOSA A 11 pp246 (94). Here you can
find an analytical expression for the foveal MTF as a function of
both pupil diameter and normalized spatial frequency.
MTF(sf,p)=(1-C1+C2*p)*exp[-A1*exp(A2*p)*sf]
+(C1-C2*p)*exp[-B1*exp(B2*p)*sf]
where A1=3.53; A2=0.43; B1=1.69; B2=0.28; C1=0.48; C2=0.037
and 2<p<8 mm; sf=sf0/sflim with sf0<50 cpd and sflim being
the cut-off frequency for wavelegth 632 nm.
I hope this will be useful to you.
Cheers
Angeles
### Units (Nilsson)
From: Thomy Nilsson <NILSSON@upei.ca>
Dear Brian,
Thanks for the list. Thought of 2 more:
retinal illuminance = incident luminance X pupil area in mm sqd X .0016
Note that the above has a correction from Judd's constant of .004. (See
Applied Optics 1983, p.3462)
When dealing with a round stimulus which is the only scource of light, one
can calculate its luminance from a measure of illuminace as
obtained with a simple light meter:
luminance (cd/m sqd) = illuminance (lumen/m sqd) / Pi X (sine A/2)sqd
where A is the visual angle of the stimulus' diameter.
Regards,
Thomy
Color (Varner)
From DVarner@swri.edu Fri Jan 13 11:31:29 1995
Hi Brian--
I happened to read your posting of the important numbers for vision.
Although I don't do color anymore (or vision either for that matter), I'm
still Dottie and Leo's student. So I'd like to add:
Unique blue-- about 475 nm
Unique green-- about 500 nm
Unique yellow-- about 575 nm
With these three numbers and a knowledge of the basic color-opponency
relationships, a person can relate spectral wavelengths to color appearance.
denise
Philosophy, numbers, New York running commentary (Graham)
<nvg@psych.columbia.edu> (Norma Graham)
Jan. 14, 1995
Dear Brian,
(1) Re your numbers: To second the point made by one of your
repondents. Most numbers that are useful to me have a limited range of
applicability, and theis range if frequently difficult to succinctly
express.
However, some idea of a number's magnitude is very frequently a
whole lot better than no idea. And I very much enjoyed your list of
replies. And I hope you can manage some numbers of this sort in your book.
And it is great that your book may be getting done!!
(2) Not mentioned in your email summary of other replies you
received is one I do find useful (not only quantitatively but also it
reminds me of the qualitative relationship). Namely
0.5 db per period is the loss of contrast sensitivity moving away from the
fovea (relatively independent of spatial frequency, at moderate temporal
frequencies, and for photopic luminances, etc. etc. and so forth).
How good is this number? Asking how the loss depends on number of periods
is a whole lot more useful than asking how it depends on distance in visual
angle. Actually, there is a loss of between 0.5 and 1.0 log units contrast
sensitivity (10 to 20 db) in thirty periods in John's and my data (Fig. 4
from Robson and Graham, 1981, reprinted as Fig. 5.6 in Graham, Visual
Pattern Analyzers, 1989). Also see see Fig. 13.5 (section 13.4.3.in the
book) showing that this "number-of-periods" rule of thumb is consistent
with size-scaled models of the retinal inhomogeneity.
(3) A related point, made with regard to Charles Anderson's window
of attention, in particular with regard to his using grating sensitivity as
one support for it: Figuring out the window of attention from the number
of cycles over which grating sensitivity increases is problematic unless
you get the grating someplace where there is sufficient homogeneity. And
if you get someplace where there is uniformity, grating sensitivity
increases out to at least 30 (or 60) than 10 or 15 cycles. And we did not
test higher. Of course, it may only be a factor of 2 (or 4) different
from his estimate, but I am not sure that it wouldn't increase even out to
greater distances (for conditions where the observers have reason to spread
their window of attention as wide as possible). See FIg. 3 from Robson and
Graham, 1981 reprinted as Fig. 5.7 of Graham, Visual Pattern Analyzers,
1989.)
(4) Of course another number which appears here and elsewhere is
that the amount by which sensitivity would be predicted if you double the
number of things involved. A notice one response dealt with it in the
binocular situation. But you might try something more general. Is it dealt
with in the text of the book anyway?
(5) I find that remembering contrast sensitivity at the peak of
the sensitivity vs. spatial or temporal frequency in typical conditions is
something between 100 and 1000 is useful. I usually say 300 for moderate
temporal frequencies and moderate photopic but this really does depend on a
lot. (Again, however, compared to total ignorance, this is a very useful
number!!)
### A challenge (Sivak)
From: "Dr. J. Sivak" <jsivak@sciborg.uwaterloo.ca>
To Brian Wandell,
The giant squid eye cannot be 19 feet in diameter ( see P. Cavanagh).
Even 19 inches sounds big.
### A response (Cavanagh)
From: "Patrick Cavanagh" <patrick@burrhus.harvard.edu>
Hmm, well I can't vouch for the accuracy of the eyeball size estimate but the
source is "How Animals See" from Facts on File -- I forget the author's name.
She has been criticized before for errors but the giant squid is an elusive
character (as opposed to the giant squid axon) and I have no idea how to check
her information.
Patrick
### And a resolution (Cavanagh/Backus)
From: "Patrick Cavanagh" <patrick@burrhus.harvard.edu>
> Dear Professor Cavanagh:
>
> I have "How Animals See" (picked it up for \$4, says the sticker still on
> it). "Squid," in the index, leads to page 9, which, I regret to report,
> gives 370 cm (12 ft), not 19 ft, as the diameter of the giant squid's eye.
>
> Sandra Sinclair, 1985
> How Animals See: Other Visions of Our World
> New York: Facts On File Publications.
What cruel tricks time plays on memory. I had remembered the giant squid eye
from "How Animals See" as being 19 feet, perhaps I did the conversion wrong.
Nonetheless, it is a big eye.
I have been searching for a copy of "How Animals See" for the last 4 years,
bookstores, out-of-print books services, everywhere but no luck. Please buy a
copy for me if you see another one.
> Re: the bilateral symmetry of the body is responsible for the apparent
> left-right, without up-down, reversal in a mirror. I give you this
> creature in return: shaped like an upright wine bottle, the creature is
> capable of spinning itself about the vertical axis by means of a bent arm
> that sticks out from the side of the bottle, with a jet nozzle at its end.
> Male arms are bent one way, and female arms the other, so that males and
> females spin in opposite directions. In the mirror, a female would see not
> a left-right reversal of herself (though such an identification could be
> made), but surely herself as she would look were she male.
>
> Sincerely,
> Ben Backus
> UC Berkeley graduate student (Marty Banks' lab)
Mmm, a real gender bender. But why make her spin? If she stops spinning and
looks at her nozzle organ in the mirror, it will be bent in the male direction
(a left-right reversal to us but an embarrassing male-female reversal to her).
But rather than a wine bottle body which has an obvious top, imagine a perfectly
cylindrical body. Now in the mirror "she" could see herself as a male or "she"
could equally well see herself as an upside down female. Which would you pick?
On the other hand, so to speak, up and down would have to be labeled
before the gender-specific nozzle bend could be established so there
still isn't true up-down symmetry. If sex were designated by, say, the
color of the body, the mirror images would only produce an up-down
reversal and their species would have lost an endless source of good
jokes (eg. No, I don't like your theory but turn over and I'll kiss
you).
Patrick
### Handsome units (Kooijman)
From: "a.c.kooijman" <a.c.kooijman@med.RUG.NL>
Dear Brian,
I have a paper in front of me with some handsome numbers and
relations on light units:
some examples:
*
1 photopic lumen = 1 scotopic lumen for monochromatic light of 555 nm
'lumen' can be replaced by:
lux, candela, nit, troland, apostilb, footlambert (though you have to
avoid this non-metric unit)
* in 'free' viewing to a scene or surface:
retinal illuminance [in troland] =
luminance of the scene [in candela/sq m] * pupil area [in sq mm]
* in a maxwellian view condition:
place a lux meter at 10 cm behind the focus of the maxwellian focus
(thus about in the middle of the brains of the subject)
retinal illuminance [in troland] =
10E-4 * the illumination on the lux meter (full illuminated
sensor)
best wishes
Aart
### Philosophy (Nelson)
From: jnelson@ln.nimh.nih.gov (Jerry Nelson)
17 Jan 95
Dr. Wandell,
What an outpouring you got in response to a request for useful
constants in vision. The wonderful replies included much more
-- people's pet relationships, perhaps their pet ideas.
Is their an unsatisfied need here? Very good people will
complain privately of their inability to pursue questions they
consider fundamental. At meetings some very important
intellectual work is done outside the formal constructs
intended to house them. There is a richness and creativity in
all of us which is ill-housed by the usual professional
structures.
What did you tap? It was clever to ask for only one number.
That kept replies down to lengths under 3 long paragraphs.
You tapped a playfulness. And acceptance was contingent on
how HELPFUL TO OTHERS the contribution might be -- zero
pomposity here.
But without review the contributions were uneven, and without
a ritualized structure (INTRO, METHODS, etc) the format is
chaos -- charming, but a lot of editing work for you and your
book.
Anyway, what fun. After your book is over, fight postpartum
letdown by starting a MODERATED newsgroup -- entries come to
you, and if a single glance can't quickly reveal humor or
helpfulness, then it doesn't go in. Even in mean times,
American science has a mutual helpfulness hard to match
elsewhere.
Thanks again, good luck with the text.
--jerry
### Angles, eye movements, and more (R. Boynton)
From: ps45@sdcc12.UCSD.EDU (robert m boynton)
I was off CV-net for a while for some reason, and missed
the request for magic numbers. I would like to add
a few.
The rule of thumb varies remarkably from one submitter to
another, doesn't it? I really liked Al Ahumada's modal
number of eyes per subject.
Anyway, here are mine:
.3 mm on the retina = 1 degree of visual angle
number of eye movements per second during very intensive visual search: 3 to 5
Nodal point of eye is 7 mm behind corneal vertex
Axial length of the eye is about 1 inch or 25 cm
Index of refraction of eye media (roughly) 4/3
Wavelengths of peak sensitivity: 505 and 555 nm
Maximum density of cones at foveal center: About 130,000,000/mm^2
Time for complete dark adaptation after exposure to a bright
light: 30 to 45 minutes
Cheers,
### Receptors, units, behavior (Ross)
From: jr@psy.uwa.edu.au (John Ross)
Brian: Some more (approximate) numbers, all useful I find.
1. Number of rods per (human) eye 120m
Number of cones per eye 6m
Number of fibres/optic nerve 1m
Ratio of receptors to fibres = 125:1
2. 30'' of arc = minimum separable; closest spacing of cones (approx.)
3. 100ms = critical duration for Bloch's law (can be less); period of
frequency at which flicker and motion are seen best.
4. 3 or 4 = number of cone types.
5. 4 = Jevons number: maximum number of objects countable without
error at a glance.
6. 4 or 5 log units: improvement in sensitivity with dark adaptation.
7. 7 = Miller magic number; minimum number of 'stations' for best
apparent motion.
8. 500 = best contrast sensitivity.
John Ross
### Optics, fun, retina, many (Tyler)
From: cwt@skivs.ski.org (Chris Tyler)
Dear Brian,
Well, I see you received a goodly supply of numbers. Some others that I
have come across are:
The human eye is exactly the same size as a quarter.
The rod-free, capillary-free foveola is 1/2 deg in diameter, same as
the sun, the moon, and the pinky fingernail at arm's length.
One deg is about .3 mm (300 u) on the retina.
Diameter of smallest cone or rod is 1 u, packed at up to 300 per deg.
Diameter of largest human cone is 10 u, as large as those of any amphibian.
There are 10^8 photopigment molecules per rod. Each one can be
photoisomerized in 100 femtosec. They are packed into 800 disks
Density of cones falls with E^-2/3 out to 20 deg eccentricity (not
reciprocally).
There are about 4000 cones in the central (foveolar) half degree.
The same number in a 5 deg diameter patch at 40 deg eccentricity.
The range of visible light is 1 octave at around 1/2 the diameter of the
smallest cone.
Each human striate cortex is 4 x 8 cm, making a total of 64 sq cm.
There are about 6400 hypercolumns therein.
You can think of this area as spanning 8 octants around the visual
field and 8 doublings from 1/2 to 64 degrees eccentricity.
The viewing distance at which an object is projected at veridical size
on the foveal representation is 20 cm (I think; you can work it out
for your own best foveal magnification estimate).
Foveal cone detection threshold occurs at an average of 1 quantum per
cone for extended stimuli.
A comment on Dave Burr's law of Psychometric functions. Mayer and I
found that the slope (beta) varied from 2-6 across observers. It is
well known that it falls to 1 for suprathreshold discrimination tasks.
Optimal contrast threshold is 1/4%.
Visual apparatus needs to be engineered to 1 part per thousand.
The bandwidth of the visual system is 1 teraHerz (e.g., the raster
bandwidth for a panoramic wall panel that looked like a full-field
dynamic view of the real world).
The number of possible brain states that we can experience in a lifetime is
10^10^10 (a googol).
The Ferry-Porter Law. In the absence of flicker adaptation, CFF
increases directly with log I over 6 log units. (25 Hz per decade in
the periphery, 10 Hz per decade in foveola)
The maximum CFF ever reported is 107 Hz from my right eye at 300,000
Td.
The finest Vernier acuity reported was .8 arc sec in Dennis Levi's
right eye
Tyler's Law of disparity scaling: The maximum disparity for depth
resolution scales linearly with stimulus size. Horizontal binocular
fusion follows the same law.
The best stereoacuity reported is 2 arc sec. Contrary to the limited
statements in most textbooks, this is sufficient to discriminate the
distance of 2 miles from infinity (e.g., distant mountains) or resolve
the depth of a human ovum (100 u) at 10 cm.
The vertical horopter is tilted at an average of two degrees, just
enough to ensure that it falls in the plane of the ground for an erect
observer who is fixating anywhere in the ground plane. Pettigrew
showed that the tilt was 20 deg for the cat and ground owl, and zero
for the tree owl, so that the horopter fell in the ground plane for
all four species.
At a pupil diameter of 3.6 cm, the pupil area is 10 cm, so for normal
viewing conditions, Trolands = 10 x cd/m2.
Speaking of units, it should be arc sec, arc min and Td for
consistency with the physics and astronomy communities. All units
from proper names are capitalized (e.g., Hz, V, etc).
0.1 log unit is 27%
The standard error of a Poisson process (such as light) is equal to
the square root of its mean, which is why taking more readings
improves S/N ratio. This fact is unknown to most undergraduates, in
my experience.
The eyes are half-way down the head - a fact known to most artists.
If the sun's light was blocked, it would be eight minutes before the
moon went dark.
I am happy to dig out the references for any of these that interest you, so
let me know if you want them. Christopher
### Adaptation (Thompson)
From: P Thompson <pt2@unix.york.ac.uk>
I like your useful numbers. I might suggest adding that visual
aftereffects reach their maximum 'strength' after an adaptation period of
about 60 seconds.
Peter
### Visual sensitivity (Klein)
From: klein@adage.Berkeley.EDU (Stanley Klein)
Brian, here are some numbers from Dennis Levi and Stan Klein,
The following numbers are give or take a factor of 2 (or so) to keep the
numbers simple:
Detection thresholds:
edge - about 1%
line (a pair of adjacent opposite polarity edges) - about 1 %min
dipole (a pair of adjacent opposite polarity lines)- about 1%min^2
quadrupole (a pair of adjacent opposite polarity dipoles)- about 1%min^3
1 Td is about 1 absorbed photon per cone per integration time (E100 msec).
The best hyperacuity is equivalent to a retinal distance of one tenth of the
wavelength of red light (beat that, Patrick).
One can localize a peripheral point to within one tenth of a hypercolumn.
One hypercolumn is .1 times eccentricity
The best temporal hyperacuity for vernier acuity of moving targets (no
tracking) is about 1 msec.
The Weber fraction is crudely about 10% for almost anything (except for
position (or size) where one must remember to also divide by 2pi). Too bad
the presence of multiple channels makes this number difficult to use.
For a monochromatic display to be able to present perceptually lossless
images without halftoning it must be able to display about 100 bits/min^2.
The perceptual information in a monochromatic image is up to about 20
bits/min^2.
People are quite happy with monochromatic images compressed to about 1
bit/m^2.
One can derive Levick's formula (blur angle=pupil size times diopters of
defocus) from the following geometric construction. 1) Focus at infinity. 2)
Look at something at a finite distance. 3) The linear blur projected onto the
object will be the size of the viewer's pupil (about 3 mm) independent of the
object distance.
Many psychophysical functions vary with eccentricity according to
Th =k(E+E2)
where Th is the threshold, k is the slope, E is the stimulus eccentricity,
and E2 is the eccentricity at which the threshold is twice the foveal value:
For resolution and contrast senssitivity, E2 is about 2-3 degrees.
For position judgements E2 is about .5 -1 degree.
For absolute motion, E2 is about 5-10 degrees.
What's your favorite E2?
We like the position judgement E2 and are starting to call it L2 (the E2
characteristic of the local sign). It is the value that we expect to be similar
to the anatomical cortical magnification factor.
What a great idea to do this.
Stan
### Genetics (Piantanida)
From: Tom Piantanida <pianta@unix.sri.com>
Brian, Here are a few more numbers for your survey.
As a rule of thumb, for image stabilization to occur, spatial noise needs
to be 1 minarc or less, and latency needs to be 1 msec or less.
The genes encoding rhodopsin and cyanopsis have five exons, those
encoding chlorolabe and erythrolabe have six.
The opsin proteins of rhodopsin and cyanolabe consist of 348 amino
acids, those of chlorolabe and erythrolabe consist of 364.
The gene that encodes rhodopsin is on chromosome 3 (3q21- 3qter), the
cyanolabe gene is on chromosome 7 (7q22-7qter), and the genes for
chlorolabe and erythrolabe are on the X chromosome (Xq28).
Panum's fusional area can be as large as about 2 degrees under
stabilized-image conditions.
When the fusional limit is exceeded, diplopia occurs, but fusion can
be reestablished at disparities close to 2/3 of the fusional limit
(not 6minarc as Fender and Julesz reported.)
This is a great idea.
Tom Piantanida
### Units, references, sensitivity (Meese)
From: tim@vis.psg.bham.ac.uk (Tim Meese)
Here are a few numbers for your list, if your still collecting them (they
are mainly obvious, but I find them useful):
1) I prefer to report stimulus contrast in dB (20*log(c)), though my
referees often seem to prefer percent or proportion! The conversion is
easy however: an increase of 6dB represents a doubling of contrast (or
whatever other variable you may have converted to dB). In absolute terms
0dB is a contrast of 1% (if c is in the range 0 to 100%), or 1 (ie. 100%,
if c is in the range 0 to 1).
2) Useful stimulus spacing for detection/discrimination experiments etc:
about 2 or 3 dB
3) An approximation I find useful: To convert from the spread (SD) of a
cummulative Gaussian to the spread (logits) of a logistic curve
(1/(Exp[-L]+1)), multiply the former by 0.6. Graphical analysis shows this
to be a fair approximation, and it has been reported in a foot note by
Taylor and Creelman (1967; JASA, 41, pp782-)
4) There are some more good numbers relating alternative measures of
Gaussian widths in Norma Graham's book (p50).
5) Some undergraduates have problems in visualising large numbers. An easy
way of making 1,000,000 seem small is to imagine (or build) a cube with
sides of 10 cm (ie. something quite small that can easily be held in one
hand). However, this cube is also 1,000,000 cubic mm.
6) Percentage of 2AFC trials in which you guess: 0%
(the feedback beeps must be wrong)
sincerely
Tim Meese
### Binocularity (Pietsch)
From: PO2::"reading@ucs.indiana.edu" 9-JAN-199
My favorite one involves the enhancement of visual performance gained by
going binocular. It applies to absolute light detection thresholds,
brightness discrimination, critical flicker frequency, and visual acuity.
The gain in sensitivity is always [more or less] equal to the square root
of 2! You can arrive at this result through sampling theory,probability
summation, or areal summation [an adaptation of Piper's law to binocular
vision]. One of the implications of this approach is that a three eyed
individual should enjoy an enhancement equal to the square root of three!
[as suggested by the late Fergus Campbell}
***
PIETSCH, P. AND SCHNEIDER, C. W., Two-eyed Versus One-eyed Salamanders: Does
Binocularity Enhance the Optically Evoked Skin Blanching Reactions of
Ambystoma Larvae? PHYSIOL BEHAV 48(2) 357-359, 1990. A wide variety of
visual functions show increases attributable to binocularity, and the question
pursued here was whether a second eye enhances the visually stimulated skin
blanching reaction of the larval salamander. Dermal melanin spots (melansome
aggregations within dermal melanophores), which contract or expand to lighten
or darken the skin reflectance, were measured in eyeless (controls), one-eyed
and two-eyed Ambystoma punctatum larvae after chronic exposure of subjects to
a white background (i. e., stimulus conditions for maximum blanching). The
eyeless subjects showed no blanching (thus remained dark) in white cups, and
they exhibited melanin spots 7 or 8 times the size those of the other two
groups. All one-eyed or two-eyed subjects exhibited blanching reactions;
planometric comparison revealed a significantly larger melanin spot area for
one-eyed than for two-eyed animals; i. e., the binocular condition permitted
greater contraction of the pigment spots than did the monocular condition.
Analytical data compared favorably with independently ascertained pigmentation
indices. The results indicate that a second eye quantitatively elevates the
blanching maximum of a larval salamander.
IF YOU'RE INTERESTED, I'LL SEND YOU A REPRINT OF THE ENTIRE ARTICLE.
PAUL PIETSCH
INDIANA UNIVERSITY, BLOOMINGTON, IN 47405 USA
### Philosophy (Cohn)
From: tecohn@mindseye.berkeley.edu (Theodore E. Cohn)
Who could have guessed that numbers would be so much fun? I've been
thinking a bit about numbers and it occurs to me there are two categories of
same. The first category includes those which represent pschyological (or
anatomical or physiological) constants such as the number of foveal cones
per square mm. The other category includes numbers we use to divide up
continua into easy-to-describe pieces. I heard one of the latter cited the
other day...its the distance beyond which we accept the view that image
disparity doesn't contribute to depth perception. (a ref is C. Schor and M.
Flom "Relative value of stereopsis as a function of viewing distance, Am J
Optom 46:805-809, 1968). With respect to the second category, universal
agreemnt on the value of a given number would neither be achievable nor
expected. In the case of the former, agreement could be reached to within
the irreducible measurement error implicit in operations used to define the
number.
Ted
### Numbers to facts (Ross)
From: jr@psy.uwa.edu.au (John Ross)
Brian:
I would like to add a few more numbers to the beginning of my list:
1 = the number concurrent perceptual interpretations of visual
input (as in rivalry, ambiguous figures, but more generally scene
interpretations)
2 = Ahumada's modal number (of eyes), but also of receptor systems
(rods and cones), major pathways (magno and parvo), and dimensionality of
retinal images.
3 (or 4?) = dimensionality of perceived space.
John Ross
### Sensitivity, eye movement, color (G. Boynton)
From: boynton@blue.stanford.edu (Geoff Boynton)
One quantum is sufficient to activate a rod. (Cornsweet, Hecht et al)
Average saccade size is about 10' of arc. (Ratliff, F., and
Riggs. L.A. (1950) Involuntary motions of the eye during monocular
fixation. J Exptl. Psychol 40 687-701
Pupil size ranges from 3mm under 2.5 log cd/mm^2 to 8mm in darkness
(Wyszeki \& Stiles, Color science. New York: Wiley, 1982)
There are 11 basic color names (Boynton, Robert M. Olson, Conrad
X. Vision Research. 1990 Vol 30(9) 1311-1317.)
### Neural waves (Drosler)
From: <Jan.Drosler@psychologie.uni-regensburg.de>
The velocity of signal propagation in the retina (in radial direction )
is one degree of visual angle per millisecond.
It was measured behaviorally by letting the observer synchronize two
LEDs flashing at identical periods, one of which is fixated in the
center of the fovea. The temporal lead of the peripheral LED necessary
for a judgement of synchrony permits the calculation of the velocity
stated above. It appears to be constant over a range of zero to 55
degrees of visual angle. The standard error is 0.46 ms at all
meridians tested.
Reference: Arch. Psychol. 131, 249 -266, 1979.
### Optics/units (Knoblauch)
From: Ken Knoblauch <Ken.Knoblauch@cismibm.univ-lyon1.fr>
Has anybody suggested that 1.13 mm pupil diameter (artificial
probably) should give you 1 troland = 1 cd/m^2?
### Field of view (Grigsby)
From: Scott Grigsby <SGRIGSBY@FALCON.AAMRL.WPAFB.AF.MIL>
total extent of human field-of-view under optimal conditions: ~200 deg
extent of overlap of the two eye fields: ~120 deg
extent of FOV that does binocular processing: ~40 deg
(see Grigsby and Tsou (1994), Vision Research 34, 2841-2848)
extent of eye movement before a compensating head movement: 20 deg
(Robinson (1979), Human Factors 21, 343-352)
1 microcandle in the plane of the pupil = 1 troland
### Cortex (Shadlen)
From: mike@monkeybiz.Stanford.EDU (Mike Shadlen)
I don't know; but here are some other numbers
in a cubic mm of cortex there are
10^5 neurons
10^9 synapses
2 miles of axons
These and several other fun numbers can be found in Stevens CF, 'What
form should a cortical theory take' in Large-Scale Neuronal Theories
of the Brain, Koch C and Davis JL eds., MIT press, Cambridge, Mass.,
1994, pp.239-255.
### Saccades (Groh)
From: groh@monkeybiz.Stanford.EDU (Jennifer M. Groh)
Peak saccade velocities in humans can reach 700+ deg/sec, though
in my experience tenured faculty are slower by ~250 deg/sec. You
can cite
RHS Carpenter, Movements of the Eyes, Pion, London, 1988
p. 70
Bahill AT, Clark MR, Stark L, 1975, Glissades--eye movements generated
by mismatched components of the saccadic motoneuronal control signal,
Mathematical Biosciences 26:303-318
as a specific reference on the "main sequence", which is the
amplitude-duration-velocity relationship of saccadic eye movements.
Another rule of thumb for you:
duration of saccades > 5 deg = 20-30 ms + 2 ms per degree.
This also comes from Carpenter.
### Rod circuit sensitivity (Robson)
From: From: John RobsonI have a good number for you. It is 1 and is the number of
photoisomerisations per rod required to saturate the retinal rod circuit.
### Number of Colors
From: Mark Fairchild> From Wandell to Fairchild:
>
> How many discriminable colors
> are there, and how did people count them? No important reason for asking,
> but I find people are always asking me about this and I always give them
> a vague answer I don't like.
>
I get asked that question alot also and I also give vague answers. I
guess I usually get away with something like "it depends on the viewing
conditions." I did a little digging and found on page 21 of Hunt's
Measuring Colour 2nd Ed. a line saying that 10 million was one reliable
estimate and he referenced Judd and Wyszecki. I went to Judd &
Wyszecki, Color in Business Science and Industry, 3rd Ed. and found them
saying " since about 10 million surface colors can be distinguished by
the normal human eye ..." on page 388. They however gave no direct
reference. That section was talking about the ISCC-NBS method of color
naming so there might be something in the paper that describes that
(Judd and Kelly, Method of Designating Colors, J. Res. Netl. Bur. Std.,
23 355 (1939).). I'm not going to look that one up. I guess if 10
million is a good enough estimate for Judd, Wyszecki, and Hunt, then
it's good enough for me.
From shev@midway.uchicago.edu Fri Oct 25 08:15:55 1996
Mark and Brian, Thanks for replying to my inquiry. The following FYI.
A conservatively low estimate of the number of discriminable colors,
ignoring luminance, is 6,000. I arrived at this using two different
approaches.
APPROACH #1
1. 10,000,000 colors with luminance included (Hunt, Judd, Wyszecki)
2. 1% Weber fraction from 1 to 1,000,000 trolands (implying ~1620
discriminable luminance steps). This assumption is the conservative
part, leading to a low-end estimate.
So, 10,000,000/1620 = 6,173.
The number is 10,800 if the 1% Weber fraction is applied over 1-10,000 td.
The number is 21,500 if the Weber fraction is 2% over 1-10,000 td.
APPROACH #2
This approach takes into consideration that the 10,000,000 colors are
*surface* colors but ignores overall level of illumination.
1. 10,000,000 surface colors (Hunt, Judd, Wyszecki), ignoring luminance.
2. Assume incremental and decremental contrast are *each* discriminable
with a 1% Weber fraction over a 100:1 contrast range (implying
2x463=926 discriminable contrast steps). Again, this is conservative.
So, 10,000,000/(926) = 10,799.
Another approach is (1) to take discrimination around the spectral locus
(using wavelength discrim data) and along the non-spectral purple line,
and then (2) to count the number of JNDs from each discrminable point on
the locus to EE white. This requires some pretty heavy assumptions in the
model of JNDs and will lead to an inflated estimate, because discriminable
colors at the spectrum locus may not be discriminable when desaturated.
I may try it as an exercise.
..Steve
### Location of Optic Nerve Head (Blindspot)
From wsimpson@io.uwinnipeg.ca Fri Jan 10 13:02:15 1997
Dear Brian,
I just looked at your list, which includes
Retina
5. ...size of optic nerve head (optic disc/blind spot)
BUT I always need to look up its location whenever I use a stimulus
outside the fovea. Maybe you could add the location? 15 deg nasal.
Best wishes,
Bill
### Visual angle of a pencil line
From: "Robert A. (Bob) Morris"
I just worked this out for a course I'm teaching on art and vision: A
line drawn by a number two pencil subtends about 1/5 degree when
viewed at 18", and so falls approximately at the peak of adult human
grating acuity.
I wonder if crayon line widths fall at the peak of 3-year old acuity?
Bob Morris
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https://www.wps.com/academy/how-to-split-first-and-last-name-in-excel-using-formula-quick-tutorials-1863889/?chan=web_help_center
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How to split first and last name in excel using formula
July 22, 2022
1.5K Views
0
A free Office suite fully compatible with Microsoft Office
A free Office suite fully compatible with Microsoft Office
WPS Excel has become a significant tool in the world of business. It permits its users to acquire various programming skills. It serves as a convenient platform in accessing analytical and computing features. Excel holds a prime importance in organizing and sorting data, managing finances and graphing. Excel plays a key role in compiling data which results in compacting of information at one place.
Although Excel is quite handy but it contains some technicalities. With proper skills and knowledge of different operations of Excel these technicalities can be handled easily . One of the main problem that is faced by a number of excel users is splitting of first and last name of full name. In this article we have penned down some easy and simple ways to split name. Following are some techniques that would prove to be fruitful in solving the mentioned problem.
How to spilt first and last name in excel using formula
Using formula for splitting full name is an appropriate and simple method. The following formula separates the full name in first and second name. The given steps will guide you to separate name by using formula
1. So the first step is to extract the first name. For this purpose we will use the formula =LEFT(A2,SEARCH( ,A2)-1)
2. Select the full name then enter the formula given above in second cell i.e. column B.
3. In column B we will get the first name.
1. Now for the second name we will use the formula =RIGHT(A2,LEN(A2)-SEARCH( ,A2,1))
2. Now we will again select the full name and will enter the formula in third cell i.e. column C.
3. In column C we will get the second name.
How to separate first and last name by comma
In splitting first and last name with comma we are going to imply the same steps mentioned previously. The only difference is a slight change in the formula that is the usage -2 instead of -1 as we have to remove both comma and space. For getting first name the formula will be =LEFT(A2,SEARCH( ,A2)-2)
For extracting second name, we will use the formula=RIGHT(A2,LEN(A2)-SEARCH( ,A2,1))
How to separate middle name from full name
There are some names which consists of a middle name. To separate the middle name we will use a different formula that is =MID(A2,SEARCH( ,A2,1)+1,SEARCH( ,A2,SEARCH( ,A2,1)+1)-SEARCH( ,A2,1))
In this way you will be able to separate middle name from the full name.
How to split name by using flash fill
There is another technique of splitting name that is the usage of flash fill method.
1. In this method, first we will click data button then we will click flash fill which will activate the features of flash fill.
2. Then we will create a column in front of the column containing the original names.
3. After that we will type the part of the name that we want to extract.
4. Then we will type that part of name in the second cell. Automatically, the names will populate in all the other cells.
Did you get the concept of splitting full name in excel? To access more features of Word Document, PowerPoint slides and excel Spreadsheets, you can follow WPS Academy.
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## The problem of local minima
If the search area is so large that the quadrant monotonic model does not apply to the entire search area then the algorithm might converge on a local minimum. A local minimum is a candidate block location where the neighbouring blocks have greater distortion but the candidate block is not the best of the entire search area. Thus the block is only the best within its own locality and a better block is available elsewhere in the search area.
Mountain climber using the quadrant monotonic assumption to reach the peak. On the left is a contour map where the quadrant monotonic assumption holds over the entire search area with the corresponding profile on the right. The climber succeeds in finding the peak because the assumption holds.
The quadrant monotonic assumption and the problem of local minima are analogous to a mountain climber making the assumption that from an arbitrary starting position he will eventually arrive at the highest point as long as he continues to climb up (Illustrated above). This is of course true if there is only one peak. If there is more than one peak, however, the climber might arrive at a point where moving in any direction results in a drop in altitude and mistakenly conclude that he has reached the highest peak in the range. (Illustrated below).
In this illustration the mountain climber again uses the quadrant monotonic assumption to reach the peak. On the left is the contour map showing that the quadrant monotonic assumption does not hold over the entire search area. The climber fails because the quadrant monotonic assumption does not hold.
Local minima are common in images and the local minima are typically not as good as the global minimum. This is a serious problem for sub-optimal block matching algorithms.
[Sub-optimal BMAs part 1]
[Sub-optimal BMAs part 2]
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https://byjus.com/questions/calculate-the-compound-interest-on-rs-200-for-1-5-years-at-10-per-annum-compounded-semiannually/
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# Calculate the compound interest on Rs.200 for 1.5 years at 10% per annum compounded semiannually.
Compound Interest = Principal(1+Rate/100)^Time -Principal
or
CI = P [(1+r)^t – 1]
Here Rate = 10/2 = 5. t = 3, coz it is compounded semiannually.
Substituting the values in the above equation, we can get the answer as
Rs 31.25/-
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http://math.stackexchange.com/questions/46577/why-is-the-complex-locus-of-mathrmargz-theta-a-half-line-and-not-a?answertab=votes
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# Why is the **complex locus** of $\mathrm{arg}(z)=\theta$ a half line, and not a line?
I know that the locus of $\mathrm{arg}(z)=\theta$ is a half line with angle $\theta$, but I'm not sure why?
I can start the proof: $$z=x+iy$$ $$\theta=\mathrm{arg}(z)=\arctan\left(\frac{y}{x}\right)$$ $$\tan(\theta)=\frac{y}{x}$$ $$y=x\cdot \tan(\theta)$$ Which tells me that the locus is a line with gradient $\tan(\theta)$ passing through $(0,0)$, but I know that it should be a half line with gradient $\tan(\theta)$ starting at $(0,0)$.
Why is this?
-
$\arg(z)$ is not simply $\arctan(y/x)$, because that would make $\arg(1) = \arg(-1) = 0$. Rather, $\arg(z)$ is the real number $\theta$ such that $z = \lvert z \rvert \cos \theta + i \lvert z \rvert \sin \theta$. – Rahul Jun 20 '11 at 22:19
I see. I only had the specific case where x > 0. – david4dev Jun 20 '11 at 23:02
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0
# What is 9 divided by four fifths in simplest form?
Wiki User
2017-02-04 11:56:27
11¼. Division by a fraction is mathematically equivalent to multiplication by its reciprocal.
9 divided by four fifths in simplest form is 45 over 4
Wiki User
2017-02-04 11:56:27
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# Voltmeter Are Used?
## How does a voltmeter work?
Voltmeter works on the principle of Ohm’s law, which states that the voltage across a resistance is directly proportional to the current passing through it. In order to implement it in real time, we form the construction of a galvanometer, such that a coil is suspended in a magnetic field.
## What is called voltmeter?
A voltmeter, also known as a voltage meter, is an instrument used for measuring the potential difference, or voltage, between two points in an electrical or electronic circuit. Some voltmeters are intended for use in direct current (DC) circuits; others are designed for alternating current (AC) circuits.
## What is voltmeter class 8?
A voltmeter is a device (instrument) used for measuring electric potential difference between two points in a circuit. For this purpose, it is put in parallel with that branch of circuit, at the ends of which the potential difference is to be measured.
## Where is voltmeter placed?
Voltmeters are always placed in parallel with the circuit element being analyzed, and positions 3 and 4 are equivalent because they are connected with wires (and potential is always the same anywhere in an ideal wire).
## How many types of voltmeter are there?
There are two kinds of voltmeters. One kind has a needle, or “pointer”, that points to a number that tells the number of volts.
## Do voltmeters need batteries?
Multimeters indicate the presence of, and measure the quantity of, electrical properties such as voltage, current, and resistance. If your multimeter is digital, it will require a small battery to operate. If it is analog, it does not need a battery to measure voltage. Some digital multimeters are autoranging.
## What is voltmeter with diagram?
A voltmeter is an instrument used for measuring electric potential difference between two points in an electric circuit. Analog voltmeters move a pointer across a scale in proportion to the voltage measured; digital voltmeters give a numerical display of voltage by use of an analog-to-digital converter.
## How do you read a voltmeter?
Hence, the voltmeter reading would be V=IR = 0.
## What is range of voltmeter?
The range of practical voltmeter ranges between 1000 to 3000 volts.
## Why do we convert galvanometer into voltmeter?
A galvanometer is a device which is used to detect small electric current flowing in the circuit and Voltmeter is an instrument used to measure the potential difference across the two ends of a circuit element.
## What is difference between voltage and current?
Voltage is the difference in charge between two points. Current is the rate at which charge is flowing.
## What is a DC shunt?
Shunt Series Accuenergy’s line of DC current shunts are specially built to measure electrical DC current systems. The principle use of shunts is to measure electrical currents based on the small voltage drop created across a high precision resistor placed in series with the load.
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OpenKattis
Kattis Set 09
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## Kattis Set 09
#### End
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The end is near!
Contest is over.
Not yet started.
Contest is starting in -306 days 11:47:15
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# Problem GBus Planning
Picture by AutoPhoto via Wikimedia Commons, cc by
An elementary school class is going on a road trip by bus to a computer factory. However, the driver is really tired of having to drive fighting kids all day, so she suggested that the class should split up and go to the factory in groups. She has already observed which kids dislike each other and are likely to fight, so she wants to make sure they do not end up in the same group. Of course, driving everyone one-by-one is a waste of time and money, so the driver wants to minimize the number of groups she has to drive. In addition, the bus is pretty small, and may only fit up to $c$ kids at a time.
You are to write a program which helps her with the task of making these groups. Given the number of kids and their enemies, find the minimum number of groups required, as well as a division of the class into this minimum number of groups
## Input
The first line contains three integers $n$, $k$, and $c$ ($1 \le n \le 17$, $0 \le k \le \frac{n(n-1)}{2}$ and $1 \le c \le n$) – the number of kids, pairs of enemies, and the capacity of the bus, respectively. Then follow $n$ lines with the kids’ names. Each name consists solely of the characters A-Z and a-z, is non-empty, and at most $10$ characters long. Then follow $k$ lines, each containing a pair of space-separated names indicating a pair of kids that dislike each other. No pair of names appears twice, and no kid is their own enemy.
## Output
On the first line, output the minimum number of groups, followed by one line per group containing the names of the children in that group (separated by spaces).
Sample Input 1 Sample Output 1
2 0 1
Alice
Bob
2
Alice
Bob
Sample Input 2 Sample Output 2
3 2 3
Alice
Charlie
Bob
Alice Charlie
Bob Charlie
2
Alice Bob
Charlie
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# Elaboration of equipotential surfaces of planets using biorthogonal expansions
2016;
: pp. 39– 43
Authors:
1
Lviv Polytechnic National University
2
Lviv polytechnic National University
3
Department of Cartography and Geospatial Modeling of Lviv Polytechnic National University
Purpose. Using known and fixed Earth potential, presented asthe biorthogonal expansion, to culculate the geoid surface, which describes the actual shape of the planet. The external gravitational field is generally described by the series of spherical functions. Since the geoid is determined with the help of such functions, a question arises converning the identity to define the shape, moreover its several points does not belong to the region of convergence. Methodology and results. We consider representation of potential by convergent series everywhere, which makes it possible to find the geoid without specifying the location of points on the surface, although the geoid heights calculation is carried out by various relations. According to the known function of the mass distribution of the Earth, represented by the second degree polynomial, internal and external potential of elliptical planet are defined and the equipotential surfaces are found. Calculated values via these formulas and their degree of coincidence was analyzed. Defined in two ways surfaces do not coincide with each other because the difference in the values of the radius-vector amouts up to ten meters. So, when applying biorthogonal expansions of higher orders in constructing equipotential surfaces based on information about the external gravitational field it is necessary to take into account characteristics of expansion. Originality. Method of determining the shape of the Earth using the biorthogonal expansions of mass distribution function is proposed. This representation is characterized by a convergence for considered series and gives the opportunity to build digital models of the geoid (volumetric or as an isolines map). Practical significance. The results of numerical experiments, described in the article, led to the conclusion about the possibility of determining the equipotential surfaces that adequately describe the physical surface of the planet not only of the second but higher orders using biorthogonal expansions only with additional investigations. Calculation of geoid heights with high accuracy opens the way to observe many regional and local geodynamic phenomena, such as the movement of tectonic plates, and high accuracy leveling using GPS technology can solve a number of geodetic problems.
1. Antonov V. A, Timoshkova Ye.I., Kholshevnikov K.V. Vvedenie v teoriu Njutonovskogo potenciala. (Glav.red. Phiz.-mat. Lit.), 1988, 272 p.
2. Balmino D. Predstavlenie potenciala Zemli s pomoschiu sovokupnosti tochechnyh mass, nahodiashchihsia vnutri Zemli. Ispolzovanie iskustviennyh suputnikov dlia geodezii. Moscow: Mir, 1975, pp. 175–183.
3. Zagrebin D. V. Vvedenie v teoreticheskuiu gravimetriu. Leningr. Otd. Izd. Nauka, 1976, 292 p.
4. Marchenko A. N. Approksimacia globalnogo, regio-nalnogo I lokalnogo gravitacionnogo polia Zemli sistemoi potencialov necentralnyh multipolei. Tr. II Orlovskoi konferencii "izuchenie Zemli kak planet metodami astronomii, geodezii I geophiziki". Kyiv: Naukova dumka, 1982, pp. 56–59.
5. Marchenko A. N. Hilbertovy prostranstva funkcii, harmonicheskih vne sphery Bjerhamera, globalnaia funkcia anomalnogo polia. Kyiv: Nauka, 1983, 22 p. (Rukopis dep. v UkrNIINTI, № 292 Ук-Д83).
6. Meshcheriakov G. A. Zadachi teorii potenciala i obobshchenaia Zemlia. Мoscow: Nauka, 1991, 216 p. (Glav.red. Phiz.-mat. Lit.).
7. Moritc G. Sovremennaia phizicheskaia geodezia. Мoscow: Nedra, 1983, 392 p.
8. Pellinen L. P. Vysshaia geodezia(teoreticheskaia geodezia . Мoscow: Nedra, 1978, 264 p.
9. Fys M. M., Fotca R. S., Sohor A. R., Volos V. O. Metod znahodjennia gustyny rozpodilu mas planet z urahuvanniam stoksovyh stalyh do chetvertogo stepenia. Geodynamics. Lviv, Lviv Polytechnic Publishing House, 2008, vol. 1(7), pp. 25–34.
10. Fys M. M., Gubar Yu. P., Pokotylo I. Ya. Netradyciinyi metod pobudovy poverhon' rivnia planety (geoida, selenoida, aeroida) za ii zovnishnim gravitaciinym polem. Zbirnyk naukovyh prats konferencii "Suchasni dosjaghnennja gheodezychnoji nauky ta vyrobnyctva v Ukraini". Lviv, 1997, pp. 39–42.
11. Fys M. M. Pro odyn class neortogonalnyh dla elipsoida harmoniinyh funkcii. Zbirnyk naukovyh prats zahidnogo geodezychnogo tovarystva "Suchasni dosjaghnennja gheodezychnoji nauky ta vyrobnyctva" [Modern achievements in geodetic science and industry]. Lviv, 2006, vol. І (11, pp. 126–130.
12. Fys M. M. O shodimosti v sredniem biortogonalnyh riadov vnutri elipsoida. Differencialnye uravnienia i ih prilogenia. Lviv: "Vyshcha shkola", 1983, Vol. 172, 2 p.
13. Cherniaha P. G., Fys M .M., Golubinka Yu. I., Yurkiv M. I. Porivniannia odnogo klasu harmonichnyh ta kuliovyh funkcii pry predstavlenni potencialu planet. Zbirnyk naukovyh prats zahidnogo geodezychnogo tovarystva "Suchasni dosjaghnennja gheodezych¬noji nauky ta vyrobnyctva" [Modern achievements in geodetic science and industry]. Lviv, 2014, vol. II (28), pp. 19–23.
14. Konopliv A., Banerdt W., Sjogren W. Venus gravity: 180 th degree and order model. Icarus, 1999, no. 139, pp. 3–18.
15. Konopliv A., Asmar S., Yuang D. Resent gravity models as a result of the Lunar Prospector mission. Icarus, 2001, no. 150, pp. 1–18.
16. Kraup T. A contribution to the mathematical foundation of physical geodesity. Danish Geodetic Institute, 1969, Copenhagen, vol. 44.
17. Lerch F. I. Model improvement using GEOS-3 (GEM 9 and 10). J. Geophys. Res. 1979, no. 138, pp. 3897–3916.
18. Marchenko A. N. Parameterization of the Earth's Gravity Field: Point and Line Singularities. Published by Lviv Astronomical and Geodetic Society. Lviv, Ukraine, 1998, 210 p.
19. Moritz G. Fundamental geodetic constant. Proceedings of the IAG XVII Gener. Assemb. IUGG/IAG, 1979, p. 24.
20. Pavlis N. K, Holmes S. A., Kenyon S. C. An Earth Gravitational Model to degree 2160: EGM2008. EGU General Assembly. Geophysical Reaseach Abstracts. 2008, no. 10, p. 2.
21. Yung D. N., Sjogren W., Konopliv A. S. Gravity field of Mars: 75 degree and order model. Geophys. Res. 2001, no. 10, pp. 23377–2340.
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# Set Theory - The in operator
The in relationship indicates that an object belongs to a set (equivalently, that the set contains the object).
## Notation
$$\Large \in$$
## Example
$$\Large 1 \in \{1,2,3,4\}$$
Discover More
Set Theory - Cross Product (Cartesian product)
The cartesian product is a set operator between two set A and B that produces a set of all pairs (a, b) where: (ie a in A) and (ie b in B) See See
Set Theory - Operator (Operations)
operator in Set Theory that applies to set structure (set, bag) A data operator produce another set. All relational algebra operator are set-operator. A relationship operator produce...
Set Theory - Relationship
The relationships between set structure is: given by a mathematical function and is described by the set operators. You can visualize them with venn diagram.
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# If a and b are positive numbers, is a < b?
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If a and b are positive numbers, is a < b? [#permalink]
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Updated on: 17 Jun 2017, 02:10
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If a and b are positive numbers, is a < b?
(1) a < b/2 + 2
(2) a < b/2 -2
_________________
we shall fight on the beaches,
we shall fight on the landing grounds,
we shall fight in the fields and in the streets,
we shall fight in the hills;
we shall never surrender!
Originally posted by bkpolymers1617 on 17 Jun 2017, 01:45.
Last edited by Bunuel on 17 Jun 2017, 02:10, edited 1 time in total.
Renamed the topic and edited the question.
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Re: If a and b are positive numbers, is a < b? [#permalink]
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17 Jun 2017, 02:45
Given a and b are positive numbers, we need to find out if a < b?
(1) a < b/2 + 2
Equation can be re-written as a-2 < b/2
For values a = b = 10 statement a<b is false as a=b
Similarly, for values a = 3,b=3.2 statement a<b is true
Hence, not sufficient.
(2) a < b/2 - 2
Equations can be rewritten as a+2 < b/2
If a=b=any value, the equation a+2 < b/2 will not stand.
If a>b for any positive number, the equation a+2 < b/2 will not stand.
Only when a<b, this statement stands.
Hence this statement alone is sufficient.(Option B)
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Re: If a and b are positive numbers, is a < b? [#permalink]
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17 Jun 2017, 03:09
If a and b are positive numbers, is a < b?
(1) a < b/2 + 2. If b = 2 and a = 2, then answer is NO but if b = 4 and a = 1, then the answer is YES. Not sufficient.
(2) a < b/2 - 2. Notice that b/2 is less than b, so b/2 - 2 is even less. So, even if b/2 - 2 is greater than a, then so must be b alone. Sufficient.
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Re: If a and b are positive numbers, is a < b? [#permalink]
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17 Jun 2017, 03:18
S1 -> if b = 2 then a < 3 so it can be 2 but if b = 4, a < 6 so it can be 5 and can be 1. Insufficient
S2 -> b = 2 then a < -1, if b = 4 a < 0. Sufficient.
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Re: If a and b are positive numbers, is a < b? [#permalink]
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17 Jun 2017, 07:38
Top Contributor
bkpolymers1617 wrote:
If a and b are positive numbers, is a < b?
(1) a < b/2 + 2
(2) a < b/2 - 2
Target question: Is a < b?
Given: a and b are positive numbers
Statement 1: a < b/2 + 2
This statement doesn't FEEL sufficient, so I'll TEST some values.
There are several values of a and b that satisfy statement 1. Here are two:
Case a: a = 0 and b = 2, in which case a < b
Case b: a = 2.5 and b = 2, in which case a > b
Since we cannot answer the target question with certainty, statement 1 is NOT SUFFICIENT
Aside: For more on this idea of plugging in values when a statement doesn't feel sufficient, read my article: http://www.gmatprepnow.com/articles/dat ... lug-values
Statement 2: a < b/2 - 2
Since b is POSITIVE, we know that b/2 < b
We can also say that b/2 - 2 < b/2
We can COMBINE these inequalities to get: a < b/2 - 2 < b/2 < b
If we ignore the middle parts, we see that a < b
Since we can answer the target question with certainty, statement 2 is SUFFICIENT
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Re: If a and b are positive numbers, is a < b? [#permalink]
### Show Tags
15 Oct 2018, 06:17
Bunuel wrote:
If a and b are positive numbers, is a < b?
(1) a < b/2 + 2. If b = 2 and a = 2, then answer is NO but if b = 4 and a = 1, then the answer is YES. Not sufficient.
(2) a < b/2 - 2. Notice that b/2 is less than b, so b/2 - 2 is even less. So, even if b/2 - 2 is greater than a, then so must be b alone. Sufficient.
Dear brunel
Any tips in deciding the numbers to test the inequalities? i am taking too much in agreeing on the numbers to test the DS questions. thanks in advance
Re: If a and b are positive numbers, is a < b? [#permalink] 15 Oct 2018, 06:17
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# Heat Of Fusion
Jason Jarrell
Chemistry II
12-8-99
Heat of Fusion
Objective: The objective of this experiment is to find the heat of fusion of water by using a calorimeter. The calorimeter will be used to melt ice in water to find the heat of fusion.
Theory: Heat of fusion is known to be the amount of heat that it takes to allow one mole of a substance to turn from solid to liquid. The heat of fusion of water is known to be 80 cal/mol. This experiment will use a calorimeter with distilled water and ice to find an experimental value of heat of fusion of water. Equations used in this experiment will be LFM1 = M2CT where LF is the heat of fusion, M1 is the mass of the substance being melted (the ice), M 2 is the mass of the substance that is doing the melting (the water), C is the specific heat of the substance (water is one calorie per gram per degree), and T is the change in temperature.
Procedure: The following materials will be needed for this experiment, a cup, a thermometer, distilled water, ice, and a scale. The first step is to find the mass of the cup without anything in it. After that fill it about half full with distilled water, and find the mass of that. After the mass of the cup and water is measured, put about five or six normal blocks of ice in to the cup. Use the thermometer to stir the mixture. Every 15 seconds record the temperature of the mixture. When the temperature has leveled off, use a fork to pick out any blocks of ice that are left.
Data & Calculations:
Mass of cup: 3.39 g
Mass of cup and water: 169.17 g
Mass of cup, water and ice: 208.11 g
Mass of water: 165.78g
Mass of ice: 38.94 g
Change in temperature: 18.9 C
(38.94 g)X = (165.78 g)(1 cal/gram C)(18.9 C)
38.94 g X = 3133.242 cal
X = 80.46 cal/g
80.46 – 80 = .6 % error
80
Conclusion & Discussion: This experiment was good to show how find an experimental value of heat of fusion of a substance. Some of the possible sources of error could have been:
There could have been human error in reading equipment.
There could have been equipment error in measuring.
Some extra water could have come out when digging out the ice.
The ice was tap water and not distilled.
Some water could have been splashed out when stirring.
Some heat could have been lost into the cup.
Some heat could have been lost to the environment.
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Home > Mom's Life > Family Finances > Understanding Finances > Calculating Your Personal Profits and Losses
Calculating Your Personal Profits and Losses
In This Article:
Page 4
How Much House Payment Can You Afford?
As I've said, I'm no financial expert, but here's one thing I do know: Many lenders are happy to give you enough financial rope to hang yourself, especially for a mortgage. If you default on the loan, they'll just take the house back, keep the money you paid, and start fresh with a new borrower. It's up to you to stay within your means and only borrow what you know you can afford, which might not be the entire amount they're willing to give you.
"But my lender seems so knowledgeable," you might say. "He showed me a formula based on my income that proves I can afford the house I want." Some lenders use the 28/36 qualifying ratio to calculate how much to lend you. The numbers are percentages of your gross income: either 28% or 36% of what you earn before taxes or any other deductions. Lenders following this ratio will allow you to commit a maximum of 28% of your gross income for housing expenses or a maximum of 36% of that gross for housing expenses plus your other recurring debt. However, there is variation among lenders in what constitutes housing expenses and recurring debt and also in how strictly this ratio is applied. Furthermore, some lenders use a different calculation altogether. And, just to make it more confusing, a budget counselor is likely to give you a very different opinion: Many will suggest that you calculate what you can afford based on your net – not your gross – income and their allowable percentages will be more conservative.
The critical fact to keep in mind when deciding how much to borrow is this: Just because a lender will give it to you doesn't mean it's a good idea. You – not the lender – are the only one who can decide what's in your best interests.
Using Your Data to Organize a Budget
By now, all of this microscopic examination of your financial data might have inspired you to create a budget. We're getting very close to the point where I, the professional organizer, wish you well and hand you over to your financial advisor. But, in the spirit of organizing your data without judging it, I can offer you one more chart before you go.
When you create and follow a budget, you apply organizing skills where there once were none. Budgeting is a form of project management, which is a form of organizing, meaning it's just another process improvement to be made and has no reflection on your character or your worth as a person.
Next: Page 5 >>
More on: Family Finances
Reproduced from Organize Your Personal Finances in No Time, by Debbie Stanley, by permission of Pearson Education. Copyright © 2005 by Que Publishing. Please visit Amazon to order your own copy.
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# Bar, Ball, Cone
Geometry Level 2
A chocolate shop sells its products in 3 different shapes: a cylindrical bar, a spherical ball, and a cone. These 3 shapes are of the same height and radius, as shown in the picture. Which of these choices would give you the most chocolate?
$\text{ I. A full cylindrical bar } \hspace{.4cm} \text{ or } \hspace{.45cm} \text{ II. A ball plus a cone }$
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# Why does friction cause a car to turn?
I've had a lot of difficulty conceptually understanding the physics of how a car turns on an unbanked curve, so I'm hoping you could help me out. When a car is moving in uniform circular motion, we know that $|\vec{a}| = \frac{v^2}{r}$, and the direction of acceleration is towards the radius of the circle about which the car is moving. Drawing a free body diagram for the car shows that there are only three forces acting on it: gravity $(\vec{F_g})$, the normal force $(\vec{F_n})$, and friction $(\vec{F_f})$. Since gravity and the normal force negate each other, the car isn't accelerating in the $y$ direction. Because it is in uniform circular motion we know it is accelerating in the $x$ direction, and summing up the forces in this direction yields $$\vec{F_{net x}}=m\vec{a}=\vec{F_f}$$ which implies that the centripetal acceleration is due to the frictional force.
What I am having difficulty understanding is why this intuitively makes sense. I've read some other people's answers on this question but I haven't found anything satisfactory. In particular, many people talk about how wheels "are pushing the pavement to the left or right", and this causes the pavement to exert a force on the car wheels by Newton's third law, but this hasn't made sense to me.
Another way of putting this might be that I don't understand why friction should be directed inwards towards the center of the circle about which one is turning. I would expect that, since the wheels have been turned, that friction would be directed in the opposite direction of where the car is moving to prevent the car from continuing to move forward and skidding on the road.
I hope this makes sense, thanks.
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How hard is it to push a car (in neutral, no brakes, level ground) in the direction the wheels are pointed, compared to moving it sideways under the same conditions? – DJohnM Oct 6 '13 at 20:39
I had fun trying to make this as intuitive as possible. I hope I've succeeded without doing the physics of the situation much injustice.
When a car is driving straight ahead, the plane in which the wheels are rotating is aligned with the direction of movement. Another way of saying this is that the rotation axis is perpendicular to the momentum vector $\vec{p}=m\vec{v}$ of the car. So the friction merely makes it harder for the car to move, which is part of the reason why you need to put your foot on the gas pedal to maintain a constant speed. At the same time, the friction is what allows you to maintain that constant speed because the rotating tires sort of grab onto the ground, which is the intuitive picture of friction. The tires grab the ground and pull/push it backwards beneath themselves, as you would do when dragging yourself over the floor (if it had handles to grab onto). Those grabbing and pulling/pushing forces are what keeps you going.
Things change when the wheels are turned. The plane in which they are rotating now is at an angle with the direction of motion. Alternatively but equivalently, we could say the rotation axis now makes an angle with the momentum vector of the car. To see how friction then makes the car turn, think again in terms of the wheels grabbing onto the ground. The fact that they now make an angle with the direction of motion, means the force the tires are exerting is also at an angle with the direction of motion - or equivalently, the momentum vector.
Now, a force is a change in momentum$^1$ and so (because the wheels are part of the rigid body that is a car) this force will change the direction of the car's momentum vector until it is aligned with the exerted force. Imagine dragging yourself forward on a straight line of handles on the floor and then suddenly grabbing hold of a handle slightly to one side instead of the one straight ahead. You'll steer yourself away from the original direction in which you were headed.
$^1$ Mathematically: $$\vec{F}=\frac{d\vec{p}}{dt}$$
-
The key IMO is that the "effective friction" of a wheel that is free to rotate is very different in different directions. Moving inline with the wheel the effective friction is low because the wheel can rotate. Moving perpendicular to the wheel the effective friction is simply the friction between wheel surface and ground which is high.
If the wheel is paralell to the direction of motion then it exerts a small force slowing the car down. If it is perpendicular to the car then it exerts a much larger force slowing the car down.
To see what happens when we turn the wheels digonally lets resolve the cars veolicty into components parlell and perpendicular the the wheel. Bot of these components will point diagonally forwards. That means that their coorespond frictional forces point digonally backwards.
due to the higher effective friction the component of the frictional force perpendicular to the wheel will be much stronger than the one paralell to the wheel. So lets only only consider that one from now on. if the front of the wheel is pointing right then the perpendicular frictional force will point backwards and right.
Resolving that force back to the car we see a force pointing backwards and a force pointing right.
If the car was 4 way symetrical and all wheels turned in the same direction then the car would just start moving diagonally but the wheels on a car don't do that. only the front wheels turn so the front of the car is pushed right causing the car to turn right.
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If we want to keep a body in a circular orbit a central force is needed. If that is intuitive enough then,when you turn the wheel you are effectively changing the direction of the driving force ( if we are travelling at uniform speed this force is still there though it cancels off exactly with total frictional forces ) changing the force direction will not change the velocity direction instantly. if you make only one angle change of the driving force, it will take a finite amount of time to get the vehicle adjusted to the new angle. So it is the Inertia that gives this apparent slipping. Circular orbit is bit different, you keep on changing the direction of the driving force, now since inertia makes the vehicle slip all the time the Frictional force is active. But the maximum frictional force is limited by mu*R , R = mg , If the central force needed for our intended orbit is less than that my*R then vehicle sustains minimum slip, but if it is insufficient, slipping and turning occur successively until we reach a larger orbit with a higher Radius of curvature, that will make the central force lesser ( since it is mv^2/r ) so that it could be sustained by mu* R ( mu is the coefficient of friction ) Alternatively V could be made smaller so that the frictional force could sustain necessary central force. Main thing there are no two forces centrifugal and centripetal, It just makes this whole thing confusing. The thing is that central force that is needed for a particular orbit and a speed. It has to come from some external means.
P.S. Elaborate further on a sudden direction change to a vehicle which is traveling at a uniform speed. As soon as we turn it you can resolve the velocity in two directions. one along the new direction and perpendicular to that direction. Now the perpendicular component will push the vehicle outward, with a friction force in the direction opposite to slip, Because of that 'perpendicular velocity' will decrease. However if this remains the same, the constant speed of the vehicle in the new direction does not increase back to v because the driving force is barely sufficient to keep the velocity constant i.e. No acceleration. Because of the turn kinetic energy is lost. The key here is Inertia.
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## protected by Qmechanic♦Mar 20 '14 at 17:33
Thank you for your interest in this question. Because it has attracted low-quality or spam answers that had to be removed, posting an answer now requires 10 reputation on this site (the association bonus does not count).
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Molecule Simulator Step By Step Instructions:
Standards:
M5P4 Students will make connections among mathematical ideas and to other disciplines.
S5P2 Students will explain the difference between a physical change and a chemical change.
a. Investigate physical changes by separating mixtures and manipulating paper to demonstrate examples of physical change.
b. Recognize that the changes in state of water (water vapor/steam, liquid, ice) are due to temperature differences and are examples of physical changes.
c. Investigate the properties of a substance before, during, and after a chemical reaction to find evidence of change.
Requirements:
1. Has at least 10 molecule sprites
2. Each molecule moves and bounces off edge and each other
3. Each molecule moves faster and slower depending on the energy variable.
Extras:
1. Molecules "stick" closer together when energy is lower.
2. Molecules move farther apart when energy is higher.
3. Label sprites show the states of matter (solid, liquid, gas) at different energy levels.
Finished Sample:
Step By Step Instructions:
Open Scratch and Draw Water Molecule:
1. Open Scratch
2. Delete the Cat
3. Draw a Water Molecule
a. Click on "Draw a New Sprite"
b. Draw Three Circles
c. Use Flood Fill to fill the Circles
d. Label with Text Tool (O, H, H)
4. Label the Water Molecule (You always have to name your sprites)
Add Scripts for Molecule to Move and Bounce:
5. Create the Scripts for the Water Molecule (Test, Run, Fix)
a. Basic Movement Script:
b. Simulate Molecule "Wiggle"
c. Test and Run
d. Bouncing off other Molecules
e. Run and Test
Add Variables for Speed and Size:
6. Create Variables for Speed and Size (What is a Variable?)
a. Click on Variables
b. Click "Make a variable"
c. Type "Temperature" -> Click "OK"
d. Click "Make a variable"
e. Type "Size" -> Click "OK"
7. Double Click on Sliders for Temperature and Speed
8. Put the Variable Blocks into the Scripts
a. Put "Temperature" into the Move 5 steps block
b. Add "set size to" Command
c. Put "Size" variable block into "set size to" command
9. Run and Test
Make More Molecules:
10. Make more molecules!
a. Use the stamp tool and make one more Molecule
b. Run and Test
c. Make lots of Molecules
11. Save Your Work. Click "File -> Save"
(Scratch 1.4 automatically saves to a Folder it creates called "Scratch Projects" in your home directory.)
12. Finished Project:
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## Welcome to the Treehouse Community
Want to collaborate on code errors? Have bugs you need feedback on? Looking for an extra set of eyes on your latest project? Get support with fellow developers, designers, and programmers of all backgrounds and skill levels here with the Treehouse Community! While you're at it, check out some resources Treehouse students have shared here.
### Looking to learn something new?
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# collection and control
Now that we have the while loop set up, it's time to compute the sum! Using the value of counter as an index value, retrieve each value from the array and add it to the value of sum.
For example: sum = sum + newValue. Or you could use the compound addition operator sum += newValue where newValue is the value retrieved from the array.
while.swift
```let numbers = [2,8,1,16,4,3,9]
var sum = 0
var index = 0
while index += numbers {
print([index])
index++
}
```
So what we are trying to do here in task 1 is set up a while loop. The way the english of the task reads helps you understand how. They say "while counter is less than the length of the array" and hint you can use the count method on the array to calculate that.
The second task tells us to add each array item to the sum variable. Again it hints we can use the counter as the index number for the array item.
So then your code would look something like this.
```let numbers = [2,8,1,16,4,3,9]
var sum = 0
var counter = 0
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0
# What is 6 percent as a fraction in simplest form?
Updated: 4/28/2022
Wiki User
12y ago
50/3, or 16 2/3
Lvl 10
2y ago
Wiki User
12y ago
it is 3/50
Earn +20 pts
Q: What is 6 percent as a fraction in simplest form?
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Related questions
### Convert 24 percent in to a fraction in simplest form?
24% as a fraction in its simplest form is 6/25
### What is 600 percent as a fraction in a simplest form?
It is 6, which is an integer, not a fraction.
### 60 percent as a fraction in simplest?
60% as a fraction is 6/10 and the simplest form is 3/5
3/50
3 / 50
6/25
6/25
### What is 6 percent as simplified fraction?
It is: 6% = 0.06 as a decimal and 3/50 as a fraction in its simplest form
89/100
### How do you write 6 percent as fraction?
6/100 or in simplest form, 3/50
6/25
### What is 6 percent as a fraction and acute in simplest form?
6% = 6/100, which can be simplified, if required. The phrase "acute fraction" is not recognised.
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posted by Heather
Write two 5-digit addends whose sum would give an estimate of 60,000
1. Steve
21,849 + 38,932 ≈ 20,000+40,000
2. Aileen
70,000
3. Kristina
30,000+30,000
4. David
bbb
5. Toni
Write two 5- digit addends whose sum would give an estimate of 60,000
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6. ### math
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http://blog.mrwaddell.net/archives/751
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This post is a little personal venting along with some serious math. I am in an Algebra 2 training today, and the instructor is doing some long division for finding slant asymptotes. Okay, nothing wrong with that, but I ask why we aren’t using synthetic division when we just finished using synthetic division in the previous example when the WHOLE room (minus 2 pp) yells it me, “Because it can’t be done.” When I insist it can work, they challenge me to put it on the board, they just don’t believe it. The session ends before I can.
The frustrating thing is, the entire room of math teachers with the exception of 2 others are completely and totally wrong. At least 2 other people had my back.
As evidence, see this paper by
And, just to prove to myself that Pat is correct, (and he is, don’t doubt it) I tried a problem side by side with the long division I know works:
Really, it works. I especially like the fact that synthetic division does not always need a 1 in front of the leading term.
Oops, I just did another synthetic division that is impossible!
I think I need to do four more so that I can be like the White Queen and do six impossible things before breakfast.
And yes, I am getting to the training tomorrow early so these can be on the board when everyone walks in. Don’t tell me something is impossible unless it truly is. Calling something impossible because you don’t want to think or learn about it is kind of like saying the quadratic equation y = x2 – x + 7 has no solutions. It has solutions, they just aren’t nice and easy like you are used to.
—————–
Special thanks to Pat, who introduced me to this in 2009 in his blog.
### 5 Responses to “Never tell me it can’t be done”
1. […] I read Glenn’s post Never Tell Me It Can’t Be Done. Again, we get so fixated on the process we’ve been ‘taught’ or even used to […]
2. Thank you for this post. I can’t say that I’ve ever questioned when you can or can’t use synthetic division so I really appreciate seeing what is possible.
3. Glen, I herd it was impossible for years before I sat down and derived this approach. In math, “it’s impossible!” requires the same standard of proof as ” it’s possible.”
By the way, I have a little on history of terms from stats, z, t, etc on a page at http://pballew.net/etyindex.html and a couple of stats relate blogs you might enjoy. http://pballew.blogspot.com/2011/09/standard-deviation-as-distance.html for example.. just search stats… thanks for an interesting blog
4. Thank you Pat. I will dive into the history of z, t etc today.
5. […] never say never to a kid, they’ll prove you wrong! Math teachers will too – see Never tell me it can’t be done.) They were still struggling – so I was trying to think of a new method. Then the […]
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A207419 T(n,k)=Number of nXk 0..1 arrays avoiding 0 0 1 and 0 1 0 horizontally and 0 1 1 and 1 0 1 vertically 7
2, 4, 4, 6, 16, 6, 9, 36, 36, 9, 14, 81, 92, 81, 14, 21, 196, 241, 241, 196, 21, 31, 441, 720, 742, 720, 441, 31, 46, 961, 1889, 2760, 2760, 1889, 961, 46, 68, 2116, 4719, 8465, 13260, 8465, 4719, 2116, 68, 100, 4624, 12102, 24317, 50139, 50139, 24317, 12102 (list; table; graph; refs; listen; history; text; internal format)
OFFSET 1,1 COMMENTS Table starts ..2....4.....6.....9.....14......21.......31........46........68........100 ..4...16....36....81....196.....441......961......2116......4624......10000 ..6...36....92...241....720....1889.....4719.....12102.....30414......74588 ..9...81...241...742...2760....8465....24317.....73405....214117.....601411 .14..196...720..2760..13260...50139...175903....656508...2338562....7964440 .21..441..1889..8465..50139..220882...891720...3919354..16153614...62844340 .31..961..4719.24317.175903..891720..4094666..21072638..99746042..440141222 .46.2116.12102.73405.656508.3919354.21072638.129528304.721835038.3728749264 LINKS R. H. Hardin, Table of n, a(n) for n = 1..611 EXAMPLE Some solutions for n=4 k=3 ..1..0..0....0..1..1....0..0..0....1..0..0....1..0..1....0..1..1....0..1..1 ..0..0..0....1..1..1....0..1..1....0..1..1....1..0..1....1..0..0....0..0..0 ..0..1..1....0..0..0....0..0..0....0..0..0....1..0..1....0..0..0....1..0..0 ..0..0..0....0..0..0....1..0..0....0..0..0....1..0..0....0..0..0....0..0..0 CROSSREFS Column 1 is A038718(n+2) Column 2 is A207069 Sequence in context: A208164 A207960 A207564 * A208007 A207123 A207442 Adjacent sequences: A207416 A207417 A207418 * A207420 A207421 A207422 KEYWORD nonn,tabl AUTHOR R. H. Hardin Feb 17 2012 STATUS approved
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Last modified May 17 19:26 EDT 2021. Contains 343988 sequences. (Running on oeis4.)
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Updated 2015-09-02 22:51:11 by pooryorick
started by Theo Verelst
When dealing with the solution of (2d order) differential equations, electrical circuits, drawing circles, making musical waves or fourier analysis, sine waves or sine values as function of lets say x are essential.
Lets first draw a sine wave with a reasonable graphical accuracy, make sure you have a (preferably scrollable) Tk canvas to work on, and that the tcl variable mc contains the path to that canvas.
This routine, called with the number of x-steps draws 2 periods of a sine wave, with 1:1 x:y scale ratio, and rouding of y coordinates by truncation:
```proc drawsine { {n 256} } {
global mc
set pi 3.1415926535
\$mc del gr
for {set i 1} {\$i < 2*\$n} {incr i} {
\$mc create line [expr 100+\$i-1] \
[expr 100+\$n-\$n*sin(2*\$pi*(\$i-1)/\$n)] \
[expr 100+\$i] \
[expr 100+\$n-\$n*sin(2*\$pi*(\$i)/\$n)] \
-tag gr
}
}
drawsine 256```
Now when we make the y coordinate quantized per 10 of the above pixel widths
```proc drawsine_yq { {n 256} } {
global mc
set pi 3.1415926535
\$mc del gr
for {set i 1} {\$i < 2*\$n} {incr i} {
\$mc create line [expr 100+\$i-1] \
[expr 100+\$n-\$n*sin(2*\$pi*10*int((\$i-1)/10)/\$n)] \
[expr 100+\$i] \
[expr 100+\$n-\$n*sin(2*\$pi*10*int((\$i)/10)/\$n)] \
-tag gr
}
}```
```proc drawsine_yq { {n 256} } {
global mc
set pi 3.1415926535
\$mc del gr
for {set i 1} {\$i < 2*\$n} {incr i} {
\$mc create line [expr 100+\$i-1] \
[expr 100+\$n-\$n*sin(2*\$pi*10*int((\$i-1)/10)/\$n)] \
[expr 100+\$i] \
[expr 100+\$n-\$n*sin(2*\$pi*10*int((\$i-1)/10)/\$n)] \
-tag gr
}
}
drawsine_yq 256```
the graph looks like above, depending on whether the vertical connection lines are drawn or not.
Alternatively, we can clearly (because even with double accuracy floating point numbers, computer accuracy isn't infinite) quantize the x coordinates, and then interpolate somehow:
```proc drawsine_xq { {n 256} } {
global mc
set pi 3.1415926535
\$mc del gr
for {set i 1} {\$i < 2*\$n} {incr i} {
\$mc create line [expr 100+10*int((\$i-1)/10)] \
[expr 100+\$n-\$n*sin(2*\$pi*(\$i-1)/\$n)] \
[expr 100+10*int((\$i-1)/10)] \
[expr 100+\$n-\$n*sin(2*\$pi*\$i/\$n)] \
-tag gr
}
}```
Note that in the case of Y quantisation there are equal distances of 10 between all y values in the graph, and for X quantisation, the X coordinates are rounded to a fixed grid of 10 wide spacing (the images have been shrunk by a factor of 2).
Now if we take a non simple related x and y quantisation and an unclear (though well defined) interpolation strategy, in this case straight line through the sampled points, and make a fairly course quantisation (allow only a few values on the x and y axis where everythingis rounded to), results become maybe artistic, but not very scientifically usable:
```proc drawsine_xyq { {n {256}} } {
global mc
set pi 3.1415926535
\$mc del gr
set qx 30
set qy 18
for {set i 1} {\$i < 2*\$n} {incr i} {
\$mc create line [expr 100+\$qx*int((\$i-1)/\$qx)] \
[expr 100+\$n-\$n*sin(2*\$pi*\$qy*int((\$i-1)/\$qy)/\$n)] \
[expr 100+\$qx*int((\$i)/\$qx)] \
[expr 100+\$n-\$n*sin(2*\$pi*\$qy*int((\$i)/\$qy)/\$n)] \
-tag gr
}
} ```
Remember, these values aren't rounded, and no attempt is made to make the quantized patterns come as close as possible to the sine wave by any measure such as max or average distance or integral of absolute difference, or of course least square sense or so.
Things get even more hard to follow when sampling and signal reconstruction are also in the signal path.
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http://talkstats.com/threads/t-distribution.65895/
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| 67,216,913
| 9,118
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T distribution
Hasan
New Member
1. True or False: For a t distribution with 12 degrees of freedom, the area between 2.6810 and 2.1788 is 0.970.
2. True or False: For a t distribution with 12 degrees of freedom, the area between -2.6810 and 2.1788 is 0.970.
| 84
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| 2.578125
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| 0.892922
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https://logosconcarne.com/2020/05/04/secret-code-ii/
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| 424,872,423
| 40,419
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Secret Code II
‘X’ is for… ??
I’ve written about secret codes before. The Pigpen cipher was pretty simplistic, almost more a child’s game, but the Playfair cipher was a bit more interesting. That one came from a murder mystery novel by Dorothy L. Sayers. Today I thought I’d show you another simple code from another mystery novel.
1829 2125 0038 2226 1600 2125 0027 1722 4200 1829 1600 4219 2518 1617 1900 4122 3916 3200 3700 4019 0025 2016 0028 1724 2718 2241 4400 2118 0020 2516 2500 1829 1600 1819 2316 1517 2118 1617 0038 2226 1643 0015 2921 3829 0030 2025 1800 2425 2521 2841 2500 4120 4240 1617 2500 1822 0018 2916 0031 1619
Stop! See if you can decode the above before continuing.
You just kept on reading, didn’t you. Well, honestly, I would have too.
What the encoded paragraph says is:
“This code is from the mystery novel X by Sue Grafton. It uses the Typewriter Code, which just assigns numbers to the keyboard keys in order.”
And just number the letters left-to-right, top-to-bottom. So Q=1, W=2, etc. up to M=26.
That makes this a simple substitution cipher — very easy to break given enough of a sample.
As described in the novel, it’s a restricted system: Just letters, uppercase only. No punctuation. No digits (write them: “ONE SIX” etc.). It doesn’t even encode spaces.
For the sample above, I’ve used the code mainly as described, but I added spaces (space=0) because it was an easy fix (and because mushing together words sucks).
§
There is a simple wrinkle in that all number codes are two digits, so 4 is 04, and pairs of numbers are joined together. Use a final 00 to fill out the last pair if necessary.
That’s all there is to it.
It’s not a good code, but it’s not a super obvious one either. If you figure out E=3, it doesn’t mean F=4, as it would in a simple decoder ring version of a substitution code. This at least scrambles the order of the letters.
The crucial thing (and the primary difficulty), as always, lies in both you and the message recipient sharing the code key. Using the keyboard trick means no code books; one creates the key as needed.
We can expand the cipher to include punctuation keys, even the Return key for line endings. It just requires both sides agree on the numbering scheme, whatever it is. (It can be backwards, for instance.)
We can even double up to get upper and lower case: Q=1, q=2, W=3, w=4, E=5, e=6, and so on. This makes the code ever so slightly harder to break by introducing more codes and mapping each letter to two numbers.
Simple and trivial (and not at all secure) but useful to keep something important (and brief) from casually prying eyes.
§ §
The code is really just to kick off another Mystery Monday post about two favorite authors: Sue Grafton and Sara Paretsky.
They make a pair that both compares and contrasts.
Two female authors with long-running well-known series about female private eyes.
Grafton gave us Kinsey Millhone, working in “Santa Teresa” (a fictional Santa Barbara), California, and Paretsky gave us V.I. Warshawski, working in Chicago.
Both PIs have their own business and work alone. Both also live alone and have difficulty sustaining romantic relationships. Both have a retirement-age male neighbor who is a close friend, often helpful personally or with a case, but who is also sometimes a hindrance. Both have prickly irascible personalities.
That said, the two women are entirely different.
§
I’ll let Ms Millhone describe herself:
“My name is Kinsey Millhone. I’m a private detective by trade, doing business as Millhone Investigations. I’m female, thirty-eight years old, twice divorced, and childless, a status I maintain with rigorous attention to my birth-control pills.”
Later on the same page she adds: “I’m miserly and cheap.”
The quote is from the penultimate book in the series, simply titled “X”. The series ends with the last book Grafton wrote before she died: “Y” Is for Yesterday.
The Kinsey Millhone series is also called the Alphabet Mystery series because Grafton followed a pretty cool pattern with her titles:
1. “A” Is for Alibi
2. “B” Is for Burgler
3. “C” Is for Corpse
You get the idea. Part of the fun was thinking ahead to what crime-related word Grafton would pick for upcoming books. I’ve been following her since the early 1980s — certainly by her second book if not the first.
“X” is unique in the series for not being for anything. (Grafton couldn’t come up with a good crime word for X. Instead, the book tries to use every possible X word it can in the text. It’s kind of a hoot.)
All the books follow a general arc of Kinsey investigating a case, doing a lot of leg-work, interviewing people, chasing clues, finally putting it all together, and then caught in some life-threatening situation at the end.
They are not murder mysteries; Kinsey is usually hired for the more ordinary private eye stuff: locating a missing person, for example. Often what she discovers ends up being life-threatening for her, hence the final showdown scene.
They’re extremely readable. Great books for travel or the beach or when you just want a really comfortable detective story. They’re great for escape because Grafton keeps Kinsey in her 30s in 1980s “Santa Teresa” throughout the series, which ultimately only covers a period of a few years.
As with most book series starting long ago, the early books are skinny, and the later books are fat. Someone never returned my paperback of “A” but “B” is 210 pages. My paperback copy of “X” is 498 pages.
I bought the ebook of “A” as well as “Y” so I have her first and last books electronically. There will never be a “Z” or any other letters. (Readers long wondered what she’d do after “Z”. Quit? Double letters? We’ll never know.)
One thing I like about Grafton: She made it explicit in her will: Kinsey Millhone dies with me. There will never be “Sue Grafton’s Kinsey Millhone” books, which I really respect. Major props on that one, Ms Grafton. Kudos!
§ §
Before I move on to Paretsky, as an aside, Lawrence Block has a really fun murder mystery series about Bernie Rhodenbarr, the gentleman burglar.
Bernie, a book-lover who owns a used book store as cover, only steals from those who have so much they can (and really, in Bernie’s opinion, ought to) share their wealth. Bernie’s dedicated to helping them do it (albeit without their knowledge or permission).
The problem is, all too often, during his perfectly innocent burglary, there’s a murder, and his only hope of clearing his name is catching the real murderer.
It’s a droll comedy series; lots of fun.
Point is, Bernie has a BFF, Carolyn, a lesbian soulmate who owns a pet-grooming salon. They regularly drink and chat together, and one of their pastimes is guessing Sue Grafton titles beyond “Z”. Bernie loves a good pun (as do I).
I no longer have those books, but two I recall were “AA” Is for Battery, and “DD” Is for Busted. (There was one about “F” Is for Train, too.)
Not only will we never know what Grafton would have done after; we’ll never even know what “Z” would have been.
§ §
Sara Paretsky’s Victoria Iphigenia Warshawski (“Vic” or “V. I.” but god help you if you say “Vicky”) is from the tough-as-nails breed of PI.
Both parents are dead (true of Kinsey, too). Dad was a tough Chicago cop; Vic still has friends on the force who remember him and who saw her grow up. Her dad’s best friend, Bobby Mallory, now a Chicago PD Lieutenant, is very unhappy with VI’s career choice.
VI has a good friend, Dr. “Lotty” Herschel, who also isn’t too happy with the career choice, especially as she often finds herself patching Vic up.
As with Grafton, these are also not murder mysteries. VI mainly specializes in financial crime and other white collar issues. But, as with Millhone, she tends to stumble over much larger crimes no one knew about.
Also as with Millhone, things often get dicey at the end, although Warshawski sometimes gets banged up long before the final scene (hence the tch-tch patching up by Lotty).
The contrast of locale, sunny little “Santa Teresa” versus big city Chicago, is very much reflected in the tone of the books. The Paretsky books are slightly darker and edgier.
Vic is slightly edgier than Kinsey, although neither take any BS.
§
I do have a gripe about a pattern these books follow that both surprises and annoys me.
VI stumbles onto some big crime that know one suspected was going on. And no one ever believes her. Not even her friends — they’re always dismissing what she says.
But it always turns out she’s right, so at what point does their light bulb go on? As far as readers know, VI is right every time. Why wouldn’t her friends figure she’s on to something this time?
(I do think Paretsky is making a pointed statement about men not believing, or even listening to, women, and having seen it in real life, I get it. But it’s still annoying how no one ever learns to believe her. Idiots.)
§ §
I’ve known these private eyes for decades and really enjoy the stories. If you like private eye stories, you can’t go wrong with Grafton and Paretsky.
I mentioned recently that Tony Hillerman is my second favorite mystery author. Going by what I’ve followed and bought over the years, Sue Grafton and Sara Paretsky are tied for third place (with Dorothy L. Sayers and Rex Stout breathing down their necks).
One of these days I’ll tell you about first place.
Stay typewritten, my friends!
The canonical fool on the hill watching the sunset and the rotation of the planet and thinking what he imagines are large thoughts. View all posts by Wyrd Smythe
7 responses to “Secret Code II”
• SelfAwarePatterns
Block wrote a book on writing that I read many years ago. I think it was the first one on writing I read. I still have it lying around somewhere. I don’t remember it being all that great as far as writing techniques, but it covered the basics and served to lift the air of mystery about how it works.
He described his own writing process, which is largely discovery and hammering our five pages a day, something which seemed imminently doable to me at the time. (The fact that I haven’t done it to any great extent shows it isn’t as easy as I thought back then.)
• Wyrd Smythe
I can’t say Block is at the top of any of my lists, although I did enjoy his Bernie Rhodenbarr stories. I used to have a couple books of unrelated short stories he wrote, too. Crime stories mostly. They were okay.
Robert Parker (who wrote the Spenser series) also supposedly wrote five pages a day. Long hand on legal pads. Every day without fail. And supposedly he never revised, never plotted, just wrote. Or so I heard long ago, and who knows what embellishments my mind has added over the years.
I guess when it comes to (fiction) writing, you just have to make yourself do it. It’s that way sometimes with blog writing. I realized long ago I didn’t really have any good (fiction) stories to tell, and I’ve recently realized I just don’t like people enough to write fiction. (My whole approach seems more oriented in trying to educate them…. and I can just hear Dr. Phil’s voice,… “So, how’s that been working out for you?” 😮 )
• SelfAwarePatterns
One of the risks with discovery writing is you might have to end up making large scale structural changes. Some writers take pride in never making those kinds of changes, but typically some of their stories are less satisfying than they otherwise could have been. The writer might bring enough other stuff (interesting settings, characters, ideas, etc) that you don’t mind, but not always.
I’m with you on the need to make yourself do it, particularly on blogging. I try to push myself to write at least one post a week. If I didn’t do that, I’d probably never post.
I don’t know that you have to like people all that much in order to write fiction. Some writers have a pretty dismal view of humanity. I think of writers like Richard K. Morgan, the author of the Altered Carbon series, whose misanthropy permeates those books. It’s a view that can lead to pretty grim stories, although it can also be flipped and used for dark comedy.
On educating people, I think, like persuasion, it has to be seen as a long term game, one that will sometimes bear fruit, but often won’t. It’s very rare that I see someone change their mind during a discussion. It virtually never happens. (When it does happen, it’s usually because the person wasn’t that committed to whatever view they started with.) When I’ve successfully convinced someone of something, I usually only found out about it long after the fact, and then only if the conversation ended on amicable terms.
• Wyrd Smythe
All I can say is that my writing definitely benefits from as much going over as I can stand. That said, I have banged out posts that didn’t suck.
Maybe if one has a strong writing voice and a good sense of where the plot goes it can work. Parker certainly had a specific writing voice. I have no idea how he went about building a plot. I would figure any mystery writer would have to have a plot outline, otherwise where’s the mystery?
The misanthropy thing for me is Yin-Yang; there’s a fascination, too. The misanthropy is mostly a product of my experiences. And it’s a good point, I could express that in fiction. I think the recent realization has to do with where we are these days — we’re going backwards, and I’ve become disengaged. It’s probably more the first thing I mentioned; I just don’t have any stories that burn in me to tell.
I think I’ve mentioned there are a lot of teachers (and some preachers) in my family tree. There can be a persuasive element to teaching, but persuading someone towards knowledge is, as you suggest, very much an uphill battle. Education should be about feeding the thirst to learn.
Debate is a whole other mode and it’s true that, especially these polarized days, one doesn’t change minds so much as maybe plant thought seeds. These days it’s rare for most people to even go so far as being able to say, “Yeah, I see your point, it’s valid, but my view differs.” It seems more about separate corners and refusing to grant any validity to opposing arguments.
It does take a strong ego, a confidence that one is usually right, to be able to admit being wrong, or even that the other guy has a valid point. It’s easy to see a slippery slope where admitting to one point leads to another and down you go. (It doesn’t seem many people value truth and facts quite as much as not being wrong.)
The irony is that, at least in my experience, it’s our mistakes and being wrong that teach us the most. Those lessons always seem more well-remembered to me than the slow slog of learning. Like touching a hot stove. Don’t do that again!
• SelfAwarePatterns
I’m with you on writing being improved by going over it again. The amount of passes I need to make through a post seems to grow as the size of the post increases. I can typically nail out a short 300-500 word post without too much editing. But once it grows past 1000 words, the amount of editing increases. It’s not exponential, but it feels more than linear.
One of the nice things about writing a story is you always have the ability to go back and make relatively minor edits in the earlier portions that can serve to effectively foreshadow later events. At least with modern word processing, it’s easy, and I suspect every author does it, whether or not they’ll admit it. It was a lot more effort in the typewriter era.
I don’t know that people today are really that much more argumentative than they’ve been historically. The book, How to Win Friend and Influence People, was published in 1935. Many of its examples range from the Civil War through the early 1930s, but the way people act in those examples seems very resonant with how they act today. What seems to be different today is that it all happens much faster, and it’s therefore much easier for it to cascade out of control.
Yes, admitting when we’re wrong can be hard. It’s even harder when the person on the other side is being nasty. Carnegie was right, you can’t win an argument. It’s still sometimes fun to have them. 🙂
• Wyrd Smythe
Interesting point about the growth of complexity with size. Due, perhaps, to increasing interconnections between aspects of the text, so there’s more that has to sync up? I do find I have to focus on coherence as a post gets more complex.
I’m sure I’ve mentioned the thing about how it always requires N+1 proof readings to catch all errors, where N is the actual number performed. Regardless of how large you make N, this rule applies.
I go back far enough to have used a typewriter to write. Software editors are a whole world better. Spell check alone is worth its weight.
“I don’t know that people today are really that much more argumentative than they’ve been historically.”
Oh, hell no! We haven’t changed as a species in a good long time.
What’s different is the amplification factor social media provides. Combined with the sheer numbers of people involved these days. For any weird fringe view, there will still be a ton of people in that group. Social media amplifies that, and our thirst for something new feeds the amplification. Cascades, as you say, out of control.
Combine that with a mainstream pulling away from facts and science, and it’s a recipe for a really bad stew.
“Yes, admitting when we’re wrong can be hard.”
Polarization is a pretty easy trap to fall into, especially when facts and the dialectic are devalued as they are currently. We stick with our perceived tribes, especially in insecure times.
Those who are deeply into science can be a little better at processing unwelcome facts, but even those who place the highest reverence on facts are still human. I think the mark of a true intellect is that those seeds do sprout (assuming they were good seeds). At some point, if one is true to the ideals of knowledge, a good argument eventually breaks through.
At least it does for me. Maybe I’m atypical. 😉 (Actually, I know that’s true!) 😀
• Secret Code III | Logos con carne
[…] days back in 2015 and more recently about other codes from books. See The Playfair Cipher and Secret Code II. And now you see why this is Secret Code […]
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## Wednesday, May 24, 2017
### diamond square fractals
I have previously worked on fractal automata which produce animating patterns in 2D and 3D. A special case of this is where only the larger scale affects the smaller scales, giving static images. One shortcoming with these automata and images is that they have square or cubic symmetry, so features are just at right angles, which gets a bit dull, this was one reason I made this cubic modification for 2D images, giving a hexagonal symmetry.
A tweet by someone called R4_Unit showed a different way to achieve a similar goal. Use the diamond-square algorithm which is used for plasma fractals and hilly height fields, but use automata rules rather than random offsets. R4_Unit applied this in 3D, and has since submitted his code here, and a user called Softology has added his own implementation to his Visions of Chaos software. In 2D the diamond-square algorithm gives a pseudo octagonal symmetry, quite similar to automata rule type 7 that I made in the 2nd link above which gave nice results:
But with a rule that is more complicated than diamond-square. The 3D diamond-square is actually cube-irregular_octahedron-octahedron and doesn't have nearly the simple symmetry of diamond-square, so here I investigate what the results look like in 2D.
For 4 parent points there are 2^4 parent combinations and the central point is either black or white for each, giving 2^16 rule sets. However, once you constrain it to square symmetry, this drops down to only 64 rule sets, which I draw using a 4x4 start grid:
0001
0010
0011
1111
And for the symmetric start grid:
0000
0110
0110
0000
five along one up seems like the 'nicest' in both cases.
For more than two colours the number of combinations explodes. One way to contain that is to demand 'bit symmetry' which in the two-colour case means the rule sets acts the same if black and white are swapped. Unfortunately it is not possible to have a two-colour bit symmetric rule with this diamond-square approach, because there are an even number of identical parents. If two parents are white and two are black then there is no child colour that acts the same if black and white are reversed.
The generalised 'bit symmetry' for n colours means the rules are the same if the n colours are permuted in any way. Then for three colours it is possible to have bit symmetric rules, and there are only three of them:
Overall the rule is to choose the most popular colour of the parent. If this is a tie, then choose the least popular. The choice occurs when there are three parent corners of a single colour. Do you choose the colour there is none of, one of or three of among the parents? The above shows each respectively.
The third is the simpler rule (as it is subsumed by the first criteria), and gives more solid results. Here is a close up of the only complex feature it seems to generate:
These teardrop shapes are all identical. While they are polygonal, each protrusion seems to have an infinite number of edges, in fact it is a tree. So each teardrop is a tree.
In fact, these are the only three bit-symmetric rule sets possible for the diamond-square algorithm for any number of colours, and none exist in 3D at all (assuming cubic symmetry), so they are quite unique.
However, a square lattice is not the only scalable lattice, the only other I know is the triangular lattice. And in this lattice bit-symmetric automata occur only with four colours, and there are only two of them:
The rule is the same as for the square lattice case, but the choice is when two of your parents are the same colour, do you choose the colour there is one of or two of? Again, the more solid results come from the latter case. It is quite a wonderful shape, with almost rounded looking blobs that belie its triangular lattice frame. These used a randomised 4x4(ish) start pattern, and here is higher resolution with an 8x8(ish) start pattern:
The rule for this and the teardrop above are incredibly simple, pick the most popular parent colour, if it is a tie then pick the least popular.
Here are three simplest cases:
The apparent fifth colour is a nowhere-dense mix of the four colours and primarily takes the shape of a modified Koch snowflake, by contrast the dense colours are all finite sided polygons.
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# In wind turbine with induction generator system, I keep getting NAN EM torque. How to fix this?
2 views (last 30 days)
Bayan Hussein on 15 Mar 2022
Commented: Bayan Hussein on 23 Mar 2022
I am trying to simulate an induction generator-based wind turbine system connected to an infinite bus through 2 transmission lines. I am using double squirrel cage generator and the input is the speed in perunit. I compute the speed using a small block for the swing equation Tem-Tm=Jdw/dt (and Wm=Wg) but for now I am fizing it to 1.09. The EM torque from the generator gives me NAN no matter what I change. Does anyone know how to fix this issue?
Peter O on 16 Mar 2022
So you're generating, fixing the rotor speed of the machine to 1.09 per unit and supplying the machine from an infinite 3-phase bus (leaving just torque and currents as unknowns), but the output you're getting back is NaN? What are the parameters of the machine? To confirm, the infinite bus is running at 1.0 PU, with its frequency set to 1.0?
Bayan Hussein on 23 Mar 2022
Hello Peter O. First, I would like to thank you so much for your answer. What I'm trying to do is simulate a wind-turbine with a double squirrel cage generator. I am supposed to work with the speed rather than the mechanical torque. Thus, I am using the swing equation to generate this speed in a loop-kind-of block. However, right now I am facing the issue of the NaN EM torque so the speed itself is not being calculated. I tried to fix it to 1.09 to see what is the issue but I still get the same NaN problem.
The machine parameters are as follows
I am supposed to work with an infinite bus so I added the 3ph source to emulate that. However, I kept at as swing because I dont know what to set it as. Is its voltage and frequency to 1.0 pu with these parameters?
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Home > Standard Deviation > Calculate Standard Deviation Error
# Calculate Standard Deviation Error
## Contents
Create an account EXPLORE Community DashboardRandom ArticleAbout UsCategoriesRecent Changes HELP US Write an ArticleRequest a New ArticleAnswer a RequestMore Ideas... National Center for Health Statistics (24). Advertisement Autoplay When autoplay is enabled, a suggested video will automatically play next. The confidence interval of 18 to 22 is a quantitative measure of the uncertainty – the possible difference between the true average effect of the drug and the estimate of 20mg/dL. useful reference
Rating is available when the video has been rented. Todd Grande 22,962 views 9:33 Understanding Standard Error - Duration: 5:01. The standard error (SE) is the standard deviation of the sampling distribution of a statistic,[1] most commonly of the mean. When to use standard error? http://www.radford.edu/~biol-web/stats/standarderrorcalc.pdf
## How To Calculate Standard Error Of The Mean
If there is no change in the data points as experiments are repeated, then the standard error of mean is zero. . . The formula to calculate Standard Error is, Standard Error Formula: where SEx̄ = Standard Error of the Mean s = Standard Deviation of the Mean n = Number of Observations of Show more Language: English Content location: United States Restricted Mode: Off History Help Loading...
All Rights Reserved. Standard Error Example Retrieved 17 July 2014. Gurland and Tripathi (1971)[6] provide a correction and equation for this effect. Edwards Deming.
The standard deviation of the age for the 16 runners is 10.23, which is somewhat greater than the true population standard deviation σ = 9.27 years. Calculate Standard Error From Standard Deviation In Excel As will be shown, the standard error is the standard deviation of the sampling distribution. Sign in to make your opinion count. Tips Calculations of the mean, standard deviation, and standard error are most useful for analysis of normally distributed data.
## Standard Error Example
Flag as... http://www.wikihow.com/Calculate-Mean,-Standard-Deviation,-and-Standard-Error American Statistical Association. 25 (4): 30–32. How To Calculate Standard Error Of The Mean This information is referred to as a sample. Formula For Calculating Standard Error Compare the true standard error of the mean to the standard error estimated using this sample.
In other words, it is the standard deviation of the sampling distribution of the sample statistic. http://freqnbytes.com/standard-deviation/calculate-a-standard-deviation-from-a-standard-error.php Standard Error of the Estimate A related and similar concept to standard error of the mean is the standard error of the estimate. The following expressions can be used to calculate the upper and lower 95% confidence limits, where x ¯ {\displaystyle {\bar {x}}} is equal to the sample mean, S E {\displaystyle SE} Show more unanswered questions Ask a Question Submit Already answered Not a question Bad question Other If this question (or a similar one) is answered twice in this section, please click Convert Standard Deviation Standard Error
Flag as... Method 3 The Standard Deviation 1 Calculate the standard deviation. Standard error of the mean It is a measure of how precise is our estimate of the mean. #computation of the standard error of the mean sem<-sd(x)/sqrt(length(x)) #95% confidence intervals of this page If you are interested in the precision of the means or in comparing and testing differences between means then standard error is your metric.
It is usually calculated by the sample estimate of the population standard deviation (sample standard deviation) divided by the square root of the sample size (assuming statistical independence of the values Calculate Variance Standard Deviation Because of random variation in sampling, the proportion or mean calculated using the sample will usually differ from the true proportion or mean in the entire population. set.seed(20151204) #generate some random data x<-rnorm(10) #compute the standard deviation sd(x) 1.144105 For normally distributed data the standard deviation has some extra information, namely the 68-95-99.7 rule which tells us the
## They may be used to calculate confidence intervals.
Learn More . Flag as... Flag as duplicate Thanks! Calculate Median Standard Deviation The larger the sample, the smaller the standard error, and the closer the sample mean approximates the population mean.
The sample mean will very rarely be equal to the population mean. Next, consider all possible samples of 16 runners from the population of 9,732 runners. However, the mean and standard deviation are descriptive statistics, whereas the standard error of the mean describes bounds on a random sampling process. http://freqnbytes.com/standard-deviation/calculate-standard-error-from-standard-deviation-excel.php Full list of contributing R-bloggers R-bloggers was founded by Tal Galili, with gratitude to the R community.
Add up all the numbers and divide by the population size: Mean (μ) = ΣX/N, where Σ is the summation (addition) sign, xi is each individual number, and N is the Loading... Repeating the sampling procedure as for the Cherry Blossom runners, take 20,000 samples of size n=16 from the age at first marriage population. Similar Worksheets Calculate Standard Deviation from Standard Error How to Calculate Standard Deviation from Probability & Samples Worksheet for how to Calculate Antilog Worksheet for how to Calculate Permutations nPr and
Because these 16 runners are a sample from the population of 9,732 runners, 37.25 is the sample mean, and 10.23 is the sample standard deviation, s. Larger sample sizes give smaller standard errors As would be expected, larger sample sizes give smaller standard errors. Sign in to report inappropriate content. Consider a sample of n=16 runners selected at random from the 9,732.
All the R Ladies One Way Analysis of Variance Exercises GoodReads: Machine Learning (Part 3) Danger, Caution H2O steam is very hot!! The sample standard deviation s = 10.23 is greater than the true population standard deviation σ = 9.27 years. Loading... This often leads to confusion about their interchangeability.
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Age of the Universe: Exercises
At this point, you should open a blank document in your word processor and record your answers to these questions. Copy each question into the journal and type your response below in a different color or font. Be sure to write in complete sentences and express your ideas so that others can understand. Save your work and keep this window open throughout this study so you can easily record other questions and answers. Questions for thought: 1. What do you think of when you think of the universe? 2. What do you know about the age of the universe? 3. How do you think scientists might obtain an estimate of the age of the universe? Write a brief response to these questions in your journal.
The information that follows will be used in your exploration of the age of the universe. A list of questions at the end of the section will help this exploration.
Notation: Basic units Because astronomical distances are so large, astronomers like to use a large measure, a parsec. One parsec (pc) is approximately 3.26 light years, or 3.1 x 1013 kilometers (km). (One light year is the distance that light travels in one year.) In this study, the distance d of a galaxy from Earth is given in Megaparsecs (Mpc) Mega means million, so 1 Mpc = 1 x 106 pc. 1 Mpc = 3.1 x 1019 km
The objectives and assumptions of this study are listed below. These will appear in each of the Kéyah Math Studies.
Objectives
To determine:
1. a linear model for distance of a galaxy from the earth against the velocity of recession;
2. an estimate for the age of the universe.
Assumption
The relation between distance and velocity is linear.
d(x103) v(x103) Constructing the Hubble Diagram The data points to the left are taken from a plot of redshift z of receding galaxies against their distance d from Earth (Galaxies and Cosmology, Jones and Lambourne, page 243) with the second row indicating redshift x velocity of light = velocity of recession of galaxy = v The numbers in the distance column give distance d of the galaxy from Earth in Megaparsecs x 103 (Mpc); the velocity is kilometer per sec x 103. So, for example, a galaxy which is 0.028 x 103 Mpc from Earth is receding at a velocity of 2.7 x 103 km/sec. Linear Models In this part, you will find a linear function to express the relationship between distance and velocity. Specifically you will: * express velocity as a function of distance from the earth, * find the Hubble constant, and * determine how fast the Whirlpool Galaxy is receding from us. For this demonstration module, instructions are provided in italics and solutions are in blue italics. 0.028 2.7 0.076 4.2 0.108 10.5 0.136 14.1 0.153 10.5 0.226 13.2 0.283 19.8 0.359 28.2 0.363 20.7 0.408 29.4 0.438 31.8 0.472 44.4 0.476 32.1 0.476 37.2 0.493 33 0.556 34.5 0.639 46.5
Problem Set (Round off to three decimal places for this work.)
The applets needed to complete this problem are provided in the text for this Introductory module. Usually, you will need to follow these steps. (You may choose to use your graphing calculator to do this work-ask your instructor.)
* Scroll down to "Tool Chest" on the menu to the left of the screen.
* Click "linear regression" (since we are looking for a linear function).
The linear regression applet will open in a separate screen.
Question 1: Plot the points corresponding to the data in Table 1. The first coordinate is distance from the earth; denote this by d (units are 103 Mpc). The second coordinate is velocity of recession of galaxy, units 103 km/sec. You can copy each column from the table and paste it into the appropriate column of the applet.
Enter the data provided in Table 1into the columns on the applet screen; enter the d value in the "x" column, then enter the v value in the "y" column. You can copy each column from the table and paste it into the appropriate column of the applet.
Click "Plot", you should see the screen below.
Question 2: Now find the line of best fit. (The resulting graph is called a Hubble Diagram.)
Click "analyze" on the applet containing the plotted data and see the screen as shown.
The equation of the line is shown in lower right window of the applet screen:
y = 70.345x + 0.737.
Question 3: What is the slope of this line? What are the units?
The slope is indicated by the constant m, it represents the change in y divided by the change in x. The units for the numerator are the units for velocity, 103 km/sec; the units for the denominator are units for distance, in this case 103 Mpc; thus the units for the slope are (km/sec)/Mpc.
m = 70.345 ( km/sec)/Mpc.
Your answer is the Hubble constant, denoted by H0. Here's what it means: if a galaxy is 10 Mpc away from us, it is receding at the rate of 10 x 70.345 km/sec = 703.45 km/sec; if a galaxy is 15 Mpc away, it is receding at the rate of 15 x 70.345 km/sec = 1055.175km/sec; etc.
Question 4: The Whirlpool Galaxy is approximately 30 x 106 light years away from us. (One light year is the distance that light travels in one year.) How fast is it receding from us? We need to use the same units for all measurements, so converting 30 x 106 light years to Mpc, we get 9.2 Mpc.
Now use the Hubble constant to answer the question.
The galaxy is receding at the rate of 9.2 x 70.345 = 647.174 km/sec. (This calculation can be done using MathPad or your calculator).
Information: We can use the Hubble constant to approximate the age of the universe. The modified equation from #1 (ignoring the very small constant, assuming that when d = 0, then v = 0, and rounding the Hubble Constant to one decimal place), v = 70.3d, is known as Hubble's Law.
Question 5: A Black Hole, Perseus A, has been detected traveling away from us at the rate of 1631 km/s. How far away is it?
This could actually be solved easily by dividing 1631 by 70.3, however we will use the Plot-Solve Applet to show how it can solve equations. The equation we need to solve is
1631 = 70.3d,
for the distance d.
Click "Tool Chest" on the left menu, and then click "Plot-Solve." You'll see the screen below.
Now click "New Function" and type in the left side of the equation, 1631; then click "Plot."
Next, click "New Function" and type in the right side of the equation (remember to use * for multiplication, and use x for d), 70.3*x; then click "Plot again".
The screen at this point doesn't look very interesting. The reason is that the graph of y = 1631 is outside the range of the screen, so you'll need to change the y-range. Enter 1600 for "YMin" and 1800 for "YMax"; this range includes 1631, which is what we want to see.
We also want to see the solution, so you'll need to change the x-range also: let's guess and enter 10 for "XMin" and 30 for "XMax."
Now, click "Plot" again. You should see the screen shown below. The solution to our equation is the x-coordinate of the intersection of the two graphs.
Place the point of the arrow on your cursor right on the intersection of the two graphs and click "Plot." The x-coordinate can be read in the window to the right of the screen, you should see
x = 23.2.
So, the black hole is 23.2 Mpc away from Earth.
Question 6: Recall the well-known formula for distance, distance = (rate of speed) x (time traveled), or d = vt, and substitute this into the Hubble Law for d. Simplify the result and solve for time, t.
v = 70.3d
v = 70.3(vt)
Cancel the v from both sides to get
1 = 70.3t
Solve for t to get
t = 1/70.3
In order to keep the same units and express the age of the universe in years, the Hubble constant, H0 = 70.3 km/s (per Mpc), expressed in Mpc/yr (per Mpc) is 72.5 x 10-12 Mpc/yr (per Mpc).
Question 6: Using this value for the Hubble constant (replace 70.3) and the answer to Question 6, compute t. This answer is a very good approximation to the current age of the universe.
t = 1/(72.5 x 10-12) = 13.8 x 109 years, or 13.8 billion years
(Note the units here: the Mpc's cancel leaving only years.)
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# Topic: Re: How do you calculate the power required for helical impeller mixer?
posed by
You left out one essential piece of information needed to calculate the power, and that is the rotational speed. So here are a the power requirements for a few appropriate speeds: 60 rpm 1,400 W 50 rpm 936 W 40 rpm 599 W 30 rpm 337 W The viscous power number for the impeller is 312. Viscous power number is Power divided by all the following (Viscosity times Rotational Speed squared times Impeller Diameter cubed). Of course the units of the variables must be consistent to obtain a dimensionless relationship.
The answers by this expert are based on the best available interpretation of the information provided. The consequences of the application of this information are the responsibility of the user. If clarification is needed, please submit a further question.
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# maths
posted by .
Hello I'm stuck on this question here it is x² + 3x - 6.75 = 0 So what does x equal to ?
• maths -
x + 3x - 6.75 = 0
4x = 6.75
x = 1.6875
You should ALWAYS proofread your problem after posting to make sure it makes sense.
I have no idea what the ? means.
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FutureStarr
16 Out of 40 As a Percentage OR
## 16 out of 40 as a percentage
via GIPHY
Not every professor is a 16 out of 40 favorite, but not every professor is a professor. There’s no shortage of content online, but we’ve taken a look at what we think are the best content and blog posts online to give you a list of your next brilliant ideas or to provide some motivation. If you’re looking for one last tip to help make your sale, maybe you’ll find a new tool here.
### Percentage
via GIPHY
This percentage calculator is a tool that lets you do a simple calculation: what percent of X is Y? The tool is pretty straightforward. All you need to do is fill in two fields, and the third one will be calculated for you automatically. This method will allow you to answer the question of how to find a percentage of two numbers. Furthermore, our percentage calculator also allows you to perform calculations in the opposite way, i.e., how to find a percentage of a number. Try entering various values into the different fields and see how quick and easy-to-use this handy tool is. Is only knowing how to get a percentage of a number is not enough for you? If you are looking for more extensive calculations, hit the advanced mode button under the calculator.
Other than being helpful with learning percentages and fractions, this tool is useful in many different situations. You can find percentages in almost every aspect of your life! Anyone who has ever been to the shopping mall has surely seen dozens of signs with a large percentage symbol saying "discount!". And this is only one of many other examples of percentages. They frequently appear, e.g., in finance where we used them to find an amount of income tax or sales tax, or in health to express what is your body fat. Keep reading if you would like to see how to find a percentage of something, what the percentage formula is, and the applications of percentages in other areas of life, like statistics or physics. (Source: www.omnicalculator.com)
### Value
via GIPHY
Due to inconsistent usage, it is not always clear from the context what a percentage is relative to. When speaking of a "10% rise" or a "10% fall" in a quantity, the usual interpretation is that this is relative to the initial value of that quantity. For example, if an item is initially priced at \$200 and the price rises 10% (an increase of \$20), the new price will be \$220. Note that this final price is 110% of the initial price (100% + 10% = 110%).
Calculating the percentage of a known value is quite straightforward. In a numerical reasoning test, these questions tend to require you to identify/manipulate the relevant information in order to use the formula. Graphs or tables will often be used to present the information, such as the one below. For example, Shimmy’s friend Elissa also bought a house in 2010. She recently had it valued and found it was now worth £230,000. This is 80% greater than what she originally paid for it. To work out the original value of the house, you need to use the following process: (Source: www.wikijob.co.uk)
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# how to calculate 1% of erc20 18 decimal token?
I am building a system that should calculate the 1% of the tokens sent to it. I am thinking that it is ok to deal with large numbers but when it comes to smaller numbers like 1 token or 2 token [not the complete token], how can we calculate 1% of them? The token is 18 decimal places. I am looking for a chunk of code snippet for calculating 1% of the tokens value entered.
• Tokens and Ether units (such as wei) do not have direct connection unless you give them explicit price in a crowdsale or so. If you only look at a token contract it's impossible to determine the value of one token. So it is a bit unclear to me what you are asking. – Lauri Peltonen Jul 2 '19 at 11:28
• thanks for a response. so let say a wallet send WHOLE 1 Token to contract and contract has to calculate 1% of it in transfer function, so complete token will be this (1000000000000000000) and then we will calculate 1% of it. Now if someone send only one chunk of the token, that is instead of 1 with 18 decimals, he sent only 1 token then how can we calculate 1% of it? – Fariha Abbasi Jul 2 '19 at 11:32
• If somebody sends only 1 (aka 1 wei), then you don't quite have a choice, and 1% of it is 0 wei. In fact, even 99% of it would be 0 wei. However, please note that 1 wei is typically considered an extremely small amount (that's why the "wei-resolution" of `1e18` has been introduced to begin with). So it's not likely for anyone to transfer this amount, and even if you give them less than the 1% that they're entitled to get, it's not going to make much of a difference. – goodvibration Jul 2 '19 at 11:41
For the sake of this answer let me invent two new terms:
1) Display amount [of tokens]. The amount of tokens you typically see in wallets and exchanges. Can usually be a decimal number.
2) Absolute amount [of tokens]. The amount used in technical context and is always an integer. The amount given for example to ERC20's `transfer` function.
I think you are a bit confused about the topic of token decimal places. In absolute units 1 token is always the minimum anyone can own, transfer and receive - regardless of how many decimal places the token has. So if you call an ERC20 contract's `transfer` method the parameter you give it is always an integer and you are transferring full tokens.
However for display purposes 1 "absolute amount of tokens" is not the same thing as 1 "display amount of tokens". When displaying the amount of tokens the decimal places are taken into consideration. So if you have 18 decimal places then 1 display token = 1 000 000 000 000 000 000 absolute tokens and to issue a transaction with the `transfer` function for 1 display token you give it 1 000 000 000 000 000 000 as a parameter.
Usually people talk about display amounts of tokens but technical people may talk about absolute amount of tokens.
You can read more clarifications for example here: Decimals on ERC20 Tokens and here: What denomination should I issue my ERC20 token by?
I'll try and give a complete answer to the OP, building upon the very good generic explanations given by other answers.
To make things very clear here, let's talk about the case of a token defined with ZERO decimals. This effectively means that one token cannot be divided further. In Lauri Peltonen's wording it means "display amount = absolute amount".
So if you want to calculate a fraction of ONE indivisible token, well you can't. Solidity's integer division will simply truncate the decimal part of the result, and it will simply return zero if the result is below 1. Even trying to take 1% of 99 tokens will fail, in the sense that it will yield zero fees. Also, all fees will come in 100tokens-steps, so 800 to 899 tokens yield 8 and 900 yields 9, etc.
There are 3 ways to take this:
1. Ignore it, because for example the token value is very low and a transaction for less than 1000 tokens would cost more gas than the value of the tokens. Also, a loss of a fraction of a token in fees per transfer is negligible. This is often the case when there are 18 decimals.
2. Disallow it. You detect cases where fees are zero and revert the tx, effectively forcing users to transfer at least 100 tokens.
3. Work around it in a way that makes sense to your business/Dapp, e.g. by forcing a round up or by keeping track of the remainder.
If you want to keep track of the lost decimal part, you can always use the "precision" trick, and here's an example of how to treat this case:
``````contract feeBurningToken is ERC20Token {
using SafeMath for uint;
uint256 constant PRECISION = 10 ** 3;
mapping(address => uint256) unpaidPreciseFees; //stored e.g. in milliTokens
function transfer(address to, uint256 amount) {
uint256 fee = feesToBurn(amount);
super.transfer(bonfire, fee);
super.transfer(to, amount - fee);
}
function transferFrom(address from, address to, uint256 amount) {
uint256 fee = feesToBurn(amount);
super.transferFrom(from, bonfire, fee);
super.transferFrom(from, to, amount - fee);
}
function feesToBurn(uint256 amount) public returns (uint256 fee){
uint256 fee = amount / 100;
uint256 preciseFee = amount.mul(PRECISION) / 100;
unpaidPreciseFees[destination] += preciseFee.mod(PRECISION);
//whenever unpaidFees get to 1000 or over, one full token is due,
//so we can now add it to the fees on this transfer.
if (unpaidPreciseFees[destination] >= PRECISION) {
unpaidPreciseFees[destination] -= PRECISION;
fee++;
}
return fee;
}
}
``````
Depending on your context and how you tweak things, you might need more SafeMath in your code.
Obviously there are many ways of treating the remainder, you could for example do the contrary of the code I proposed, and round up on the first tx, then keep track of the userCredit, spending it to cover the millitokens part of the next transfer to the same destination. Btw it is more of a business decision than a developer's responsibility imho.
NOTES:
Notice this code puts an additional limit on the size of transfers, so go easy on the value chosen for precision, esp. if your token has a very large supply. A token with 18 decimals, with a precision of 18 decimals will still be able to transfer up to around 10**(77-18-18) = 10**41 "display tokens" before the multiplication overflows and reverts the tx.
• thanks for a detailed response, it is helpful. However, I need to burn 1% of the tokens on every transfer, so what you suggest shall I do when the 1% of the transfer tokens results into value less than 1 ? – Fariha Abbasi Jul 5 '19 at 7:38
• The code I suggested will save and accumulate all unburned token fractions from any transfer, and burn one more token each time those accumulated fractions amount to one full token. This should ensure that your token accounting is consistent over time, and fair to each user. – blackscale Jul 5 '19 at 8:01
• `value / 100` if you need to round result down,
• `(value + 99) / 100` to round up,
• `(value + 50) / 100` to round to nearest.
The result will be rounded to the 1/10^18 of the whole token, so precision will be quite good in all three cases.
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You are on page 1of 2
# A. Read the story. Answer the questions.
## 1 Where did Tom and Peter meet?
2 What did they want to play?
3 Did the boys play basketball?
Yesterday afternoon, I invited Peter to go to the
playground to play basketball with me. We met there
at 5.00 oclock. There were some older boys there
playing football. We wanted to join them, but they
didnt want to play with us. So Peter and I decided to
find some other place to play. We went to a nearby
park. When we arrived there we saw some kids
fighting. They were of our age. We tried to help, but
we couldnt. One of the boys even punched me so
hard that I fell down. Peter started to call for help.
There were some older kids in the park too, but they
didnt want to help. I somehow managed to get out of
the park and I saw my father. He didnt see me, so I
## E. Complete these sentences with the verbs in
brackets. Use the simple past. Write the
negatvie forms, too.
Example: Mum opened (open) the door for us.
Mum didnt open the door for us.
1 That event ____________ (happen) lon ago.
That event _______ _______ long ago.
2 They ____________ (walk) to school together
yesterday.
They _______ _______ to school together
yesterday.
3 Dad ____________ (try) to fix the light.
Dad _______ _______ to fix the light.
F. Complete the sentences rom the story.
1 We _____ there at five oclock.
2 Tom _____ his father.
## B. Complete the sentences from the story.
Grammar note
1 Tom _____ Peter to go to the playground with him.
2 We _____ to help, but we couldnt.
3 Peter _____ to call for help.
C. Complete the table with the verbs from the
story.
This is the past simple tense, affirmative
regular verbs. What do we add to regular verbs
to make simple past affirmative?
Most verbs + -ed
punch - _________
want - _________
start - _________
Verbs ending in e + -d
invite - _________
decide - _________
arrive - _________
manage - _________
Verbs ending y after a try - _________
consonant remove y
D. Read and compare. How do we make
negative forms?
1 Peter started to call for help.
1 a Peter didnt start to call for help.
2 I tried to help.
2 a I dint try to help.
*We use ___________ + infinitive (verb without ed
/ -d) to make the negative of the past simple of
regular verbs.
## These are irregular verbs. There is no rule to make
these verbs. You need to learn the form for each verb:
bend - bent
break broke
bring - brought
fall - fell
fly - flew
get got
go - went
hear heard
keep kept
lose lost
meet - met
sell - sold
shoot - shot
sleep slept
see - saw
take - took
tell - told
## G. Complete the sentences with the correct
simple past of the verbs in brackets.
1 I _____ (lose) my pen on the bus.
2 We _____ (sell) our car last week.
3 The baby _____ (sleep) right through the night.
4 Peter _____ (get) a watch for his birthday.
H. Past simple irregular verbs negative
forms.
a) Complete the sentence from the story.
I saw my father, but he ______ ______ me.
*To make the neagtive of the past simple tense of
irregular verbs, we use didnt and the infinitive
(first form) of the verb.
## b) Make these sentences negative.
Example:
He broke the window yesterday.
He didnt break the window yesterday .
1 I heard a noise in the night.
__________________________________.
2 He brought his pet mouse to school.
__________________________________.
3 My book fell off the desk.
__________________________________.
SIMPLE PAST QUESTIONS
## 3 A: _____________________ (your friends / want)
to help?
B: Yes, ___________
## D. Match the sentences to the picures.
1 The wicked Queen gave Snow White a poisoned
apple.
3 Pinocchios nose grew longer every time he told a
lie.
4 Dinosaurs lived millions of years ago.
a.
## A. Read these two sentences. How do we make
questions? Complete the rules.
1 He watched TV all day.
1 a Did he watch TV all day?
2 He went to London.
2 a Did he go to London?
*To make questions in the past simple we use _____
and the infinitive. We put Did ____________ the
subject.
B. Make questions.
b.
Example:
She tried to help.
Did she try to help
## 1 He fell off a horse yesterday.
_____________________________?
2 She bought a new computer last week.
_____________________________?
3 They invited me to their party.
_____________________________?
4 She studied French when she was in France.
_____________________________?
c.
1 - ___
2 - ___
3 - ___
## C. Make questions an give short answers.
Example:
A: _______________________ (you/see) him
yesterday?
B: Yes, ___________.
A: Did you see
B:Yes, I did
.
him yesterday?
## 1 A: _____________________ (Tom / break) that
window?
B: Yes, ___________ .
2 A: _____________________ (you and Sue / arrive)
yesterday?
B: No, ___________ .
## *We use the simple past tense to talk about things
that happened in the past. The simple past tense is
also used to talk about things that happened in stories
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1. Newbie
Join Date
Oct 2011
Posts
26
## syntactic function
Help me, please, to decide on the function of the words "to examine" and "to ptotect" in the sentence:
"The police has two major responsibilities: to examine the victim, and to protect all the evidence."
I believe they are predicate, but I was tought that infinitive cannon function as a predicate with few exceptions.
2. ## Re: syntactic function
Originally Posted by marin123
Help me, please, to decide on the function of the words "to examine" and "to ptotect" in the sentence:
"The police have two major responsibilities: to examine the victim, and to protect all the evidence."
I believe they are predicate, but I was taught that infinitive cannot function as a predicate with few exceptions.
Would it make it easier to understand if it were expressed it as:
"The two main functions of the police are to examine the victim and to protect all the evidence"?
3. Newbie
Join Date
Oct 2011
Posts
26
## Re: syntactic function
Surely that would be easier)))). But I have a fixed sentence, where I must define the syntactic function of the infinitives "to examine" and "to ptotect" in the given sequence, without any alternations
4. Member
Join Date
Sep 2010
Posts
268
## Re: syntactic function
police / has / responsibilities = to examine/ victim . . . and . . to protect/evidence."
These may be a predicate nominatives.
The use of the colon makes this structure unique.
Last edited by susiedq; 01-Mar-2012 at 15:21.
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Join Date
Oct 2011
Posts
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## Re: syntactic function
I see then an attribute here. what exect responsibilities? - responsibilities to examine and protect.
6. Newbie
Join Date
Feb 2012
Posts
27
## Re: syntactic function
No predicate. "two mayor responsibilities" is the object of 'have'. The coordinated infinitive phrases function as a restrictive apposition - it defines the object, names the object.
7. Newbie
Join Date
Feb 2012
Posts
27
## Re: syntactic function
Originally Posted by susiedq
police / has / responsibilities = to examine/ victim . . . and . . to protect/evidence."
These may be a predicate nominatives.
Absurd!
8. ## Re: syntactic function
Incidentally, the sentence is incorrect. It should be "The police have ..."
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# What is the probability of selecting a black card from a deck of cards?
## What is the probability of selecting a black card from a deck of cards?
Therefore, the probability of drawing a black card or a 6= (26+2)/52 = 28/52 = 7/13. What is the probability of getting a spade card or picture card, if a card is drawn from a deck of 52 playing cards?
## What is the probability of drawing a black ace from a deck of 52 cards?
A deck has 52 cards (without jokers). There are four aces, only two of them are black. So there are 2/52 black aces in a deck. Thus, 2 out of 52 is the probability of drawing a black ace from the deck.
## What is the probability of choosing an ace or a black card?
However, The probability of getting a black card or an ace [which we may denote as P(black or ace)] is not P(black) + P(ace) since the former is 28/52 (there are 26 black cards and 2 red aces) while the latter is 26/52 + 4/52.
READ: Is there a Costco in Laughlin?
## What is the probability that a card selected at random from a standard deck of 52 cards is an ace or a heart?
Ace or heart probability is 16/52=4/13. A deck of ordinary cards is shuffled and 13 cards are dealt.
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FFT Example
```import matplotlib.pyplot as plt
import plotly.plotly as py
import numpy as np
# Learn about API authentication here: https://plot.ly/python/getting-started
# Find your api_key here: https://plot.ly/settings/api
Fs = 150.0; # sampling rate
Ts = 1.0/Fs; # sampling interval
t = np.arange(0,1,Ts) # time vector
ff = 11.5 # frequency of the signal
y = np.sin(2*np.pi*ff*t)
n = len(y) # length of the signal
k = np.arange(n)
T = n/Fs
frq = k/T # two sides frequency range
frq = frq[range(n/2)] # one side frequency range
Y = np.fft.fft(y)/n # fft computing and normalization
Y = Y[range(n/2)]
fig, ax = plt.subplots(2, 1)
ax[0].plot(t,y)
ax[0].set_xlabel('Time')
ax[0].set_ylabel('Amplitude')
ax[1].plot(frq,abs(Y),'r') # plotting the spectrum
ax[1].set_xlabel('Freq (Hz)')
ax[1].set_ylabel('|Y(freq)|')
```
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MATLAB and Simulink resources for Arduino, LEGO, and Raspberry Pi
# Making bigger matrix with smaller matrixes elements.
Asked by Ayob on 21 Apr 2013
I'm assembling a bigger matrix with elements of other smaller matrixes. in my method I build a connecting vector for example C=[1 5 6 7 8 9] when in each element like for example C(2)=5, 2 is a local dimension in smaller matrix and 5 is a global dimension in main matrix which is assembled.
If I want to use easy code writing I should use this kind of coding:
``` %for i=1:size(C,2)
for j=1:size(C,2)
B(C(i),C(j))=A(i,j)```
but I'm trying to use Matlab special features in matrix calculations to put A with local dimension in B with global dimension. because Matlab features are fast and need less code writing.
Note: A and B are n*n and m*m matrixes.
Image Analyst on 21 Apr 2013
First of all, let's see if this is what you meant (since your code does not run as-is, and I tried to fix it but am not sure I got what you intended):
``` C=[1 5 6 7 8 9]
A = magic(9); % Some sample data.
for i=1:size(C, 1)
for j=1:size(C, 2)
B(C(i),C(j))=A(i,j);
end
end
B```
Results in command window:
```C =
1 5 6 7 8 9
B =
47 0 0 0 58 69 80 1 12
```
Ayob on 21 Apr 2013
Yes,you are right.But A and B are both matrixes not vectors.
Image Analyst on 21 Apr 2013
Please take note that A is a 2D 9 by 9 matrix. And B ended up as a 1D row vector because your loop over i only went from 1 to size(C,1), in other words from 1 to 1 so it did just one row. That was your code, not mine, though you commented it out. That's why I asked you if my attempts to fix your code were correct, which you said it was.
## Products
No products are associated with this question.
Answer by Matt J on 21 Apr 2013
` B(C,C)=A;`
Matt J on 21 Apr 2013
You're confusing people by writing
```%for i=1:size(C,2)
```
with the '%', as if you have commented this line out. Is there a loop over i or not? If yes, why do you have the '%'? And if i runs from 1 to size(C,2)=4 then A needs to have 4 rows in order for A(i,j) to be accessible when i=4.
Ayob on 21 Apr 2013
Clear and complete answer by Matt J.
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## Reflection: Developing a Conceptual Understanding Inequalities - Section 3: Instruction & Teacher Modeling
This reflection speaks to the importance of ensuring students understand the verbiage behind inequalities.
Developing a Conceptual Understanding: Inequalities In Words
# Inequalities
Unit 4: One-Step Equations & Inequalities
Lesson 8 of 15
## Big Idea: Inequalities NOT Equations! Understanding the meaning and significance of inequalities.
Print Lesson
4 teachers like this lesson
Standards:
Subject(s):
Math, greater than, Inequalities (Algebra), less than, Expressions (Algebra), greater than or equal to, less than or equal to
70 minutes
### Michelle Braggs
##### Similar Lessons
###### Algebraic Expressions and the Real-World
Big Idea: Expressions can be used to represent a mathematical or real-world problem in an abstract way using numbers and symbols to make meaning of and understand problems.
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Resources(18)
New Haven, CT
Environment: Urban
###### Show What You Know about Expressions, Equations, & Inequalities
6th Grade Math » Expressions, Equations, & Inequalities
Big Idea: What are the similarities and differences between expressions, equations, and inequalities? What do students understand? What gaps do they have in their understanding? Students review and take the quiz.
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Resources(12)
Somerville, MA
Environment: Urban
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• anonymous
Find equations for the horizontal or oblique asymptotes, if any, for each of the following rational functions. #1) F(x)=5/x-1 #2)F(x)= x^2-4/x+2 #3) F(x)= x+2/x^2-9 #4) F(x)=3x^2+1/x-2 Can you please show your work so I can understand how you got the answer. Thank you! I will definitely award medal
Mathematics
• Stacey Warren - Expert brainly.com
Hey! We 've verified this expert answer for you, click below to unlock the details :)
SOLVED
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Simulating propagation of separated wave modes in general anisotropic media, Part I: qP-wave propagators
Next: 1.2 Two-layer TI model Up: 1. Simulating propagation of Previous: 1. Simulating propagation of
### 1.1 Homogeneous VTI model
For comparison, we first appply the original anisotropic elastic wave equation to synthesize wavefields in a homogeneous VTI medium with weak anisotropy, in which , , , and . Figure 4a and 4b display the horizontal and vertical components of the displacement wavefields at 0.3 s. Then we try to simulate propagation of separated wave modes using the pseudo-pure-mode qP-wave equation (equation 22 in its 2D form). Figure 4c and 4d display the two components of the pseudo-pure-mode qP-wave fields, and Figure 4e displays their summation, i.e., the pseudo-pure-mode scalar qP-wave fields with weak residual qSV-wave energy. Compared with the theoretical wavefront curves (see Figure 4f) calculated on the base of group velocities and angles, pseudo-pure-mode scalar qP-wave fields have correct kinematics for both qP- and qSV-waves. We finally remove residual qSV-waves and get completely separated scalar qP-wave fields by applying the filtering to correct the projection deviation (Figure 4g).
Elasticx,Elasticz,PseudoPurePx,PseudoPurePz,PseudoPureP,WF,PseudoPureSepP
Figure 4.
Synthesized wavefields in a VTI medium with weak anisotropy: (a) x- and (b) z-components synthesized by original elastic wave equation; (c) x- and (d) z-components synthesized by pseudo-pure-mode qP-wave equation; (e) pseudo-pure-mode scalar qP-wave fields; (f) kinematics of qP- and qSV-waves; and (g) separated scalar qP-wave fields.
Then we consider wavefield modeling in a homogeneous VTI medium with strong anisotropy, in which , , , and . Figure 5 displays the wavefield snapshots at 0.3 s synthesized by using original elastic wave equation and pseudo-pure-mode qP-wave equation respectively. Note that the pseudo-pure-mode qP-wave fields still accurately represent the qP- and qSV-waves' kinematics. Although the residual qSV-wave energy becomes stronger when the strength of anisotropy increases, the filtering step still removes these residual qSV-waves effectively.
Elasticx,Elasticz,PseudoPurePx,PseudoPurePz,PseudoPureP,WF,PseudoPureSepP
Figure 5.
Synthesized wavefields in a VTI medium with strong anisotropy: (a) x- and (b) z-components synthesized by original elastic wave equation; (c) x- and (d) z-components synthesized by pseudo-pure-mode qP-wave equation; (e) pseudo-pure-mode scalar qP-wave fields; (f) kinematics of qP- and qSV-waves; and (g) separated scalar qP-wave fields.
Simulating propagation of separated wave modes in general anisotropic media, Part I: qP-wave propagators
Next: 1.2 Two-layer TI model Up: 1. Simulating propagation of Previous: 1. Simulating propagation of
2014-06-24
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# Solved – Using Kolmogorov–Smirnov test
I am trying to use the Kolmogorov–Smirnov test. I managed to calculate the difference between the empirical points and the theoretical distribution \$D\$ (following Wikipedia). But then I am a bit confused about the test:
1. Is the null hypothesis that the empirical data is or is not distributed according to the theoretical distribution?
2. How do I determine the critical level \$alpha\$ when I have \$D\$? I can use this class to calculate the Kolmogorov Smirnov distribution.
I think I have all the ingredients, but I am not sure how to put them together.
Contents
1) The null hypothesis is that the data is distributed according to the theoretical distribution.
2) Let $$N$$ be your sample size, $$D$$ be the observed value of the Kolmogorov-Smirnov test statistic, and define $$lambda = D(0.12 + sqrt{N} + 0.11 / sqrt{N})$$. Then the p-value for the test statistic is approximately:
$$Q = 2 sum_{j=1}^{infty}(-1)^{j-1}exp{-2j^2lambda^2}$$
Obviously you can't calculate the infinite sum, but if you sum over 100 values or so this will get you very, very, very close. This approximation is quite good even for small values of $$N$$, as low as 5 if I recall correctly, and gets better as $$N$$ increases. Note, however, that @whuber in comments proposes a better approach.
This is a perfectly reasonable alternative to the Shapiro-Wilk test I suggested in answer to your other question, by the way. Shapiro-Wilk is more powerful, but if your sample size is in the high hundreds, the Kolmogorov-Smirnov test will have quite a bit of power too.
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# Is there a “closed” form of sequence $u_{n+1}=\frac{u_n^2}{u_n+1}$
Let $u_0=1$ and $u_{n+1}=\frac{u_n^2}{u_n+1}, \forall n\in \mathbb{N}$.
a) Find the formula of $u_n$?
b) Calculate the limit $\displaystyle\varlimsup_{n\rightarrow \infty} (u_n)^{\frac{1}{n}}$.
-
In a), what is the meaning of the formula? Note that $x_n=1/u_n$ is such that $x_{n+1}=x_n^2+x_n$ and that quadratic functions are notoriously difficult to iterate, as witnessed by the Julia and Mandelbrot sets. – Did Feb 29 '12 at 17:40
For b), note that $u_{n}\leqslant2^{-2^{n-1}}$ for every $n\geqslant1$. – Did Feb 29 '12 at 17:41
It helps to look at $v_n = \frac{1}{u_n}$ instead. Then the recurrence becomes $v_0 = 1$ and $v_{n+1} = v_n (v_n + 1)$ which is easier to understand. – TMM Feb 29 '12 at 18:43
That $v_{n+1}=v_n(v_n+1)$ is wrong. It is $v_{n+1}=v_n+1/v_n$ – GEdgar Feb 29 '12 at 22:33
@GEdgar: Really? – Did Mar 2 '12 at 18:49
For a "formula"
Using the methods in the paper: http://www.fq.math.ca/Scanned/11-4/aho-a.pdf, and as mentioned in the comments to the OEIS sequence: https://oeis.org/A007018 I believe we can show that
$$\frac{1}{u_n} = \text{largest even integer less than } k^{2^n}$$ for $n \ge 1$.
where $k$ is a constant (definable as a limit of a sequence defined in terms of $\{u_n\}$, see that paper).
Basically, the methods in the paper can be used to show that
$$\frac{1}{u_n} \lt k^{2^n} \lt \frac{1}{u_n}+2$$
Since $\frac{1}{u_n}$ is an even integer for $n\ge 1$, the result follows.
-
Hint for b):
Show that $u_n\rightarrow0$ (it's easy to show that the sequence is positive, bounded below by 0, and is decreasing. So, it converges and its limit satisfies the formula $L=L^2/(L+1)$).
Then use the fact that, for a sequence $(a_n)$ of positive numbers, if $\lim\limits_{n\rightarrow\infty} {a_{n+1}\over a_n}$exists, then $\lim\limits_{n\rightarrow\infty} \root n\of{a_n}$ exists and the two limits are equal.
This may be a "sledgehammer", though...
Incidentally: The first few terms of the sequence sequence are $1, {1\over2}, {1\over2\cdot3}, {1\over 6\cdot 7}, {1\over 42\cdot43}, {1\over 1806\cdot 1807}, \ldots$. The denominators listed here, when entered into OEIS, gives an interesting result (see sequence A100016 at the bottom).
-
It's only seen by some as a sledgehammer because the theorem about the two limits is relatively unknown. I don't know why many people don't know it, considering people know of the Root and Ratio tests and this theorem is quite similar material. +1 by the way. – Ragib Zaman Feb 29 '12 at 17:25
Dear David Mitra. Please give me a proof of your lats statement. If $\displaystyle\lim_{n\rightarrow\infty}\frac{a_{n+1}}{a_n}$ exists, then $\displaystyle\lim_{n\rightarrow\infty}a_n^{\frac{1}{n}}$ exists and two limits are equal. – impartialmale Mar 17 '12 at 21:53
@impartialmale See Lemma 3 in these notes of Pete L. Clark's. I think the proof is also in Rudin's (baby) analysis text. You should be able to find a proof in most Real Analysis texts. – David Mitra Mar 17 '12 at 22:12
Taking Didier's comment, write in fact $z_n = (1/u_n)+(1/2)$ to get recurrence $$z_{n+1} = z_n^2+\frac{1}{4}$$ and then consult the literature about the Mandelbrot set. It is known that the recurrence $z_{n+1} = z_n^2+c$ has "closed form" (in a certain precise sense$^1$) if and only if $c=0$ or $c=-2$. Thus, in this case $c=1/4$ it has no "closed form".
$^1$Eremenko, page 663, in: L. Rubel, "Some research problems about algebraic differential equations, II" Ill. J. Math. 36 (1992) 659--680.
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Copyright © University of Cambridge. All rights reserved.
## 'Escape from Planet Earth' printed from http://nrich.maths.org/
### Show menu
Steven from City of Sunderland College sent us in a wonderful, complete solution to this problem, which we recommend reading in full to any budding problem solver. The main points are as follows:
The gravitational potential energy of a cannon ball of mass $m$ at a distance $r$ from the centre of a planet of mass $M$ is
$$V = -\frac{GMm}{r}\;.$$
The kinetic energy of a cannon ball launched at speed $v$ is
$$KE = \frac{1}{2}mv^2\;.$$
Suppose that a cannon ball just escapes the pull of a planet and makes it to infinity. At this point, both its potential and kinetic energies will be zero. Thus, the initial kinetic and potential energies must sum to zero. So, if launched from a planet of radius $R$ we must have
$$\frac{GMm}{R} = \frac{1}{2}mv^2\;.$$
This gives the escape velocity $v$ as
$$v =\sqrt{\frac{2GM}{R}}\;.$$
Putting in the numbers for Earth gives
$$v=\sqrt{\frac{2\times 6.674\times 10^{-11}\times 5.9763\times 10^{24}}{6.378\times 10^6}}=11.2\textrm{ km s}^{-1}\;.$$
For the moon, Jupiter and the sun the escape velocity changes by the relative change in the factor $\sqrt{\frac{M}{R}}$. For the moon, Jupiter and the sun these are $0.2122$, $5.32$ and $55.26$, giving rise to escape velocities of
Moon: $2.37\textrm{ km s}^{-1}$,
Jupiter: $59.5\textrm{ km s}^{-1}$,
Sun: $619\textrm{ km s}^{-1}$.
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# 15-344/Homework Assignment 7
(diff) ← Older revision | Latest revision (diff) | Newer revision → (diff)
This assignment is due at the tutorials on Thursday November 19. Here and everywhere, neatness counts!! You may be brilliant and you may mean just the right things, but if the teaching assistants will be having hard time deciphering your work they will give up and assume it is wrong.
Solve problems 8, 10, 15, 20, 24, 28, and 29 in section 5.3 and problems 1, 3, 16, 39, 40, 63, and 65 in section 5.4, but submit only your solutions of the underlined problems.
Dror's notes above / Students' notes below
Homework Assignment 7 Solution
15 How many 8-digit sequences are there involving exactly six different digits?
A) We have 2 cases: 1 digit repeated three times (for example, 11123456) and 2 digits repeated (for example, 11223456)
For the first case, there are $\binom{10}{6}$ ways to choose 6 digits from 10 digit choice. Then there are $\frac{8!}{3!}$ arrangements involving a triple digit and and $\binom{6}{1}$ different triple digits possible (111,222 and so on).So, in total, we have: $\binom{10}{6} \binom{6}{1}\frac{8!}{3!}$
Similarly, for the second case, we have: $\binom{10}{6} \binom{6}{2}\frac{8!}{2!2!}$
The sum of the two cases gives the answer:
$\binom{10}{6}(\binom{6}{1}\frac{8!}{3!}+ \binom{6}{2}\frac{8!}{2!2!})$
24 How many arrangements of the word PREPOSTEROUS are there in which the five vowels are consecutive?
A)
We have P-2, R-2, E-2, O-2, S-2, T-1,U-1
The vowels are OOUEE and its variations.
The number of all the possible variations of OOUEE is: $\frac{5!}{2!2!}$
The remaining 7 letters can be arranged in $\frac{7!}{2!2!2!}$ ways.
Also, there are 8 different positions for the vowel group in the word. For example, OOUEE_ _ _ _ _ _ _ and _ _ _ _ _ _ _ OOUEE
In total, $8\frac{5!}{4}\frac{7!}{8} = 151200$
28 How many ways are there to place nine different rings on the four fingers of your right hand if a) the order of the rings on a finger does not matter b) the order of the rings on a finger is considered
A) a) For each ring, we have a choice of being on 1 - 4th finger. So, we have $4^9$ possible such ways.
A) b) First, without considering the different types of the rings, consider the distribution of the nine rings to four fingers: $\binom{n+k-1}{k-1} = \binom{9+4-1}{4-1} = \binom{12}{3}$ If we consider the fact that the rings are all different, we have the permutations $9!$ So, in total, $9!\binom{12}{3}$ ways.
39 How many non-negative integer solutions are there to the equation $2x_{1} + 2x_{2} + x_{3} + x_{4} = 12$.
A) This can be viewed as a special type of distribution problems.
First we think of all the possible pairs of $x_{1}, x_{2}$. For example:
$x_{1} = 0, x_{2} = 0 \to x_{3}+x_{4} = 12$
$x_{1} = 1, x_{2} = 0 \to x_{3}+x_{4} = 10$ (two such cases since we can interchange the role of x1 and x2)
and so on.
In total, we have:
$\binom{12+2-1}{2-1} + 2\binom{10+2-1}{2-1} + 3\binom{8+2-1}{2-1} ... + 7\binom{0+2-1}{2-1} = 140 ways$
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Filters
138 Hits in 1.8 sec
### On a special class of boxicity 2 graphs [article]
Sujoy Kumar Bhore, Dibyayan Chakraborty, Sandip Das, Sagnik Sen
2016 arXiv pre-print
We define and study a class of graphs, called 2-stab interval graphs (2SIG), with boxicity 2 which properly contains the class of interval graphs. ... We provide a matrix characterization for a subclass of 2SIG graph. ... Moreover, our graph class is based on local structures of boxicity 2 graphs in some sense. Thus, the study of this class may help us understand the structure of boxicity 2 graphs in a better way. ...
### Poset boxicity of graphs
W.T Trotter, Douglas B West
1987 Discrete Mathematics
In this paper, a special class of posets is used to show that the poset boxicity of a graph on n points is at most O(log log n). ... Seheinerman defined the poset boxicity of a graph G to be the minimum t such that G is the intersection graph of intervals in some t-dimensional poset. ... First, we use a special class of posets to show that the poset boxicity of a graph on n points is always at most O(log log n). consider Q. ...
### Boxicity and Maximum degree [article]
L. Sunil Chandran, Mathew C. Francis, Naveen Sivadasan
2006 arXiv pre-print
The concept of boxicity finds applications in various areas such as ecology, operation research etc. We show that for any graph G with maximum degree Δ, (G) < 2 Δ^2 + 2. ... For a graph G, its boxicity (G) is the minimum dimension d, such that G is representable as the intersection graph of (axis--parallel) boxes in d--dimensional space. ... In fact, given any ∆, it is not difficult to construct graphs of boxicity Ω(∆) on arbitrarily large number of vertices, using a construction given by Roberts [10] . ...
### Boxicity and maximum degree
L. Sunil Chandran, Mathew C. Francis, Naveen Sivadasan
2008 Journal of combinatorial theory. Series B (Print)
The concept of boxicity finds applications in various areas such as ecology, operation research etc. We show that for any graph G with maximum degree ∆, box(G) ≤ 22 + 2. ... For a graph G, its boxicity box(G) is the minimum dimension d, such that G is representable as the intersection graph of (axis-parallel) boxes in d-dimensional space. ... There have been many attempts to estimate or bound the boxicity of graph classes with special structure. In his pioneering work, F. S. ...
### Boxicity and Treewidth [article]
2005 arXiv pre-print
In particular, we show that, if the boxicity of a graph is b >= 3, then there exists a simple cycle of length at least b-3 as well as an induced cycle of length at least floor of (log(b-2) to the base ... Our result leads to various interesting consequences, like bounding the boxicity of many well known graph classes, such as chordal graphs, circular arc graphs, AT-free graphs, co--comparability graphs ... Consequences on Special Classes of Graphs Chordal Graphs Let C be a cycle in a graph G. A chord of C is an edge of G joining two nodes of C which are not consecutive. ...
### An upper bound for Cubicity in terms of Boxicity
L. Sunil Chandran, K. Ashik Mathew
2009 Discrete Mathematics
The boxicity of any graph G, box(G) is the minimum positive integer b such that G can be represented as the intersection graph of axis-parallel on the real line. ... Any graph of treewidth tw has cubicity at most (tw + 2) log 2 n . ... A prime example of a graph class defined in this way is the class of interval graphs. ...
### Characterization of the graphs with boxicity ⩽2
Martin Quest, Gerd Wegner
1990 Discrete Mathematics
In this paper we will give a combinatorial characterization of the graphs with b(G)s2, called boxicity 2-graphs, by means of the arrangement of zeros and ones in special matrices attached to the graph. ... Following Roberts [4] the boxicity b(G) of a graph G is defined as the smallest d such that G is the intersection graph of boxes in Euclidean d-space, i.e. parallelepipeds with edges parallel to the coordinate ... Note, that the class of boxicity 2-graphs contains the class of interval graphs. In the following the set of vertices of a graph G is called V(G), the set of edges E(G). ...
### An upper bound for Cubicity in terms of Boxicity [article]
L. Sunil Chandran, K. Ashik Mathew
2006 arXiv pre-print
The boxicity of any graph G, box(G) is the minimum positive integer b such that G can be represented as the intersection graph of axis parallel b-dimensional boxes. ... A b-dimensional cube is a Cartesian product R_1 × R_2× ... × R_b, where each R_i (for 1 ≤ i ≤ b) is a closed interval of the form [a_i,a_i+1] on the real line. ... A prime example of a graph class defined in this way is the class of interval graphs. Definition 1. ...
### Parameterized and Approximation Algorithms for Boxicity [article]
Abhijin Adiga and Jasine Babu and L. Sunil Chandran
2014 arXiv pre-print
Extending the same idea in one of our algorithms, we also get an O(n√( n)/√( n)) factor approximation algorithm for computing boxicity and an O(n ( n)^3/2/√( n)) factor approximation algorithm for computing ... Boxicity of a graph G(V, E), denoted by box(G), is the minimum integer k such that G can be represented as the intersection graph of axis parallel boxes in R^k. ... FPT Algorithm to Compute Boxicity of F + k 1 e − k 2 e Graphs In this section, we give a proof of Theorem 2. Let G(V, E) be a F + k 1 e − k 2 e graph on n vertices, where k 1 + k 2 = k. ...
### Geometric Representation of Graphs in Low Dimension Using Axis Parallel Boxes
L. Sunil Chandran, Mathew C. Francis, Naveen Sivadasan
2008 Algorithmica
In most cases, the first step usually is computing a low dimensional box representation of the given graph. Deciding whether the boxicity of a graph is at most 2 itself is NP-hard. ... A number of NP-hard problems are either polynomial time solvable or have much better approximation ratio on low boxicity graphs. ... Acknowledgement We thank the anonymous referee for carefully reading the paper and pointing out a serious mistake in the last section of the first version of the paper. ...
### The cubicity of hypercube graphs
2008 Discrete Mathematics
The parameter boxicity generalizes cubicity: the boxicity box(G) of a graph G is defined as the minimum dimension k such that G is representable as the intersection graph of axis-parallel boxes in k-dimensional ... For a graph G, its cubicity cub(G) is the minimum dimension k such that G is representable as the intersection graph of (axis-parallel) cubes in k-dimensional space. on the real line.) ... Acknowledgement The first author's work was partially supported by a DST grant SR/S3/EECE/62/2006. ...
### Geometric Representation of Graphs in Low Dimension [chapter]
2006 Lecture Notes in Computer Science
In most cases, the first step usually is computing a low dimensional box representation of the given graph. Deciding whether the boxicity of a graph is at most 2 itself is NP-hard. ... A number of NP-hard problems are either polynomial time solvable or have much better approximation ratio on low boxicity graphs. ... Researchers have also tried to bound the boxicity of graph classes with special structure. Scheinerman [18] showed that the boxicity of outer planar graphs is at most 2. ...
### On the Boxicity and Cubicity of Hypercubes [article]
2006 arXiv pre-print
In this paper, we show that cub(H_d) = Θ(d/ d).The parameter boxicity generalizes cubicity: the boxicity box(G) of a graph G is defined as the minimum dimension k such that G is representable as the intersection ... For a graph G, its cubicity cub(G) is the minimum dimension k such that G is representable as the intersection graph of (axis--parallel) cubes in k--dimensional space. ... There have also been attempts to estimate or bound the boxicity of graph classes with special structure. Scheinerman [13] showed that the boxicity of outer planar graphs is at most 2. ...
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# bzoj 1832: [AHOI2008]聚会
1 #include<bits/stdc++.h>
2 #define N 500005
3 #define M 10000005
4 #define LL long long
5 #define inf 0x3f3f3f3f
6 using namespace std;
7 inline int ra()
8 {
9 int x=0,f=1; char ch=getchar();
10 while (ch<'0' || ch>'9') {if (ch=='-') f=-1; ch=getchar();}
11 while (ch>='0' && ch<='9') {x=x*10+ch-'0'; ch=getchar();}
12 return x*f;
13 }
14 struct node{
15 int to,next;
16 }e[N<<1];
18 int fa[N][20],deep[N],n,m;
20 void dfs(int x)
21 {
22 for (int i=1; i<=19; i++)
23 if (deep[x]>=(1<<i))
24 fa[x][i]=fa[fa[x][i-1]][i-1];
25 else break;
27 {
28 if (e[i].to==fa[x][0]) continue;
29 fa[e[i].to][0]=x;
30 deep[e[i].to]=deep[x]+1;
31 dfs(e[i].to);
32 }
33 }
34 int lca(int x, int y)
35 {
36 if (deep[x]<deep[y]) swap(x,y);
37 int t=deep[x]-deep[y];
38 for (int i=0; (1<<i)<=t ; i++)
39 if (t&(1<<i)) x=fa[x][i];
40 for (int i=19; i>=0; i--)
41 if (fa[x][i]!=fa[y][i])
42 x=fa[x][i],y=fa[y][i];
43 if (x==y) return x;
44 return fa[x][0];
45 }
46 int lca_cost(int x, int y)
47 {
48 int sum=0;
49 if (deep[x]<deep[y]) swap(x,y);
50 int t=deep[x]-deep[y];
51 for (int i=0; (1<<i)<=t ; i++)
52 if (t&(1<<i)) x=fa[x][i],sum+=(1<<i);
53 for (int i=19; i>=0; i--)
54 if (fa[x][i]!=fa[y][i])
55 x=fa[x][i],y=fa[y][i],sum+=(1<<i)*2;
56 if (x==y) return sum;
57 return sum+2;
58 }
59 int main()
60 {
61 n=ra(); m=ra();
62 for (int i=1; i<n; i++)
63 {
64 int x=ra(),y=ra();
65 insert(x,y); insert(y,x);
66 }
67 dfs(1);
68 for (int i=1; i<=m; i++)
69 {
70 int x=ra(),y=ra(),z=ra();
71 int ans1=lca_cost(x,y)+lca_cost(lca(x,y),z);
72 int ans2=lca_cost(x,z)+lca_cost(lca(x,z),y);
73 int ans3=lca_cost(y,z)+lca_cost(lca(y,z),x);
74 int ans=min(min(ans1,ans2),ans3);
75 if (ans==ans1) printf("%d ",lca(x,y));
76 else if (ans==ans2) printf("%d ",lca(x,z));
77 else if (ans==ans3) printf("%d ",lca(y,z));
78 printf("%d\n",ans);
79 }
80 }
posted @ 2017-02-23 20:26 ws_ccd 阅读(52) 评论(0编辑 收藏
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ummm could you please be more specific?
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## Thursday, November 5, 2015
### Multiple Representations: Midpoint and Endpoint Socrative Practice
One value I push and is a common thread throughout my lessons is the idea of multiple representations. I think, far too much in education, we teach kids "the easiest way", which may be the easiest way for you, or for most kids, but it could very well be the most difficult method for others. I make every attempt to analyze math topics from multiple perspectives.
If I am a strong algebra student/equation solver, how would I best understand this concept?
If I am a strong visual student/grapher, how would I best understand this concept?
If I am a strong tactile student/manipulative-user, how would I best understand this concept?
I know lots of philosophies put emphasis on catering to the different types of learners (visual, kinesthetic, auditory), but that's not exactly what I'm trying to do here. You see, I am an auditory learner - very auditory. In fact, a specialist "tested" me and said I was the highest auditory result he'd ever seen. Evidently I'm weird - nobody that knows me well is shocked. But you see, it's a beautiful thing. In a way, we are all weird. We all learn differently. Some of us are about the minutiae - detail people. Some of us are big picture visionaries. How do these types best learn? I don't think we necessarily can have a perfectly ironed-out answer here. However, we should give students choices in how they can approach a problem
When I taught midpoint and endpoint this year, I stressed using reasoning and these multiple approaches to arrive at an answer. Students were taught how to find midpoint and endpoint (1) algebraically, (2) graphically, and (3) on a number line with marker manipulatives. Some kids heavily preferred the algebraic approach. Some initially avoided the algebra like the plague. The great thing is - by the end of the lesson set - students saw how all the methods were interwoven. Because of students' strength in one approach and the interconnectivity, these initial strengths eventually translated into a gradual strengthening of their approaches in other, more weak, methods.
To facilitate this type of practice, I used Socrative (I'm a huge Socrative fan) and a strategically assembled worksheet.
Socrative Code (if you'd like to use this activity): SOC-18350734
Here is the handout. For each problem kids had to (1) specify if the problem was an endpoint or midpoint problem (I know this is obvious, but this question seems to help the kids focus), (2) prove the calculation algebraically, (3) prove the solution on number lines, and (4) prove the solution on the coordinate plane.
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Tarzan swinging on a rope -- find the acceleration
• isukatphysics69
In summary,Tarzan swings from a cliff at the end of a 20.0m vine that hangs from a high tree limb. Immediately after Tarzan steps off the cliff, the tension in the vine is 719N and makes an angle of 29.0deg with the vertical. Tarzan's weight is 822N.The magnitude of Tarzan's acceleration immediately after he steps off the cliff is 0.3m/s^2.
isukatphysics69
Homework Statement
Tarzan swings from a cliff at the end of a 20.0m vine that hangs from a high tree limb. Immediately after Tarzan steps off the cliff, the tension in the vine is 719N and makes an angle of 29.0deg with the vertical. Tarzan's weight is 822N.
What is the magnitude of Tarzan's acceleration immediately after he steps off the cliff?
netforce = m*a
The Attempt at a Solution
netforce =m*a
netforcex = 719cos71
netforcey = 719sin71-822
netforce= sqrt((719cos71)^2+(719sin71-822)^2))
a = 273.88/822
a=0.3m/s^2not sure what i am doing wrong here! >=[
isukatphysics69 said:
a = 273.88/822
Weight is not the same as mass.
In general, always write out and check your units in all steps. It will help you identify mistakes such as, in this case, getting a dimensionless acceleration.
Orodruin said:
Weight is not the same as mass.
In general, always write out and check your units in all steps. It will help you identify mistakes such as, in this case, getting a dimensionless acceleration.
i see i made a stupid mistake there i meant 273/83.87 now i am getting 3.2 and its still wrong
isukatphysics69 said:
netforce= sqrt((719cos71)^2+(719sin71-822)^2))
What is the result of this computation?
Orodruin said:
What is the result of this computation?
273.8758934
isukatphysics69 said:
273.8758934
I suggest you redo that math.
Edit: Also, again, units! The units here should be Newton. Otherwise it is not a force.
Orodruin said:
I suggest you redo that math.
Edit: Also, again, units! The units here should be Newton. Otherwise it is not a force.
I put it into the calculator and got the same thing, i have the calculator on degree mode do you have yours on radian?
isukatphysics69 said:
I put it into the calculator and got the same thing, i have the calculator on degree mode do you have yours on radian?
What is 90-29?
Orodruin said:
What is 90-29?
omg i am so STUPID! >=[
So the correct solution is what? There is an easier way of solving this problem, but full solutions are not allowed until the OP shows that they have solved the problem so I cannot tell you until you show your full correct solution.
Orodruin said:
So the correct solution is what? There is an easier way of solving this problem, but full solutions are not allowed until the OP shows that they have solved the problem so I cannot tell you until you show your full correct solution.
netforce= sqrt((719cos61)^2+(719sin61-822)^2)) = 18.21643467
Edit: this is wrong also, there is something fundamental i am doing wrong here. i think i am supposed to be using v^2/R
isukatphysics69 said:
netforce= sqrt((719cos71)^2+(719sin71-822)^2)) = 18.21643467
Edit: this is wrong also, there is something fundamental i am doing wrong here. i think i am supposed to be using v^2/R
First of all, that is still wrong and it still has 71 in it so it is unclear how you changed it. Second, just as he steps off the ledge, his velocity is zero and all acceleration is tangential. The radial acceleration (which is v^2/r) is zero.
Orodruin said:
First of all, that is still wrong and it still has 71 in it so it is unclear how you changed it. Second, just as he steps off the ledge, his velocity is zero and all acceleration is tangential. The radial acceleration (which is v^2/r) is zero.
Sorry i am so tired i will try again and come back in 10 minutes i am screwing up big time on this problem and in LIFE! BRB
I have no idea what i am doing wrong here whose going to be online tomorrow?? my school is closed again because of the storm and there's no tutoring and my stupid self needs help! >=[ i will get my phd in physics one day tarzan not going to stop me!
Your original idea is workable. You are just not putting numbers into your calculator correctly. Double check everything.
isukatphysics69
I FIGURED IT OUT ITS 4.75 M/S^2!
I FIGURED IT OUT ITS 4.75 M/S^2!
Now i am trying to find the angle of acceleration with the vine... i did tan^-1(-193/348) and got -29 degrees but marked wrong what am i doing wrong here
FIGURED IT OUT I WAS DOING ANGLE WITH X AXIS NOT VINE NOW I UNDERSTAND ITS -90 DEGREES I AM SO STUPID
Last edited:
What is -193 and 348? Consequently, what is the angle that you are computing?
Orodruin said:
What is -193 and 348? Consequently, what is the angle that you are computing?
I edit post bro see above now I'm working on another problem thank you tho
So here is the easier way of doing things.
Completely ignore the given forces, they are essentially useless for the solution to the problem. Since the vine has a fixed length, the motion must be along a circle and the vine takes up any force in the radial direction. This means that any force component in the radial direction will be canceled by the tension in the vine. Only the force component in the tangential direction remains and it is equal to the mass multiplied by the projection of the gravitational acceleration on the tangential direction. Since the acceleration is force/mass, the acceleration will just be the projection of the gravitational acceleration onto the tangential direction, i.e., ##g\sin(29^\circ) = 4.75~{\rm m/s}^2##.
Since the vine is radial and the motion tangential, the angle between them must be a right angle.
isukatphysics69
1. What is the formula for calculating acceleration?
The formula for acceleration is a = (vf - vi) / t, where a is acceleration, vf is final velocity, vi is initial velocity, and t is time.
2. How is acceleration related to Tarzan swinging on a rope?
Acceleration is the change in velocity over time. In the case of Tarzan swinging on a rope, his velocity changes as he swings back and forth, so acceleration is a key factor in his movement.
3. Can you measure Tarzan's acceleration while he is swinging on a rope?
Yes, Tarzan's acceleration can be measured using a device called an accelerometer. This device can track changes in velocity and calculate acceleration.
4. Is Tarzan's acceleration constant while swinging on a rope?
No, Tarzan's acceleration is not constant while swinging on a rope. It changes as he moves back and forth, and can also be affected by external factors such as wind resistance and the weight of his body.
5. How does Tarzan's acceleration affect his speed while swinging on a rope?
Tarzan's acceleration directly affects his speed while swinging on a rope. The greater the acceleration, the faster he will swing. When his acceleration is zero, he will reach the top of his swing and momentarily stop before changing direction.
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# Can Money Be Tight If The Fed Does Nothing?
There seems to be a lot of debate in the blogosphere about how to evaluate whether monetary policy is too tight. Specifically, Arnold Kling notes:
According to Sumner, we know that the Fed tightened because we know that real interest rates on TIPS rose. The real rate is the nominal rate minus the expected rate of inflation. Because the nominal rate did not rise, the increase in the real rate came from a big drop in expected inflation. So we know that the Fed tightened because expectation inflation dropped, and we know that expected inflation dropped because the Fed tightened. That’s too circular for my taste.
Scott and I hold somewhat similar views, but I don’t want to focus on whether Scott’s logic is correct; rather I want to focus on the first sentence of Arnold’s comment. Specifically, I want to isolate the phrase “we know that the Fed tightened”. The phrase is important because it asks whether the relative tightness of monetary policy requires that the Fed’s behavior has changed. I intend to argue that monetary policy can be too tight even if the Fed does nothing.
Recall the equation of exchange:
MV = Py
where M is the money supply, V is velocity, P is the price level, and y is real output. Now, let’s make the modification that I like to use by viewing the money supply as the product of the monetary base (B) and the money multiplier (m):
mBV = Py
Now let’s suppose that the Federal Reserve is targeting nominal income. Assuming that the money multiplier and velocity are constant, the Fed can adjust the monetary base to target nominal income. Another way of thinking about this is that movements in the money supply offset corresponding movements in velocity such that nominal income remains stable. Thus, the policy can be seen as one that seeks to achieve monetary equilibrium (reductions in velocity — increases in money demand — are offset by increases in the money supply). Given the severe informational problems with predicting the behavior of velocity and the money multiplier, the nominal income target serves as a measuring stick to which monetary equilibrium is being maintained.
Given this framework, consider what happens to the money multiplier and to velocity when Lehman Brothers fails and Paulson and Bernanke testify to Congress that the situation is dire. Individuals restrain from consumption and thus velocity falls. In addition, fears about the banking system cause banks to increase their demand for reserves and individuals to increase their demand for currency relative to deposits; each of which reduce the money multiplier. In this case, with m and V falling, nominal income will fall as a result of falling velocity and multiple deposit destruction (thereby maintaining the equality).
Note that the Fed has really done nothing in this instance. However, I would consider monetary policy to be tight because the Fed could (should) increase the monetary base to offset the changes in velocity and the money multiplier thereby stabilizing nominal income. In other words, the Fed could increase currency and bank reserves into the system to meet the corresponding increase in demand.
In actuality, this is precisely what happened beginning in September 2008 when Lehman collapsed and Paulson and Bernanke gave their testimony. David Beckworth’s excellent graph shows that the sudden drop off in the money multiplier and velocity corresponding to this period.
Thus, it would seem that monetary policy became tight without the Fed doing anything. (Of course, not to be outdone by secular forces, the Fed did start paying interest on bank reserves — thereby increasing the demand for bank reserves and reducing the money multiplier — in October).
Critics of my view would argue that the monetary base increased significantly in the aftermath of September 2008. However, as I wrote back in July:
Ultimately, the money multiplier (M1) has fallen from around 1.6 prior to the recession to .93 as of June 17. At the beginning of January 2008, the monetary base was roughly \$848 billion. Given that money multiplier, this would suggest that M1 was around \$1.356 trillion. Thus, given the current money multiplier, this would suggest that the monetary base would have to be about \$1.458 trillion today to maintain the same money supply — an increase of roughly 72%.
Of course, the 72% increase would only be sufficient keep the money supply constant and would not offset any change in velocity. In actuality, we would want the money supply to grow to increase nominal income (even if we were targeting a rate of deflation consistent with productivity growth).
Ultimately, targeting nominal income is an attempt to maintain monetary equilibrium. In this view, nominal income is the metric by which to measure whether monetary policy is too tight (or loose). What’s more, this view suggests that monetary policy can be too tight even if the Fed doesn’t “tighten.” Given the utter collapse of nominal income that characterized the crisis (nominal income growth actually became negative), I would argue that monetary policy was/is too tight and that this claim is consistent with the view outlined above.
[NOTE: For more on this view, see my earlier post.]
### 8 responses to “Can Money Be Tight If The Fed Does Nothing?”
1. Josh:
If the Fed issues 100 paper dollars to anyone who offers \$100 worth of bonds in exchange, then the fed is not being tight. It is being neutral. If the fed issues \$101 for that bond, then it is being too easy, and if the fed issues only \$99, it is being too tight.
Under a free banking system, competitive banks would always stand ready to issue \$100 in exchange for \$100 worth of stuff. The quantity of money would be dictated by the invisible hand, and not by economists fiddling with tautological equations. We can only hope the fed will try to emulate a free banking system.
2. Your statements “Bernanke gave their testimony” and “monetary policy became tight without the Fed doing anything” contradict themselves. :-)
3. sumnerbentley
Josh, Nice post. I would add that by Kling’s logic money was easy in the early 1930s because the Fed cut rates sharply, and money was tight during the German hyperinflation. So nominal rates simply cannot be correct indicator of whether policy has been loosened or tightened.
4. nice post, good to see a pretty refreshing POV
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https://support.office.com/en-us/article/N-function-a624cad1-3635-4208-b54a-29733d1278c9?CTT=5&origin=HP010079186&CorrelationId=881ee48e-296c-4c2c-bf41-84ba307071e5&ui=en-US&rs=en-US&ad=US
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N function
# N function
This article describes the formula syntax and usage of the N function in Microsoft Excel.
## Description
Returns a value converted to a number.
## Syntax
N(value)
The N function syntax has the following arguments:
• Value Required. The value you want converted. N converts values listed in the following table.
If value is or refers to N returns A number That number A date, in one of the built-in date formats available in Microsoft Excel The serial number of that date TRUE 1 FALSE 0 An error value, such as #DIV/0! The error value Anything else 0
## Remarks
• It is not generally necessary to use the N function in a formula, because Excel automatically converts values as necessary. This function is provided for compatibility with other spreadsheet programs.
• Excel stores dates as sequential serial numbers so they can be used in calculations. By default, January 1, 1900 is serial number 1, and January 1, 2008 is serial number 39448 because it is 39,448 days after January 1, 1900.
## Example
Copy the example data in the following table, and paste it in cell A1 of a new Excel worksheet. For formulas to show results, select them, press F2, and then press Enter. If you need to, you can adjust the column widths to see all the data.
Data 7 Even TRUE 4/17/2011 Formula Description Result =N(A2) Because A2 contains a number, the number is returned. 7 =N(A3) Because A3 contains text, 0 is returned. 0 =N(A4) Because A4 is the logical value TRUE, 1 is returned. 1 =N(A5) Because A5 is a date, the serial number of the date is returned (varies with the date system used). 40650 =N("7") Because "7" is text, 0 is returned. 0
Applies To: Excel 2013, Excel 2007, Excel 2010, Excel Starter, Excel Online, Excel 2016 for Mac
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Vous êtes sur la page 1sur 13
# Lecture No.
9
MEASUREMENT OF FLUID FLOW
9.1 General types o lo! "eters
Obstruction meter. Fluid meters that belong to this type indicate a flow
by a change in pressure.
Full bore meter (Venturi meter, orifice meter, flow nozzle)
Insertion meter (Pitot tube)
Variable area meter. An area meter is one in which the pressure drop is
constant and the reading is dependent upon a ariable flow area.
!otameter
Others" #agnetic field, orte\$ shedding, turbine meter etc.
9.# \$entur% "eter
Venturi meter is a full%bore obstruction meter with conerging and dierging
sections. It consists of a tube with a constricted throat which causes the
increase of fluid elocity at the e\$pense of pressure. &he throat is followed
by a gradually dierging section where the elocity is decreased with an
increase in pressure and slight friction losses. A schematic of Venturi meter
is shown in Figure '.(.
Figure '.(. )chematic of a Venturi meter
*(
\$eloc%ty at t&e t&roat 'u
#
()
( )
c
+ c
, ,
+ P g -. -.
u +g /
( (
('.()
Mass lo! rate 'm
0( an* +eloc%ty %n t&e "a%n l%ne 'u
1
()
( ( ( + + +
m u A u A
0
('.+)
For incompressible fluid flow"
+ +
( + +
(
A
u u
A
('.1)
where"
.
## is the Venturi coefficient2
3
+
(
diameter of throat 4
diameter of main line 4
2
%P is the pressure drop, P
(
5 P
+
2
is the density of the fluid2
/ is the differential head, (
P
/
)2
- is a correction factor.
I"portant notes)
If 6 7.+8, then
,
( ( .
Values of .
"
.
.
7.'9 for 4
(
3 + to 9 inches
.
7.'' for 4
(
: 9 inches
.
## 3 7.'9, for /erschel type with !e : +77,777
*+
Values of -"
For incompressible fluids (i.e., li;uids)" - 3 (
For compressible fluids (i.e., gases)"
( )
( )
(
, +
(
(
+
+
(
, + +
( (
P
( (
P
P
-
P
P P
( ( (
P P
1
_
1
1
,
_ 1
]
1
,
_ _
1
1
, ,
1
]
('.,)
where"
p
.
.
.
)ee also E,uat%on 1-.#1 and F%/ure 1-.10 of Perry<s .h= /andboo>
(*
th
ed.)
&he permanent pressure loss across Venturi meter is dependent on
and the discharge cone angle () in the dierging section"
3 8 5 *
o
(7 to (8? of %P
: (8
o
(7 to 17? of %P
9.1 Or%%ce "eter
A sharp% or s;uare%edged orifice is a clean%cut s;uare%edged hole with
straight walls perpendicular to the flat upstream face of a thin plate faced
crosswise of a channel.&he stream issuing from such an orifice attains its
minimum cross section (vena contracta) at a distance downstream of the
orifice which aries with .
*1
Figure '.+. )chematic of a sharp%edged orifice
\$eloc%ty at t&e or%%ce)
( )
c o o
+ c
, ,
+ P g -. -.
u +g /
( (
('.8)
where" .
o
is the orifice discharge coefficient
- is a correction factor.
@ther parameters are as defined in )ection '.+.
I"portant notes)
Values of .
o
3 f(!e, )"
+.8
7.+( 9
o 7.*8
. 7.8'8' 7.71(+ 7.(9, '(.*(
!e
+ + ('.A)
=;uation '.A has been plotted to gie F%/ure 9.1 below (or F%/ure
1-.#- of Perry<s .h= /andboo>, *
th
ed.)
*,
Figure '.1. .oefficient of discharge for s;uare%edged circular (Figure
(7%+7 of Perry<s .h= /andboo>, *
th
ed.)
Values of -"
For compressible fluid (i.e., gases)"
,
+
(
7.,( 7.18 P
- ( (
P
_ +
,
('.*)
)ee also F%/ure 1-.10 of Perry<s .h= /andboo> (*
th
ed.)
9.2 3%tot tu4e
A pitot tube is an insertion meter that measures local elocity (i.e., elocity at
a point). Applications include finding the elocity of a moing craft such as a
boat or an airplane. A similar deice, called Pitot static tube shown in Figure
'.,, is used to measure the elocity at different radial positions in a pipe.
*8
A
B
Figure '.,. Pitot tube with sidewall static tap (Fig. (7%8 Perry<s .h=
/andboo>, *
th
ed.)
\$eloc%ty at A '%.e.5 !&ere t&e t%p %s locate*(5 u
o
)
o c
u . +g / ('.9)
where"
. is the correction coefficient
I"portant notes)
Values of ."
. 3 ( t 7.7( for simple Pitot tubes
. 3 7.'9 to ( for Pitot static tubes
For gases at elocities : +77 ft s
%(
"
(
B A
o c
B B
P P
u . +g (
( P
1
_
1
1
,
1
]
('.')
Cenerally, for compressible fluids"
( )
c A B
o
+
,
o
B
g P P
u
#a +
( #a ...
, +,
_
+ + +
,
('.(7)
For =;uations '.' and '.(7, P
A
and P
B
are impact and static pressures,
respectiely.
*A
If the Pitot static tube is inserted at the middle of a tube or pipe, the
elocity measured is ma\$imum (u
o
3 u
ma\$
). &o determine the aerage
elocity in the tube or pipe, we use Figure '.8.
Figure '.8. Velocity ratio ersus !eynolds number for smooth circular
pipes (Figure (7%* of Perry<s .h= /andboo>, *
th
ed.)
9.6 Rota"eter
&he rotameter (Figure '.A) consists of a solid float or plummet that is free to
moe inside a gradually tapered ertical glass tube. &he fluid flows upward
and the flow rate is indicated by the e;uilibrium position reached by the float
which can be read from the adDacent scale usually etched on the glass tube.
Balance of forces at e;uilibrium (i.e., steady%state)"
( )
4 C B f f f f f
c c c
g g g
F F F V V V
g g g
('.(()
where"
F
4
drag force resulting from form and s>in friction for flow around
the float
F
B
buoyant force acting so as to raise the float
F
C
graity force acting downward on a float
V
f
olume of the float
**
f
density of the float
density of the fluid
Figure '.A. )chematic representation of a rotameter
=nergy balance between ( and + (see Figure '.A)
+ +
+ ( + (
c
P P u u
F 7
+g
+ + ('.(+)
where"
F summation of drag or friction losses
.ontinuity e;uation
( ( + +
m u) u )
0
('.(1)
.ombining e;uations '.(+ and '.(1
+
+ (
( c +
+
) P
u ( +g F
)
_ _
+
, ,
('.(,)
( ( c +
(
+
+
P
u . +g
)
(
)
('.(8)
*9
(
F
B
F
4
F
C
+
+
(
Flow
4rag force (F
4
)
( ) ( )
+
4 f f f f
F P ) P . ) ('.(A)
where"
%P
f
pressure drop acting on the top of the float
+
f
. fraction of the ma\$imum pressure drop (%P) that is not
recoered
.ombining e;uations '.((, '.(8 and '.(A
( )
f f
( !
+
(
f +
+
+V g
u .
)
) (
)
,
('.(*)
where"
(
!
f
.
.
.
rotameter coefficient
f ( +
) ) ) (rotameter tapers gradually)
( )
( )
f f
+
( !
f ( +
+gV
)
u .
) ) )
+
('.(9)
#ass flow rate
( )
f f
! +
f
+gV
m . )
)
0
('.(')
Important notes"
Proper design of rotameter float will ma>e .
!
constant oer wide range
of !e.
For a constant%density fluid in a single rotameter, the terms within the
s;uare root symbol of e;uation '.(9 are practically constant and
independent of flow rate2 hence,
*'
E
( ! +
u . ) ('.+7)
&he preceding e;uation shows that the elocity is e;ual to a constant
multiplied by the minimum cross section for flow ()
+
).
9.0 Notc&es an* We%rs
Fotch An opening in the side of the tan> or reseroir which e\$tends
aboe the surface of the fluid. It is used to measure discharge
flow rate.
Geir A notch on a lager scale, usually found in riers. It maybe
sharp%crested but may also hae a substantial depth in the
direction of flow. It can be used to measure flow rate or raise
water leels.
T&e General We%r E,uat%on
Gith reference to Figure '.*,
Velocity through the strip" u +gh ('.+()
4ischarge through the strip" H Au b h +gh
H /
7 7
H b +gh h
/
(I +
7
H +g bh dh
('.++)
Figure '.*. =lemental strip of flow through a notch
97
Rectan/ular !e%r
Figure '.9. !ectangular weir
b 3 B
( 1
/
+ +
theo
7
+
H B +g h dh B +g/
1
('.+1)
1
+
actual d
+
H . +g/
1
('.+,)
\$.notc& !e%r
Figure '.'. V%notch or triangular weir geometry
( ) b + / h tan
+
('.+8)
( )
(
/
+
theo
7
H + +gtan / h h dh
+
9(
/
1 8
+ +
theo
7
+
H + +gtan /h h
+ 8
1 _
1
,
]
8
+
theo
9
H +gtan /
(8 +
('.+A)
8
+
actual d
9
H . +gtan /
(8 +
('.+*)
7UESTIONS 8 3RO9LEMS)
(. =\$plain how a Pitot tube measures the speed of a boat.
+. =\$plain the principle of ariable area meters.
1. A Pitot tube is inserted into the center of an air duct 7.9 m in diameter.
A pressure gage attached to the Pitot tube reads %P 3 ' FIm
+
. &he
Pitot tube coefficient is 7.'9+. .alculate the mass flow rate of air, at a
temperature of 18
o
. and a pressure of (7( >Pa.
,. A Venturi meter with an entrance diameter of 7.1m and a throat
diameter of 7.+m is used to measure the olume of gas flowing
through a pipe. &he discharge coefficient of the meter is 7.'A.
Assuming the specific weight of the gas to be constant at ('.A+ FIm
1
,
calculate the olume flowing when the pressure difference between
the entrance and the throat is measured as 7.7A m on a water J%tube
manometer.
8. A Venturi meter is used for measuring flow of water along a pipe. &he
diameter of the Venturi throat is two%fifths the diameter of the pipe.
&he inlet and throat are connected by water%filled tubes to a mercury
J%tube manometer. &he elocity of flow along the pipe is found to be
+.8 /
mIs, where / is the manometer reading in m of mercury.
4etermine the loss of head between inlet and throat of the Venturi
when / is 7.,' m. &he relatie density of mercury is (1.A.
A. Brine (specific graity (.(9) is flowing though a 97%mm pipe at a
ma\$imum rate of 7.7(8 m
1
Is. In order to measure the flow rate, a
9+
sharp%edged orifice, connected to simple J%tube manometer is to be
1'7 mm /g. Ghat size orifice should be installedK
*. A closed tan> has a 7.7+8%m diameter orifice in one of its ertical
sides. &he tan> contains oil to a depth of 7.A( m aboe the centre of
the orifice and the pressure in the air space aboe the oil is maintained
at (1*97 FIm
+
aboe atmospheric. 4etermine the discharge from the
orifice. &he coefficient of discharge of the orifice is 7.A( and the
relatie density of oil is 7.'.
9. A Venturi meter is fitted in a horizontal pipe of 7.(8 m diameter to
measure the flow of water which may be anything up to +,7 m
1
Ih. &he
pressure head at the inlet for this flow is (9 m aboe atmospheric and
the pressure head at the throat must not be less than * m below
atmospheric. Between the inlet and the throat there is an estimated
frictional loss of (7? of the difference in pressure head between these
points. .alculate the minimum allowable diameter for the throat.
'. 4educe an e\$pression for the discharge of water oer a right%angled
sharp edged V%notch, gien that the coefficient of discharge is 7.A(.
A rectangular tan> (A m by Am has the same notch in one of its
short ertical sides. 4etermine the time ta>en for the head, measured
from the bottom of the notch, to fall from (8 cm to *.8 cm.
(7.4eelop a formula for the discharge oer a '7%degree V%notch weir in
terms of head aboe the bottom of the V.
A channel coneys 177 litersIsec of water. At the outlet end
there is a '7%degree V%notch weir for which the coefficient of discharge
is 7.89. At what distance aboe the bottom of the channel should the
weir be placed in order to ma>e the depth in the channel (.17 mK Gith
the weir in this position what is the depth of water in the channel when
the flow is +77 litersIsecK
91
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Math 6308 Ordinary differential Equations II. Spring 2018
Schedule: 1:30 - 2:45 pm TR Classroom: Skiles 255 Office: Skiles 102A Phone: 404-894-4750 Email: zengch@math.gatech.edu Webpage: www.math.gatech.edu/~zengch Office hours: 10:05 - 11am W & 12:30 - 1:25pm R
Syllabus
Instructor: Chongchun Zeng
Prerequisites: MATH 2552&4541, or permission of the instructor
Participation: Attendence in classes is mandatory. Students are expected to read the materials before and/or after the classes. It is suggested that you start to work on homework problems right after they are assigned.
Main reference book: Ordinary differential equations with applications, Carmen Chicone, 2nd edition, Springer. (Homework will mainly come from this book.)
Materials: In this semester, we will develop the qualitative theory for systems of nonlinear ordinary differential equations. After a brief review of the material covered in Math 6307, the main subjects include invariant manifold theory, bifurcation theory, normal forms, periodic orbits and homoclinic orbits. The topics are among
• Invariant manifolds: existence, uniqueness, smoothness, and other properties of invariant manifolds and invariant foliations.
• Basic bifurcations: saddle-node, transcritical, pitch-fork, and Hopf bifurcations.
• Periodic orbits: Poincare map, persistence, period-doubling, and periodic orbtis in Hamiltonian systems.
• Homoclinic orbits: Melnikov method, Smale-Birkhoff theorem, etc.
• 1 take-home final exam 40%: `open-textbook' and `open-notes', must be entirely independent work.
• 4 homework, 15% each: homework will be assigned at the end of each lecture and then put on this webpage in the below. You may discuss the homework problems with other students in this class, but you should write down the solutions and complete the homework independently. On each collecting day, the homework is due in the classroom at the beginning of the lecture. The corresponding assignment includes all the problems assigned in previous weeks, but after the last assignment. Some selected problems from each assignment will be graded and the score will be given based on both the graded problems and the completion of the whole assignment. The exact homework collecting dates are:
1/25(Thu.), 2/15(Thu.), 3/8(Thu.), 4/5(Thu.)
Homework assignments:
• Homework Assignment #1 (due Thu. 1/25):
• Assigned on Tue. 1/9: P159, 2.30; P163, 2.34; P193, 2.86, 2.87
• Assigned on Thu. 1/11: P228, 3.2; P233, 3.5, 3.7
• Homework Assignment #2 (due Thu. 2/15):
• Homework Assignment #3 (due Thu. 3/8)
• Assigned on Tue. 2/13: Page 347, 4.4, 4.5
• Assigned on Thu. 2/22: Page 558, 8.9
• Homework Assignment #4 (due Thu. 4/5)
• Assigned on Tue. 3/6: Page 564, 8.18, 8.19
• Assigned on Tue. 3/13: P373, 5.2
• Assigned on Tue. 3/27: P382, 5.9
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# Tutor profile: Chelsey B.
Chelsey B.
Geotechnical Engineer
## Questions
### Subject:Geometry
TutorMe
Question:
Given a circle of radius 10 cm that has a hole in it that has a diameter of 10 cm, determine the area of the circle minus the hole.
Chelsey B.
The area of a circle is determined using the equation Area=Pi*radius squared (A=Pi*r^2). To determine the area of the circle minus the area of the hole we will do Area=(Pi*10^2)-(Pi*5^2)=235.5. Note that for the smaller circle the problem statement gave us the diameter, so we had to divide that by 2 to get the radius of the hole.
### Subject:Basic Math
TutorMe
Question:
If 1+2=3 2+3=5 3+4=7 What is the next logical equation in the list
Chelsey B.
Evaluating the patterns, we can see that each equation is a number X + the number X+1 and the first number in the equation is equal to the number X+1 from the previous equation. So applying that logic the next logical equation would be 4+5=9.
### Subject:Civil Engineering
TutorMe
Question:
What is the best method for determining the settlement of a layer of cohesionless soil?
Chelsey B.
There are several methods for determining the settlement of soils. It is important to understand how each method works and what soil types it applies to. For cohesionless soils such as sands, the Hough Method is a good method for evaluating the settlement.
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https://mapleprimes.com/users/Lonely/answers
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## 400 Reputation
16 years, 308 days
## assuming 1 < x < 2...
solve((1+x)^(p+1) > 2^p*(x^p+x), p)
assuming 1 < x < 2
## for what value the function is increasin...
how to show
f(x) =
x >= 1
it looks like the (i am not sure):
df/dx > 0 for 0 < m <=3
df/dx < 0 for m < 0
df/dx < 0 for m > 0
## there was a misprint how to...
there was a misprint
how to see it :
p*(int(f(x), x = 0 .. infinity)) = sum(int(f(x), x = n .. n+p), n = 0 .. infinity)+sum((p-n)*(int(f(x), x = n-1 .. n)), n = 1 .. p-1)
## how to prove this with maple...
e < (1+1/n)^((1/8)*(n^(1/3)+(n+1)^(1/3))^3)
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The Question:
BLYs Method
HSAs Method
SMAs Method
Would have done it similarly but without introducing the variable B, so all in terms of R
R : 4R
R+7 : 3(R+7)
We know the Bs were unchanged so
4R = 3(R+7)
Giving us R=21
And then B is 3(21+7)=84
## One thought on “The Bead Problem”
1. BLY says:
I like SMA’s approach of focusing on the unchanged and leaving in terms of r; although I’d use x to represent number of beads (any colour) since I’m predicting that pupils will be confused with blue beads represented with r.
HSA’s approach is really neat and demonstrates some pretty smooth thinking. 💭 BLY
Like
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# nLab compactly generated triangulated category
Contents
### Context
#### Homological algebra
homological algebra
Introduction
diagram chasing
### Theorems
#### Stable homotopy theory
stable homotopy theory
Introduction
# Contents
#### Compact objects
objects $d \in C$ such that $C(d,-)$ commutes with certain colimits
# Contents
## Definition
###### Definition
Let $\mathcal{T}$ be a triangulated category with coproducts. Then $\mathcal{T}$ is compactly generated if there is a set $\mathcal{G}$ of objects of $\mathcal{T}$ such that
1. Whenever $X$ is an object such that $\mathcal{T}(\Sigma^m G, X) = 0$ for all $G \in \mathcal{G}$ and $m \in \mathbb{Z}$, then $X = 0$.
2. All objects in $\mathcal{G}$ are compact i.e. for all $G \in \mathcal{G}$ and for every family $\{ X_i \mid i \in I\}$ of objects of $\mathcal{T}$
$\coprod_{i \in I} \mathcal{T}(G, X_i) \to \mathcal{T}\Big(G,\coprod_{i \in I} X_i\Big)$
is an isomorphism.
## Properties
###### Proposition
If $\mathcal{T}$ is a triangulated category with coproducts, then a set of objects $\mathcal{G}$ satisfies the two conditions of def. if and only if the smallest localizing subcategory of $\mathcal{T}$ that contains $\mathcal{G}$ is $\mathcal{T}$ itself.
This is Lemma 2.2.1. of (Schwede-Shipley).
Brown representability theorem holds in compactly generated triangulated categories.
###### Theorem
If $\mathcal{T}$ is a compactly generated triangulated category and $H : \mathcal{T}^\mathrm{op} \to Ab$ is a cohomological functor, then $H$ is representable.
## References
• Stefan Schwede, Brooke Shipley, Stable model categories are categories of modules. Topology 42 (2003), no. 1, 103–153.
• Amnon Neeman, Triangulated categories, Annals of Mathematics Studies, 148. Princeton University Press, 2001.
Last revised on February 10, 2014 at 06:54:30. See the history of this page for a list of all contributions to it.
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# Tagged with sum of two squares and cube …
## #math trivia #264 solution
#math trivia for #September21: #264 is the sum of a square and a cube, and also of two squares and a cube. What are they? — Burt Kaliski Jr. (@modulomathy) September 22, 2013 Answer: The square and cube are 256 = 162 and 8 = 23. The two squares and cube are 4 = 22, … Continue reading
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Question DetailsNormal
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MM305 Unit 7 Project | Complete Solution
Question posted by
Name:
Directions:
• In this project you will be using the States Data filethat is found in Doc Sharing under the Instructor-Graded Projects:
• Steps for accessing the States Data file:
1. Open your Excel with PhStat2
2. Click File—Open
3. Select the States Data file
• Not all questions require the use of technology or the States Date file.
• You may insert your answers, including any charts, graphs, or output, on this document.
• Be sure to put your name on this document and save it to your computer.
1. Even though independent gasoline stations have been having a difficult time, Susan Solomon has been thinking about starting her own independent gas station. Susan’s problem is to decide how large her station should be. The annual returns will depend on both the size of the station and a number of marketing factors related to oil industry and demand for gasoline. After careful analysis, Susan developed the following table:
Size of Gasoline Station
Good Market (\$)
Fair Market (\$)
Poor Market (\$)
Small
\$70,000
\$30,000
-\$30,000
Medium
\$110,000
\$50,000
-\$40,000
Large
\$170,000
\$70,000
-\$50,000
1. Develop a decision table for this decision.
2. What is the Maximax decision?
3. What is the Maximin decision?
4. What is the criterion of realism decision? Use α = 0.6.
5. Develop an Opportunity Loss Table
6. What is the Minimax Regret Decision?
2. Data collected on the yearly demand for 50-pound bags of fertilizer at Sunshine Garden Supply are shown in the following table.
Year
Demand for Fertilizer (1,000s of Bags)
1
4
2
7
3
5
4
5
5
10
6
7
7
8
8
9
9
11
10
14
11
15
Use Excel QM to:
1. Develop a three-year moving average to forecast sales in year 12.
2. Develop a 3-year weighted average to predict demand in year 12, in which sales in the most recent year is given a weight of 2 and sales in the two years prior to that are each given a weight of 1.
3. Develop a regression/trend line to estimate the demand for fertilizer in year 12.
4. Based on the three forecasts you have created, which forecast is the most accurate?
3. Kaplan College has decided to "wire" its campus. The first stage in this effort is to install the "backbone," i.e., to connect all the buildings. The table below gives the distances between the various buildings on campus in hundreds of feet.
Distances in Hundreds of Feet
From
To
Building 1
Building 2
Building 3
Building 4
Building 5
Building 6
Building 1
3
7
5
5
4
Building 2
5
2
6
6
Building 3
5
4
4
Building 4
5
3
Building 5
4
Building 6
1. How should the buildings be connected to minimize the total length of cable?
2. What length of cable is required?
4. The following represents the distances in miles from a warehouse (node 1) to various cities in Montana. The major outlet store is located at node 7.
From Node
To Node
Distance
1
2
40
1
4
100
2
3
20
2
4
30
3
4
60
3
5
40
3
6
20
4
5
70
4
7
50
5
6
50
5
7
80
6
7
40
1. Find the shortest route and distance from Node 1 to Node 7.
5. The network of a city sewer system and their capacities are shown below. Remember that the arc has both capacity and reverse capacity. For example, row 1 is the flow from node 1 to node 2 and row 2 is the reverse flow from node 2 to node 1. There are eight branches in this network.
From Node
To Node
Fluid Flow
1
2
150
2
1
100
1
3
0
3
1
150
1
4
300
4
1
300
1
5
150
5
1
100
2
4
300
4
2
200
3
4
250
4
3
300
3
5
300
5
3
250
4
5
100
5
4
0
Determine the maximum flow (in hundreds of gallons of water per minute) from node 1 to node 5
6. The governor of Michigan believes that the state can improve the state’s crime rate if the state can reduce the college debt carried by its citizens and if they can increase the percent of the population covered by health insurance.
1. Using the States Data Set and PHStat, create the multiple regression prediction equation.
2. Predict the crime rate for Michigan if the college debt were \$25,000 and the percent not covered by insurance was 10?
7. A concessionaire for the local ballpark has developed a table of conditional values for the various alternatives (stocking decisions) and states of nature (size of crowd).
Stocking Decision
Large Crowd (\$)
Average Crowd (\$)
Small Crowd (\$)
Large Inventory
\$22,000
\$12,000
-\$2,000
Average Inventory
\$15,000
\$12,000
\$6,000
Small Inventory
\$9,000
\$6,000
\$5,000
If the probabilities associated with the states of nature are 0.30 for a large crowd, 0.50 for an average crowd, and 0.20 for a small crowd, determine:
1. The alternative that provides the greatest expected monetary value (EMV).
2. The expected value of perfect information (EVPI).
• Save your project in a location that you will remember and with your full name. When you are ready to submit your project, click on the Dropbox and complete the steps below:
• Click the link that says, “Submit an Assignment.”
• Click the “Add Attachments” button.
• Make sure that you save a copy of your submitted project.
\$ 30.00
[Solved] MM305 Unit 7 Project | Complete Solution
• This Solution has been Purchased 5 time
• Submitted On 08 Feb, 2015 08:45:07
If the probabilities associated with the states of nature are 0.30 fo...
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2. What is the latest possible time that Activity D may be started without delaying the completion of the project?
We could start Activity D at T=10 weeks, rather than the early start tome of t= 6wee...
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;************************************************************* ; regrid_12.ncl ; ; Concepts illustrated: ; - Interpolating from a fixed grid to a lower resolution using conservative remapping ; - Computing dispersion statistics ; - Drawing a cylindrical equidistant map ;************************************************************* load "\$NCARG_ROOT/lib/ncarg/nclscripts/csm/gsn_code.ncl" load "\$NCARG_ROOT/lib/ncarg/nclscripts/csm/gsn_csm.ncl" load "\$NCARG_ROOT/lib/ncarg/nclscripts/csm/contributed.ncl" begin ;********************************************* ; Read global grid [180W to 180E] ; ELEV: [LAT | 5401] x [LON | 10801] *includes* cyclic pt ; We do *not* want to read the cyclic point. ;********************************************* diri = "/Users/shea/Data/netCDF/" fili = "ETOPO2_GLOBAL_2_ELEVATION.nc" ; NGDC f = addfile(diri+fili,"r") Z = short2flt( f->ELEV(:,0:10799) ) ; Do not read the cyclic pt printVarSummary(Z) ; Z: min=-10654 max=8593 ;********************************************* ; areal average conservative remapping ;********************************************* NLATR = 180 MLONR = 360 LATR = latGlobeFo(NLATR, "LATR", "latitude" , "degrees_north") LONR = lonGlobeFo(MLONR, "LONR", "longitude", "degrees_east") ; 0-360 LONR = (/LONR-180/) ; -180 to 180 LONR&LONR = LONR ZR = area_conserve_remap_Wrap (Z&LON,Z&LAT,Z, LONR,LATR, False) ; 5.2.0 ;ZR = area_hi2lores_Wrap (Z&LON,Z&LAT,Z, True, 1.0, LONR,LATR, False) ; 5.1.0 ZR@long_name = ZR@long_name +": 1x1 grid" printVarSummary(ZR) ; ZR (1x1): min=-10439.7 max=5713.99 ;********************************************* ; Calculate assorted statistics and print ; stat_dispersion is a bit slow for *large* variables ;********************************************* opt = True opt@PrintStat = True ;statZ = stat_dispersion(Z , opt ) ; bit slow (5401,10800) statZR = stat_dispersion(ZR, opt ) ;********************************************* ; Plot ;********************************************* wks = gsn_open_wks ("ps", "regrid") gsn_define_colormap(wks, "BlAqGrYeOrReVi200") setvalues NhlGetWorkspaceObjectId() "wsMaximumSize": 500000000 end setvalues res = True res@gsnMaximize = True res@gsnAddCyclic = False ; dataset allready has cyclic pt res@gsnSpreadColors = True res@cnFillMode = "RasterFill" res@cnLinesOn = False ; Turn off contour lines res@cnFillOn = True ; Turn on contour fill res@cnLevelSelectionMode = "ManualLevels" ; set manual contour levels res@cnMinLevelValF = -5000 ; set min contour level res@cnMaxLevelValF = 5000 ; set max contour level res@cnLevelSpacingF = 500 res@lbLabelAutoStride = True res@mpCenterLonF = 0 res@mpOutlineOn = True res@mpFillOn = False plot = gsn_csm_contour_map_ce(wks,Z , res) plot = gsn_csm_contour_map_ce(wks,ZR, res) end
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https://www.scribd.com/document/76445992/In-Soo-Baek-and-Sang-Hung-Lee-Hausdorff-Dimension-of-Some-Specified-Perturbed-Cantor-Set
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# J. Korean Math. Soc. 31 (1994), No. 4, pp.
539–544
HAUSDORFF DIMENSION OF SOME SPECIFIC PERTURBED CANTOR SET IN SOO BAEK
AND
SANG HUN LEE
1. Introduction We [1] investigated the Hausdorff dimension and the packing dimension of a certain perturbed Cantor set whose ratios are uniformly bounded. In this paper, we consider a specific perturbed Cantor set whose ratios are not necessarily uniformly bounded but satisfy some other conditions. In fact, in the hypothesis, only the condition of the uniform boundedness of ratios on the set is substituted by a “∗-condition". We use energy theory related to Hausdorff dimension in this study while we [1] used Hausdorff density theorem to find the Hausdorff dimension of some perturbed Cantor set. In the end, we give an example which explains aforementioned facts. 2. Preliminaries We recall the definition of perturbed Cantor set of [1]. Let Iφ =[0,1]. We can obtain the left subinterval Iσ,1 and the right subinterval Iσ,2 of Iσ deleting middle open subinterval of Iσ inductively for each σ ∈ {1, 2}n , where n = 0, 1, 2, · · · . Consider E n = ∪σ ∈{1,2}n Iσ . Then {E n } is a decreasing sequence of closed sets. For each n, we fix | Iσ,1 | / | Iσ |= an+1 and | Iσ,2 | / | Iσ |= bn+1 for all σ ∈ {1, 2}n ,where | I | denotes the diameter of I. We call F = ∞ E n a perturbed Cantor set. n=0 We recall the s-dimensional Hausdorff measure of F : H s (F) = lim Hδs (F),
δ→0
Received June 17, 1993. Revised July 10, 1994. Supported in part by TGRC-KOSEF and the Basic Science Research Institute program, Ministry of Education, Korea, 1994, project NO BSRI-94-1401
540
In Soo Baek and Sang Hun Lee
where Hδs (F) = inf{ ∞ | Un |s : {Un }∞ is a δ-cover of F}, and the n=1 n=1 Hausdorff dimension of F : dim H (F) = sup{s > 0 : H s (F) = ∞} (= inf{s > 0 : H s (F) = 0})(see[2]). We note that if {an } and {bn } are given, then a perturbed Cantor set F is determined, vice versa. We are now ready to study the ratio geometry of the perturbed Cantor set. 3. Main results In this section, F means a perturbed Cantor set determined by {an } and {bn }. We introduce ∗-condition which plays an important role in the energy theory related to Hausdorff dimension. Let s ∈ (0, 1). F is said to satisfy ∗-condition for s, if for each δ > 0 there exist positive < 1 and N such that k (ais + bis ) k [max(ai , i=N i=1 bi )]δ < (1 − )k−N for all k ≥ N . Before going into the main theorem, it is fruitful to know some properties of the ratios {an }, {bn } of a perturbed Cantor set F. LEMMA 1. Let {an }, {bn } be sequences in (0,1) and dn = 1 − (an + bn ) > 0 for each n. Then
s s (1) there exists unique sn such that ann + bnn = 1 for each n = 1, 2, · · · . (2) 0 ≤ s ≤ s ≤ 1, where s = lim infn→∞ sn , s = lim supn→∞ sn . (3) if lim inf n→∞ an > 0 and lim infn→∞ bn > 0, then lim supn→∞ s s (an + bn ) = 1. s s (4) if lim inf n→∞ n log(ak + bk ) is finite for some s , then s ≤ s ≤ s k=1 . s s (5) if lim inf n→∞ n log(ak + bk ) is finite for some s , then there is k=1 some B < ∞ such that lim inf n→∞ n log(ais +bis ) < B for any k. i=k s s (6) if lim supn→∞ (an + bn ) ≤ 1 for some s and lim infn→∞ dn > 0, then F satisfies the ∗-condition for s .
Proof. (1) - (5) follow immediately from the definitions of lim inf and lim sup. For (6), we can find d > 0 such that an + bn ≤ 1 − d for all n ≥ N1 for some N1 since lim infn→∞ dn > 0. Given δ > 0, we choose
Hausdorff dimension of some specific perturbed Cantor set
541
0 < < 0 such that (1 − d)δ < 1 − 2 . For such > 0, there is N ≥ s s N1 such that an + bn < 1 + for all n ≥ N . Therefore for all k ≥ N , k (ais + bis ) k [max(ai , bi )]δ < k (ais + bis ) k (1 − d)δ < i=N i=1 i=N i=1 k (1 + )(1 − 2 ) < k (1 + )( 1− ) < (1 − )k−N . i=N i=N 1+
s s THEOREM 2. If lim infn→∞ n log(ak + bk ) is finite for some s and k=1 lim infn→∞ dn > 0 and F satisfies the ∗-condition for s , then dim H (F) = s.
Proof. We define a set function µ by µ(Iσ ) = lim inf
n→∞ I ∈Iσ ∩E n
|I |s
for each σ ∈ {1, 2}k , where k = 1, 2, · · · . Clearly µ(Iσ ) = µ(Iσ,1 ) + µ(Iσ,2 ) for all σ . Then µ is extended to a mass distribution on F (see [2,Proposition1.7]). For x, y ∈ F, we denote x∧ y by the interval that both x and y belong to a common interval of E n with greatest integer n. Since lim infn→∞ dn > 0, we find d > 0 such that ak + bk ≤ 1 − d for all k ≥ N1 for some N1 . Then for σ ∈ {1, 2}k where k ≥ N1 , the gap of Iσ,1 and Iσ,2 is not less than d|Iσ |. Let 0 < t < s. By slight abuse of notation, we write I ∈ E k to mean that the interval I is one of the basic intervals Iσ of E k , where σ ∈ {1, 2}k . If x∧ y ∈ E k and k ≥ N1 , then for σ ∈ {1, 2}k , |x − y|−t dµ(x)dµ(y) ≤2d
−t x∧ y=Iσ |Iσ |−t µ(Iσ,1 )µ(Iσ,2 ) n→∞
=2d −t |Iσ |−t lim inf
I ∈Iσ,1 ∩E n
|I |s lim inf
n→∞ I ∈Iσ,2 ∩E n
|I |s
≤2d −t |Iσ |−t lim inf
n→∞ I ∈Iσ,1 ∩E n ,I ∈Iσ,2 ∩E n n
|I |s |I |s
=2d
−t
lim inf
n→∞ i=k+2 n
(ais + bis )2 |Iσ |−t |Iσ,1 |s |Iσ,2 |s (ais + bis )2 |Iσ |2s−t .
i=k+2
≤2d −t lim inf
n→∞
542
In Soo Baek and Sang Hun Lee
Using Lemma 1 (5), for σ ∈ {1, 2}k where k ≥ N1 , we obtain |x − y|−t dµ(x)dµ(y)
x∧ y=Iσ
≤2d
−t
B|Iσ |2s−t for some B < ∞.
Since F satisfies the ∗-condition for s, there exist positive < 1 and N ≥ N1 such that k (ais + bis ) k [max(ai , bi )]s−t < (1 − )k−N for all k ≥ N . i=N i=1 Therefore for all k ≥ N , we have |I |2s−t =
I ∈E k I ∈E k k
|I |s |I |s−t {(ais + bis )[max(ai , bi )]s−t }
i=1
<α(1 − )k−N , where α =
N −1 s i=1 (ai
+ bis ). Thus for k ≥ N , we have |x − y|−t dµ(x)dµ(y)
x∧ y=I ∞
∞ k=N I ∈E k
≤2d −t B
|I |2s−t α
k=N I ∈E k
≤2d −t B . Since dn > 0 for each n, d ∗ = min1≤k≤N dk exists. If x∧ y ∈ E k and k < N , then we similarly obtain |x − y|−t dµ(x)dµ(y)
x∧ y=Iσ ∗ −t ≤2(d ) |Iσ |−t µ(Iσ,1 )µ(Iσ,2 )
≤2(d ∗ )−t B|Iσ |2s−t
Hausdorff dimension of some specific perturbed Cantor set
543
for σ ∈ {1, 2}k . Thus
N −1 k=0 I ∈E k x∧ y=I N −1
|x − y|−t dµ(x)dµ(y) |I |2s−t
k=0 I ∈E k N −1
≤2(d ) B ≤2(d ) B ≤2(d ) B2 . Hence
∗ −t k=0 N ∗ −t
∗ −t
2k
|x − y|−t dµ(x)dµ(y)
F ∞ F
=
k=0 I ∈E k N −1 x∧ y=I
|x − y|−t dµ(x)dµ(y) |x − y|−t dµ(x)dµ(y)
x∧ y=I
=
k=0 I ∈E k ∞
+
k=N I ∈E k x∧ y=I
|x − y|−t dµ(x)dµ(y) α < ∞.
≤2(d ∗ )−t B2 N + 2d −t B
Since µ is a mass distribution on F, dim H (F) ≥ s (see [2,Theorem 4.13 (a)]). s Now dim H (F) ≤ s follows easily from the fact that lim infn→∞ n log(ak + k=1 s bk ) < ∞ and lim infn→∞ dn > 0 .
s s COROLLARY 3. If lim inf n→∞ n log(ak +bk ) is finite, lim infn→∞ dn > k=1 s s 0 and lim supn→∞ (an + bn ) ≤ 1, then dim H (F) = s.
Proof. It follows immediately from Theorem 2 and Lemma 1 (6).
544
In Soo Baek and Sang Hun Lee
EXAMPLE 4. Consider a sequence {n k }∞ of natural numbers such that k=1 n k+1 − n k > k + 1 for all k. Let 0 < < 1 and bn k = 1 − for all k. 4 We also take a decreasing sequence {an k }∞ such that 0 < an k ≤ 2 with k=1
2 limk→∞ an k = 0, and log(an2k + bn k ) < 0. Let bn k +i = 1 , 1 ≤ i ≤ k for 4 1 each k. For each k and 1 ≤ i ≤ k , we find an k +i such that log{(an k +i ) 2 + 2 (bn k +i ) 2 } = δkk , where δk = − log(an2k + bn k ). We put an = bn = 1 if 4 lim n = n k , n k + 1, · · · , n k + k, where k = 1, 2, · · · . Then √ infn→∞ an = 0, 3−3 and lim infn→∞ dn > 0. In fact, dn ≥ min( 2 , 1 − { 1 + ( 4 6 )2 }) > 0 for all 4 1 2 n. Clearly lim supn→∞ (an2 + bn ) = 1 and lim infn→∞ n log(ak2 + bk2 ) = k=1 1 log(1 − ) 2 . Using Corollary 3, we obtain dim H (F) = 1 . 2 1 1 1 1 1 1 1 1
References
[1] I. S. Baek, Dimensions of the perturbed Cantor set, Real Analysis Exchange 19(1) (19931994), 269-273. [2] K. J. Falconer, Fractal geometry, (John Wiley and Sons. 1990).
SANG HUN LEE Department of Mathematics Kyungpook National University Taegu 702-701, Korea IN SOO BAEK Department of Mathematics Pusan University of Foreign Studies Pusan 608-738, Korea
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https://www.doorsteptutor.com/Exams/KVPY/Stream-SA/Mathematics/Questions/Topic-Coordinate-Geometry-9/Part-1.html
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# Coordinate Geometry (KVPY (Kishore Vaigyanik Protsahan Yojana) Stream-SA (Class 11) Math): Questions 1 - 4 of 14
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## Question number: 1
» Coordinate Geometry » Angles Between Two Lines
Appeared in Year: 2011
MCQ▾
### Question
In the adjoining figure AB = 12 cm, CD = 8 cm, BD = 20 cm; ∠ ABD = ∠ AEC = ∠ EDC = 90 0. If BE = x, then,
### Choices
Choice (4) Response
a.
x has only one value and
b.
x has two possible values whose difference is 4
c.
x has two possible values whose sum is 28
d.
x cannot be determined with the given information
## Question number: 2
» Coordinate Geometry » Equations of Internal and External Bisectors of Angles
Appeared in Year: 2011
MCQ▾
### Question
Let ABC be a triangle with ∠ B = 90 0. Let AD be the bisector of ∠ A with D on BC. Suppose AC = 6 cm and the area of the triangle ADC is 10 cm 2. Then the length of BD in cm is equal to
### Choices
Choice (4) Response
a.
b.
c.
d.
## Question number: 3
» Coordinate Geometry » Distance Formula
Appeared in Year: 2011
MCQ▾
### Question
A frog is presently located at the origin (0,0) in the XY-plane. It always jumps from a point with integer coordinates to a point with integer coordinates moving a distance of 5 units in each jump. What is the minimum number of jumps required for the frog to go from (0,0) to (0,1)?
### Choices
Choice (4) Response
a.
4
b.
9
c.
2
d.
3
## Question number: 4
» Coordinate Geometry » Section Formula
Appeared in Year: 2011
MCQ▾
### Question
A piece of paper in the shape of a sector of a circle (see Fig. 1) is rolled up to form a right-circular cone (see Fig. 2). The value of the angle is?
### Choices
Choice (4) Response
a.
b.
c.
d.
f Page
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3. Do a seat belt experiment. Impulse is change in momentum. Momentum, Impulse, Momentum and Impulse, Conservation of Momentum in One Dimension, Elastic Collisions, Inelastic Collisions, Energy-Momentum Problems MS-PS2-1. Every moving object has momentum. Determine the impulse. But there are other ways to think about momentum! We will examine some physics in action in the real world. What is impulse. LAB 2: Understanding Momentum. Car manufacturers use this idea and design crumple zones into cars, such that the car has a momentum is existing. , HS-PS2-2. When there is a car crash, the car, its contents and the passengers decelerate rapidly. Impulse = change in momentum . 5. 4. WORKSHEET When a light car and a massive truck collide, momentum conservation requires that Gives us an idea of how hard it is to accelerate a moving object. a [m/s] is the acceleration of an object; F [N] is the force acting on an object; m [kg] is the mass of an object. The impulse and energy that is usually witnessed in the car impulse have a symbol of J or Imp. During a collision, the motion of the driver and passenger carries them towards the windshield. The process of minimizing an impact force can be approached from the definition of the impulse of force: . We review their content and use your feedback to keep the quality high. Subjects: Basic Principles, Physical Science, Physics. F = force F = ma m =mass a =acceleration F = mv v =change in velocity = tttime or rate Ft = impulse Ft = mvmv =change in momentum Observe the graph of force versus time. m/s.Impulse and momentum have the same units; when an impulse is applied to an object, the momentum of What is momentum. Determine the impulse. Gnome Garden.
Forces and Motion. Drag Racing Club. Impulse: The process of transferring momentum is called impulse. Impulse and Momentum. Spinning rear wheels lift from the ground. Path Painter. Impact Force Equation is written as such. There's two main concepts here: impulse (change in momentum) and pressure (force over some area). In a car crash, you decelerate from some initial velocity which you were driving at to rest. Regardless of any specifics of the crash, if you start with some $v_i$, your change in momentum will be the same. 10 Fly Car Stunt. Air-bags in Motor Vehicles. Luke Henderson Physics Videos. 2 m/s to +1 Simple Machines Videos & Response Sheets [Note: use the video links on the right, NOT the links inside the Solve momentum/impulse problems involving the application of kinematic equations Bell Ringer-Write a response to the following prompt: As head of NIHS, identify three recommendations for improving crash worthiness Impulse Practice Force (F) is equal to the change in momentum (P) over the change in time (t). The van is about to crash head-on into a wall at 50km/h. Picture a car crash on the highway. If an impact stops a moving object, then the change in momentum is a fixed quantity, and extending the time of the collision will decrease the time average of the impact force by the same factor. The time here refers to the experience of the crash test dummy inside the car. impact force. The momentum of the car is: 20000 kg m/s 3 Conservation of Momentum Notes PDF Hitting a rocket with a projectile Calculate the momentum (in kg m/s) of the ostrich I am trying to figure out why they said that the right answer is 2? An airbag is another way of minimizing force in a car accident, in this case by reducing the time over which the occupants move forward toward the dashboard or wind-shield. And the change in momentum (P) is also equal to the impulse (J). The impulse is the same for each car. 1. Between 70 and 80 ms B. The change in momentum of one colliding vehicle is accompanied by an equal and opposite change of momentum of the other vehicle. A 2000kg car driving at 50km/h has a lot of momentum. F = m * v / (2 * d) Where F is the avg. IMPULSE We discussed above the factors changing momentum which are mass and velocity. B. interaction between two objects. Momentum = (mass) x (velocity) (Momentum equals mass times velocity.) A car or truck might look a long way away but it can get to you fast and hit you hard. It is also equal to the average force times the change in time. P = m x v. Momentum is the quantity of movement and the quantity of matter travelled and the velocity at which it travels is multiplied.
The car hits a wall and rebounds (moving west) with a speed of 0.100 m/s. Lets take an example of a car travelling a steep road. Title: Momentum and Impulse Author: bill nye Last modified by: Marilyn Baker Created Date: 5/3/2005 3:05:11 PM How hard is it to stop a moving object? Extending the time of collision will reduce the average impact force. Because F x t=m x delta v, impulse is the change in momentum. Instinct causes the driver to stiffen his legs against the impending crash.
Reflection form wear your seatbelt. If your system is subject to some non-negligible net external force then its overall momentum will change by Newton's second law. product of force and time. Crumple zones, crush zones, or crash zones are a structural safety feature used in vehicles, mainly in automobiles, to increase the time over which a change in velocity (and consequently momentum) occurs from the impact during a collision by a controlled deformation; in recent years, it is also incorporated into trains and railcars.. Crumple zones are designed to increase
In particular, we will focus upon. The train car on the left, mass m 1, is moving with speed V o when it collides with a stationary car of mass m 2. Distinguish between impulse and momentum. Momentum = (Mass x Velocity) or. 2. ACT 1: Momentum practice CPO Watch an updated version at https://youtu.be/2XKOzibVqJgWhat happens to vehicles and their occupants in crashes is determined by science. At a stoplight, a large truck (3000 kg) collides with a motionless small car (1200 kg). The truck comes to an instantaneous stop; the car slides straight ahead, coming to a stop after sliding 10 meters. LAB 5: Egg Drop. Lesson 3: Energy, Momentum, and Understanding Car Crashes Many of us have lost students to violent motor vehicle crashes. CRASH COURSE ACTIVITY Understanding Car Crashes Video 11 Crash Course Definitions impulse:product of force and time interval during which the force acts;impulse equals change in momentum, Ft=(mv) impact: qualitative term for force T E A C H E R L E S S O 3N Here, m is mass and v is velocity. A large force is required for an object to lose momentum quickly (such as a car stopping quickly when it hits a stationary object). Word Document File; Video notes to go along with the youtube video (link on page) that discusses momentum and impulse as it relates to car crashes. This is known as the impulse-momentum change theorem. A larger impulse means that a greater force is experienced by the occupants of the cars. Blog. They experience great forces because of the change in momentum which can maybe our impulse and momentum calculator would be more useful. In this part of Lesson 1, we will examine some real-world applications of the impulse-momentum change theorem. a property of a moving object that depends on the object's mass and velocity. If two things crash together, their total momentum is the same before and after the collision. ) Both can be applied d to a car crash, the momentum before impact is equal to after impact , and the EKE before the impact is equal to PEe+EKE+heat during impact. The Role of Mass And Energy.
Using the impulse formula, J=force X time, the impulse would be -3,330 Ns. When people run, walk, etc. d is the distance traveled at the time of collision. The car would have the maximum momentum if a bike and a car ride down the street at the same velocity. Relate Impulse and Momentum to collision of objects (eg. Conservation of Momentum When two or more bodies act on one another in a collision, the total momentum of the bodies remains constant. The impulse - momentum relationship is important to understand safety in car crashes. where. Intuitively, it seems obvious that the collision time is the same for both cars, but it is only true for objects traveling at ordinary speeds. Simple example using the impulse-momentum theorem to calculate the force on a driver in a car crash with different assumptions about the distance over which the driver comes to a stop. Example $$\PageIndex{4}$$: analyzing a car crash. 1. a = F / m,. The car hits a wall and rebounds (moving west) with a speed of 0.100 m/s. - The hood crumples, rises and strikes the windshield. Wheels are fast, and traffic can go really fast much faster than the faster person can run. Newton explained the relationship between crash forces and inertia in his circle one 1st 2nd 3rd Law of Motion. m/s 2 Calculate the cars new momentum if its velocity is doubled. Why are cars designed to be destroyed in a collision? If the impact lasts 0.03s, then calculate the average force acting on the van during the crash. If the tennis ball was at rest before it was hit, its final momentum is equal to the impulse, 1.4 kg m/s. Principle of Angular Impulse and Momentum (Learn to solve any problem) GCSE Physics - Momentum Part 1 of 2 - Conservation of Momentum Principle #59 Principle of Work and Energy (Learn to solve any problem) Conservation of Linear MomentumCollisions: Crash Course Physics #10 Momentum Change of momentum and Impulse. C. measure of an objects inertia in motion. Technically, yes, but applying the concept of momentum and impulse, you will know that the change in momentum will occur over a short period of time, therefore, the hitting force is greater, making your car crash (and stop) completely. 1 Answer. p 2 F t p 1 You can see now that the balls final momentum is the sum of the ini-tial momentum and the impulse. 5. LAB 3: Web lab impulse and momentum. Momentum (P) is equal to mass (M) times velocity (v). Using the definition of impulse, the change in momentum of car 1 is given by p 1 = F 1 t, where F 1 is the force on car 1 due to car 2, and t is the time the force acts (the duration of the collision). Boyd Science. the effect of collision time upon the amount of force an object experiences, and. You can figure out something's momentum by multiplying its mass by its velocity: momentum = mv. Ex 6.2 In terms of impulse and momentum, why are padded dashboards safer in automobiles? The conservation of momentum applies to an isolated system. Video is just over 22 minutes long. Video Notes: Understanding car crashes. Drifting Mania. During a crash, there is change in momentum (which product of . Put another teddy in another toy car without a seat belt. Explain your answer. In many cases, an object needs to be brought to rest from a certain initial velocity. After the collision, the car does not rebound, the final velocity thus being zero. In this case the car-wall system is acted on by a net external force (the contact force of the wall with the ground). The physics car crash one! Find the change in the momentum of the car if the mass of the car is 4 tons. Set the velocity of the car to -30 m/s. Crash Test Dummy (Momentum, Impulse) | Physics | CK-12 Exploration Series (The negative sign simply indicates the car is moving toward the left. Experiment with the number of discs, masses, and initial conditions Now you can perform the classic momentum lab with all the same calculations, but without the inconvenient physical air track and photogates The volume is essentially zero During an elastic collision, the total momentum in both the i direction and the j 1. Endless Racer 3D (Unity) Run Race 3D 2. m is the mass of an object. 2. Impulse = (force) (time) Impulse = change in momentum. In most of the case mass is constant and for momentum change velocity changes. 4.
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https://iuee.eu/en/a-recipe-calls-for-6-cups-of-water-and-4-cups-of-flour-a-what-is-the-ratio-of-water-to-flour-b.6246783.html
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InDumire799
32
# A recipe calls for 6 cups of water and 4 cups of flour. a) what is the ratio of water to flour b) if the recipe is increased to use 6 cups of flour, how much water should be used c) If the recipe is decreased to use 2 cups of water, how much flour should be used NEED HELP ASAP
Jordanstarnes
A) 6:4B) 9 cups of water for 6 cups of flourC) 3 cups of flour for 2 cups of waterHope it helps! :)
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http://www.chegg.com/homework-help/street-crossing-consider-child-waiting-street-corner-gap-tra-chapter-4.5-problem-66a-solution-9780321421326-exc
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# Calculus with Applications (9th Edition) View more editions Solutions for Chapter 4.5 Problem 66AProblem 66A: Street Crossing Consider a child waiting at a street corner ...
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Chapter: Problem:
Street Crossing Consider a child waiting at a street corner for a gap in traffic that is large enough so that he can safely cross the street. A mathematical model for traffic shows that if the expected waiting time for the child is to be at most 1 minute, then the maximum traffic flow, in cars per hour, is given by
where x is the width of the street in feet. Find the maximum traffic flow and the rate of change of the maximum traffic flow with respect to street width for the following values of the street width. Source: An Introduction to Mathematical Modeling.
a. 30 ft
b. 40 ft
STEP-BY-STEP SOLUTION:
Chapter: Problem:
Corresponding Textbook
Calculus with Applications | 9th Edition
9780321421326ISBN-13: 0321421329ISBN:
Alternate ISBN: 9780321422743, 9780321450319, 9780321455697, 9780321518897, 9780321563361, 9780321570109, 9780558840297
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https://www.archivemore.com/how-do-you-calculate-height-from-the-length-of-a-femur/
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How do you calculate height from the length of a femur?
To calculate the estimated height based on the person’s femur, first measure the femur in centimeters. If the subject is female, multiply the length by 2.47 and add 54.1 to arrive at the approximate height. If the subject is male, multiply by 2.32 and add 65.53.
How is stature calculated from bones?
Stature estimation is obtained from measurements of long bones; namely the humerus, femur, and tibia. If these bones are unavailable, the ulna, radius, and fibula can also provide a good range for the expected height of an individual.
How long should my tibia be?
The average length of the male tibia on the right was 36.45 cm.; on the left, 36.48 cm. In females the right tibia measured on an average 34.5 cm.; the left, 34.6 cm. The scale of length in the male varies from 31.0 to 45.5 cm., and in the female from 28.0 to 39.0 cm.
What is femur length in ultrasound?
Femur length measures the longest bone in the body and reflects the longitudinal growth of the fetus. Its usefulness is similar to the biparietal diameter (BPD, which is the diameter between the 2 sides of the head.
What if femur length is short?
Fetuses with shorter-than-expected femur length have been found to be at higher risk for skeletal dysplasia, otherwise known as dwarfism. 2 This is different from short stature, which is a height that is three or more standard deviations below the mean for age but is proportional.
What is normal femur length in pregnancy?
The foetal femoral length growth presents a characteristic appearance between the 12th and 42nd pregnancy week. In the 12th pregnancy week it is 11 mm on the average, 33 m in the 20th, 58 mm in the 30th, and 76 mm at birth.
How accurate is femur length in ultrasound?
It is well established that ultrasound measurement of femur length and biparietal diameter are comparably accurate estimators of gestational age when obtained in the first half of pregnancy. Both estimators, however, become less accurate later in pregnancy.
Does femur length indicate height?
The femur is the most commonly used bone for estimating the stature. It was also noted that the femur length was approximately 4 times the height of the person. This study showed that the femur length can be used for estimating the height of the individuals.
How can I improve my fetal femur length?
Amongst a mass of evidence in literature, only Chang et al. (14) established the role of milk on growth of femur length of foetus, showing that higher milk intake is related to better growth of femur length of foetus.
When should I be concerned about femur length?
In general your height and that of your husband will decide whether your baby will have long or short femurs. In general if the discrepancy is more than three weeks should you be worried. (meaning at 22 weeks the femur length is less than 19).
What is considered short femur in fetus?
A short femur was considered isolated when both the abdominal circumference and estimated fetal weight measured greater than the 10th percentile for gestational age. All gestational age-specific biometry values were determined by standards derived by Hadlock. IUFD was defined as fetal death ≥20 weeks’ gestation.
What causes a short femur?
In most cases, researchers suspect congenital short femur is caused by a disruption during early prenatal development, which can be caused randomly, or as a result of an outside force such as an infection or trauma.
Can you detect dwarfism on an ultrasound?
How Is Dwarfism Diagnosed? Most pregnant women have a prenatal ultrasound to measure the baby’s growth at around 20 weeks. At that stage, features of achondroplasia aren’t yet noticeable.
How soon can dwarfism be detected?
Disproportionate dwarfism is usually apparent at birth or early in infancy. Proportionate dwarfism may not be diagnosed until later in childhood or the teenage years if your child isn’t growing at an expected rate.
Is dwarfism picked up in pregnancy?
If one or both parents have a family history of a condition that causes short stature, their baby can be tested for this during pregnancy (prenatal diagnosis).
Can you tell if a baby is a little person?
Doctors are able to diagnose most cases of achondroplasia even before birth by doing an ultrasound in the later stages of pregnancy. The ultrasound can show if a baby’s arms and legs are shorter than average and if the baby’s head is larger.
4 feet 10 inches
Does small baby mean small person?
Babies who are born small for their gestational age (SGA) tend to remain small as children. This study tracked SGA babies past their adolescent growth . . . Babies who are born small for their gestational age (SGA) tend to remain small as children.
Is being a little person a recessive gene?
One is recessive, which means you inherit two mutated genes (one from each parent) to have the condition. The other is dominant. You only need one mutated gene — from either parent — to have the disorder. Other risk factors for dwarfism include a hormone deficiency or malnutrition.
How is dwarfism a dominant trait?
Achondroplasia is inherited in an autosomal dominant pattern, which means one copy of the altered gene in each cell is sufficient to cause the disorder. About 80 percent of people with achondroplasia have average-size parents; these cases result from new mutations in the FGFR3 gene.
Is achondroplasia dominant or recessive?
When achondroplasia is inherited, it is inherited in an autosomal dominant manner. Over 80 percent of individuals who have achondroplasia have parents with normal stature and are born with achondroplasia as a result of a new (de novo) gene alteration (mutation).
What body systems are affected by achondroplasia?
Achondroplasia is the most common form of skeletal dysplasia, occurring in about one in every 40,000 births. Achondroplasia impairs the growth of bone in the limbs and causes abnormal growth in the spine and skull.
How long is the average lifespan of a person with achondroplasia?
According to the NHGRI, most people with achondroplasia have a normal life span. However, there’s a slightly increased risk of death during the first year of life. There may also be an increased risk of heart disease later in life.
Can achondroplasia be prevented?
Currently, there is no way to prevent achondroplasia, since most cases result from unexpected new mutations. Doctors may treat some children with growth hormone, but this does not significantly affect the height of the child with achondroplasia. In some very specific cases, surgeries to lengthen legs may be considered.
Why is achondroplasia not curable?
Currently there are no treatments able to reverse achondroplasia, which is caused by mutations in a gene — called FGFR3 — that result in the excess production of proteins that slow bone growth, nor are there ways to treat the genetic culprit itself.
How does achondroplasia affect everyday life?
Affected patients experience various orthopedic and neurological complications and might face multiple medical and non-medical challenges in their daily life [5,6,7,8]. Adult patients reported physical and mental impairments as well as lower quality of life and lower self-esteem than healthy relatives [9, 10].
What is the prognosis for achondroplasia?
Survival. Most of those with achondroplasia will have a normal or near normal life expectancy. However, there is an increased risk for premature death [107,108,109] related not only to sudden unexpected deaths in infancy (see below) but also, it appears, to cardiovascular complications in mid-adult life [108].
What limitations does a person with achondroplasia have?
limited joint flexibility and arthritis. lower back pain or leg numbness. recurring ear infections and risk of hearing loss.
Are the femur and tibia the same length?
No, the femur and tibia are not the same lengths. While they are the two longest bones in the body, the femur is between three and four inches longer…
Is femur or tibia longer?
The femur is the single bone of the thigh. The patella is the kneecap and articulates with the distal femur. The tibia is the larger, weight-bearing bone located on the medial side of the leg, and the fibula is the thin bone of the lateral leg.
0.80
Why is my femur longer than my tibia?
Structural LLD occurs when either the thigh bone (femur) or the shin bone (tibia) is shorter in one leg than in the other. The condition typically presents at birth, but it can also happen as a child grows. Some potential causes of structural LLD include: Bone injuries: Bone breaks can slow down bone growth in one leg.
What is a normal leg length?
In a nutshell:
MINUS (short) STANDARD (normal)
Inside leg length 75 cm | 29.5 inches 81 cm | 32 inches
Length with large hem 80 cm | 31.5 inches 86 cm | 34 inches
What leg length is regular women’s?
Women’s leg lengths
Length cm inches
Short 76 30
Regular 81 32
Long 86 34
Tall 91 36
How do I know my leg size?
Place a flat object between your legs at crotch height like a tall, thin book, a level, or a ruler to use as a reference. Then, use a tape measure to measure form the floor to the top of your reference object. You can use your inseam measurement to help you choose trousers with the right leg length for you.
Is a 32 leg regular or long?
These are inside leg measurements and are a guide to fit only. The garment lengths will vary according to the style….Jeans & Trousers inside leg length.
Size To fit Inside leg length
Inches CM
Regular 32 81
Long 34 86
What length are M&S regular trousers?
All Size Guides
Trouser leg lengths To fit inside leg *Actual trouser inside leg length will vary according to style
Short 29
Standard (Medium) 31
Long 33
Is M size 10 or 12?
M means medium, which translates roughly to women’s size 8 to 10.
What size is medium in marks?
Dual Size
Small Medium Large
36-38 40-42 44-46
34-36 38-40 42-44
82-87 92-97 102.5-108
65-70 75-80.5 86-92
Is a size 16 a medium or large?
HOW TO MEASURE?
Global Size UK USA
Small 8-10 4-6
Medium 12 8
Large 14-16 10-12
X-Large 18 14
Can I get measured for a bra at M&S?
Our online bra fitters will do their best in all conditions. However, you’ll get the most out of your appointment if you wear your best-fitting bra, have stable internet connection and join your appointment from a quiet, private space. If possible, we’d suggest having a tape measure to hand, too.
Are M&S doing bra measuring?
Our priority is helping our customers shop with confidence and we have extensive measures in place across our stores to ensure the safety & wellbeing of everyone. Whilst our expert teams can still advise you on finding the perfect style and size, our bra fitting service remains paused in some stores.
Where can I get my bra fitted?
Big shops that carry a line of intimates like Bloomingdale’s, Nordstrom, and Macy’s, are known to have bra fitting experts who can help you out with getting measured and picking the perfect comfy bra for you.
What cup size is 32 inches?
US Bra Size Tables
US Inches Centimeters
Bra Size Band, Under Bust Cup, Over Bust
32C 28-30 in 86-89 cm
34C 30-32 in 91-94 cm
36C 32-34 in 97-99 cm
Why do you add 4 inches to band size?
It was designed as a method for measuring way back in the 1950s when bras were made from silk and satin and did not stretch. The 4 inches extra allowed breathing room, something which in 2015 is completely redundant as all bras now have elastic and stretch in them.
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# Boyle's Law Lab Report
767 Words4 Pages
Purpose/Objective The purpose of this experiment is to test the validity of Boyle’s Law. The objective is to determine the relationship between the pressure and volume of a confined gas. Materials Pencil Lab notebook Pressure sensor Syringe Chemicals Air Data and Results Volume (mL) Pressure (kPa) P/V PxV 5.0 mL 192.64 kPa 38.53 kPa/mL 963.20 kPa*mL 7.5 mL 136.13 kPa 18.13 kPa/mL 1020.96 kPa*mL 10.0 mL 104.54 kPa 10.45 kPa/mL 1045.40 kPa*mL 12.5 mL 83.50 kPa 6.68 kPa/mL 1043.75 kPa*mL 15.0 mL 70.51 kPa 4.70 kPa/mL 1057.65 kPa*mL 17.5 mL 60.37 kPa 3.45 kPa/mL 1056.48 kPa*mL 20.0 mL 53.22 kPa 2.66 kPa/mL 1064.40 kPa*mL Boyle’s Law Plot Sample Calculations P/V: 53.22 kPa/20.0 mL = 2.66 kPa/mL PxV: 53.22 kPa x 20.0 mL = 1064.40 kPa*mL Discussion and Comments This experiment was a simple and expedient method of proving Boyle’s law by demonstrating that pressure and volume have an indirect relationship. When the volume was doubled from 5.0 mL to 10.0 mL the pressure was nearly halved from 192.64 kPa to 104.54 kPa. When the volume is halved from 20.0 mL to 10.0 mL the pressure is almost doubled from 53.22 kPa to 104.54 kPa. The relationship between the pressure and volume of a confined gas is inverse. The line of the graph is curved not straight. This indicates an indirect relationship. The amount of the gas and the temperature are assumed to be constant in this experiment. PV=k. Pressure is inversely proportional to volume. The errors in this experiment were systematic due to not properly attaining the correct amount of gas in the syringe. For example, when the amount of gas was doubled from 5.0 mL to 10.0 mL, the pressure is halved and the reading should have been 96.32 kPa as opposed to 104.54 kPa. This experiment correlated with the material present in our textbook in
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# So, I need help figuring out how much canned food to give my dogAugust 6, 2018 4:01 PM Subscribe
I am going to be giving my elderly lab dog (12 years old) a sensitive stomach dog food (4health brand at Tractor Supply). I need to figure out how much to give her, I feed her twice a day morning and evening.
She weights 68 lbs, the calorie content on the can says:
Metabolizable Energy (ME) 1,102 kcal ME/kg; 412 kcal/can
So, how many calories are in a can of food - 1102 or 412 ?
Also could use some suggestions on how many calories to give her a day. She is not very active, not as much as when she was younger of course.
There are 412 calories in a can. Is there no weight chart to guide you on the can or the pack of cans? I'd feed one in the morning and one at night and then adjust if she's gaining or losing weight. We weigh weekly when we're establishing feeding amounts for a specific dog; metabolism varies.
posted by DarlingBri at 4:25 PM on August 6, 2018 [1 favorite]
I'm guessing the can is less than 1 kg, it's likely 0.37386 kg (which is 412 / 1102). As I read it, there are 412 kcal per can, and 1102 kcal per kilogram.
posted by smcameron at 4:25 PM on August 6, 2018 [1 favorite]
412 calories (kcal) per can. The other figure is for a kilogram.
A dog of around 70lb with low activity probably needs about three of those cans (around 1200 calories a day). Most dogs need more calories in the winter than in the summer.
As DarlingBri advises, weigh your dog every week and see how it goes. Adjust calories to keep your dog at a healthy weight.
posted by pipeski at 4:30 PM on August 6, 2018
The food cans I get say the same thing, it's 412 calories per can. My dog weighs the same as yours and I give him 4 cans a day, but he is 4 and active. I would give your dog 3 cans and weigh her to see how it's going. You should also factor in treats and so on into the overall calorie intake.
posted by ananci at 10:00 PM on August 6, 2018
IDK but for the last three years I have been giving my same size dog two cups high quality dog food, add water, add half a cup canned pumpkin, microwave for 1 & 1/2 minutes, add 1/2 cup yogurt, serve.
The pumpkin and yogurt have been great for his stomach. And he loves it. The microwaving part is essential.
I also give him some canned fish but that is controversial because it has salt in it but I think it is the right thing for him. He is thriving.
And then I add or subtract the kibble part of the recipe according to his weight gain or loss.
posted by cda at 3:10 AM on August 7, 2018
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# Compute volume from mesh Mathematical demonstration
## Recommended Posts
Hi,
I start a new topic from an older because I have a tiebreaker !
the older topic : [url="http://www.gamedev.net/topic/312027-is-it-possible-to-calculate-the-volume-for-a-mesh/"]http://www.gamedev.n...ume-for-a-mesh/[/url]
My question is: How can we prove the formula V= sum(each tetrahedron volume with the origin)
[source lang="cpp"]
float volume = 0;
int* index = I;
for (i = 0; i < T; i++)
{
Vector3 v0 = V[*index++];
Vector3 v1 = V[*index++];
Vector3 v2 = V[*index++];
volume += Dot(v0,Cross(v1,v2));
}
volume /= 6;
[/source]
I know I must use 3D integrals and Fourier moments but I don't know how I can do that !
Please someone give me some clues about a possible math proof ?
PS: I already try to find this formula with the gauss theorem as said in the older topic but I never find the result but an approximate result! I obtain the formuma here but with some addition in more...
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This formula is a simple result of the divergence theorem.
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but I don't see how it is a "simple" result of the divergence theorem because you don't compute with element surface but with tetrahedron where the base is each triangle of the parametrized surface and the common point is the origin of the 3d space (0,0,0).
The divergence theorem said that the volume is equal to the outward flux of the surface, isn't it ?
Here we don't use surface but volume of tetrahedron composed with origin ?
Could you explain me with math if it's possible ?
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Alright.
Take the vector field: F=(x-x0)/3, x and x0 are in R^3.
Its divergence with respect to x is 1.
So: Volume = Int[Div(F)*dV] = Int[<F,dS>] = Sum[A_i * (d_i - <n_i,x_0>)]/3
Where: A_i = Area of triangle. With its plane equation <n_i,x>=d_i, with unit normal.
If you choose x0=0, then you get Sum[A_i*d_i/3]
For unit n_i, d_i is exactly the distance of the plane from the origin. So what is written in the sum is exactly the volume of the tetrahedron. Edited by max343
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Hum, I'm really sorry but I don't understand your demonstration.
If you take vector field F (x,y,z) -> (x-x0,0,0)/3 the divergence is 1/3, not 1 ??
Moreover I don't understand how you insert your inner product d_i - <n_i,x0> and directly the area of a triangle after the gauss theorem
=> Int[<F,dS>] = Sum[A_i * (d_i - <n_i,x_0>)]/3 ??
One more time thanks !
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x and x0 are points in R^3 and not coordinates.
As for the second question:
F*dS is a two-form defined over the boundary of the volume, while <F,n>*dA is the corresponding two-form defined over subset of R^2.
F*dS = <F,n>*dA = (<x,n> - <x0,n>)/3*dA = (d - <x0,n>)/3*dA
Now, d - <x0,n>/3 is a constant within the triangle, so the integral is: A*(d - <x0,n>)/3.
I've been abusing notation here a bit. It makes little sense to do an inner product between 1-forms and vectors, but it should be fairly obvious what I meant in <~,n>.
Does this make more sense? Edited by max343
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Ok well,
I don't know it was possible to use points in the function F ? I believed it was just vector field ?
I have a last question (in my mind easy but I want to be sure) : I use only vertices in the code above (v0 v1 v2).
Or your variable d is a distance not a point.
How can we say that A*(d-<x0,n>)3 == 1/2 (P1^P2).d/3 where d must be a vertice ? (n is not obligatory (1,1,1)?)
Moreover if i use the formule of area it's also distance and not vertices....
Many thanks Edited by joohooo
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Well, vector fields assign vectors to points. So it's only natural to use points in them.
As for 'd', then yes, it's the distance from the plane to the origin when unit normal is used. Since during the integration 'x' lies on the plane, <x,n> is constant and it equals exactly to 'd'.
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Yes So it doesn't prove my formula with vertices ??
The aim of this formula is to not use distance but just vertices to gain time !
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I haven't been following the thread, but the code posted in the first post seems fine to me. Instead of "Dot(v0,Cross(v1,v2))" I would have written "Determinant(v0,v1,v2)", but other than that it's exactly how I would have written it.
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[quote name='joohooo' timestamp='1354205457' post='5005341']
Yes So it doesn't prove my formula with vertices ??
The aim of this formula is to not use distance but just vertices to gain time !
[/quote]
Why not?
For tetrahedron: A*d/3 == Det(v1-v0,v2-v0,v3-v0)/6, where 'A' is the area of some triangle and 'd' is the distance from the opposing vertex to the plane of the triangle. In the Det formula v1,v2,v3 define the same triangle, while v0 is the opposing vertex. Edited by max343
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[quote name='max343' timestamp='1354207366' post='5005355']
[quote name='joohooo' timestamp='1354205457' post='5005341']
Yes So it doesn't prove my formula with vertices ??
The aim of this formula is to not use distance but just vertices to gain time !
[/quote]
Why not?
For tetrahedron: A*d/3 == Det(v1-v0,v2-v0,v3-v0)/6, where 'A' is the area of some triangle and 'd' is the distance from the opposing vertex to the plane of the triangle. In the Det formula v1,v2,v3 define the same triangle, while v0 is the opposing vertex.
[/quote]
One more time thanks !
But ... I have one more question?? (yes again)
I agree with you for the Det but in the proof we have already use the origin for x0 ! Or in my formula the origin will be your V0 in the det formula as the opposing vertex for each triangle in the mesh. It's not a wrong thing to use the origin for the X0 in F and for the opposing vertex. In my mind , it's look like a ruse... no ?
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I'm not sure whether I got you right, but why is it a bad thing to use the origin? You can choose x0 to be whatever you want, and as long as it's constant this formula will work.
I'll try to rephrase. In the vector field F you can choose x0 as any point in R^3. After some manipulations we can interpret the result as the sum of volumes of multiple tetrahedrons that are comprised of the faces of the mesh and have one vertex in common, that is x0.
Geometrically this makes a lot of sense for sphere-like (or more generally convex) meshes, just choose x0 to be somewhere within the object and the formula should come trivially to you without the divergence theorem. For arbitrary oriented manifolds (or when x0 is not within the enclosed volume) this is not so clear, but the divergence theorem takes care of that. Edited by max343
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Yep,
I understand your last topic and that's why I decided to use this theorem (it's a very interesting thing).
But my worry is completely in the proof because we use the origin in the vector field (make a lot of sense to simplify the formula) AND I use the origin for one of my vertice (opposing vertex) in the Det formula to obtain the dot product of a cross product of three points (after simplification)...
Or in the formula we obtain A*d/3 where d == <x,n> because (<x,n>-<x0,n>)/3==<x,n>/3. To simplify the det formula I use x (opposing vertex)=Origin=xo. And it's here my problem....
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This is not a problem since this is exactly the same point.
Let's recap it by steps:
2. Afterwards, you choose a vector field whose divergence is 1. This vector field is defined up to a constant x0.
3. As any x0 is good, you choose one (for instance the origin) and obtain the flux integral by the divergence theorem.
4. Once finished computing the flux integral, [b]you choose to interpret the formula[/b] as the sum of volumes of tetrahedrons that share the vertex x0.
5. You rewrite the formula from step 4, with an equivalent formulation that uses Det. Again, this is just a change in interpretation.
In steps 2-5 x0 is fixated and chosen exactly once. Hence there's no problem.
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Two students took the same courses last year. Each received the following final grades. In evaluating the grades, student 2 claims that it is a tie, while student 1 says he has a better grade. Who...
consider the following dataset (3,3,3,3,4,4,4,4,5,6,6,6,8,8,9,10,11,12,13)(I already put it in order). Find: a. Mode, median, mean b. The minimum, maximum, three quartiles, IQR, variance,...
From our class, (20 females, 15 males), four students are to be chosen to form a committee. In how many different ways can the student be chosen such that there would be at least one female?
A hospital conducts a survey of patients' satisfaction with its service. Each survey respondent ranks the 3 items with which he/she is most satisfied with the hospital service. The items to be ranked...
2. Over the past five years, the public health graduates’ mean GPA is 2.75. The chair is interested in finding out whether graduates this year have made any significant improvement in GPA. So, she...
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https://oneconvert.com/unit-converters/power-converter/watt-to-centiwatt
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# Convert watt to centiwatt
## How to Convert watt to centiwatt
To convert watt to centiwatt , the formula is used,
$\mathrm{1W}=100\mathrm{cW}$
where the Watt to cW value is substituted to get the answer from Power Converter.
1 Watt
=
100 cW
1 cW
=
0.01 Watt
Example: convert 15 Watt to cW:
15 Watt
=
15
x
100 cW
=
1500 cW
## watt to centiwatt Conversion Table
watt (Watt) centiwatt (cW)
0.01 Watt 1 cW
0.1 Watt 10 cW
1 Watt 100 cW
2 Watt 200 cW
3 Watt 300 cW
5 Watt 500 cW
10 Watt 1000 cW
20 Watt 2000 cW
50 Watt 5000 cW
100 Watt 10000 cW
1000 Watt 100000 cW
### Popular Unit Conversions Power
The most used and popular units of power conversions are presented for quick and free access.
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https://testbook.com/question-answer/an-ammeter-has-a-current-range-of-0-5-a-and-its--62aaf843c85ba2b6501f57ae
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# An ammeter has a current range of 0 - 5 A, and its internal resistance is 0.4 Ω. In order to change the range to 0 - 50 A we need to add a resistance of _______
This question was previously asked in
MAHATRANSCO AE Electrical 2017 Official Paper
View all MAHATRANSCO AE Papers >
1. 0.4 Ω in series with the meter,
2. 1.0 Ω in series with the meter
3. 0.044 Ω in parallel with the meter
4. 0.055 Ω in parallel with the meter
Option 3 : 0.044 Ω in parallel with the meter
Free
MAHATRANSCO AE Electrical Full Test 1
2.8 K Users
130 Questions 150 Marks 120 Mins
## Detailed Solution
Concept:
We can extend the range of the ammeter by keeping a shunt resistance as shown:
R= internal resistance of the coil
Rsh = Shunt resistance
I = Required full-scale range
I= Full scale deflection of current
As the two resistances, Rm and Rsh are in parallel, the voltage drop across the resistance will be equal.
$${I_m}{R_m} = \left( {I - {I_m}} \right){R_{sh}}$$
$${R_m} = \left( {\frac{I}{{{I_m}}} - 1} \right){R_{sh}}$$
$${R_{sh}} = \frac{{{R_m}}}{{\left( {\frac{I}{{{I_m}}} - 1} \right)}}$$
$${R_{sh}} = \frac{{{R_m}}}{{\left( {m - 1} \right)}}$$
Where $$m = \frac{I}{{{I_m}}}$$
‘m’ is called multiplying power
Calculation:
Given:
Meter resistance (Rm) = 0.4 Ω
Full scale deflection current (Im) = 5 A
Required full scale reading (I) = 50 A
∴ The required shunt resistance that should be connected in parallel/shunt will be:
$${R_{sh}} = \frac{{{R_m}}}{{\left[ {\frac{I}{{{I_m}}} - 1} \right]}}$$
m = 50/5 = 10
$$R_{sh}=\frac{0.4}{9}=0.044\ \ \Omega$$
Important Points
• To increase the ranges of ammeter, we need to connect a small shunt resistance in parallel with ammeters.
• To increase the ranges of a voltmeter, we need to connect a high series of multiplier resistance in series with voltmeters.
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https://dsp.stackexchange.com/questions/53317/compare-two-signals-find-times-when-they-are-affected-by-the-same-processes
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# Compare two signals, find times when they are affected by the same processes
I have two (noisy) signals $$S_1(t)$$ and $$S_2(t)$$. They contain information on three distinct events occurring in a physical system. Event $$A$$ affects only $$S_1(t)$$, event $$B$$ affects both $$S_1(t)$$ and $$S_2(t)$$, and event $$C$$ affects only $$S_2(t)$$. I want to identify the sequence of events in the system by analyzing the signals.
Is there some generic way to identify if two strings of data contain correlated events and find the timing of said events?
More specifically, my signals take the form $$$$\textrm{d}S_i (t + \textrm{d}t) = S_i^{\textrm{true}}(t)\textrm{d}t + \textrm{d}W(t),$$$$ where $$S_i^{\textrm{true}}$$ is the noiseless value of the signal calculated from the state of the system at time $$t$$, $$\textrm{d}W(t)$$ is a Wiener element (uncorrelated Gaussian noise), and the statement is to be read as an update formula. The l.h.s. is what is available to me. In case of an event the true values change sign in the following manner: \begin{align*} A &\Rightarrow\ S_1^{\textrm{true}} \to -S_1^{\textrm{true}}, \\ B &\Rightarrow\ S_1^{\textrm{true}} \to -S_1^{\textrm{true}} \textrm{ and } S_2^{\textrm{true}} \to -S_2^{\textrm{true}},\\ C &\Rightarrow\ S_2^{\textrm{true}} \to -S_2^{\textrm{true}}. \end{align*} The way I am trying to do things right now: I am using a running average to get rid off some of the noise, then I find the second derivatives of the smoothed signals. If the true value changes, there will be a peak in the second derivative. I then determine whether there is a peak in both signals or just one at the given time. However, this does not work very well, as the derivatives pick up the noise, too.
Here is a plot of my raw and running window averaged signals: I have no experience with signal processing. Is there a canonical procedure? If the question is too broad, where can I find recipes to deal with issues as such?
You can model your system as a hypothesis testing problem. For example, let $$M_1$$ and $$M_2$$ denote the two systems which are correlated. Let the observed signal in $$M_1$$ and $$M_2$$ be given as \begin{align} y_1(t) = s_1(t) + n_1(t) \\ y_2(t) = s_2(t) + n_2(t), \end{align} respectively, where $$n_i(t) \sim \mathcal{N}(0, \sigma_i^2)$$, for $$i = \{1,2\}$$, and $$s_i(t)$$, for $$i = \{1,2\}$$, is the original signal. Since the two systems are correlated, you can model them as \begin{align} \begin{bmatrix} y_1(t) \\ y_2(t) \end{bmatrix} = \begin{bmatrix} 1 & \rho \\ \rho & 1 \end{bmatrix} \begin{bmatrix} s_1(t) \\ s_2(t) \end{bmatrix} + \begin{bmatrix} n_1(t) \\ n_2(t) \end{bmatrix}, \end{align} where $$-1 < \rho < 1$$ denotes the correlation constant between the two systems. Now, we can model the system as a hypothesis testing problem as follows: \begin{align} \mathcal{H}_0: \rho = 0, &\quad \mbox{uncorrelated model} \\ \mathcal{H}_1: \rho \ne 0, &\quad \mbox{correlated model}. \end{align} If you assume that $$s_1(t)$$ and $$s_2(t)$$ are known, then find the maximum likelihood estimate (MLE) of $$\rho$$ under $$\mathcal{H}_1$$. If you assume $$s_1(t)$$ and $$s_2(t)$$ are unknown, then use generalized likelihood ratio test, i.e., find the MLE of $$s_i(t)$$ under each hypothesis and replace $$s_i(t)$$ with its MLE.
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https://www.coursehero.com/file/p2hr130i/Please-use-this-module-with-care-Do-not-put-unnecessary-marks-on-any-part-of/
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# Please use this module with care do not put
• 23
Course Hero uses AI to attempt to automatically extract content from documents to surface to you and others so you can study better, e.g., in search results, to enrich docs, and more. This preview shows page 3 - 6 out of 23 pages.
Please use this module with care. Do not put unnecessary marks on anypart of this SLM. Use a separate sheet of paper in answering the exercises andtests. And read the instructions carefully before performing each task.If you have any questions in using this SLM or any difficulty in answeringthe tasks in this module, do not hesitate to consult your teacher or facilitator.Thank you.What I Need To KnowThis module was designed and written with you in mind. It is here tohelpyou solve problems involving permutation of identical objects andcircularpermutation. The scope of this module permits it to be used inmany
different learning situations. The lessons are arranged to follow thestandardsequence of the course but the pacing in which you read andcomprehendthe contents and answer the exercises in this module willdepend on yourability.After going through this module, you are expected to be able todemonstrate understanding of key concepts of permutation. Specifically, youshould be able to:1) illustrate the permutation of identical objects and circularpermutation; and2) solve problems involving permutation of identical objects andcircular permutation.1CO_Q3_Mathematics 10_Module 2What I KnowAre you ready? You are tasked to answer the following questionsbefore we proceed with our lesson. Do not worry, we only want to know howknowledgeable you are with the topics that we will be discussing in thismodule.DIRECTION: Read and analyze each item carefully. Write the letter of the
correct answer on the blank before the item number.___ 1. Which concept is involved when the letters of the word “LOVE”taken 4 at a time equals 24?
___ 2. In how many ways can the letters of the word “SWIMMING” bearranged?
___ 3. There are three identical green flags, three identical white flags,andtwo identical red flags. Using all eight flags, how manysignals canbe made?
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self
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Grammatical person
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http://mathhelpforum.com/discrete-math/139112-inclusion-exclusion-problem.html
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# Thread: Inclusion - Exclusion Problem
1. ## Inclusion - Exclusion Problem
Hi.
I am having some difficulty with this inclusion-exclusion problem, and I would love some help.
************************************************** ******
Determine the number of integer solutions to X1 + X2 + X3 + X4 = 19, where -5 ≤ Xi ≤ 10, for all 1 ≤ i ≤ 4.
What I have done is create another equation to solve, since we already know how to solve non-negative integer solutions, C(n + r - 1, r), but I don't know whether I have done this step right.
Let Y1 + Y2 + Y3 + Y4 = 24, where Xi ≤ 15, Yi = Xi + 5 for all 1 ≤ i ≤ 4. IS THIS RIGHT?
If this is right, I am pretty sure I know how to go on from here, but I am unsure of this particular step.
Thanks in advance
2. Originally Posted by purakanui
Determine the number of integer solutions to X1 + X2 + X3 + X4 = 19, where -5 ≤ Xi ≤ 10, for all 1 ≤ i ≤ 4.
What I have done is create another equation to solve, since we already know how to solve non-negative integer solutions, C(n + r - 1, r), but I don't know whether I have done this step right.
Let Y1 + Y2 + Y3 + Y4 = 24, where Xi ≤ 15, Yi = Xi + 5 for all 1 ≤ i ≤ 4.
Your approach is correct. The difficulty is that upper limit of 10.
This is one case where generating functions are useful.
Expand the expression $\displaystyle \left( {\sum\limits_{k = 0}^{15} {x^k } } \right)^4$. The coefficient of $\displaystyle x^{24}$ is the answer.
But if you must use inclusion/exclusion then the answer is:
$\displaystyle \binom{24+4-1}{4-1}-\left( {\sum\limits_{k = 16}^{24} {4 \cdot \binom{26-k}{2}} } \right)$
We remove all solutions in which a variable is at least 16.
3. ## Thanks for your help
I have used an online expander (the coefficient is 2265, do you blame me!) which is the same answer I got from the other approach (the inclusion exclusion way).
The Y1 + Y2 + Y3 + Y4 = 24 was wrong because of Yi = Xi + 5, however there are 4 Values, so I should of added (5 x 4) = 20, so Y1 + ... + Y4 = 39.
Thanks for your help, we are just covering generating functions now, and it's good to see both sides.
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https://everything.explained.today/Nil_ideal/
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# Nil ideal explained
In mathematics, more specifically ring theory, a left, right or two-sided ideal of a ring is said to be a nil ideal if each of its elements is nilpotent.[1] [2]
The nilradical of a commutative ring is an example of a nil ideal; in fact, it is the ideal of the ring maximal with respect to the property of being nil. Unfortunately the set of nil elements does not always form an ideal for noncommutative rings. Nil ideals are still associated with interesting open questions, especially the unsolved Köthe conjecture.
## Commutative rings
In commutative rings, the nil ideals are better understood than in noncommutative rings, primarily because in commutative rings, products involving nilpotent elements and sums of nilpotent elements are both nilpotent. This is because if a and b are nilpotent elements of R with an=0 and bm=0, and r is any element of R, then (a·r)n = an·rn = 0, and by the binomial theorem, (a+b)m+n=0. Therefore, the set of all nilpotent elements forms an ideal known as the nilradical of a ring. Because the nilradical contains every nilpotent element, an ideal of a commutative ring is nil if and only if it is a subset of the nilradical, and so the nilradical is maximal among nil ideals. Furthermore, for any nilpotent element a of a commutative ring R, the ideal aR is nil. For a noncommutative ring however, it is not in general true that the set of nilpotent elements forms an ideal, or that a·R is a nil (one-sided) ideal, even if a is nilpotent.
## Noncommutative rings
The theory of nil ideals is of major importance in noncommutative ring theory. In particular, through the understanding of nil rings—rings whose every element is nilpotent—one may obtain a much better understanding of more general rings.[3]
In the case of commutative rings, there is always a maximal nil ideal: the nilradical of the ring. The existence of such a maximal nil ideal in the case of noncommutative rings is guaranteed by the fact that the sum of nil ideals is again nil. However, the truth of the assertion that the sum of two left nil ideals is again a left nil ideal remains elusive; it is an open problem known as the Köthe conjecture.[4] The Köthe conjecture was first posed in 1930 and yet remains unresolved as of 2022.
## Relation to nilpotent ideals
The notion of a nil ideal has a deep connection with that of a nilpotent ideal, and in some classes of rings, the two notions coincide. If an ideal is nilpotent, it is of course nil. There are two main barriers for nil ideals to be nilpotent:
1. There need not be an upper bound on the exponent required to annihilate elements. Arbitrarily high exponents may be required.
2. The product of n nilpotent elements may be nonzero for arbitrarily high n.
Clearly both of these barriers must be avoided for a nil ideal to qualify as nilpotent.
In a right artinian ring, any nil ideal is nilpotent. This is proven by observing that any nil ideal is contained in the Jacobson radical of the ring, and since the Jacobson radical is a nilpotent ideal (due to the artinian hypothesis), the result follows. In fact, this has been generalized to right noetherian rings; the result is known as Levitzky's theorem. A particularly simple proof due to Utumi can be found in .
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## Algebra: A Combined Approach (4th Edition)
$\frac{x^{2}y^{2}}{49}$
Based on the power of a quotient rule, we know that $(\frac{a}{c})^{n}=\frac{a^{n}}{c^{n}}$ (where $n$ is a positive integer and $a$ and $c$ are real numbers). Therefore, $(\frac{xy}{7})^{2}=\frac{x^{2}y^{2}}{7^{2}}=\frac{x^{2}y^{2}}{49}$.
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