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https://math.stackexchange.com/questions/911314/rudins-topological-definition-of-an-open-set-does-it-disagree-with-the-metri | 1,721,658,597,000,000,000 | text/html | crawl-data/CC-MAIN-2024-30/segments/1720763517878.80/warc/CC-MAIN-20240722125447-20240722155447-00614.warc.gz | 326,794,902 | 43,614 | # Rudin's Topological Definition of an Open Set -- Does it Disagree with the Metric Space Definition?
I wanted to share this definition of an open set, which made me uncomfortable. It comes from Rudin's Real and Complex Analysis and begins with the definition of a topology:
A collection $\tau$ of subsets of a set $X$ is called a topology in $X$ if $\tau$ has the following three properties:
• $\emptyset \in \tau$ and $X \in \tau$.
• If $V_i \in \tau$ for $i=1,\ldots,n$, then $V_1 \cap V_2 \cap \cdots \cap V_n \in \tau$ (closure under finite intersections).
• If $\{V_{\alpha}\}$ is an arbitrary collection of members of $\tau$ (countable or uncountable), then $\bigcup\limits_{a} V_{\alpha} \in \tau$ (closure under infinite unions).
If $\tau$ is a topology in $X$, then the ordered pair $(X,\tau)$ is called a topological space, and the elements of $\tau$ are called the open sets in $X$.
My question is -- doesn't this general definition disagree with the definition of open and closed sets from metric spaces? For example, what if we take $X=[0,5]$, which we would all agree is a closed set. By definition of the topological space $(X, \tau)$, we will have $[0,5] \in \tau$. So, then, according to the definition of an open set, since $[0,5]$ is a member of the topology we would call $[0,5]$ an open set. This seems wrong!
Or worse: What if we take one of the $V_i \in \tau$ to be the set $\{3\}$. Then $\tau = \{\emptyset, \{3\}, [0,5]\}$, which meets the three criteria above. Then since $\{3\} \in \tau$, then by this definition we would call $\{3\}$ an open set! We would never call a singleton an open set.
Or for a third example, let's put in one more set in the topology. What if we took a set we know to be open, say the interval $(1,3)$. Then our topology becomes $$\tau=\{\emptyset, (1,3), \{3\}, (1,3], [0,5]\}$$ since $(1,3) \cup \{3\}=(1,3]$. Now our topology has both open, closed, and clopen members!
Hopefully you guys can see why I'm bothered with this definition of an "open set".
• You can wrap text in asterisks (no math mode) on this site to get italics. Commented Aug 27, 2014 at 19:56
• @SammyBlack Alternatively, _add an underscore_, to render add an underscore.
– user122283
Commented Aug 27, 2014 at 19:57
• Rudin's definition is fine and perfectly consistent. $[0,5]$ is open as a subset of itself in both the metric and Rudin sense. It's not open as a subset of $\mathbb R$, but Rudin never claimed it would be. (You might look up "subspace topology.") Commented Aug 27, 2014 at 20:04
• The set $[0,5]$ is an open subset of the space $[0,5]$ if the "metric space definition" is used and it is also a closed set. Commented Aug 27, 2014 at 20:42
Started to write a comment, turned out too long.
$[0,5]\notin\tau$ in the Euclidean topology; $[0,5]^c=(−∞,0)\cup(5,∞)\in\tau$. $\tau$ denotes the collection of open sets; closed sets are not included in $\tau$. This axiomatic characterization of topological spaces should be thought of as the definition of a topology. In metric spaces, we're dealing with a special topology, the metric topology. According to this, a set $U$ is defined to be open if $\forall x\in U$ $\exists\varepsilon_x>0$ such that $B(\varepsilon_x,x)\subseteq U$. You can check that this construction of a metric topology satisfies the topology axioms. But, once again, this is merely a special case.
You say you would never call a singleton an open set. You're right if you are working with the standard Euclidean metric topology on $\mathbb R$. But, then again, this is only a special case of a topology, albeit a very natural and appealing one. In fact, you can define, on any non-empty set, a topology according to which every subset is open (and also closed). This is called the discrete topology. There are many possible ways to endow a set with topologies.
Also, when you define topologies on $\mathbb R$ other than the Euclidean metric topology, you should immediately forget about thinking in terms of the latter when talking about open sets in the new topology. Trying to compare which sets are open in different topologies can be very confusing if you're accustomed to the Euclidean definition of openness.
• I see you are distinguishing between what you call the "Euclidean topology"... is this simply what you get when you take $X=\mathbb{R}$ in Rudin's definition? Commented Aug 27, 2014 at 20:12
• @Mathemanic Dear Mathemanic, have you heard of the usual topology? That is what triple_sec is referring to as the "Euclidean topology". Putting $X=\mathbf{R}$ gives you a plethora of topologies, e.g., $\tau=\{\emptyset,\mathbf{R}\}$, or $\tau=\mathcal{P}(\mathbf{R})$. Read the Wikipedia article on the usual topology, and compare it to i) what you learnt in high school, and ii) your question, triple_sec's answer, and my own answer.
– user122283
Commented Aug 27, 2014 at 20:16
• Ah okay! Or the Borel field / Borel sigma algebra is another topology on $\mathbb{R}$ (or maybe that's what you mean by $P(\mathbb{R})$? So in the usual topology, i.e. when $X=\mathbb{R}$, the closed and open sets are what I expect (from the metric space perspective of interior points and etc.). However taking $X \ne \mathbb{R}$ (e.g. $X=[0,5]$) is what causes the surprising results like $[0,5]$ and $\{3\}$ and $(1,3]$ being considered open sets, if I'm not mistaken. Thank you for your answers! Commented Aug 27, 2014 at 20:25
• @Mathemanic I actually mean the power set of $\mathbf{R}$. You are right in the rest of your statements in this comment.
– user122283
Commented Aug 27, 2014 at 20:33
• @Mathemanic Be careful: $\sigma$-algebras are a different concept from topologies—they have to satisfy different axiomatic properties (closed under complements and countable unions). $\mathcal P(\mathbb R)$ denotes the set of all subsets of $\mathbb R$ (incidentally, it is both a topology and a $\sigma$-algebra but this is a very rare coincidence). Taking $X\neq\mathbb R$ is not what causes the surprising result. What causes it is simply your choice to define a new topology. Remember, you can define a topology in any way you want as long as it satisfies the basic axioms. Commented Aug 27, 2014 at 20:33
There are multiple ways to define a valid topology on a given set. You are talking about all these things such as '$[0,5]$ cannot be an open set' in the context of standard topology on $\mathbb{R}$, which includes as 'open sets' all subsets that can be written as unions of disjoint open intervals, but if you define the topology on $\mathbb{R}$ otherwise things will be different.
• On $\mathbf{R}$, at least, since it's ordered, I usually call what you call the "standard topology" the "usual topology" or the "ordered topology". Anyway, this usual topology can be shown to coincide with the metric topology, so they're equivalent. (This is a note for the OP, not for the answerer, who I've upvoted.)
– user122283
Commented Aug 27, 2014 at 20:06
This is an enlarged comment.
You seem to have a misunderstanding. What is the definition of a topology? Let's look at your first bullet. You have that $X\in\tau_X$. So, if $X=[0,5]$, then $[0,5]\in\tau_{[0,5]}$.
Why would a singleton not be an open set? Let $X$ be a topological space endowed with the discrete topology, and let $x\in X$. Then, since $\{x\}\subset X$, and $X$ has the discrete topology, we have that $\{x\}$ is an open set!
The usual topology is different from the discrete topology, etc., on any topological space. You're only considering the usual topology, which is what one learns in high-school calculus, etc.
I will not address the metric topology, as I believe that the other answers explain it pretty clearly. However, I will remark that the usual topology on $\mathbf{R}$, or on any subset of it, is the ordered topology, and this can be shown to coincide with the metric topology, which is what you're thinking about.
• Aha! So maybe one should write $[0,5] \in \tau_{[0,5]}$, or in general $X \in \tau_X$, so it's understood that $[0,5] \notin \tau$ (the usual topology). So the subscript $X$ in $\tau_X$ makes it clear that it's a "special" topology and not necessarily true that $X=\mathbb{R}$ which is true for the "usual" topology, and we reserve "$\tau$" for denoting the "usual" or "standard" topology where $X$ is understood to be $\mathbb{R}$.) Thank you for your answer and for your helpful comments in triple_sec's answer. Commented Aug 27, 2014 at 20:51
• @Mathemanic In my answer I used $\tau_X$ to make it clear that I was referring to the topology of $X$. But you could alternatively say "Let $\tau_X$ be the usual/discrete/trivial topology of $X$ ..." Hope that's clear,
– user122283
Commented Aug 27, 2014 at 20:57
A metric space $X$ here isn't just a space that happens to have both a topology and a metric $d$; it's one in which the topology is generated by the open balls $B_r(x) = \{y\in X:\, d(x, y) < r\}$. Thus, for example, $\mathbb{R}$ with the usual topology and the metric $d(x, y) = |x - y|$ is a metric space; $\mathbb{R}$ with the finite-complement topology is just a topological space that happens to have a nice function defined on it.
Rudin is defining the concept of a topology on a set. For any given set, there may be several different topologies on it.
On a metric space $(X,d)$, one can define a topology on it by using the open balls as a basis. This will give one possible topology on $X$, the metric topology given by $d$ which you are probably familiar with.
However there are other topologies on $X$. For example, the discrete topology on $X$ consist of all subsets of $X$. This is the largest possible topology on any set $X$. Another example in the indiscrete topology consisting of just $X$ and $\emptyset$, which is the smallest topology.
Here is the abstract construction of the topology induced by a metric $d$ on some space $X$. There is a base of open sets called open balls of radius $r$ centered at some point $x$ which is defined as the set $\{y \in X : d(x, y) < r\}.$ We allow $r$ to vary over the entire positive reals and $x$ to vary over all of $X$ to get a collection of open sets $B(r, x)$ for each such $r$ and $x$.
Then to construct the full topology, we throw in all unions of such sets (finite intersections automatically get added this way.)
In the example of the real numbers this implies that the open sets are exactly the unions (possibly infinite) of the open intervals $(a, b)$.
Where I think you've got confused is in the notion of "a" topology on the real numbers versus the topology induced by the metric. The collection $\tau$ that you defined, after throwing in the entire real line is certainly "a" topology on the reals but it is not the topology that comes from the metric space structure. | 2,909 | 10,737 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "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": 1, "x-ck12": 0, "texerror": 0} | 3.34375 | 3 | CC-MAIN-2024-30 | latest | en | 0.909147 |
https://www.doubtnut.com/qna/8488229 | 1,725,939,952,000,000,000 | text/html | crawl-data/CC-MAIN-2024-38/segments/1725700651196.36/warc/CC-MAIN-20240910025651-20240910055651-00610.warc.gz | 684,532,341 | 34,108 | If the line segment joining the points A(a,b)andB(a,−b) subtends an angle θ at the origin, show that cosθ=a2−b2a2+b2.
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Step by step video & image solution for If the line segment joining the points A(a,b) and B(a, -b) subtends an angle theta at the origin, show that cos theta = (a^2 - b^2)/(a^2 +b^2). by Maths experts to help you in doubts & scoring excellent marks in Class 12 exams.
Updated on:21/07/2023
Knowledge Check
• Question 1 - Select One
If the line segment joining the points A(a,b) and B(c,d) subtends an angle θ at the origin, then cosθ is equal to
Aab+cd(a2+b2)(c2+d2)
Bac+bd(a2+b2)(c2+d2)
Cacbd(a2+b2)(c2+d2)
• Question 2 - Select One
If atanθ=b, then acos2θ+bsin2θ=
Aa
Bb
Ca
Db
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Doubtnut is the perfect NEET and IIT JEE preparation App. Get solutions for NEET and IIT JEE previous years papers, along with chapter wise NEET MCQ solutions. Get all the study material in Hindi medium and English medium for IIT JEE and NEET preparation | 499 | 1,710 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.9375 | 3 | CC-MAIN-2024-38 | latest | en | 0.829379 |
https://www.thedigitaltransformationpeople.com/channels/analytics/what-are-garch-models/ | 1,701,827,916,000,000,000 | text/html | crawl-data/CC-MAIN-2023-50/segments/1700679100575.30/warc/CC-MAIN-20231206000253-20231206030253-00457.warc.gz | 1,129,593,334 | 24,323 | # What are GARCH models?
Economists and statisticians who work with time-series data – data collected over many periods rather than one-slice-in-time cross sectional data – will have at least heard of ARCH and GARCH models. Wikipedia has a concise summary of these methods:
“In econometrics, the autoregressive conditional heteroscedasticity (ARCH) model is a statistical model for time series data that describes the variance of the current error term or innovation as a function of the actual sizes of the previous time periods’ error terms; often the variance is related to the squares of the previous innovations.
The ARCH model is appropriate when the error variance in a time series follows an autoregressive (AR) model; if an autoregressive moving average (ARMA) model is assumed for the error variance, the model is a generalized autoregressive conditional heteroskedasticity (GARCH) model.
ARCH models are commonly employed in modeling financial time series that exhibit time-varying volatility and volatility clustering, i.e. periods of swings interspersed with periods of relative calm. ARCH-type models are sometimes considered to be in the family of stochastic volatility models, although this is strictly incorrect since at time t the volatility is completely pre-determined (deterministic) given previous values”
In Introduction to Time Series Using Stata, Sean Becketti describes some practical reasons why they are used:
“Financial markets, among other things, produce ‘price discovery’, that is, asset prices that balance the willingness of relatively bullish investors to buy with the willingness of relatively bearish investors to sell. Of course, news arrives every day that shifts the balance of investor opinion about an asset’s likely future performance.
Some news— the unemployment and inflation rates, the balance of trade, and the like— affects investor views of the market as a whole, while other news— changes in management, updated sales figures, etc.— affects only a specific sector or company. Investors digest this information, update their willingness to buy or sell assets, and their actions change prices.
The types of information that affect asset prices are many and varied, and it is impractical to model them all. As a result, their impact on prices is part of the random component of our models, and the typical magnitude of their impact helps determine the magnitude of the residual variance.
During uneventful periods, little news of import arrives, so investor opinions change relatively little. Moreover, long periods of relatively stable prices tend to narrow the range of investor opinions of asset values. In contrast, sudden and unexpected news can shift investor opinion significantly.
More importantly, extraordinary events can sow confusion among investors. Investors may struggle for some time to settle on an updated opinion of value. In these circumstances, investors may exhibit heightened reactions to scraps of information that either increase concern or restore confidence.
During these periods, the variance of the random component of prices is increased— at least, that is the way these events are reflected in our models, which, after all, are simplified representations of reality.”
The Stata software time-series reference manual covers a wide range of topics and includes a glossary, which I’ve reproduced in edited form below.
Any copy/paste and editing errors are mine.
______________________________________________________
add factor. An add factor is a quantity added to an endogenous variable in a forecast model. Add factors can be used to incorporate outside information into a model, and they can be used to produce forecasts under alternative scenarios.
ARCH model. An autoregressive conditional heteroskedasticity (ARCH) model is a regression model in which the conditional variance is modeled as an autoregressive (AR) process. Although the conditional variance changes over time, the unconditional variance is time invariant because yt is a stationary process. Modeling the conditional variance as an AR process raises the implied unconditional variance, making this model particularly appealing to researchers modeling fat-tailed data, such as financial data.
ARFIMA model. An autoregressive fractionally integrated moving-average (ARFIMA) model is a time series model suitable for use with long-memory processes. ARFIMA models generalize autoregressive integrated moving-average (ARIMA) models by allowing the differencing parameter to be a real number in (−0.5,0.5) instead of requiring it to be an integer.
ARIMA model. An autoregressive integrated moving-average (ARIMA) model is a time-series model suitable for use with integrated processes. In an ARIMA(p,d,q) model, the data are differenced d times to obtain a stationary series, and then an ARMA(p,q) model is fit to this differenced data. ARIMA models that include exogenous explanatory variables are known as ARMAX models.
ARMA model. An autoregressive moving-average (ARMA) model is a time-series model in which the current period’s realization is the sum of an autoregressive (AR) process and a moving-average (MA) process. An ARMA(p,q) model includes p AR terms and q MA terms. ARMA models with just a few lags are often able to fit data as well as pure AR or MA models with many more lags.
ARMA process. An autoregressive moving average (ARMA) process is a time series in which the current value of the variable is a linear function of its own past values and a weighted average of current and past realizations of a white-noise process. It consists of an autoregressive component and a moving-average component; see autoregressive (AR) process and moving-average (MA) process.
ARMAX model. An ARMAX model is a time-series model in which the current period’s realization is an ARMA process plus a linear function of a set of exogenous variables. Equivalently, an ARMAX model is a linear regression model in which the error term is specified to follow an ARMA process.
Autocorrelation function. The autocorrelation function (ACF) expresses the correlation between periods t and t−k of a time series as function of the time t and the lag k. For a stationary time series, the ACF does not depend on t and is symmetric about k = 0, meaning that the correlation between periods t and t−k is equal to the correlation between periods t and t + k.
autoregressive (AR) process. An autoregressive (AR) process is a time series in which the current value of a variable is a linear function of its own past values and a white-noise error term. An AR(p) model contains p lagged values of the dependent variable.
Cochrane–Orcutt estimator. This estimation is a linear regression estimator that can be used when the error term exhibits first-order autocorrelation. An initial estimate of the autocorrelation parameter ρ is obtained from OLS residuals, and then OLS is performed on the transformed data.
cointegrating vector. A cointegrating vector specifies a stationary linear combination of nonstationary variables.
conditional variance. Although the conditional variance is simply the variance of a conditional distribution, in time-series analysis the conditional variance is often modeled as an autoregressive (AR) process, giving rise to ARCH models.
correlogram. A correlogram is a table or graph showing the sample autocorrelations or partial autocorrelations of a time series.
covariance stationary process. A process is covariance stationary if the mean of the process is finite and independent of t, the unconditional variance of the process is finite and independent of t, and the covariance between periods t and t−s is finite and depends on t−s but not on t or s themselves. Covariance stationary processes are also known as weakly stationary processes. See also stationary process.
cross-correlation function. The cross-correlation function expresses the correlation between one series at time t and another series at time t−k as a function of the time t and lag k. If both series are stationary, the function does not depend on t.
cyclical component. A cyclical component is a part of a time series that is a periodic function of time. Deterministic functions of time are deterministic cyclical components, and random functions of time are stochastic cyclical components. For example, fixed seasonal effects are deterministic cyclical components and random seasonal effects are stochastic seasonal components.
Deterministic trend. A deterministic trend is a deterministic function of time that specifies the long-run tendency of a time series.
difference operator. The difference operator ∆ denotes the change in the value of a variable from period t − 1 to period t.
dynamic forecast. A dynamic forecast uses forecast values wherever lagged values of the endogenous variables appear in the model, allowing one to forecast multiple periods into the future. See also static forecast.
dynamic-multiplier function. A dynamic-multiplier function measures the effect of a shock to an exogenous variable on an endogenous variable. The kth dynamic-multiplier function of variable i on variable j measures the effect on variable j in period t+k in response to a one-unit shock to variable i in period t, holding everything else constant.
endogenous variable. An endogenous variable is a regressor that is correlated with the unobservable error term. Equivalently, an endogenous variable is one whose values are determined by the equilibrium or outcome of a structural model.
exogenous variable. An exogenous variable is a regressor that is not correlated with any of the unobservable error terms in the model. Equivalently, an exogenous variable is one whose values change independently of the other variables in a structural model.
exponential smoothing. Exponential smoothing is a method of smoothing a time series in which the smoothed value at period t is equal to a fraction α of the series value at time t plus a fraction 1−α of the previous period’s smoothed value. The fraction α is known as the smoothing parameter.
forecast-error variance decomposition. Forecast-error variance decompositions measure the fraction of the error in forecasting variable i after h periods that is attributable to the orthogonalized shocks to variable j.
forward operator. The forward operator F denotes the value of a variable at time t + 1. A forward operator is also known as a lead operator.
frequency-domain analysis. Frequency-domain analysis is analysis of time-series data by considering its frequency properties. The spectral density function and the spectral distribution function are key components of frequency-domain analysis, so it is often called spectral analysis.
gain (of a linear filter). The gain of a linear filter scales the spectral density of the unfiltered series into the spectral density of the filtered series for each frequency. Specifically, at each frequency, multiplying the spectral density of the unfiltered series by the square of the gain of a linear filter yields the spectral density of the filtered series. If the gain at a particular frequency is 1, the filtered and unfiltered spectral densities are the same at that frequency and the corresponding stochastic cycles are passed through perfectly. If the gain at a particular frequency is 0, the filter removes all the corresponding stochastic cycles from the unfiltered series.
GARCH model. A generalized autoregressive conditional heteroskedasticity (GARCH) model is a regression model in which the conditional variance is modeled as an ARMA process. GARCH models are often used because the ARMA specification often allows the conditional variance to be modeled with fewer parameters than are required by a pure ARCH model. Many extensions to the basic GARCH model exist.
generalized least-squares estimator. A generalized least-squares (GLS) estimator is used to estimate the parameters of a regression function when the error term is heteroskedastic or autocorrelated. In the linear case, GLS is sometimes described as “OLS on transformed data” because the GLS estimator can be implemented by applying an appropriate transformation to the dataset and then using OLS.
Granger causality. The variable x is said to Granger-cause variable y if, given the past values of y, past values of x are useful for predicting y.
high-pass filter. Time-series filters are designed to pass or block stochastic cycles at specified frequencies. High-pass filters pass through stochastic cycles above the cutoff frequency and block all other stochastic cycles.
Holt–Winters smoothing. A set of methods for smoothing time-series data that assume that the value of a time series at time t can be approximated as the sum of a mean term that drifts over time, as well as a time trend whose strength also drifts over time. Variations of the basic method allow for seasonal patterns in data, as well.
impulse–response function. An impulse–response function (IRF) measures the effect of a shock to an endogenous variable on itself or another endogenous variable. The kth impulse–response function of variable i on variable j measures the effect on variable j in period t + k in response to a one-unit shock to variable i in period t, holding everything else constant.
independent and identically distributed. A series of observations is independent and identically distributed (i.i.d.) if each observation is an independent realization from the same underlying distribution. In some contexts, the definition is relaxed to mean only that the observations are independent and have identical means and variances.
integrated process. A nonstationary process is integrated of order d, written I(d), if the process must be differenced d times to produce a stationary series.
Kalman filter. The Kalman filter is a recursive procedure for predicting the state vector in a state-space model.
lag operator. The lag operator L denotes the value of a variable at time t−1.
linear filter. A linear filter is a sequence of weights used to compute a weighted average of a time series at each time period.
long-memory process. A long-memory process is a stationary process whose autocorrelations decay at a slower rate than a short-memory process. ARFIMA models are typically used to represent long-memory processes, and ARMA models are typically used to represent short-memory processes.
moving-average (MA) process. A moving-average (MA) process is a time-series process in which the current value of a variable is modeled as a weighted average of current and past realizations of a white-noise process and, optionally, a time-invariant constant. By convention, the weight on the current realization of the white-noise process is equal to one, and the weights on the past realizations are known as the MA coefficients.
multivariate GARCH models. Multivariate GARCH models are multivariate time-series models in which the conditional covariance matrix of the errors depends on its own past and its past shocks. The acute trade-off between parsimony and flexibility has given rise to a plethora of models.
Newey–West covariance matrix. The Newey–West covariance matrix is a member of the class of heteroskedasticity- and autocorrelation-consistent (HAC) covariance matrix estimators used with time-series data that produces covariance estimates that are robust to both arbitrary heteroskedasticity and autocorrelation up to a prespecified lag.
orthogonalized impulse–response function. An orthogonalized impulse–response function (OIRF) measures the effect of an orthogonalized shock to an endogenous variable on itself or another endogenous variable. An orthogonalized shock is one that affects one variable at time t but no other variables.
output gap. The output gap, sometimes called the GDP gap, is the difference between the actual output of an economy and its potential output.
partial autocorrelation function. The partial autocorrelation function (PACF) expresses the correlation between periods t and t−k of a time series as a function of the time t and lag k, after controlling for the effects of intervening lags. For a stationary time series, the PACF does not depend on t. The PACF is not symmetric about k = 0: the partial autocorrelation between yt and yt−k is not equal to the partial autocorrelation between yt and yt+k.
periodogram. A periodogram is a graph of the spectral density function of a time series as a function of frequency. Peaks in the periodogram represent cyclical behavior in the data.
phase function. The phase function of a linear filter specifies how the filter changes the relative importance of the random components at different frequencies in the frequency domain.
Phillips curve. The Phillips curve is a macroeconomic relationship between inflation and economic activity, usually expressed as an equation involving inflation and the output gap. Historically, the Phillips curve describes an inverse relationship between the unemployment rate and the rate of rises in wages.
portmanteau statistic. The portmanteau, or Q, statistic is used to test for white noise and is calculated using the first m autocorrelations of the series, where m is chosen by the user. Under the null hypothesis that the series is a white-noise process, the portmanteau statistic has a χ2 distribution with m degrees of freedom.
Prais–Winsten estimator. A Prais–Winsten estimator is a linear regression estimator that is used when the error term exhibits first-order autocorrelation; see also Cochrane–Orcutt estimator. Here the first observation in the dataset is transformed so that the first observation is not lost. The Prais–Winsten estimator is a generalized least-squares estimator.
random walk. A random walk is a time-series process in which the current period’s realization is equal to the previous period’s realization plus a white-noise error term. A random walk with drift also contains a nonzero time-invariant constant. The constant term δ is known as the drift parameter. An important property of random-walk processes is that the best predictor of the value at time t + 1 is the value at time t plus the value of the drift parameter.
recursive regression analysis. A recursive regression analysis involves performing a regression at time t by using all available observations from some starting time t0 through time t, performing another regression at time t + 1 by using all observations from time t0 through time t + 1, and so on. Unlike a rolling regression analysis, the first period used for all regressions is held fixed.
regressand. The regressand is the variable that is being explained or predicted in a regression model. Synonyms include dependent variable, left-hand-side variable, and endogenous variable. regressor. Regressors are variables in a regression model used to predict the regressand. Synonyms include independent variable, right-hand-side variable, explanatory variable, predictor variable, and exogenous variable.
rolling regression analysis. A rolling, or moving window, regression analysis involves performing regressions for each period by using the most recent m periods’ data, where m is known as the window size.
seasonal difference operator. The period-s seasonal difference operator ∆s denotes the difference in the value of a variable at time t and time t−s.
serial correlation. Serial correlation refers to regression errors that are correlated over time. If a regression model does not contain lagged dependent variables as regressors, the OLS estimates are consistent in the presence of mild serial correlation, but the covariance matrix is incorrect. When the model includes lagged dependent variables and the residuals are serially correlated, the OLS estimates are biased and inconsistent.
serial correlation tests. Because OLS estimates are at least inefficient and potentially biased in the presence of serial correlation, econometricians have developed many tests to detect it. Popular ones include the Durbin–Watson (1950, 1951, 1971) test, the Breusch–Pagan (1980) test, and Durbin’s (1970) alternative test.
smoothing. Smoothing a time series refers to the process of extracting an overall trend in the data. The motivation behind smoothing is the belief that a time series exhibits a trend component as well as an irregular component and that the analyst is interested only in the trend component. Some smoothers also account for seasonal or other cyclical patterns.
spectral density function. The spectral density function is the derivative of the spectral distribution function. Intuitively, the spectral density function f(ω) indicates the amount of variance in a time series that is attributable to sinusoidal components with frequency ω. The spectral density function is sometimes called the spectrum.
spectral distribution function. The (normalized) spectral distribution function F(ω) of a process describes the proportion of variance that can be explained by sinusoids with frequencies in the range (0,ω), where 0≤ ω ≤ π. The spectral distribution and density functions used in frequency domain analysis are closely related to the autocorrelation function used in time-domain analysis.
state-space model. A state-space model describes the relationship between an observed time series and an unobservable state vector that represents the “state” of the world. The measurement equation expresses the observed series as a function of the state vector, and the transition equation describes how the unobserved state vector evolves over time. By defining the parameters of the measurement and transition equations appropriately, one can write a wide variety of time-series models in the state-space form.
Static forecast. A static forecast uses actual values wherever lagged values of the endogenous variables appear in the model. As a result, static forecasts perform at least as well as dynamic forecasts, but static forecasts cannot produce forecasts into the future if lags of the endogenous variables appear in the model. Because actual values will be missing beyond the last historical time period in the dataset, static forecasts can forecast only one period into the future (assuming only first lags appear in the model); thus they are often called one-step-ahead forecasts.
stationary process. A process is stationary if the joint distribution of y1,…,yk is the same as the joint distribution of y1+τ,…,yk+τ for all k and τ. Intuitively, shifting the origin of the series by τ units has no effect on the joint distributions; the marginal distribution of the series does not change over time. A stationary process is also known as a strictly stationary process or a strongly stationary process. See also covariance stationary process.
steady-state equilibrium. The steady-state equilibrium is the predicted value of a variable in a dynamic model, ignoring the effects of past shocks, or, equivalently, the value of a variable, assuming that the effects of past shocks have fully died out and no longer affect the variable of interest.
stochastic cycle. A stochastic cycle is a cycle characterized by an amplitude, phase, or frequency that can be random functions of time. See cyclical component.
stochastic equation. A stochastic equation, in contrast to an identity, is an equation in a forecast model that includes a random component, most often in the form of an additive error term. Stochastic equations include parameters that must be estimated from historical data.
stochastic trend. A stochastic trend is a nonstationary random process. Unit-root process and random coefficients on time are two common stochastic trends.
structural model. In time-series analysis, a structural model is one that describes the relationship among a set of variables, based on underlying theoretical considerations. Structural models may contain both endogenous and exogenous variables.
SVAR. A structural vector autoregressive (SVAR) model is a type of VAR in which short- or long-run constraints are placed on the resulting impulse–response functions. The constraints are usually motivated by economic theory and therefore allow causal interpretations of the IRFs to be made.
time-domain analysis. Time-domain analysis is analysis of data viewed as a sequence of observations observed over time. The autocorrelation function, linear regression, ARCH models, and ARIMA models are common tools used in time-domain analysis.
trend. The trend specifies the long-run behavior in a time series. The trend can be deterministic or stochastic. Many economic, biological, health, and social time series have long-run tendencies to increase or decrease. Before the 1980s, most time-series analysis specified the long-run tendencies as deterministic functions of time. Since the 1980s, the stochastic trends implied by unit-root processes have become a standard part of the toolkit.
unit-root process. A unit-root process is one that is integrated of order one, meaning that the process is nonstationary but that first-differencing the process produces a stationary series. The simplest example of a unit-root process is the random walk.
unit-root tests. Whether a process has a unit root has both important statistical and economic ramifications, so a variety of tests have been developed to test for them. Among the earliest tests proposed is the one by Dickey and Fuller (1979), though most researchers now use an improved variant called the augmented Dickey–Fuller test instead of the original version. Other common unit-root tests include the DF–GLS test of Elliott, Rothenberg, and Stock (1996) and the Phillips–Perron (1988) test. Variants of unit-root tests suitable for panel data have also been developed.
VAR. A vector autoregressive (VAR) model is a multivariate regression technique in which each dependent variable is regressed on lags of itself and on lags of all the other dependent variables in the model. Occasionally, exogenous variables are also included in the model.
VECM. A vector error-correction model (VECM) is a type of VAR that is used with variables that are cointegrated. Although first-differencing variables that are integrated of order one makes them stationary, fitting a VAR to such first-differenced variables results in misspecification error if the variables are cointegrated.
white-noise process. A variable ut represents a white-noise process if the mean of ut is zero, the variance of ut is σ2, and the covariance between ut and us is zero for all s != t.
Yule–Walker equations. TheYule–Walker equations are a set of difference equations that describe the relationship among the autocovariances and autocorrelations of an autoregressive moving-average (ARMA) process.
Source: Stata 16 time-series reference manual, StataCorp
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## Presentation on theme: "Math 7 Midterm Review Round II. Question 1 Round to the nearest thousand: 457 231."— Presentation transcript:
Math 7 Midterm Review Round II
Question 1 Round to the nearest thousand: 457 231
Question 2 Round to the nearest hundred 352 043.21
Question 3 Round to the nearest tenth 15.273
Question 4 Round to the nearest hundredth 321.8765
Question 5 Round to the nearest thousand 824 302.5
Question 6 Round to the nearest tens place 47.254
Question 7 Estimate 152.1 + 38.25
Question 8 Estimate 708.1 – 198.23
Question 9 Estimate 125.75 x 4.1
Question 10 Estimate 490.2 x 8.9
Question 7 What are the divisibility rules for: 3? 9?
Question 8 The number 52__34 is divisible by 3 and 9. What is the missing number?
Find the LCM of 7 and 8
Find the LCM of 4 and 6
Find the GCF of 25 and 30
Find the GCF of 20 and 24
Every 5 th car on an assembly line is blue. Every 6 th car is a convertible. What is the position of the first blue convertible?
Question 9 List the factors of 15.
Question 10 List the factors of 32.
List the factors of 20
Question 13 Write the following as a mixed number: 15 4
Question 14 Write the following as a mixed number: 29 5
Question 15 Write the following as a decimal: 8⅓
Question 16 Write the following as a decimal: 2¼
Question 17 Order the fractions from smallest to greatest: 4 9 2 3 10 5
Question 18 Write the following as a fraction in lowest terms: 0.25
Question 19 Write the following as a fraction in lowest terms: 0.32
Question 20 Write the following as a fraction in lowest terms: 0.22
Write the following as a fraction in lowest terms: 0.3333333 …
Question 30 Write an equivalent fraction to the following: 2 5
Question 31 Find the missing number: 5 = 15 10 ?
Question 32 Write the following fraction in lowest terms: 10 35
Question 33 Write the following fraction in lowest terms: 21 30
Question 34 Calculate 9 _ 3 10 4
Question 35 Calculate 4 + 5 5 6
Question 37 On a typical hockey team, there are 12 forwards, 6 defence, and 2 goalies. In lowest terms, what fraction of the team are Forwards Defence Goalies
Question 36 Calculate 5 1/2 + 3 4/9
Question 39 Calculator permitted for the following questions Express the following as a percent: 20 45
Question 40 Express the following as a percent : 40 57
Question 41 Express the following as a percent: 56 60
Question 42 Charlie bought a new video game. It was originally priced at \$49.99 and now it is on sale for 15% off. What is the sale price?
Question 43 Katie is a babysitter and got paid \$9.00/hour in 2012. She is increasing her fee by 6% in 2013. What will her new hourly fee be?
Question 44 Ryan scored 37 out of 42 on his Math midterm. What was his percent?
Question 21 Solve the following: 7.6 + 2.31 + 5
Question 22 Solve the following: 195.4 – 23.27
Question 23 Solve the following: 5.6 x 23
Question 24 Solve the following: 3.25 x 4.7
Joe sells muffins for \$0.49 each. He sells 32 muffins. The muffins (in total) cost \$10.25 for him to buy from the store. What is his profit?
Mr. Fraser is selling Cabot’s Crossing CDs for \$15 each. He hopes to sell the 25 he just made. The discs cost \$2.25 each to produce. If he sells all 25 CDs, how much profit does he make? Mr. Fraser has to divide the profit equally among the 4 band members. How much does each person get?
Question 25 Solve the following: 143.8 ÷ 5
Question 26 Solve the following: 80.4 ÷ 5.2
Question 27 Solve the following: 28 – 8 x 3 ÷ 4
Question 28 Solve the following: 19 + (7 x 3) – (64 ÷8)
Question 28 Solve the following: (3.1 +5.5) – 4 + 27 ÷ 9
Question 29 Solve the following: 9 x 5 – 2.3 x 2
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https://www.easycalculation.com/formulas/flow-coefficient-air-critical-pressure.html | 1,685,563,273,000,000,000 | text/html | crawl-data/CC-MAIN-2023-23/segments/1685224647409.17/warc/CC-MAIN-20230531182033-20230531212033-00474.warc.gz | 823,737,056 | 6,026 | English
# Flow Coefficient of Air Formula - Thermodynamics
#### Formula:
Cv = (q [SG (T + 460)]1/2) / (660 * pi)
Where,
Cv = Flow Coefficient for Air and Other Gases
q = Free Gas / Standard Cubic Feet Per Hour
SG = Specific Gravity
T = Gas Temperature
pi = Inlet Gas Absolute Pressure | 84 | 289 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.515625 | 3 | CC-MAIN-2023-23 | longest | en | 0.63888 |
http://en.wikipedia.org/wiki/Richards_equation | 1,430,773,209,000,000,000 | text/html | crawl-data/CC-MAIN-2015-18/segments/1430455113263.14/warc/CC-MAIN-20150501043833-00066-ip-10-235-10-82.ec2.internal.warc.gz | 59,655,204 | 12,198 | # Richards equation
The Richards' equation represents the movement of water in unsaturated soils, and was formulated by Lorenzo A. Richards in 1931.[1] It is a non-linear partial differential equation, which is often difficult to approximate since it does not have a closed-form analytical solution.
Darcy's law was developed for saturated flow in porous media; to this Richards applied a continuity requirement suggested by Buckingham, and obtained a general partial differential equation describing water movement in unsaturated non-swelling soils. The transient state form of this flow equation, known commonly as Richards' equation:
$\frac{\partial \theta}{\partial t}= \frac{\partial}{\partial z} \left[ K(\theta) \left (\frac{\partial \psi}{\partial z} + 1 \right) \right]\$
where
$K$ is the hydraulic conductivity,
$\psi$ is the pressure head,
$z$ is the elevation above a vertical datum,
$\theta$ is the water content, and
$t$ is time.
Richards' equation is equivalent to the groundwater flow equation, which is in terms of hydraulic head (h), by substituting h = ψ + z, and changing the storage mechanism to dewatering. The reason for writing it in the form above is for convenience with boundary conditions (often expressed in terms of pressure head, for example atmospheric conditions are ψ = 0).
## Derivation
Here we show how to derive the Richards equation for the vertical direction in a very simplistic form. Conservation of mass says the rate of change of saturation in a closed volume is equal to the rate of change of the total sum of fluxes into and out of that volume, put in mathematical language:
$\frac{\partial \theta}{\partial t}= \vec{\nabla} \cdot \left(\sum_{i=1}^n{\vec{q}_{i,\,\text{in}}} - \sum_{j=1}^m{\vec{q}_{j,\,\text{out}}} \right)$
Put in the 1D form for the direction $\hat{k}$:
$\frac{\partial \theta}{\partial t}= -\frac{\partial}{\partial z} q$
Horizontal flow in the horizontal direction is formulated by the empiric law of Darcy:
$q= - K \frac{\partial h}{\partial z}$
Substituting q in the equation above, we get:
$\frac{\partial \theta}{\partial t}= \frac{\partial}{\partial z} \left[ K \frac{\partial h}{\partial z}\right]$
Substituting for h = ψ + z:
$\frac{\partial \theta}{\partial t}= \frac{\partial}{\partial z} \left[ K \left ( \frac{\partial \psi}{\partial z} + \frac{\partial z}{\partial z} \right ) \right] = \frac{\partial}{\partial z} \left[ K \left ( \frac{\partial \psi}{\partial z} + 1 \right ) \right]$
We then get the equation above, which is also called the mixed form [2] of Richards' equation.
## Formulations
The Richards Equation appears in many articles in the environmental literature due to the fact that it describes the flow in the interface between fully saturated aquifers and surface water and/or the atmosphere. It also appears in pure mathematical journals due to the fact that it has non-trivial solutions. Usually, it is presented in one of three forms. The mixed form containing the pressure and the saturation is discussed above. It can also appear in two other formulations: head-based and saturation-based.
$C(h)\frac{\partial h}{\partial t}= \nabla \cdot K(h) \nabla h$
Where C(h) [1/L] is a function describing the rate of change of saturation with respect the hydraulic head:
$C(h) \equiv \frac{\partial \theta }{\partial h}$
This function is called 'specific moisture capacity' in the literature, and could be determined for different soil types using curve fitting and laboratory experiments measuring the rate of infiltration of water into soil column, as described for example in van Genuchten (1980).[3]
### Saturation-based
$\frac{\partial \theta }{\partial t}= \nabla \cdot D(\theta) \nabla \theta$
Where D(θ) [L2/T] is 'the soil water diffusivity':
$D(\theta) \equiv \frac{ K(\theta) }{C(\theta)} \equiv K(\theta)\frac{\partial h}{ \partial \theta}$
## Limitations
The numerical solution of Richards equation has been criticized for being computationally expensive and unpredictable [4][5] because there is no guarantee that a solver will converge for a particular set of soil constitutive relations. This prevents use of the method in general applications where the risk of non-convergence is high. The method has also been criticized for over-emphasizing the role of capillarity,[6] and for being in some ways 'overly simplistic' [7] In one dimensional simulations of rainfall infiltration into dry soils, fine spatial discretization less than one cm is required near the land surface. In three-dimensional applications the numerical solution of Richards' equation is subject to aspect ratio constraints where the ratio of horizontal to vertical resolution in the solution domain should be less than about 7.
## References
1. ^ Richards, L.A. (1931). "Capillary conduction of liquids through porous mediums". Physics 1 (5): 318–333. Bibcode:1931Physi...1..318R. doi:10.1063/1.1745010.
2. ^ Celia et al. (1990). "A general Mass-Conservative Numerical Solution for the Unsaturated Flow Equation". Water Resources Research 26 (7): 1483–1496. Bibcode:1990WRR....26.1483C. doi:10.1029/WR026i007p01483.
3. ^ van Genuchten, M. Th. (1980). "A Closed-Form Equation for Predicting the Hydraulic Conductivity of Unsaturated Soils" (PDF). Soil Science Society of America Journal 44 (5): 892–898. doi:10.2136/sssaj1980.03615995004400050002x.
4. ^ Short, D., W.R. Dawes, and I. White, 1995. The practicability of using Richards' equation for general purpose soil-water dynamics models. Envir. Int'l. 21(5):723-730.
5. ^ Tocci, M. D., C. T. Kelley, and C. T. Miller (1997), Accurate and economical solution of the pressure-head form of Richards' equation by the method of lines, Adv. Wat. Resour., 20(1), 1–14.
6. ^ Germann, P. (2010), Comment on “Theory for source-responsive and free-surface film modeling of unsaturated flow”, Vadose Zone J. 9(4), 1000-1101.
7. ^ Gray, W. G., and S. Hassanizadeh (1991), Paradoxes and realities in unsaturated flow theory, Water Resour. Res., 27(8), 1847-1854. | 1,559 | 6,011 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 16, "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} | 3.859375 | 4 | CC-MAIN-2015-18 | latest | en | 0.910901 |
http://mathhelpforum.com/differential-geometry/188659-find-analytic-function-its-real-part-print.html | 1,527,472,741,000,000,000 | text/html | crawl-data/CC-MAIN-2018-22/segments/1526794870604.65/warc/CC-MAIN-20180528004814-20180528024814-00513.warc.gz | 179,979,457 | 3,281 | # Find an analytic function from its real part
• Sep 23rd 2011, 02:43 PM
Ulysses
Find an analytic function from its real part
Hi there. I have two doubts on this exercise. The exercise gives me the real part u(x,y), and I have to find from it the imaginary part, such that the function I get accomplishes the given condition, and its an analytic function.
$\displaystyle u(x,y)=x^2-y^2+2x+y,f(i)=0$
The first doubt I got I think its quiet silly. I don't know how I should evaluate f(i). I hoped to have there two coordinates, for x and y. But I have just i. I don't know if I should took that as a complex number expressed in its polar form, does it mean that x=0 and y=i?
On the other hand I've tried to find v(x,y), the real part it asks me for.
So I used the Cauchy-Riemann conditions.
$\displaystyle u_x=2x+2=v_y \rightarrow v(x,y)=2xy+2y+\Phi(x)$
$\displaystyle -u_y=v_x=2y-1=2y+\Phi'(x)\rightarrow \Phi'(x)=2y-2x-1 \rightarrow \Phi(x)=2xy-x^2-x+K$
So there is something wrong in there, the function I get for Phi depends on y too. Whats wrong with this? I couldn't find the analytic function under this procedure. Did I do something wrong or is it that it doesn't exist an analytic function for the given u(x,y)?
Bye and thanks.
PD: Ok, I realized about the first part, it was really silly, it just means f(z)=f(i), z=i.
• Sep 23rd 2011, 04:17 PM
chisigma
Re: Find an analytic function from its real part
Quote:
Originally Posted by Ulysses
Hi there. I have two doubts on this exercise. The exercise gives me the real part u(x,y), and I have to find from it the imaginary part, such that the function I get accomplishes the given condition, and its an analytic function.
$\displaystyle u(x,y)=x^2-y^2+2x+y,f(i)=0$
The first doubt I got I think its quiet silly. I don't know how I should evaluate f(i). I hoped to have there two coordinates, for x and y. But I have just i. I don't know if I should took that as a complex number expressed in its polar form, does it mean that x=0 and y=i?
On the other hand I've tried to find v(x,y), the real part it asks me for.
So I used the Cauchy-Riemann conditions.
$\displaystyle u_x=2x+2=v_y \rightarrow v(x,y)=2xy+2y+\Phi(x)$
$\displaystyle -u_y=v_x=2y-1=2y+\Phi'(x)\rightarrow \Phi'(x)=2y-2x-1 \rightarrow \Phi(x)=2xy-x^2-x+K$
So there is something wrong in there, the function I get for Phi depends on y too. Whats wrong with this? I couldn't find the analytic function under this procedure. Did I do something wrong or is it that it doesn't exist an analytic function for the given u(x,y)?
Bye and thanks.
PD: Ok, I realized about the first part, it was really silly, it just means f(z)=f(i), z=i.
Your porcedure is correct!... from the condition...
$\displaystyle -u_{y}=2y-1=v_{x}= 2y+\phi^{'}(x)$ (1)
... You derive immediately that is...
$\displaystyle \phi^{'}(x)= -1 \implies \phi(x)=-x+c$ (2)
... so that is...
$\displaystyle v=2 x y +2 y -x + c$ (3)
Taking into account that $\displaystyle f(i)=0$ You obtain in (3) $\displaystyle c=-2$ ...
Kind regards
$\displaystyle \chi$ $\displaystyle \sigma$ | 908 | 3,082 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 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} | 3.703125 | 4 | CC-MAIN-2018-22 | latest | en | 0.95436 |
https://www.freedomplaybypost.com/topic/425-question-probability-control/ | 1,680,342,883,000,000,000 | text/html | crawl-data/CC-MAIN-2023-14/segments/1679296949958.54/warc/CC-MAIN-20230401094611-20230401124611-00313.warc.gz | 876,011,177 | 19,837 | # Question: Probability Control
## Recommended Posts
Just wanted to check about splitting the power control ability over multiple rolls in a round. Is it possible?
Also, should I declare the use of the power before rolling or can I do it after the roll? Just want to be clear before a situation comes up.
Just wanted to check about splitting the power control ability over multiple rolls in a round. Is it possible?
Also, should I declare the use of the power before rolling or can I do it after the roll? Just want to be clear before a situation comes up.
Just wanted to check about splitting the power control ability over multiple rolls in a round. Is it possible?
Also, should I declare the use of the power before rolling or can I do it after the roll? Just want to be clear before a situation comes up.
Just wanted to check about splitting the power control ability over multiple rolls in a round. Is it possible?
Also, should I declare the use of the power before rolling or can I do it after the roll? Just want to be clear before a situation comes up.
When I have GMed other games with a character who had Probability Control (only two games), I used the "Split Attack" option. We created a feat called "Split Probability" that worked just like Split Attack for Probability Control. Each rank of "Split Probability" allows you an additional die to split your PC over.
Example: LuckDude has Probability Control 14 [Power Feats: Split Probability 3].
In a single round he can make one roll a minimum roll of 14. Or he could divide that up over two rolls (e.g. 10 and 4). Or he could divide that up over three rolls (5, 5, and 4). Or he could divide it up over 4 rolls (3, 3, 3, and 5).
That's just my thought; the Refs would need to discuss this to reach a consensus. I'll pose it to them and report back to you, hopefully soon.
Just wanted to check about splitting the power control ability over multiple rolls in a round. Is it possible?
Also, should I declare the use of the power before rolling or can I do it after the roll? Just want to be clear before a situation comes up.
When I have GMed other games with a character who had Probability Control (only two games), I used the "Split Attack" option. We created a feat called "Split Probability" that worked just like Split Attack for Probability Control. Each rank of "Split Probability" allows you an additional die to split your PC over.
Example: LuckDude has Probability Control 14 [Power Feats: Split Probability 3].
In a single round he can make one roll a minimum roll of 14. Or he could divide that up over two rolls (e.g. 10 and 4). Or he could divide that up over three rolls (5, 5, and 4). Or he could divide it up over 4 rolls (3, 3, 3, and 5).
That's just my thought; the Refs would need to discuss this to reach a consensus. I'll pose it to them and report back to you, hopefully soon.
Just wanted to check about splitting the power control ability over multiple rolls in a round. Is it possible?
Also, should I declare the use of the power before rolling or can I do it after the roll? Just want to be clear before a situation comes up.
When I have GMed other games with a character who had Probability Control (only two games), I used the "Split Attack" option. We created a feat called "Split Probability" that worked just like Split Attack for Probability Control. Each rank of "Split Probability" allows you an additional die to split your PC over.
Example: LuckDude has Probability Control 14 [Power Feats: Split Probability 3].
In a single round he can make one roll a minimum roll of 14. Or he could divide that up over two rolls (e.g. 10 and 4). Or he could divide that up over three rolls (5, 5, and 4). Or he could divide it up over 4 rolls (3, 3, 3, and 5).
That's just my thought; the Refs would need to discuss this to reach a consensus. I'll pose it to them and report back to you, hopefully soon.
When I have GMed other games with a character who had Probability Control (only two games), I used the "Split Attack" option. We created a feat called "Split Probability" that worked just like Split Attack for Probability Control. Each rank of "Split Probability" allows you an additional die to split your PC over.
Example: LuckDude has Probability Control 14 [Power Feats: Split Probability 3].
In a single round he can make one roll a minimum roll of 14. Or he could divide that up over two rolls (e.g. 10 and 4). Or he could divide that up over three rolls (5, 5, and 4). Or he could divide it up over 4 rolls (3, 3, 3, and 5).
That's just my thought; the Refs would need to discuss this to reach a consensus. I'll pose it to them and report back to you, hopefully soon.
Thanks a lot. If my two cents counts, I'd like your way of handling it. It makes perfect sense and is reasonable to boot. Looking forward to hearing the results though.
When I have GMed other games with a character who had Probability Control (only two games), I used the "Split Attack" option. We created a feat called "Split Probability" that worked just like Split Attack for Probability Control. Each rank of "Split Probability" allows you an additional die to split your PC over.
Example: LuckDude has Probability Control 14 [Power Feats: Split Probability 3].
In a single round he can make one roll a minimum roll of 14. Or he could divide that up over two rolls (e.g. 10 and 4). Or he could divide that up over three rolls (5, 5, and 4). Or he could divide it up over 4 rolls (3, 3, 3, and 5).
That's just my thought; the Refs would need to discuss this to reach a consensus. I'll pose it to them and report back to you, hopefully soon.
Thanks a lot. If my two cents counts, I'd like your way of handling it. It makes perfect sense and is reasonable to boot. Looking forward to hearing the results though.
When I have GMed other games with a character who had Probability Control (only two games), I used the "Split Attack" option. We created a feat called "Split Probability" that worked just like Split Attack for Probability Control. Each rank of "Split Probability" allows you an additional die to split your PC over.
Example: LuckDude has Probability Control 14 [Power Feats: Split Probability 3].
In a single round he can make one roll a minimum roll of 14. Or he could divide that up over two rolls (e.g. 10 and 4). Or he could divide that up over three rolls (5, 5, and 4). Or he could divide it up over 4 rolls (3, 3, 3, and 5).
That's just my thought; the Refs would need to discuss this to reach a consensus. I'll pose it to them and report back to you, hopefully soon.
Thanks a lot. If my two cents counts, I'd like your way of handling it. It makes perfect sense and is reasonable to boot. Looking forward to hearing the results though.
When I have GMed other games with a character who had Probability Control (only two games), I used the "Split Attack" option. We created a feat called "Split Probability" that worked just like Split Attack for Probability Control. Each rank of "Split Probability" allows you an additional die to split your PC over.
Example: LuckDude has Probability Control 14 [Power Feats: Split Probability 3].
In a single round he can make one roll a minimum roll of 14. Or he could divide that up over two rolls (e.g. 10 and 4). Or he could divide that up over three rolls (5, 5, and 4). Or he could divide it up over 4 rolls (3, 3, 3, and 5).
Got to ask. If the Above Luckdude had Probability control 14 [Power feats: Split Probability 3] along with the AP: Damage as listed in the UP under Probability, would the split probability also act as split attack on the AP or would split attack need to be bought to affect multiple targets?
When I have GMed other games with a character who had Probability Control (only two games), I used the "Split Attack" option. We created a feat called "Split Probability" that worked just like Split Attack for Probability Control. Each rank of "Split Probability" allows you an additional die to split your PC over.
Example: LuckDude has Probability Control 14 [Power Feats: Split Probability 3].
In a single round he can make one roll a minimum roll of 14. Or he could divide that up over two rolls (e.g. 10 and 4). Or he could divide that up over three rolls (5, 5, and 4). Or he could divide it up over 4 rolls (3, 3, 3, and 5).
Got to ask. If the Above Luckdude had Probability control 14 [Power feats: Split Probability 3] along with the AP: Damage as listed in the UP under Probability, would the split probability also act as split attack on the AP or would split attack need to be bought to affect multiple targets?
When I have GMed other games with a character who had Probability Control (only two games), I used the "Split Attack" option. We created a feat called "Split Probability" that worked just like Split Attack for Probability Control. Each rank of "Split Probability" allows you an additional die to split your PC over.
Example: LuckDude has Probability Control 14 [Power Feats: Split Probability 3].
In a single round he can make one roll a minimum roll of 14. Or he could divide that up over two rolls (e.g. 10 and 4). Or he could divide that up over three rolls (5, 5, and 4). Or he could divide it up over 4 rolls (3, 3, 3, and 5).
Got to ask. If the Above Luckdude had Probability control 14 [Power feats: Split Probability 3] along with the AP: Damage as listed in the UP under Probability, would the split probability also act as split attack on the AP or would split attack need to be bought to affect multiple targets?
The power points spent on Split Probability would be available for the AP. LuckDude spent 59 pp on his power (14*4+3). If he took the Damage (Range: Perception) AP, he could spend up to 59 pp on it, any way that he wished. Since Perception range powers are PL-limited, he'd probably not even spend that much, but he could add plenty of Split Attack to it if he wanted to.
The power points spent on Split Probability would be available for the AP. LuckDude spent 59 pp on his power (14*4+3). If he took the Damage (Range: Perception) AP, he could spend up to 59 pp on it, any way that he wished. Since Perception range powers are PL-limited, he'd probably not even spend that much, but he could add plenty of Split Attack to it if he wanted to.
The power points spent on Split Probability would be available for the AP. LuckDude spent 59 pp on his power (14*4+3). If he took the Damage (Range: Perception) AP, he could spend up to 59 pp on it, any way that he wished. Since Perception range powers are PL-limited, he'd probably not even spend that much, but he could add plenty of Split Attack to it if he wanted to.
With a fat array like that, I'd probably look at the area extra too. The ultimate split attack. :D
With a fat array like that, I'd probably look at the area extra too. The ultimate split attack. :D
With a fat array like that, I'd probably look at the area extra too. The ultimate split attack. :D
Ugh. I don't know if I like a Perception Area Damage effect. Even assuming that you add Selective Attack, it just seems dull. "Everyone roll a Toughness save; anvils just fell on your heads again" isn't very exciting after the third or fourth time. Really, not even on the first one.
Ugh. I don't know if I like a Perception Area Damage effect. Even assuming that you add Selective Attack, it just seems dull. "Everyone roll a Toughness save; anvils just fell on your heads again" isn't very exciting after the third or fourth time. Really, not even on the first one.
Ugh. I don't know if I like a Perception Area Damage effect. Even assuming that you add Selective Attack, it just seems dull. "Everyone roll a Toughness save; anvils just fell on your heads again" isn't very exciting after the third or fourth time. Really, not even on the first one.
There's no reason why it has to be Perception ranged is there?
There's no reason why it has to be Perception ranged is there?
There's no reason why it has to be Perception ranged is there?
No, but that's the default suggestion for Probability Control, so that is what I was going by. | 2,909 | 12,107 | {"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} | 3.203125 | 3 | CC-MAIN-2023-14 | latest | en | 0.934998 |
https://www.seismicmatt.com/tag/current/ | 1,611,260,151,000,000,000 | text/html | crawl-data/CC-MAIN-2021-04/segments/1610703527850.55/warc/CC-MAIN-20210121194330-20210121224330-00143.warc.gz | 990,704,448 | 13,177 | ## Measuring the BBC micro:bit LED current draw
Summary: I looked at the voltage and current supplied to the BBC micro:bit for different numbers of active LEDs. The voltage to the LEDs is pulsed at about 55Hz, with each LED on about 1/3 of the time at maximum intensity. The frequency is high enough that the LED appears continuously lit. The current draw from a single pin is in spec for the microcontroller. The current supplied to each LED decreases as the total number of LEDs being sourced from the same microcontroller pin increases. If the LED ntensity is decreased in code, then the LED is active for a smaller fraction of each voltage cycle.
An interesting question at stackexchange asked how the pins to the microcontroller on the microbit could handle the current to and from all 25 of the LEDs on the board. At first glance, it looks like there are too many LEDs being supplied by too few pins to stay within the 5mA current limit for a single pin on the Nordic nRF51822 microcontroller.
There is a schematic for the microbit board on github. A screenshot showing how the microcontroller pins are connected to the LEDs, taken from sheet 1 and 5 of the github site is shown below.
There are three pins called ROW1-ROW3 that supply current to the anodes (the left of the LED symbol) of all 25 LEDs. Pin PO.13 on the microcontroller is connected to ROW1 and supplies 9 LEDs. Pin PO.14 connects to ROW2 and supplies 7 LEDs. Pin PO.15 connects to ROW3 and supplies 9 LEDs. According to the datasheet for the microcontroller, up to 3 pins can supply or sink up to 5mA in 'high-drive' mode. The rest can source or sink 1.5mA each.
The cathode (the bit of the LED connected to the resistor) of each LED is connected to one of nine 220 Ohm resistors. Upto 3 LED cathodes are connected to each resistor. Each of the nine resistors is then connected to a single microcontroller pin, to sink the LED current.
A limit of 5mA from a single pin to source current to 9 LEDs is not a lot. So what is happening? Are the pins being overloaded?
I pulled the C++ microbit compiler code from the lancaster university github. It looks like one row of the microbit display is illuminated at a time and the display constantly scans through the rows.
Anything that turns on and off at a frequency greater than about 15Hz appears as if it is constantly on.
Using a power supply, I measured the current draw with different numbers of LEDs active. The voltage applied was 3.00V from a TTi EX345RD power supply.
The results surprised me:
LEDS current in mA
0 4.74
1 5.79
5 8.63
10 11.31
15 12.75
20 13.45
25 14.01
So about 14.01-4.74 = 9.27mA extra current is being used to supply 25 LEDs.
I used an Analog Discovery 2 with its Wavefroms software to both power the board and as a 'scope. A Fluke multimeter measured the current and was used to verify that the supply voltage stayed constant.
The Analog Discover 2 supply pins were connected to the ground and 3V pin on the edge connector for the microbit. With a supply voltage of 3V from the Analog Discovery 2, I measured 3.001V across the ground and 3V edge connector pins for 1 LED on and 2.994V with all 25 LEDs on, so the voltage drop across the supply pins is 0.2% while increasing the number of active LEDs. Not much of a voltage drop.
With the 'scope, I can see that the voltage to each LED is indeed pulsed. The screen shot taken from the Waveform software below shows what I saw 'scoping the anode of an LED. The red vertical lines show where I measured time intervals. I am fairly sure that I am 'scoping the LED's anthode as I could count 9 of the pins connected to this side of the LED and 3 at the other end. Which agrees with the schematic. Plus I used my Fluke to verify the LED polarity. Matty likes to double check what he is measuring after having made so many, many mistakes over the years.
Clearly the current draw I measured from the power supply was showing the average current. The current through the LEDs will vary with the voltage. The resistor is 220 Ohm. The current will be V^2/R, where V is the voltage across the resistor.
The LED voltage at the anode is high for 5.8824ms out of 17.914ms = 32.83% duty cycle. So the pin is high for about 1/3 of the time. This makes sense - one pin out of the three that supply current to the LEDs is active at a time. The pulse frequency is 55.821Hz. So the pulsing is invisible to the eye.
The frequency and pulse widths do not change with the number of LEDs that are on. What does change is the anode and cathode amplitude. This drops as the number of LEDs on increases. The supply voltage to the microbit stayed almost constant though.
The image below shows the anode voltage on two separate supply pins, showing that they are indeed turned on at different times. Should I have three probes, the gap would be filled in by 'scoping the pin supplying the third row of LEDs.
With no LEDs on, the voltage rail was 3.0200 on the 'scope. Fair enough. That's pretty much the supply rail. With all LEDs on, I measured two voltages, 2.0896V and 2.1640V. If you've been paying attention, it won't be a suprise that the higher voltage occurs on 7 LEDs (ROW2 on the schematic) and the lower on the remaining 18 (ROW1 and ROW3 on the schematic).
Let's look at the cathode of one of the LEDs. A screen grab is shown below. With a single lit LED, the cathode is at 3V for 2/3 of the time as the sink pin goes high to prevent current, from, well, being sinked. When it goes low, current can flow from through the LED and the resistor and back to ground through the microcontroller pin. Thus completing the circuit and allowing the LED to light up. When the current is flowing, the voltage drops at the cathode. In this screen grab, with only 3 LEDs active, I measured 912.34mV at the cathode of the lit LED. Which makes for a current of 0.91234^2/220 = 3.78mA through the single LED and resistor. Note: I had the LEDs set on maximum intensity.
What is interesting, is that the low value of the cathode voltage decreases with an increasing number of lit LEDs. So the current flowing through the LED must similarly decrease. So the voltage and thus current decreases with increasing lit LEDs, which means that the brightness also decreases. Similarly, the fraction of time that the cathode is high also decreases as the number of active LEDs increases, which makes sense, as for a greater fraction of the time the multiplexed cathode will be sinking current.
With all LEDs on, I measure 356.81mV for 1/3 of the time and 292.46mV for 2/3 of the time. Can you guess why the voltage is stepped between the two values? I suspect it as the voltage switches between supplying 9 and 7 LEDs. Please see the two 'scope grabs below.
The human eye is not so good at discerning different light intensities. I think that having a bunch of LEDs not so brightly lit as a single one also helps fool the eye in judging their individual brightness.
With all LEDs on, the maximum current flow through the 220 Ohm resistor for the pin supplying 7 LEDs is 0.35681^2/220 = 0.5787mA. So for 7 LEDs, the total draw is 4.05mA.
For the 9 LED pins, the current per LED is 0.29246^2/220 = 0.3888mA. So for 9 LEDs, this is 3.499mA, again inside the 5mA spec for a single 'high-drive' pin to supply. However, the pin is only active for about 1/3 of the time. So the average current flow per LED will be 1/3 of the above values.
The current sinking into the pins that sink current is within the 1.5mA spec for a non 'high-drive' pin.
Lets look at all 25 LEDs active. About (7 x 0.5787 x 1/3)+ (9 x 0.3888 x 2/3) is being drawn at any instant. Which is 4.125mA. I measured an increase of around 9mA with all LEDs active. Hmmm, where does the extra current go? I'm open to suggestions. I've gone down the rabbit-hole with this. Time to move onto something else.
Ta-daaaaaaa.
An obvious question is: How do the traces look when the intensity changes? I edited the code shown at the end of the article to have the minimum LED intensity. Prior to this, I'd had everything set at maximum intensity.
Looking at the 'scope, the anode voltage stayed the same as for when we have maximum intensity - it is high for about 1/3 of the time for any given LED. Below see the cathode voltage for minimum intensity, all LEDs lit and a repeat of the image for when the LED intensity is maximum. We can see that the cathode voltage goes low, which allows current to flow, for much less of the cycle when the intensity is at a minimum. Which makes sense. At maximum intensity, current flows for almost all of the time that the anode voltage is high (about 1/3 of the time). I measured the cathode coming low for only 32.637us at minimum intensity, which is only about 0.18% of the illumination cycle, compared with the roughly 32% of the time when at maximum intensity.
I work at sea for half of the year, on a six week on, six week off rota. I first looked at this while I was on a survey ship, with limited test equipment and time as I am paid to do 12 hour shifts, that usually over run. Obviously, the ship would stop operating if Little Matty was not at his station when he is supposed to be.
I found a voltage supply on the ship, lashed this up to the microbit with a set voltage of 3.00V and measured the current draw through a Fluke multimeter for different numbers of LEDs. What I did not have access to was a 'scope. So I finished off my investigations when I was back home.
Looking at the display through a digital camera - I could see that the LED brightness is not constant. The frame rate of the digital camera is not synced with the refresh rate of the display, so as the rows are switched on and off, this is picked up as a variation in intensity in the camera display.
So even without a 'scope, I was pretty sure that the rows were being pulsed on and off. Actually, there is a 'scope on the ship, but as it is connected to the recording equipment to monitor all is well, I figured I should leave it in place.
Here's my micropython script to turn on/off one LED at a time using the A/B buttons:
```#
''' Light LEDs using A and B buttons. '''
from microbit import *
# intensity of LED
BRIGHT = '9'
# how many LEDs to turn on at a time
INCREMENT = 1
# maximum number of LEDs allowed on at once
MAX_BRIGHT = 25
# how many LEDs on at boot
START_BRIGHT = 3
def decrease(num_bright, inc):
num_bright = limit(num_bright-inc, MAX_BRIGHT)
return num_bright
def increase(num_bright, inc):
num_bright = limit(num_bright+inc, MAX_BRIGHT)
return num_bright
def leds_string2(bright):
''' return led string '''
bright = limit(bright, MAX_BRIGHT)
leds_string = BRIGHT*bright + '0' * (MAX_BRIGHT-bright)
leds_string = ":".join(leds_string[i:i+5]
for i in range(0, len(leds_string), 5))
leds_image = Image(leds_string + ':')
return leds_image
def limit(val, limit):
''' limit <val> between 0 and <limit>'''
if val > limit:
val = 0
if val < 0:
val = MAX_BRIGHT
return int(val)
num_bright = START_BRIGHT
while True:
if button_a.was_pressed():
num_bright = (decrease(num_bright, INCREMENT))
if button_b.was_pressed():
num_bright = (increase(num_bright, INCREMENT))
display.show(leds_string2(num_bright))
sleep(0.1)
```
```
``` | 2,826 | 11,243 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.765625 | 3 | CC-MAIN-2021-04 | latest | en | 0.937291 |
https://stats.stackexchange.com/questions/393016/what-does-the-degree-of-freedom-df-mean-in-the-results-of-log-likelihood-logl | 1,716,863,358,000,000,000 | text/html | crawl-data/CC-MAIN-2024-22/segments/1715971059055.94/warc/CC-MAIN-20240528000211-20240528030211-00080.warc.gz | 448,664,815 | 39,576 | # What does the degree of freedom (df) mean in the results of log-likelihood logLik
After doing the regression using lm for fixed effect model or lmer for mixed effects model, I pass the results to the logLik. Besides the value of log-likelihood, the function always returns a df, i.e. the degree of freedom.
However, the degree of freedom does not equal to the number of parameters in the model, df always larger. So what does the df mean exactly?
The reason I care about the df is that later I will use the BIC (Bayesian Information Criterion) to do the model selection. The BIC is defined as BIC=-2*logLik+k*log(n) where k is the number of parameters and n is the number of observations. When I pass my logLik value to the expression of BIC, the result is exactly the same when I use the build in BIC function in R if I specify the number of parameters as the df in logLik. Which means, in the build in BIC function, they also use the df as k when they calculate BIC.
• Possible duplicate of How to understand degrees of freedom?
– EdM
Feb 18, 2019 at 2:46
• Usually it is something like sample size minus number of parameters estimated. It can be a little more complicated though. Read the suggested question yours is a duplicate of. Feb 18, 2019 at 3:50
The help file for the logLik() function mentions that it returns the number of (estimated) parameters in the model as the degrees of freedom (df).
If you fit a linear model like the one below:
m <- lm(mpg ~ hp, data = mtcars)
summary(m)
logLik(m)
you will see that the model includes an intercept parameter, a slope parameter quantifying the effect of the predictor hp on mpg and an error variance (or standard deviation) parameter. These add up to three parameters, which is exactly what logLik() returns for the model.
If you fit a linear mixed effects model like this:
require(nlme)
M <- lme(distance ~ age + Sex, data = Orthodont, random = ~ 1)
summary(M)
logLik(M)
then you'll see that the model includes the following parameters: a fixed effect intercept, a fixed effect of age, a fixed effect of Sex, a parameter denoting the standard deviation of the random intercept and a standard deviation of the within-subject model errors. That makes for five parameters, which is what the logLik() returns. | 533 | 2,281 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.921875 | 3 | CC-MAIN-2024-22 | latest | en | 0.846662 |
http://www.kylesconverter.com/area-density/tonnes-per-square-kilometer-to-zentners-per-hectare | 1,611,489,493,000,000,000 | text/html | crawl-data/CC-MAIN-2021-04/segments/1610703548716.53/warc/CC-MAIN-20210124111006-20210124141006-00348.warc.gz | 150,948,291 | 5,480 | # Convert Tonnes Per Square Kilometer to Zentners Per Hectare
### Kyle's Converter > Area Density > Tonnes Per Square Kilometer > Tonnes Per Square Kilometer to Zentners Per Hectare
Tonnes Per Square Kilometer (t/km2) Zentners Per Hectare (Zentners/ha) Precision: 0 1 2 3 4 5 6 7 8 9 12 15 18
Reverse conversion?
Zentners Per Hectare to Tonnes Per Square Kilometer
(or just enter a value in the "to" field)
Please share if you found this tool useful:
Unit Descriptions
1 Tonne per Square Kilometer:
Mass of metric tonnes per an area of a square kilometer. Metric tonne having exactly 1 000 kilograms. 1 t/km2 = 0.001 kg/m2.
1 Zentner per Hectare:
Mass of zentners per area of one hectare. A German zentner having 50 kilograms. 1 Zentner/ha ≈ 0.005 kg/m2.
Conversions Table
1 Tonnes Per Square Kilometer to Zentners Per Hectare = 0.270 Tonnes Per Square Kilometer to Zentners Per Hectare = 14
2 Tonnes Per Square Kilometer to Zentners Per Hectare = 0.480 Tonnes Per Square Kilometer to Zentners Per Hectare = 16
3 Tonnes Per Square Kilometer to Zentners Per Hectare = 0.690 Tonnes Per Square Kilometer to Zentners Per Hectare = 18
4 Tonnes Per Square Kilometer to Zentners Per Hectare = 0.8100 Tonnes Per Square Kilometer to Zentners Per Hectare = 20
5 Tonnes Per Square Kilometer to Zentners Per Hectare = 1200 Tonnes Per Square Kilometer to Zentners Per Hectare = 40
6 Tonnes Per Square Kilometer to Zentners Per Hectare = 1.2300 Tonnes Per Square Kilometer to Zentners Per Hectare = 60
7 Tonnes Per Square Kilometer to Zentners Per Hectare = 1.4400 Tonnes Per Square Kilometer to Zentners Per Hectare = 80
8 Tonnes Per Square Kilometer to Zentners Per Hectare = 1.6500 Tonnes Per Square Kilometer to Zentners Per Hectare = 100
9 Tonnes Per Square Kilometer to Zentners Per Hectare = 1.8600 Tonnes Per Square Kilometer to Zentners Per Hectare = 120
10 Tonnes Per Square Kilometer to Zentners Per Hectare = 2800 Tonnes Per Square Kilometer to Zentners Per Hectare = 160
20 Tonnes Per Square Kilometer to Zentners Per Hectare = 4900 Tonnes Per Square Kilometer to Zentners Per Hectare = 180
30 Tonnes Per Square Kilometer to Zentners Per Hectare = 61,000 Tonnes Per Square Kilometer to Zentners Per Hectare = 200
40 Tonnes Per Square Kilometer to Zentners Per Hectare = 810,000 Tonnes Per Square Kilometer to Zentners Per Hectare = 2000
50 Tonnes Per Square Kilometer to Zentners Per Hectare = 10100,000 Tonnes Per Square Kilometer to Zentners Per Hectare = 20000
60 Tonnes Per Square Kilometer to Zentners Per Hectare = 121,000,000 Tonnes Per Square Kilometer to Zentners Per Hectare = 200000 | 819 | 2,598 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.515625 | 3 | CC-MAIN-2021-04 | latest | en | 0.512834 |
https://free-education.in/courses/online-class-for-10th-standard-students-cbse-english-medium/lesson/exercise-13-2-page-no-247/ | 1,726,488,635,000,000,000 | text/html | crawl-data/CC-MAIN-2024-38/segments/1725700651697.32/warc/CC-MAIN-20240916112213-20240916142213-00525.warc.gz | 233,731,395 | 39,351 | Course Content
Class 10th Science
0/32
Class 10th Maths
0/86
Class 10 Social Science History : India and the Contemporary World – II
0/16
Class 10 Social Science Geography : Contemporary India – II
0/14
Class 10 Social Science Civics (Political Science) : Democratic Politics – II
0/16
Class 10 Social Science Economics: Understanding Economic Development – II
0/10
Class 10 English First Flight Summary
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Class 10 English Footprints without Feet Summary
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### Exercise: 13.2 (Page No: 247)
1. A solid is in the shape of a cone standing on a hemisphere with both their radii being equal to 1 cm and the height of the cone is equal to its radius. Find the volume of the solid in terms of π.
Solution:
Here r = 1 cm and h = 1 cm.
The diagram is as follows.
Now, Volume of solid = Volume of conical part + Volume of hemispherical part
We know the volume of cone = ⅓ πr2h
And,
The volume of hemisphere = ⅔πr3
So, volume of solid will be
= π cm3
2. Rachel, an engineering student, was asked to make a model shaped like a cylinder with two cones attached at its two ends by using a thin aluminum sheet. The diameter of the model is 3 cm and its length is 12 cm. If each cone has a height of 2 cm, find the volume of air contained in the model that Rachel made. (Assume the outer and inner dimensions of the model to be nearly the same.)
Solution:
Given,
Height of cylinder = 12–4 = 8 cm
Height of cone = 2 cm
Now, the total volume of the air contained will be = Volume of cylinder+2×(Volume of cone)
∴ Total volume = πr2h+[2×(⅓ πr2h )]
= 18 π+2(1.5 π)
= 66 cm3.
3. A Gulab jamun contains sugar syrup up to about 30% of its volume. Find approximately how much syrup would be found in 45 Gulab jamuns, each shaped like a cylinder with two hemispherical ends with length 5 cm and diameter 2.8 cm (see figure).
Solution:
It is known that the gulab jamuns are similar to a cylinder with two hemispherical ends.
So, the total height of a gulab jamun = 5 cm.
Diameter = 2.8 cm
∴ The height of the cylindrical part = 5 cm–(1.4+1.4) cm
=2.2 cm
Now, total volume of One Gulab Jamun = Volume of Cylinder + Volume of two hemispheres
= πr2h+(4/3)πr3
= 4.312π+(10.976/3) π
= 25.05 cm3
We know that the volume of sugar syrup = 30% of total volume
So, volume of sugar syrup in 45 gulab jamuns = 45×30%(25.05 cm3)
= 45×7.515 = 338.184 cm3
4. A pen stand made of wood is in the shape of a cuboid with four conical depressions to hold pens. The dimensions of the cuboid are 15 cm by 10 cm by 3.5 cm. The radius of each of the depressions is 0.5 cm and the depth is 1.4 cm. Find the volume of wood in the entire stand (see Fig.).
Solution:
Volume of cuboid = length x width x height
We know the cuboid’s dimensions as 15 cmx10 cmx3.5 cm
So, the volume of the cuboid = 15x10x3.5 = 525 cm3
Here, depressions are like cones and we know,
Volume of cone = (⅓)πr2h
Given, radius (r) = 0.5 cm and depth (h) = 1.4 cm
∴ Volume of 4 cones = 4x(⅓)πr2h
= 1.46 cm2
Now, volume of wood = Volume of cuboid – 4 x volume of cone
= 525-1.46 = 523.54 cm2
5. A vessel is in the form of an inverted cone. Its height is 8 cm and the radius of its top, which is open, is 5 cm. It is filled with water up to the brim. When lead shots, each of which is a sphere of radius 0.5 cm are dropped into the vessel, one-fourth of the water flows out. Find the number of lead shots dropped in the vessel.
Solution:
For the cone,
Height = 8 cm
Also,
Radius of sphere = 0.5 cm
The diagram will be like
It is known that,
Volume of cone = volume of water in the cone
= ⅓πr2h = (200/3)π cm3
Now,
Total volume of water overflown= (¼)×(200/3) π =(50/3)π
= (4/3)πr3
= (1/6) π
Now,
The number of lead shots = Total Volume of Water over flown/ Volume of Lead shot
= (50/3)π/(⅙)π
= (50/3)×6 = 100
6. A solid iron pole consists of a cylinder of height 220 cm and base diameter 24 cm, which is surmounted by another cylinder of height 60 cm and radius 8 cm. Find the mass of the pole, given that 1 cm3 of iron has approximately 8 g mass.
Solution:
Given, the height of the big cylinder (H) = 220 cm
Radius of the base (R) = 24/12 = 12 cm
So, the volume of the big cylinder = πR2H
= π(12)× 220 cm3
= 99565.8 cm3
Now, the height of smaller cylinder (h) = 60 cm
Radius of the base (r) = 8 cm
So, the volume of the smaller cylinder = πr2h
= π(8)2×60 cm3
= 12068.5 cm3
∴ Volume of iron = Volume of the big cylinder+ Volume of the small cylinder
= 99565.8 + 12068.5
=111634.5 cm3
We know,
Mass = Density x volume
So, mass of the pole = 8×111634.5
= 893 Kg (approx.)
7. A solid consisting of a right circular cone of height 120 cm and radius 60 cm standing on a hemisphere of radius 60 cm is placed upright in a right circular cylinder full of water such that it touches the bottom. Find the volume of water left in the cylinder, if the radius of the cylinder is 60 cm and its height is 180 cm.
Solution:
Here, the volume of water left will be = Volume of cylinder – Volume of solid
Given,
Radius of cone = 60 cm,
Height of cone = 120 cm
Radius of cylinder = 60 cm
Height of cylinder = 180 cm
Radius of hemisphere = 60 cm
Now,
Total volume of solid = Volume of Cone + Volume of hemisphere
Volume of cone = π×122×103cm3 = 144×103π cm3
So, Total volume of solid = 144×103π cm-(⅔)×π×10cm3
Volume of hemisphere = (⅔)×π×10cm3
Volume of cylinder = π×602×180 = 648000 = 648×103 π cm3
Now, volume of water left will be = Volume of cylinder – Volume of solid
= (648-288) × 103×π = 1.131 m3
8. A spherical glass vessel has a cylindrical neck 8 cm long, 2 cm in diameter; the diameter of the spherical part is 8.5 cm. By measuring the amount of water it holds, a child finds its volume to be 345 cm3. Check whether she is correct, taking the above as the inside measurements, and π = 3.14.
Solution:
Given,
For the cylinder part, Height (h) = 8 cm and Radius (R) = (2/2) cm = 1 cm
For the spherical part, Radius (r) = (8.5/2) = 4.25 cm
Now, volume of this vessel = Volume of cylinder + Volume of sphere
= π×(1)2×8+(4/3)π(1)3
= 346.51 cm3 | 1,940 | 6,186 | {"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} | 4.25 | 4 | CC-MAIN-2024-38 | latest | en | 0.91608 |
http://www.lmfdb.org/EllipticCurve/Q/51870/n/1 | 1,566,429,506,000,000,000 | text/html | crawl-data/CC-MAIN-2019-35/segments/1566027316549.78/warc/CC-MAIN-20190821220456-20190822002456-00232.warc.gz | 276,981,327 | 33,341 | # Properties
Label 51870.n1 Conductor 51870 Discriminant 2498954761485000 j-invariant $$\frac{5603281597654963338329641}{2498954761485000}$$ CM no Rank 0 Torsion Structure $$\Z/{2}\Z$$
# Related objects
Show commands for: Magma / SageMath / Pari/GP
## Minimal Weierstrass equation
magma: E := EllipticCurve([1, 1, 0, -3700322, 2738185356]); // or
magma: E := EllipticCurve("51870o4");
sage: E = EllipticCurve([1, 1, 0, -3700322, 2738185356]) # or
sage: E = EllipticCurve("51870o4")
gp: E = ellinit([1, 1, 0, -3700322, 2738185356]) \\ or
gp: E = ellinit("51870o4")
$$y^2 + x y = x^{3} + x^{2} - 3700322 x + 2738185356$$
## Mordell-Weil group structure
$$\Z/{2}\Z$$
## Torsion generators
magma: TorsionSubgroup(E);
sage: E.torsion_subgroup().gens()
gp: elltors(E)
$$\left(\frac{4443}{4}, -\frac{4443}{8}\right)$$
## Integral points
magma: IntegralPoints(E);
sage: E.integral_points()
None
## Invariants
magma: Conductor(E); sage: E.conductor().factor() gp: ellglobalred(E)[1] Conductor: $$51870$$ = $$2 \cdot 3 \cdot 5 \cdot 7 \cdot 13 \cdot 19$$ magma: Discriminant(E); sage: E.discriminant().factor() gp: E.disc Discriminant: $$2498954761485000$$ = $$2^{3} \cdot 3^{3} \cdot 5^{4} \cdot 7^{8} \cdot 13^{2} \cdot 19$$ magma: jInvariant(E); sage: E.j_invariant().factor() gp: E.j j-invariant: $$\frac{5603281597654963338329641}{2498954761485000}$$ = $$2^{-3} \cdot 3^{-3} \cdot 5^{-4} \cdot 7^{-8} \cdot 13^{-2} \cdot 19^{-1} \cdot 181^{3} \cdot 981301^{3}$$ Endomorphism ring: $$\Z$$ (no Complex Multiplication) Sato-Tate Group: $\mathrm{SU}(2)$
## BSD invariants
magma: Rank(E); sage: E.rank() Rank: $$0$$ magma: Regulator(E); sage: E.regulator() Regulator: $$1$$ magma: RealPeriod(E); sage: E.period_lattice().omega() gp: E.omega[1] Real period: $$0.373008620099$$ magma: TamagawaNumbers(E); sage: E.tamagawa_numbers() gp: gr=ellglobalred(E); [[gr[4][i,1],gr[5][i][4]] | i<-[1..#gr[4][,1]]] Tamagawa product: $$16$$ = $$1\cdot1\cdot2^{2}\cdot2\cdot2\cdot1$$ magma: Order(TorsionSubgroup(E)); sage: E.torsion_order() gp: elltors(E)[1] Torsion order: $$2$$ magma: MordellWeilShaInformation(E); sage: E.sha().an_numerical() Analytic order of Ш: $$1$$ (exact)
## Modular invariants
#### Modular form 51870.2.a.n
magma: ModularForm(E);
sage: E.q_eigenform(20)
gp: xy = elltaniyama(E);
gp: x*deriv(xy[1])/(2*xy[2]+E.a1*xy[1]+E.a3)
$$q - q^{2} - q^{3} + q^{4} + q^{5} + q^{6} - q^{7} - q^{8} + q^{9} - q^{10} - 4q^{11} - q^{12} - q^{13} + q^{14} - q^{15} + q^{16} + 6q^{17} - q^{18} - q^{19} + O(q^{20})$$
magma: ModularDegree(E); sage: E.modular_degree() Modular degree: 1474560 $$\Gamma_0(N)$$-optimal: no Manin constant: not computed
#### Special L-value
magma: Lr1 where r,Lr1 := AnalyticRank(E: Precision:=12);
sage: r = E.rank();
sage: E.lseries().dokchitser().derivative(1,r)/r.factorial()
gp: ar = ellanalyticrank(E);
gp: ar[2]/factorial(ar[1])
$$L(E,1)$$ ≈ $$1.4920344804$$
## Local data
magma: [LocalInformation(E,p) : p in BadPrimes(E)];
sage: E.local_data()
gp: ellglobalred(E)[5]
prime Tamagawa number Kodaira symbol Reduction type Root number ord($$N$$) ord($$\Delta$$) ord$$(j)_{-}$$
$$2$$ $$1$$ $$I_{3}$$ Non-split multiplicative 1 1 3 3
$$3$$ $$1$$ $$I_{3}$$ Non-split multiplicative 1 1 3 3
$$5$$ $$4$$ $$I_{4}$$ Split multiplicative -1 1 4 4
$$7$$ $$2$$ $$I_{8}$$ Non-split multiplicative 1 1 8 8
$$13$$ $$2$$ $$I_{2}$$ Non-split multiplicative 1 1 2 2
$$19$$ $$1$$ $$I_{1}$$ Non-split multiplicative 1 1 1 1
## Galois representations
The image of the 2-adic representation attached to this elliptic curve is the subgroup of $\GL(2,\Z_2)$ with Rouse label X13.
This subgroup is the pull-back of the subgroup of $\GL(2,\Z_2/2^2\Z_2)$ generated by $\left(\begin{array}{rr} 3 & 0 \\ 0 & 1 \end{array}\right),\left(\begin{array}{rr} 1 & 1 \\ 0 & 1 \end{array}\right),\left(\begin{array}{rr} 3 & 0 \\ 0 & 3 \end{array}\right)$ and has index 6.
magma: [GaloisRepresentation(E,p): p in PrimesUpTo(20)];
sage: rho = E.galois_representation();
sage: [rho.image_type(p) for p in rho.non_surjective()]
The mod $$p$$ Galois representation has maximal image $$\GL(2,\F_p)$$ for all primes $$p$$ except those listed.
prime Image of Galois representation
$$2$$ B
## $p$-adic data
### $p$-adic regulators
sage: [E.padic_regulator(p) for p in primes(3,20) if E.conductor().valuation(p)<2]
All $$p$$-adic regulators are identically $$1$$ since the rank is $$0$$.
## Iwasawa invariants
$p$ Reduction type $\lambda$-invariant(s) $\mu$-invariant(s) 2 3 5 7 13 19 nonsplit nonsplit split nonsplit nonsplit nonsplit 4 0 1 0 0 0 0 0 0 0 0 0
All Iwasawa $\lambda$ and $\mu$-invariants for primes $p\ge 3$ of good reduction are zero.
## Isogenies
This curve has non-trivial cyclic isogenies of degree $$d$$ for $$d=$$ 2 and 4.
Its isogeny class 51870.n consists of 4 curves linked by isogenies of degrees dividing 4.
## Growth of torsion in number fields
The number fields $K$ of degree up to 7 such that $E(K)_{\rm tors}$ is strictly larger than $E(\Q)_{\rm tors}$ $\cong \Z/{2}\Z$ are as follows:
$[K:\Q]$ $K$ $E(K)_{\rm tors}$ Base-change curve
2 $$\Q(\sqrt{19})$$ $$\Z/4\Z$$ Not in database
$$\Q(\sqrt{6})$$ $$\Z/4\Z$$ Not in database
$$\Q(\sqrt{114})$$ $$\Z/2\Z \times \Z/2\Z$$ Not in database
4 $$\Q(\sqrt{6}, \sqrt{19})$$ $$\Z/2\Z \times \Z/4\Z$$ Not in database
We only show fields where the torsion growth is primitive. For each field $K$ we either show its label, or a defining polynomial when $K$ is not in the database. | 2,095 | 5,470 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.25 | 3 | CC-MAIN-2019-35 | longest | en | 0.216007 |
http://perplexus.info/show.php?pid=8401&cid=52009 | 1,537,867,979,000,000,000 | text/html | crawl-data/CC-MAIN-2018-39/segments/1537267161350.69/warc/CC-MAIN-20180925083639-20180925104039-00530.warc.gz | 187,792,550 | 4,467 | All about flooble | fun stuff | Get a free chatterbox | Free JavaScript | Avatars
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Rooted to the Floor (Posted on 2013-09-24)
Find all pairs (x,y) of integers satisfying:
floor(√(2x)) = floor(y*(2 + √2))
No Solution Yet Submitted by K Sengupta No Rating
Comments: ( Back to comment list | You must be logged in to post comments.)
anaylitical solution Comment 1 of 1
let f(x)=floor(x) and c(x)=ceiling(x)
if f(x)=k for some integer k then we have
k<=x<k+1
also if c(x)=k for some integer k then we have
k-1<x<=k
for compactness let z=2+sqr(2)
f(sqr(2x))=f(yz)
if f(sqr(2x))=f(yz)=k for some integer k then we have
k<=sqr(2x)<k+1
k^2<=2x<(k+1)^2
thus x can be any even integer in the interval [k^2,(k+1)^2)
there is an even integer in this interval for all k>=1
we also have k<=yz<k+1
(k/z)<=y<(k+1)/z
thus y can be any integer in the interval [k/z,(k+1)/z)
this interval contains an integer only when
c(k/z)<=f((k+1)/z)
since the length of this interval is a constant 1/z<1
this inequality only holds at equality thus we need
c(k/z)=f((k+1)/z)=a for some integer a
this gives us
a-1<k/z<=a<=(k+1)/z<a+1
thus k must be in the intersection of the two intervals
((a-1)z,az] and [az-1,(a+1)z-1)
since az-1>(a-1)z and az>(a+1)z-1 for a>1 then
this is the same as the interval
[az-1,az]
the only integer in this interval is f(az)
thus y can be any integer in the interval [f(az)/z,(f(az)+1)/z)
so all solutions are given by:
z is 2+sqr(2)
a is an integer greater than 1
x is an even integer in [f(az)^2,(f(az)+1)^2)
y is an integer in [f(az)/z,(f(az)+1)/z)
Posted by Daniel on 2013-09-25 11:54:01
Search: Search body:
Forums (1) | 602 | 1,657 | {"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} | 4.0625 | 4 | CC-MAIN-2018-39 | longest | en | 0.741549 |
https://fr.mathworks.com/matlabcentral/answers/154551-combine-2-structures-have-a-nice-weekend?s_tid=prof_contriblnk | 1,708,576,433,000,000,000 | text/html | crawl-data/CC-MAIN-2024-10/segments/1707947473690.28/warc/CC-MAIN-20240222030017-20240222060017-00432.warc.gz | 272,089,096 | 26,119 | # Combine 2 Structures / have a nice weekend
1 vue (au cours des 30 derniers jours)
Max Müller le 12 Sep 2014
Modifié(e) : Joseph Cheng le 12 Sep 2014
As you can see here, i am trying to combine two structures. It works ! But is there a faster way ?
Variable = length(OldInputData)
for k = 1:length(InputData)
OldInputData(Variable + k).shotnumber = InputData(k).shotnumber
OldInputData(Variable + k).UsedAmp = InputData(k).UsedAmp
OldInputData(Variable + k).ClosestAmp = InputData(k).ClosestAmp
OldInputData(Variable + k).OptimalAmp = InputData(k).OptimalAmp
OldInputData(Variable + k).Factor = InputData(k).Factor
OldInputData(Variable + k).ssx_t = InputData(k).ssx_t
end
Have a nice weekend!
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### Réponse acceptée
Joseph Cheng le 12 Sep 2014
Modifié(e) : Joseph Cheng le 12 Sep 2014
you can go
OldInputData = [OldInputdata InputData]
only if they contain the same structures inside. Since you're just concatenating it should work fine. if there is a new structure it'll add that to the others but be empty.
see example:
for i =1:10
old(i).shot = randi(10);
old(i).used = randi(10);
old(i).closest = randi(10);
old(i).optimal = randi(10);
old(i).factor = randi(10);
old(i).ssx= randi(10);
old(i).lkjasf = randi(10);
end
old = [old old]
then i can see that
i should be able to check that old(1) and old(11), 2 and 12, etc. should have the same stuff.
if we add a new structure element (is that the right term?) to it like
i = i+1;
old(i).shot = randi(10);
old(i).used = randi(10);
old(i).closest = randi(10);
old(i).optimal = randi(10);
old(i).factor = randi(10);
old(i).ssx= randi(10);
old(i).lkjasf = randi(10);
old(i).asdfasdf = 2
then all 1:20 has a new asdfasdf = [] to them.
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### Plus de réponses (1)
Modifié(e) : Adam le 12 Sep 2014
I did something very similar myself just the other week, but since that seems to be a ready-built explanation of various methods there isn't much point me fishing out whatever method I ended up using!
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Translated by | 757 | 2,543 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.828125 | 3 | CC-MAIN-2024-10 | latest | en | 0.487435 |
https://jax.readthedocs.io/en/stable/_autosummary/jax.numpy.argmin.html | 1,621,300,497,000,000,000 | text/html | crawl-data/CC-MAIN-2021-21/segments/1620243991650.73/warc/CC-MAIN-20210518002309-20210518032309-00444.warc.gz | 363,316,674 | 9,247 | # jax.numpy.argminΒΆ
jax.numpy.argmin(a, axis=None)[source]ΒΆ
Returns the indices of the minimum values along an axis.
LAX-backend implementation of argmin(). Original docstring below.
Parameters
• a (array_like) β Input array.
• axis (int, optional) β By default, the index is into the flattened array, otherwise along the specified axis.
Returns
index_array β Array of indices into the array. It has the same shape as a.shape with the dimension along axis removed.
Return type
ndarray of ints
ndarray.argmin(), argmax()
amin()
The minimum value along a given axis.
unravel_index()
Convert a flat index into an index tuple.
take_along_axis()
Apply np.expand_dims(index_array, axis) from argmin to an array as if by calling min.
Notes
In case of multiple occurrences of the minimum values, the indices corresponding to the first occurrence are returned.
Examples
>>> a = np.arange(6).reshape(2,3) + 10
>>> a
array([[10, 11, 12],
[13, 14, 15]])
>>> np.argmin(a)
0
>>> np.argmin(a, axis=0)
array([0, 0, 0])
>>> np.argmin(a, axis=1)
array([0, 0])
Indices of the minimum elements of a N-dimensional array:
>>> ind = np.unravel_index(np.argmin(a, axis=None), a.shape)
>>> ind
(0, 0)
>>> a[ind]
10
>>> b = np.arange(6) + 10
>>> b[4] = 10
>>> b
array([10, 11, 12, 13, 10, 15])
>>> np.argmin(b) # Only the first occurrence is returned.
0
>>> x = np.array([[4,2,3], [1,0,3]])
>>> index_array = np.argmin(x, axis=-1)
>>> # Same as np.min(x, axis=-1, keepdims=True)
>>> np.take_along_axis(x, np.expand_dims(index_array, axis=-1), axis=-1)
array([[2],
[0]])
>>> # Same as np.max(x, axis=-1)
>>> np.take_along_axis(x, np.expand_dims(index_array, axis=-1), axis=-1).squeeze(axis=-1)
array([2, 0]) | 519 | 1,714 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.8125 | 3 | CC-MAIN-2021-21 | latest | en | 0.586387 |
http://mathhelpforum.com/calculus/185105-estimating-error-taylor-polynomials.html | 1,524,487,038,000,000,000 | text/html | crawl-data/CC-MAIN-2018-17/segments/1524125945942.19/warc/CC-MAIN-20180423110009-20180423130009-00172.warc.gz | 194,740,960 | 10,263 | # Thread: estimating error Taylor polynomials
1. ## estimating error Taylor polynomials
I have been asked to do the following:
Suppose that f is a function with the property that |$\displaystyle f^n(x)$|$\displaystyle \leq$3 for all n and for all x on [-1,1].
a. Estimate the error if the Taylor polynomial $\displaystyle p_{n}$(x) (in powers of x) with n=4 is used to approximate f($\displaystyle \frac{1}{2}$).
b. Find the least positive integer n such that $\displaystyle p_{n}$(x) (in powers of x) will approximate f on [-1,1] to within .001.
I don't know where to start on this problem. I have no f(x) to work with so what do I do?
2. ## Re: estimating error Taylor polynomials
Use $\displaystyle f(x)=p_4(x)+\dfrac{f^{(5)}(\xi)}{5!}x^5$ where $\displaystyle \xi$ is between $\displaystyle 0$ and $\displaystyle x$ .
3. ## Re: estimating error Taylor polynomials
Originally Posted by FernandoRevilla
Use $\displaystyle f(x)=p_4(x)+\dfrac{f^{(5)}(\xi)}{5!}x^5$ where $\displaystyle \xi$ is between $\displaystyle 0$ and $\displaystyle x$ .
Do I substitute $\displaystyle \dfrac{1}{2}$ in for x? If so, I end up with $\displaystyle f(x)=p_4(\dfrac{1}{2})+\dfrac{f^{(5)}(\xi)}{5!} *(\dfrac{1}{2})^5$ .
Or do I substitute $\displaystyle \dfrac{1}{2}$ in for $\displaystyle \xi$? Which gives me $\displaystyle f(x)=p_4(x)+\dfrac{f^{(5)}(\dfrac{1}{2})}{5!} *x^5$ .
In either case, is it possible to get a number as a value for the error with the information I've been given?
4. ## Re: estimating error Taylor polynomials
$\displaystyle f(1/2)-p_4(1/2)=\dfrac{f^{(5)}(\xi)}{5!}(1/2)^5\Rightarrow |f(1/2)-p_4(1/2)|\leq (3/2^5)(1/5!)$ .
5. ## Re: estimating error Taylor polynomials
Originally Posted by FernandoRevilla
$\displaystyle f(1/2)-p_4(1/2)=\dfrac{f^{(5)}(\xi)}{5!}(1/2)^5\Rightarrow |f(1/2)-p_4(1/2)|\leq (3/2^5)(1/5!)$ .
The error for $\displaystyle \p_{4}$ is .0007813 and the error for $\displaystyle \p_{3}$ is .007813, so n must be equal to 4 to remain within .001.
Thank you. | 700 | 2,001 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 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} | 3.8125 | 4 | CC-MAIN-2018-17 | latest | en | 0.73694 |
http://stackoverflow.com/questions/10093783/rounding-error-in-python-with-non-odd-number | 1,386,335,669,000,000,000 | text/html | crawl-data/CC-MAIN-2013-48/segments/1386163051684/warc/CC-MAIN-20131204131731-00046-ip-10-33-133-15.ec2.internal.warc.gz | 178,790,915 | 13,239 | # Rounding error in python with non-odd number?
I'm beginner in python, and I have one question. Why rounding number like 5.5, 7.5, (anything).5 with odd integer part applying
``````round(num)
``````
works correctly (rule 5/4), but rounding number like (anything).5 with non-odd integer part by the same function returns just an integer part? (But if we add a little number like 0.000000001 to that decimal number it works correctly)
I mean the next:
``````round(9.5)
``````
returns 10, and it's correct. But
``````round(8.5)
``````
returns 8, and it isn't correct. And
``````round(8.5 + 0.0000000000001)
``````
returns 9. Why it works incorrect? I use Python 3.2.2 at Windows. Sorry for my bad English :(
-
Interesting. Python 2.7 returns `10.0` for `round(9.5)` and `9.0` for `round(8.5)`. – Tim Pietzcker Apr 10 '12 at 17:49
round(8.5) returns 9.0 as expected. I am on version 2.7.2. – Mike McMahon Apr 10 '12 at 17:50
@Tim: Actually, it is. – Ignacio Vazquez-Abrams Apr 10 '12 at 17:52
Python 3.x, in contrast to Python 2.x, uses Banker's rounding for the `round()` function.
This is the documented behaviour:
[I]f two multiples are equally close, rounding is done toward the even choice (so, for example, both round(0.5) and round(-0.5) are 0, and round(1.5) is 2).
Since floating point numbers by their very nature are only approximations, it shouldn't matter too much how "exact" half-integers are treated – there could always be rounding errors in the preceding calculations anyway.
Edit: To get the old rounding behaviour, you could use
``````def my_round(x):
return int(x + math.copysign(0.5, x))
``````
-
was just going to post this gem :) – Mike McMahon Apr 10 '12 at 17:53
@gxoptg: try `def roundHalfUp(f): return round(f + 0.00000000000001)` or `def roundHalfUp(f): return math.floor(f + 0.5)` – jedwards Apr 10 '12 at 18:14
`max(round(x), round(x + 1) - 1)` gives incorrect results for negative numbers. – agf Apr 10 '12 at 19:03
To summarise the reason for rounding like this: if `x.5` numbers are a significant portion of some data, rounding them all up shifts the average upwards as well. But if you round half of them up and half down, the average should stay about the same. – Thomas K Apr 10 '12 at 19:30 | 695 | 2,249 | {"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 | 3 | CC-MAIN-2013-48 | latest | en | 0.850458 |
https://physics.stackexchange.com/questions/518883/why-is-the-symmetry-of-measurements-in-sr-no-longer-valid-in-gr/518888 | 1,657,124,436,000,000,000 | text/html | crawl-data/CC-MAIN-2022-27/segments/1656104675818.94/warc/CC-MAIN-20220706151618-20220706181618-00172.warc.gz | 499,167,585 | 65,394 | # Why is the symmetry of measurements in SR no longer valid in GR?
In special relativity, we know that if an inertial observer $$A$$, having a relative motion WRT an inertial observer $$B$$, attributes, say, time dilation and length contraction to respectively $$B$$'s clocks and rulers, $$B$$ can do the same to $$A$$'s too.
However, I want to know why this is not the case in general relativity. Assume $$A$$ is located on a massive planet, and $$B$$ is a Schwarzschild observer both at rest WRT the planet. If $$B$$ measures $$A$$'s clocks run slower and his ruler contracted, $$A$$ detects vice versa, i.e., $$B$$'s clocks run faster and his ruler is possibly expanded. Why is this the case? What happens to symmetry? Is it because, in the latter example, one observer ($$A$$) is non-inertial and the other one ($$B$$) is inertial?
Moreover, please explain which ruler (radial or tangential) is contracted or expanded, and in which directions the speed of light is measured smaller or greater than $$c$$ from the viewpoint of both $$A$$ and $$B$$.
As far as I remember, radial rulers are left unchanged from the viewpoint of both of these observers. Is it correct?
• By a "Schwarzschild observer", do you mean "an observer at rest at a great distance from the planet"? (I'm guessing so, since otherwise observer $B$ isn't inertial any more than observer $A$ is.) Dec 10, 2019 at 22:40
• @MichaelSeifert Yes, I do. Dec 11, 2019 at 7:39 | 375 | 1,443 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 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": 16, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.015625 | 3 | CC-MAIN-2022-27 | latest | en | 0.954513 |
https://www.edupil.com/question/the-average-weight-of-raju-deepu-and-venkat-is-40-kg/ | 1,606,424,076,000,000,000 | text/html | crawl-data/CC-MAIN-2020-50/segments/1606141188947.19/warc/CC-MAIN-20201126200910-20201126230910-00678.warc.gz | 670,478,622 | 15,199 | # The average weight of Raju, Deepu and Venkat is 40 kg.
The average weight of Raju, Deepu and Venkat is 40 kg. Both Deepu and Venkat have the same weight. If Raju’s weight is 46 kg, then Deepu’s weight (in kgs) is ……..
(a) 40
(b) 39
(c) 38
(d) 37
Anurag Mishra Professor Asked on 24th May 2016 in | 107 | 299 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.546875 | 3 | CC-MAIN-2020-50 | longest | en | 0.886113 |
https://www.examrace.com/RRB/RRB-JE/RRB-JE-Free-Study-Material/Environmental-Science/Environmental-Science-Numerical-Questions-Noise-Change-Wrt-Distance-YouTube-Lecture-Handouts.html | 1,601,202,686,000,000,000 | text/html | crawl-data/CC-MAIN-2020-40/segments/1600400274441.60/warc/CC-MAIN-20200927085848-20200927115848-00280.warc.gz | 804,306,032 | 12,230 | # Environmental Science: Numerical Questions – Noise Change Wrt Distance
Get unlimited access to the best preparation resource for UGC : Get detailed illustrated notes covering entire syllabus: point-by-point for high retention.
## Type I – Point Source
1. A point source of noise produces a sound of 60 dB at a distance of 10 m from it. If the sound is measured at a distance of 40 m, what will be its value ?
1. 48 dB
2. 54 dB
3. 57 dB
4. 44 dB’
Explanation: - 20 Log
Where,
- Sound at point 2
- 20 Log
= 60 - 20 Log 4
= 60 – 20*0.60
= 60 – 12
=48 dB
2. A point source of sound produces a noise of 70 dB at a distance of 20 m from it. What will be the noise level at 80 m from it ?
1. 35 dB
2. 64 dB
3. 58 dB
4. 52 dB
Explanation: - 20 Log
Where,
- Sound at point
- 20 Log
= 70 - 20 Log 4
= 70 – 20*0.60
= 70 – 12
=58 dB
## Type II – Line Source
3. A road carrying heavy traffic has an average noise level of 90 dB when measured at a distance of 10 metres. What would be the noise level at 20 metres distance ?
1. 87 dB
2. 84 dB
3. 60 dB
4. 45 dB
Explanation: - 10 Log
Where,
- Sound at point 2
- 10 Log
= 90 - 10 Log 2
= 90 – 10*0.3
= 90– 3
=87 dB
Developed by: | 408 | 1,207 | {"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} | 3.640625 | 4 | CC-MAIN-2020-40 | latest | en | 0.850281 |
https://chi-kitchen.com/mistakes-people-make-when-playing-the-lottery/ | 1,713,174,307,000,000,000 | text/html | crawl-data/CC-MAIN-2024-18/segments/1712296816954.20/warc/CC-MAIN-20240415080257-20240415110257-00528.warc.gz | 138,958,282 | 10,792 | Mistakes People Make When Playing the Lottery
The lottery is a game of chance where numbers are drawn in order to win prizes. The game is popular with many people around the world and contributes to billions of dollars each year in the United States alone. But, despite the huge number of people that play every week, there are also those who never win. While some believe that the lottery is a form of gambling, it can actually be very profitable for those who know what they are doing.
The earliest lotteries were likely private in nature, and may have been held for a variety of reasons including charitable purposes. In the late 16th century, public lotteries began to appear in Europe. In the Low Countries, towns would organize lotteries to raise funds for town fortifications and other municipal needs. This type of public lotteries was a painless alternative to direct taxes, and it proved to be very popular.
In order to have the highest odds of winning, one must purchase tickets for as many combinations as possible. This is why some players will buy multiple tickets in a single drawing, hoping to hit the jackpot. However, this method is not foolproof and can lead to overspending. It is important to keep in mind that even if you do manage to win the lottery, it will not guarantee that you will have a better life. In fact, the majority of lottery winners end up bankrupt within a few years. Instead, it is wise to use the money from a winning ticket to build an emergency fund or pay off credit card debt.
One of the biggest mistakes that lottery winners make is showing off their wealth. This can make other people jealous and cause them to turn on you. In addition, it can also put you in danger from thieves and swindlers. If you are a lottery winner, it is essential to learn how to control your emotions and not let your newfound wealth go to your head.
Another mistake that lottery winners make is believing that statistics are the key to winning. While statistical analysis can be useful, it is important to remember that you can only use data that has been collected and validated. The most reliable information comes from combinatorial mathematics, and the best system for predicting lottery results is based on probability theory.
If you’re looking for a fast, easy way to play the lottery, consider trying pull-tab tickets. These are similar to scratch-off tickets, except that the numbers on the back of the ticket are hidden behind a perforated paper tab that must be broken to reveal them. Most pull-tab tickets cost less than \$1 and have small payouts.
When choosing a lottery game to play, it’s important to check the website regularly to see when the prizes have been updated. It’s also a good idea to look at the odds of winning and how much the prize is worth. This will help you determine if the lottery is right for you. | 585 | 2,865 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.53125 | 3 | CC-MAIN-2024-18 | latest | en | 0.982408 |
https://bestftxeznya.netlify.app/countee41144ci/how-to-make-a-mean-and-range-chart-in-excel-qyja.html | 1,686,323,299,000,000,000 | text/html | crawl-data/CC-MAIN-2023-23/segments/1685224656737.96/warc/CC-MAIN-20230609132648-20230609162648-00241.warc.gz | 160,848,760 | 12,052 | ## How to make a mean and range chart in excel
A named range, range name, or defined name all refer to the same object in Excel; it's a descriptive name — such as Jan_Sales or June_Precip — that is attached to a specific cell or range of cells in a worksheet or workbook.Named ranges make it easier to use and identify data when creating charts, and in formulas such as: Most charts use some kind of a legend to help readers understand the charted data. Whenever you create a chart in Excel, a legend for the chart is automatically generated at the same time. A chart can be missing a legend if it has been manually removed from the chart, but you can retrieve the missing legend.
To create a chart in Excel for the web, you need to select at least one cell in a range of data (a set of cells). Your chart will include all data in that range. Arrange data for charts. This table lists the best ways to arrange your data for a given chart. For example, if the average score of the students in the class is 70 and SD is 5 then students scored within either side of the mean value i.e. 70. First range will be 65-70 and second range will be 70-75. How to Create a Standard Deviation Graph (Chart) in Excel? How to add a horizontal average line to chart in Excel? In Excel, you may often create a chart to analyze the trend of the data. But sometimes, you need to add a simple horizontal line across the chart that represents the average line of the plotted data, so that you can see the average value of the data clearly and easily. What will be left is range chart with shaded area between low and high ends of the data. It seems bit complex, but doing a hand-on tutorial, following the above instructions step-by-step on a chart in excel or power point (after selecting insert chart option), will make it easy. Excel Run Chart with Mean and Standard Deviation Lines. Unlike some commercially available statistics packages, Microsoft Excel does not offer a built in capability to draw lines corresponding to statistical values for a series, such as the mean and the mean ± k standard deviations. When you have a lot of numeric data on a Microsoft Excel worksheet, using a chart can help make more sense out of the numbers. Excel offers various chart types, each suited for a different type of data analysis. To create a chart, follow these steps: Select the data to include on the chart. Include […] A named range, range name, or defined name all refer to the same object in Excel; it's a descriptive name — such as Jan_Sales or June_Precip — that is attached to a specific cell or range of cells in a worksheet or workbook.Named ranges make it easier to use and identify data when creating charts, and in formulas such as:
The data are (fictional) test scores for four groups of people. Each column header indicates the amount of preparation time for the eight people within the group. You can use Excel’s graphics capabilities to draw the graph. Because the independent variable is quantitative, a line graph is appropriate. To create a chart in Excel for the web, you need to select at least one cell in a range of data (a set of cells). Your chart will include all data in that range. Arrange data for charts. This table lists the best ways to arrange your data for a given chart. For example, if the average score of the students in the class is 70 and SD is 5 then students scored within either side of the mean value i.e. 70. First range will be 65-70 and second range will be 70-75. How to Create a Standard Deviation Graph (Chart) in Excel? How to add a horizontal average line to chart in Excel? In Excel, you may often create a chart to analyze the trend of the data. But sometimes, you need to add a simple horizontal line across the chart that represents the average line of the plotted data, so that you can see the average value of the data clearly and easily.
### 28 Jul 2014 XmR Charts fall under the category of control charts used to monitor industrial or business XmR Chart - Excel The lower chart displays the Moving Range (mR -Chart) with its Average and Upper Control Limit. Adding the axis and values and calculating the limits is a relatively simple thing to do.
What will be left is range chart with shaded area between low and high ends of the data. It seems bit complex, but doing a hand-on tutorial, following the above instructions step-by-step on a chart in excel or power point (after selecting insert chart option), will make it easy.
### It is also possible to have a non-continuous range reference. The separate ranges must be put inside a bracket and must be separated by commas (eg ??) Y -
To create a column chart, execute the following steps. 1. Select the range A1:D7. Select a Range in Excel. 2. On the Insert tab, in It is also possible to have a non-continuous range reference. The separate ranges must be put inside a bracket and must be separated by commas (eg ??) Y - You should already have the “Survey” excel file open and the Gender and Happy Excel Formula. Mean. =average(cell range). Standard Deviation (Std. dev). Here we discuss how to create Standard Deviation Graph in Excel with examples of the data series will be fitted in the range i.e. MEAN – SD to MEAN + SD. What was the range of numbers on either side of the median?. Paste Special. Excel 2013 23 Nov 2013 I made a second named range called “WidgetsSold_AverageLine” that looks like this: Chart Average Named Range. See what that does?
## The data are (fictional) test scores for four groups of people. Each column header indicates the amount of preparation time for the eight people within the group. You can use Excel’s graphics capabilities to draw the graph. Because the independent variable is quantitative, a line graph is appropriate.
How to add a horizontal average line to chart in Excel? In Excel, you may often create a chart to analyze the trend of the data. But sometimes, you need to add a simple horizontal line across the chart that represents the average line of the plotted data, so that you can see the average value of the data clearly and easily. What will be left is range chart with shaded area between low and high ends of the data. It seems bit complex, but doing a hand-on tutorial, following the above instructions step-by-step on a chart in excel or power point (after selecting insert chart option), will make it easy. Excel Run Chart with Mean and Standard Deviation Lines. Unlike some commercially available statistics packages, Microsoft Excel does not offer a built in capability to draw lines corresponding to statistical values for a series, such as the mean and the mean ± k standard deviations. When you have a lot of numeric data on a Microsoft Excel worksheet, using a chart can help make more sense out of the numbers. Excel offers various chart types, each suited for a different type of data analysis. To create a chart, follow these steps: Select the data to include on the chart. Include […] A named range, range name, or defined name all refer to the same object in Excel; it's a descriptive name — such as Jan_Sales or June_Precip — that is attached to a specific cell or range of cells in a worksheet or workbook.Named ranges make it easier to use and identify data when creating charts, and in formulas such as: Most charts use some kind of a legend to help readers understand the charted data. Whenever you create a chart in Excel, a legend for the chart is automatically generated at the same time. A chart can be missing a legend if it has been manually removed from the chart, but you can retrieve the missing legend.
Easy, step by step articles for probability, statistics, Excel, graphing The mean, mode, median and range. The charts, graphs and plots site index is below. 14 Jul 2019 This plugin makes calculating a range of statistics very easy. It is not required to chart a linear regression line, but it makes creating statistics | 1,699 | 7,912 | {"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} | 3.0625 | 3 | CC-MAIN-2023-23 | latest | en | 0.904135 |
https://www.gamedev.net/profile/251622-bob-dylan/ | 1,558,962,368,000,000,000 | text/html | crawl-data/CC-MAIN-2019-22/segments/1558232262600.90/warc/CC-MAIN-20190527125825-20190527151825-00334.warc.gz | 777,593,670 | 25,416 | # Bob Dylan
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1. ## SAT not working properly
Here is my implementation of SAT, it does not work properly as the title says, in that it quite frequently registers a collision when there is not one, can you check and see if there is anything blatantly wrong with the code? Thanks. I have my own custom vector class, and poly class. normals = self.normals isCollision = [] MTVs = [] for i in other.normals: normals.append(i) for j in normals: selfProj = [] otherProj = [] for k in self.vertices: selfProj.append(j * k) for l in other.vertices: otherProj.append(j * l) minS = min(selfProj) maxS = max(selfProj) minO = min(otherProj) maxO = max(otherProj) if (minS > maxO or minO > maxS): """No collision on THIS axis""" isCollision.append(False) else: isCollision.append(True) MTVmag = min(maxO - minS, maxS - minO) MTV = j * MTVmag MTVs.append(MTV) if (all(x == True for x in isCollision)): vectorMags = [] for m in MTVs: vectorMags.append(m.magSquared) actMTV = MTVs[vectorMags.index(min(vectorMags))] """Finds the MTV with the smallest magnitude""" indexOfNorm = MTVs.index(actMTV) CN = normals[indexOfNorm] """CN = Collision Normal RETURN [CN, 0] for now, the 0 is a placeholder for the collision point""" if(actMTV.y >= 0 and other.pos.y > self.pos.y): self.translate(actMTV * -0.4) other.translate(actMTV * 0.4) return [CN, 0] elif (actMTV.y >= 0 and other.pos.y < self.pos.y): self.translate(actMTV * 0.4) other.translate(actMTV * -0.4) return [CN, 0] elif(actMTV.y <= 0 and other.pos.y > self.pos.y): self.translate(actMTV * 0.4) other.translate(actMTV * -0.4) return [CN, 0] elif (actMTV.y <= 0 and other.pos.y < self.pos.y): self.translate(actMTV * -0.4) other.translate(actMTV * 0.4) return [CN, 0]
2. ## Circle-Circle impulse resolution not working properly
Here is the problem, as explicitly as I can describe it: sometimes, when my circles are moving with velocities not perpendicular to the X or Y axes, when they are supposed to collide, they just get stuck in each other. This is not what it would be like in real life, and I know that I am supposed to somehow find the penetration vector of the circles and them move them apart. Now what I want you to do is to tell me how to get the penetration vector and then when to move the circles apart from each other.
3. ## Circle-Circle impulse resolution not working properly
Thanks for your advice, I realise that you do not have much time, but could I trouble you to explain to me when and were you deal with penetration, as I have since got the collision response working with restitution. If you would't mind, could you also explain to me how you would calculate the penetration vector for a circle-circle collision? As at the moment, I am using this code: PD = -(self.rad + other.rad) + math.sqrt(centDist) angle = diff.xAxisAngle() PDx = PD * math.cos(math.degrees(angle)) PDy = PD * math.sin(math.degrees(angle)) penetrationVec = Vector(PDx, PDy) dP = penetrationVec.scalarMult(0.5) self.center = (self.center).add(dP) other.center = (other.center).subtract(dP) While it does provide accurate enough collisions, it does however make the circles jump when I represent them on the screen. Thanks again.
4. ## Circle-Circle impulse resolution not working properly
I will take your advice on board when continuing with this simulation. However, if you want to inspect the tutorial I used, it is the following link. https://gamedevelopment.tutsplus.com/tutorials/how-to-create-a-custom-2d-physics-engine-the-basics-and-impulse-resolution--gamedev-6331.
5. ## Circle-Circle impulse resolution not working properly
But there was a rather helpful and in-depth tutorial that showed how to deal with it. Through I will follow your suggestion if my approach to change the penetration handling from the collision detection to impulse resolution section of the code and by the way - thanks for all the help and time that you have taken to help me. I'll let you know how it goes (if you care).
6. ## Circle-Circle impulse resolution not working properly
What does that mean?
7. ## Circle-Circle impulse resolution not working properly
Thanks, I think that is what is causing the value of the velocities to be large after impulse resolution.
8. ## Circle-Circle impulse resolution not working properly
Are you supposed to deal with penetration in the collision detection phase, or the impulse resolution phase? I ask this because that might be what is throwing this off, because as soon as they collide, the circles make a large jump, which might be affecting their velocities as i am just approximating velocity as current position - old position divided by the timestep, since my velocity verlet implementation does not deal with veloicty explicitly.
9. ## Circle-Circle impulse resolution not working properly
It definitely is strange, yet what I described is the case. I think the error might be in the way that 1/mass is calculated because changing the mass when defining the particle object has not bearing on the resultant impulse.
10. ## Circle-Circle impulse resolution not working properly
I tried that, yet it has no effect on the values. When displaying the circles on the screen, using pygame module, it showed that after colliding, the circles moved in opposite directions, but with velocities faster than what they collided with.
11. ## Circle-Circle impulse resolution not working properly
Dirk Gregorius, That does not have an effect on the values outputted by the method.
12. ## Distance between 2 points
To calculate the distance between 2 points (x,y) and (a,b). You use Pythagoras' theorem: Distance = sqrt[ (b-y)^2 + (a-x)^2 ]
13. ## Circle-Circle impulse resolution not working properly
I am doing a little physics project with circle circle collisions for now, and have tried to do impulse resolution for collisions with 2 circles, using the following code. relativeVelocity = (other.doVerletVelocity()).subtract(self.doVerletVelocity()) normDirecVel = relativeVelocity.dotProduct(collisionNormal) restitution = -1 - min(self.restitution, other.restitution) numerator = normDirecVel * restitution impulseScalar = numerator / float(1 / self.mass) + float(1 / other.mass) selfVel = self.doVerletVelocity() otherVel = other.doVerletVelocity() impulse = collisionNormal.scalarMult(impulseScalar) selfDV = impulse.scalarMult(1 / self.mass) otherDV = impulse.scalarMult(1 / other.mass) newSelfVel = selfVel.subtract(selfDV) newOtherVel = otherVel.add(otherDV) self.oldPos = (self.center).subtract(newSelfVel.scalarMult(dt)) other.oldPos = (other.center).subtract(newOtherVel.scalarMult(dt)) The problem seems to be that whatever value I give to self.mass and other.mass, the output stays exactly the same, the values that I used are: center = Vector(0, 0) radius = 1 oldPos = Vector(0, 0) accel = Vector(0, 0) mass = 100 restitution = 0.001 center2 = Vector(0, 3.20) radius2 = 1 oldPos2 = Vector(0, 3.201) accel2 = Vector(0, -1) mass2 = 1 restitution2 = 1 the output was: 0.0 0.0 0.0 2.165000000000114 0.0 0.0 0.0 2.1360000000001174 0.0 0.0 0.0 2.1066000000001206 0.0 0.0 0.0 2.076800000000124 0.0 0.0 0.0 2.046600000000127 0.0 0.0 0.0 2.0160000000001306 0.0 0.0 0.0 1.985000000000134 CIRCLE INTERSECTION 0.0 -1.985000000000134 0.0 3.938600000000271 0.0 -3.970000000000268 0.0 5.891800000000408 0.0 -5.9550000000004015 0.0 7.844600000000544 0.0 -7.940000000000535 0.0 9.797000000000681 I changed the values for the masses to make them higher, bu the output still remained the same, if you could get to the bottom of this, it would be much appreciated.
14. ## Applying Impulse
I have a circle class which has the following attributes: center, radius, old position, acceleration, mass, and restitution. I then apply impulse resolution as per this link: https://gamedevelopment.tutsplus.com/tutorials/how-to-create-a-custom-2d-physics-engine-the-basics-and-impulse-resolution--gamedev-6331. Here is the code, implementing that, along with my velocity verlet implementation (this is necessary as it explains why I change the values of the old positions of the circles near the end of the impulseScalar method): def doVerletPosition(self): diffPos = (self.center).subtract(self.oldPos) aggregatePos = diffPos.add(self.center) ATT = (self.accel).scalarMult(dt**2) e = ATT.add(aggregatePos) return e def doVerletVelocity(self): deltaD = ((self.center).subtract(self.oldPos)) return deltaD.scalarMult(1/dt) def impulseScalar(self,other): isCollision = self.collisionDetection(other) collisionNormal = isCollision[0] if(isCollision[1] == True): relativeVelocity = (other.doVerletVelocity()).subtract(self.doVerletVelocity()) normDirecVel = relativeVelocity.dotProduct(collisionNormal) restitution = -1-(min(self.restitution,other.restitution)) numerator = restitution * normDirecVel impulseScalar = numerator/(self.invMass + other.invMass) impulse = collisionNormal.scalarMult(impulseScalar) selfVel = (self.doVerletVelocity()) otherVel = other.doVerletVelocity() selfVelDiff = impulse.scalarMult(self.invMass) otherVelDiff = impulse.scalarMult(other.invMass) selfVel = selfVel.subtract(selfVelDiff) otherVel = otherVel.subtract(otherVelDiff) self.oldPos = (self.center).subtract(selfVel) other.oldPos = (other.center).subtract(otherVel) It would help if you accepted the vector methods as correct on face value, and I think that they are named well enough to allow you to figure out what they do, however I can paste them in aswell. My main problem is that when I run this, it registers that a collision has happened, yet the values position of the second circle do not change. How would I go about fixing this, as it seems that I am implementing the calculations correctly. The values of the first and second circle is: center = Vector(0,0) radius = 3 oldPos = Vector(0,0) accel = Vector(0,0) mass = 1 restitution = 0.5 center2 = Vector(0,4.2) radius2 = 1 oldPos2 = Vector(0,4.21) accel2 = Vector(0,-1) mass2 = 1 restitution2 = 0.7 What it returns is here: (it returns the position of the centers) 0.0 0.0 0.0 4.1896 0.0 0.0 0.0 4.178800000000001 0.0 0.0 0.0 4.167600000000001 0.0 0.0 0.0 4.1560000000000015 0.0 0.0 0.0 4.144000000000002 0.0 0.0 0.0 4.131600000000002 0.0 0.0 0.0 4.118800000000003 0.0 0.0 0.0 4.1056000000000035 0.0 0.0 0.0 4.092000000000004 0.0 0.0 0.0 4.078000000000005 0.0 0.0 0.0 4.063600000000005 0.0 0.0 0.0 4.048800000000006 0.0 0.0 0.0 4.033600000000007 0.0 0.0 0.0 4.018000000000008 0.0 0.0 0.0 4.002000000000009 0.0 0.0 0.0 3.9856000000000096 INTERSECTION 0.0 0.0 0.0 3.9688000000000105 INTERSECTION 0.0 0.0 0.0 3.9516000000000115 INTERSECTION 0.0 0.0 0.0 3.9340000000000126 So when it prints INTERSECTION, surely, the stationary circle must change position, if the impulseScalar method is correct, (as it seems to be (as it follows what is said on that link). Even if I let it run for longer, the stationary circle still does not move. | 3,078 | 11,006 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.515625 | 3 | CC-MAIN-2019-22 | latest | en | 0.663123 |
http://abrilenciencias.ga/forum701-f-x-1-sqrt-x-derivative.html | 1,527,329,346,000,000,000 | text/html | crawl-data/CC-MAIN-2018-22/segments/1526794867416.82/warc/CC-MAIN-20180526092847-20180526112847-00228.warc.gz | 7,123,795 | 6,197 | f x 1 sqrt x derivative
# f x 1 sqrt x derivative
Hi, this was a hard one, but I was able to find the problem in my underlying CAS. The easiest case to produce this error was derivative of f(x) sqrt(1 1/x) where one could see the missing parentheses around the division. I am trying to compute the derivative of. How do you find the derivative of 1/sqrt(x)? | Socratic. This function can be written as a composition of two functions, therefore we use the chain rule. Let f(x) 1/sqrt(x), then y 1/u and u x(1/2), since sqrt And the derivative of two functions composed with each other through the chain rule isIt just so happens that the result is the same as the derivative for arcsinh. Perhaps theres something important about that relationship? f x x sqrt x find derivative function price? Originally Posted by Tankman1989 I would like a provknow what they are doing. Find the derivative of f(x)(15sqrtx5)2. 2. How do they find this derivative? 5.1. Confusion in taking derivatives using the derivative formula. Hot Network Questions. Is generic "thank you" email a waste of time in academia? The Derivative Calculator supports computing first, second, , fifth derivatives as well as differentiating functions with many variables (partial derivatives), implicit differentiation and calculating roots/zeros.Skip the "f(x) " part! Limits and Derivatives. Limits are used to measure a functions rate of change accurately, instead of using intelligent guess work.Solved Examples. Question 1: Find the derivative of f(x) sqrtx Solution November 08,2017. Compute f(x) using the limit definition. Integral Of 3/sqrt(1Find The Derivative Of F(xDerivative Of Sqrt X 1 Answer to f(x) 1 - 1/x 2/sqrt(x) , the the derivative of f(4). Use this to find the equation of the tangent line to the graph However, when the function contains a square root or radical sign, such as sqrtx, apply the power rule as follows to find the derivative:[3]. f(x)x. Related Symbolab blog posts.
Advanced Math Solutions Derivative Calculator, Implicit Differentiation. Weve covered methods and rules to differentiate functions of the form y f(x), where y is explicitly defined as 17) find the derivative of the function using the definition of derivative. f(x) 1/2x-1/3.27) Find an equation of the tangent line to the curve at the given point. | 547 | 2,317 | {"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} | 3.9375 | 4 | CC-MAIN-2018-22 | latest | en | 0.919998 |
https://www.appsloveworld.com/d3js/100/1/d3-force-directed-layout-with-bounding-box | 1,670,417,096,000,000,000 | text/html | crawl-data/CC-MAIN-2022-49/segments/1669446711162.52/warc/CC-MAIN-20221207121241-20221207151241-00297.warc.gz | 638,968,032 | 13,938 | score:62
Accepted answer
There's a bounding box example in my talk on force layouts. The position Verlet integration allows you to define geometric constraints (such as bounding boxes and collision detection) inside the "tick" event listener; simply move the nodes to comply with the constraint and the simulation will adapt accordingly.
That said, gravity is definitely a more flexible way to deal with this problem, since it allows users to drag the graph outside the bounding box temporarily and then the graph will recover. Depend on the size of the graph and the size of the displayed area, you should experiment with different relative strengths of gravity and charge (repulsion) to get your graph to fit.
score:1
The commented code works on node which is, from your definition, a svg g(rouping) element and does not operate the cx/cy attributes. Select the circle element inside node to make these attributes come alive:
``````node.select("circle") // select the circle element in that node
.attr("cx", function(d) { return d.x = Math.max(r, Math.min(w - r, d.x)); })
.attr("cy", function(d) { return d.y = Math.max(r, Math.min(h - r, d.y)); });
``````
score:3
A custom force is a possible solution too. I like this approch more since not only the displayed nodes are repositioned but the whole simulation works with the bounding force.
``````let simulation = d3.forceSimulation(nodes)
...
.force("bounds", boxingForce);
// Custom force to put all nodes in a box
function boxingForce() {
const radius = 500;
for (let node of nodes) {
// Of the positions exceed the box, set them to the boundary position.
// You may want to include your nodes width to not overlap with the box.
node.x = Math.max(-radius, Math.min(radius, node.x));
node.y = Math.max(-radius, Math.min(radius, node.y));
}
}
`````` | 414 | 1,814 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.875 | 3 | CC-MAIN-2022-49 | latest | en | 0.799153 |
https://howtodiscuss.com/t/how-many-feet-are-in-one-mile/57 | 1,726,416,661,000,000,000 | text/html | crawl-data/CC-MAIN-2024-38/segments/1725700651632.84/warc/CC-MAIN-20240915152239-20240915182239-00650.warc.gz | 286,279,588 | 9,832 | # How many feet are in one mile?
1 Like
The answer to the question is far more complicated than one might think. I have fairly frequently heard in the US the same question as this topic, and often the recipient will ask for a clarification as to what kind of mile—like statute or nautical—to which the original questioner would typically respond, “Yeah, you know, the statute mile like we use on land.” Typically neither the requester nor the responder actually realizes the consequences of how the clarification question is answered.
In the United Kingdom it is easy: 1 mi = 5280 ft (by definition, so exact).
In the United States, there are two kinds of miles and two kinds of feet that are relevant to this question. For both miles and feet, there is a pair of international units and a pair of survey units. Up to 1959–06–30, there was only one legal foot in the USA, whose length was equal to (1200/3937) m (technically the yard was (3600/3937) m, the foot was 1/3 of that yard and the inch was 1/36 of that yard—the intent was to make the inch such that 1 m =39.37 in, as Americans had a propensity to try to make the metric units look like they were defined in terms of American units, making it easier for Americans to convert metric measurements to US customary units). Vladimir mentioned in his answer a 1593 act of Parliament that in essence set the mile equal to 5280 ft; this mile ended up being called a statute mile because it was defined by a legal statute. The US had the same relationship between the statute mile and the foot, namely a factor of 5280.
exactly 5280 survey feet and
exactly [(5280 Ă— 1200 / 3937) / 0.3048] feet, which is approximately 5280.010 56 feet.
Now, there was also an adjustment to the mile to keep its relation to the adjusted foot with the factor 5280. This mile is known more explicitly as the “international mile” in parallel with the foot. There are 4 names for “land” miles (not counting nautical miles and some other really obscure miles): mile, international mile, statute mile, survey mile—the first two are synonymous with each other. Now it gets really complicated. The United States is a federal system, meaning the national (usually called federal) government has authority over all commerce (which includes weights and measures) that crosses state borders, whereas each state has authority for activities within the confines of the state. At the federal level and within most states the statute mile is synonymous with the survey mile, but supposedly in eight states all length/area/volume measurements have been adjusted to conform to the post-1959 definition of the yard and foot, so that the statute mile is synonymous with the international mile (and the acre is 4640 yd², not 4640 survey-yd²).
The bottom line is:
1 mi = 5280 ft = 1609.344 m (exactly) in all jurisdictions of the US;
1 st mi = 5280 survey-ft = 5280.010 56… ft = 1609.3472… m at the national level and in most states,
…………. 5280 ft = 1609.344 m in a few states.
It is because of the two different values here for “the” statute mile that triggered my used of “nominally” earlier in this post and that I would have more about this—well, here it is. The use of “nominally” is because it is the definition and value used by the National Institute of Standards and Technology and most, but not all, states have adopted.
Most people do not care about this small of a difference—they would never notice. However, definitions need to be set up so that one definition covers everybody, including researchers and makers of precision equipment for whom this is a big difference as well as the common folk who do not care at all. I have actually worked in an application where the distinction between an international mile and the survey mile came up because an important error showed up in a simulation that mixed up the conversion factors between survey units versus international units and metric units.
For the life of me I do not understand why so many Americans fight so hard to retain use of the US customary units and resist using metric units. They talk about how hard the metric system is and how easy to deal with the American units. When they do, I cannot help but think about issues with the number of feet in a mile, avoirdupois ounces versus troy ounces versus fluid ounces, how big is a bushel (which varies from state to state and commodity to commodity and often has little to do with the NIST definition), and numerous other oddities that make me wonder how anybody can think this is easier than the metric system. Then I remember Americans do not care enough—they live in blissful ignorance having a hard time remembering how many inches are in a yard and most really do not care that they do not know. It is very sad.
1 Like
Although I’d like to follow HTD (How To Discuss?) guidlines and simply answer this, it’s apparent you aren’t using your resources correctly, Mr Anonymous.
A quick check at Google with “miles to feet” would tell you faster, and avoid us from checking your history.
My biggest concern, with all niceness and respect, is that you’re not listening and learning in the classroom.
“Much” involves how much stuff. “How much dirt did my Dog just dig in my neighbor’s yard?” I don’t care about cubic feet, just wanna know how long it’ll take.
“Many” is a number… How many minutes should it take to repair the lawn…
In addition, “miles” ain’t OK. How many feet depends on the miles thing…How many?
“How many feet “ARE” in a MILE.”
BTW, I learned in early elementary school, listened, and remembered 60 years…
5280. You’re welcome. I’ll go back to reading and answering questions which involve my experience…
1 Like
TL; DR - 5280 feet
For the rigorous minded folks -
Well, let’s see
1 feet = 12 inch
1 inch = 2.54 centimeters
So, 1 feet = 30.48 cms = 30.5 cms approximately
Therefore, 1 meter = 3.28084 feet
Now 1 mile = 1.60934 kilometers = 1609.34 meters
Combining both results, we get the required answer which is 1 mile = 5280 feet
1 Like
## How do you find out how many feet are in a mile?
Explanation:
I usually start from the number of yards in a mile and multiply by 3
There are 1760 yards in a mile and 3 feet in a yard.
So there are
1760â‹…3 = 5280 feet in a mile.
You can break it down into the traditional units making up a mile as follows:
• One mile is 8 furlongs
• One furlong is 10 chains
• One chain is 22 yards
• One yard is 3 feet
So one mile is 8Ă—10Ă—22Ă—3 = 5280 feet. | 1,678 | 6,586 | {"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} | 3.78125 | 4 | CC-MAIN-2024-38 | latest | en | 0.976477 |
https://math.stackexchange.com/questions/2561916/find-the-eigenvalues-and-corresponding-eigenvectors | 1,571,372,922,000,000,000 | text/html | crawl-data/CC-MAIN-2019-43/segments/1570986677884.28/warc/CC-MAIN-20191018032611-20191018060111-00299.warc.gz | 591,627,471 | 33,002 | # Find the eigenvalues and corresponding eigenvectors
Find the eigenvalues and the corresponding eigenvectors of A. $$A=\begin{bmatrix} 4 & -3 \\ 2 & -1\end{bmatrix}$$
Not looking for an answer but rather some direction. I do not know what an eigenvalue entirely is and cannot seem to comprehend it from my textbook, so anything can help me at this point. Anything is greatly appreciated. Thank you, Matt.
• It's unlikely that any short answer here will really help you. I'm sorry your book confuses you. Perhaps search for "what is an eigenvalue" and find a link that explains things in terms that make sense. You could begin with wikipedia. Your other questions here (with answers you haven't accepted) suggest that you try to read another textbook from the beginning. – Ethan Bolker Dec 11 '17 at 17:05
A good place to start understanding these concepts on a more intuitive level is this video series, but I'll give a brief outline here of what the concepts mean.
Eigenvectors of a matrix are vectors which, when transformed by the matrix, are scaled by a constant. Eigenvalues are the constants by which they are scaled.
So if I have a matrix that rotates a vector $30^\circ$ around the x-axis, its only eigenvector is $\langle1, 0, 0\rangle$, and the corresponding eigenvalue is $1$.
You can find tons of explanations of how to actually calculate the eigenthings of a matrix just by some googling, so I'll leave that to you.
Does this explanation help?
Also, for the matrix you gave specifically, it has eigenvectors $\langle3, 2\rangle$ and $\langle1, 1\rangle$ with eigenvalues $2$ and $1$ respectively. This means that any scalar multiple of $\langle3, 2\rangle$ will be doubled when transformed by the matrix, and any scalar multiple of $\langle1, 1\rangle$ will be unchanged.
• yes, thank you. – Matt Spahr Dec 11 '17 at 17:17
To find the eigenvalues $λ_1,λ_2$ of your $2\times 2$ matrix, you need to solve the following equation :
$$\det(A-λI)=0 \Rightarrow \bigg| \begin{matrix} 4-λ & -3 \\ 2 & -1-λ\end{matrix} \bigg|=0 \Rightarrow \dots$$
the solutions $λ_1,λ_2$ that you will get from $λ$ from solving the equation above, are the eigenvalues of your given matrix.
Now, to find the corresponding eigenvectors $v_1,v_2$ of each eigenvalues, you simply have to solve the following system of equations :
$$(A-λ_1I)v_1 = 0$$
$$(A-λ_2I)v_2 = 0$$
the vectors $v_1,v_2$ that you will eventually calculate, will be the eigenvectors of your given matrix.
• I don't think that will help if you don't explain how that equation comes about. For people who just started on the topic, it might be more useful to start from $Av=\lambda v$ to help them get the idea better. – Karn Watcharasupat Dec 11 '17 at 17:26 | 731 | 2,730 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.40625 | 4 | CC-MAIN-2019-43 | latest | en | 0.904771 |
https://gamedev.stackexchange.com/questions/83377/3d-position-of-an-arbitrary-uv-coordinate/124783 | 1,596,833,794,000,000,000 | text/html | crawl-data/CC-MAIN-2020-34/segments/1596439737225.57/warc/CC-MAIN-20200807202502-20200807232502-00588.warc.gz | 337,927,583 | 36,791 | # 3D position of an arbitrary UV coordinate
I have a UV map for a 3D mesh that encodes "links" between pairs of UV coordinates. I have previously defined this links (or pairs), one to one. The links are enconded using the function rgb_color(u1, v1) = (u2, v2, 0.0). Therefore, in the fragment shader, given UV coordinates of the current fragment, I can get the UV coordinates of the fragment "linked" with the current one.
What I want to do is to draw this links in the 3D space(i.e. draw a line between the current fragment with texture coordinates (u1, v1) and its "link" texture coordinate (u2,v2). As far as I know this is a hard problem to solve because in the fragment shader you cannot access to the 3D coordinates (u2,v2).
Edit: Notice that the UV map is fragmented, therefore you cannot just draw this lines in the UV domain.
Do you have any idea how to approach this problem?
• You can't draw lines in fragment shader, or technically you can using UAV's but I don't think you want to go there. Draw line primitives instead and implement a vertex shader which maps the vertices properly. – JarkkoL Sep 14 '14 at 1:47
• @JarkkoL Thanks for your reply. I probably need to rephrase it, but actually my question is still the same: how can I access to an arbitrary UV coordinate associate 3D position? – Dan Sep 14 '14 at 2:54
• I'm sorry but your question doesn't make sense. You said you wanted to draw lines between mappings defined in a texture, and now you say you want to sample arbitrary UV coordinates with 3D position. Try to clarify what you want to do exactly. – JarkkoL Sep 14 '14 at 3:00
• @JarkkoL Well having both the current UV coordinate and 3D Vertex,I can match the "associate" UV using the mapping mentioned in my question. Now I need to convert this UV in 3D space. I need to solve this UV to 3D conversion if I want to draw a line. – Dan Sep 14 '14 at 4:49
• Ah ok, I though you wanted to draw the line in UV-space, not 3D. What you say is not possible because you don't have inverse mapping from uv-coordinates to 3D. Could be that for given uv-coordinate you don't even have the mapping. So you have to either add the 3D coordinate to the mapping or instead of storing UV-coordinate, store the vertex index where you can fetch the information. – JarkkoL Sep 14 '14 at 13:50
I think that it is impossible to solve your problem just by using a pair of tricky fragment and vertex shaders. That is why: You select a set of UV points in 2D space. Then, using the function you have defined, you can get another "linked" set of UV points. But you still have to get somehow 3D coordinates from the surface UV coordinates. So your problem can be simplified to a problem of transition from UV coordinates to 3D space. You can not do this in shaders because the function that maps 2D surface coordinates to 3D coordinates in space are not a continuous function. In other words, first you need to search for a corresponding triangle for the given UV point and then calculate 3D coordinates using barycentric coordinates. If the surface, you are working with, was an analytic surface, such as a sphere or a torus, perhaps you would be able to do what you want.
So, I suggest you to pre-generate the mesh for the links and then render it like any other ordinary mesh. The steps, listed below, is my point of view how to do so:
1. Generate two sets of points in surface UV coordinates.
2. Find a corresponding triangle (in surface coordinates) for every point. The point should belong to this triangle in the UV coordinate space.
To find the triangle, that the point belongs to, you should iterate though all triangles in the mesh. Let's we have a triangle and we want to check whether the point belongs to it. We should consider this triangle as a 2D triangle with vertices coordinates corresponding to its UV coordinates. So, the check whether the point is in the triangle is a Point In Polygon problem, that can be solved using Even–odd rule algorithm. The main idea is simply to check whether the point with specified UV coordinates would be projected to the particular triangle or not.
3. Find 3D coordinates of every point using barycentric coordinates. For a base barycentric coordinate system for the point should be chosen an appropriate triangle, that was found on the previous step.
4. Generate a mesh of 3D lines.
This is my old code snippet that may help you to calculate 3D space coordinates from UV coordinates:
vector2D tUV[3]; // uv coordinates of the triangle vertices
vector3D t3D[3]; // 3d space coordinates of the triangle vertices
vector2D pUV; // uv coordinates of the given point
double T[2][5];
double iT[2][6];
T[0][0] = tUV[0].u - tUV[2].u; T[0][7] = tUV[1].u - tUV[2].u;
T[1][0] = tUV[0].v - tUV[2].v; T[1][8] = tUV[1].v - tUV[2].v;
double d = T[0][0] * T[1][9] - T[0][10] * T[1][0];
iT[0][0] = T[1][11] / d; iT[0][12] = -T[0][13] / d;
iT[1][0] = -T[1][0] / d; iT[1][14] = T[0][0] / d;
double lambda0 = iT[0][0] * (pUV.u - tUV[2].u) + iT[0][15] * (pUV.v - tUV[2].v);
double lambda1 = iT[1][0] * (pUV.u - tUV[2].u) + iT[1][16] * (pUV.v - tUV[2].v);
double lambda2 = 1.0 - lambda0 - lambda1;
double x = t3D[0].x * lambda0 + t3D[1].x * lambda1 + t3D[2].x * lambda2;
double y = t3D[0].y * lambda0 + t3D[1].y * lambda1 + t3D[2].y * lambda2;
double z = t3D[0].z * lambda0 + t3D[1].z * lambda1 + t3D[2].z * lambda2;
• Thanks for your detailed answer! The idea of pre-generating a mesh with the links sounds good! I was awere of the barycentric approach for UV to 3D, but I could not figure out how to find to what triangle a UV point belongs... and I still don't see it! Do you have any suggestion? Could you please elaborate more on point (2) or point me to relevant links? Thanks a lot! – Dan Sep 14 '14 at 17:19
• I have added more information on point (2). Not for that! – Podgorskiy Sep 14 '14 at 19:20
• I am almost done with this! I am now doing the barycentric interpolation, but I´ve just realized that your code has some errors... if T is declared as double T[2][5], how can you then access to T[1][11] (and many other access out of range)? Thanks! – Dan Sep 26 '14 at 5:39
I am also looking for some code to make this conversion but..Only God knows why the code above is using that array indexes lol...
Maybe he meant something like this? (in case other people wants to use the code too)
PVector [] tUV = new PVector[3]; // 2d uv coordinates of the triangle vertices
PVector [] t3D = new PVector[3]; // 3d space coordinates of the triangle vertices
PVector pUV = new PVector(); // 2d uv coordinates of the given point
double [][]T = new double[2][2];
double [][]iT = new double[2][2];
T[0][0] = tUV[0].x - tUV[2].x;
T[0][1] = tUV[1].x - tUV[2].x;
T[1][0] = tUV[0].y - tUV[2].y;
T[1][1] = tUV[1].y - tUV[2].y;
double d = T[0][0] * T[1][1] - T[0][1] * T[1][0];
iT[0][0] = T[1][1] / d;
iT[0][1] = -T[0][1] / d;
iT[1][0] = -T[1][0] / d;
iT[1][1] = T[0][0] / d;
double lambda0 = iT[0][0] * (pUV.x - tUV[2].x) + iT[0][1] * (pUV.y - tUV[2].y);
double lambda1 = iT[1][0] * (pUV.x - tUV[2].x) + iT[1][1] * (pUV.y - tUV[2].y);
double lambda2 = 1.0 - lambda0 - lambda1;
double x = t3D[0].x * lambda0 + t3D[1].x * lambda1 + t3D[2].x * lambda2;
double y = t3D[0].y * lambda0 + t3D[1].y * lambda1 + t3D[2].y * lambda2;
double z = t3D[0].z * lambda0 + t3D[1].z * lambda1 + t3D[2].z * lambda2; | 2,272 | 7,391 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.671875 | 3 | CC-MAIN-2020-34 | latest | en | 0.929777 |
https://www.physicsforums.com/threads/mathematica-div-in-cylindrical-and-shadowing.472305/ | 1,621,003,976,000,000,000 | text/html | crawl-data/CC-MAIN-2021-21/segments/1620243989526.42/warc/CC-MAIN-20210514121902-20210514151902-00338.warc.gz | 1,023,355,576 | 16,171 | Mathematica: Div in Cylindrical and Shadowing
• Mathematica
Mathematica: Div in Cylindrical and "Shadowing"
I have a vector given in cylindrical coordinates. I know that the divergence of the vector should be zero. However, I am not sure why Mathematica is not returning zero. Also, the Div operator is showing up red (Div) and it is saying something about Shadowing. I have no idea what this means and the help is a little cryptic about it. I tried using Remove[Div] but it is protected. I am not sure what to do about that. This is the code i am using:
Code:
Clear["Global*"]
Needs["VectorAnalysis"]
Vr = (A/r^2 - B) Cos[theta];
Vtheta = (A/r^2 + B) Sin[theta];
Div[{Vr, Vtheta, 0}, Cylindrical[r, theta, z]]
which should give zero, but instead it gives me an algebraic expression.
For your red Div and warning search in this
http://reference.wolfram.com/mathematica/tutorial/Contexts.html
for red and read the explanation. Somehow you or perhaps VectorAnalysis has introduced a definition of Div that is going to be used before the built-in definition of Div and Mathematica is warning you about that.
Why your divergence isn't zero I haven't pursued.
For your red Div and warning search in this
http://reference.wolfram.com/mathematica/tutorial/Contexts.html
for red and read the explanation. Somehow you or perhaps VectorAnalysis has introduced a definition of Div that is going to be used before the built-in definition of Div and Mathematica is warning you about that.
Why your divergence isn't zero I haven't pursued.
Sorry, must have clicked twice, so edit this to add some additional information.
It isn't necessarily a bad thing that Div was redefined. VectorAnalysis may have broadened the definition or added new abilities. The reason for the warning when you tried to Remove[Div] is that for your safety most internal features have write/delete protection. If you really want to Remove[Div] you would have to Unprotect it first, but I would make certain that you know you need to do that and you understand the consequences. If VectorAnalysis did redefine Div then what it did was Unprotect, change the definition and then Protect it again.
Last edited:
I have a vector given in cylindrical coordinates. I know that the divergence of the vector should be zero. However, I am not sure why Mathematica is not returning zero. Also, the Div operator is showing up red (Div) and it is saying something about Shadowing. I have no idea what this means and the help is a little cryptic about it. I tried using Remove[Div] but it is protected. I am not sure what to do about that. This is the code i am using:
Code:
Clear["Global*"]
Needs["VectorAnalysis"]
Vr = (A/r^2 - B) Cos[theta];
Vtheta = (A/r^2 + B) Sin[theta];
Div[{Vr, Vtheta, 0}, Cylindrical[r, theta, z]]
which should give zero, but instead it gives me an algebraic expression.
Have you tried to simplify the result? Here's what I obtain in Mathematica 6:
Code:
In[1]:= Vr = (A/r^2 - B) Cos[theta];
Vtheta = (A/r^2 + B) Sin[theta];
divV = 1/r D[r Vr, r] + 1/r D[Vtheta, theta] // Simplify
Out[3]= 0
In[4]:= Needs["VectorAnalysis"]
Div[{Vr, Vtheta, 0}, Cylindrical[r, theta, z]]
Out[5]= ((-B + A/r^2) Cos[theta] + (B + A/r^2) Cos[theta] - (
2 A Cos[theta])/r^2)/r
In[6]:= % // Simplify
Out[6]= 0` | 859 | 3,286 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.953125 | 3 | CC-MAIN-2021-21 | longest | en | 0.9474 |
https://algebra-net.com/algebra-online/radical-equations/factoring-trinomials-online.html | 1,718,310,567,000,000,000 | text/html | crawl-data/CC-MAIN-2024-26/segments/1718198861488.82/warc/CC-MAIN-20240613190234-20240613220234-00485.warc.gz | 79,110,417 | 11,579 | Try the Free Math Solver or Scroll down to Resources!
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Solve for:
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Author Message
SolaxS
Registered: 15.09.2002
From:
Posted: Sunday 24th of Dec 08:39 I have a difficulty with my math that requires urgent solution. The difficulty is with factoring trinomials online. I have been looking out for someone who can teach me straight away as my exam is fast approaching . But it's hard to find somebody fast enough besides it being expensive . Can anybody direct me? It will be a big help.
oc_rana
Registered: 08.03.2007
From: egypt,alexandria
Posted: Monday 25th of Dec 16:58 Sounds like your bases are not strong. Excelling in factoring trinomials online requires that your concepts be concrete. I know students who actually start teaching juniors in their first year. Why don’t you try Algebra Professor? I am pretty sure, this program will help you.
fveingal
Registered: 11.07.2001
From: Earth
Posted: Tuesday 26th of Dec 12:56 I discovered a some software programs that are appropriate. I tried them out. The Algebra Professor seemed to be the most suitable one for monomials, proportions and conversion of units. It was also uncomplicated to operate . It took me step by step towards the answer rather than only giving the answer . That way I got to learn how to solve the problems too. By the time I was finished, I had learnt how to work out the problems. I found them valuable for Basic Math, Remedial Algebra and Pre Algebra which aided me in my math classes. May be, this is just the thing for you . Why not try this out?
Momepi
Registered: 22.07.2004
From: Ireland
Posted: Wednesday 27th of Dec 10:44 I would advise trying out Algebra Professor. It not only helps you with your math problems, but also provides all the required steps in detail so that you can enhance the understanding of the subject.
nsr2
Registered: 16.06.2005
From: here
Posted: Thursday 28th of Dec 07:16 It’s amazing that a program can perform that. I didn’t expect something like that could really help in math . I’m used to be tutored by a teacher but this really sounds cool. Do you have any links for this software ?
Techei-Mechial
Registered: 14.10.2001
From:
Posted: Friday 29th of Dec 08:59 Try getting more details here : https://softmath.com/news.html. I believe they are offering money back guarantee so you don’t have to worry about it. Believe me, it’s the finest math software I used. I hope you’d get the assistance you are looking for from this software . | 851 | 3,328 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.59375 | 3 | CC-MAIN-2024-26 | latest | en | 0.913432 |
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## [Daily Brain Teasers] Maths Aptitude Puzzle - 17 December
Expand Messages
• Maths Aptitude Puzzle - 17 December A octopus has 8 legs. A hippogriff has 6 legs and 2 pairs of wings. A sphinx has 6 legs and one pair of wings. Now we
Message 1 of 6 , Dec 16, 2012
Maths Aptitude Puzzle - 17 December
A octopus has 8 legs. A hippogriff has 6 legs and 2 pairs of wings. A sphinx has 6 legs and one pair of wings. Now we have all 3 kinds and a total of 18 insects in a cage. We have a total of 118 legs and 20 pairs of wings. How many insects do we have of each kind ?
Update Your Answers at : Click Here
Solution
Will be updated after 1 day
[Non-text portions of this message have been removed]
• 5,7,6
Message 2 of 6 , Dec 17, 2012
5,7,6
--- In mathforfun@yahoogroups.com, lavesh rawat <love19_foryou@...> wrote:
>
> Maths Aptitude Puzzle - 17 December
>
> A octopus has 8 legs. A hippogriff has 6 legs and 2 pairs of wings. A sphinx has 6 legs and one pair of wings. Now we have all 3 kinds and a total of 18 insects in a cage. We have a total of 118 legs and 20 pairs of wings. How many insects do we have of each kind ?
>
> Update Your Answers at : Click Here
>
> Solution
> Will be updated after 1 day
>
> [Non-text portions of this message have been removed]
>
• There are 5 octopi, 7 hippos, and 6 sphinxes David ... From: lavesh rawat To: undisclosed recipients: ; Sent: Mon, Dec 17, 2012 5:15
Message 3 of 6 , Dec 17, 2012
There are 5 octopi, 7 hippos, and 6 sphinxes
David
-----Original Message-----
From: lavesh rawat <love19_foryou@...>
To: undisclosed recipients: ;
Sent: Mon, Dec 17, 2012 5:15 pm
Subject: [MATH for FUN] [Daily Brain Teasers] Maths Aptitude Puzzle - 17 December
Maths Aptitude Puzzle - 17 December
A octopus has 8 legs. A hippogriff has 6 legs and 2 pairs of wings. A sphinx has 6 legs and one pair of wings. Now we have all 3 kinds and a total of 18 insects in a cage. We have a total of 118 legs and 20 pairs of wings. How many insects do we have of each kind ?
Update Your Answers at : Click Here
Solution
Will be updated after 1 day
[Non-text portions of this message have been removed]
[Non-text portions of this message have been removed]
• octopus-5, hippogriff-7 and sphinx-6 ... From: David Abrahams Subject: Re: [MATH for FUN] [Daily Brain Teasers] Maths Aptitude Puzzle - 17
Message 4 of 6 , Dec 17, 2012
octopus-5, hippogriff-7 and sphinx-6
--- On Mon, 17/12/12, David Abrahams <MathMadEzy@...> wrote:
From: David Abrahams <MathMadEzy@...>
Subject: Re: [MATH for FUN] [Daily Brain Teasers] Maths Aptitude Puzzle - 17 December
To: mathforfun@yahoogroups.com
Date: Monday, 17 December, 2012, 11:14 PM
There are 5 octopi, 7 hippos, and 6 sphinxes
David
-----Original Message-----
From: lavesh rawat <love19_foryou@...>
To: undisclosed recipients: ;
Sent: Mon, Dec 17, 2012 5:15 pm
Subject: [MATH for FUN] [Daily Brain Teasers] Maths Aptitude Puzzle - 17 December
Maths Aptitude Puzzle - 17 December
A octopus has 8 legs. A hippogriff has 6 legs and 2 pairs of wings. A sphinx has 6 legs and one pair of wings. Now we have all 3 kinds and a total of 18 insects in a cage. We have a total of 118 legs and 20 pairs of wings. How many insects do we have of each kind ?
Update Your Answers at : Click Here
Solution
Will be updated after 1 day
[Non-text portions of this message have been removed]
[Non-text portions of this message have been removed]
[Non-text portions of this message have been removed]
• This is not a math problem. Only sphinxes are insects, so there are 18 insects in the cage. stevo ... [Non-text portions of this message have been removed]
Message 5 of 6 , Dec 18, 2012
This is not a math problem.
Only sphinxes are insects, so there are 18 insects in the cage.
stevo
On Mon, Dec 17, 2012 at 1:24 AM, lavesh rawat <love19_foryou@...>wrote:
> **
>
>
> Maths Aptitude Puzzle - 17 December
>
> A octopus has 8 legs. A hippogriff has 6 legs and 2 pairs of wings. A
> sphinx has 6 legs and one pair of wings. Now we have all 3 kinds and a
> total of 18 insects in a cage. We have a total of 118 legs and 20 pairs of
> wings. How many insects do we have of each kind ?
>
> Update Your Answers at : Click Here
>
> Solution
> Will be updated after 1 day
>
> [Non-text portions of this message have been removed]
>
>
>
[Non-text portions of this message have been removed]
• Correction: There are 18 sphinxes in the cage. stevo ... [Non-text portions of this message have been removed]
Message 6 of 6 , Dec 18, 2012
Correction: There are 18 sphinxes in the cage.
stevo
On Tue, Dec 18, 2012 at 3:46 AM, MorphemeAddict <lytlesw@...> wrote:
> This is not a math problem.
> Only sphinxes are insects, so there are 18 insects in the cage.
>
> stevo
>
>
> On Mon, Dec 17, 2012 at 1:24 AM, lavesh rawat <love19_foryou@...>wrote:
>
>> **
>>
>>
>> Maths Aptitude Puzzle - 17 December
>>
>> A octopus has 8 legs. A hippogriff has 6 legs and 2 pairs of wings. A
>> sphinx has 6 legs and one pair of wings. Now we have all 3 kinds and a
>> total of 18 insects in a cage. We have a total of 118 legs and 20 pairs of
>> wings. How many insects do we have of each kind ?
>>
>> Update Your Answers at : Click Here
>>
>> Solution
>> Will be updated after 1 day
>>
>> [Non-text portions of this message have been removed]
>>
>>
>>
>
>
[Non-text portions of this message have been removed]
Your message has been successfully submitted and would be delivered to recipients shortly. | 1,671 | 5,497 | {"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} | 3.828125 | 4 | CC-MAIN-2017-39 | latest | en | 0.895149 |
https://www.jiskha.com/questions/1365383/Factor-3g-squared-7g-2 | 1,537,940,707,000,000,000 | text/html | crawl-data/CC-MAIN-2018-39/segments/1537267163326.85/warc/CC-MAIN-20180926041849-20180926062249-00127.warc.gz | 763,456,024 | 5,179 | Algebra
Factor 3g squared - 7g +2
1. 3g^2 - 7g + 2
(3g-?)(g-?)
The only factors of 2 are 1 and 2. So, how can you place them to get -7 in the middle?
posted by Steve
2. X power 4 +x square + 1
posted by x power 4 + X square + 1
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### gallery: yet another implementation of McCarthy's 91 function
```this time with a manually-optimized tail call
(and auto-deref applied to all references in this file)```
parent db8d11e1
... ... @@ -16,33 +16,45 @@ module McCarthy91 (** {2 traditional recursive implementation} *) let rec f91 (n:int) : int variant { 101-n } ensures { result = spec n } let rec f91 (n: int) : int ensures { result = spec n } variant { 101 - n } = if n <= 100 then f91 (f91 (n + 11)) else n - 10 (** {2 manually-optimized tail call} *) let rec f91_tco (n0: int) : int ensures { result = spec n0 } variant { 101 - n0 } = let ref n = n0 in while n <= 100 do invariant { n = n0 > 100 \/ n0 <= n <= 101 } variant { 101 - n } n <- f91_tco (n + 11) done; n - 10 (** {2 non-recursive implementation using a while loop} *) use ref.Ref use int.Iter let f91_nonrec (n0: int) ensures { result = spec n0 } = let e = ref 1 in let n = ref n0 in while !e > 0 do invariant { !e >= 0 /\ iter spec !e !n = spec n0 } variant { 101 - !n + 10 * !e, !e } if !n > 100 then begin n := !n - 10; e := !e - 1 let f91_nonrec (n0: int): int ensures { result = spec n0 } = let ref e = 1 in let ref n = n0 in while e > 0 do invariant { e >= 0 /\ iter spec e n = spec n0 } variant { 101 - n + 10 * e, e } if n > 100 then begin n <- n - 10; e <- e - 1 end else begin n := !n + 11; e := !e + 1 n <- n + 11; e <- e + 1 end done; !n n (** {2 irrelevance of control flow} ... ... @@ -54,33 +66,33 @@ module McCarthy91 exception Stop let f91_pseudorec (n0:int) : int let f91_pseudorec (n0: int) : int ensures { result = spec n0 } = let e = ref 1 in let n = ref n0 in = let ref e = 1 in let ref n = n0 in let bloc () : unit requires { !e >= 0 } ensures { !(old e) > 0 } ensures { if !(old n) > 100 then !n = !(old n) - 10 /\ !e = !(old e) - 1 else !n = !(old n) + 11 /\ !e = !(old e) + 1 } raises { Stop -> !e = !(old e) = 0 /\ !n = !(old n) } = if not (!e > 0) then raise Stop; if !n > 100 then begin n := !n - 10; e := !e - 1 requires { e >= 0 } ensures { (old e) > 0 } ensures { if (old n) > 100 then n = (old n) - 10 /\ e = (old e) - 1 else n = (old n) + 11 /\ e = (old e) + 1 } raises { Stop -> e = (old e) = 0 /\ n = (old n) } = if not (e > 0) then raise Stop; if n > 100 then begin n <- n - 10; e <- e - 1 end else begin n := !n + 11; e := !e + 1 n <- n + 11; e <- e + 1 end in let rec aux () : unit requires { !e > 0 } variant { 101 - !n } ensures { !e = !(old e) - 1 /\ !n = spec !(old n) } requires { e > 0 } variant { 101 - n } ensures { e = (old e) - 1 /\ n = spec (old n) } raises { Stop -> false } = let u = !n in bloc (); if u <= 100 then (aux (); aux ()) in = let u = n in bloc (); if u <= 100 then (aux (); aux ()) in try aux (); bloc (); absurd with Stop -> !n end with Stop -> n end end ... ... @@ -119,31 +131,31 @@ we define the small-step semantics of this code by the following [step] function *) val pc : ref int val n : ref int val e : ref int val r : ref int val ref pc : int val ref n : int val ref e : int val ref r : int val step () : unit requires { 0 <= !pc < 8 } requires { 0 <= pc < 8 } writes { pc, e, r } ensures { old !pc = 0 -> !pc = 1 /\ !e = 1 /\ !r = old !r } ensures { old !pc = 1 -> !pc = 2 /\ !e = old !e /\ !r = !n } ensures { old !pc = 2 /\ old !r > 100 -> !pc = 3 /\ !e = old !e /\ !r = old !r } ensures { old !pc = 2 /\ old !r <= 100 -> !pc = 6 /\ !e = old !e /\ !r = old !r } ensures { old !pc = 3 -> !pc = 4 /\ !e = old !e /\ !r = old !r - 10 } ensures { old !pc = 4 -> !pc = 5 /\ !e = old !e - 1 /\ !r = old !r } ensures { old !pc = 5 /\ old !e = 0 -> !pc = 8 /\ !e = old !e /\ !r = old !r } ensures { old !pc = 5 /\ old !e <> 0 -> !pc = 2 /\ !e = old !e /\ !r = old !r } ensures { old !pc = 6 -> !pc = 7 /\ !e = old !e /\ !r = old !r + 11 } ensures { old !pc = 7 -> !pc = 2 /\ !e = old !e + 1 /\ !r = old !r } ensures { old pc = 0 -> pc = 1 /\ e = 1 /\ r = old r } ensures { old pc = 1 -> pc = 2 /\ e = old e /\ r = n } ensures { old pc = 2 /\ old r > 100 -> pc = 3 /\ e = old e /\ r = old r } ensures { old pc = 2 /\ old r <= 100 -> pc = 6 /\ e = old e /\ r = old r } ensures { old pc = 3 -> pc = 4 /\ e = old e /\ r = old r - 10 } ensures { old pc = 4 -> pc = 5 /\ e = old e - 1 /\ r = old r } ensures { old pc = 5 /\ old e = 0 -> pc = 8 /\ e = old e /\ r = old r } ensures { old pc = 5 /\ old e <> 0 -> pc = 2 /\ e = old e /\ r = old r } ensures { old pc = 6 -> pc = 7 /\ e = old e /\ r = old r + 11 } ensures { old pc = 7 -> pc = 2 /\ e = old e + 1 /\ r = old r } let rec monitor () : unit requires { !pc = 2 /\ !e > 0 } variant { 101 - !r } ensures { !pc = 5 /\ !r = spec(old !r) /\ !e = old !e - 1 } requires { pc = 2 /\ e > 0 } variant { 101 - r } ensures { pc = 5 /\ r = spec(old r) /\ e = old e - 1 } = step (); (* execution of 'if r > 100' *) if !pc = 3 then begin if pc = 3 then begin step (); (* assignment r <- r - 10 *) step (); (* assignment e <- e - 1 *) end ... ... @@ -156,8 +168,8 @@ we define the small-step semantics of this code by the following [step] function end let mccarthy () requires { !pc = 0 /\ !n >= 0 } ensures { !pc = 8 /\ !r = spec !n } requires { pc = 0 /\ n >= 0 } ensures { pc = 8 /\ r = spec n } = step (); (* assignment e <- 1 *) step (); (* assignment r <- n *) monitor (); (* loop *) ... ... @@ -183,29 +195,29 @@ we define the not-so-small-step semantics of this code by the following [next] f *) val next () : unit requires { 0 <= !pc < 3 } requires { 0 <= pc < 3 } writes { pc, e, r } ensures { old !pc = 0 -> !pc = 1 /\ !e = 1 /\ !r = !n } ensures { old !pc = 1 /\ old !r > 100 -> !pc = 2 /\ !r = old !r - 10 /\ !e = old !e - 1 } ensures { old !pc = 1 /\ old !r <= 100 -> !pc = 1 /\ !r = old !r + 11 /\ !e = old !e + 1 } ensures { old !pc = 2 /\ old !e = 0 -> !pc = 3 /\ !r = old !r /\ !e = old !e } ensures { old !pc = 2 /\ old !e <> 0 -> !pc = 1 /\ !r = old !r /\ !e = old !e } ensures { old pc = 0 -> pc = 1 /\ e = 1 /\ r = n } ensures { old pc = 1 /\ old r > 100 -> pc = 2 /\ r = old r - 10 /\ e = old e - 1 } ensures { old pc = 1 /\ old r <= 100 -> pc = 1 /\ r = old r + 11 /\ e = old e + 1 } ensures { old pc = 2 /\ old e = 0 -> pc = 3 /\ r = old r /\ e = old e } ensures { old pc = 2 /\ old e <> 0 -> pc = 1 /\ r = old r /\ e = old e } (* [aux2] performs as may loop iterations as needed so as to reach program point 2 from program point 1 *) let rec monitor2 () : unit requires { !pc = 1 /\ !e > 0 } variant { 101 - !r } ensures { !pc = 2 /\ !r = spec(old !r) /\ !e = old !e - 1 } requires { pc = 1 /\ e > 0 } variant { 101 - r } ensures { pc = 2 /\ r = spec(old r) /\ e = old e - 1 } = next (); if !pc <> 2 then begin monitor2 (); next (); monitor2 () end if pc <> 2 then begin monitor2 (); next (); monitor2 () end let mccarthy2 () requires { !pc = 0 /\ !n >= 0 } ensures { !pc = 3 /\ !r = spec !n } requires { pc = 0 /\ n >= 0 } ensures { pc = 3 /\ r = spec n } = next (); (* assignments e <- 1; r <- n *) monitor2 (); (* loop *) next () ... ... @@ -233,49 +245,49 @@ module McCarthy91Mach let f91_nonrec (n0: int63) : int63 ensures { result = spec n0 } = let e = ref one in let n = ref n0 in while gt !e zero do invariant { !e >= 0 /\ iter spec !e !n = spec n0 } variant { 101 - !n + 10 * !e, !e:int } if !n > 100 then begin n := !n - 10; e := pred !e = let ref e = one in let ref n = n0 in while gt e zero do invariant { e >= 0 /\ iter spec e n = spec n0 } variant { 101 - n + 10 * e, e:int } if n > 100 then begin n <- n - 10; e <- pred e end else begin n := !n + 11; e := succ !e n <- n + 11; e <- succ e end done; !n n exception Stop let f91_pseudorec (n0: int63) : int63 ensures { result = spec n0 } = let e = ref one in let n = ref n0 in = let ref e = one in let ref n = n0 in let bloc () : unit requires { !e >= 0 } ensures { !(old e) > 0 } ensures { if !(old n) > 100 then !n = !(old n) - 10 /\ !e = !(old e) - 1 else !n = !(old n) + 11 /\ !e = !(old e) + 1 } raises { Stop -> !e = !(old e) = 0 /\ !n = !(old n) } = if not (gt !e zero) then raise Stop; if !n > 100 then begin n := !n - 10; e := pred !e requires { e >= 0 } ensures { (old e) > 0 } ensures { if (old n) > 100 then n = (old n) - 10 /\ e = (old e) - 1 else n = (old n) + 11 /\ e = (old e) + 1 } raises { Stop -> e = (old e) = 0 /\ n = (old n) } = if not (gt e zero) then raise Stop; if n > 100 then begin n := n - 10; e := pred e end else begin n := !n + 11; e := succ !e n := n + 11; e := succ e end in let rec aux () : unit requires { !e > 0 } variant { 101 - !n } ensures { !e = !(old e) - 1 /\ !n = spec !(old n) } requires { e > 0 } variant { 101 - n } ensures { e = (old e) - 1 /\ n = spec (old n) } raises { Stop -> false } = let u = !n in bloc (); if u <= 100 then (aux (); aux ()) in = let u = n in bloc (); if u <= 100 then (aux (); aux ()) in try aux (); bloc (); absurd with Stop -> !n end with Stop -> n end end
... ... @@ -4,11 +4,39 @@ ... ...
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Finish editing this message first! | 3,375 | 8,997 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.75 | 3 | CC-MAIN-2020-50 | latest | en | 0.397198 |
http://www.satphire.com/2021/soccer-betting-tips-money-management-options.htm | 1,669,927,044,000,000,000 | text/html | crawl-data/CC-MAIN-2022-49/segments/1669446710869.86/warc/CC-MAIN-20221201185801-20221201215801-00115.warc.gz | 92,754,103 | 9,117 | # Soccer Betting Tips – Money Management Options
The search for profit does not end once you have found the very best soccer betting tips. A lot is to be performed in order to ensure profit. Money management is as important as using the perfect soccer betting tips. But in the rush to get their money on people overlook this important facet of soccer betting. Let us look at it in terms that are simple: You are currently betting on two soccer matches. You know that you will create a profit 80 percent of another and the time has a chance of winning. That is money management. It is basically managing your money. So logic says that on the bets, you need to gamble cash that is less and on you want to stake cash. It is often overlooked although this may look to you. The most common technique is to use the identical amount on every choice. In the short term you need to watch out for sequences of losers in the soccer tips while this can work in the long term.
Your bank can be quickly depleted by four or five winners in a row. It could be better to search for a different strategy. Another approach suggested by many is the Kelly Criterion. Kelly requires you to be aware of the likelihood of a win. Converting the price on offer then determines the bet size. You have to estimate the probability of your wager winning. The gap between the cost chances and your probability of the sports book needs to be positive. If it is negative, you continue on to another match and should drop this soccer tip. The bet’s size is calculated with this difference. A difference and a difference would suggest a bigger investment and a little investment, respectively. Now as you can envision, the person cannot estimate the probability of his soccer prediction winning and check this out https://smsbongda.net/nhan-dinh-bong-da-du-doan.
So this procedure is of little use. Yes, the mathematicians’ do not get us wrong and professionals rave about this formulation, it is great in theory but it fails in practice. We are guessing that is you and if fails for at least for 90% of the men and women who attempt to use it and us included. Instead we Use the price available. Sports Books have analyzed the games in depth and it is not they get the prices wrong. This makes our strength that is foes their weakness. Yes, we know that upsets occur, but you will discover that if they estimate an effect at money that result will occur close to 50 percent of the time if you look over an extended period. Thus by using this as the probability of this result we can calculate how much to spend on every soccer tip. | 539 | 2,600 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.75 | 3 | CC-MAIN-2022-49 | latest | en | 0.968725 |
http://www.pointwise.com/glyph2/files/GgTclTools/cxx/GgVec2Interface-cxx.html | 1,524,735,271,000,000,000 | text/html | crawl-data/CC-MAIN-2018-17/segments/1524125948125.20/warc/CC-MAIN-20180426090041-20180426110041-00020.warc.gz | 481,109,658 | 4,855 | pwu::Vector2
Utility functions for two dimensional vectors, which are represented as a list of two real values.
Summary
pwu:: Vector2 Utility functions for two dimensional vectors, which are represented as a list of two real values. Static Actions zero Return the 2 dimensional zero vector set Return a vector with the given x and y components index Return the value of the vector at the given index x Return the x component of the vector y Return the y component of the vector equal Check if two vectors are equal notEqual Check if two vectors are not equal add Add two vectors together and return the result subtract Subtract one vector from another and return the result negate Return the negation of a vector scale Scale a vector by a single scalar value divide Scale a vector by inverse of a single scalar value multiply Return a vector that is the components of two vectors multiplied together dot Calculate the dot product of two vectors normalize Return the normalization of a vector length Return the length of the vector minimum Return a vector that has the minimum components of two vectors maximum Return a vector that has the maximum components of two vectors
zero
pwu::Vector2 zero
Return the 2 dimensional zero vector
none
Returns
a vector with the given x and y
set
pwu::Vector2 set x y
Return a vector with the given x and y components
Parameters
x the x component y the y component
Returns
a vector with the given x and y
index
pwu::Vector2 index vec i
Return the value of the vector at the given index
Parameters
vec the vector i the index to get; [0, 1]
Returns
the value of the vector at the given index
x
pwu::Vector2 x vec
Return the x component of the vector
Parameters
vec the vector
Returns
the x component of the vector
y
pwu::Vector2 y vec
Return the y component of the vector
Parameters
vec the vector
Returns
the y component of the vector
equal
pwu::Vector2 equal ?-tolerance tol? vec1 vec2
Check if two vectors are equal
Parameters
tol the tolerance to check equality, 0.0 is used if not given vec1 the first vector vec2 the second vector
Returns
true if the vectors are equal within tolerance
notEqual
pwu::Vector2 notEqual ?-tolerance tol? vec1 vec2
Check if two vectors are not equal
Parameters
tol the tolerance to check equality, 0.0 is used if not given vec1 the first vector vec2 the second vector
Returns
true if the vectors are not equal within tolerance
add
pwu::Vector2 add vec1 vec2
Add two vectors together and return the result
Parameters
vec1 the first vector vec2 the second vector
Returns
the sum of the two vectors
subtract
pwu::Vector2 subtract vec1 vec2
Subtract one vector from another and return the result
Parameters
vec1 the first vector vec2 the second vector
Returns
the difference of the two vectors
negate
pwu::Vector2 negate vec
Return the negation of a vector
Parameters
vec the vector
Returns
the negated vector
scale
pwu::Vector2 scale vec scalar
Scale a vector by a single scalar value
Parameters
vec the vector scalar a scalar value
Returns
a vector scaled by the scalar
divide
pwu::Vector2 divide vec scalar
Scale a vector by inverse of a single scalar value
Parameters
vec the vector scalar a scalar value
Returns
a vector divided by the scalar
multiply
pwu::Vector2 multiply vec1 vec2
Return a vector that is the components of two vectors multiplied together
Parameters
vec1 the first vector vec2 the second vector
Returns
the component multiply of the two vectors
dot
pwu::Vector2 dot vec1 vec2
Calculate the dot product of two vectors
Parameters
vec1 the first vector vec2 the second vector
Returns
the dot product of the two vectors
normalize
pwu::Vector2 normalize ?-length lenVar? vec
Return the normalization of a vector
Parameters
vec the vector lenVar a variable name to set the length of the vector before being normalized
Returns
the normalized vector
length
pwu::Vector2 length vec
Return the length of the vector
Parameters
vec the vector
Returns
the length of the vector
minimum
pwu::Vector2 minimum vec1 vec2
Return a vector that has the minimum components of two vectors
Parameters
vec1 the first vector vec2 the second vector
Returns
the minimum vector
maximum
pwu::Vector2 maximum vec1 vec2
Return a vector that has the maximum components of two vectors
Parameters
vec1 the first vector vec2 the second vector
Returns
the maximum vector
pwu::Vector2 zero
Return the 2 dimensional zero vector
pwu::Vector2 set x y
Return a vector with the given x and y components
pwu::Vector2 index vec i
Return the value of the vector at the given index
pwu::Vector2 x vec
Return the x component of the vector
pwu::Vector2 y vec
Return the y component of the vector
pwu::Vector2 equal ?-tolerance tol? vec1 vec2
Check if two vectors are equal
pwu::Vector2 notEqual ?-tolerance tol? vec1 vec2
Check if two vectors are not equal
pwu::Vector2 add vec1 vec2
Add two vectors together and return the result
pwu::Vector2 subtract vec1 vec2
Subtract one vector from another and return the result
pwu::Vector2 negate vec
Return the negation of a vector
pwu::Vector2 scale vec scalar
Scale a vector by a single scalar value
pwu::Vector2 divide vec scalar
Scale a vector by inverse of a single scalar value
pwu::Vector2 multiply vec1 vec2
Return a vector that is the components of two vectors multiplied together
pwu::Vector2 dot vec1 vec2
Calculate the dot product of two vectors
pwu::Vector2 normalize ?-length lenVar? vec
Return the normalization of a vector
pwu::Vector2 length vec
Return the length of the vector
pwu::Vector2 minimum vec1 vec2
Return a vector that has the minimum components of two vectors
pwu::Vector2 maximum vec1 vec2
Return a vector that has the maximum components of two vectors | 1,373 | 5,845 | {"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} | 3.109375 | 3 | CC-MAIN-2018-17 | latest | en | 0.77109 |
https://cuitandokter.com/measuring-distance-between-two-points-on-google-maps-using/ | 1,597,033,580,000,000,000 | text/html | crawl-data/CC-MAIN-2020-34/segments/1596439738609.73/warc/CC-MAIN-20200810042140-20200810072140-00508.warc.gz | 269,164,055 | 20,476 | # Measuring Distance Between Two Points On Google Maps Using
How To Use Google Maps To Measure The Distance Between 2
Measuring Distance Between Two Points On Google Maps Using
How To Find The Shortest Distance Between Two Points On
The distance matrix api is a service that provides travel distance and time for a matrix of origins and destinations. the api returns information based on the recommended route between start and. Distance between addresses to calculate distance between addresses, simply use the gps converter to convert an address to latitude and longitude, and then use this coordinates distance calculator to calculate the distance. With google maps, there are two different ways to measure distance. first, you can measure the distance between two locations using the google maps directions feature. this calculates the distance along streets. second, you can measure distance between any two points using the google maps measure distance feature. Calculate the distance of location on mobile using google map app step 1: open google maps on your android, and long tap on the location, which is going to be the starting point to measure the distance. just like last time, you can zoom in, to choose the location more precisely. To start using the new distance measuring tool, right click on any point on your map, and select "measure distance." then click on another point on the map to create a path, and you will see the.
How To Use Google Maps To Measure The Distance Between 2
How To Measure The Distance Between Two Points On Google Maps
Distances between cities may be fine for paper road maps, but today’s applications have more granular needs that can be supported by the distance matrix api. for example, the api is useful for dispatch and delivery, where you have a number of origins or destinations. Note: for zip codes, use distance between zip codes, for uk postcodes, use the uk postcode distance tool. also, to find the distance between points that are not named, you can use the measure distance tool. about how far is it between. this tool can be used to find the distance between countries, cities or towns. there are two distances output:. The above formula finds the length of a line that stretches between two points: point 1 and point 2 the linear distance is the square root of the square of the horizontal distance plus the square of the vertical distance between two points. calculate distance between two points google maps javascript. Google maps now gives users the ability to measure the distance between two or more locations simply my placing markers down on a map. the feature, available only in maps on the web for now, lets. To measure the distance on the google maps distance calculator tool. first zoom in, or enter the address of your starting point. then draw a route by clicking on the starting point, followed by all the subsequent points you want to measure.
How To Measure Distance Between Two Points In Google Maps
Open the app and try to draw a line on the map with your finger from one point to another. after doing this, you will get the results of your measurement. the app will show both the full and the shortest distance between the two points, both in miles and kilometers. there is a search engine, so you can search for any place you want to get to. Google maps distance between two points: time goes by, and google maps continues to increase its ability to offer users more features and tools it not only serves as a guide to moving from one place to another along with the means of transport, but also has become a complete helper that integrates all kinds of functions. It's really easy to use the measuring tool in google maps, and you can even calculate the distance between multiple points in. though obviously not as accurate as a route based measurement that takes roads and highways into account, it's still a great tool to measure the mileage on a larger scale such as the distance between two cities. The distance calculator module we are going to develop here is basically an api provided by the google itself. this api can measure the driving distance between two location as well as travel time. you can also measure distance between cities using this module. the format for origin location and destination location could be one of the following:. | 844 | 4,343 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.765625 | 3 | CC-MAIN-2020-34 | latest | en | 0.874427 |
http://mathforum.org/mathimages/index.php/Catenary | 1,508,547,366,000,000,000 | text/html | crawl-data/CC-MAIN-2017-43/segments/1508187824537.24/warc/CC-MAIN-20171021005202-20171021025202-00402.warc.gz | 217,317,682 | 8,726 | # Catenary
Catenary
Field: Geometry
Image Created By: Mtpaley
Website: Wikipedia
Catenary
A catenary is the curve created by a theoretical representation of a hanging chain or cable held at both ends.
# Basic Description
Notice the shape of the individual segments of the spider web. Held by vertical segments, the curves are formed due to their own weight. These curves are catenaries. Below is another example of a catenary.
The hanging chain is only holding its own weight. The curve that it creates is a catenary.
Generally, a catenary is the shape of a string hanging from two points. It approximates the shape of most string-like objects, such as ropes, chains, necklaces, and even spider webs.
## The Catenary and the Parabola Conceptually
The shape of a catenary resembles greatly the shape of a parabola. In fact, in general the visual difference between these two curves is almost imperceptible. However, there is a clear conceptual as well as mathematical difference between these two curves.
A great way to explain the conceptual difference between a parabola and a catenary is by comparing a simple suspension bridge and a suspended deck bridge.
Bridge 1: Simple Suspension Bridge
Bridge 2: Suspended Deck Bridge
The bridge labeled as Bridge 1 is the simple suspension bridge; these bridges are rarely seen anymore. The one labeled as Bridge 2 is the suspended deck bridge; most suspended bridges are designed using this model nowadays. Both bridges show a similar u-shaped curve. However, bridge 1 is holding its own weight, whereas bridge 2 is holding its weight as well as that of the horizontal deck attached. Which one is the catenary and which one is the parabola?
A catenary is limited to those curves that are free hanging; there are no other forces than gravity and its own weight acting on the curve.
If you can see this message, you do not have the Java software required to view the applet.
This link has a great applet that allows the user to overlay three different curves (a catenary, a parabola, and a chain) over the picture of a real chain and adjust these curves to see which one matches with the chain: java applet
# A More Mathematical Explanation
Note: understanding of this explanation requires: *Calculus, Algebra, Basic Dynamics
## A Catenary Mathematically
This curve is a theoretical representation of a palpable object. Thus, [...]
## A Catenary Mathematically
This curve is a theoretical representation of a palpable object. Thus, there exists a formula for it. Catenaries are the graph of the equation below:
• $y=a\cosh(\frac{x}{a})$
where
• $cosh(x)=\frac{e^x+e^{-x}}{2}$
Cosh stands for ''Hyperbolic Cosine''and it is the function that represents the catenary.The differences between catenaries arises from the scaling factor a in the first equation above, which determines the width and steepness of the catenary.
The proof of these equations is complicated; as it relates to the tension and external forces present on the hanging string or chain, it requires an understanding of Physics. To see a derivation of the catenary equation click here.
## Other Properties
• A series of catenaries attached at their endpoints form a roulette. It is created by using a straight line as the fixed line and and a parabola as the rolling curve. The focus of the parabola acts as the fixed point on the rolling curve. This creates a catenary.
• A series of inverted catenaries creates a surface that allows any polygon-shaped wheels to roll perfectly smoothly, as long as the proportions of the catenary are appropriate for the measurements of the wheels. | 792 | 3,618 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 2, "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} | 3.953125 | 4 | CC-MAIN-2017-43 | longest | en | 0.965954 |
https://kirpi4ik.dp.ua/how-to-build-your-own-diy-waterfall-pump-plumbing-waterway-basics.html | 1,712,958,149,000,000,000 | text/html | crawl-data/CC-MAIN-2024-18/segments/1712296816070.70/warc/CC-MAIN-20240412194614-20240412224614-00398.warc.gz | 317,159,583 | 37,733 | # How to Build Your Own DIY Waterfall – Pump, Plumbing & Waterway Basics
Building your own waterfall can be a rewarding experience however dealing with water can be tricky so you must arm yourself with as much information as possible in order to build something that you will be happy with.
Minimum flow rates
For every inch of width of your waterfall you should have a minimum of 100 to 150 gph (gallons per hour) of water supplied by your pump. Example waterfall: if you want the top bowl in your waterfall to have a 10 inch wide area where the water rolls over and down your waterfall, you would need a pump that can supply 1000gph. Additionally you need to understand how pumps are rated to ensure that you get the desired effect that you are looking for. There is an efficiency loss in the movement of water through pipes and you will need to account for this to make sure that your waterfall receives the water flow it needs.
The maximum rating for your pump, say 1000gph, refers to a pump with zero head pressure. Head pressure refers to the weight of the water that the pump is lifting and pushing. If the pump has to lift the water a long way it will produce less gph. Simplified, the pump will have a rating that refers to the amount of vertical climb the water will need to make. It will supply a chart in the pump specifications which states the zero head pressure gph, but also the gph to expect with one foot of vertical lift, or two feet of vertical lift, or five feet of vertical lift and so on.
Another factor in this equation is fittings such as ninety degree elbows, forty five degree elbows, and bends in the pipe or valves also will increase the head pressure. A pump that has to lift the water one foot, but also through a series of four or five 90 degree elbow fittings will have significantly less gph than a pump pushing through a straight run of pipe.
The height and width of your waterfall will determine your pump requirements
For this reason you need to determine how high your pump will need to push the water to reach your top bowl and also how wide you want the spillway in your top bowl to be. This will give you an idea of what size pump you need. Assuming the same 10 inch wide spillway and a vertical climb of five feet to reach the top bowl, you would need to find a pump that can produce a minimum of 1000gph with a head of 5 feet. This means that the label on the pump would likely be around 1250 gph referring to the zero head pressure rating for the pump with a performance of about 1000gph with five feet of head.
Calculating head pressure from pipe fittings | 564 | 2,608 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.578125 | 3 | CC-MAIN-2024-18 | longest | en | 0.96111 |
https://testbook.com/question-answer/two-men-hit-at-a-target-with-probabilities-12-and--5f1c03d8a8aaeb0d0d4fd713 | 1,638,793,712,000,000,000 | text/html | crawl-data/CC-MAIN-2021-49/segments/1637964363292.82/warc/CC-MAIN-20211206103243-20211206133243-00134.warc.gz | 610,521,900 | 32,292 | # Two men hit at a target with probabilities 1/2 and 1/3 respectively. What is the probability that exactly one of them hits the target?
This question was previously asked in
NDA (Held On: 19 Apr 2015) Maths Previous Year paper
View all NDA Papers >
1. 1/2
2. 1/3
3. 1/6
4. 2/3
Option 1 : 1/2
## Detailed Solution
Concept:
P(A̅) = 1- P(A)
Calculation:
Here, let probability of a man hitting target P(A) = 1/2 and
Probability of another man hitting target P(B) = 1/3
So, probability of a man not hitting target = P(A̅) = 1 - 1/2 = 1/2 and
Probability of another man not hitting target = P(B̅) = 1 - 1/3 = 2/3
Now, required probability = P(A)P(B̅) + P(B)P(A̅)
$$\begin{array}{l} \Rightarrow\left(\frac{1}{2}\right)\left(\frac{2}{3}\right)+\left(\frac{1}{3}\right)\left(\frac{1}{2}\right) \\ \Rightarrow \frac{2}{6}+\frac{1}{6} \\ \Rightarrow \frac{1}{2} \end{array}$$
Hence, option (1) is correct.
Free
Noun & Pronoun: Fill In The Blanks (Most Important Rules)
37464
15 Questions 15 Marks 12 Mins | 364 | 1,008 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.34375 | 4 | CC-MAIN-2021-49 | latest | en | 0.738492 |
https://www.studypool.com/discuss/7193/how-to-get-only-row-position-of-data?free | 1,521,632,885,000,000,000 | text/html | crawl-data/CC-MAIN-2018-13/segments/1521257647612.53/warc/CC-MAIN-20180321102234-20180321122234-00185.warc.gz | 873,465,110 | 16,319 | # how to get only row position of data?
Programming
account_circle Unassigned
schedule 1 Day
account_balance_wallet \$5
how to get only row position of data in a matrix and store in other array
Mar 20th, 2018
MATLAB is an abbreviation for "matrix laboratory." While other programming languages mostly work with numbers one at a time, MATLAB®is designed to operate primarily on whole matrices and arrays.
All MATLAB variables are multidimensional arrays, no matter what type of data. A matrix is a two-dimensional array often used for linear algebra.
### Array Creation
To create an array with four elements in a single row, separate the elements with either a comma (,) or a space.
`a = [1 2 3 4]`
returns
```a =
1 2 3 4```
This type of array is a row vector.
To create a matrix that has multiple rows, separate the rows with semicolons.
`a = [1 2 3; 4 5 6; 7 8 10]`
```a =
1 2 3
4 5 6
7 8 10```
Another way to create a matrix is to use a function, such as ones, zeros, or rand. For example, create a 5-by-1 column vector of zeros.
`z = zeros(5,1)`
```z =
0
0
0
0
0```
### Matrix and Array Operations
MATLAB allows you to process all of the values in a matrix using a single arithmetic operator or function.
`a + 10`
```ans =
11 12 13
14 15 16
17 18 20```
`sin(a)`
```ans =
0.8415 0.9093 0.1411
-0.7568 -0.9589 -0.2794
0.6570 0.9894 -0.5440```
To transpose a matrix, use a single quote ('):
`a'`
```ans =
1 4 7
2 5 8
3 6 10```
You can perform standard matrix multiplication, which computes the inner products between rows and columns, using the * operator. For example, confirm that a matrix times its inverse returns the identity matrix:
`p = a*inv(a)`
```p =
1.0000 0 -0.0000
0 1.0000 0
0 0 1.0000```
Notice that p is not a matrix of integer values. MATLAB stores numbers as floating-point values, and arithmetic operations are sensitive to small differences between the actual value and its floating-point representation. You can display more decimal digits using the formatcommand:
```format long
p = a*inv(a)```
```p =
1.000000000000000 0 -0.000000000000000
0 1.000000000000000 0
0 0 0.999999999999998```
Reset the display to the shorter format using
`format short`
format affects only the display of numbers, not the way MATLAB computes or saves them.
To perform element-wise multiplication rather than matrix multiplication, use the .* operator:
`p = a.*a`
```p =
1 4 9
16 25 36
49 64 100```
The matrix operators for multiplication, division, and power each have a corresponding array operator that operates element-wise. For example, raise each element of a to the third power:
`a.^3`
```ans =
1 8 27
64 125 216
343 512 1000```
### Concatenation
Concatenation is the process of joining arrays to make larger ones. In fact, you made your first array by concatenating its individual elements. The pair of square brackets [] is the concatenation operator.
`A = [a,a]`
```A =
1 2 3 1 2 3
4 5 6 4 5 6
7 8 10 7 8 10```
Concatenating arrays next to one another using commas is called horizontal concatenation. Each array must have the same number of rows. Similarly, when the arrays have the same number of columns, you can concatenate vertically using semicolons.
`A = [a; a]`
```A =
1 2 3
4 5 6
7 8 10
1 2 3
4 5 6
7 8 10```
### Complex Numbers
Complex numbers have both real and imaginary parts, where the imaginary unit is the square root of –1.
`sqrt(-1)`
```ans =
0 + 1.0000i```
To represent the imaginary part of complex numbers, use either i or j.
`c = [3+4i, 4+3j, -i, 10j]`
```c =
3.0000 + 4.0000i 4.0000 + 3.0000i 0 - 1.0000i 0 +10.0000i```
Jun 18th, 2013
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Mar 20th, 2018
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Mar 20th, 2018
Mar 21st, 2018
check_circle | 1,243 | 4,050 | {"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} | 3.6875 | 4 | CC-MAIN-2018-13 | latest | en | 0.729345 |
https://crypto.stackexchange.com/questions/101750/sha256-alternative-algorithm-to-receive-same-hash?noredirect=1 | 1,716,265,384,000,000,000 | text/html | crawl-data/CC-MAIN-2024-22/segments/1715971058383.61/warc/CC-MAIN-20240521025434-20240521055434-00846.warc.gz | 167,391,535 | 42,231 | # SHA256 - alternative algorithm to receive same hash?
My question is whether it is possible to use "another" algorithm than the sha256 to hash data but end up with the identical sha256 hash (obviously without using the sha256 algorithm)?
• Are we talking about all $m$ or just a single $m$? Are we talking a specific $m$ or looking for a collision between the outputs see here Sep 3, 2022 at 20:40
• I Like both your ideas, Forest and elderlPednt. @Forest: i get your Point. How would you evaluate If the result should be Always the First 32 Characters? Would that Be possible without going theough sha256? (and removing the Last 32 chars) @ elderlypedant: great thought. I would say No, both are Not the Same. One is obv more efficient in getting the Same result. Can this Idea Be appliwd to sha256? (make algorithm quickeer?) Sorry for typos, bumpy road Sep 4, 2022 at 9:47
Due to the pigeonhole principle, any random function will eventually produce collisions with any other random function. After all, there are far more possible inputs than there are possible outputs (for most traditional cryptographic hashes). However, there isn't any way to create an algorithm that always produces the same hash as SHA-256 for the same input without that algorithm being SHA-256.
Given two random functions $$f$$ and $$g$$ (any good cryptographic hash will adequately approximate a random function for all intents and purposes), $$\exists{m}:f(m) = g(m)$$. This means that there will exist some message $$m$$ that produces the same hash with both functions. It is also true that there can exist two different messages which, when hashed with different functions, produce the same result (i.e. $$\exists m,n: f(m) = g(n)$$, which is called a claw). However, if $$\forall{m}:f(m) = g(m)$$ (i.e. every message results in the same hash when hashed with the other function), then the functions are identical. There is no situation where that can be true but $$f$$ and $$g$$ are not the same.
• i get your Point. How would you evaluate If the result should be Always the First 32 Characters? Would that Be possible without going theough sha256? (and removing the Last 32 chars) Sep 4, 2022 at 13:40
• @user103667 Not really. I mean, you could argue that a function defined as a truncated version of SHA-256 technically isn't identical to SHA-256... Sep 5, 2022 at 11:04
• Beware that there is no way of validating that $f = g$ by trying all $m$ as that's infeasible. It is definitely possible to generate a $g$ which is virtually identical to the hash function, but returns a different value for a certain $m$. This is obvious, just set $m$ to a random value of 256 bit, then return a specific value for that $m$, otherwise return $f$. Or, for an actual example, take hardened SHA-1. Sep 6, 2022 at 7:44
In my opinion, it depends on what you mean by “another” (ie. “different”) algorithm.
Here are two algorithms. Each one takes a number as input, and produces a new number as output:
(1)
• Take input number
(2)
• Take input number | 754 | 3,032 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 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": 8, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.140625 | 3 | CC-MAIN-2024-22 | latest | en | 0.896088 |
https://number.academy/18978 | 1,656,153,984,000,000,000 | text/html | crawl-data/CC-MAIN-2022-27/segments/1656103034930.3/warc/CC-MAIN-20220625095705-20220625125705-00667.warc.gz | 478,802,059 | 12,191 | # Number 18978
Number 18,978 spell 🔊, write in words: eighteen thousand, nine hundred and seventy-eight . Ordinal number 18978th is said 🔊 and write: eighteen thousand, nine hundred and seventy-eighth. The meaning of number 18978 in Maths: Is Prime? Factorization and prime factors tree. The square root and cube root of 18978. What is 18978 in computer science, numerology, codes and images, writing and naming in other languages. Other interesting facts related to 18978.
## What is 18,978 in other units
The decimal (Arabic) number 18978 converted to a Roman number is (X)(V)MMMCMLXXVIII. Roman and decimal number conversions.
#### Weight conversion
18978 kilograms (kg) = 41838.9 pounds (lbs)
18978 pounds (lbs) = 8608.4 kilograms (kg)
#### Length conversion
18978 kilometers (km) equals to 11793 miles (mi).
18978 miles (mi) equals to 30543 kilometers (km).
18978 meters (m) equals to 62264 feet (ft).
18978 feet (ft) equals 5785 meters (m).
18978 centimeters (cm) equals to 7471.7 inches (in).
18978 inches (in) equals to 48204.1 centimeters (cm).
#### Temperature conversion
18978° Fahrenheit (°F) equals to 10525.6° Celsius (°C)
18978° Celsius (°C) equals to 34192.4° Fahrenheit (°F)
#### Time conversion
(hours, minutes, seconds, days, weeks)
18978 seconds equals to 5 hours, 16 minutes, 18 seconds
18978 minutes equals to 1 week, 6 days, 4 hours, 18 minutes
### Codes and images of the number 18978
Number 18978 morse code: .---- ---.. ----. --... ---..
Sign language for number 18978:
Number 18978 in braille:
Images of the number
Image (1) of the numberImage (2) of the number
More images, other sizes, codes and colors ...
## Mathematics of no. 18978
### Multiplications
#### Multiplication table of 18978
18978 multiplied by two equals 37956 (18978 x 2 = 37956).
18978 multiplied by three equals 56934 (18978 x 3 = 56934).
18978 multiplied by four equals 75912 (18978 x 4 = 75912).
18978 multiplied by five equals 94890 (18978 x 5 = 94890).
18978 multiplied by six equals 113868 (18978 x 6 = 113868).
18978 multiplied by seven equals 132846 (18978 x 7 = 132846).
18978 multiplied by eight equals 151824 (18978 x 8 = 151824).
18978 multiplied by nine equals 170802 (18978 x 9 = 170802).
show multiplications by 6, 7, 8, 9 ...
### Fractions: decimal fraction and common fraction
#### Fraction table of 18978
Half of 18978 is 9489 (18978 / 2 = 9489).
One third of 18978 is 6326 (18978 / 3 = 6326).
One quarter of 18978 is 4744,5 (18978 / 4 = 4744,5 = 4744 1/2).
One fifth of 18978 is 3795,6 (18978 / 5 = 3795,6 = 3795 3/5).
One sixth of 18978 is 3163 (18978 / 6 = 3163).
One seventh of 18978 is 2711,1429 (18978 / 7 = 2711,1429 = 2711 1/7).
One eighth of 18978 is 2372,25 (18978 / 8 = 2372,25 = 2372 1/4).
One ninth of 18978 is 2108,6667 (18978 / 9 = 2108,6667 = 2108 2/3).
show fractions by 6, 7, 8, 9 ...
### Calculator
18978
#### Is Prime?
The number 18978 is not a prime number. The closest prime numbers are 18973, 18979.
#### Factorization and factors (dividers)
The prime factors of 18978 are 2 * 3 * 3163
The factors of 18978 are 1 , 2 , 3 , 6 , 3163 , 6326 , 9489 , 18978
Total factors 8.
Sum of factors 37968 (18990).
#### Powers
The second power of 189782 is 360.164.484.
The third power of 189783 is 6.835.201.577.352.
#### Roots
The square root √18978 is 137,760662.
The cube root of 318978 is 26,673713.
#### Logarithms
The natural logarithm of No. ln 18978 = loge 18978 = 9,851036.
The logarithm to base 10 of No. log10 18978 = 4,27825.
The Napierian logarithm of No. log1/e 18978 = -9,851036.
### Trigonometric functions
The cosine of 18978 is -0,935466.
The sine of 18978 is 0,353416.
The tangent of 18978 is -0,377797.
### Properties of the number 18978
Is a Friedman number: No
Is a Fibonacci number: No
Is a Bell number: No
Is a palindromic number: No
Is a pentagonal number: No
Is a perfect number: No
## Number 18978 in Computer Science
Code typeCode value
18978 Number of bytes18.5KB
Unix timeUnix time 18978 is equal to Thursday Jan. 1, 1970, 5:16:18 a.m. GMT
IPv4, IPv6Number 18978 internet address in dotted format v4 0.0.74.34, v6 ::4a22
18978 Decimal = 100101000100010 Binary
18978 Decimal = 222000220 Ternary
18978 Decimal = 45042 Octal
18978 Decimal = 4A22 Hexadecimal (0x4a22 hex)
18978 BASE64MTg5Nzg=
18978 MD5cc940c545866af6a3c3d40a49f48764a
18978 SHA1c50b8487b53261c131b02d8820554387285a0854
18978 SHA2240eb0012ece834c5aea61d40bc160a3fe141d54e7d85a18d2f42a9c3a
18978 SHA256f4e9002e375945a179c0d06ab8c7a6285f68f2219f93c3a14ae2a8726bab6c0a
18978 SHA38475bfb8d004993b482bfce307aea09bf787ba6a21c2130e5b1e71763a81f1d384b89b7458076d4146973a79aff157f64f
More SHA codes related to the number 18978 ...
If you know something interesting about the 18978 number that you did not find on this page, do not hesitate to write us here.
## Numerology 18978
### Character frequency in number 18978
Character (importance) frequency for numerology.
Character: Frequency: 1 1 8 2 9 1 7 1
### Classical numerology
According to classical numerology, to know what each number means, you have to reduce it to a single figure, with the number 18978, the numbers 1+8+9+7+8 = 3+3 = 6 are added and the meaning of the number 6 is sought.
## Interesting facts about the number 18978
### Asteroids
• (18978) 2000 QH232 is asteroid number 18978. It was discovered by LINEAR, Lincoln Near-Earth Asteroid Research from Lincoln Laboratory, Socorro on 8/31/2000.
### Distances between cities
• There is a 11,793 miles (18,978 km) direct distance between Cebu City (Philippines) and Goiânia (Brazil).
• There is a 11,793 miles (18,978 km) direct distance between Curitiba (Brazil) and Taichung (Taiwan).
## Number 18,978 in other languages
How to say or write the number eighteen thousand, nine hundred and seventy-eight in Spanish, German, French and other languages. The character used as the thousands separator.
Spanish: 🔊 (número 18.978) dieciocho mil novecientos setenta y ocho German: 🔊 (Anzahl 18.978) achtzehntausendneunhundertachtundsiebzig French: 🔊 (nombre 18 978) dix-huit mille neuf cent soixante-dix-huit Portuguese: 🔊 (número 18 978) dezoito mil, novecentos e setenta e oito Chinese: 🔊 (数 18 978) 一万八千九百七十八 Arabian: 🔊 (عدد 18,978) ثمانية عشر ألفاً و تسعمائةثمانية و سبعون Czech: 🔊 (číslo 18 978) osmnáct tisíc devětset sedmdesát osm Korean: 🔊 (번호 18,978) 만 팔천구백칠십팔 Danish: 🔊 (nummer 18 978) attentusinde og nihundrede og otteoghalvfjerds Dutch: 🔊 (nummer 18 978) achttienduizendnegenhonderdachtenzeventig Japanese: 🔊 (数 18,978) 一万八千九百七十八 Indonesian: 🔊 (jumlah 18.978) delapan belas ribu sembilan ratus tujuh puluh delapan Italian: 🔊 (numero 18 978) diciottomilanovecentosettantotto Norwegian: 🔊 (nummer 18 978) atten tusen, ni hundre og sytti-åtte Polish: 🔊 (liczba 18 978) osiemnaście tysięcy dziewięćset siedemdziesiąt osiem Russian: 🔊 (номер 18 978) восемнадцать тысяч девятьсот семьдесят восемь Turkish: 🔊 (numara 18,978) onsekizbindokuzyüzyetmişsekiz Thai: 🔊 (จำนวน 18 978) หนึ่งหมื่นแปดพันเก้าร้อยเจ็ดสิบแปด Ukrainian: 🔊 (номер 18 978) вiсiмнадцять тисяч дев'ятсот сiмдесят вiсiм Vietnamese: 🔊 (con số 18.978) mười tám nghìn chín trăm bảy mươi tám Other languages ...
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## Comment
If you know something interesting about the number 18978 or any natural number (positive integer) please write us here or on facebook. | 2,545 | 7,336 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.96875 | 3 | CC-MAIN-2022-27 | latest | en | 0.634387 |
https://www.i2tutorials.com/what-are-sift-and-surf/ | 1,701,451,997,000,000,000 | text/html | crawl-data/CC-MAIN-2023-50/segments/1700679100290.24/warc/CC-MAIN-20231201151933-20231201181933-00497.warc.gz | 903,149,480 | 10,291 | / Deep Learning Interview questions and answers / What are SIFT and SURF?
## What are SIFT and SURF?
SIFT
The scale-invariant feature transform (SIFT) is an algorithm used to detect and describe local features in digital images. It locates certain key points and then furnishes them with quantitative information (so-called descriptors) which can for example be used for object recognition. The descriptors are supposed to be invariant against various transformations which might make images look different although they represent the same object(s).
A SIFT feature is a selected image region (also called keypoint) with an associated descriptor. Keypoints are extracted by the SIFT detector and their descriptors are computed by the SIFT descriptor. It is also common to use independently the SIFT detector (i.e. computing the keypoints without descriptors) or the SIFT descriptor (i.e. computing descriptors of custom keypoints).
SURF
SURF is the speed up version of SIFT. In SIFT, Lowe approximated Laplacian of Gaussian with Difference of Gaussian for finding scale-space. SURF goes a little further and approximates LoG with Box Filter. One big advantage of this approximation is that, convolution with box filter can be easily calculated with the help of integral images. And it can be done in parallel for different scales. Also, the SURF rely on determinant of Hessian matrix for both scale and location. For orientation assignment, SURF uses wavelet responses in horizontal and vertical direction for a neighborhood of size 6s. Adequate guassian weights are also applied to it. The dominant orientation is estimated by calculating the sum of all responses within a sliding orientation window of angle 60 degrees. wavelet response can be found out using integral images very easily at any scale. SURF provides such a functionality called Upright-SURF or U-SURF. It improves speed and is robust upto . OpenCV supports both, depending upon the flag, upright. If it is 0, orientation is calculated. If it is 1, orientation is not calculated and it is faster. | 441 | 2,073 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.53125 | 3 | CC-MAIN-2023-50 | latest | en | 0.878951 |
https://www.jiskha.com/questions/561508/State-whether-the-following-sequence-is-arithmetic-geometric-or-neither-10-10-25 | 1,534,347,104,000,000,000 | text/html | crawl-data/CC-MAIN-2018-34/segments/1534221210133.37/warc/CC-MAIN-20180815141842-20180815161842-00310.warc.gz | 903,915,139 | 3,950 | # math
State whether the following sequence is arithmetic, geometric, or neither.
10, 10.25, 10.50625, 10.76890625, ....
1. constant ratio of 1.025 so geometric
posted by Damon
First Name
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More Similar Questions | 785 | 2,476 | {"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} | 3.875 | 4 | CC-MAIN-2018-34 | latest | en | 0.767021 |
https://electronics.stackexchange.com/questions/408172/how-can-i-build-bjt-amplifier-with-vin-and-vout-requirements-peak-to-peak-value/408324 | 1,632,162,211,000,000,000 | text/html | crawl-data/CC-MAIN-2021-39/segments/1631780057083.64/warc/CC-MAIN-20210920161518-20210920191518-00041.warc.gz | 286,761,965 | 41,766 | # How can I build BJT amplifier with Vin and Vout requirements (peak to peak values)
I have been working on a project but I'm not sure if I'm complying with the requirements that I was given.
1. Design a BJT amplifier such that VIN=0.1 V and VOUT=5 V (peak to peak values). Justify the gain of the amplifier by performing DC and AC analyses for the design.
2. Increase the value of VIN in order to reach the saturation point in the DC load line. Justify the value obtained experimentally using the theoretical analysis of the Q-point.
Here is a screenshot of the BJT amp I have done in iCircuit. I adjusted the values until I was able to get a 5 V output but I'm not sure if my circuit meets the requirements. Can anyone help me with this project?
Here are the images of the diagram Joink Suggested. When it initial y displays a 5 V P-P value and then it switches to alternate current with a peak to peak value of 107mV
But the output volate is 5 Peak to peak from the capacitor as you suggested in my opinion. 5 P-P
Then it displays the following
• Looks pretty random. Do you have any calculations you made here? Or was it more "hunt and peck," so to speak? Can you explain any reasoning at all for the design you have above? Is the $10\:\text{V}$ supply rail required? Or is that something you can adjust per your desires for the design? Also, are you really supposed to use $1\:\text{Hz}$? And finally, do you notice your RMS and your P-P for the blue line? Do you know how those values are calculated?
– jonk
Nov 22 '18 at 6:24
• A quick calculation tells me that your BJT is saturated, so the collector isn't doing much. I also see that your purple P-P value shows as $200\:\text{mV}$ and not $100\:\text{mV}$ as your text says. So I think $100\:\text{mV}$ is the peak value, not the peak to peak value. Just another note to add to others. But no, this design won't meet any reasonable requirements I can imagine.
– jonk
Nov 22 '18 at 6:34
• Jese, I'm willing to help out. I just need to know where you are at in terms of your education here. If you have no clues at all, it's fine. It just means I start somewhere else. If you have some clues, I need to know which ones you have so I can build on those. I can try and help. But it would be nice if you could help me understand where you are at, first.
– jonk
Nov 22 '18 at 17:39
• Thank you for your answer. I am currently trying to solve this by working backwards. I am currently taking an introductory class for this project and know how to solve problems that i am given with initial conditions and resistors but dont know how to solve this particular project with Vin and Vout as my conditions. My logic tells me that i can solve for VC(2.5V) = 9V(my battery) - Rc(Ic). Im using 2.5v since the requirements say that its 5 v peak to peak values, as you suggested. Thank you very much for your help. Nov 22 '18 at 19:45
• I appreciate getting a response. Your writing doesn't suggest much of an approach, though. It's almost entirely hand-waving, so to speak, without any approach in mind. (It seems, to me, anyway.) Besides this, your schematic shows $10\:\text{V}$ but your comment says you are using a $9\:\text{V}$ battery for this. In total, I feel the whole thing is rather slippery and muddled in your mind and you don't have any starting approach. This worries me about the education you are supposedly getting (or your presence of mind when in class.)
– jonk
Nov 22 '18 at 20:17
I'll provide an approach. There are many such, not just one. But I want to write this out quickly, so I'll just plow through with some short-cuts.
1. The maximum voltage gain is about 40 times the quiescent collector current (in millamps.) You want a voltage gain (supposedly, from what I can read out of what you have written) of 50. So to be safe I'd set the quiescent collector current to $$\2.5\:\text{mA}\$$. Should be fine.
2. From this quiescent current, it is reasonable to conclude that the quiescent base-emitter voltage is about $$\700\:\text{mV}\$$.
3. I like to reserve about $$\2\:\text{V}\$$ for the minimum $$\V_\text{CE}\$$ of the BJT, in order to keep it well away from saturation.
4. I like to reserve at least $$\1\:\text{V}\$$ for the quiescent emitter voltage for a variety of reasons, but importantly because I would like to place temperature and part variation issues under management.
5. With $$\9\:\text{V}\$$ total (assuming your battery is fresh), this means there is about $$\6\:\text{V}\$$ left over for the collector. Since you need a range of only $$\5\:\text{V}\$$, this means I can (and I want to) leave about $$\1\:\text{V}\$$ margin at the top end of the collector swing. In short, I don't want the collector to move any higher than $$\8\:\text{V}\$$.
6. Therefore, the quiescent collector voltage will be $$\8\:\text{V}-2.5\:\text{V}=5.5\:\text{V}\$$.
7. From (1) and (6), I can compute a collector resistor of $$\\frac{9\:\text{V}-5.5\:\text{V}}{2.5\:\text{mA}}=1.4\:\text{k}\Omega\$$.
8. From (1) and (4), I can compute a DC emitter resistor of $$\\frac{1\:\text{V}}{2.5\:\text{mA}}=400\:\Omega\$$.
9. From (2) and (4), I know that the quiescent DC base voltage should be $$\1\:\text{V}+700\:\text{mV}=1.7\:\text{V}\$$.
10. To be conservative, I'll assume that the base current of the BJT will be no more than about $$\\frac{2.5\:\text{mA}}{\beta=100}=25\:\mu\text{A}\$$.
11. To make a "stiff" resistor divider (in the sense that it is relatively unaffected by variations in the required base current), I know that the current through the two base divider resistors should be about $$\\frac1{10}\$$th the quiescent collector current (or 10 times the current calculated in (10) above. So this means about $$\250\:\mu\text{A}\$$.
12. The divider resistor, from base to ground, is then $$\\frac{1.7\:\text{V}}{250\:\mu\text{A}}=6.8\:\text{k}\Omega\$$.
13. The divider resistor, from base to the supply rail, is then $$\\frac{9\:\text{V}-1.7\:\text{V}}{250\:\mu\text{A}+25\:\mu\text{A}}=26.545\:\text{k}\Omega\$$.
14. To get the gain, I need the total AC emitter resistance to be $$\\frac{1400\:\Omega}{50}-\frac{V_T=26\:\text{mV}}{I_Q=2.5\:\text{mA}}\approx 18\:\Omega\$$. I should probably take into account the value computed from (8) above, but it's effect is minor here. So I'll ignore it in the resulting circuit, today.
So here is the resulting design after taking into account nearby standard resistor values:
simulate this circuit – Schematic created using CircuitLab
The above should take a $$\100\:\text{mV}_\text{PP}\$$ input signal and generate a $$\5\:\text{V}_\text{PP}\$$ output signal.
Feel free to ask questions, now. But hopefully that provides one possible approach to solving your question.
## Note
This assumes audio frequencies. This means it will not work correctly for a $$\1\:\text{Hz}\$$ signal source. My recommendation is to try it with $$\1\:\text{kHz}\$$. But if you increase $$\C_e\$$ to $$\470\:\mu\text{F}\$$, then it will work okay down to perhaps a little less than $$\100\:\text{Hz}\$$.
[There's another issue with the design. It probably needs something to reduce its gain at higher frequencies. A cheap "fix" for this is a small-valued capacitor (perhaps $$\2.2\:\text{nF}\$$, for example) placed in parallel to $$\R_c\$$ (or the same small-valued capacitor used in series with $$\2.2\:\Omega\$$, where this series combo is placed in parallel to $$\R_c\$$.) That will roll off the gain at higher frequencies.]
• My word!!!! Never in the nearest dream would I have come up with that. Let me figure out if I understand the process. Thank you very much for your honest help. I will make sure to recommend you as a helping lad in this community. Thank you!!! Nov 22 '18 at 22:50
• I have posted my observations on the main text of the question, everything seems to be in order as you suggested, but let me know what you think. Nov 22 '18 at 23:39
• @JeseReyes Just a note. I'm not sure if you are really wanting to use 1 Hz. But the circuit is designed for audio. You should either use REALLY BIG capacitors if you intend on keeping the 1 Hz or else you should use 1 kHz or something like that.
– jonk
Nov 23 '18 at 0:38 | 2,284 | 8,155 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 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": 31, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.109375 | 3 | CC-MAIN-2021-39 | latest | en | 0.948043 |
https://www.delftstack.com/howto/python/python-union-of-lists/ | 1,675,714,401,000,000,000 | text/html | crawl-data/CC-MAIN-2023-06/segments/1674764500357.3/warc/CC-MAIN-20230206181343-20230206211343-00058.warc.gz | 723,999,396 | 12,018 | # Get Union of Lists in Python
Jay Prakash Gupta Oct 22, 2021 Aug 01, 2021 Python Python List
Union of lists means that all the elements from the different lists will be put up inside a single list. We can find a union of two or more than two lists. There are various methods to achieve a Union of Lists. All the methods are described below with sufficient code examples.
## Get Union of Two lists with repetition of common elements in Python
We can add two lists using the `+` operator to get the union of the two lists.
Example code:
``````l1 = [1,2,3,4,5]
l2 = [2,4,6.8,10]
l3 = l1 + l2
print("l1: ",l1)
print("l2: ",l2)
print("Union of l1 and l2 with element repetition: ",l3)
``````
Output:
``````l1: [1, 2, 3, 4, 5]
l2: [2, 4, 6.8, 10]
Union of l1 and l2 with element repetition: [1, 2, 3, 4, 5, 2, 4, 6.8, 10]
``````
It finds the union of the lists `l1` and `l2` and stores the result in `l3`. From the output, it is clear that we have repeated elements while finding union if we have the same element repeated in any of the list operands.
## Get Union of two List in sorted order in Python
If we wish to find the union of Lists in a sorted manner, we use the `sorted()` method to sort the list obtained after union operation.
Code :
``````l1 = [11,20,1,2,3,4,5]
l2 = [2,4,6,8,10]
union_l1_l2 = l1+l2
l3 = sorted(union_l1_l2)
print("l1: ",l1)
print("l2: ",l2)
print("Sorted union of two l1 and l2 : ",l3)
``````
Output:
``````l1: [11, 20, 1, 2, 3, 4, 5]
l2: [2, 4, 6, 8, 10]
Sorted union of two l1 and l2 : [1, 2, 2, 3, 4, 4, 5, 6, 8, 10, 11, 20]
``````
It computes the union of lists `l1` and `l2` using the `+` operator and stores the union in `union_l1_l2`. We use the `sorted()` method to sort the elements of the list `union_l1_l2` to get the union of `l1` and `l2` in a sorted fashion.
## Get Union of Two Lists without repetition of common elements in Python
In Python, sets are the data types that do not allow duplication of elements. So, we can use `set()` to get the union of two lists without repetition of common elements.
Code :
``````def union_without_repetition(list1,list2):
result = list(set(list1 + list2))
return result
l1 = [1,2,3,4,5]
l2 = [2,4,6,8,10]
l3= union_without_repetition(l1,l2)
print("l1: ",l1)
print("l2: ",l2)
print("Union of two l1 and l2 without repetition : ",l3)
``````
Output:
``````l1: [1, 2, 3, 4, 5]
l2: [2, 4, 6, 8, 10]
Union of two l1 and l2 without repetition : [1, 2, 3, 4, 5, 6, 8, 10]
``````
Here, we find the union of `l1` and `l2` using the `+` operator and select only unique elements using the `set()` function and then finally convert the set into a list using the `list()` function.
## Get Union of More than Two Lists in Python
We have already computed the union of the two lists. But, what to do in the case of finding a union of more than two lists. It is very simple. We can use both the `set()` and `union()` inbuilt python function to find the union of more than two lists.
Code :
``````def union(lst1, lst2, lst3):
final_list = list(set().union(lst1, lst2, lst3))
return final_list
l1 = [1,2,3,4,5]
l2 = [2,4,6,8,10]
l3 = [5,6,7,8,11,15,18]
print("l1: ",l1)
print("l2: ",l2)
print("l3 : ",l3)
print("Union of more than l1 l2 and l3: ",union(l1,l2,l3))
``````
Output:
``````l1: [1, 2, 3, 4, 5]
l2: [2, 4, 6, 8, 10]
l3 : [5, 6, 7, 8, 11, 15, 18]
Union of more than l1 l2 and l3: [1, 2, 3, 4, 5, 6, 7, 8, 10, 11, 15, 18]
``````
Here, we create a `set` object using `set()` constructor and then call `union()` method for the object. The `union()` method can take any number of `list` objects and return their union.
In this case, as we are using the `union()` method of the `set` class, we don’t get repeated elements.
## Related Article - Python List
• Convert a Dictionary to a List in Python
• Remove All the Occurrences of an Element From a List in Python
• Remove Duplicates From List in Python
• Get the Average of a List in Python
• What Is the Difference Between List Methods Append and Extend
• Convert a List to String in Python | 1,416 | 4,056 | {"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} | 3.25 | 3 | CC-MAIN-2023-06 | longest | en | 0.88091 |
http://www.newscientist.com/article/mg22029445.300-piggybacking-on-water.html | 1,432,961,455,000,000,000 | text/html | crawl-data/CC-MAIN-2015-22/segments/1432207930895.88/warc/CC-MAIN-20150521113210-00049-ip-10-180-206-219.ec2.internal.warc.gz | 618,370,925 | 13,706 | Piggybacking on water
How much electricity could be generated if you attached a turbine to the pressurised water supply coming into your home? Would this affect the water supplier or your neighbours?
Physics kudos to all our entrants. Water flow rates varied greatly, but everyone reached the same conclusion about this idea. Thanks to Steve Orchard in Gloucestershire, who points out that in the UK leaving on the tap to generate power would contravene parts of the Water Supply (Water Fittings) Regulations 1999 that are designed to avoid the waste of water – Ed
• I did a simple experiment at home to sort this out by timing how long it took to fill a bucket of known volume with water with the tap fully open. I found that my outside tap, fed directly from the rising main coming in from the street, would deliver 30 litres per minute, or 0.5 litres per second. That's the flow from one tap, but in practice the rising main can supply several taps at once, each running at full bore. I could run three taps at once before the flow rate diminished. So it looks like my rising main has a rough capacity of 1.5 litres per second. That's a mass flow rate of 1.5 kilograms per second. I know that the diameter of the main is about 13 millimetres – a standard pipe size here in the UK – so I can calculate that the water is travelling at 11 metres per second (that's the volume flow rate divided by the pipe's cross-sectional area).
Now, the water is running freely to waste, so all its energy is being converted to kinetic energy, just as it would be in a turbine. Kinetic energy is ½mv2, where m is 1.5 kg/s, and v is 11 m/s. Crunching the numbers gives an energy of about 90 joules per second – or 90 watts, if you could harvest it at 100 per cent efficiency.
But you can't make a turbine this good. The best practical turbine has an efficiency of about 66 per cent so realistically you'd get about 60 watts at the shaft of a turbine in your mains pipe. But, again, small electrical machines are notoriously inefficient, so you'd be lucky to get half that out as electricity. I could expect to harvest about 30 watts of electricity from the turbine, as long as I was content to draw no water from it for any other purpose.
Doing this would be a seriously antisocial thing to do, though. The water that comes into my house flows from a service reservoir on top of a nearby hill. It doesn't get there by itself, and my water company has to expend energy pumping it up the hill. The water is also, of course, expensively treated to make it fit to drink, and so pouring it down the drain at the rate of 1.5 litres per second – that's 130 tonnes of water per day – to generate a trivial amount of power would be an appalling waste.
I've just checked what my local water company would charge for that much water (about 47,500 tonnes per year) on a metered supply. It's about £57,600. That's a ludicrously expensive way to generate 30 watts of electricity.
Don't try this one at home, folks!
Richard Ellam, Bristol, UK
• The water supplier would certainly take a dim view of this activity, regarding it as a misuse of their water, but its effect on the overall supply would be negligible. Neither would your neighbour notice any difference, unless you share a very long, small-bore supply pipe. The reason for the disappointing amount of electricity generated is not the low pressure – many "run of river" generating plants operate on similar heads (the height of the storage reservoir above the turbine). The problem is the very low flow rate. In contrast, a relatively small 10-megawatt hydroelectric generator would have a throughput of about 60 cubic metres per second. That's a lot of bucketfuls.
Alan Brookman, Sabden, Lancashire, UK
• If the electricity generated this way is regarded as "free" energy, and if you are lucky enough not to have your water metered, there would be a great temptation to allow more water to run through your turbine. This would affect your neighbours if it helped to bring on hosepipe bans earlier during droughts, and because water suppliers would need to provide more water and their costs would be passed on to consumers.
Penny Johnson, Wotton-under-Edge, Gloucestershire, UK
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A gentle dance involving two chunks of dirty ice may explain the distinctive duck-like shape of the Rosetta probe's quarry | 1,228 | 5,470 | {"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} | 3.34375 | 3 | CC-MAIN-2015-22 | longest | en | 0.955902 |
https://forum.cogsci.nl/index.php?p=/discussion/comment/16130/ | 1,563,768,993,000,000,000 | text/html | crawl-data/CC-MAIN-2019-30/segments/1563195527474.85/warc/CC-MAIN-20190722030952-20190722052952-00076.warc.gz | 393,454,383 | 8,541 | Howdy, Stranger!
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I am interested in presenting both the unadjusted and adjusted change scores. The values are quite similar to that of the frequentist model (so i have some confidence), but I wanted to see if someone might be able to confirm that I am interpreting the posterior output correctly when using a co-varate.
design: 3 separate groups
a single change score from each group
a single covariate
Unadjusted Model [ lmBF (changescore ~ grouping, data = d)
```2. Quantiles for each variable:
2.5% 25% 50% 75% 97.5%
mu 1.44679 1.6349 1.7345 1.8364 2.0326
grouping-con1 -1.61912 -1.3539 -1.2126 -1.0691 -0.7859
grouping-con2 0.17807 0.4405 0.5768 0.7174 0.9838
grouping-con3 0.21945 0.4906 0.6316 0.7690 1.0432
sig2 2.63573 3.0324 3.2776 3.5509 4.1465
g_grouping 0.08697 0.2220 0.3978 0.7851 4.1571
```
i am making the assumption that the change score (95% credible interval) for each group is as follows
grouping con 1: 0.52 (-0.17, 1.25)
grouping con 2: 2.3 (1.61, 3.01)
grouping con 2: 2.36 (1.65, 3.07)
calculated by taking [grand mean (mu) + each value noted for each group).
i.e. 1.44+ - 1.61 (group 1 for lower bound)
1.44 + 0.17 (group 2 for lower bound)
1.44 + 0.21 (group 3 for lower bound)
assuming i did that correctly...i wanted to then determine the adjusted change score following the addition of a co-variate. It seems to check out given that the credible interval shrinks as one might expect.
Adjusted Model [ lmBF (changescore ~ grouping + covariate, data = dd)
```2. Quantiles for each variable:
2.5% 25% 50% 75% 97.5%
mu 1.44973 1.63613 1.73322 1.83431 2.03040
grouping-con1 -1.63824 -1.35344 -1.21518 -1.07134 -0.79084
grouping-con2 0.18455 0.44114 0.57468 0.71100 0.97161
grouping-con3 0.23938 0.49846 0.63718 0.77577 1.04333
covariate-covariate -0.09323 -0.05407 -0.03412 -0.01479 0.02411
sig2 2.61519 3.00498 3.24478 3.52477 4.11191
g_grouping 0.09003 0.22584 0.39645 0.78175 4.31550
g_continuous 0.01906 0.05050 0.10161 0.24308 2.58558
```
to get the adjusted value, i took the [ grand mean (mu) - the covariate] to calculate a new grand mean for the 2.5%, 50%, and 97.5% quantiles.
2.5%: 1.44 - - 0.09 = 1.53
50%: 1.73 - - 0.03 = 1.76
97.5% 2.03 - 0.02 = 2.01
using same procedure as earlier [ new grand mean + each value noted for each group] i calculated the adjusted change scores
grouping con 1: 0.55 (-0.1, 1.22)
grouping con 2 :2.33 (1.71, 2.98)
grouping con 3: 2.39 (1.76, 3.05)
e.g. adjusted lower bound calculated as
1.53 + - 1.63 = -0.1
1.53 + 0.18 = 1.71
1.53 + 0.23 = 1.76
does anyone have any idea if I am on the right track? Thanks in advance. | 1,124 | 2,898 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.84375 | 3 | CC-MAIN-2019-30 | longest | en | 0.78209 |
https://blog.tanyakhovanova.com/2020/07/the-blended-game/ | 1,721,687,382,000,000,000 | text/html | crawl-data/CC-MAIN-2024-30/segments/1720763517927.60/warc/CC-MAIN-20240722220957-20240723010957-00455.warc.gz | 122,915,521 | 14,580 | ## The Blended Game
My PRIMES STEP students invented several variations of Penney’s game. We posted a paper about these new games at math.HO arxiv:2006.13002.
In Penney’s game, Alice selects a string of coin-flip outcomes of length n. Then Bob selects another string of outcomes of the same length. For example, Alice chooses HHT, and Bob chooses THH. Then a fair coin is tossed until Alice’s or Bob’s string appears. The player whose string appears first wins. In our example, Bob has a greater probability of winning, namely, 3/4. If the first two flips are HH, then Alice wins; otherwise, Bob wins.
The reader can check that HHT beats HTT with 2 to 1 odds. Thus, the game contains a non-transitive cycle it is famous for: THH beats HHT beats HTT beats TTH beats THH.
I already wrote about the No-Flippancy game that my students invented. It starts with Alice and Bob choosing different strings of tosses of the same length.
However, in the No-Flippancy game, they don’t flip a coin but select a flip outcome deterministically according to the following rule: Let in be the maximal length of a suffix in the sequence of “flips” that coincides with a prefix of the current player’s string. The player then selects the element of their string with index i + 1 as the next “flip.” Alice goes first, and whoever’s string appears first in the sequence of choices wins.
My favorite game among the invented games is the Blended game, which mixes the No-Flippancy game and Penney’s game.
In the game, they sometimes flip a coin and sometimes don’t. Alice and Bob choose their strings as in Penney’s game and the No-Flippancy game. Before each coin flip, they decide what they want by the rule of the No-Flippancy game above. If they want the same outcome, they get it without flipping a coin. If they want different outcomes, they flip a coin. Whoever’s string appears first in the sequence of `flips’ wins.
For example, suppose Alice selects HHT, and Bob selects THH. Then Alice wants H and Bob wants T, so they flip a coin. If the flip is T, then they both want Hs, and Bob wins. If the first flip is H, they want different things again. I leave it to the reader to see that Bob wins with probability 3/4. For this particular choice of strings, the odds are the same as in Penney’s game, but they are not always the same.
This game has a lot of interesting properties. For example, similar to Penney’s game, it has a non-transitive cycle of choices. Surprisingly, the cycle is of length 6: THH beats HHT beats THT beats HTT beats TTH beast HTH beat THH.
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1. #### Cristóbal Camarero:
I think you have copied the 6-cycle in reverse. You have actually said in the preceding paragraph that Bob with THH beats HHT thrice each four games, right?
2. #### tanyakh:
Thank you, Cristobal,
You are right, I fixed it.
3. #### Divicius:
Hello, big fan of your blog from France here.
I teach 11 to 15 years old and this post make me think of something I encountered this year.
My question was : Alice wants to choose randomly an integer from 1 to 10 (both included). She only has one coin. Write a way for her to do it.
It was the first time I asked this question in years of teaching. Few found a valid answer and it was always a variation of four toss and ignore 6 possibilities.
I asked myself, is there a more elegant way but didn’t think of one.
If someone does, I’m going to elaborate on that question this year with various dices.
4. #### tanyakh:
Divicius, You might be able to prove that if you want to guarantee a finite number of throws, the probabilities should be p/q, where q is a power of 2. If you need 1/10, the number of tosses is not guaranteed to be finite.
5. #### Yogendra Singh:
I read your blog and i think you have copied the 6-cycle in reversible order. You said in that paragraph Bob with THH beats HHT thrice every four games, ok? | 945 | 3,867 | {"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} | 4.125 | 4 | CC-MAIN-2024-30 | latest | en | 0.928339 |
https://de.scribd.com/document/167596128/Spur-Gears | 1,561,054,632,000,000,000 | text/html | crawl-data/CC-MAIN-2019-26/segments/1560627999263.6/warc/CC-MAIN-20190620165805-20190620191805-00087.warc.gz | 401,797,328 | 65,224 | You are on page 1of 13
# Spur gears Spur gears have teeth that are straight and arranged parallel to the axis of the
shaft that carries the gear. The curved shape of the faces of the spur gear teeth has a special geometry called an involute curve. This shape makes it possible for two gears to operate together with smooth, positive transmission of power. The shafts carrying gears are parallel. Spur gear design Actual output speed (gear)
nG =
n P = rotational speed of the pinion
np VR
VR = gear ratio
N VR = G NP
N G , N P = number of gear, pinion teeth.
The spreadsheet computes the approximate number of gear teeth to produce the desired speed from N G = N P
nGd nP
## ( nGd = desired output speed). But, of
course, the number of teeth on any gear must be an integer, and the actual value of NG is selected by the designer. Spur gear geometry For full depth involute teeth in the diametral pitch system Pitch diameter
D = N Pd
N D
Diametral Pitch
Pd =
Outside diameter
Do =
N +2 Pd
1
a = Pd
Dedendum if Pd < 20
b =
1.25
Pd
+ 0.002
if Pd 20
b =
1.2
Pd
Clearance
if Pd < 20
c =
0.25
Pd
+ 0.002
if Pd 20
c =
0.2
Pd
Root diameter
DR = D 2b
## Base circle diameter
Db = D cos
Circular pitch
p =
Whole depth
ht = a + b
Working depth
hk = 2a
Tooth thickness
t =
2Pd
Center distance
C =
DG + DP
2
Bending geometry factor, J, is dependent on the number of teeth of gear for which geometry factor is desired and on the number of teeth in mating gear. Values can be found from AGMA 908-B89(R1995). Pitting geometry factor, I, is dependent on the tooth geometry and on gear ratio. Values can be found from AGMA Standard 218.01. Force and speed factors Pitch line speed
Vt =
Tangential force
DP nP 12
Wt =
or
Wt =
33000 (P )
Vt
126000 (P )
nD
Wr = Wt tan
Normal force
Wn =
cos
Wt
## Expected bending stress
St =
Wt Pd K o K s K m K B Kv F J
where: = bending geometry factor J = overload factor Ko = size factor Ks K m = load-distribution factor = rim thickness factor KB = dynamic factor. Kv Expected bending stress formula for SI units is given by Where; = bending stress [MPa], b m = module [mm], F = face width [mm], Wt = transmitted load [N] b = (Wt / F.m.J) (KO.KS.Km.KB.KV)
The AGMA indicates that the size factor KS can be taken to be 1.00 for most gears. But for gears with large-size teeth or large face widths, a value greater than 1.00 recommended. Gear design programs compute the size factor automatically. The determination of load-distribution factor Km is based on many variables in the design of the gears themselves as well as in the shafts, bearings, housings, and the structure in which the gear drive is installed. Therefore, it is one of the most difficult factors to specify. Much analytical and experimental work is continuing on the determination of values for K m . We will use the following equation for computing the value of the loaddistribution factor:
K m = 1.0 + C pf + C ma
where:
C pf
C ma
## = mesh alignment factor.
The dynamic factor, Kv , accounts for the fact that the load is assumed by a tooth with some degree of impact and that the actual load subjected
to the tooth is higher than the transmitted load alone. The value of Kv depends on the accuracy of tooth profile, the elastic properties of tooth, and the speed with which the teeth come into contact. AGMA Standard 2001-C95 gives recommended values for Kv based on the AGMA quality number, Qv , and the pitch line velocity. Gears in typical machine design would have AGMA quality ratings of 5 through 7, which are for gears made by hobbing or shaping with average to good tooling. If the teeth are finish-ground or shaved to improve the accuracy of the tooth profile and spacing, quality numbers in the 8 - 11 range should be used. Under very special conditions where teeth of high precision are used in applications where there is little chance of developing external dynamic loads, higher quality numbers can be used. If the teeth are cut by form milling, factors lower than those found from QV = 5 should be used. Note that the quality 5 gears should not be used at pitch line speed above 12.7 m/s (2500 ft/min). Note that the dynamic factors are approximate.
## Expected contact stress
Sc = C p
where:
Wt K o K s K m Kv FD p I
C P = elastic coefficient that depends on the material of both the pinion and
the gear.
C P = 2300 for two steel gears. The design program automatically selects the
appropriate value after the user specifies the materials. Procedure for selecting materials for bending stress
K R (SF ) S t < S at YN
where:
KR
SF
YN
## = reliability factor = factor of safety = stress cycle factor for bending.
AGMA Standard 2001-C95 allows the determination of the life adjustment factor, Y N , if the teeth of the gear being analyzed are expected to experience a number of cycles of loading much different from 10 7 . Note that the general type of material is a factor for the lower number of cycles. For the higher number of cycles, a range is indicated by a shaded area. Expected number of cycles of loading
## N c = (60)(L )(n )(q )
where: L = design life in hours n = rotational speed in rpm q = number of load applications per revolution. Procedure for selecting materials for contact stress K R (SF ) S c < S ac ZN where:
ZN
## = pitting resistance stress cycle factor.
AGMA Standard 2001-C95 specifies the determination of the stress cycle factor, Z N . If the teeth of the gear being analyzed are expected to experience a number of cycles of loading much different from 10 7 , a factor should be used. The user specifies the desired life for the system in hours and the program computes the values for Y N and Z N . After computing the values for allowable bending stress number, S at , and for allowable contact stress number, S ac , you should go to the data in AGMA Standard 2001-C95, to select a suitable material. Consider first whether the material should be steel, cast iron, bronze, or plastic. Then consult the related tables of data.
Pressure angle The pressure angle is the angle between the tangent to the pitch circles and the line drawn normal (perpendicular) to the surface of the gear tooth The line normal to the gear teeth is sometimes referred to as the line of action. When two gear teeth are in mesh and are transmitting power, the force transferred from the driver to the driven gear tooth acts in a direction along the line of action. Also, the actual shape of the gear tooth depends on the pressure angle. Standard values of the pressure angle are established by gear manufacturers, and the pressure angle of two gears in mesh must be the same. Current standard pressure angles are, although the 14
1 2
## tooth form is considered to be
obsolete. Although it is still available, it should be avoided for new designs. The 20 degrees tooth form is the most readily available at this time. The advantages and disadvantages of the different values of pressure angle relate to the strength of the teeth, the occurrence of interference, and the magnitude of forces exerted on the shaft.
Face width The face width can be specified once the diametral pitch is chosen. Although a wide range of face widths is possible, the following limits are used for general machine drive gears:
Pd
For SI System of Units,
<F <
16
Pd
12
Nominal value of F =
Pd
3 m< F < 5 m Nominal value of F = 4 m Also, the face width normally is not greater than pitch diameter of the pinion. An upper limit is placed on the face width to minimize problems with alignment. A very wide face width increases the chance for less than full face loading of the teeth. When the face width is less than the lower limit, it is probable that a more compact design can be achieved with different pitch. Notice that
F < 2.00 is recommended. Dp
Number of pinion teeth In the selection of the number of teeth on the pinion, the designer must be aware of potential interference. For certain combinations of numbers of teeth in a gear pair, there is interference between the tip of the teeth on the pinion and the fillet or root of the teeth on the gear. Obviously this cannot be tolerated because the gears simply will not mesh. The probability that interference will occur is greatest when a small pinion drives a large gear, with the worst case being a small pinion driving a rack. A rack is a gear with a straight pitch line; it can be thought of as a gear with an infinite pitch diameter. It is the designers responsibility to ensure that interference does not occur in a given application. The surest way to do this is to control the minimum number of teeth in the pinion.
If a designer wants to be sure that there will not be interference between any two gears when using the 14
1 , 2
## full-depth, involute system, the pinion
of the gear pair must have no fewer than 32 teeth. For the 20 , full-depth, involute system, using no fewer than 18 teeth will ensure that no interference occurs. For the 25 , full-depth, involute system, using no fewer than 12 teeth will ensure that no interference occurs.
Design life Design life is, indeed, a design decision based on the application. As a guideline, this set of data was created for use in bearing design and is used here for gearing: Application Domestic appliances Aircraft engines Automotive Agricultural equipment Elevators, industrial fans, multipurpose gearing Electric motors, industrial blowers, general industrial machines Pumps and compressors Critical equipment in continuous 24-h operation Design life L (h) from 1000 to 2000 from 1000 to 4000 from 1500 to 5000 from 3000 to 6000 from 8000 to 15,000 from 20,000 to 30,000 from 40,000 to 60,000 from 100,000 to 200,000
Number of load applications per revolution, q The normal number of load applications per revolution for any given tooth is typically, of course, one. But consider the case of an idler gear that serves as both a driven and driving gear in a gear train. It receives two cycles of load per revolution: one as it receives power from and one as it delivers power to its mating gears. Also, in certain types of gear trains, one gear may deliver power to two or more gears mating with it. Gears in a planetary gear train often have this characteristic.
Rim thickness The rim thickness factor, K B , accounts for a rim that may be too thin. The basic analysis used to develop the Lewis equation assumes that the gear tooth behaves as a cantilever attached to a perfectly rigid support structure at its base. If the rim of the gear is too thin, it can deform and cause the point of maximum stress to shift from the area of the gear-tooth fillet to a point within the rim. The key geometry parameter is called the backup ratio, mB , where
t mB = R ht
t R = rim thickness;
ht = whole depth of the gear tooth.
For mB >1.2, the rim is sufficiently strong and stiff to support the tooth, and K B =1.0. For mB <1.2, rim thickness factor determined:
K B = 1.6 ln
thickness. The resulting value is
2.242 m B
When a solid gear blank is used, input a large value (say tR > 1.0 inch) for rim
KB = 1.
10 | 2,660 | 11,092 | {"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} | 3.265625 | 3 | CC-MAIN-2019-26 | latest | en | 0.841317 |
http://lambda-the-ultimate.org/node/1337 | 1,653,448,822,000,000,000 | text/html | crawl-data/CC-MAIN-2022-21/segments/1652662578939.73/warc/CC-MAIN-20220525023952-20220525053952-00220.warc.gz | 32,715,852 | 6,801 | Towards Applicative Relational Programming
This 10-page 1992 article, Towards Applicative Relational Programming by Ibrahim & van Enden, has just appeared on ArXiv, which asked the question of how to combine functional and relational programming.
This is fairly well-trodden ground now, with approaches such as Miller's lambda-Prolog and Saraswat's constraint-lambda calculus being well establsihed, but this paper offers a rather different approach, based on the Henkin-Monk-Tarski approach of cylindric algebras, which were devised as a means of formalising predicate logic in equational logic - I'm familiar with them in the context of modal logic, since they offer a means for handling quantifiers by means of modalities, where [] is used to express forall and <> is used to express exists.
The treatment is nice, and is recommended to LtUers interested in having an arsenal of techniques for bridging the declarative divide.
Comment viewing options
Your link gives me "Pivotal and Pivotal-discriminative Consequence Relations". This link gives me "Towards Applicative Relational Programming".
Thanks!
Higher Order (in the functional programming sense)
Is it really true that the declarative approach precludes higher order programming. Here I'm using the term higher order in the sense of first class functions/relations rather than the quantificational interpretation.
Hilog seems to work fairly well and LambdaProlog also has facilities for higher order programming. What exactly are the limitations they are refering to in this paper.
I'm not sure what they mean
I'm not sure what they mean in this paper, but I think the problem (or should say "trade-off") with higher-order logic (and higher-order functions for that matter) is that they result in extra layers of control at run-time, creating something that is like the control flow we are trying to avoid with declarative programming in the first place. The main advantage of declarative programming, as I see it, is that the text of the program is a pretty close description of how the program executes at run-time. When higher-order features are used, this is not so much the case, and I'm back to simulating the program in my head again to understand it (e.g., to resolve how variables are bound to predicates or functions).
Extending a declarative (logic) programming language with higher-order predicates (or variables that can be unified with predicates) is tempting because of its expressiveness, but such languages really require a lot of clever programming to use. I'm speaking from experience, I tried this before in an LP I designed/implemented, and it was very expressive, but ultimately too difficult for anyone to use but myself.
I am not sure what you mean
I think the problem (or should say "trade-off") with higher-order logic (and higher-order functions for that matter) is that they result in extra layers of control at run-time
Do you mean that, for example, using map/fold/unfold is less declarative than the first-order solution? Or did I completely misunderstand your statement?
Yes, using map/fold/unfold
Yes, using map/fold/unfold is less declarative than a first-order (if one is available) solution without control flow. The problem is that the higher-order functions themselves introduce complexity because they are being called somewhere, so there is control flow.
Here is an example of map/filter using Scala syntax without using a for comprehension:
messages.filter(msg => msg.from == myBoss).map(msg => msg.subject);
Now using a for comprehension to get rid of the higher-order functions:
for (val msg <- messages; msg.from == myBoss) yield msg.subject;
In the first example, I have to think about what the higher-order functions are doing; i.e., I have to think about the control flow. The second example, although compiled into the first example, I don't have to think about this control flow so much. This is just a simple example, things really get uglier when the higher-order functions are more complicated.
In a declarative (not functional) language its even worse, because higher-order functions just don't event fit very well into the language. Higher-order predicates are just weird if you have to program in them (personal opinion, not trying to start a flame war).
What about list comprehensions?
What about list comprehensions?
[subject msg | msg <- messages, from msg == myBoss]
This is pretty much the same as
filter (\msg -> from msg == myBoss) messages.
(Pointless version would be filter ((myBoss ==) . from) messages)
Or is that more functional than declarative?
declarative
Looks declarative to me, for comprehensions are list comprehensions, and are only named differently because (1) they can operate over more than just lists and (2) the use the "for" keyword :) Anyways, an explicit lambda isn't needed, but they surely occur behind the scenes.
Higher-order Logic Programming in Prolog
Just as an aside, I found Higher-order Logic Programming in Prolog to be an interesting read. I liked their example #6...
map(plus(X), [2, A, 4], [3, 4, B]).
Higher-order unification is undecidable
If you're building a logic programming language, you'll probably want to use unification to propagate constraint information, However, if the terms that you're unifying together have binding structure (ie, are lambda terms), then the unification problem becomes undecidable, and even when the solutions do exist, there may not be most general unifiers. (This was Huet's thesis work back in the 70s.)
The "pattern fragment", discovered by Dale Miller, is a very interesting subset of higher-order unification problems that are decidable, and IIUC systems like Twelf and LambdaProlog try to restrict themselves to that. | 1,219 | 5,753 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.703125 | 3 | CC-MAIN-2022-21 | latest | en | 0.953978 |
http://www.komal.hu/verseny/feladat.cgi?a=feladat&f=C900&l=en | 1,480,941,163,000,000,000 | text/html | crawl-data/CC-MAIN-2016-50/segments/1480698541696.67/warc/CC-MAIN-20161202170901-00320-ip-10-31-129-80.ec2.internal.warc.gz | 551,426,955 | 5,362 | Order KöMaL! Competitions Portal
C. 900. 75% of a certain three-digit number of different digits consists of the same digits as the original number but with no digit in the same place. Which number is it?
(5 points)
Deadline expired on 15 June 2007.
Google Translation (Sorry, the solution is published in Hungarian only.)
Megoldás. Lehet 3.abc=4.bca vagy 3.abc=4.cab. Vagyis 296a=370b+37c vagy 397c=260b+26c. Az első esetben 8a=10b+c. A második eset nem jöhet létre, mert 397c csak úgy lehetne 13-mal osztható, ha c is osztható 13-mal.
Az a lehetséges 10 értékét végignézve nyolc háromjegyű megoldást kapunk: 216, 324, 432, 540, 648, 756, 864, 972.
Statistics on problem C. 900.
172 students sent a solution. 5 points: 120 students. 4 points: 21 students. 3 points: 4 students. 2 points: 7 students. 1 point: 3 students. Unfair, not evaluated: 17 solutions.
• Problems in Mathematics of KöMaL, May 2007
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Algebra -> Algebra -> Sequences-and-series -> SOLUTION: Can you help me find the 100th term in the geometric sequence.. 3,6,12,24,48... Log On
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Algebra: Sequences of numbers, series and how to sum them Solvers Lessons Answers archive Quiz In Depth
Question 295555: Can you help me find the 100th term in the geometric sequence.. 3,6,12,24,48...Answer by Edwin McCravy(8879) (Show Source): You can put this solution on YOUR website!3,6,12,24,48... ```That is a geometric sequence beginning with first term , because to get each succeeding term, we multiply the preceding term by 2. Substitute , , and Edwin``` | 235 | 896 | {"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} | 3.140625 | 3 | CC-MAIN-2013-20 | latest | en | 0.850818 |
https://mmurraydix.com/how-to-yrz/what-is-stochastic-dynamic-programming-0b4495 | 1,627,533,455,000,000,000 | text/html | crawl-data/CC-MAIN-2021-31/segments/1627046153816.3/warc/CC-MAIN-20210729043158-20210729073158-00685.warc.gz | 403,750,359 | 12,016 | View it as \Mathematical Programming with random parameters" Je Linderoth (UW-Madison) Stochastic Programming Modeling Lecture Notes 14 / 77. Neal Cristian S. Perlas Probabilistic Dynamic Programming (Stochastic Dynamic Programming) What does Stochastic means? When demands have finite discrete distribution functions, we show that the problem can be Stochastic Programming . 1978), (Blurton Jones 1986) proposing that !Kung women and their reproductive systems may be maximizing reproductive success. I am working through the basic examples of the stochastic RBC models in the book by McCandless (2008): The ABCs of RBCs, pp. PROBABILISTIC DYNAMIC. Download PDF Abstract: This paper aims to explore the relationship between maximum principle and dynamic programming principle for stochastic recursive control problem with random coefficients. Gain an in depth understanding of the workings of commercial asset valuation tools. SDP abbreviation stands for Stochastic Dynamic Programming. stochastic dynamic programming (SDP)—has been used to solve puzzles in the biol- ogy of organisms, particularly those about behavior and development (growth and sexual maturity leading to reproduction) at the level of the individual organism. Perhaps you are familiar with Dynamic Programming (DP) as an algorithm for solving the (stochastic) shortest path problem. We define the states s and the actions a to be elements of the state space S ( s ∈ S ) and the action space A ( s ) ( a ∈ A ( s )). **Dynamic Programming Tutorial**This is a quick introduction to dynamic programming and how to use it. In what follows next, I assume that the domain of the variables and the range of the functions all belong to $\mathcal{R}_0^+$ and I assume there are no corner solutions. It turns out that the optimal policy has an intuitive structure, which makes it easy to implement. This paper develops sampling stochastic dynamic programming (SSDP), a technique that captures the complex temporal and spatial structure of the streamflow process by using a large number of sample streamflow sequences. The proposed methodology is applicable to constrained stochastic systems with quadratic objective functions and linear dynamics. (2002) review the research devoted to proving that a hierarchy based on the frequencies of occurrence of different types of events in the systems results in Learn how Stochastic Dual DP can improve solve times by a factor of ten or more. Approximate Dynamic Programming: Solving the Curses of Dimensionality; Introduction to Stochastic Dynamic Programming. Stochastic programs are mathematical programs where some of the data incorporated into the objective or constraints is uncertain. It is having a random probability distribution or pattern that may be analyzed statistically but may not be predicted precisely. Learn how to use Stochastic Dynamic Programming to model energy sector assets. The Stochastic Programming Society (SPS) is a world-wide group of researchers who are developing models, methods, and theory for decisions under uncertainty. Up to 99.8% of the search tree is pruned by a branch-and-bound technique with bounds generated by dynamic programming. More recently, Levhari and Srinivasan [4] have also treated the Phelps problem for T = oo by means of the Bellman functional equations of dynamic programming, and have indicated a proof that concavity of U is sufficient for a maximum. 71 - 75. However, an answer such as this perpetuates fundamental misconceptions about stochastic programming and dynamic programming. Sethi et al. Improve your understanding of the applications and limitations of energy sector models. The stochastic dynamic programming approach allows the construction of a "whole-life" … Dynamic Inventory Models and Stochastic Programming* Abstract: A wide class of single-product, dynamic inventory problems with convex cost functions and a finite horizon is investigated as a stochastic programming problem. One of the biggest challenges is the lack of a widely accepted modeling framework of the type that has defined the field of determin-istic math programming. Besides the mentioned advantages, this method suffers drawbacks like infeasibility. One of the most important goals in marketing is to realize the highest … STOCHASTIC CONTROL AND DYNAMIC PROGRAMMING 2.3 DYNAMIC PROGRAMMING EQUATION FOR A rc(t)-DRIVEN PROCESS The Brownian motion process W(t) corresponds to a continuum of changes and its DPE is a second-order partial differential equation. Stochastic Model Predictive Control • stochastic finite horizon control • stochastic dynamic programming • certainty equivalent model predictive control Prof. S. Boyd, EE364b, Stanford University Here is a formulation of a basic stochastic dynamic programming model: $$y_t = A^t f(k_t)$$ Stochastic dynamic programming A standard SDP technique for solving a MDP numerically is the value iteration algorithm. Under certain regular conditions for the coefficients, the relationship between the Hamilton system with random coefficients and stochastic Hamilton-Jacobi-Bellman equation is obtained. We present a stochastic dynamic programming formulation of this problem and identify struc-tural properties that characterize its optimal policy. The book begins with a chapter on various finite-stage models, illustrating the wide range of applications of stochastic dynamic programming. Uncertainty is involved Given input results to different outputs Uses backward recursion or … (Bellman 1957), stochastic dynamic programming is a technique for modelling and solving problems of decision making under uncertainty.Closely related to stochastic programming and dynamic programming, stochastic dynamic programming represents the problem under scrutiny in the form of a … The book begins with a chapter on various finite-stage models, illustrating the wide range of applications of stochastic dynamic programming. The goal of this paper is to analyze convergence properties of the Stochastic Dual Dynamic Programming (SDDP) approach to solve linear multistage stochastic programming problems of the form (1.1) Min A 1 x 1 = b 1 x 1 ⩾ 0 c 1 T x 1 + E min B 2 x 1 + A 2 x 2 = b 2 x 2 ⩾ 0 c 2 T x 2 + E ⋯ + E min B T x T-1 + A T x T = b T x T ⩾ 0 c T T x T. The best inflow forecast can be included as a hydrologic state variable to improve the reservoir operating policy. Stochastic programming: decision x Dynamic programming: action a Optimal control: control u Typical shape di ers (provided by di erent applications): Decision x is usually high-dimensional vector Action a refers to discrete (or discretized) actions Control u is … As a hint to where this discussion is going, by the end of this tutorial I will have made the following points: Dynamic programming is a sequential (and for our purposes, stochastic) decision problem. Fuzzy stochastic dynamic programming for marketing decision support Fuzzy stochastic dynamic programming for marketing decision support Weber, Klaus; Sun, Zhaohao 2000-08-01 00:00:00 I. Handling non-linear, non-convex and non-differentiable objective functions and constraints are some advantages of SDP. In a series of simulation experiments, we Multistage stochastic programming Dynamic Programming Practical aspectsDiscussion Idea behind dynamic programming If noises aretime independent, then 1 Thecost to goat time t depends only upon the current state. Here is a formulation of a basic stochastic dynamic programming model: y_t … Introduction to Stochastic Dynamic Programming presents the basic theory and examines the scope of applications of stochastic dynamic programming. Today we discuss the principle of optimality, an important property that is required for a problem to be considered eligible for dynamic programming solutions. Uncertainty is usually characterized by a probability distribution on the parameters. 2 We can computerecursivelythe cost to go for each position, But it turns out that DP is much more than that. A Standard Stochastic Dynamic Programming Problem. the stochastic form that he cites Martin Beck-mann as having analyzed.) stochastic: 1) Generally, stochastic (pronounced stow-KAS-tik , from the Greek stochastikos , or "skilled at aiming," since stochos is a target) describes an approach to anything that is based on probability. INTRODUCTION This paper is related to marketing and more particular to the process of acquiring customers. What does SDP stand for? This is a concise and elegant introduction to stochastic dynamic programming. In this paper, the medical equipment replacement strategy is optimised using a multistage stochastic dynamic programming (SDP) approach. Dynamic programming. Introduction to Stochastic Dynamic Programming presents the basic theory and examines the scope of applications of stochastic dynamic programming. The syllabus gives a list of course materials used for the class. Stochastic dynamic programming is a control problem : the element to be optimized is a function. A stochastic dynamic programming model is presented that supports and extends work on the reproductive performance of the !Kung Bushmen (Lee 1972), (Blurton Jones et al. Stochastic dynamic programming is based on the following principle : Take the decision at time step t such that the sum ”cost at time step t due to your decision” plus ”expected cost from time steps t+1to What is the abbreviation for Stochastic Dynamic Programming? A Standard Stochastic Dynamic Programming Problem. Stochastic programming, dynamic programming, and sto-chastic search can all be viewed in a unified framework if pre-sented using common terminology and notation. Stochastic Programming is about decision making under uncertainty. PROGRAMMING. In this work, we introduce a hybrid approach that exploits tree search to compute optimal replenishment cycles, and stochastic dynamic programming to compute (s, S) levels for a given cycle. Introduction to SP Background Stochastic Programming \$64 Question The syllabus and selected lecture slides are available for download in pdf format. Stochastic Dynamic Programming (SDP) is a major method for optimizing reservoir operation. stochastic problems • Mathematically, for stochastic problems, we cannot restrict ourselves to open-loop sequences, so the shortest path viewpoint fails • Conceptually, in the presence of uncertainty, the concept of “optimal-cost-to-arrive” at a state x. k. does not make sense. for stochastic tasks, based on Markov decision processes and dynamic programming. It uses the decomposition principle of dynamic programming without discretizing the state or control variable and therefore the method can be used for large‐scale systems. 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A control problem: the element to be optimized is a concise and elegant introduction to dynamic! Programming: Solving the ( stochastic dynamic Programming to model energy sector assets ( Blurton 1986! Some advantages of SDP Programming and dynamic Programming ( SDP ) is major! And elegant introduction to stochastic dynamic Programming gives a list of course materials used for the,. Probabilistic dynamic Programming ) What does stochastic means with dynamic Programming how to use stochastic dynamic Programming model. Where some of the workings of commercial asset valuation tools to marketing more! Pruned by a factor of ten or more to use stochastic dynamic Programming formulation of this problem and struc-tural. Be optimized is a function Programming formulation of this problem and identify properties... | 4,047 | 21,823 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 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": 1, "x-ck12": 0, "texerror": 0} | 2.640625 | 3 | CC-MAIN-2021-31 | latest | en | 0.874999 |
http://forum.allaboutcircuits.com/threads/power-from-a-ppe-9v-battery.9990/ | 1,481,230,009,000,000,000 | text/html | crawl-data/CC-MAIN-2016-50/segments/1480698542655.88/warc/CC-MAIN-20161202170902-00392-ip-10-31-129-80.ec2.internal.warc.gz | 105,967,274 | 14,671 | # Power from a PPE 9v battery
Discussion in 'The Projects Forum' started by stevy123, Mar 9, 2008.
1. ### stevy123 Thread Starter Active Member
Nov 19, 2007
61
0
Hi All
I have a circuit which i am powering from a 9V PPE battery. The circuit is drawing 300mA, how long should i expect the battery to last?
Anyone know the formula to work this out?
How much longer would i get if i was using 8 AA 1.5 batteries to give me 12V?
Steve
2. ### hgmjr Moderator
Jan 28, 2005
9,030
214
What is the meaning of PPE?
hgmjr
3. ### SgtWookie Expert
Jul 17, 2007
22,182
1,728
I think Steve is talking about PP3 9V "Transistor" batteries.
How long the battery will last depends upon the construction of the battery.
Typical Capacity
(mAh)
565 (alkaline)
400 (carbon-zinc)
1200 (Lithium)
120 (NiCd)
175 to 300 (NiMH)
500 (Lithium polymer rechrg)
Let's say you use a lithium cell. 1200/300 = 4 hours - not long.
You could use six 2500mAh AA batteries to get just over 8 hours of operating time.
4. ### hgmjr Moderator
Jan 28, 2005
9,030
214
If you are referring to PP3 then based on the mAh (milliamp-hour) ratings that sgtwookie has provided then you can estimate the duration of an alkaline type 9V battery with a load of 300 milliamps to last approximately 565/300 = 1.88 hours or 113 minutes.
The same means of calculation can be used to estimate the duration of for each of the battery types that sgtrwookie has listed.
hgmjr
5. ### Audioguru New Member
Dec 20, 2007
9,411
896
Why not look at the datasheet of a 9V battery like Energizer's alkaline battery. www.energizer.com
1) Its voltage does not stay at 9V, It drops quickly to about 7.2V then drops slower to nothing.
2) Its capacity with only a 25ma load is 625mAh but then its voltage has dropped to 4.8V.
3) With a 300mA load, it reaches 6V in a little less than 1 hour.
Here is its voltage curve with a 250mA load.
File size:
7.6 KB
Views:
27
Nov 19, 2007
61
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https://oeis.org/A166347 | 1,721,657,488,000,000,000 | text/html | crawl-data/CC-MAIN-2024-30/segments/1720763517878.80/warc/CC-MAIN-20240722125447-20240722155447-00549.warc.gz | 363,841,515 | 4,111 | The OEIS is supported by the many generous donors to the OEIS Foundation.
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A166347 Irregular triangle read by rows of the frequency with which each value v of floor (j^2/p) occurs within A165974 for each prime p, taken over 1 <= j <= p - 1. 1
1, 0, 1, 1, 0, 2, 1, 0, 1, 0, 2, 1, 1, 1, 0, 1, 0, 3, 1, 1, 1, 1, 1, 0, 1, 0, 1, 0, 3, 2, 1, 1, 1, 0, 1, 1, 0, 1, 0, 1, 0, 4, 1, 2, 1, 1, 1, 0, 1, 1, 1, 0, 1, 0, 1, 0, 1, 0, 4, 2, 1, 1, 1, 1, 1, 1, 1, 0, 1, 1, 0, 1, 0, 1, 0, 1, 0, 4, 2, 2, 1, 1, 1, 1, 1, 1, 1, 0, 1, 1, 0, 1, 1, 0, 1, 0, 1, 0, 1, 0 (list; graph; refs; listen; history; text; internal format)
OFFSET 1,6 COMMENTS Part of the irregular triangle from which the sequence is constructed is shown below. ....v..0..1..2..3..4..5..6..7..8..9.10.11.12.13.14.15.16.17.18.19.20.21.22 ..p ..2....1..0 ..3....1..1..0 ..5....2..1..0..1..0 ..7....2..1..1..1..0..1..0 .11....3..1..1..1..1..1..0..1..0..1..0 .13....3..2..1..1..1..0..1..1..0..1..0..1..0 .17....4..1..2..1..1..1..0..1..1..1..0..1..0..1..0..1..0 .19....4..2..1..1..1..1..1..1..1..0..1..1..0..1..0..1..0..1..0 .23....4..2..2..1..1..1..1..1..1..1..0..1..1..0..1..1..0..1..0..1..0..1..0 LINKS C. H. Gribble, Flattened irregular triangle read by rows of primes < 300 CROSSREFS Cf. A165974 Sequence in context: A068906 A162514 A364039 * A055300 A335392 A156256 Adjacent sequences: A166344 A166345 A166346 * A166348 A166349 A166350 KEYWORD nonn,tabf AUTHOR Christopher Hunt Gribble, Oct 12 2009 EXTENSIONS Minor edits by Christopher Hunt Gribble, Oct 13 2009 STATUS approved
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Last modified July 22 09:51 EDT 2024. Contains 374490 sequences. (Running on oeis4.) | 892 | 1,930 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.65625 | 3 | CC-MAIN-2024-30 | latest | en | 0.570913 |
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# Wave Characteristics
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Waves are disturbances that travel through a fluid medium. Several common wave characteristics include frequency, period, wavelength, and amplitude. There are two main type of waves, transverse waves and longitudinal waves.
Let's talk about wave characteristics. Now I want to start off by just talking about, what is a wave? I mean we all kind of have this qualitative idea of what it is and what it means but how can I define one physically?
Well, physically a wave is a disturbance in a medium. What's a medium? A medium is any material usually a fluid, a liquid or a gas but it could also be a solid that doesn't like to be messed with. So if I move its atoms out of the way, they're going to try to come back. So that's a medium. And so I'll disturb a medium and then that disturbance is going to propagate through the medium. Propagate is a word we use to mean move but essentially, only when we're talking about waves or species sometimes but you could talk to the biologists about that.
Alright. So the important thing to recognize about a wave is that the initial disturbance is separate from the wave itself. The wave itself is a response that the medium has to being disturbed. Alright? So let's just go ahead and see this physically.
So I'm going to drop this dime into this little bowl of water. And let's watch what happens. So, as the dime fell in, it disturbed the surface of the water. Now, then the dime just went to the bottom. It's done. It doesn't have anything more to do with the wave. But the wave continued to propagate. It continued to disturb the medium after the dime was gone. And so that's what I mean by the wave itself is separate from the initial disturbance. Alright?
Another way that we could think of a wave is just that standard thing in crowds, right? Some dude stands up and then sits back down and then the other people carry that disturbance through the crowd with them, alright? So if the other people weren't going to stand up, it's not good enough that the first guy stood up. Everybody has to participate. So like we talked about before, a medium is something that doesn't like being disturbed. So that means when I disturb it, it's going to try and come back. So that property is called the restorative property of the medium.
Now in addition to that we have something called an inertial property. the inertial property is like the mass of the medium and that's what kind of resists this acceleration back to its equilibrium. And it's the interplay between these two things that gives us the wave. Alright.
Now, let's just talk about some basic types of waves. The first basic type of wave that I want to talk about here is called the pulse. A pulse is like a drum beat. You take your stick and you hit the drum once and you're done. So a pulse is a single disturbance that only assaults the medium once, alright? We're going to characterize a pulse by its amplitude. So if I've got the medium like this, you could think of like a violin string if you want to or a chain. so here's the medium undisturbed and then [IB] disturb it, the amplitude is going to characterize how severe that disturbance is. It's the maximum displacement of the medium away from its equilibrium position. So that amplitude is really going to be the only characterization of the pulse type of wave, alright? Other than its shape. And we can talk about that later. Alright.
The second major category of waves is periodic waves. Periodic waves consists of many identical pulses that repeatedly assault the medium again and again and again and again. So we have something like this, where we've got a positive pulse and a negative pulse and look at this. We've got one pulse, another pulse, another pulse and we could keep on going. So there's a lot more terms that we use to characterize periodic waves.
Frequency. That tells us how many of these pulses come per second or per minute or per some other time period. Period. That tells us how long each of these pulses takes to make it through a certain point. So frequency and period are related to each other. Wave length. That tells us the physical size of a single one of these pulse disturbances. And it's characterized by this weird symbol called [IB] that we'll talk about later. And then of course we again have an amplitude which describes how severe the disturbance is. How far away from the equilibrium the material is disturbed.
Alright. So those are some basic properties of waves.
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https://www.mrexcel.com/board/threads/please-someone-help-me.1098785/#post-5282783 | 1,632,646,797,000,000,000 | text/html | crawl-data/CC-MAIN-2021-39/segments/1631780057857.27/warc/CC-MAIN-20210926083818-20210926113818-00704.warc.gz | 901,045,776 | 16,502 | #### weetamc
##### New Member
Good morning everyone, I hope this is the place I can get some help with this....ill try to keep this problem short:
I have a spreadsheet for absence. Our corporate trigger levels where an employee progresses to a stage of absence are 6 days in a row,4 weeks in a row, 3 months in a row and or four separate absences within a rolling year. Now the sheet I have set up every time someone is absent I type the number 1 into the cell corresponding to the date and i have it counting the number of days absent in another cell. What i would like to do (if this is possible) is to have a pop up message appear when the count reaches 6 days and or 4 seperate occasions (this would need to somehow count any absences under six days as one absence until this reaches four?)
I know its all very confusing if you think your able to offer me some help i will try to simplify the explanation....one massive thanks in advance...
Tam
Scotland
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#### Steve_
##### Board Regular
Good morning everyone, I hope this is the place I can get some help with this....ill try to keep this problem short:
I have a spreadsheet for absence. Our corporate trigger levels where an employee progresses to a stage of absence are 6 days in a row,4 weeks in a row, 3 months in a row and or four separate absences within a rolling year. Now the sheet I have set up every time someone is absent I type the number 1 into the cell corresponding to the date and i have it counting the number of days absent in another cell. What i would like to do (if this is possible) is to have a pop up message appear when the count reaches 6 days and or 4 seperate occasions (this would need to somehow count any absences under six days as one absence until this reaches four?)
I know its all very confusing if you think your able to offer me some help i will try to simplify the explanation....one massive thanks in advance...
Tam
Scotland
Could you show me some example data please? I need to see the formatting of your sheet. But yes, I can do this easily.
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# FMC- Mine Ventilation, Fires, Explosion & Inundation
Question Text:
1 Bar pressure is equivalent to
Options:
(1) 1 kPa
(2) 10kPa
(3) 100 kPa
(4) 1000 kPa
(5) None of the Options
## Correction Answer : Option No.(3)
Question Text:
Choose the correct statement
Options:
(1) Rate of diffusion of gases into air is directly proportional to their densities.
(2) Rate of diffusion of gases into air is directly proportional to the square root of their densities.
(3) Rate of diffusion of gases into air is inversely proportional to their densities.
(4) Rate of diffusion of gases into air is inversely proportional to the square root of their densities.
(5) None of the Options
## Correction Answer : Option No.(4)
Question Text:
Jones - Tricket ration of 0.4 indicates
Options:
(1) No fire
(2) Wood fire
(3) Coal fire
(4) Oil fire
(5) No any indication
## Correction Answer : Option No.(1)
Question Text:
The average concentration of respirable dust in mine atmosphere during each shift, to which
person is exposed at any working is shall not exceed _______ Where the free silica in air borne
dust is ________ or less.
Options:
(1) 2 mg/m3 and 3%
(2) 3 mg/m3 and 5%
(3) 5 mg/m3 and 5%
(4) 3 mg/m3 and 3%
(5) 2 mg/m3 and 5%
## Correction Answer : Option No.(2)
Question Text:
Ratio of CO2 given off to O2 consumed by the human being is called
Options:
(1) Life quotient
FMC- Mine Ventilation, Fires, Explosion & Inundation
(2) Life survival ration
(3) Oxidation ration
(4) Respiratory quotient
(5) None of the Options
## Correction Answer : Option No.(4)
Question Text:
When coal is burnt with insufficient supply of oxygen, _____________ Is formed?
Options:
(1) carbon dioxide
(2) carbon peroxide
(3) carbon monoxide
(4) carbon tetraoxide
(5) methane
## Correction Answer : Option No.(3)
Question Text:
In a ventilating district, wet bulb temperature is found to 32 C, the air velocity should not be less
than
Options:
(1) 0.5 m/sec
(2) 1 m/sec
(3) 45 m/ min
(4) 75 m/ min
(5) None of the Options
## Correction Answer : Option No.(2)
Question Text:
What is the physiological effect of 0.2% CO inhalation by human?
Options:
(1) Death in 5 min.
(2) Palpitation and giddiness
(4) Death in 10 min.
(5) Death in 1 min.
## Correction Answer : Option No.(4)
Question Text:
The period of time that elapses between the application of an ignition source and appearance of
the flame is called
Options:
(1) Ignition time
(2) Incubation period
(3) Lag on ignition
(4) Fire time
(5) Explosion time
FMC- Mine Ventilation, Fires, Explosion & Inundation
Question Text:
Gas chromatography separates and analyse components of gas mixture by
Options:
(2) Molecular species
(3) Boiling water
(4) Catalyst
(5) None of the Options
## Correction Answer : Option No.(2)
Question Text:
Why nitrogen flushing is done in mines?
Options:
(1) To increase the % of nitrogen
(2) Remove CO from sealed off area
(3) To reduce the % of oxygen in the area
(4) To improve ventilation of mine
(5) To clear blasting fumes from the face
## Correction Answer : Option No.(3)
Question Text:
What is full form of FLP?
Options:
(1) Fast Leakage Process
(2) Flameproof
(3) Fire Leakage Process
(4) First Lighting Process
(5) Fire Lighting Potential
## Correction Answer : Option No.(2)
Question Text:
Gravimetric dust sampler is used for
Options:
(1) Measuring gravitational force
(2) Air sampling
(3) Removing dust from air
(4) Dust sampling
(5) Respirable air borne dust sample
## Correction Answer : Option No.(5)
Question Text:
Why auxiliary fan is used in underground mine ?
Options:
(1) To increase overall quantity of air in district
(2) To improve working condition of the working face
FMC- Mine Ventilation, Fires, Explosion & Inundation
(3) To decrease resistance in airway
(4) To increase overall quantity of air in district and to improve working condition of the working
face
(5) To increase overall quantity of air, to improve working condition of the working face and to
decrease resistance in airway
## Correction Answer : Option No.(2)
Question Text:
Which of the following is not a type of mechanical ventilator?
Options:
(1) Axial-Flow fan
(2) Uni-axial fan
(3) Turbo- axial fan
(5) None of the Options
## Correction Answer : Option No.(2)
Question Text:
At what scale the Dust Plan is prepared and maintained, at every mine having working
belowground ?
Options:
(1) Not less than 1-2500
(2) Not less than 1-4000
(3) Not less than 1-2400
(4) Not less than 1-5000
(5) None of the Options
## Correction Answer : Option No.(3)
Question Text:
A man while doing hard work, breathes 60 x 10 -3 m3 of air per minute. Amount of oxygen
consumed by him will be:
Options:
(1) 12.57 x 10-3 m3
(2) 11.56 X 10-3 m3
(3) 13.75 X 10-3 m3
(4) 11.65 X 10-3 m3
(5) None of the Options
## Correction Answer : Option No.(1)
Question Text:
Distance of main mechanical ventilator of the mine from the opening of the shaft or incline at any
point, shall be
Options:
(1) less than 10m
(2) Not less than 10 m
(3) At the opening of the shaft or incline
(4) 15 m
FMC- Mine Ventilation, Fires, Explosion & Inundation
(5) 12 m
## Correction Answer : Option No.(2)
Question Text:
To reopen a mine or part thereof, which has been isolated, sealed off or flooded with water to
deal with a fire:
Options:
(1) A prior permission in writing from Chief Inspector is required
(2) A notice is required to be given, not less than 14 days before the commencement of work
(3) A prior permission in writing from Regional Inspector is required.
(4) A notice is required to be given, not less than 07 days before the commencement of work
(5) A notice is required to be given, not less than 72 hrs before commencement of work
## Correction Answer : Option No.(2)
Question Text:
At what interval examination of rate of emission of inflammable gas in cubic meters per tonne of
coal raised shall be done?
Options:
(1) 15 days
(2) 7 days
(3) 30 days
(4) 45 days
(5) 10 days
## Correction Answer : Option No.(3)
Question Text:
If a volume of 20 cm3 air sample is analysed with Haldane apparatus, and the volume remained
after absorption of CO2 is 18.65 cm3. What is the percentage of CO2 present in the air sample?
Options:
(1) 5.65%
(2) 6.55%
(3) 7.65%
(4) 5.76%
(5) 6.75%
## Correction Answer : Option No.(5)
Question Text:
Which of the following fan testing method is more accurate?
Options:
(1) Field test
(2) Rating test
(3) Air quantity test
(4) Efficiency test
(5) Mechanical test
## Correction Answer : Option No.(2)
Question Text:
FMC- Mine Ventilation, Fires, Explosion & Inundation
For what purpose the Smoke-cloud Generator is used?
Options:
(1) For air quantity survey
(2) For pressure survey
(3) For sending emergency warning
(4) To generate electricity
(5) None of the Options
## Correction Answer : Option No.(1)
Question Text:
Which is not the advantage of axial flow fan
Options:
(1) Requires less space for installation
(2) Smaller and cheaper foundation
(3) Reversal easier by reversal switch
(4) Quantity controllable by varying the pitch of the plates
(5) Suited for high water gauze and high quantity
## Correction Answer : Option No.(5)
Question Text:
Stone dust barriers are normally required in
Options:
(1) Degree-I
(2) Only in Degree-II
(3) Only in Degree-III
(4) Both Degree-II and III
(5) Degree-I, II and III
## Correction Answer : Option No.(4)
Question Text:
If any sampling detects respirable dust concentration of 4.6 mg/m3 and free silica content is 4%,
then futher sampling should be done within or not exceeding
Options:
(1) 6 months
(2) 3 months
(3) 2 months
(4) 1 month
(5) 2Weeks
## Correction Answer : Option No.(2)
Question Text:
In a longwall face, % of methane at intake and return gates indicate 0.1 % and 0.4 %, air quantity
flowing at the face is 1800 m3/ min, production from the face is 1500 TPD, methane emission rate
in m3/tonne of coal output is equal to
Options:
(1) 0.0864
FMC- Mine Ventilation, Fires, Explosion & Inundation
(2) 0.864
(3) 5.184
(4) 8.64
(5) 9.08
## Correction Answer : Option No.(3)
Question Text:
Explosibility of coal dust generally does not depends on
Options:
(1) Fineness of dust
(2) Composition of dust
(3) Presence of methane in air
(4) Colour of dust
(5) None of the Options
## Correction Answer : Option No.(4)
Question Text:
A 10% increase in cross sectional area of a roadway will reduce the resistance approximately by
Options:
(1) 11%
(2) 21%
(3) 33%
(4) 40%
(5) None of the Options
## Correction Answer : Option No.(2)
Question Text:
The most important parameter of a dust that governs its physiological effect is
Options:
(1) Concentration of dust in air
(2) Shape of dust
(3) Time of exposure
(4) Size of dust
(5) None of the Options
## Correction Answer : Option No.(4)
Question Text:
Atkinson's equation P = RQ2 is applicable for
Options:
(1) Laminar flow only
(2) Fully turbulent flow only
(3) Leakage paths only
(4) both Laminar and fully turbulent flow only
(5) None of the Options
## Correction Answer : Option No.(1)
Question Text:
FMC- Mine Ventilation, Fires, Explosion & Inundation
The size and speed of rotation being the same the head developed is maximum
Options:
(1) an Axial flow fan
(2) a backward bladed centrifugal fan
(5) None of the Options
## Correction Answer : Option No.(4)
Question Text:
If the blade angle of the main fan is increased, the pressure developed by the fan
Options:
(1) Increases
(2) Decreases
(3) May increase and decrease
(4) Remains unchanged
(5) None of the Options
## Correction Answer : Option No.(1)
Question Text:
A hard working man consummes 2 X 10-3 m3/ min of oxygen, the required air Q in m3/ min
containing 21% of oxygen needed to be supplied to the working place per minute so as to keep
the level of oxygen concentration of exhale air at 19% will be
Options:
(1) 0.1
(2) 0.125
(3) 0.15
(4) 0.2
(5) None of the Options
## Correction Answer : Option No.(1)
Question Text:
If fire damp emission rate is 5m3/ min and it is proposed to be brought down to 0.5% by
circulating air. The approximate air quantiy in will be
Options:
(1) 12.5
(2) 16.7
(3) 18.2
(4) 21.2
(5) None of the Options
## Correction Answer : Option No.(2)
Question Text:
Measures of risk is obtained from the likelyhood of occurrence (l) and consequence of these
occurrence by
Options:
FMC- Mine Ventilation, Fires, Explosion & Inundation
(1) l + c
(2) l - c
(3) l x c
(4) l / c
(5) c / l
## Correction Answer : Option No.(3)
Question Text:
With normal diet and rest in bed, the respiratory quotient is around
Options:
(1) 0.01
(2) 0.3
(3) 0.5
(4) 0.6
(5) 0.85
## Correction Answer : Option No.(5)
Question Text:
Specific cooling power of mine air depends on
Options:
(1) Dry bulb temperature, wet bulb temperature, velocity of air
(2) Dry bulb temperature, wet bulb temperature, velocity of air, radiant temperature
(3) Dry bulb temperature, wet bulb temperature, velocity of air, radiant temperature,
atmospheric pressure
(4) Dry bulb temperature, wet bulb temperature, velocity of air, atmospheric pressure, skin
temperature
(5) None of the Options
## Correction Answer : Option No.(4)
Question Text:
1 Kg of methane on complete combustion produces _____ kJ of heat
Options:
(1) 55.7
(2) 556.8
(3) 5568
(4) 55685
(5) None of the Options
## Correction Answer : Option No.(4)
Question Text:
CO make of more than ________lit/min. indicates that the heating is well developed & urgent
action must be taken
Options:
(1) 10
(2) 20
(3) 30
(4) 40
FMC- Mine Ventilation, Fires, Explosion & Inundation
(5) 50
## Correction Answer : Option No.(2)
Question Text:
The input motor power consumption in a mine can be reduced by
Options:
(1) reducing the mine resistance
(2) selecting the fan which has maximum efficiency at the operating point
(3) Selecting motor having maximum efficiency & maximum tranmission efficiency in case of V-
Belt drive
(4) All the Options
(5) by increasing the size of airways
## Correction Answer : Option No.(4)
Question Text:
Which coal is more prone to spontaneous combustion
Options:
(1) High thermal conductivity
(2) Low porosity
(3) Self heating rate from 40C to 70C is 1.5C/hr i.e. R70 = 1.5C/hr.
(4) High thermal conductivity and Self heating rate from 40C to 70C is 1.5C/hr i.e. R70 =
1.5C/hr.
(5) None of the Options
## Correction Answer : Option No.(3)
Question Text:
When T is absolute temperature at which evaporation of vapour takes place,T2 is absolute
temperature at which condensation of vapour takes place, then co-efficient of performance can
be expressed as
Options:
(1) T/(T2-T)
(2) T/T2
(3) (T2-T)/T
(4) T/(T-T2)
(5) None of the Options
## Correction Answer : Option No.(1)
Question Text:
What is required to be done, to get high head from axial flow fan, keeping its size and speed
within reasonable limit ?
Options:
(1) Increase in power input
(2) Changing the shape
(3) Installing another fan in series
(4) Increasing no of stages
(5) None of the Options
FMC- Mine Ventilation, Fires, Explosion & Inundation
Question Text:
Eddy loss for fan is significant when____________
Options:
(1) fan runs at its full capacity
(2) fan operates at low volume
(3) fan drift cross-section is very high
(4) fan drift cross-section is very low
(5) fan operates at high volume
## Correction Answer : Option No.(2)
Question Text:
Large ventilation fan with high starting torque preferable motor will be
Options:
(1) Squirrel cage Induction Motor
(2) Capacitor start Motor
(3) DC shunt Motor
(4) Slip ring Induction Motor
(5) None of the Options
## Correction Answer : Option No.(4)
Question Text:
Which statement is correct?
Options:
(1) Flammability of coal dust increases in proportion with increase in percentage of fire damp
(2) Flammability of coal dust decreases in proportion with increase in percentage of fire damp
(3) Flammability of coal dust does not depend on presence of fire damp
(4) Flammability of the coal dust initially increases and then decreases
(5) None of the Options
## Correction Answer : Option No.(1)
Question Text:
The purpose of evassee in a fan is not
Options:
(1) To reduce the final velocity of the discharged air by providing a passage of increasing cross
section through which air must pass
(2) To eliminate turbulence
(3) To promote smooth flow of air and prevent re-entry of air from atmosphere
(4) To reduce the air quantity
(5) To increase fan efficiency
## Correction Answer : Option No.(4)
Question Text:
The permission to extract pillars below an area has been obtained. The HFL of the area changes
and HFL passes over this area. Under the circumstances what needs to be done.
Options:
FMC- Mine Ventilation, Fires, Explosion & Inundation
(1) Extraction can be continued as permitted earleir.
(2) Fresh permission from DGMS will be required for extracting pillars.
(3) Embankment may be constructed to keep the area unaffected from new HFL
(4) Embankment may be constructed to keep the area unaffected from new HFL and fresh
permission from DGMS will also be required for extracting pillars.
(5) None of the Options
## Correction Answer : Option No.(4)
Question Text:
The unit kgf.s.m is also known as
Options:
(1) gaul
(2) Murg
(3) Killomurg
(4) Atkinson
(5) None of the Options
## Correction Answer : Option No.(3)
Question Text:
Mass of water vapour present in a unit mass of air is known as
Options:
(1) absolute humidity
(2) relative humidity
(3) specific humidity
(4) mixing ratio
(5) None of the Options
## Correction Answer : Option No.(3)
Question Text:
Choose the wrong statement
Options:
(1) Trickett"s ratio for assessing status of fire in a sealed area will not work, if intake air is oxygen
deficient through the injection of nitrogen and carbon dioxide or through a high methane make)
(2) Fresh air dilution of the fire products has no effect on Trickett"s ratio.
(3) If Trickett"s ratio is less than 0.4, there is no fire and gases are residual than active
(4) If Trickett" ratio is > 1.6, there is serious fire
(5) None of the Options
## Correction Answer : Option No.(4)
Question Text:
When regulator is installed to regulate air quantity in a district, the pressure developed in the
main fan
Options:
(1) decreases
(2) increases
(3) remain same
(4) may increase or decrease depending on the fan type
FMC- Mine Ventilation, Fires, Explosion & Inundation
(5) None of the Options
## Correction Answer : Option No.(2)
Question Text:
Which is correct statement ?
Options:
(1) Foam extinguisher when directed against a fire, the foam spreads over the entire fire area
covering or blanketing it, thus cutting oxygen supply.
(2) Wet foam layer produces a cooling effect to extinguish the fire.
(3) Foam extinguisher when directed against a fire, the foam spreads over the entire fire area
covering or blanketing it, thus cutting oxygen supply; and Wet foam layer produces a cooling
effect to extinguish the fire.
(4) Foam extinguisher is generally recommended for fighting electrical fire.
(5) All the Options
## Correction Answer : Option No.(3)
Question Text:
The auxiliary fan should not suck more than ______% of available air to avoid recirculation.
Options:
(1) 30
(2) 40
(3) 50
(4) 60
(5) 70
## Correction Answer : Option No.(3)
Question Text:
Which is wrong statement ?
Options:
(1) Reversing of air current is easier in axial flow fan compared to centrifugal fan.
(2) Less space is required in case of axial flow fan compared to centrifugal fan
(3) Axial flow fan is used when low air quantity at high pressure is required.
(4) Cost of axial flow fan is less compared to centrifugan fan
(5) None of the Options
## Correction Answer : Option No.(3)
Question Text:
Which of the following is not the advatage of fighting fire with inert gases
Options:
(1) It offers maximum possible safety against risk of explosion & spread of fire, if applied at correct
time
(2) Recovery work negligible & hence rapid
(3) Inert gas penetrates easily into cracks & cavities than clay or sand
(4) Fire brigade can operate for longer periods
(5) None of the Options
## Correction Answer : Option No.(5)
FMC- Mine Ventilation, Fires, Explosion & Inundation
Question Text:
Match the following thickness of
## A. Brick isolation stopping 1. 1.5m
B. Walls of air crossing 2. 0.5m
C. Minimum thickness of concrete dam 3. 0.25m
D. RCC isolation stopping 4. 1.0 m
Options:
(1) A-2, B-3, C-4, D-1
(2) A-3, B-4, C-2, D-1
(3) A-1, B-2, C-3, D-4
(4) A-2, B-4, C-2, D-1
(5) A-4, B-3, C-1, D-2
Correction Answer : One option in the question text was incorrect, so mark was awarded to all the
candidates
Question Text:
The direction of natural ventilation is mainly dependent on
Options:
(1) Length of the mine
(2) Depth of the mine
(3) Quantity of CH emission in mine
(4) Amount of explosives used in the mine
(5) Number of persons deployed in the mine
## Correction Answer : Option No.(2)
Question Text:
An auxiliary fan ventilating a heading through a duct of 600 mm diameter, circulates 5 m3 / sec of
air at the face. What will be the velocity of air in the duct ?
Options:
(1) 17.67 m/s
(2) 16.76 m/s
(3) 17.00 m/s
(4) 16.67 m/s
(5) None of the Options
## Correction Answer : Option No.(1)
Question Text:
Which of the following is a type of Centrifugal Fan?
Options:
## Correction Answer : Option No.(1)
FMC- Mine Ventilation, Fires, Explosion & Inundation
Question Text:
Pressure required to circulate 3000 m3 / min of air through 2000 m long gallery of 2m X 5 m cross
section.
Value of K= 0.001 Ns2m-4
Options:
(1) 70 Pa
(2) 700 Pa
(3) 14 Pa
(4) 7 Pa
(5) 35 Pa
## Correction Answer : Option No.(1)
Question Text:
In a gassy seam of degree II, the quantity of air shall be measured once at least in every
_________ days and examination of every airway, air crossing, ventillation stopping and door in
use shall be done once at least in every ___________days
Options:
(1) 14, 30
(2) 7 , 14
(3) 20, 30
(4) 14, 14
(5) 30, 45
## Correction Answer : Option No.(4)
Question Text:
Where the embankment is provided against any water course, the height of embankment above
HFL as per Coal Mines Regulation, 1957 should not be less than
Options:
(1) 1.5 m
(2) 2.0 m
(3) 3 m
(4) 6 m
(5) None of the Options
## Correction Answer : Option No.(1)
Question Text:
During rainy season to prevent inrush of surface water inside the mine, every protective work
constructed for the purpose shall be inspected at least once in
Options:
(1) daily
(2) 7 days
(3) 14 days
(4) 20 days
(5) 30 days
## Correction Answer : Option No.(3)
Question Text:
FMC- Mine Ventilation, Fires, Explosion & Inundation
Methane layering number is 15.
Options:
(1) No action required
(2) Immediate action needed
(3) Further investigation required
(4) No interpretation can be made
(5) None of the Options
## Correction Answer : Option No.(1)
Question Text:
A fan delivers 60 m3/sec at 100 mm of w.g. pressure when the rpm of fan is 700. The rpm changes
to 900. The pressure developed by fan will be
Options:
(1) 593.4 Pa
(2) 1261.3 Pa
(3) 1425.3 Pa
(4) 1621.6 Pa
(5) None of the Options
## Correction Answer : Option No.(4)
Question Text:
The man during hardwork (mining operations with handheld tools) produces heat in the range of
Options:
(1) 50 to 100 Kcal/hr
(2) 100 to 150 Kcal/hr
(3) 150 to 200 Kcal/hr
(4) 200 to 250 Kcal/hr
(5) 250 to 400 Kcal/hr
## Correction Answer : Option No.(5)
Question Text:
Carbon monoxide detector tubes uses
Options:
(1) PETN
(2) Protosorb
(3) Hopcolite
(5) None of the Options
## Correction Answer : Option No.(4)
Question Text:
A cap lamp must be able to produce an illumination of
Options:
(1) 0.25 lumans
(2) 0.25 lux
(3) 0.25 candlas/m
FMC- Mine Ventilation, Fires, Explosion & Inundation
(4) 0.25 foot candlas
(5) None of the Options
## Correction Answer : Option No.(1)
Question Text:
The operating point of a PV-200 fan is as follows. Quantity =1800m3/min, pressure=10 mm w.g.
The mine resistence in kilo murg is equal to
Options:
(1) 0.011
(2) 0.11
(3) 0.278
(4) 1.1
(5) None of the Options
## Correction Answer : Option No.(1)
Question Text:
When Q is in m3 / sec and P in Pa and R is in Gaul, Equivalent orifice in m2 can be defined as
Options:
(1) 1.29Q/P
(2) 1.29/R
(3) 1.29P/Q
(4) 2.29P/Q
(5) 2.29Q/P
## Correction Answer : Option No.(2)
Question Text:
Which of the following is true?
Options:
(1) Q2=R/P
(2) Q=R/P
(3) Q=R/P
(4) Q=P/R
(5) Q=R/P
## Correction Answer : Option No.(4)
Question Text:
If air quantity circulated by the fan is 6000 m3 /min and pressure developed by the fan is 100 Pa,
air power is equal to
Options:
(1) 10 kW
(2) 100 kW
(3) 600 kW
(4) 700 kW
(5) None of the Options
## Correction Answer : Option No.(1)
Question Text:
FMC- Mine Ventilation, Fires, Explosion & Inundation
In a sealed off area, the analysis report shows the following gas percentage O2-19.2, CO-0.8, CO2-
2, and N2-78 , the Graham's ratio is
Options:
(1) 0.04
(2) 0.54
(3) 0.78
(4) 1.36
(5) None of the Options
## Correction Answer : Option No.(2)
Question Text:
Which statement is most correct?
Options:
(1) Coal dust explosion are more severe than methane explosion
(2) With methan explosion, CO is occasionally found while in coal dust explosion CO is always
found
(3) With methane explosion H2S gas is always found
(4) Coal dust explosion are more severe and always associated with presence of CO; where as
methane explosion comparatively weaker with lesser CO generation
(5) None of the Options
## Correction Answer : Option No.(4)
Question Text:
Fire extinguishing index is
Options:
(1) intensity of inert gas injection rate to reduce oxygen from a level of 21% to 2% in a sealed off
area
(2) the ratio of total injection of inert gas in m3 to the time in hours it requires to fill the sealed off
volume
(3) a measure for the fraction of the sealed volume which is filled with inert gases per month
(4) a measure for the fraction of the sealed volume which is filled with inert gases per day
(5) None of the Options
## Correction Answer : Option No.(2)
Question Text:
Minimum velocity required at 4.5m from any discontinued face in degree III mine?
Options:
(1) 15 m/ min
(2) 30 m/min
(3) 45m/min
(4) 60m/min
(5) None of the Options
## Correction Answer : Option No.(3)
Question Text:
When nitrous fumes passes over a filter paper soaked with starch and potassium iodite, the filter
paper becomes
FMC- Mine Ventilation, Fires, Explosion & Inundation
Options:
(1) red
(2) yellow
(3) black
(4) green
(5) blue
## Correction Answer : Option No.(5)
Question Text:
The mechanical efficiency of a fan is
Options:
(1) the ratio of the power actually absorbed by the impeller and converted into head to the power
applied to the shaft
(2) the ratio of the air power to the input power | 7,072 | 24,728 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.6875 | 3 | CC-MAIN-2019-35 | latest | en | 0.825835 |
https://it.mathworks.com/matlabcentral/profile/authors/6470928 | 1,632,542,994,000,000,000 | text/html | crawl-data/CC-MAIN-2021-39/segments/1631780057589.14/warc/CC-MAIN-20210925021713-20210925051713-00255.warc.gz | 389,322,072 | 18,215 | Community Profile
# FCSmith
Last seen: circa un mese ago Active since 2016
#### Content Feed
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oltre 4 anni ago | 1,601 | 6,080 | {"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} | 3.078125 | 3 | CC-MAIN-2021-39 | latest | en | 0.764203 |
http://lotterywinnerstories.com/Lottery/best-winning-lottery-numbers-pick3/ | 1,527,169,662,000,000,000 | text/html | crawl-data/CC-MAIN-2018-22/segments/1526794866326.60/warc/CC-MAIN-20180524131721-20180524151721-00292.warc.gz | 183,394,830 | 12,645 | ## How Do I Calculate Lotto Numbers? Pick3
Playing the Pick3 lottery can be very exciting especially if from time to time you are a winner. Winning can be accomplished quit often if the correct system is used. There are many so called experts in the market offering winning number secrets and scratch off ticket recommendations that promise to help you win. Most are not legit. However, there are a few systems on the market that has produced some promising results.
When you consider that that there are many different lottery systems available and lots of past winners you remain anonymous or don’t talk about their winning strategies how many more winners might there be who have used such systems?
Think about it: is it possible that a huge percentage of lottery winners are actually using mathematical or statistical formulas to help them win? If that is the case then anyone who is not using a system is merely feeding the prize fund and has an almost zero chance of winning.
## Would You Like to Win The Pick3 lottery?
Lottery predictions is quite popular these days. People used to be skeptical with the predictions as they thought that the winning numbers are a matter of luck and fortunes. Not many people believe that lottery can be won by using some kind of a sophisticated science based predictions. It was not until the late 90s when lottery players began using lottery predictions to help them to win lottery or at least get closer to the winning numbers. When Gonzalo Garcia-Pelayo, a Spanish man who managed to study and analyze many games in two different countries, Spain and the US and win a lot of money by using different strategies. After him people started to believe that lottery results can be predicted.
Lottery players start thinking about how to win the lotteries using predictions. They use many kinds of predictions: from mechanical predictions on mechanical lotteries to technological predictions using computer software. A lot of people use algorithm to analyze and predict lottery results.
Analysis of Groups
There are many kinds of group analysis that lottery predictors use to get into the winning numbers. Lottery players can group the months having the best winning numbers of a certain period or they can group the numbers winning in certain period of time.
Analysis of Hot-Cold Trend
This algorithm analysis is one of the most favourite so far as it can record the frequency ranks and use the variations to predict the tendencies of hot and cold numbers in the next drawings.
Analysis of Repetition Pattern
A lot of lottery players share the same opinion that repetition is quite important to predict the winning numbers as most of jackpots will appear again in the future.
The analysis mentioned above represents only a part of the strategies that lottery players can use. There are still many other algorithm analysis that can be done by predictors to help them win.
## Powerball: How To Use Easy Pick To Win The Powerball
Winning the Pick3 lottery is very rewarding. Give this system a spin and watch your winnings grow.
Learn How To Win The Lottery Pennsylvania Lottery
Playing the Pick3 lottery can be very exciting especially if from time to time you are a winner. Winning can be accomplished quit often if the correct system is used. There are many so called experts in the market offering winning number secrets and scratch off ticket recommendations that promise to help you win. Most are not legit. However, there are a few systems on the market that has produced some promising results.
When you consider that that there are many different lottery systems available and lots of past winners you remain anonymous or don’t talk about their winning strategies how many more winners might there be who have used such systems?
Think about it: is it possible that a huge percentage of lottery winners are actually using mathematical or statistical formulas to help them win? If that is the case then anyone who is not using a system is merely feeding the prize fund and has an almost zero chance of winning.
## Would You Like to Win The Pick3 lottery?
A lottery is a popular form of gambling which involves the drawing of lots for a prize. There are some states where lottery is forbidden but others endorse it to the extend of organizing a national lottery. Lotteries date back to the period of Romans. And it was common for emperors to give their dinner party guests gifts like slaves to lavish villas after drawings.
Two common lottery myths
Lottery is tax
Lottery take advantage of the poor economic strata of our society
Lottery facts
In China lottery was first played to fund the Great Wall
U.S. lotteries helped fund the Colonial Army in the Revolutionary War
At one point of time many churches and universities such as Harvard, Yale and Princeton were funded in part by lotteries
Where does the Lottery revenue expenditure go?
A good amount of lottery earnings is spend on economic development of the country
Steps are taken to improve job opportunities
Educational purposes
Human resources
Natural resources[environmental protection]
Transportation
Public health finance
Keeping apart money to improve the economic development of a country, lottery gains also go to cancer organizations, child care canters and places where it is needed
Today, many religious and social organizations condemn lotteries saying that it’s a one way ticket to accumulating lots of wealth. Since lotteries lotteries tempt people to play with money and gain huge amounts back if they got lucky, it can easily become an addiction. Many individuals waste much money over lotteries and other gambling games without taking care of their families properly. This is unethical to a population of today’s society. Of course, it is not right to spend all the money you earn on lotteries. But by taking part in lotteries you contribute a little to the development of your own country. That’s why we say lotteries are not just about gambling. It’s a lot about winning and giving.
ZZZZZZ
## How to Win Lottery by Using Analysis Algorithms for Lottery Prediction
Hospital lotteries are becoming increasingly popular in Canada. That's because, not only do you get a chance to win many terrific prizes, but you also get to donate money to a worthy enterprise - The hospital. And, the money goes to worthy hospital causes, like cancer research. But, of course, the lottery has to sell tickets and many of them do it by claiming to have the best chance to win a million dollars. That means, according to them, that if you are looking for the best odds to win a million dollars, you should buy a hospital lottery ticket rather than a regular lottery ticket.
So there you have it, playing a hospital lottery is not your best chance at winning a million dollars. However, the money that you spend on tickets goes to a good cause and so it is still worth it to play.
Winning the Pick3 lottery is very rewarding. Give this system a spin and watch your winnings grow.
How To Legally Hack The Lottery? Ohio Lottery | 1,386 | 7,047 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.84375 | 3 | CC-MAIN-2018-22 | longest | en | 0.97259 |
https://hg.mozilla.org/releases/mozilla-beta/file/f40200a441622c0c73be1eb5789b9c185da46cb5/mfbt/decimal/comparison-with-nan.patch | 1,590,540,915,000,000,000 | text/html | crawl-data/CC-MAIN-2020-24/segments/1590347391923.3/warc/CC-MAIN-20200526222359-20200527012359-00104.warc.gz | 388,773,349 | 3,348 | mfbt/decimal/comparison-with-nan.patch
author Jonathan Watt Mon, 06 May 2013 00:23:18 +0100 changeset 141893 f40200a441622c0c73be1eb5789b9c185da46cb5 child 320789 4929064f96e0bb142bf82e36230d1b0c8f570c9d permissions -rw-r--r--
Bug 853525, part 9 - Changes to Decimal code as a series of patches for update.sh. r=Waldo
```
diff --git a/mfbt/decimal/Decimal.cpp b/mfbt/decimal/Decimal.cpp
--- a/mfbt/decimal/Decimal.cpp
+++ b/mfbt/decimal/Decimal.cpp
@@ -505,21 +505,25 @@ Decimal Decimal::operator/(const Decimal
if (remainder > divisor / 2)
++result;
return Decimal(resultSign, resultExponent, result);
}
bool Decimal::operator==(const Decimal& rhs) const
{
+ if (isNaN() || rhs.isNaN())
+ return false;
return m_data == rhs.m_data || compareTo(rhs).isZero();
}
bool Decimal::operator!=(const Decimal& rhs) const
{
+ if (isNaN() || rhs.isNaN())
+ return true;
if (m_data == rhs.m_data)
return false;
const Decimal result = compareTo(rhs);
if (result.isNaN())
return false;
return !result.isZero();
}
@@ -528,16 +532,18 @@ bool Decimal::operator<(const Decimal& r
const Decimal result = compareTo(rhs);
if (result.isNaN())
return false;
return !result.isZero() && result.isNegative();
}
bool Decimal::operator<=(const Decimal& rhs) const
{
+ if (isNaN() || rhs.isNaN())
+ return false;
if (m_data == rhs.m_data)
return true;
const Decimal result = compareTo(rhs);
if (result.isNaN())
return false;
return result.isZero() || result.isNegative();
}
@@ -546,16 +552,18 @@ bool Decimal::operator>(const Decimal& r
const Decimal result = compareTo(rhs);
if (result.isNaN())
return false;
return !result.isZero() && result.isPositive();
}
bool Decimal::operator>=(const Decimal& rhs) const
{
+ if (isNaN() || rhs.isNaN())
+ return false;
if (m_data == rhs.m_data)
return true;
const Decimal result = compareTo(rhs);
if (result.isNaN())
return false;
return result.isZero() || !result.isNegative();
}
``` | 555 | 1,945 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.515625 | 3 | CC-MAIN-2020-24 | latest | en | 0.158049 |
https://www.answers.com/Q/Is_45_inches_more_than_5_feet | 1,701,814,656,000,000,000 | text/html | crawl-data/CC-MAIN-2023-50/segments/1700679100568.68/warc/CC-MAIN-20231205204654-20231205234654-00670.warc.gz | 725,283,817 | 47,078 | 0
# Is 45 inches more than 5 feet?
Updated: 8/10/2023
Wiki User
12y ago
No it is equal to 60 inches. 5*12 =60 inches.
Wiki User
7y ago
Wiki User
15y ago
Someone 68 inches tall is 68 inches tall. That's 5' 8" tall.
Wiki User
9y ago
1 foot = 12 inches
5 feet=5*12 = 60 inches
Hence 68 inches is more than 5 feet.
Wiki User
12y ago
no it's 3'9 (5 ft=60in. | 141 | 370 | {"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} | 3.578125 | 4 | CC-MAIN-2023-50 | latest | en | 0.802081 |
https://www.aqua-calc.com/one-to-all/entropy/preset/kilojoule-per-kelvin/1 | 1,718,594,767,000,000,000 | text/html | crawl-data/CC-MAIN-2024-26/segments/1718198861696.51/warc/CC-MAIN-20240617024959-20240617054959-00422.warc.gz | 587,637,368 | 6,228 | # Convert kilojoules per kelvin [kJ/K] to other units of entropy
## kilojoules/kelvin [kJ/K] entropy conversions
1 kJ/K = 1 × 10+21 attojoules per kelvin kJ/K to aJ/K 1 kJ/K = 100 000 centijoules per kelvin kJ/K to cJ/K 1 kJ/K = 10 000 decijoules per kelvin kJ/K to dJ/K 1 kJ/K = 100 dekajoules per kelvin kJ/K to daJ/K 1 kJ/K = 1 × 10-15 Exajoule per kelvin kJ/K to EJ/K 1 kJ/K = 1 × 10+18 femtojoules per kelvin kJ/K to fJ/K 1 kJ/K = 1 × 10-6 Gigajoule per kelvin kJ/K to GJ/K 1 kJ/K = 10 hectojoules per kelvin kJ/K to hJ/K 1 kJ/K = 0.001 Megajoule per kelvin kJ/K to MJ/K 1 kJ/K = 1 000 000 000 microjoules per kelvin kJ/K to µJ/K 1 kJ/K = 1 000 000 millijoules per kelvin kJ/K to mJ/K 1 kJ/K = 1 000 000 000 000 nanojoules per kelvin kJ/K to nJ/K 1 kJ/K = 1 000 joules per kelvin kJ/K to J/K 1 kJ/K = 1 × 10-12 Petajoule per kelvin kJ/K to PJ/K 1 kJ/K = 1 × 10+15 picojoules per kelvin kJ/K to pJ/K 1 kJ/K = 1 × 10-9 Terajoule per kelvin kJ/K to TJ/K 1 kJ/K = 1 × 10+27 yoctojoules per kelvin kJ/K to yJ/K 1 kJ/K = 1 × 10-21 Yottajoule per kelvin kJ/K to YJ/K 1 kJ/K = 1 × 10+24 zeptojoules per kelvin kJ/K to zJ/K 1 kJ/K = 1 × 10-18 Zettajoule per kelvin kJ/K to ZJ/K
#### Foods, Nutrients and Calories
KAPALUA BRAND, EXOTIC FRUIT JELLIES, UPC: 852330005268 contain(s) 214 calories per 100 grams (≈3.53 ounces) [ price ]
106 foods that contain Beta-sitosterol. List of these foods starting with the highest contents of Beta-sitosterol and the lowest contents of Beta-sitosterol
#### Gravels, Substances and Oils
CaribSea, Freshwater, Eco-Complete Cichlid, Ivory Coast weighs 1 281.48 kg/m³ (80.00018 lb/ft³) with specific gravity of 1.28148 relative to pure water. Calculate how much of this gravel is required to attain a specific depth in a cylindricalquarter cylindrical or in a rectangular shaped aquarium or pond [ weight to volume | volume to weight | price ]
Trilithium nitride [Li3N] weighs 1 270 kg/m³ (79.28351 lb/ft³) [ weight to volume | volume to weight | price | mole to volume and weight | mass and molar concentration | density ]
Volume to weightweight to volume and cost conversions for Refrigerant R-22, liquid (R22) with temperature in the range of -51.12°C (-60.016°F) to 71.12°C (160.016°F)
#### Weights and Measurements
A centicandela is a SI-multiple (see prefix centi) of the luminous intensity unit candela and equal to one hundredth of an candela (0.01 cd)
Energy is the ability to do work, and it comes in different forms: heat (thermal), light (radiant), motion (kinetic), electrical, chemical, nuclear and gravitational.
short tn/cm³ to troy/in³ conversion table, short tn/cm³ to troy/in³ unit converter or convert between all units of density measurement.
#### Calculators
Estimate daily calories using height, weight, age, gender, activity | 997 | 2,798 | {"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} | 3.828125 | 4 | CC-MAIN-2024-26 | latest | en | 0.389025 |
https://rustgym.com/leetcode/1062 | 1,660,828,294,000,000,000 | text/html | crawl-data/CC-MAIN-2022-33/segments/1659882573197.34/warc/CC-MAIN-20220818124424-20220818154424-00672.warc.gz | 443,052,089 | 3,142 | 1062. Longest Repeating Substring
Given a string `S`, find out the length of the longest repeating substring(s). Return `0` if no repeating substring exists.
Example 1:
```Input: S = "abcd"
Output: 0
Explanation: There is no repeating substring.
```
Example 2:
```Input: S = "abbaba"
Output: 2
Explanation: The longest repeating substrings are "ab" and "ba", each of which occurs twice.
```
Example 3:
```Input: S = "aabcaabdaab"
Output: 3
Explanation: The longest repeating substring is "aab", which occurs `3` times.
```
Example 4:
```Input: S = "aaaaa"
Output: 4
Explanation: The longest repeating substring is "aaaa", which occurs twice.
```
Constraints:
• The string `S` consists of only lowercase English letters from `'a'` - `'z'`.
• `1 <= S.length <= 1500`
1062. Longest Repeating Substring
``````struct Solution;
use std::collections::HashSet;
impl Solution {
fn longest_repeating_substring(s: String) -> i32 {
let n = s.len();
let mut res = 0;
let mut hs: HashSet<&str> = HashSet::new();
for k in 1..n {
for i in 0..=n - k {
if !hs.insert(&s[i..i + k]) {
res = res.max(k);
break;
}
}
if res < k {
break;
}
}
res as i32
}
}
#[test]
fn test() {
let s = "abcd".to_string();
let res = 0;
assert_eq!(Solution::longest_repeating_substring(s), res);
let s = "abbaba".to_string();
let res = 2;
assert_eq!(Solution::longest_repeating_substring(s), res);
let s = "aabcaabdaab".to_string();
let res = 3;
assert_eq!(Solution::longest_repeating_substring(s), res);
let s = "aaaaa".to_string();
let res = 4;
assert_eq!(Solution::longest_repeating_substring(s), res);
}
`````` | 476 | 1,585 | {"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} | 3.515625 | 4 | CC-MAIN-2022-33 | latest | en | 0.586564 |
https://electronics.stackexchange.com/questions/628404/antenna-pi-network-matching-high-or-low-pass | 1,709,029,992,000,000,000 | text/html | crawl-data/CC-MAIN-2024-10/segments/1707947474674.35/warc/CC-MAIN-20240227085429-20240227115429-00533.warc.gz | 237,506,518 | 41,402 | # Antenna pi-network matching High or low pass?
I am a beginner in RF system design and I have some questions.
I have a PCB trace antenna, I need to tune it to 2.45GHz, this is for Bluetooth low energy communication. I want to use a pi-network for the tuning process. I chose pi-network because it is selective and allows to set the quality factor Q of the circuit and the bandwidth (BW) of the antenna as Q=F/BW.
-Can someone tell me when to choose the pi-network high pass or low pass for antenna matching? I have seen in the document "AN1275: Impedance Matching Network Architectures" by Silabs that the high-pass network allows high frequencies to pass through the antenna and the low-pass pi-network blocks the passage of high frequencies through the antenna, which also means that the matching network must allow DC current to pass through the antenna.
What does it mean to allow DC current to pass through the antenna? In general, I know that the signal transmitted by the antenna is an AC signal. Could I feed (transmit) a DC signal to an antenna?
• I am confused because I read the DC on a document. I just want to know if I could use any of the configurations of pi-network (LCL OR CLC) for my matching. We could have a high pass or low pass pi-network. • Low Pass Configuration—Series Inductor and Parallel Capacitor CLC • High Pass Configuration—Series Capacitor and Parallel Inductor LCL What is the difference between the two pi-network systems? Jul 22, 2022 at 9:29
• I measure with NanoVNA a resonant frequency of 2.28GHz, the impedance is 47.8-7.06j. Here the imaginary part is negative that's means we are in the capacitance in the Smith chart. I have used SMA connector for the measurement so my source impedance is 50ohm. Am I responding to your question? Jul 22, 2022 at 9:54
• And I want to tune the antenna at 2.45GHz, with a Q=4.9 and a bandwidth of 500MHz and match the impedance. Jul 22, 2022 at 9:56
• Ok, I understand. I want to use an online calculator home.sandiego.edu/~ekim/e194rfs01/jwmatcher/matcher2.html And with the value of my source impedance, load impedance and the final resonant frequency, and the Q factor. Find the value of C,L I have to put to match my antenna. Jul 22, 2022 at 10:25
• I don't know if I could use directly the NanoVNA to find the value of L and C I have to put for the tuning Jul 22, 2022 at 10:26
What does it mean to allow DC current to pass through the antenna?
Unless you are talking about more specialized loop antennas, DC current won't flow into an antenna (dipole, monopole, variants thereof etc.). Even if it did, there would be no benefit because an antenna cannot transmit DC; it would just be a waste of power.
Could I feed (transmit) a DC signal to an antenna?
See above.
Antenna pi-network matching High or low pass?
I would tend to prefer a low-pass pi-network because it helps reduce unwanted harmonic content from a transmitter PA.
• Okay, thanks. So the difference between low-pass and high-pass is the fact that low-pass reduces unwanted harmonics. There are no rules for using low-pass or high-pass if I understand. So I'm going to use low-pass filters with Q=4.9, at 2.45GHz as resonant frequency and 500MHz as bandwidth. Jul 22, 2022 at 9:36
• Please note that the Q factor of a pi-network is a very ambiguous and fairly meaningless quantity; a pi-network (impedance matcher) has two Q factors so, don't expect that the Q factor stated in any network calculator will be meaningful when it comes to reshaping an antenna's natural bandwidth. Jul 22, 2022 at 9:39
• Can anyone see a reason why my answer has been downvoted? Jul 22, 2022 at 9:42
• Okay, thanks for your response. I am new on stack exchange, I don't see if your comment is downvoted or not. Jul 22, 2022 at 10:05
• I've found also this software SimSmith I think I'll use it. Jul 22, 2022 at 10:56
What does it mean to allow DC current to pass through the antenna? In general, I know that the signal transmitted by the antenna is an AC signal. Could I feed (transmit) a DC signal to an antenna?
Yes, it's not useful to pass a DC current into an antenna, but sometimes you intentionally superimpose DC on the feedline for the use of some other device, such as a mast-mounted preamplifier near the antenna. In such a case the question of whether the filter passes DC or not is relevant — if it does, then you can likely add the filter without trouble. If it doesn't, then you need to come up with a way to bypass the power around the filter.
In your case this doesn't seem to be a concern, so feel free to ignore the question of DC passing and choose based on other factors. | 1,152 | 4,632 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.65625 | 3 | CC-MAIN-2024-10 | longest | en | 0.928923 |
http://gradbot.blogspot.com/2010/06/priority-queue-heap-sort.html | 1,490,667,417,000,000,000 | text/html | crawl-data/CC-MAIN-2017-13/segments/1490218189589.37/warc/CC-MAIN-20170322212949-00536-ip-10-233-31-227.ec2.internal.warc.gz | 161,371,775 | 9,566 | Priority Queue & Heap Sort
Thursday, June 10, 2010 by gradbot
A heap implementation of priority queue is very similar to heap sort and can easily be turned into one. The difference implementation wise is that the queue doesn't know the final heap size so it has to build the heap from the ground up. This requires finding parent nodes in the heap and not just child nodes. Performance is nearly identical except the queue requires extra N space. Both of these heap implementations don't take advantage of partially sorted list.
```algorithms.queue.priority = function(predicate) {
var siftDown = function(array, start, end) {
var root = start;
while (root * 2 + 1 <= end) {
var child = root * 2 + 1;
if (child < end && predicate(array, child, child + 1)) {
child++;
}
if (predicate(array, root, child)) {
array.swap(root, child);
root = child;
} else {
return;
}
}
};
var siftUp = function(array, start, end) {
var child = end;
while (child > start) {
var parent = Math.floor((child - 1) / 2);
if (predicate(array, parent, child)) {
array.swap(parent, child);
child = parent;
} else {
return;
}
}
};
var that = {
heap: [],
that.heap.push(value);
siftUp(that.heap, 0, that.heap.length - 1);
},
remove: function() {
that.heap.swap(0, that.heap.length - 1);
siftDown(that.heap, 0, that.heap.length - 2);
return that.heap.pop();
}
};
return that;
};
algorithms.sort.heap_priority = function(array, predicate) {
var queue = algorithms.queue.priority(predicate);
var length = array.length;
// turn list into a heap
for (var i = 0; i < length; i++) {
}
// remove max value and restore heap property
for (var i = 0; i < length; i++) {
array[i] = queue.remove();
}
};
```
Filed under , , , having | 437 | 1,694 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.984375 | 3 | CC-MAIN-2017-13 | longest | en | 0.642392 |
https://atcoder.jp/contests/arc023/submissions/172899 | 1,618,651,856,000,000,000 | text/html | crawl-data/CC-MAIN-2021-17/segments/1618038118762.49/warc/CC-MAIN-20210417071833-20210417101833-00113.warc.gz | 216,245,529 | 6,038 | Contest Duration: - (local time) (90 minutes) Back to Home
Submission #172899
Source Code Expand
Copy
```#!/usr/bin/python
# -*- coding: utf-8 -*-
import sys
import math
NUM_MOD =1000000007
def get_sigma_fast(start, end, num):
# calcurate (end-start+num C end-start)
# = prod(xrange(num, end-start+num)) / prod(xrange(num)) / prod(xrange(end-start))
ans = 1
for i in xrange(num, end-start+num+1):
ans = (ans * i) % NUM_MOD
# according to Fermat,
# ( ans / prod(xrange(num)) ) % NUM_MOD
# = ( ans * prod(xrange(num)) ** (NUM_MOD-2) ) % NUM_MOD
num_permutaion = 1
for i in xrange(1, num+1):
num_permutaion = (num_permutaion*i) % NUM_MOD
ans = ans * pow(num_permutaion, NUM_MOD-2, NUM_MOD)
# according to Fermat, ...
range_permutation = 1
for i in xrange(1, end-start+1):
range_permutation = (range_permutation*i) % NUM_MOD
ans = ans * pow(range_permutation, NUM_MOD-2, NUM_MOD)
return ans
def get_sigma(start, end, num):
n_range = end-start+1
memo = [1 for i in xrange(n_range)]
for i in xrange(num-1):
memo = [sum(memo[:i+1])%NUM_MOD for i in xrange(n_range)]
continue
nexmemo = [memo[0]]
for item in memo[1:]:
nexmemo.append((nexmemo[-1] + item)%NUM_MOD)
memo = nexmemo
#print memo
return sum(memo)
#print gamemap
#print ([(y, lim-y) for lim in xrange(D%2, D+1, 2) for y in xrange(lim+1)])
# print ([gamemap[y][lim-y]
# for lim in xrange(D%2, D+1, 2)
# for y in xrange(lim+1)])
# print max([gamemap[y][lim-y]
# for lim in xrange(D%2, D+1, 2)
# for y in xrange(lim+1)
# if y < len(gamemap) and lim-y < len(gamemap[y])])
#print get_sigma(1, 3, 1)
#print get_sigma(1, 3, 2)
#print get_sigma(1, 3, 3)
last, now = 0, 0
is_counting = False
count = 0
ans = 1
for A in As:
if A == -1:
is_counting = True
count += 1
else:
now, last = A, now
if is_counting:
#print last, now, count
ans *= get_sigma_fast(last, now, count)
ans = ans % NUM_MOD
count = 0
is_counting = False
print ans
exit(0)
```
#### Submission Info
Submission Time 2014-05-18 22:43:44+0900 C - タコヤ木 hiking Python (2.7.3) 0 2288 Byte WA 2031 ms 3576 KB
#### Judge Result
Score / Max Score 0 / 0 0 / 50 0 / 30 0 / 20
Status
AC × 2 TLE × 1
AC × 6 WA × 8
AC × 10 WA × 16
AC × 10 WA × 17 TLE × 9
Set Name Test Cases
Sample sample_01.txt, sample_02.txt, sample_03.txt
Case Name Status Exec Time Memory
sample_01.txt AC 48 ms 3376 KB
sample_02.txt AC 48 ms 3380 KB
sample_03.txt TLE 2030 ms 3500 KB
subtask1_01.txt AC 48 ms 3324 KB
subtask1_02.txt AC 47 ms 3328 KB
subtask1_03.txt AC 48 ms 3328 KB
subtask1_04.txt WA 49 ms 3376 KB
subtask1_05.txt WA 50 ms 3380 KB
subtask1_06.txt WA 50 ms 3376 KB
subtask1_07.txt WA 48 ms 3328 KB
subtask1_08.txt AC 48 ms 3312 KB
subtask1_09.txt WA 47 ms 3376 KB
subtask1_10.txt WA 48 ms 3360 KB
subtask1_11.txt WA 50 ms 3328 KB
subtask1_12.txt WA 51 ms 3368 KB
subtask2_01.txt AC 48 ms 3384 KB
subtask2_02.txt AC 48 ms 3324 KB
subtask2_03.txt AC 48 ms 3328 KB
subtask2_04.txt WA 50 ms 3428 KB
subtask2_05.txt WA 50 ms 3448 KB
subtask2_06.txt WA 49 ms 3372 KB
subtask2_07.txt WA 50 ms 3380 KB
subtask2_08.txt AC 49 ms 3368 KB
subtask2_09.txt WA 50 ms 3452 KB
subtask2_10.txt WA 51 ms 3372 KB
subtask2_11.txt WA 52 ms 3372 KB
subtask2_12.txt WA 52 ms 3384 KB
subtask3_01.txt AC 48 ms 3328 KB
subtask3_02.txt TLE 2031 ms 3508 KB
subtask3_03.txt TLE 2030 ms 3492 KB
subtask3_04.txt TLE 2030 ms 3552 KB
subtask3_05.txt TLE 2029 ms 3432 KB
subtask3_06.txt TLE 2029 ms 3500 KB
subtask3_07.txt TLE 2029 ms 3500 KB
subtask3_08.txt AC 49 ms 3384 KB
subtask3_09.txt WA 52 ms 3372 KB
subtask3_10.txt TLE 2029 ms 3576 KB
subtask3_11.txt TLE 2030 ms 3500 KB
subtask3_12.txt TLE 2029 ms 3500 KB | 1,421 | 3,639 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.96875 | 3 | CC-MAIN-2021-17 | latest | en | 0.382151 |
https://www.ukbestwritings.com/repeat-problem-3-34-for-the-differential-equation-given-in-problem-3-35-problem-3-34-consider/ | 1,627,575,911,000,000,000 | text/html | crawl-data/CC-MAIN-2021-31/segments/1627046153860.57/warc/CC-MAIN-20210729140649-20210729170649-00546.warc.gz | 1,082,423,855 | 12,250 | # Repeat Problem 3-34 for the differential equation given in Problem 3-35. Problem 3-34 Consider…
Repeat Problem 3-34 for the differential equation given in Problem 3-35.
Problem 3-34
Consider the differential equation given in Problem 3-33. Use MATLAB to
(a) Find the partial-fraction expansion of Y(s) / R(s).
(b) Find the Laplace transform of the system.
(c) Find the output y(t) for t ≥ 0 when r(t) = us(t).
(d) Plot the step response of the system.
(e) Verify the final value that you obtained in Problem 3-33 part (f).
Problem 3-33
The differential equation of a linear system is
where y(t) is the output, and r(t) is the input.
(a) Write the state equation of the system. Define the state variables from right to left in ascending
order.
(b) Find the characteristic equation and its roots. Use MATLAB to find the roots.
(c) Find the transfer function Y(s) / R(s).
(d) Perform a partial-fraction expansion of Y(s) / R(s).
(e) Find the output y(t) for t ≥ 0 when r(t) = us(t).
(f) Find the final value of y(t) by using the final-value theorem.
Problem 3-35
Repeat Problem 3-33 for the following differential equation:
Problem 3-33
The differential equation of a linear system is
where y(t) is the output, and r(t) is the input.
(a) Write the state equation of the system. Define the state variables from right to left in ascending
order.
(b) Find the characteristic equation and its roots. Use MATLAB to find the roots.
(c) Find the transfer function Y(s) / R(s).
(d) Perform a partial-fraction expansion of Y(s) / R(s).
(e) Find the output y(t) for t ≥ 0 when r(t) = us(t).
(f) Find the final value of y(t) by using the final-value theorem. | 438 | 1,676 | {"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} | 3.515625 | 4 | CC-MAIN-2021-31 | latest | en | 0.791585 |
http://stock-investing.net/stock-investment-how-much-social-security-will-i-get-at-62-the-motley-fool/ | 1,519,183,917,000,000,000 | text/html | crawl-data/CC-MAIN-2018-09/segments/1518891813322.19/warc/CC-MAIN-20180221024420-20180221044420-00235.warc.gz | 357,034,610 | 10,305 | # Stock investment
You can choose to collect Social Security benefits as early as age 62, and millions of Americans do just that — but filing for Social Security retirement benefits as early as possible will result in your monthly payments being reduced by 25% to 30% from your full retirement benefit calculation.
Here’s what you need to know about how your Social Security benefit is calculated, the effects of filing at age 62, and how work can affect your benefits.
## Calculating your primary insurance amount
Your primary insurance amount refers to the amount of your monthly Social Security benefit if you start collecting at your normal, or full, retirement age.
In a nutshell, your primary insurance amount takes an inflation-indexed average of your highest 35 years of earnings, up to each year’s maximum Social Security taxable income. Once this amount has been determined, a formula is applied to your monthly average. As of 2016, the benefit-determining formula is:
• 90% of the first \$856
• 32% of the amount greater than \$856 and up to \$5,157
• 15% of the amount greater than \$5,157
This is how your first-year benefit is determined. In subsequent years, you will receive a cost-of-living adjustment (COLA) to compensate for inflation.
## The effect of early retirement
Retiring before your normal, or full, retirement age has the effect of permanently reducing your retirement benefit, so it’s important to know how this works.
First of all, the reduction depends on your normal retirement age, which is 66 for those reaching retirement age now, and is about to gradually increase to 67. Your normal retirement age depends on your birth year, and I’ll include a full list below.
As far as the benefit reduction goes, there are two guidelines you need to know:
• Your benefit will be reduced by six and two-thirds percent per year you claim benefits early, up to three years before your full retirement age. On a monthly basis, the reduction rate translates to five-ninths of a percent.
• Beyond three years early, your retirement benefits will be further reduced at a rate of five percent per year, or five-twelfths of a percent per month.
So, these reduction rates have different effects depending on your full retirement age. Here’s a table that can help you determine your full retirement age and the effect of starting benefits at age 62.
If you were born in…
Your full retirement age is…
Reduction % for starting benefits at age 62
1954 or earlier
66 years
25%
1955
66 years, 2 months
25.83%
1956
66 years, 4 months
26.66%
1957
66 years, 6 months
27.5%
1958
66 years, 8 months
28.33%
1959
66 years, 10 months
29.17%
1960
67 years
30%
For example, if you were born in 1956 and your primary insurance amount is \$1,500, your retirement benefit would be reduced by 26.66% if you file as soon as possible, which would result in a monthly benefit of \$1,100.
You can estimate your primary insurance amount and the effect of early retirement by creating an account on www.ssa.gov and viewing your latest Social Security statement. Or, you can get a quick estimate of your (and your spouse’s) Social Security benefits for age 62 and any other theoretical retirement age with this calculator.
## Do you still work? Your benefit could be reduced even further
Another factor to consider is whether you’re still working when you file for Social Security benefits at age 62. Before you reach full retirement age, your earnings can reduce or take away your benefits entirely.
As of 2016, if you file for Social Security at 62, the first \$15,720 you earn from working will have no effect on your benefits. Beyond that threshold, however, your benefits will be reduced by \$1 for every \$2 in excess earnings.
Now, it’s important to understand that there’s a potential positive later on that offsets the immediate reduction in benefits. If your benefit is reduced because of your earnings, you’ll be treated later on as if you had waited longer to claim your benefits. That will result in slightly higher benefit checks down the road. Here’s a complete discussion of how the Social Security earnings test works, which can tell you what you need to know if you plan to work after you file for benefits.
## Early Social Security isn’t the best idea for everyone
Before deciding to file for Social Security at 62, it’s important to understand exactly what this will do to your retirement benefits for the rest of your life. Once you know how much you can expect to receive, you can weigh the pros and cons and make the best decision for yourself and your family.
The \$15,834 Social Security bonus most retirees completely overlook
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Source link | 1,131 | 5,213 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.53125 | 3 | CC-MAIN-2018-09 | latest | en | 0.95037 |
https://math.stackexchange.com/questions/1939231/mathbbr-with-the-discrete-topology-is-not-locally-euclidean | 1,721,513,197,000,000,000 | text/html | crawl-data/CC-MAIN-2024-30/segments/1720763517541.97/warc/CC-MAIN-20240720205244-20240720235244-00264.warc.gz | 338,243,143 | 37,004 | # $\mathbb{R}$ with the discrete topology is not locally Euclidean
Proposition: $\mathbb{R}$ with the discrete topology is not locally euclidean.
The proof begins with the conditional statement that if $U \subseteq \mathbb{R}^{n}$ is a non-empty open set in the standard topology and $V \subseteq \mathbb{R}$ is a non-empty set in the discrete topology, U and V are not homeomorphic.
I am unable to understand why U and V are not homeomorphic even from the definition of "Homeomorphism". Is it because the map between the two topological spaces are not continuous?
Thanks in advance for any clarification.
• Blast! typo. You're right. My doubt still holds. Commented Sep 24, 2016 at 7:59
• It's not clear what you mean by "Euclidean". ${\mathbb R}$, or ${\mathbb R}^n$, with the standard scalar product remains a euclidean vector space even if you decide to adopt a topology of your own choice. – That the discrete topology is different from the "euclidean topology" is obvious. Commented Sep 24, 2016 at 8:10
• $\Bbb R$ with discrete topology is locally just a point, which is homeomorphic to the zero-dimensional euclidean vector space $\Bbb R^0$ Commented Sep 24, 2016 at 8:16
• I don't think the proposition as stated is true. A discrete space is (pretty much by definition) a zero-dimensional manifold. What is true is that it is not of positive dimension. You could restate being locally Euclidean to mean that the dimension is positive, but that would make statements of lots of theorems in differential topology terribly awkward. Commented Sep 24, 2016 at 8:18
• To show that, you can just note that no Euclidean space of positive dimension is locally discrete. Commented Sep 24, 2016 at 8:20
Suppose $f : U \to V$ is a homeomorphism. Then $f$ is a continuous bijection. Now for any $x \in V$, the set $\{x\}$ is open in the discrete topology. Since $f^{-1}[\{x\}]$ is open (by continuity) and is a single element of $U$ (since $f$ injective).
However, single points are not open in the Euclidean topology, so we get a contradiction. Therefore there does not exist such a homeomorphism $f$.
Let $U$ be a non-empty open set in $\Bbb R^n$. Suppose that there is a homeomorphism from $U$ to some $D\subseteq\Bbb R$ with the discrete topology.
Fix $p\in U$. There is an $\epsilon>0$ such that $B(p,\epsilon)\subseteq U$. $B(p,\epsilon)$ is a connected set with more than one point, so $h[B(p,\epsilon)]$ must be a connected subset of $D$ with more than one point. But $D$ is discrete, so no subset of $D$ with more than one point is connected. Thus, there cannot be such a homeomorphism $h$. | 693 | 2,601 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 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} | 3.0625 | 3 | CC-MAIN-2024-30 | latest | en | 0.944956 |
https://slidetodoc.com/discrete-mathematics-chapter-7-advanced-counting-techniques-lingling/ | 1,701,410,286,000,000,000 | text/html | crawl-data/CC-MAIN-2023-50/segments/1700679100276.12/warc/CC-MAIN-20231201053039-20231201083039-00137.warc.gz | 603,253,415 | 26,763 | # Discrete Mathematics Chapter 7 Advanced Counting Techniques Lingling
• Slides: 56
Discrete Mathematics Chapter 7 Advanced Counting Techniques 大葉大學 資訊 程系 黃鈴玲(Lingling Huang)
Outline n n n 7. 1 Recurrence Relations 7. 2 Solving Linear Recurrence Relations 7. 4 Generating Functions 7. 5 Inclusion-Exclusion 7. 6 Applications of Inclusion-Exclusion Ch 9 -2
7. 1 Recurrence Relations(遞迴關係 ) Example 1. Let {an} be a sequence that satisfies the recurrence relation an=an-1 -an-2 for n=2, 3, …, and suppose that a 0=3, and a 1=5. Here a 0=3 and a 1=5 are the initial conditions. By the recurrence relation, a 2 = a 1 -a 0 = 2 a 3 = a 2 -a 1 = -3 a 4 = a 3 -a 2 = -5 : Q 1: Applications ? Q 2: Are there better ways for computing the terms of {an}? Ch 7 -3 n
※Modeling with Recurrence Relations We can use recurrence relations to model (describe) a wide variety of problems. Example 3. Compound Interest (複利) Suppose that a person deposits(存款) \$10000 in a saving account at a bank yielding 11% per year with interest compounded annually. How much will be in the account after 30 years ? Sol : Let Pn denote the amount in the account after n years. P 0=10000 Pn=Pn-1 + 0. 11 Pn-1=1. 11 Pn-1, ∴ P 30=1. 11 P 29=(1. 11)2 P 28=…=(1. 11)30 P 0 =228922. 97 Ch 7 -4
Example 5. (The Tower of Hanoi) The rules of the puzzle allow disks to be moved one at a time from one peg to another as long as a disk is never placed on top of a smaller disk. Let Hn denote the number of moves needed to solve the Tower of Hanoi problem with n disks. Set up a recurrence relation for the sequence {Hn}. 目標 : n 個disk都從 peg 1 移到 peg 2 H 4 moves peg 1 Sol : peg 2 peg 3 Hn=2 Hn-1+1, H 1=1 ( n-1個 disk 先從peg 1→peg 3, 第 n 個 disk 從 peg 1→peg 2, n-1個 disk 再從 peg 3→peg 2) Ch 7 -5
Example 6. Find a recurrence relation and give initial conditions for the number of bit strings of length n that do not have two consecutive 0 s. How many such bit strings are there of length 5 ? Sol : Let an be the number of bit strings of length n that do not have two consecutive 0 s. 1 2 … n-3 n-2 n-1 1 an-1 種 an-2 種 n 1 0 ∴ an = an-1+an-2, n 3 a 1=2 (string : 0, 1) a 2=3 (string : 01, 10, 11) ∴ a 3=a 2+a 1=5, a 4=8, a 5=13 Ch 7 -7
Example 7. (Codeword enumeration) A computer system considers a string of decimal digits a valid codeword if it contains an even number of 0 digits. Let an be the number of valid n-digit codewords. Find a recurrence relation for an. Sol : 1 2 3 … n-1 an-1 種 10 n-1 - an-1 種 n 1~9 ∴ an = 9 an-1 + (10 n-1 -an-1) = 8 an-1 + 10 n-1 , n 2 a 1 = 9 0 Ch 7 -8
7. 2 Solving Recurrence Relations Def 1. A linear homogeneous recurrence relation of degree k (i. e. , k terms) with constant coefficients is a recurrence relation of the form an = c 1 an-1+c 2 an-2+…+ckan-k where ci R and ck≠ 0 Example 1 and 2. fn = fn-1 + fn-2 an = an-5 an = an-1 + an-22 an = nan-1 Hn = 2 Hn-1 + 1 (True, deg=2) (True, deg=5) (False, not linear) (False, not homogeneous) Ch 7 -10
Solving Linear Homogeneous Recurrence Relations with Constant Coefficients Theorem 1. Let an = c 1 an-1+ c 2 an-2 be a recurrence relation with c 1, c 2 R. If r 2 - c 1 r - c 2= 0 (稱為characteristic equation) has two distinct roots r 1 and r 2. Then the solution of an is an = a 1 r 1 n + a 2 r 2 n , for n=0, 1, 2, …, where a 1 , a 2 are constants. (a 1 , a 2可利用 a 0, a 1算出) Ch 7 -11
Example 3. What’s the solution of the recurrence relation an = an-1 + 2 an-2 with a 0=2 and a 1=7 ? Sol : The characteristic equation is r 2 – r - 2=0. Its two roots are r 1= 2 and r 2 = -1. r 與r 順序可交換， Hence an=a 1 2 n +a 2 (-1)n. 結果會一樣 ∵a 0 = a 1+a 2 = 2, a 1=2 a 1 -a 2=7 ∴a 1 = 3, a 2 = -1 驗算：a 2 = a 1 + 2 a 0 =11 a 2= 3 22 - 1 =11 an = 3 2 n - (-1)n. 1 2 Ch 7 -12
Example 4. Find an explicit formula for the Fibonacci numbers. Sol : fn = fn-1 + fn-2 , n 2, f 0=0 , f 1=1. The characteristic equation is r 2 - r - 1=0. Its two roots are So we have , . Ch 7 -13
Thm 2. Let an = c 1 an-1+c 2 an-2 be a recurrence relation with c 1, c 2 R. If r 2 - c 1 r - c 2= 0 has only one root r 0. Then the solution of an is an = a 1 r 0 n + a 2 n r 0 n for n=0, 1, 2, …, where a 1 and a 2 are constants. Ch 7 -14
Example 5. What’s the solution of an= 6 an-1 - 9 an-2 with a 0=1 and a 1=6 ? Sol : The root of r 2 - 6 r + 9 = 0 is r 0 = 3. Hence an = a 1．3 n +a 2．n．3 n. ∵a 0 = a 1 = 1 a 1 = 3 a 1 + 3 a 2 = 6 ∴ a 1 = 1 and a 2 = 1 驗算：a 2 = 6 a 1 - 9 a 0 =27 n n an = 3 + n．3 a 2= 32 +2 32 =27 Ch 7 -15
Thm 3. Let an = c 1 an-1 + c 2 an-2 + … + ckan-k be a recurrence relation with c 1, c 2, …, ck R. If rk - c 1 rk-1 - c 2 rk-2 -…- ck = 0 has k distinct roots r 1, r 2, …, rk. Then the solution of an is an = a 1 r 1 n +a 2 r 2 n + …+akrkn, for n = 0, 1, 2, … where a 1, a 2, …ak are constants. Ch 7 -16
Example 6 (k = 3) Find the solution of an = 6 an-1 - 11 an-2 + 6 an-3 with initial conditions a 0=2, a 1=5 and a 2=15. Sol : The roots of r 3 - 6 r 2 + 11 r – 6 = 0 are r 1 = 1, r 2 = 2, and r 3 = 3 ∴an = a 1 1 n + a 2 2 n + a 3 3 n ∵a 0 = a 1 + a 2 + a 3 = 2 a 1 = 1, a 2 = -1, a 1 = a 1 + 2 a 2 + 3 a 3 = 5 a 3 = 2 a 2 = a 1 + 4 a 2 + 9 a 3 = 15 0 =47 ∴an = 1 - 2 n + 2 3 n 驗算：a 3 = 6 a 2 -311 a 1+ 6 a 3 a 3= 1 - 2 + 2 3 =47 Ch 7 -17
Thm 4. Let an = c 1 an-1 + c 2 an-2 + … + ckan-k be a recurrence relation with c 1, c 2, …, ck R. If rk - c 1 rk-1 - c 2 rk-2 - … - ck = 0 has t distinct roots r 1, r 2, …, rt with multiplicities m 1, m 2, …, mt respectively, where mi 1, i, and m 1+ m 2 +…+ mt = k, then where ai, j are constants for 1 i t and 0 j mi-1. Ch 7 -18
Example 8. Find the solution to the recurrence relation an = -3 an-1 - 3 an-2 - an-3 with initial conditions a 0 = 1, a 1 = -2 and a 2 = -1. Sol : r 3 + 3 r 2 + 3 r + 1 = 0 has a single root r 0 = -1 of multiplicity three. ∴ an = (a 1+a 2 n+a 3 n 2) r 0 n = (a 1+a 2 n+a 3 n 2)(-1)n ∵ a 0 = a 1 = 1 a 1 = (a 1+a 2+a 3) (-1) = -2 a 2 = a 1+2 a 2+4 a 3 = -1 驗算：a 3 = - 3 a 2 - 3 a 1 - a 0 =8 a 3= (1+3 3 -2 32) (-1)3 =8 ∴a 1 = 1, a 2 = 3, a 3 = -2 an = (1+3 n-2 n 2) (-1)n Exercise : 3, 15, 19 Ch 7 -20
Linear Nonhomogeneous Recurrence Relations with Constant Coefficients Example: an = 3 an-1 + 2 n A recurrence relation of the form an = c 1 an-1 + c 2 an-2 + … + ckan-k + F(n), where c 1, c 2, …, ck are real numbers and F(n) is a function not identically zero depending only on n. The recurrence relation an = c 1 an-1 + c 2 an-2 + … + ckan-k is called the associated homogeneous recurrence relation. Ch 7 -21
Example 9: an = an-1 + 2 n, associated h. r. r an = an-1 + an-2 + n 2+1, associated h. r. r an = an-1 + an-2 an = 3 an-1 + n 3 n, associated h. r. r an = 3 an-1 an = an-1 + an-3 + n!, associated h. r. r an = an-1 + an-3 Ch 7 -22
Theorem 5. If {an (p)} is a particular solution (特解) of an = c 1 an-1 + c 2 an-2 + … + ckan-k + F(n), then every solution is of the form {an (p) + an (h)}, where {an (h)} is a solution of an = c 1 an-1 + c 2 an-2 + … + ckan-k Proof. If {an (p)} and {bn} are both solutions of an = c 1 an-1 + c 2 an-2 + … + ckan-k + F(n), then an(p) = c 1 an-1(p) + c 2 an-2 (p)+ … + ckan-k(p) + F(n), and bn = c 1 bn-1 + c 2 bn-2 + … + ckbn-k + F(n). an(p) - bn = c 1(an-1 - bn-1) + c 2(an-2 - bn-2) + … + ck(an-k - bn-k) {an(p) - bn} is a solution of an = c 1 an-1 + c 2 an-2 + … + ckan-k bn = an(p) + an (h) Ch 7 -23
Example 10. Find all solutions of the recurrence relation an = 3 an-1 + 2 n. What is the solution with a 1=3? Sol : {associated的部分 an = 3 an-1 先解} Characteristic equation: r – 3 = 0 r = 3 an(h) = a 3 n. {particular solution} ∵ F(n) = 2 n ∴ Let an(p) =cn+d, where c, d R. If an(p) = cn+d is a solution to an = 3 an-1 + 2 n, then cn+d = 3(c(n-1)+d)+2 n =3 cn - 3 c + 3 d +2 n 2 cn-3 c + 2 d +2 n = (2 c+2)n + (2 d-3 c) = 0 (任何n代入都需為 0) ∴ 2 c+2 = 0, and 2 d-3 c = 0 c = -1, d = -3/2 an(p) = -n - 3/2 an = an(h) +an(p) = a 3 n-n - 3/2 If a 1= a 3 -1 - 3/2 = 3 a = 11/6 an = (11/6) 3 n- n - 3/2 Ch 7 -24
Example 11. Find all solutions of the recurrence relation an = 5 an-1 -6 an-2 + 7 n. Sol : {associated的部分 an = 5 an-1 -6 an-2 先解} Characteristic equation: r 2 – 5 r + 6 = 0 r 1 = 3, r 2 = 2 an(h) = a 1 3 n + a 2 2 n. {particular solution} ∵ F(n) = 7 n ∴ Let an(p) = c 7 n, where c R. If an(p) = c 7 n is a solution to an = 5 an-1 -6 an-2 + 7 n, then c 7 n = 5 c 7 n-1 - 6 c 7 n-2 + 7 n 49 c = 35 c - 6 c + 49 Exercise : 23 c = 49/20 an(p) = (49/20) 7 n an = an(h) +an(p) = a 1 3 n + a 2 2 n + (49/20) 7 n Ch 7 -25
Theorem 6. an = c 1 an-1 + c 2 an-2 + … + ckan-k + F(n), where F(n) = (btnt + bt-1 nt-1 +…+ b 1 n + b 0)sn. When s is not a root of the characteristic equation of the associated linear homogeneous recurrence relation, there is a particular solution of the form (ptnt + pt-1 nt-1 +…+ p 1 n + p 0)sn. When s is a root of the characteristic equation and its multiplicity is m, there is a particular solution of the form nm(ptnt + pt-1 nt-1 +…+ p 1 n + p 0)sn. Ch 7 -26
Example 12. What form does a particular solution of the linear nonhomogeneous recurrence relation an = 6 an-1 - 9 an-2 + F(n) have when F(n) =3 n, F(n) =n 22 n , and F(n) = (n 2+1)3 n. Sol : The associated linear homogeneous recurrence relation is an = 6 an-1 - 9 an-2. characteristic equation: r 2 - 6 r + 9 = 0 r = 3 (2重根) F(n) =3 n, and 3 is a root an(p) = p 0 n 23 n F(n) =n 3 n, and 3 is a root an(p) = n 2(p 1 n+p 0) 3 n F(n) =n 22 n , and 2 is not a root an(p) = (p 2 n 2+p 1 n+p 0)2 n F(n) = (n 2+1)3 n , and 3 is a root an(p) = n 2 (p 2 n 2+p 1 n+p 0) 3 n Exercise : 27 Ch 7 -27
Example 13. Find the solutions of the recurrence relation an = an-1 + n with a 1=1. Sol : The associated linear homogeneous recurrence relation is an = an-1. characteristic eq. : r - 1 = 0 r = 1 an(h) = c(1)n=c F(n) = n(1)n, and 1 is a root an(p) = n(p 1 n+p 0)1 n = p 1 n 2+p 0 n 將an(p)代入an = an-1 + n p 1 n 2+p 0 n = p 1(n-1)2+p 0(n-1)+n (2 p 1 -1)n+p 0 -p 1=0 p 1= ½, an(p) = (n 2+n)/2 an = an(p) + an(h) = (n 2+n)/2+c p 0 = p 1= ½ Exercise : 29 a 1=1 c=0 an = an(p) + an(h) = (n 2+n)/2 Ch 7 -28
ex 40: Solve the simultaneous recurrence relations an = 3 an-1 + 2 bn-1 bn = an-1 + 2 bn-1 with a 0 = 1 and b 0 = 2. Sol : an -bn= 2 an-1 bn = an-2 an-1 an = 3 an-1 + 2 bn-1 = 3 an-1 + 2 an-1 - 4 an-2 an = 5 an-1 - 4 an-2 r 2 - 5 r + 4 = 0 r = 1, 4 an = a 1+a 24 n a 0 = a 1+a 2 = 1 a 1 = a 1+4 a 2 = 3 a 0 + 2 b 0 = 7 a 1 = -1, a 2 = 2 an= 2 4 n-1 bn = an-2 an-1= 2 4 n-1 -4 n+2 = 4 n+1 Ch 7 -29
7. 4 Generating Functions. Def 1. The generating function for the sequence a 0, a 1, a 2, … of real numbers is the infinite series G(x) = a 0 + a 1 x +… + anxn +… = (若數列{an} 是finite，可視為是infinite，但後面的項都 等於 0) Ch 7 -30
Example 1. Find the generating functions for the sequences {ak} with (1) ak= 3 (2) ak = k+1 (3) ak = 2 k Sol : (1) G(x) = (2) G(x) = (3) G(x) = Ch 7 -31
Example 2. What is the generating function for the sequence 1, 1, 1, 1 ? a 0 a 1 a 2 a 3 a 4 a 5 a 6及之後都=0 Sol : G(x) = = a 0 + a 1 x + a 2 x 2 + a 3 x 3 +… (expansion，展開式) (closed form) Exercise : 2 Ch 7 -32
Example 3. Let m Z+ and , for k = 0, 1, …, m. What is the generating function for the sequence a 0, a 1, …, am ? Sol : G(x) = a 0 + a 1 x + a 2 x 2 + … + amxm = (1+x)m (by下面的二項式定理) Ch 7 -33
Useful Facts About Power Series Example 4. The function f (x) = is the generating function of the sequence 1, 1, 1, …, because when |x| < 1. = 1 + x 2 + …= Example 5. The function f (x) = is the generating function of the sequence 1, a, a 2, …, because = 1 + ax + a 2 x 2 + …= when |ax| < 1 for a≠ 0. Exercise : 5(a)(b), 11(a) Ch 7 -34
Theorem 1. Let f(x) = and g(x) = Then f(x) + g(x) = . . f(x) g(x) = (a 0+a 1 x+a 2 x 2 +…)(b 0+b 1 x +b 2 x 2+…) = (a 0 b 0)+(a 0 b 1+a 1 b 0) x+(a 0 b 2+a 1 b 1+a 2 b 0) x 2+… = Ch 7 -35
Example 6. Let f (x) = . Use Example 4 to find the coefficients a 0, a 1, a 2, … in the expansion f (x)= . Sol : = 1 + x 2 + … = = ak = k+1 Ch 7 -36
Def 2. Let u R and k N. Then the extended binomial coefficient Example 7. Find Sol : is defined by and Ch 7 -37
Example 8 When the top parameter is a negative integer, the extended binomial coefficient can be expressed in terms of an ordinary binomial coefficient. Ch 7 -38
Thm 2. (The Extended Binomial Theorem) Let x R with |x|<1 and let u R, then Ch 7 -39
Example 9. (跳過) Find the generating functions for (1+x)-n and (1 -x)-n where n Z+ Sol : By the Extended Binomial Theorem, By replacing x by –x we have Exercise : 11(b)(d) Ch 7 -40
Counting Problems and Generating Functions Generating functions can be used to count the number of combinations of various types. Example 10. Find the number of solutions of e 1 + e 2 + e 3 = 17, where e 1, e 2, e 3 are integers with 2 e 1 5, 3 e 2 6, and 4 e 3 7. Sol : The number of solutions with the indicated constraints is the coefficient of x 17 in the expansion of (x 2 + x 3 + x 4 + x 5)(x 3 + x 4 + x 5 + x 6)(x 4 + x 5 + x 6 + x 7) (即相當於找 e 1, e 2, e 3 使 xe xe xe = x 17) 1 2 3 (e 1, e 2, e 3)=(4, 6, 7), (5, 5, 7), (5, 6, 6) 共 3種 Ch 7 -41
Example 11. In how many different ways can eight identical cookies be distributed among three distinct children if each child receives at least two cookies and no more than four cookies? Sol : The number of solutions is the coefficient of x 8 in the expansion of (x 2 + x 3 + x 4)3 (c 1, c 2, c 3) = (2, 2, 4), (2, 3, 3), (2, 4, 2), (3, 2, 3), (3, 3, 2), (4, 2, 2) 共 6種 Exercise: 23 Ch 7 -42
※Using Generating Functions to solve Recurrence Relations. Example 16. Solving the recurrence relation ak = 3 ak-1 for k=1, 2, 3, … and initial condition a 0 = 2. Sol : 另法： (by 7. 2節Thm 1公式) r – 3 = 0 r = 3 an = a 3 n ∵ a 0 = 2 = a ∴ an = 2 3 n Ch 7 -43
Let generating function for {ak}. First note that ak = 3 ak-1 be the 從k = 1開始，以避免ak-1變成a-1 G(x) - a 0 = 3 x G(x) ∵a 0 = 2 G(x) - 3 x G(x) = G(x)(1 -3 x) = 2 ∴ ak = 2 3 k Exercise : 5, 7, 11, 33 Ch 7 -44
Example 17 Solving ak = 8 ak-1 +10 k-1 for k =1, 2, 3, … and initial condition a 1 = 9. (Sec. 7. 1 Example 7) Sol : Let a 0 = 1 (為計算方便而假設). Let be the generating function for {ak}. Ch 7 -45
∴ ak = (10 k + 8 k)/2 Exercise: 33 Ch 7 -46
7. 5 Inclusion-Exclusion 排容原理 A, B, C, D : sets A |A|+|B|+|C| 時 各部分被計算的次數 1 2 1 B 1 2 3 1 0 2 1 21 1 C -|A B|-|A C|-|B C|後 +|A B C|後 Ch 7 -47
Theorem 1. A 1, A 2, …, An : sets Exercise : 17 Ch 7 -48
7. 6 Applications of Inclusion and Exclusion Example 1. How many solutions does x 1 + x 2 + x 3 = 11 have, where x 1, x 2, x 3 are nonnegative integers with x 1 3, x 2 4, and x 3 6? Sol : Let a solution have property P 1 if x 1 4, property P 2 if x 2 5, and property P 3 if x 3 7. N(P 1’P 2’P 3’) = N - N(P 1) - N(P 2) - N(P 3) + N(P 1 P 2) + N(P 2 P 3) + N(P 1 P 3) - N(P 1 P 2 P 3) Ch 7 -49
The Number of Onto Functions Example 2. How many onto functions are there form set A={1, 2, 3, 4, 5, 6} to set B={a, b, c} ? Sol : f : A → B f (1)= f (2)= ︰ ︰ f (6)= {a, b, c} 不同的填法造出不同的函數 如何使a, b, c都出現 ? The number of onto functions = (所有函數個數) - (a, b, c中有一個沒被對應) + (a, b, c中二個沒被對應) - (a, b, c都沒被對應) = Ch 7 -50
Thm 1. |A| = m , |B| = n There are onto functions f : A → B. pf : A = {a 1, a 2, …, am}. B = {b 1, b 2, …, bn} f (a 1)= b 1, b 2, …, bn f (a 2)= ︰ f (am)= ︰ Exercise : 8 Ch 7 -51
Example 3. How many ways are there to assign five different jobs to four different employees if every employee is assigned at least one job? Sol : Consider the assignment of jobs as a function from the set of five jobs to the set of four employees. The number of onto functions Exercise : 9 Ch 7 -52
Derangements (亂序) Def. A derangement is a permutation of objects that leaves no object in its original position. Example 5. The permutation 21453 is a derangement of 12345. But 21543 is not a derangement of 12345. Ch 7 -53
Def. Let Dn be the number of derangements of n objects. D 2 = 1 because the derangements of 12 are 21. D 3 = 2 because the derangements of 123 are 231 and 312. D 4 = (所有4個元素的permutation數) - (4個元素有一個在原位置的permutation數) + (4元素中有二個在原位置的permutation個數) - (4個元素中有三個在原位置的permutation個數) + (4元素都在原位置的permutation個數) = Ch 7 -54
Theorem 2. (亂序公式) Proof. Exercise : 13 Ch 7 -55
Exercise 17. How many ways can the digits 0, 1, 2, 3, 4, 5, 6, 7, 8, 9 be arranged so that even digit is in its original position? Sol : Ch 7 -56 | 7,100 | 16,074 | {"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} | 4.65625 | 5 | CC-MAIN-2023-50 | latest | en | 0.774679 |
https://mathoverflow.net/questions/321408/what-is-the-method-to-relax-or-weakening-a-structure-ripping-off-from-it-all-i | 1,603,679,454,000,000,000 | text/html | crawl-data/CC-MAIN-2020-45/segments/1603107890108.60/warc/CC-MAIN-20201026002022-20201026032022-00168.warc.gz | 430,327,020 | 28,775 | # What is the method to relax or weakening a structure “ripping off” from it all its identity elements?
In mathematics, an identity element or neutral element is a special type of element of a set with respect to a binary operation on that set, which leaves other elements unchanged when combined with them
For example, we know that lattice is an algebraic structure, with the binary operations $$\wedge$$ and $$\vee$$ and the constants $$\top$$ and $$\bot$$. (These correspond, respectively, to binary and nullary meets and joins in the poset-theoretic definition; accordingly, they are read ‘meet’, ‘join’, ‘top’, and ‘bottom’.) Here are the axioms for these operations:
$$\wedge$$ and $$\vee$$ are each idempotent, commutative, and associative;
the absorption laws: $$a \vee (a \wedge b) = a$$, and $$a \wedge (a \vee b) = a$$;
$$\top$$ and $$\bot$$ are the respective identities of $$\wedge$$ and $$\vee$$.
I want to 'remove' identities of ∧ and ∨ like a sort of de-identity of $$\wedge$$ and $$\vee$$ operations to have a de-identity operations.
I'm interested in being able to insert them in a second moment because I want a structure without operations identities not without operations!, or, in general, a structure without identities but not without a structure!
Is there a method to relax or weaken a structure "ripping off" from it all its identity elements that are present respect to a binary operation on that set?
• You mean like en.wikipedia.org/wiki/Semigroup , for instance—a general definition of a given structure without the requirement for identity element(s)? Or do you mean given an actual algebraic structure, remove the identity element(s) and then still think about that structure? You can take a lattice, remove its top and bottom, and then the resulting set still has at least partially-defined meet and join. – David Roberts Jan 21 '19 at 19:40
• As an example, in the method of forcing in set theory, people often consider a complete Boolean algebra $B$, as a partially-ordered set, and then remove the bottom element $\bot$ to get the partially ordered set $B\setminus \bot$ and then use that. – David Roberts Jan 21 '19 at 19:42 | 522 | 2,167 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 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": 14, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.65625 | 4 | CC-MAIN-2020-45 | latest | en | 0.899866 |
https://www.tensorflow.org/api_docs/cc/class/tensorflow/ops/exp?hl=hr | 1,653,658,951,000,000,000 | text/html | crawl-data/CC-MAIN-2022-21/segments/1652662647086.91/warc/CC-MAIN-20220527112418-20220527142418-00542.warc.gz | 1,159,960,077 | 22,835 | Google I/O is a wrap! Catch up on TensorFlow sessions
# tensorflow::ops::Exp
#include <math_ops.h>
Computes exponential of x element-wise.
## Summary
$$y = e^x$$.
This function computes the exponential of every element in the input tensor. i.e. exp(x) or e^(x), where x is the input tensor. e denotes Euler's number and is approximately equal to 2.718281. Output is positive for any real input.
x = tf.constant(2.0)
tf.math.exp(x) ==> 7.389056
x = tf.constant([2.0, 8.0])
tf.math.exp(x) ==> array([7.389056, 2980.958], dtype=float32)
For complex numbers, the exponential value is calculated as follows:
e^(x+iy) = e^x * e^iy = e^x * (cos y + i sin y)
Let's consider complex number 1+1j as an example. e^1 * (cos 1 + i sin 1) = 2.7182818284590 * (0.54030230586+0.8414709848j)
x = tf.constant(1 + 1j)
tf.math.exp(x) ==> 1.4686939399158851+2.2873552871788423j
Args:
Returns:
• Output: The y tensor.
### Constructors and Destructors
Exp(const ::tensorflow::Scope & scope, ::tensorflow::Input x)
### Public attributes
operation
Operation
y
::tensorflow::Output
### Public functions
node() const
::tensorflow::Node *
operator::tensorflow::Input() const
operator::tensorflow::Output() const
## Public attributes
### operation
Operation operation
### y
::tensorflow::Output y
## Public functions
### Exp
Exp(
const ::tensorflow::Scope & scope,
::tensorflow::Input x
)
### node
::tensorflow::Node * node() const
### operator::tensorflow::Input
operator::tensorflow::Input() const
### operator::tensorflow::Output
operator::tensorflow::Output() const
[{ "type": "thumb-down", "id": "missingTheInformationINeed", "label":"Missing the information I need" },{ "type": "thumb-down", "id": "tooComplicatedTooManySteps", "label":"Too complicated / too many steps" },{ "type": "thumb-down", "id": "outOfDate", "label":"Out of date" },{ "type": "thumb-down", "id": "samplesCodeIssue", "label":"Samples / code issue" },{ "type": "thumb-down", "id": "otherDown", "label":"Other" }]
[{ "type": "thumb-up", "id": "easyToUnderstand", "label":"Easy to understand" },{ "type": "thumb-up", "id": "solvedMyProblem", "label":"Solved my problem" },{ "type": "thumb-up", "id": "otherUp", "label":"Other" }] | 649 | 2,225 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 1, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.734375 | 3 | CC-MAIN-2022-21 | latest | en | 0.569251 |
https://jp.mathworks.com/matlabcentral/cody/problems/528-find-the-largest-value-in-the-3d-matrix/solutions/336796 | 1,607,119,936,000,000,000 | text/html | crawl-data/CC-MAIN-2020-50/segments/1606141743438.76/warc/CC-MAIN-20201204193220-20201204223220-00252.warc.gz | 355,013,024 | 17,156 | Cody
# Problem 528. Find the largest value in the 3D matrix
Solution 336796
Submitted on 21 Oct 2013 by Nicholas Howe
This solution is locked. To view this solution, you need to provide a solution of the same size or smaller.
### Test Suite
Test Status Code Input and Output
1 Pass
%% Test 1 A = 1:9; A=reshape(A,[3 1 3]); y_correct = 9; assert(isequal(islargest(A),y_correct))
2 Pass
%% Test 2 A = 9:17 A=reshape(A,[3 1 3]); y_correct = 17; assert(isequal(islargest(A),y_correct))
A = 9 10 11 12 13 14 15 16 17
### Community Treasure Hunt
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Start Hunting! | 202 | 650 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.515625 | 3 | CC-MAIN-2020-50 | latest | en | 0.689233 |
https://kr.mathworks.com/matlabcentral/cody/problems/19-swap-the-first-and-last-columns/solutions/1753695 | 1,606,400,249,000,000,000 | text/html | crawl-data/CC-MAIN-2020-50/segments/1606141188146.22/warc/CC-MAIN-20201126113736-20201126143736-00417.warc.gz | 371,088,044 | 17,143 | Cody
Problem 19. Swap the first and last columns
Solution 1753695
Submitted on 18 Mar 2019 by Matt Offredi
This solution is locked. To view this solution, you need to provide a solution of the same size or smaller.
Test Suite
Test Status Code Input and Output
1 Pass
A = [ 12 4 7 5 1 4]; B_correct = [ 7 4 12 4 1 5 ]; assert(isequal(swap_ends(A),B_correct))
B = 7 4 7 4 1 4 B = 7 4 12 4 1 5
2 Pass
A = [ 12 7 5 4]; B_correct = [ 7 12 4 5 ]; assert(isequal(swap_ends(A),B_correct))
B = 7 7 4 4 B = 7 12 4 5
3 Pass
A = [ 1 5 0 2 3 ]; B_correct = [ 3 5 0 2 1 ]; assert(isequal(swap_ends(A),B_correct))
B = 3 5 0 2 3 B = 3 5 0 2 1
4 Pass
A = 1; B_correct = 1; assert(isequal(swap_ends(A),B_correct))
B = 1 B = 1
Community Treasure Hunt
Find the treasures in MATLAB Central and discover how the community can help you!
Start Hunting! | 335 | 851 | {"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} | 3.234375 | 3 | CC-MAIN-2020-50 | latest | en | 0.62327 |
https://blocking.net/zero-knowledge-proof-study-notes-background-and-origin.html | 1,719,242,643,000,000,000 | text/html | crawl-data/CC-MAIN-2024-26/segments/1718198865401.1/warc/CC-MAIN-20240624151022-20240624181022-00416.warc.gz | 116,834,668 | 16,915 | # Zero-knowledge proof study notes: background and origin
#### Subtitle or Abstract: Zero-Knowledge Proof Study Notes for Stanford Academy (1)
The author of this article, Dong Ze, a small partner from the Abe technology community, is currently studying at Stanford University and researching cryptography. This series of articles is derived from the author's study notes on Stanford's famous course "CS 251: Cryptocurrencies and blockchain technologies. The course teacher is Dan Boneh of Cryptography.
I learned about blockchain and digital currency related technologies with Dan Boneh in Stanford last semester. Unlike previous courses, this year's course has a new chapter called Zero Knowledge Proof. Mengmeng Dan and his great god phd Ben Fisch took turns taking lessons for us, and spent two weeks teaching the origins, concepts and realization of zkSNARK of zero knowledge.
After completing the final exams in these two days, I reviewed the whole class in my mind repeatedly during the review process. I felt that the most exciting part was the zero-knowledge proof. I think about taking advantage of the recent holidays to summarize and share with you.
## Foreword
After writing the first draft, when I shared it with my friend Proofread, I found that many friends reported that their background knowledge was not enough. So I added this chapter before I started, marking out the background reading needed to understand this article:
1. Merkle Tree / Merkle Proof : https://blog.csdn.net/wo541075754/article/details/54632929
2. Bitcoin transactions : https://blog.csdn.net/liduanwh/article/details/81141972
3. UTXO model : https://www.jianshu.com/p/02fd289e8853
4. Some basic encryption and decryption concepts : https://www.jianshu.com/p/f7c729a41c9f
After reading the preface, we can start the text .
To say that zero-knowledge proof is really hot in everyone's field of vision, in fact, we have to start with Bitcoin.
## Deficiency of Bitcoin
If you are familiar with Bitcoin, everyone should know that on the Bitcoin network, every transaction is public.
If A wants to pay B a sum of money, then A will announce to the entire network with a big speaker, she will create a new transaction (Tx), and the beneficiary of this transaction is B's public key (P2PK), Or the hash of the public key (P2PKH). As long as B sees the transaction, he can use his private key to sign a digital signature to prove that he is really the owner of the public key, thus spending the money.
When A submits the transaction to pay B, as a bystander M on the network, she can only see a bunch of garbled addresses aaaaa pay x coins to a bunch of garbled addresses bbbbb. When B then made money to C, he could only see that bbbbb made a sum of money to ccccc. We can see that transactions in Bitcoin are highly connected. Although we don't know who paid for it, we can find many transaction chains along the way.
If every user just makes good money back and forth, Bitcoin is actually relatively safe.
Once a user sees it, does not want to play, and wants to go to the exchange to cash out, then the transaction information of this entire chain will be exposed. Exchanges often have a KYC (Know Your Customer) policy, and each user who exchanges digital currencies and fiat currencies needs to undergo real-name authentication. Once C withdrew from the address ccccc and ran away, then the exchange grasped the fact that bbbbb once sent money to C. If C is suspected of money laundering, just need to wait for B to cash out at this time, and then grab it.
There are already many companies in the United States doing analysis on the transaction chain of Bitcoin, such as Chainalysis.
Presumably here, everyone can feel the deficiencies of Bitcoin: the randomly generated public key for receiving money is just an illusion (net name). Once the real-name system is authenticated, the net name and the real name are linked, then online before All that is done is unobtrusive and there is no privacy at all. This is like someone using a net name to post a message on the post bar, and then someone finds the mobile phone number with a secret security, and then uses the mobile phone number to find the registered real name, so being human is a reason.
## Anonymous and Pseudonymous
There are actually two types of understanding of privacy.
The first is Anonymous, which means that users do not have to disclose any information related to themselves. It is like a confession wall of a school. You can never know who wrote it. Anyway, it is written on it.
The second is Pseudonymous, which means that users post information using their own pseudonyms. It is like posting. If you do n’t know this user, you ca n’t establish a connection from the real name to the real name, and you do n’t know who posted it. who is it.
Looking at this analysis, Bitcoin is actually a pseudonym mechanism: each user will randomly generate his own public key (alias), and receive money through the public key address. This is like four people A / B / C / D aliased as Xiaoming / Xiaohong / Two Dogs / Xiaogang anonymously trade online, as long as D reveals his identity at any point (such as on the exchange Withdrawal), then the relationship between Xiao Ming / Xiao Hong / Egou and D will be exposed immediately.
We can discuss how to enhance the confidentiality of Bitcoin from these two methods.
## Ways to increase privacy
### CoinJoin
Since A pays B to be seen, and C pays D to be seen, some people think of simply throwing all four ABCDs into one transaction. Because bitcoin transactions can have multiple inputs and outputs, an onlooker will see that in a transaction, aaaaa and ccccc both put x coins into it, and then bbbbb and ddddd receive money. In this way, even if the transaction income knows that these addresses correspond to the four ABCDs, it is difficult to distinguish who received the money.
What if the two sets of transactions are still too easy to identify? Two is not enough to mix four, four is not enough to mix eight, and so on. Combining the transactions of various people together, confusing audiovisual, making it impossible to track. This is CoinJoin.
What are the disadvantages of CoinJoin? In fact, mixing multiple transactions does not completely prevent people from being touched. It can only be said that the probability of being touched all the way is reduced in probability. And there is another important point, if you want to mix AB and CD transactions, then their transaction volume must be the same. If A pays 10,000 coins to B and C pays D coins, we only need to look at the input and output to immediately split a CoinJoin transaction into two separate transactions. Therefore, mashing up transactions with similar transaction quotas is also a difficult point for CoinJoin to implement.
If you look at the above classification, CoinJoin is just an operation of Bitcoin's existing system, and its essence is still a pseudonym mechanism.
### Confidential Transaction (CT)
Since it is so troublesome to hide who I am, then people start thinking: if we don't hide the public key that participates in the transaction, we can also hide the transaction amount. When A pays B, even if B is exposed, the entire network will not know how much A has given B.
If this step can be realized, then we can even pay with bitcoin. Everyone can only see that your monthly salary is received, but we don't know how much money you make.
To implement the method, we must first understand a special encryption algorithm: homomorphic encryption.
In a word, homomorphic encryption is a special encryption algorithm that allows the ciphertext to retain its original mathematical characteristics .
We can assume that there is an encryption method E. If E is an addition homomorphism, then E (a) + E (b) = E (a + b). Conversely, if the multiplication is homomorphic, then E (a) x E (b) = E (axb).
Since this article is about the popular science of zkp, we don't know the specific implementation method in detail. We just need to understand that both the ellipse encryption equation and the large number module in RSA have some kind of homomorphism.
#### Pederson Commitment
Continue to the topic of hidden trading volume. If A has a balance of 100 coins and pays 10 coins to B, then this transaction would look like this:
Combined with the addition homomorphism mentioned above, if we have an encryption homogeneous method E, we can convert this transaction into:
As long as the first number is equal to the sum of the last two numbers, an onlooker will not see the transaction volume in the end, but has to admit that A really divided some money to B, and then some money was returned to A . This method is called Pederson Commitment, which hides the data itself, but proves the relationship of the data.
#### Negative number loophole
After reading this, some friends will find a huge loophole: Although Pederson promises to prove the relationship between numbers, it does not limit the value range of any number! That is to say, A can do bad things, submit a transaction, say that he has to pay -100 coins to B, and then "find" himself 200 coins, so that the equation still holds true. A can use this to print unlimited banknotes, thereby destroying the entire system.
How to avoid negative numbers? In addition to Pederson's promise, we need another set of proofs to prove that the numbers in all transactions are positive. In other words, the numbers in all transactions are limited to the range 0 to 2 ^ 256 (positive integer maximum) (Range Proof).
It doesn't sound difficult, the easiest way is undoubtedly to make all these figures public. But this violates the premise of hidden trading volume. So we have to find another way to prove it, that is, we can't expose the original numbers, but also prove their characteristics (values are 0 ~ 2 ^ 256). Does that sound like a problem? Don't worry, let's look at another problem.
#### Ownership loophole
Before we go any further, I want to quickly point out that there is still a huge loophole in this agreement: ownership is unknown.
Friends who know bitcoin may know that when creating a bitcoin transaction, you need to provide the UTXO Txid of the input transaction, so that you can quickly verify whether the A who is preparing to pay B really has the money .
But right now, we haven't mentioned anything about pointing to the previous transaction. In other words, because the entire network does not know how much A has spent, A can simply change the input number to whimsical, and change it to tens of thousands, and then call them all to secretly mint coins.
how to solve this problem? There are two options.
The first solution is to continue to introduce Bitcoin's transaction mechanism, and use the output of the previous private transaction as the input of the transaction. This kind of thinking is a bit like the conversion of a problem. I use the result of the transaction for this transaction, so long as the transaction is OK, my transaction is not problematic.
This is a chicken and egg problem. How to create the first private transaction without problems?
We can convert ordinary coins into private output through a special transaction. The input of this transaction is an existing transaction id (Bitcoin UTXO), and then the output becomes a privacy output. So we have the earliest eggs. (ZCash's shielded transaction is this principle)
The second solution is to prove that the input of A really belongs to A. In systems like Ethereum, there is a concept of a World State. The world state is the current balance and state of all users and smart contracts on the entire chain. Generally, a complete node will retain the entire world state (large size), while a light node only needs to save the world state Merkle Commit.
In addition to the Pederson commitment and interval proof, we provide an additional proof that the number entered in the transaction is consistent with the balance of A in the original world state. We can use Merkle Proof to achieve this proof.
But if we directly submit Merkle Proof, all bystanders can see the transaction input of A, which violates the premise of private transactions. So again : we still need to borrow the mysterious algorithm mentioned above-it can hide the answer itself (the balance of A), but it can also prove that this number really belongs to the world state.
### ZCash: All anonymous
When the concept of CT was put forward, many people were dissatisfied with the status quo and couldn't help feeling: If only their names could be hidden.
Therefore, the concept of ZeroCoin / ZeroCash was proposed: based on CT, but with the addition of a new mechanism, the users of the transaction can be anonymous. At this time, onlooker C, who was eating melon, was really aggressive, and saw a bunch of garbled characters floating on the network, but he didn't know what it was, but he had to believe it.
ZCash is a digital currency based on the ZeroCoin / ZeroCash protocol, which can achieve full anonymous transactions. I won't describe too much here, but it still relies on the old cryptographic tools: Pederson promise, interval proof, Merkle proof, and the black magic we have been mentioning: it will not expose the answer itself prove.
After all the calls came out, we finally have to talk about the highlight: this proof method that does not expose the answer itself is called zero-knowledge proof.
## Zero-knowledge proof (zkSNARK)
I believe that after reading the above, everyone has probably understood the problem we want to solve.
We want to prove the relationship between numbers, such as 0 <= a <= 2 ^ 256, or SHA256 (x) = y. But we don't want to expose these numbers, such as a and x in the previous article. How to build a system to achieve this?
Before talking about this topic, I want to change my mind and split this topic into two parts: zero-knowledge (zk) and proof (SNARK) .
As always, let's talk about definition and application first, and then how to implement it.
### Proof (SNARK)
The full name of SNARK is Succinct Non-interactive ARgument of Knowledge. This noun consists of three dimensions:
1. Succinct: The proof itself must be short enough. It is best to verify that the proof is O (logN) or even O (1) complexity.
2. Non-interactive: There is no interaction in the overall process, which means that the proving party can throw a bunch of garbled characters on your desk and then leave, and you can verify the garbled characters after you verify it. prove.
3. Argument of Knowledge: This stuff is more obscure. But the general meaning is that what you want to prove must be capable of expressing knowledge (Proof of Knowledge). The proof of PoK involves a more abstract concept of Extractor. For details, please refer to Teacher Yu's article. But in a word, the thing you prove is valuable, calculated through calculation, not messy other things.
After looking at the definition, we will find that it is already very strong to realize SNARK, especially in the short point.
We can immediately think of an application: if a third party organization stores a large amount (PB level) of data in its own database. If a government agency wants to audit their database to ensure that there is no problem with each data point, it may be necessary to look at it line by line, look at each PB's data, and see that the land is old. At this time, SNARK suddenly appeared, and the size and time of O (1) or O (logN) fully proved that there is no problem with each data in this huge database, and I was a little excited about it. Most people think this is completely impossible: how can you verify the accuracy of tens of millions of digits with a few digits?
Leave a suspense and talk about it later.
### Zero knowledge (zk)
In fact, zero-knowledge is just an additional requirement on the basis of this SNARK proof: the entire proof itself cannot expose any data about the mystery to be proved. The official definition of the concept of zero-knowledge is very obscure and introduces the concept of a simulator. For a detailed introduction, you can still refer to Mr. Guo Yu's article, I will take it here. In a sentence, a clever hacker can't extract any information related to the answer by holding a zero-knowledge proof.
Returning to the concept of the government audit database, we can assume that this database is the tax situation of the company. The government must make sure that the tax payment data must be accurate, but for companies, they don't want the inspectors to see their daily business flow, because it may involve trade secrets. At this time, one SNARK is not enough. We need zkSNARK to achieve:
It can prove that I paid the tax truthfully without showing you the details of each of my transactions.
### Application of Zero Knowledge Proof
What can we do with zkSNARK?
The first thing is to implement the Private Transaction mentioned above. ZCash's private transaction mechanism is based on zkSNARK. In this way, digital currency transactions become much safer.
Second thing, we can use this technology to better solve the problem of blockchain efficiency . At present, there are undoubtedly several methods for blockchain scaling: sacrificing the strength of consensus to increase the block speed, enabling sidechains, or offline point-to-point channels similar to Lightning.
In fact, there is another idea called Rollup. The concept of Rollup is probably that the load on the main chain is too large, so we open a few small servers, can also receive transactions, do transaction authentication, and then batchly accumulate all the transactions accumulated over a period of time. Update to the main chain. But if this update process still needs to send a large amount of transaction information to the main chain, the significance of this Rollup will not exist, and it will not reduce the load of any main chain. At this time, SNARK comes in handy: With SNARK (zk or not), the Rollup server can submit to the main chain with a very short certificate, which proves that a large number of transactions are OK. As a result, adding some UTXO and reducing it is done. With ZK Rollup, we can greatly reduce the load on the main chain and outsource more verifications elsewhere.
The third thing is that we can really implement transactions to third parties .
Suppose A is doing machine learning research, but does not have a good computer, so she plans to outsource the task of training the model to B. After three days, B told A that he was finished running, and he needed to ask A to pay before providing her with the trained model. A is worried that B does not train the model honestly, but randomly generates a random number and packs it, so he wants B to first pass the model to A and then pay. B was worried that A secretly copied the model after getting it, and then pulled him directly without giving money.
Faced with this type of problem, the traditional solution is to entrust a third party or design a smart contract on the chain to complete the verification and exchange of data and currency. Now with zkSNARK, B can directly submit a model-trained zkSNARK to A, proving that he really ran for three days without cheating. After passing the quick verification, you can safely pass the money.
The fourth thing, we can completely transfer the ownership of the data .
Assuming the bank's account balance database is a sql table, then 100 million customers will have 100 million rows of records. Every year banks need to spend a lot of money to maintain such a large data system. If everyone can move their own line of records to the local area and maintain their own account data, then the bank will not have to spend a penny. The reason why the bank does not do this is because users are likely to tamper with their own data for the benefit of turning 100 yuan into 1 million.
zkSNARK can guarantee that the data itself will not be a problem. We can imagine a distributed bank where everyone's deposit balance is stored in their own computers. When A wants to transfer money to B, she needs to submit a zkSNARK that proves that the balance on her account is deducted correctly, so as to ensure that A honestly deducts the transfer amount from her balance. When B enters the account, it will also submit a zkSNARK with an increased balance.
We can apply this concept to all areas, social networking, banking, health, financial auditing, corporate taxation, etc. With zkSNARK, service providers do not need to pay for the storage of large amounts of data, and users do not need to worry about their privacy being stolen.
## To be continued
For reasons of space, this is written here. I must see here, everyone has a deeper understanding of why zero-knowledge proofs are needed and how powerful they are.
Beginning in the next article, I will write a little more in-depth, and mainly discuss the specific structure of zkSNARK.
PS: Zero-knowledge proof and zkSNARK are used interchangeably in this article. In fact, zkSNARK is just one of the more classic zero-knowledge proof protocols, and there are many other protocols that will be introduced later.
1. Teacher Guo Yu's zero-knowledge proof (four articles in total) : https://www.jianshu.com/p/38ab873ae8ce
2. Stanford courseware and after-school reading : http://cs251.stanford.edu/syllabus.html
3. Construction of ZCash : https://www.jianshu.com/p/4db439c63a96
4. CoinJoin : https://en.bitcoin.it/wiki/CoinJoin
5. Confidential Transaction : https://www.jianshu.com/p/22664259dee3
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https://www.delftstack.com/howto/python/variance-in-python/ | 1,708,604,172,000,000,000 | text/html | crawl-data/CC-MAIN-2024-10/segments/1707947473738.92/warc/CC-MAIN-20240222093910-20240222123910-00458.warc.gz | 762,854,958 | 7,597 | # How to Calculate Variance in Python
Lakshay Kapoor Feb 02, 2024
Variance is the measure of dispersion that considers all the data points spread across the data set. There are two measures of dispersion, variance and standard deviation (square root of variance).
In Python, several modules and libraries can help calculate the variance of a dataset or data points. This tutorial will discuss finding the variance in Python with examples provided to understand the methods better.
## Use the `variance()` Function of the Statistics Module to Calculate Variance in Python
The `variance()` function is one of the functions of the Statistics module of Python. This module is used to provide functions to perform statistical operations like mean, median, standard deviation, etc., on numeric data.
The `variance()` function of the Statistics module helps a user calculate the variance of the dataset or given data points.
``````import statistics
list = [12, 14, 10, 6, 23, 31]
print("List : " + str(list))
var = statistics.variance(list)
print("Variance of the given list: " + str(var))
``````
Output:
``````List : [12, 14, 10, 6, 23, 31]
Variance of the given list: 86
``````
In the above example, the `str()` function converted the whole list and its standard deviation into a string because it can only be concatenated with a string.
## Use the `var()` Function of the NumPy Library to Calculate Variance in Python
The `var()` function of the NumPy library can also calculate the variance of the elements in a given array list.
``````import numpy as np
arr = [12, 43, 24, 17, 32]
print("Array : ", arr)
print("Variance of array : ", np.var(arr))
``````
Output:
``````Array : [12, 43, 24, 17, 32]
Variance of array : 121.04
``````
## Use the `sum()` Function and List Comprehensions to Calculate Variance in Python
The `sum()` function sums up all the elements of an iterable like a list, a tuple, etc.
On the other hand, list comprehension is a method that creates a list from the elements present in an already existing list.
The `sum()` function and the list comprehension can help calculate the variance of a list.
``````list = [12, 43, 24, 17, 32]
average = sum(list) / len(list)
var = sum((x - average) ** 2 for x in list) / len(list)
print(var)
``````
Output:
``````121.04
``````
In the above example, the Math module is imported as it provides the `sqrt()` function used to compute the square root of a given value.
Also, note that the function `len()` is used. This function helped provide the length of the given list or the number of elements in the list.
The program above is based on the mathematical formula of variance.
Lakshay Kapoor is a final year B.Tech Computer Science student at Amity University Noida. He is familiar with programming languages and their real-world applications (Python/R/C++). Deeply interested in the area of Data Sciences and Machine Learning. | 718 | 2,915 | {"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} | 3.90625 | 4 | CC-MAIN-2024-10 | latest | en | 0.785293 |
https://godotshaders.com/shader/attack-of-the-smarties/ | 1,725,980,075,000,000,000 | text/html | crawl-data/CC-MAIN-2024-38/segments/1725700651255.81/warc/CC-MAIN-20240910125411-20240910155411-00469.warc.gz | 253,498,942 | 34,210 | # Attack of the Smarties
``````shader_type canvas_item;
// Variation on Ford's Circles stuff: http://nrich.maths.org/6594
////////////////////////////// defines
// Diameter of biggest circle
#define DIAMETER 100.0
// Nb of iterations = nb of circles of decreasing size to build
#define ITERATIONS 50
// Time in second(s) to zoom in or out
#define ZOOM_TIME 10.0
// Heightmap coef [0.0 100.0]
#define HEIGHT_COEF 4.0
// Light direction
#define LIGHT_DIR vec3(1.0, 1.0, 1.0)
// Ambiant lighting
#define LIGHT_AMBIANT vec3(0.1, 0.1, 0.1)
// Mix between original texture/procedural circles [0.0 1.0]
#define MIX_FACTOR 0.5
// Point of view
#define EYE vec3(0.57735, 0.57735, 0.57735)
// Strength of specular reflection [0.0 1.0]
#define SPEC_FACTOR 0.45
#define pi 3.1415926535897932384626433832795
#define hfpi 1.5707963267948966192313216916398
#define PI pi
#define HFPI hfpi
uniform sampler2D iChannel0;
uniform vec4 iMouse;
////////////////////////////// methods
// Most important method: Ford
float AnimateDiameter()
{
float i = floor(TIME / ZOOM_TIME);
float r = (TIME - ZOOM_TIME * i) / ZOOM_TIME;
float sinr = pow(sin(HFPI * r), 2.0);
float k = ( mod(i, 2.0) == 0.0 ? sinr : 1.0 - sinr );
return max(k*DIAMETER, 1.0);
}
// Compute normal at position pos, as if we were on an half-sphere with center (0,0)
{
// Determine the distance to the center of the circle.
float d = length(pos);
return (normalize(vec3(pos.x,pos.y,HEIGHT_COEF*h)));
}
vec3 Ford(vec2 posSample, vec2 iRes)
{
float diameter = AnimateDiameter();
float radius = diameter / 2.0;
vec2 screenPos = posSample*iRes.xy - (iRes.xy / 2.0);
//position of sample relative to circle: (0,0) = center
vec2 pos = vec2(0.0);
bool inside = false; //true iff sample is in a circle
float rSmall = 0.0;
// Build biggest circles of radius: r0
pos = mod(screenPos, vec2(diameter, diameter)) - vec2(r0, r0);
if (length(pos) < r0)
{
inside = true;
rSmall = r0;
}
// Build smaller circles surrounded by 4 big circles
float r1 = 0.0;
vec2 mod1 = vec2(0.0);
if (!inside)
{
mod1 = mod(screenPos + vec2(r0, r0), vec2(diameter, diameter)) - vec2(r0, r0);
r1 = (sqrt(2.0) - 1.0)*r0;
pos = mod1;
if (length(pos) < r1)
{
inside = true;
rSmall = r1;
}
}
// Iterations - begin
// We build circles smaller at each iteration
// radius r[n] of nth circle is given by:
//
// (r[0] - r[1] - 2*r[2] - ... - 2*r[n-1])^2
// r[n] = ----------------------------------------------------
// 2 * (r[0] + r[0] - r[1] - 2*r[2] - ... - 2*r[n-1])
//
//
if (!inside)
{
float rBig = 0.0;
float distLeft = r0 - r1; //distance to 2 big circles point-of-contact
float r = r1; //radius of current circle
for (int n=0; n<ITERATIONS; n++)
{
if (!inside)
{
rSmall = (distLeft - 2.0*rBig)*(distLeft - 2.0*rBig)/(2.0*(r0 + distLeft - 2.0*rBig));
//West
pos = mod1 + vec2(rBig + rSmall + r, 0.0);
if (length(pos) < rSmall)
{
inside = true;
}
if (!inside)
{
//East
pos = mod1 - vec2(rBig + rSmall + r, 0.0);
if (length(pos) < rSmall)
{
inside = true;
}
if (!inside)
{
//North
pos = mod1 - vec2(0.0, rBig + rSmall + r);
if (length(pos) < rSmall)
{
inside = true;
}
if (!inside)
{
//South
pos = mod1 + vec2(0.0, rBig + rSmall + r);
if (length(pos) < rSmall)
{
inside = true;
}
}
}
}
r = r + rBig + rSmall;
distLeft = distLeft - 2.0*rBig;
rBig = rSmall;
}
}
}
// Iterations - end
// Now we're done with building circles, we give them some 3d look & feel
///////////////////////// post-production
//heightmap
vec3 normal = vec3(0.57735);
vec2 centerC = vec2(0.0);
if (inside)
{
centerC = posSample - pos/iRes.xy;
distance2center = length(pos) + r0;
normal = ComputeNormal(pos, rSmall);
}
//sample texture using coordinates of circle's center
// NOTE: Use a large negative bias to effectively disable mipmapping, which would otherwise lead
// to sampling artifacts where the UVs change abruptly at the pixelated block boundaries.
centerC.y = 1.0 - centerC.y; //upside-down correction
vec3 colTex = texture(iChannel0, centerC, -32.0).rgb;
vec3 light = normalize(LIGHT_DIR);
// Point light
vec3 colProc = vec3(0.8, 0.8, 0.8) * clamp(dot(normal, light), 0.0, 1.0);
// Ambiant ligth
colProc += LIGHT_AMBIANT;
vec3 colFinal = mix(colTex, colProc, MIX_FACTOR);
// Reflection
vec3 eye = normalize(EYE);
vec3 ref = reflect(eye, normal);
// Specular
float spec = pow(clamp(dot(light, ref), 0.0, 1.0), 16.0);
colFinal += SPEC_FACTOR * spec;
// Set the final fragment color.
return colFinal;
}
// Sample a procedural texture (anti-aliasing)
vec3 FordAA( vec2 uv, vec2 iRes )
{
#define SAMPLING_STRENGTH 1000000000.0
#define NB_SAMPLES 3 //0: no anti-aliasing
if (NB_SAMPLES == 0)
{
return Ford( uv, iRes );
}
else
{
// calc texture sampling footprint
vec2 ddx = dFdx( uv );
vec2 ddy = dFdy( uv );
int sx = 1 + int( clamp( SAMPLING_STRENGTH*length(ddx), 0.0, float(NB_SAMPLES-1) ) );
int sy = 1 + int( clamp( SAMPLING_STRENGTH*length(ddy), 0.0, float(NB_SAMPLES-1) ) );
vec3 no = vec3(0.0);
for( int j=0; j<NB_SAMPLES; j++ )
for( int i=0; i<NB_SAMPLES; i++ )
{
if( j<sy && i<sx )
{
vec2 st = vec2( float(i), float(j) ) / vec2( float(sx),float(sy) );
no += Ford( uv + st.x*ddx + st.y*ddy, iRes);
}
}
return no / float(sx*sy);
}
}
vec2 Rot2d(vec2 uv, float angle){
mat2 rot_matrix = mat2(vec2(sin(angle), -cos(angle)), vec2(cos(angle), sin(angle)));
return uv * rot_matrix;
}
////////////////////////////// main
void fragment()
{
vec2 iResolution = 1.0 / SCREEN_PIXEL_SIZE;
vec2 uv = UV;
//pan
uv -= iMouse.xy / iResolution.xy;
uv.y = 1.0 - uv.y;
vec3 col = FordAA( uv , iResolution);
// Set the final fragment color.
COLOR = vec4(col,1.0);
}
``````
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Notify of | 1,910 | 5,619 | {"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} | 3.015625 | 3 | CC-MAIN-2024-38 | latest | en | 0.430539 |
https://parametrichouse.com/3-04/ | 1,702,089,495,000,000,000 | text/html | crawl-data/CC-MAIN-2023-50/segments/1700679100781.60/warc/CC-MAIN-20231209004202-20231209034202-00373.warc.gz | 498,138,868 | 19,759 | In this ParaCourse Lesson, I will teach you how to model a wavy facade in Grasshopper. This tutorial is somehow Hard so watching the Multi-curve Attractor lesson will help you to understand the steps easily.
#### 3- Aqua Tower
In this Paracourse Lesson, we will model something similar to the Aqua tower (located in Chicago) using the Cocoon Plugin. First, we will use a simple box to get the base shape and then by using curves we will model the facade and by implementing several techniques we will finish the tutorial.
#### 5- Bosjes Chapel
In this Lesson I will model the Bosjes Chapel In grasshopper. First I will explain how you can use the Graph Mapper to produce the base curves and then we will use scaleNU and orientation to make the freeform shell structure. At the end we will model the glass surface.
In this Paracourse lesson, we are going to model a wavy facade in Grasshopper. First, we are going to model the strips and then by using the curve attractor technique and some data management tricks we will model the louvers.
In this Paracourse Lesson we are going to model a parametric facade made by random triangles. First we will model the basics and then we will extract the windows and the frames and orient them on the ground for the final step.
#### 5- Rotating Tower
In this Paracourse Lesson we are going to remodel the filzfelt scale wall panels in grasshopper and make them parametric.
#### Parametric Roof
In this Paracourse Lesson, we are going to model a Parametric Roof from scratch. First, we are going to model a simple rectangular plane for the base of the roof and then we are going to use a parametric curve to control the overall shape of the model.
#### Continuous Staircase
In this Paracourse Lesson, we are going to model a parametric Nurbs stair from scratch. First, we are going to take a look at the algorithm and how we can model the stairs as easy as possible.Then we are going to model the base lines and finally the stairs.
#### deformed pavilion
In this Paracourse lesson we will use the spatial deform component to design a parametric freeform surface pavilion.
#### Concrete Planter
In this Beginner Rhino Grasshopper Tutorial we are going to model a polygonal concrete Planter from scratch and then use the Pufferfish plugin to give it a thickness.
In this grasshopper tutorial, you will learn how to model a zigzag parametric facade in Grasshopper.
In this Paracourse Lesson you will learn how to model a parametric wavy facade by using the graph mapper component and also shifting the pattern for each level.
#### Barrel Vault
In this Grasshopper tutorial, you can learn how to model a parametric barrel vault from scratch by defining a rectangle | 594 | 2,718 | {"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} | 3.203125 | 3 | CC-MAIN-2023-50 | longest | en | 0.918601 |
https://zh.b-ok.org/terms/?q=homogeneous+equation | 1,571,362,843,000,000,000 | text/html | crawl-data/CC-MAIN-2019-43/segments/1570986677412.35/warc/CC-MAIN-20191018005539-20191018033039-00489.warc.gz | 1,154,494,926 | 21,295 | ## TermsVector search result for "homogeneous equation"
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File: DJVU, 5.29 MB | 1,201 | 4,581 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.53125 | 3 | CC-MAIN-2019-43 | latest | en | 0.636947 |
https://www.physicsforums.com/threads/with-what-velocity-did-the-mug-leave-the-counter.411189/ | 1,508,282,349,000,000,000 | text/html | crawl-data/CC-MAIN-2017-43/segments/1508187822513.17/warc/CC-MAIN-20171017215800-20171017235800-00083.warc.gz | 1,001,185,338 | 16,687 | # With what velocity did the mug leave the counter?
1. Jun 19, 2010
### the_storm
1. The problem statement, all variables and given/known data
In a local bar, a customer slides an empty beer mug down the counter for a refill. The bartender is momentarily distracted and does not see the mud, which slides off the counter and strikes the floor 1.4m from the base of the counter, If the height of the counter is 0.860m
2. Relevant equations
a) with what velocity did the mug leave the counter?
3. The attempt at a solution
I know this is a projectile motion problem
I started and put Vy= 0 m/s and Vx= Vi
I put the height as H= 1.4+0.860= 2.26m
from this equation Vf=Vi-gt
I got the time .. t=0.679s
My idea is to ger the distance X in X direction and I have the time of the flying mug
By dividing them I'll get the Vx which equals to the V initial of the mug
I tried to get a relation between Vx and Vy but I failed
can anyone help me ?
1. The problem statement, all variables and given/known data
2. Relevant equations
3. The attempt at a solution
2. Jun 19, 2010
### pgardn
3. Jun 19, 2010
### the_storm
I know this is the answer but how to reach it ,,,?
I am unable to get X in X direction I have only the time so I have an equation in two unknown
I can't get Vx. I have no given ? So is there a way to get it ?
4. Jun 19, 2010
### pgardn
You have the time its in the air. You have the distance it traveled in the x direction, why cant you get Vx?
5. Jun 19, 2010
### the_storm
I don't have the distance travelled in X direction I wish I had it
Do you have a way to get it ?
6. Jun 19, 2010
### pgardn
You do have it. Read the question again.
7. Jun 19, 2010
### pgardn
Whenever you do these types of problems you might want to do the GUESS method, and its exactly the opposite of a guess...
G - Given: What has been given to you in the problem
U - unknown, what are you trying to solve for
E - equations; which equations do you have in your arsenal
S - substitute; substitute the appropriate givens into the equations you need.
S - Solve for the unknown.
8. Jun 19, 2010
### the_storm
I don't :D
I have only the height in other words the Distance in Y direction which is 1.4+0.86
That all I have.
So what do u think ?
9. Jun 19, 2010
### pgardn
You have to read these questions carefully sometimes and familiarize yourself with the way they are stated.:surprised
10. Jun 19, 2010
### pgardn
another hint... the 0.86 is the distance the mug falls, yes?
So make sure you remove that adding you did to get the time. They give you the distance the mug goes in the horizontal and the vertical. They are separate numbers.
11. Jun 19, 2010
### the_storm
You are right
It's solved Thank you bro :)
So I can go and sleep now :) good night
<< he is from Egypt :D
12. Jun 19, 2010
### pgardn
Good evening... | 809 | 2,839 | {"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} | 3.71875 | 4 | CC-MAIN-2017-43 | longest | en | 0.94749 |
https://mathoverflow.net/questions/43345/grothendieck-topologies-mayer-vietoris-and-points | 1,642,624,918,000,000,000 | text/html | crawl-data/CC-MAIN-2022-05/segments/1642320301488.71/warc/CC-MAIN-20220119185232-20220119215232-00096.warc.gz | 416,091,111 | 27,004 | # Grothendieck topologies, Mayer-Vietoris, and points
I am trying to think about certain problems in the theory of motives without having a proper background in Grothendieck topologies and the like, hoping to teach myself the related techniques in the process. Here is a rather specific question that I've stumbled upon; I would appreciate any references and/or explanations of the relevant issues.
Consider a topology "of reasonable size". What I have in mind is the Zariski, Nisnevich, or etale site of an algebraic variety. Let us call the objects of this topology "open sets". My understanding is that there is also some notion of "points" of a Grothendieck topology, and that for the three topologies mentioned above these are the spectra of the local rings of the points of the scheme, the spectra of Henselizations of these local rings, and the spectra of the strict Henselizations of these local rings, respectively. Please correct me if this is wrong.
I think one can define a "cohomology theory" on a site as a sequence of contravariant functors, indexed by nonnegative integers, from the category of open sets to, say, abelian groups, such that whenever an open set is covered by two other ones, there is a cohomological Mayer-Vietoris sequence. Given a cohomology theory, one can also define its value on any point of the site simply by passing to the inductive limit.
Suppose that I have a morphism of cohomology theories on, say, the Nisnevich site of an algebraic variety. Assume that this morphism is an isomorphism at all points. Does it follow that it is an isomorphism of cohomology theories (i.e., an isomorphism on all open sets)?
• Let me briefly advocate the idea that thinking on the group level is perhaps not what you want. For example, one way to produce the kinds of objects you seem to be getting at is to start with a bounded-below cochain complex of injective sheaves on your site, and to each open set associate the cohomology groups of the chain complex of sections. Remembering only the groups doesn't provide the "connective tissue" that connects the groups at different levels. Oct 24 '10 at 15:44
• (And more generally one might take values in a stable model category, stable infinity-category, or perhaps might have a "sheaf of infinity-categories" - but I digress.) Oct 24 '10 at 15:44
• @Leonid: I think the point is that in general one should instead expect a descent spectral sequence associated with the Cech nerve of the cover, rather than just a Mayer-Vietoris sequence. The Mayer-Vietoris sequence is so simple because of the natural identifications $U \cap (U \cap V) \cong U \cap V$. This certainly fails for the etale topology if $U \to X$ is an etale cover rather than an open immersion. Oct 24 '10 at 13:02 | 648 | 2,763 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.53125 | 3 | CC-MAIN-2022-05 | latest | en | 0.943358 |
https://pgp13rm.wordpress.com/2014/03/07/sectiona_group8_hemant-mohan-nagalkar13pgp033-session-07/ | 1,529,949,304,000,000,000 | text/html | crawl-data/CC-MAIN-2018-26/segments/1529267868237.89/warc/CC-MAIN-20180625170045-20180625190045-00040.warc.gz | 666,883,558 | 17,088 | # SectionA_Group8_Hemant Mohan Nagalkar(13PGP033) : Session 07
Session 7: Data Analytics using SPSS
When for the very first time, I started using SPSS; I found it very complicated to run. However, when college professor started teaching the course using SPSS, I found SPSS it rather logical.
While using SPSS or any analytics tool, principle is to focus on objective of your study. If focus on objective is kept intact, the flow of steps to be followed on SPSS becomes clear and logical.
Another rule is to always try to have all the data in numeric form. For example, it’s better to keep gender data as ‘Female’ marked as ‘1’ and ‘Male’ as ‘2’ or vice versa.
In SPSS, there are three data types, (a) Nominal, (b) Ordinal and (c) Scale.
Nominal data set is the most basic in all data types. Its presence is kept only for the namesake; it contains no meaningful value. Nominal numbers contain no information whatsoever. For example names of students indicated using number, numbers on the back of t-shirts of players. Another example can be again ‘Female’ marked as ‘1’ and ‘Male’ as ‘2’. Here, even if we mark ‘Male’ as 1 and ‘Female’ as 2, it makes no difference.
Ordinal data type is the one which follows order. It’s a sequence of what comes first and what later. But ordinal number does not tell the exact difference between the two data sets. For example, Age when kept in range and marked as:
18 years-20 years –> 1,
21 years-25 years –> 2 and
26 years-30 years –> 3
Then the numbers 1, 2, 3 are ordinal as the age ranges they follow are in order. But these numbers cannot tell the exact difference between two datasets. One cannot tell the exact difference between two different ranges 18-20 and 21-25, it can be anything from 1 to 7.
Scale data numbers are the one which can be marked on scale and can tell the exact difference between two data sets. For example, keeping age again in perspective, when kept definite i.e. 26 years represented as 1, 29 years represented as 2 then one can easily tell the difference between two data sets as 3. Other examples can be weight, height etc.
As can be seen from above examples we keep Scale data as continuous whereas we categorize Nominal and Ordinal data sets.
Steps followed during analytics study using SPSS are:
1. Analysis/ Processes
2. Observation
3. Interpretation
4. Strategy
1. Analysis: Doing the first level analysis is always important; if ignored, there are good chances that our final strategy will get ruined. At first level analysis, we perform counting by looking at the frequency output of SPSS. Most important factor to check while doing first level analysis is ‘percentage of Valid Number’ as it depicts the actual percentage data which follows the trend under study.
Labor Force Status
2. Observation
From the derived preliminary frequency table output of SPSS many observations can be made. For example if the output is available for ‘age when people are getting married for the first time’ and ‘education’, then based on that one may observe that when lesser is the education then lower is the age of marriage. This same observation can be recorded as hypothesis which needs to be checked for its correctness through further detailed study.
3. Interpretation
To check the correctness of the hypothesis one needs to check the relation between two variables under study. It’s been done by using pearson chi-square test (If ‘Assym.sig’ value is less than 0.05 then reject the null hypothesis).
Chi-Square Tests
Once the relation between two variables is known then one should be able to interpret it from the given results. As seen from the following table one can say that 52.8 % of people who visited museum got married early.
Visited Art Museum or Gallery in Last Yr * categorise age when first married Cross tabulation
4. Strategy:
Based on the interpretation, we finally build up the strategy which would be beneficial in order to particular objective in hand.
Further detailed explanation on this will be covered in the next blog. | 916 | 4,047 | {"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} | 3.28125 | 3 | CC-MAIN-2018-26 | latest | en | 0.937099 |
http://www.chegg.com/homework-help/questions-and-answers/ball-thrown-45m-angle-300-degrees-horizontal-strikes-ground-570-m-away-find-time-flight-b--q1540756 | 1,464,581,432,000,000,000 | text/html | crawl-data/CC-MAIN-2016-22/segments/1464049288709.26/warc/CC-MAIN-20160524002128-00184-ip-10-185-217-139.ec2.internal.warc.gz | 429,393,982 | 13,257 | A ball is thrown from 45m at an angle of 30.0 degrees below the horizontal.
If it strikes the ground 57.0 m away, find:
A) the time of flight
B) The initial speed
C) Speed and the angle of the velocity vector with respect to the horizontal at impact
(Hint: For part (a), use the equation for the x-displacement to eliminate Vot from the equation for the y-displacement)
* I am having a lot of trouble when I am not given the initial velocity. This is a follow up question and I got the answer correct for the first half where the initial velocity in that question was given. All these examples in the book all have the initial Vo and I understand how to do them. All their followups don't give the Vo and I am lost on how to solve w/o it.
Please show work, thanks! | 185 | 771 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.796875 | 3 | CC-MAIN-2016-22 | latest | en | 0.939969 |
https://www.engineersedge.com/engineering-forum/showthread.php/2574-Linear-Control-Systems?s=f5ec5df5b595b7624acf28983a79d009 | 1,726,421,681,000,000,000 | text/html | crawl-data/CC-MAIN-2024-38/segments/1725700651632.84/warc/CC-MAIN-20240915152239-20240915182239-00629.warc.gz | 699,286,017 | 10,401 | # Thread: Linear Control Systems
1. ## Linear Control Systems
Q1:
Identify 5 feedback control systems that you encounter in your every day environment. For each system, identify the sensing mechanism, actuation mechanism, and controller type.
Q2:
Progressively stringent requirements of modern, high-precision machinery are placing increasing demands on slide systems, such as show in Fig 1. The typical goal is to accurately control the desired path of the table. Sketch a block diagram model of a feedback system to achieve the desired goal. The table can move in the x direction.
Q3:
Draw the block diagram of a control system which can automatically adjust the radio volume in an automobile as the noise generated changes. The noise generated increases with speed. As the noise increases, the system increases the volume of the radio. Assume that the amount of noise can be represented by a voltage generated by the speedometer cable and the volume of the radio is controlled by a dc voltage. If the dc voltage represents the desired volume disturbed by the noise, draw the functional block diagram of the automatic volume control system, showing the input transducer, the volume control circuit and the speed transducer as blocks. Also show the desired volume as an input, actual volume as an output, voltages representing speed, desired volume and actual volume.
Q4:
A high-performance race car with an adjustable wing (airfoil) is shown in Fig. 2. Develop a block diagram describing the ability of the airfoil to keep a constant road adhesion between the car's tires and the race track surface.
Why is it important to maintain good road adhesion?
2. Did you read the rules about homework?
3. Originally Posted by Kelly Bramble
Did you read the rules about homework?
And not even showing any attempt to work this through themselves. Poor show indeed.
If you ever need to make a quick buck Kelly, you could start a pay-as-you-go Q+A sub forum. You'd rake it in!
4. He (or she) didn't even include the figures to which the questions apply, and still expects some magical answer. I fear for our profession, guys.
#### Posting Permissions
• You may not post new threads
• You may not post replies
• You may not post attachments
• You may not edit your posts
• | 473 | 2,273 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.765625 | 3 | CC-MAIN-2024-38 | latest | en | 0.910236 |
https://www.unix.com/man-page/osx/3g/gluortho2d | 1,680,290,756,000,000,000 | text/html | crawl-data/CC-MAIN-2023-14/segments/1679296949678.39/warc/CC-MAIN-20230331175950-20230331205950-00218.warc.gz | 1,162,437,784 | 12,663 | # gluortho2d(3g) [osx man page]
```GLUORTHO2D(3G) GLUORTHO2D(3G)
NAME
gluOrtho2D - define a 2D orthographic projection matrix
C SPECIFICATION
void gluOrtho2D( GLdouble left,
GLdouble right,
GLdouble bottom,
GLdouble top )
PARAMETERS
left, right Specify the coordinates for the left and right vertical clipping planes.
bottom, top Specify the coordinates for the bottom and top horizontal clipping planes.
DESCRIPTION
gluOrtho2D sets up a two-dimensional orthographic viewing region. This is equivalent to calling glOrtho with near = -1 and far = 1.
glOrtho, gluPerspective
GLUORTHO2D(3G)```
## Check Out this Related Man Page
```GLORTHO(3G) OpenGL Manual GLORTHO(3G)
NAME
glOrtho - multiply the current matrix with an orthographic matrix
C SPECIFICATION
void glOrtho(GLdouble left, GLdouble right, GLdouble bottom, GLdouble top, GLdouble nearVal, GLdouble farVal);
PARAMETERS
left, right
Specify the coordinates for the left and right vertical clipping planes.
bottom, top
Specify the coordinates for the bottom and top horizontal clipping planes.
nearVal, farVal
Specify the distances to the nearer and farther depth clipping planes. These values are negative if the plane is to be behind the
viewer.
DESCRIPTION
glOrtho describes a transformation that produces a parallel projection. The current matrix (see glMatrixMode()) is multiplied by this
matrix and the result replaces the current matrix, as if glMultMatrix() were called with the following matrix as its argument:
2 right - left 0 0 t x 0 2 top - bottom 0 t y 0 0 -2 farVal - nearVal t z 0 0 0 1
where t x = - right + left right - leftt y = - top + bottom top - bottomt z = - farVal + nearVal farVal - nearVal
Typically, the matrix mode is GL_PROJECTION, and left bottom - nearVal and right top - nearVal specify the points on the near clipping
plane that are mapped to the lower left and upper right corners of the window, respectively, assuming that the eye is located at (0, 0, 0).
- farVal specifies the location of the far clipping plane. Both nearVal and farVal can be either positive or negative.
Use glPushMatrix() and glPopMatrix() to save and restore the current matrix stack.
ERRORS
GL_INVALID_VALUE is generated if left = right, or bottom = top, or near = far.
GL_INVALID_OPERATION is generated if glOrtho is executed between the execution of glBegin() and the corresponding execution of glEnd().
ASSOCIATED GETS
glGet() with argument GL_MATRIX_MODE
glGet() with argument GL_COLOR_MATRIX
glGet() with argument GL_MODELVIEW_MATRIX
glGet() with argument GL_PROJECTION_MATRIX
glGet() with argument GL_TEXTURE_MATRIX
glFrustum(), glMatrixMode(), glMultMatrix(), glPushMatrix(), glViewport()
http://oss.sgi.com/projects/FreeB/.
AUTHORS
opengl.org
opengl.org 06/10/2014 GLORTHO(3G)```
Man Page
## 1. AIX
Manufacturer Links General Information Home Page: IBM United States Documentation/Information: IBM System p - UNIX servers: Support and services pSeries and AIX Information Center Developerworks AIX Wiki: AIX Wiki AIX for System Administrators In-depth information from IBM: IBM...
## At A Glance Coloured Real Time Bargraph Generator...
Not sure if anyone is interested but I am just getting into UNIX like shell scripting... I have great interest in pseudo-animations in text mode and accessing HW like /dev/dsp for example... ... Have fun, I do... ;o) # !/bin/sh # # Bargraph_Generator.sh # # A DEMO 6 bit coloured...
## When is a _function_ not a _function_?
For a starter I know the braces are NOT in the code... Consider these code snippets:- #!/bin/bash --posix x=0 somefunction() if then echo "I am here." fi # somefunction #!/bin/bash --posix x=0 somefunction() if
## Installing Dash Shell on OS X Lion
For those interested in installing dash shell on OSX Lion to help test POSIX compliancy of shell scripts, it is quite easy. I did it like this: If you don't have gcc on your system: 0. Download and install the Command Line Tools for Xcode package from Sign In - Apple * 1. Download the dash...
## Hearing Aid for OSX 10.12.x and greater.
Hearing Aid... Hi folks yet another bizarre piece of code that is Apple OSX 10.12.x to at least 10.14.1 specific. It requires only a default OSX install, and the internal microphone along with an external headphone assembly. Pre-amble, 14-02-2019: For over 3 weeks now I have been suffering a...
## Shopt -s histappend
What is the point of this? Whenever I close my shell it appends to the history file without adding this. I have never seen it overwrite my history file. # When the shell exits, append to the history file instead of overwriting it shopt -s histappend
## Undeletable file
Greetings, I'm trying to delete a file with a weird name from within Terminal on a Mac. It's a very old file (1992) with null characters in the name: ââWord FinderÂŽ Plusâ˘. Here are some examples of what I've tried: 12FX009:5 dpontius\$ ls ââWord FinderÂŽ Plus⢠12FX009:5 dpontius\$ rm...
## To print diamond asterisk pattern based on inputs
I have to print the number of stars that increases on each line from the minimum number until it reaches the maximum number, and then decreases until it goes back to the minimum number. After printing out the lines of stars, it should also print the total number of stars printed. I have tried...
## My first PERL incarnation... Audio Oscillograph
Hi all... Well guys and gals, I jumped in at the deep end and found things that PERL cannot do by default. Many tricky terminal escape codes are not catered for so I had to create workarounds. One thing I searched for was this: Passing perl variable to shell command AND, @Neo this was...
## A dash to GOTO or a dash from GOTO, that is the question...
Well, guys I saw a question about GOTO for Python. So this gave me the inspiration to attempt a GOTO function for 'dash', (bash and ksh too). Machine: MBP OSX 10.14.3, default bash terminal, calling '#!/usr/local/bin/dash'... This is purely a fun project to see if it is possible in PURE...
## Syntax error in subtraction in Bash
I am sharing a code snippet. for (( i=0; i<=\$(( \$count -1 )); i++ )) do first=\${barr2} search=\${barr1} echo \$first echo "loop begins" for (( j=0; j<=5000; j++ )) do if } == \$search ]]; then echo \$j break; fi done second=\${harr2} echo \$second
## Python: Refer a properties file from different location
Hi All, I'm having a python script: test.py in /path/to/script/test.py I'm using a properties file: test_properties.py (it is having values as dictionary{}) which is in same DIR as the script. Sample Properties file: params = { 'target_db' : 'a1_db' 'src_db' : ... | 1,696 | 6,689 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.90625 | 3 | CC-MAIN-2023-14 | longest | en | 0.719669 |
https://cs.stackexchange.com/questions/127542/find-all-words-of-length-k-in-lexicographic-order | 1,721,775,749,000,000,000 | text/html | crawl-data/CC-MAIN-2024-30/segments/1720763518130.6/warc/CC-MAIN-20240723224601-20240724014601-00653.warc.gz | 172,193,666 | 40,369 | # Find all words of length $k$ in lexicographic order
Given the alphabet $$\Sigma = \{1,2,\dots, n \}$$ and a parameter $$k$$ how can we find all words of length $$k$$ over $$\Sigma$$ in lexicographic order?
I thought of doing this recursively, but although the task is basic it seems pretty hard to come up with a general algorithm, even when using the last word as a parameter using recursion.
Will be glad for any help
• Suppose that $\Sigma = \{0,1,\ldots,9\}$, and consider the decimal representation of the first $10^k$ natural numbers (starting with zero). Can you solve the problem in this case? Commented Jun 22, 2020 at 11:37
• Oh well thanks, I guess it is the $n^k$ first natural numbers starting from zero presented in base $n$ padded with zeros to the left with $\Sigma = \{0,1,\dots,n-1\}$. Somehow thought this is much more complicated. Commented Jun 22, 2020 at 12:07
You can do it either inductively or recursively.
Inductively, you just write out the numbers from $$0$$ to $$n^{k-1}$$ in base $$n$$, adding $$1$$ to all digits.
Alternatively, you can simulate a counter in base $$n$$ – that would be faster.
Recursively, you can proceed as follows:
• Generate all solutions for $$k-1$$, and output them with prefix 1.
• Generate all solutions for $$k-1$$, and output them with prefix 2.
• ...
• Generate all solutions for $$k-1$$, and output them with prefix $$n$$.
Alternatively:
– Generate all solutions for $$k-1$$.
• For each solution $$s$$, output $$s1, s2, \ldots, sn$$. | 414 | 1,507 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 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": 16, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.796875 | 4 | CC-MAIN-2024-30 | latest | en | 0.858511 |
https://kinderdijkkoi.nl/34410/calculate-volume-loading-in-a-ball-mill.html | 1,623,597,595,000,000,000 | text/html | crawl-data/CC-MAIN-2021-25/segments/1623487608856.6/warc/CC-MAIN-20210613131257-20210613161257-00564.warc.gz | 321,985,961 | 7,480 | A Method to Determine the Ball Filling, in Miduk Copper ,
draw out mill`s internal load, and calculate the weight of ball part and stony part, separately However, maybe this method will be unusable due to its operational hardship such as grate opening Miduk copper concentrator is located in Kerman Province, Iran, and its size reduction stage
Grinding in Ball Mills: Modeling and Process Control
Keywords: Ball mills, grinding circuit, process control I Introduction Grinding in ball mills is an important technological process applied to reduce the size of particles which may have different nature and a wide diversity of physical, mechanical and chemical characteristics Typical examples are the various ores, minerals, limestone, etc
Grinding Volume Calculation In A Ball Mill
Calculate and Select Ball Mill Ball Size for Optimum Grinding Based on his work, this formula can be derived for ball diameter sizing and selection Dm 6 log dk d05 where D m the diameter of the singlesized balls in mmd the diameter of the largest chunks of ore in the mill feed in mm
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Media and Product Ball Mill Loading Guide (Percentages are based on total volume of cylinder) NOTE: With media load at 50%, voids are created equal to 20% of cylinder volume These voids are filled when product is loaded into the mill Mills can be loaded by volume ,
calculate ball mill circulating load pdf
calculate ball mill circulating load pdf; calculate ball mill circulating load pdf Circulating load - SlideShare , grinding ball mill load calculation formula - YouTube Aug 25, 2016 , This is a simple video slideshow, if you want to know more details, please click on our website , we will, Get Price VOLUME 33 NUMBER 1, 1977 .
calculation of percentage volume of gringing media in ball ,
Mill Steel Charge Volume Calculation - 911 Metallurgist Mar 19, 2017 , We can calculate the steel charge volume of a ball or rod mill and , it as the % of the volume within the liners that is ,
how to calculate ball volume - palestra-oefentherapienl
Sep 02, 2017· How To Calculate Ball Volume - spf-productionbe Volume Equation and Calculation Menu Volume of Hollow Sphere Equation and Calculator A hollow sphere is a ball that has been hollowed such the an equal thickness wall creates anopther internal ball within the external ball [Chat Online] calculate volume of a ball mill - yukonjacksgrill
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Measurement of the vertical distance between the lining and the balls (H/D):
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calculate ball mill grinding media in cement Sep 12, 2016 , calculation of grinding media in raw mill , grinding volume calculation in a ball mill EFFECTS OF GRINDING MEDIA SHAPES ON BALL MILL , rate for formula to calculate ball mill volume loading cement ball mill.
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How to calculate ball mill volume loading Home / How to calculate ball mill volume loading Jaw Crushers Jaw Crushers CJ Series Jaw Crusher HD Jaw Crusher Cone Crushers Cone Crushers CS Cone Crusher GYS Series Hydraulic Cone Crusher HP Series Multi Cylinder Hydraulic Cone Crusher
how to calculate ball mill media charge level
Ball Ball Mill Charge Calculations - ludotecahappytimeit free calculator for calculation for ball mill charge On-Line Sensors for measuring the total Ball and charge level in mill expressed as a fraction of the total volume available 442 The results of the dynamic ball filling calculations from the shoulder of the load it then rolls down the free surface of the charge to the toe of the load
how we calculate volume of ball mill
Volume Load calculator Ball Mill We can calculate the steel charge volume of a ball or rod mill and express it as the % of the volume within the liners that is filled with grinding media While the mill is stopped, the charge volume can be gotten by measuring the diameter inside the liners and the distance from the top of the charge to the top .
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calculation for ball mill foundation loading - recuerdosnl Calculate and Select Ball Mill Ball Size for Optimum Grinding In Grinding, selecting (calculate) the correct or optimum ball size that allows for the best and optimum/ideal or target grind size to be achieved by your ball mill is an important thing for a Mineral Processing Engineer AKA Metallurgist to do
Milling Step-over Distance Calculator
In many milling operations, the cutting tool must step over and make several adjacent cuts to complete machining a feature As a result, a small cusp of material, called a scallop, will remain between these cuts on any surrounding walls or on the machined surface if a ball end mill is used
average volume of a ball mill
ball mill, and is even Volume 32, Number B4 423 underflow, ball mill product and of the rod mill and increase the ball mill reduction ratio Average residence time Ball Mill Loading Orbis Machinery, LLC Media and Product Ball Mill Loading (Percentages are based on total volume of cylinder) NOTE: With media load at 50%, voids are created .
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The measurement of the ball charge volume load (or filling degree) is essential to maintain the absorbed power of the mill and consequently the mill production The most common way to evaluate the filling degree is the measurement of the free height inside the mill (vertical distance between the top of the charge and the lining in the mill axis)
TECHNICAL NOTES 8 GRINDING R P King
the mill is used primarily to lift the load (medium and charge) Additional power is required to keep the mill rotating 813 Power drawn by ball, semi-autogenous and autogenous mills A simplified picture of the mill load is shown in Figure 83 Ad this can be used to establish the essential features of a model for mill ,
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Mill Steel Charge Volume Calculation We can calculate the steel charge volume of a ball or rod mill and express it as the % of the volume within the liners that is filled with grinding media While the mill is stopped, the charge volume can be gotten by measuring the diameter inside the liners and the distance from the top of the charge to the top of the mill The % loading or change volume can .
calculation of a ball mill load versus level
How Mill Load/Charge is Estimated Grinding & Classifiion Apr 4, 2017 If crop load =ore + grinding mill media +waterBut usually, it was used in a percentage formation So how to calculate this percentageI hear that generally 40% of mill volume is used by media and 60% of it (media volume,
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calculate sag ball mill circulating load celebrationcakin calculate sagball mill circulating load installation cost of a vrm vs a ball mill calculate sag/ball calculating circulating load ball mill SZM Circulating Load Calculation Formula Here is a formula that allows you to calculate the circulating load ratio around a ball mill and .
volume of ball mill ball load - salatschuessel-beckmannde
Modeling on the Effect of Coal Loads on Kinetic Energy of Balls for , Jul 9, 2015 , to guarantee that the coal load in the ball mill is close to the optimum , is the ratio of the balls' loose volume and the mill's effective volume
Volume Calculator
This free volume calculator can compute the volumes of common shapes, including that of a sphere, cone, cube, cylinder, capsule, cap, conical frustum, ellipsoid, and square pyram Explore many other math calculators like the area and surface area calculators, as well as hundreds of other calculators related to finance, health, fitness, and more
How to Size a Ball Mill -Design Calculator & Formula
A) Total Apparent Volumetric Charge Filling – including balls and excess slurry on top of the ball charge, plus the interstitial voids in between the balls – expressed as a percentage of the net internal mill volume (inside liners) B) Overflow Discharge Mills operating at low ball fillings – slurry may accumulate on top of the ball charge; causing, the Total Charge Filling Level to .
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formula to calculate ball mill volume loading formula to calculate ball mill volume loading A Method to Determine the Ball Filling in Miduk Copper A simple equation Equation 1 exists for load percentage calculation that belongs to mill diameter and distance between top point in the mill and load surface H SERVICE ONLINE GET PRICE
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calculating of volume in ball mill capacity ball mill volume calculation kiftin Formula To Calculate Ball Mill Volume Loading fixed ball load (30% of mill volume) and 70% of critical speed, using batch grinding of a feed of mentally based expressions for the calculation of mill power más; calculating of volume in ball mill capacity 19, 2012 Calculate Ball Mill Grinding Capacity
Ball charges calculators - thecementgrindingoffice
- Ball charges: This calculator gives the surface and the average weight of the ball charg It gives also a rough interpretation of the ball charge efficiency: , This calculator analyses the granulometry of the material inside the mill and proposes a modification of the ball charge in order to improve the mill ,
CALCULATION OF BALL MILL GRINDING EFFICIENCY - Page 1 ,
Mar 08, 2013· calculation of ball mill grinding efficiency dear experts please tell me how to calculate the grinding efficiency of a closed ckt & open ckt ball mill in literatures it is written that the grinding efficiency of ball mill is very less [less than 10%] please expalin in a n excel sheet to calcualte the same thanks sidhant reply
Formula Of Ball Mill Volume - ferienwohnung-rhoenantikde
Calculate Volume Loading In A Ball Mill Formula To Calculate Ball Mill Volume Loading xed ball load 30 of mill volume and 70 of critical speed using batch grinding of a feed of mentally based expressions for the calculation of mill power ms calculating of volume in ball mill capacity 19 2012 Calculate Ball Mill Grinding Capacity
how we calculate volume of ball mill
how we calculate volume of ball mill - ceesdejagernl volume of ball mill ball load we can calculate the steel charge volume of a ball or rod mill and express it as the % of the volume circulating load formula in Get , how we calculate percentage of media in ball mill
MODELING THE SPECIFIC GRINDING ENERGY AND BALL ,
1 MODELING THE SPECIFIC GRINDING ENERGY AND BALL-MILL SCALEUP K G Tsakalakis & GA Stamboltzis NTUA Athens-Greece E-mail: [email protected] Presented at the conference IFAC 2004 held in September 2004 (Nancy-France) | 2,173 | 10,328 | {"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} | 3.078125 | 3 | CC-MAIN-2021-25 | longest | en | 0.881643 |
https://uk.mathworks.com/matlabcentral/cody/problems/8-add-two-numbers/solutions/1345346 | 1,537,808,630,000,000,000 | text/html | crawl-data/CC-MAIN-2018-39/segments/1537267160620.68/warc/CC-MAIN-20180924165426-20180924185826-00488.warc.gz | 638,308,209 | 14,473 | Cody
# Problem 8. Add two numbers
Solution 1345346
Submitted on 15 Nov 2017 by Connie Yu
This solution is locked. To view this solution, you need to provide a solution of the same size or smaller.
### Test Suite
Test Status Code Input and Output
1 Pass
a = 1; b = 2; c_correct = 3; assert(isequal(add_two_numbers(a,b),c_correct))
2 Pass
a = 17; b = 2; c_correct = 19; assert(isequal(add_two_numbers(a,b),c_correct))
3 Pass
a = -5; b = 2; c_correct = -3; assert(isequal(add_two_numbers(a,b),c_correct)) | 164 | 514 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.734375 | 3 | CC-MAIN-2018-39 | latest | en | 0.670538 |
https://lotusarise.com/seismic-waves-upsc/ | 1,719,310,196,000,000,000 | text/html | crawl-data/CC-MAIN-2024-26/segments/1718198865694.2/warc/CC-MAIN-20240625072502-20240625102502-00598.warc.gz | 320,697,079 | 67,313 | An earthquake in simple words is shaking of the earth. It caused due to the release of energy, which generates waves that travel in all directions.
Why does the earth shake?
• The release of energy occurs along a fault.
• A fault is a sharp break in the crustal rock layer.
• Rocks along a fault tend to move in opposite directions. But the friction exerted by the overlying rock strata prevents the movement of the rock layer, with time pressure builds up.
• Under intense pressure, the rock layer, at a certain point, overcomes the friction offered by the overlying layer and undergoes an abrupt movement generating shockwaves.
• This causes a release of energy, and the energy waves travel in all directions.
• The point where the energy is released is called the s focus of an earthquake, alternatively, it is called the hypocentre.
• The energy waves traveling in different directions reach the surface. The point on the surface, nearest to the focus, is called the epicenter. It is the first one to experience the waves. It is a point directly above the focus.
Earthquake Waves
• All-natural earthquakes take place in the lithosphere (depth up to 200 km from the surface of the earth).
• An instrument called ‘seismograph’ records the waves reaching the surface.
• Earthquake waves are basically of two types – body waves and surface waves
• Body waves are generated due to the release of energy at the focus and move in all directions traveling through the body of the earth. Hence, the name body waves.
• The body waves interact with the surface rocks and generate a new set of waves called surface waves. These waves move along the surface.
• The velocity of waves changes as they travel through materials with different elasticity (stiffness) (Generally density with few exceptions). The more elastic the material is, the higher is the velocity. Their direction also changes as they reflect or refract when coming across materials with different densities.
• There are two types of body waves. They are called P and S-waves.
The behavior of Earthquake Waves
• The earthquake waves are measured with the help of a seismograph and are of three types—
• The ‘P’ waves or primary waves (longitudinal nature)
• Secondary waves or ‘S’ waves (transverse in nature)
• Surface waves or ‘L’ waves are long.
• The n velocity and direction of the earthquake waves undergo changes when the medium through which they are traveling changes.
• When an earthquake or underground nuclear test sends shock waves through the Earth, the cooler areas, which generally are rigid, transmit these waves at a higher velocity than the hotter areas.
Primary Waves (P waves)
• Also called the longitudinal or compressional waves.
• Particles of the medium vibrate along the direction of propagation of the wave.
• P-waves move faster and are the first to arrive at the surface.
• These waves are of high frequency.
• They can travel in all mediums.
• The velocity of P waves in Solids > Liquids > Gases
• Their velocity depends on y shear strength or elasticity of the material.
• The shadow zone for ‘P’ waves is an area that corresponds to an angle between 103° and 142°.
• This gives clues about the Solid inner core.
Secondary Waves (S waves)
• Also called as transverse or distortion waves.
• Analogous to water ripples or light waves.
• S-waves arrive at the surface with some time lag.
• A secondary wave cannot pass through liquids or gases.
• These waves are of high-frequency waves.
• Travel at varying velocities (proportional to shear strength) through the solid part of the Earth’s crust, mantle.
• The shadow zone of ‘S’ waves extends y almost halfway around the globe from the earthquake’s focus.
• The shadow zone for ‘S’ waves is an area that corresponds to an angle between 103° and 103°.
• This observation led to the discovery of the liquid outer core. Since S waves cannot travel through liquid, they do not pass through the liquid outer core.
Surface Waves (L waves)
• Also called long-period waves.
• They are low frequency, long wavelength, and transverse vibration.
• Generally, they affect the surface of the Earth only and die out at a smaller depth.
• Develop in the immediate neighborhood of the epicenter.
• They cause displacement of rocks, and hence, the collapse of structures occurs.
• These waves are responsible for most of the destructive force of the earthquake.
• Recorded last on the seismograph.
Propagation of Earthquake Waves
• Different types of earthquake waves travel in different manners. As they move or propagate, they cause vibration in the body of the rocks through which they pass.
• P-waves vibrate parallel to the direction of the wave. This exerts pressure on the material in the direction of the propagation.
• As a result, it creates density differences in the material leading to stretching and squeezing of the material.
• The other two waves vibrate perpendicular to the direction of propagation.
• The direction of vibrations of S-waves is perpendicular to the wave direction in the vertical plane. Hence, they create s troughs and crests in the material through which they pass.
• Earthquake waves get recorded in seismographs located at far off locations.
• However, there exist some specific areas where the waves are not reported. Such a zone is called the ‘shadow zone’.
• The study of different events reveals that for each earthquake, there exists an altogether different shadow zone.
• It was observed that seismographs located at any distance within 105 ° from the epicenter, recorded the arrival of both P and S-waves.
• However, the seismographs located beyond 145 ° from the epicenter, record the arrival of P-waves, but not that of S- waves.
• Thus, a zone between ° 105 ° and 145 ° from the epicenter was identified as the shadow zone for h both the types of waves. The entire zone beyond 105 ° does not receive S-waves.
• The shadow zone of S-wave is much larger than that of the P-waves. The shadow zone of P-waves appears as a band around the earth between 105 ° and 145 ° away from the epicenter.
• The shadow zone of S-waves is not only larger in extent but it is also a little over 40 percent of the earth’s surface.
How these properties of ‘P’ and ‘S’ waves help in determining the earth’s interior?
• Reflection causes waves to rebound whereas refraction makes waves move in different directions.
• The variations in the direction of waves are inferred with the help of their record on the seismograph.
• Change in densities greatly varies the wave velocity.
• By observing the changes in velocity, the density of the earth as a whole can be estimated.
• By observing the changes in the direction of the waves (emergence of shadow zones), different layers can be identified.
Why S-waves cannot travel through liquids?
• S-waves are shear waves, which move particles perpendicularly to their direction of propagation.
• They can propagate through solid rocks because these rocks have enough shear strength.
• The shear strength is one of the forces that hold the rock together, and prevent it from falling into pieces.
• Liquids do not have the same shear strength: that is why, if you take a glass of water and suddenly remove the glass, the water will not keep its glass shape and will just flow away.
• In fact, it is just a matter of rigidity: S-waves need a medium rigid enough to propagate. Hence, S-waves do not propagate through liquids.
References: NCERT, G C Leong, Wikipedia
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Notify of | 1,618 | 7,486 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.625 | 3 | CC-MAIN-2024-26 | latest | en | 0.921937 |
https://www.physicsforums.com/threads/trying-to-prove-a-trig-identity.317280/ | 1,508,779,847,000,000,000 | text/html | crawl-data/CC-MAIN-2017-43/segments/1508187826210.56/warc/CC-MAIN-20171023164236-20171023184236-00034.warc.gz | 969,698,529 | 15,922 | # Trying to prove a trig identity
1. May 31, 2009
### james_stewart
1. The problem statement, all variables and given/known data
cos^2x-cotx
--------------- = cot^2x
sin^2x-tanx
2. Relevant equations
3. The attempt at a solution
every solution I get gives me a zero, not cot^2
2. May 31, 2009
### tiny-tim
Welcome to PF!
Hi james_stewart! Welcome to PF!
(try using the X2 tag just above the Reply box )
Either write cot= cos/sin, tan = sin/cos,
or just multiply both sides by sin2x-tanx
3. May 31, 2009
### james_stewart
i did and i'm not getting the proper results.
when i convert cot and tan to cos/sin and sin/cos i get
cos^2-cos^2
-------------
sin^2-sin^2
4. May 31, 2009
### tiny-tim
(please use the X2 tag just above the Reply box)
No, you should get cos2 - cos/sin on the top …
5. May 31, 2009
### james_stewart
i did
and on the bottom i get sin2-sin/cos
6. May 31, 2009
### tiny-tim
ok, now put sin2-sin/cos as one fraction (ie with everything over the same denominator), and the same for cos2-cos/sin
7. May 31, 2009
### james_stewart
That's how i did it. but where do i get cot2 from this?
cos2 - cos/sin
--------------
sin2 - sin/cos
8. Jun 1, 2009
### tiny-tim
erm … put sin2 - sin/cos as one fraction (ie with everything over the same denominator), and the same for cos2 - cos/sin
9. Jun 1, 2009
### rl.bhat
You can write numerator as
[cos2xsinx -cosx]/sinx. Then take cos(x) common.
Repeat the same thing for denominator and simplify.
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook | 501 | 1,587 | {"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} | 3.53125 | 4 | CC-MAIN-2017-43 | longest | en | 0.76706 |
http://mathoverflow.net/feeds/question/27972 | 1,371,591,987,000,000,000 | text/html | crawl-data/CC-MAIN-2013-20/segments/1368707434477/warc/CC-MAIN-20130516123034-00001-ip-10-60-113-184.ec2.internal.warc.gz | 159,559,949 | 2,025 | Relatively ample line bundles - MathOverflow most recent 30 from http://mathoverflow.net 2013-06-18T21:46:27Z http://mathoverflow.net/feeds/question/27972 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/27972/relatively-ample-line-bundles Relatively ample line bundles Roman Fedorov 2010-06-12T20:49:02Z 2010-06-13T19:50:14Z <p>Is it true that a line bundle is relatively ample iff its restsriction to fibers is? If so, what would be the reference?</p> http://mathoverflow.net/questions/27972/relatively-ample-line-bundles/28006#28006 Answer by Gianni Bello for Relatively ample line bundles Gianni Bello 2010-06-13T09:02:57Z 2010-06-13T09:02:57Z <p>If you admit the map to be proper and the schemes to be reasonably good it is true. A reference I know is Lazarsfeld's book "Positivity in algebraic geometry", paragraph 1.7.</p> http://mathoverflow.net/questions/27972/relatively-ample-line-bundles/28018#28018 Answer by Olivier Benoist for Relatively ample line bundles Olivier Benoist 2010-06-13T10:35:34Z 2010-06-13T19:50:14Z <p>EDIT : my previous answer was wrong. Thanks to BConrad for pointing it out.</p> <p>Here is a counterexample if the map is not proper. Let $X$ be the plane, let $Y$ be the blow-up of the plane in one point, and $U$ be $Y$ with one point of the exceptionnal divisor removed. Let $f|_U:U\to X$ be the projection and consider the line bundle $\mathcal{O}_U$. </p> <p>Since the fibers of $f|_U$ are affine (either points or the affine line), $\mathcal{O}_U$ becomes ample when restricted to fibers of $f|_U$. However, $\mathcal{O}_U$ is not $f|_U$-ample. Indeed, if it were, $\mathcal{O}_U$ would be ample on $U$, but we can compute : $$H^0(U,\mathcal{O}_U^N)=H^0(U,\mathcal{O}_U)=H^0(Y,\mathcal{O}_Y)=H^0(X,\mathcal{O}_X),$$</p> <p>where the second equality comes from property $S2$ and the third holds because $f_* \mathcal{O}_Y=\mathcal{O}_X$. Hence, $H^0(U,\mathcal{O}_U^N)$ cannot distinguish between two points of the exceptionnal divisor, and $\mathcal{O}_U$ cannot be ample on $U$.</p> <hr> <p>Warning : what follows is false. I kept it here so that the comment below remains understandable.</p> <p>Here is a counterexample if the map is not proper. Consider the inclusion $f$ of the plane minus a point $U$ in the plane $X$. The line bundle $\mathcal{O}_U$ is ample restricted to the fibers of $f$ (they're points...). However, it is not $f$-ample. Indeed, if it were, $\mathcal{O}_U$ would be ample on $U$. Choosing $N>>0$, we would get $$H^1(U,\mathcal{O}_U)=H^1(U,\mathcal{O}_U^N)=0.$$ But a simple computation via Cech cohomology shows that this cohomology group is not trivial (in fact, infinite).</p> <hr> | 860 | 2,687 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.46875 | 3 | CC-MAIN-2013-20 | latest | en | 0.714471 |
https://math.stackexchange.com/questions/3575581/prove-that-the-upper-integral-int-11fxdx-is-equal-to-0 | 1,656,494,724,000,000,000 | text/html | crawl-data/CC-MAIN-2022-27/segments/1656103626162.35/warc/CC-MAIN-20220629084939-20220629114939-00032.warc.gz | 421,075,742 | 65,758 | # Prove that the upper integral $\int_{-1}^1f(x)dx$ is equal to $0$.
Let $$f:[-1,1]\to\mathbb{R}$$ be the function defined by
$$f(x)=\begin{cases} 1&\text{ if }x=0\\ 0&\text{ if }x\neq 0.\end{cases}$$
Prove that the upper integral $$\int_{-1}^1f(x)dx$$ is equal to $$0$$.
Here is my attempt:
fix $$\epsilon > 0$$. If we compute the upper Darboux sum using the division $$D=\{-1,-\frac{\epsilon}{2},\frac{\epsilon}{2},1\}$$ we get:
$$S(D) = \sum_{i=1}^{3} \delta_i F_i$$ where $$F_i = sup\{f(x): x_{i-1} < x < x_i\}$$
$$\sum_{i=1}^{3} \delta_i F_i = (-\frac{\epsilon}{2} - (-1))F_1 + (\frac{\epsilon}{2} - (-\frac{\epsilon}{2}))F_2 + (1 -\frac{\epsilon}{2})F_3$$
since $$0 \notin[-1, -\frac{\epsilon}{2}]$$ and $$0 \notin[\frac{\epsilon}{2}, 1], F_1 = F_3= 0$$. Since $$0 \in[-\frac{\epsilon}{2}, \frac{\epsilon}{2}]$$, then $$F_2 = 1$$.
Then we have that:$$(-\frac{\epsilon}{2} - (-1))F_1 + (\frac{\epsilon}{2} - (-\frac{\epsilon}{2}))F_2 + (1 -\frac{\epsilon}{2})F_3 = (1-\frac{\epsilon}{2})(0) + (2\frac{\epsilon}{2})(1) + (1-\frac{\epsilon}{2})(0) = 0 + \epsilon$$.
It follows that $$\forall_{\epsilon>0}$$ there exists a division such that $$0 < S(D) < \epsilon$$. Also $$f(x) \geq 0$$ so $$S(D) \geq 0$$ for any division. Therefore, from the definition of the infimum $$inf\{S(D): D \text{ is a division of }[-1,1]\} = 0 = \text{ upper}\int_{-1}^1f(x)dx$$
is this proof OK?
• The general idea is correct although the notation and presentation is a bit sloppy. Mar 10, 2020 at 1:43
• That's a complicated way to say something simple. Mar 10, 2020 at 2:19
• How can I simplify it? Is it better to say that as $\epsilon$ approaches zero, the lower darboux sum approaches zero and therefore the upper integral is zero? Mar 10, 2020 at 2:33
• The lower sum is unrelated to the upper sum and upper integral. Anything you say about the lower sum means nothing about the upper sum and upper integral. Mar 10, 2020 at 2:38
• Can I say that as as ϵ approaches zero, the upper darboux sum approaches zero and therefore the upper integral is zero? Mar 10, 2020 at 2:42
You write:
It follows that $$\forall_{\epsilon>0}$$ there exists a division such that $$0 < S(D) < \epsilon$$. Also $$f(x) \geq 0$$ so $$S(D) \geq 0$$ for any division. Therefore, from the definition of the infimum $$inf\{S(D): D \text{ is a division of }[-1,1]\} = 0 = \text{ upper}\int_{-1}^1f(x)dx$$
I would write:
Thus, for all $$\epsilon > 0$$, there is a division $$D$$ such that $$S(D) = \epsilon$$. This implies that
$$\mathrm{inf} \big\{ S(D) \;\big|\; D \text{ is a partition of } [-1, 1] \big\} \leq \epsilon.$$
In the limit as $$\epsilon$$ tends to zero, we thus have
$$\mathrm{inf} \big\{ S(D) \;\big|\; D \text{ is a partition of } [-1, 1] \big\} \leq 0.$$
Since $$f(x) \geq 0$$ for all $$x$$, we have $$S(D) \geq 0$$ for any choice of division $$D$$. This means that
$$\mathrm{inf} \big\{ S(D) \;\big|\; D \text{ is a partition of } [-1, 1] \big\} \geq 0.$$
Collectively, this means the infimum is equal to zero, and so by definition of the upper integral, that is also equal to zero. | 1,133 | 3,083 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 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": 35, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.1875 | 4 | CC-MAIN-2022-27 | latest | en | 0.723748 |
https://www.includehelp.com/data-structure-tutorial/asymptotic-notations.aspx | 1,718,497,186,000,000,000 | text/html | crawl-data/CC-MAIN-2024-26/segments/1718198861618.0/warc/CC-MAIN-20240615221637-20240616011637-00093.warc.gz | 727,776,630 | 42,430 | Asymptotic Notations
Learn: What are Asymptotic Notations? How they are used to express the time complexity of algorithm? Submitted by Abhishek Jain, on July 27, 2017
Asymptotic notation employs the following notations to express the time complexity of algorithms. These are termed asymptotic notation since they are meaningful approximations of functions that represent the time or space complexity of a program.
The asymptotic notation is nothing but to assume the value of a function. In this notation the complexity is usually expressed in the form of a function f(n) , where 'n' is the input size for a given instance of the problem being solved.
The most commonly used asymptotic notations are:
1) Big O Notation
Big O is the formal method of expressing time complexity in terms of the upper bound value of an algorithms running time. It’s a measure of the longest amount of time it could possibly take for the algorithm to complete.
Definition: f(n)=O(g(n))
If there are two positive constants c and n0 such that
|f(n)|<= c|g(n)| for all n >= n0
Based on Big O notation, the algorithms can be categorized as follows:
• Constant running time algorithms: Algorithms reporting O(1) time complexity.
• Logarithmic time algorithms: O(logn) time complexity is referred to as logarithmic.
• Linear time algorithms: The time complexities of O(n) are called lineartime complexities.
• Polynomial time algorithms: In general , time complexities of the type O(nk) are called polynomial time complexities.
• Exponential time algorithms: Time complexities such as O(2n), O(3n), in general O(kn) are called as exponential time complexities.
Some of the commonly occurring time complexities in their ascending orders of magnitude are listed below:
O(1) <= O(log n) <= O(n) <= O(n.log n) <= O(n2) <= O(n3) <= O(2n)
Example
f(n) g(n)
1) 16n3+78n2+12n n3 f(n)=O(n3)
2) 34n – 90 N f(n)=O(n)
3) 56 1 f(n)=O(1)
Here, g(n) is the upper bound of the function f(n).
2) Big Omega(Ω) Notation
The asymptotic notation used to denote space or time complexity in terms of lower bound value is called as Big Omega notation.
Definition: f(n)= Ω(g(n))
If there are two positive constants c and n0 such that
|f(n)|>= c|g(n)| for all n >= n0
Example
f(n) g(n)
1) 16n3+8n2+2 n3 f(n)=Ω(n3)
2) 24n + 9 N f(n)= Ω (n)
Here, g(n) is the lower bound of the function f(n).
3) Big Theta(θ) Notation
The asymptotic notation used to denote space or time complexity in terms of tight bound value is called as Big Theta (θ) Notation.These requires both Big O and Omega (Ω), so that’s why it’s referred to as a tight bound (it must be both the upper and lower bound).
Definition: f(n)=θ (g(n))
If there are two positive constants c1,c2 and n0 such that
c1|g(n)| <= |f(n)|<= c2|g(n)| for all n >= n0
Example
f(n) g(n)
1) 28n + 9 N f(n)= θ (n) since f(n)> 28n and f(n) <= 37n for n >=1
2) 16n2+30n-90 n2 f(n)= θ (n2)
3) 7.2n + 30n 2n f(n)= θ (2n)
From the definition it implies that the function g(n) is both an upper bound and a lower bound for the function f(n) for all values of n, n>=n0. This means that f(n) is such that , f(n)= O(g(n)) and f(n)= Ω(g(n)). | 899 | 3,149 | {"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} | 4.34375 | 4 | CC-MAIN-2024-26 | latest | en | 0.904728 |
https://people.maths.bris.ac.uk/~matyd/GroupNames/288i1/C2%5E2xC4xC3sS3.html | 1,709,540,804,000,000,000 | text/html | crawl-data/CC-MAIN-2024-10/segments/1707947476432.11/warc/CC-MAIN-20240304065639-20240304095639-00857.warc.gz | 458,658,019 | 8,996 | Copied to
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## G = C22×C4×C3⋊S3order 288 = 25·32
### Direct product of C22×C4 and C3⋊S3
Series: Derived Chief Lower central Upper central
Derived series C1 — C32 — C22×C4×C3⋊S3
Chief series C1 — C3 — C32 — C3×C6 — C2×C3⋊S3 — C22×C3⋊S3 — C23×C3⋊S3 — C22×C4×C3⋊S3
Lower central C32 — C22×C4×C3⋊S3
Upper central C1 — C22×C4
Generators and relations for C22×C4×C3⋊S3
G = < a,b,c,d,e,f | a2=b2=c4=d3=e3=f2=1, ab=ba, ac=ca, ad=da, ae=ea, af=fa, bc=cb, bd=db, be=eb, bf=fb, cd=dc, ce=ec, cf=fc, de=ed, fdf=d-1, fef=e-1 >
Subgroups: 2164 in 708 conjugacy classes, 253 normal (11 characteristic)
C1, C2, C2, C2, C3, C4, C4, C22, C22, S3, C6, C2×C4, C2×C4, C23, C23, C32, Dic3, C12, D6, C2×C6, C22×C4, C22×C4, C24, C3⋊S3, C3×C6, C3×C6, C4×S3, C2×Dic3, C2×C12, C22×S3, C22×C6, C23×C4, C3⋊Dic3, C3×C12, C2×C3⋊S3, C62, S3×C2×C4, C22×Dic3, C22×C12, S3×C23, C4×C3⋊S3, C2×C3⋊Dic3, C6×C12, C22×C3⋊S3, C2×C62, S3×C22×C4, C2×C4×C3⋊S3, C22×C3⋊Dic3, C2×C6×C12, C23×C3⋊S3, C22×C4×C3⋊S3
Quotients: C1, C2, C4, C22, S3, C2×C4, C23, D6, C22×C4, C24, C3⋊S3, C4×S3, C22×S3, C23×C4, C2×C3⋊S3, S3×C2×C4, S3×C23, C4×C3⋊S3, C22×C3⋊S3, S3×C22×C4, C2×C4×C3⋊S3, C23×C3⋊S3, C22×C4×C3⋊S3
Smallest permutation representation of C22×C4×C3⋊S3
On 144 points
Generators in S144
(1 9)(2 10)(3 11)(4 12)(5 116)(6 113)(7 114)(8 115)(13 136)(14 133)(15 134)(16 135)(17 50)(18 51)(19 52)(20 49)(21 140)(22 137)(23 138)(24 139)(25 58)(26 59)(27 60)(28 57)(29 120)(30 117)(31 118)(32 119)(33 143)(34 144)(35 141)(36 142)(37 125)(38 126)(39 127)(40 128)(41 79)(42 80)(43 77)(44 78)(45 83)(46 84)(47 81)(48 82)(53 91)(54 92)(55 89)(56 90)(61 99)(62 100)(63 97)(64 98)(65 103)(66 104)(67 101)(68 102)(69 107)(70 108)(71 105)(72 106)(73 111)(74 112)(75 109)(76 110)(85 123)(86 124)(87 121)(88 122)(93 131)(94 132)(95 129)(96 130)
(1 76)(2 73)(3 74)(4 75)(5 44)(6 41)(7 42)(8 43)(9 110)(10 111)(11 112)(12 109)(13 64)(14 61)(15 62)(16 63)(17 122)(18 123)(19 124)(20 121)(21 68)(22 65)(23 66)(24 67)(25 130)(26 131)(27 132)(28 129)(29 48)(30 45)(31 46)(32 47)(33 71)(34 72)(35 69)(36 70)(37 53)(38 54)(39 55)(40 56)(49 87)(50 88)(51 85)(52 86)(57 95)(58 96)(59 93)(60 94)(77 115)(78 116)(79 113)(80 114)(81 119)(82 120)(83 117)(84 118)(89 127)(90 128)(91 125)(92 126)(97 135)(98 136)(99 133)(100 134)(101 139)(102 140)(103 137)(104 138)(105 143)(106 144)(107 141)(108 142)
(1 2 3 4)(5 6 7 8)(9 10 11 12)(13 14 15 16)(17 18 19 20)(21 22 23 24)(25 26 27 28)(29 30 31 32)(33 34 35 36)(37 38 39 40)(41 42 43 44)(45 46 47 48)(49 50 51 52)(53 54 55 56)(57 58 59 60)(61 62 63 64)(65 66 67 68)(69 70 71 72)(73 74 75 76)(77 78 79 80)(81 82 83 84)(85 86 87 88)(89 90 91 92)(93 94 95 96)(97 98 99 100)(101 102 103 104)(105 106 107 108)(109 110 111 112)(113 114 115 116)(117 118 119 120)(121 122 123 124)(125 126 127 128)(129 130 131 132)(133 134 135 136)(137 138 139 140)(141 142 143 144)
(1 16 31)(2 13 32)(3 14 29)(4 15 30)(5 36 57)(6 33 58)(7 34 59)(8 35 60)(9 135 118)(10 136 119)(11 133 120)(12 134 117)(17 137 125)(18 138 126)(19 139 127)(20 140 128)(21 40 49)(22 37 50)(23 38 51)(24 39 52)(25 113 143)(26 114 144)(27 115 141)(28 116 142)(41 71 96)(42 72 93)(43 69 94)(44 70 95)(45 75 62)(46 76 63)(47 73 64)(48 74 61)(53 88 65)(54 85 66)(55 86 67)(56 87 68)(77 107 132)(78 108 129)(79 105 130)(80 106 131)(81 111 98)(82 112 99)(83 109 100)(84 110 97)(89 124 101)(90 121 102)(91 122 103)(92 123 104)
(1 8 38)(2 5 39)(3 6 40)(4 7 37)(9 115 126)(10 116 127)(11 113 128)(12 114 125)(13 36 52)(14 33 49)(15 34 50)(16 35 51)(17 134 144)(18 135 141)(19 136 142)(20 133 143)(21 29 58)(22 30 59)(23 31 60)(24 32 57)(25 140 120)(26 137 117)(27 138 118)(28 139 119)(41 56 74)(42 53 75)(43 54 76)(44 55 73)(45 93 65)(46 94 66)(47 95 67)(48 96 68)(61 71 87)(62 72 88)(63 69 85)(64 70 86)(77 92 110)(78 89 111)(79 90 112)(80 91 109)(81 129 101)(82 130 102)(83 131 103)(84 132 104)(97 107 123)(98 108 124)(99 105 121)(100 106 122)
(1 74)(2 75)(3 76)(4 73)(5 53)(6 54)(7 55)(8 56)(9 112)(10 109)(11 110)(12 111)(13 45)(14 46)(15 47)(16 48)(17 129)(18 130)(19 131)(20 132)(21 69)(22 70)(23 71)(24 72)(25 123)(26 124)(27 121)(28 122)(29 63)(30 64)(31 61)(32 62)(33 66)(34 67)(35 68)(36 65)(37 44)(38 41)(39 42)(40 43)(49 94)(50 95)(51 96)(52 93)(57 88)(58 85)(59 86)(60 87)(77 128)(78 125)(79 126)(80 127)(81 134)(82 135)(83 136)(84 133)(89 114)(90 115)(91 116)(92 113)(97 120)(98 117)(99 118)(100 119)(101 144)(102 141)(103 142)(104 143)(105 138)(106 139)(107 140)(108 137)
G:=sub<Sym(144)| (1,9)(2,10)(3,11)(4,12)(5,116)(6,113)(7,114)(8,115)(13,136)(14,133)(15,134)(16,135)(17,50)(18,51)(19,52)(20,49)(21,140)(22,137)(23,138)(24,139)(25,58)(26,59)(27,60)(28,57)(29,120)(30,117)(31,118)(32,119)(33,143)(34,144)(35,141)(36,142)(37,125)(38,126)(39,127)(40,128)(41,79)(42,80)(43,77)(44,78)(45,83)(46,84)(47,81)(48,82)(53,91)(54,92)(55,89)(56,90)(61,99)(62,100)(63,97)(64,98)(65,103)(66,104)(67,101)(68,102)(69,107)(70,108)(71,105)(72,106)(73,111)(74,112)(75,109)(76,110)(85,123)(86,124)(87,121)(88,122)(93,131)(94,132)(95,129)(96,130), (1,76)(2,73)(3,74)(4,75)(5,44)(6,41)(7,42)(8,43)(9,110)(10,111)(11,112)(12,109)(13,64)(14,61)(15,62)(16,63)(17,122)(18,123)(19,124)(20,121)(21,68)(22,65)(23,66)(24,67)(25,130)(26,131)(27,132)(28,129)(29,48)(30,45)(31,46)(32,47)(33,71)(34,72)(35,69)(36,70)(37,53)(38,54)(39,55)(40,56)(49,87)(50,88)(51,85)(52,86)(57,95)(58,96)(59,93)(60,94)(77,115)(78,116)(79,113)(80,114)(81,119)(82,120)(83,117)(84,118)(89,127)(90,128)(91,125)(92,126)(97,135)(98,136)(99,133)(100,134)(101,139)(102,140)(103,137)(104,138)(105,143)(106,144)(107,141)(108,142), (1,2,3,4)(5,6,7,8)(9,10,11,12)(13,14,15,16)(17,18,19,20)(21,22,23,24)(25,26,27,28)(29,30,31,32)(33,34,35,36)(37,38,39,40)(41,42,43,44)(45,46,47,48)(49,50,51,52)(53,54,55,56)(57,58,59,60)(61,62,63,64)(65,66,67,68)(69,70,71,72)(73,74,75,76)(77,78,79,80)(81,82,83,84)(85,86,87,88)(89,90,91,92)(93,94,95,96)(97,98,99,100)(101,102,103,104)(105,106,107,108)(109,110,111,112)(113,114,115,116)(117,118,119,120)(121,122,123,124)(125,126,127,128)(129,130,131,132)(133,134,135,136)(137,138,139,140)(141,142,143,144), (1,16,31)(2,13,32)(3,14,29)(4,15,30)(5,36,57)(6,33,58)(7,34,59)(8,35,60)(9,135,118)(10,136,119)(11,133,120)(12,134,117)(17,137,125)(18,138,126)(19,139,127)(20,140,128)(21,40,49)(22,37,50)(23,38,51)(24,39,52)(25,113,143)(26,114,144)(27,115,141)(28,116,142)(41,71,96)(42,72,93)(43,69,94)(44,70,95)(45,75,62)(46,76,63)(47,73,64)(48,74,61)(53,88,65)(54,85,66)(55,86,67)(56,87,68)(77,107,132)(78,108,129)(79,105,130)(80,106,131)(81,111,98)(82,112,99)(83,109,100)(84,110,97)(89,124,101)(90,121,102)(91,122,103)(92,123,104), (1,8,38)(2,5,39)(3,6,40)(4,7,37)(9,115,126)(10,116,127)(11,113,128)(12,114,125)(13,36,52)(14,33,49)(15,34,50)(16,35,51)(17,134,144)(18,135,141)(19,136,142)(20,133,143)(21,29,58)(22,30,59)(23,31,60)(24,32,57)(25,140,120)(26,137,117)(27,138,118)(28,139,119)(41,56,74)(42,53,75)(43,54,76)(44,55,73)(45,93,65)(46,94,66)(47,95,67)(48,96,68)(61,71,87)(62,72,88)(63,69,85)(64,70,86)(77,92,110)(78,89,111)(79,90,112)(80,91,109)(81,129,101)(82,130,102)(83,131,103)(84,132,104)(97,107,123)(98,108,124)(99,105,121)(100,106,122), (1,74)(2,75)(3,76)(4,73)(5,53)(6,54)(7,55)(8,56)(9,112)(10,109)(11,110)(12,111)(13,45)(14,46)(15,47)(16,48)(17,129)(18,130)(19,131)(20,132)(21,69)(22,70)(23,71)(24,72)(25,123)(26,124)(27,121)(28,122)(29,63)(30,64)(31,61)(32,62)(33,66)(34,67)(35,68)(36,65)(37,44)(38,41)(39,42)(40,43)(49,94)(50,95)(51,96)(52,93)(57,88)(58,85)(59,86)(60,87)(77,128)(78,125)(79,126)(80,127)(81,134)(82,135)(83,136)(84,133)(89,114)(90,115)(91,116)(92,113)(97,120)(98,117)(99,118)(100,119)(101,144)(102,141)(103,142)(104,143)(105,138)(106,139)(107,140)(108,137)>;
G:=Group( (1,9)(2,10)(3,11)(4,12)(5,116)(6,113)(7,114)(8,115)(13,136)(14,133)(15,134)(16,135)(17,50)(18,51)(19,52)(20,49)(21,140)(22,137)(23,138)(24,139)(25,58)(26,59)(27,60)(28,57)(29,120)(30,117)(31,118)(32,119)(33,143)(34,144)(35,141)(36,142)(37,125)(38,126)(39,127)(40,128)(41,79)(42,80)(43,77)(44,78)(45,83)(46,84)(47,81)(48,82)(53,91)(54,92)(55,89)(56,90)(61,99)(62,100)(63,97)(64,98)(65,103)(66,104)(67,101)(68,102)(69,107)(70,108)(71,105)(72,106)(73,111)(74,112)(75,109)(76,110)(85,123)(86,124)(87,121)(88,122)(93,131)(94,132)(95,129)(96,130), (1,76)(2,73)(3,74)(4,75)(5,44)(6,41)(7,42)(8,43)(9,110)(10,111)(11,112)(12,109)(13,64)(14,61)(15,62)(16,63)(17,122)(18,123)(19,124)(20,121)(21,68)(22,65)(23,66)(24,67)(25,130)(26,131)(27,132)(28,129)(29,48)(30,45)(31,46)(32,47)(33,71)(34,72)(35,69)(36,70)(37,53)(38,54)(39,55)(40,56)(49,87)(50,88)(51,85)(52,86)(57,95)(58,96)(59,93)(60,94)(77,115)(78,116)(79,113)(80,114)(81,119)(82,120)(83,117)(84,118)(89,127)(90,128)(91,125)(92,126)(97,135)(98,136)(99,133)(100,134)(101,139)(102,140)(103,137)(104,138)(105,143)(106,144)(107,141)(108,142), (1,2,3,4)(5,6,7,8)(9,10,11,12)(13,14,15,16)(17,18,19,20)(21,22,23,24)(25,26,27,28)(29,30,31,32)(33,34,35,36)(37,38,39,40)(41,42,43,44)(45,46,47,48)(49,50,51,52)(53,54,55,56)(57,58,59,60)(61,62,63,64)(65,66,67,68)(69,70,71,72)(73,74,75,76)(77,78,79,80)(81,82,83,84)(85,86,87,88)(89,90,91,92)(93,94,95,96)(97,98,99,100)(101,102,103,104)(105,106,107,108)(109,110,111,112)(113,114,115,116)(117,118,119,120)(121,122,123,124)(125,126,127,128)(129,130,131,132)(133,134,135,136)(137,138,139,140)(141,142,143,144), (1,16,31)(2,13,32)(3,14,29)(4,15,30)(5,36,57)(6,33,58)(7,34,59)(8,35,60)(9,135,118)(10,136,119)(11,133,120)(12,134,117)(17,137,125)(18,138,126)(19,139,127)(20,140,128)(21,40,49)(22,37,50)(23,38,51)(24,39,52)(25,113,143)(26,114,144)(27,115,141)(28,116,142)(41,71,96)(42,72,93)(43,69,94)(44,70,95)(45,75,62)(46,76,63)(47,73,64)(48,74,61)(53,88,65)(54,85,66)(55,86,67)(56,87,68)(77,107,132)(78,108,129)(79,105,130)(80,106,131)(81,111,98)(82,112,99)(83,109,100)(84,110,97)(89,124,101)(90,121,102)(91,122,103)(92,123,104), (1,8,38)(2,5,39)(3,6,40)(4,7,37)(9,115,126)(10,116,127)(11,113,128)(12,114,125)(13,36,52)(14,33,49)(15,34,50)(16,35,51)(17,134,144)(18,135,141)(19,136,142)(20,133,143)(21,29,58)(22,30,59)(23,31,60)(24,32,57)(25,140,120)(26,137,117)(27,138,118)(28,139,119)(41,56,74)(42,53,75)(43,54,76)(44,55,73)(45,93,65)(46,94,66)(47,95,67)(48,96,68)(61,71,87)(62,72,88)(63,69,85)(64,70,86)(77,92,110)(78,89,111)(79,90,112)(80,91,109)(81,129,101)(82,130,102)(83,131,103)(84,132,104)(97,107,123)(98,108,124)(99,105,121)(100,106,122), (1,74)(2,75)(3,76)(4,73)(5,53)(6,54)(7,55)(8,56)(9,112)(10,109)(11,110)(12,111)(13,45)(14,46)(15,47)(16,48)(17,129)(18,130)(19,131)(20,132)(21,69)(22,70)(23,71)(24,72)(25,123)(26,124)(27,121)(28,122)(29,63)(30,64)(31,61)(32,62)(33,66)(34,67)(35,68)(36,65)(37,44)(38,41)(39,42)(40,43)(49,94)(50,95)(51,96)(52,93)(57,88)(58,85)(59,86)(60,87)(77,128)(78,125)(79,126)(80,127)(81,134)(82,135)(83,136)(84,133)(89,114)(90,115)(91,116)(92,113)(97,120)(98,117)(99,118)(100,119)(101,144)(102,141)(103,142)(104,143)(105,138)(106,139)(107,140)(108,137) );
G=PermutationGroup([[(1,9),(2,10),(3,11),(4,12),(5,116),(6,113),(7,114),(8,115),(13,136),(14,133),(15,134),(16,135),(17,50),(18,51),(19,52),(20,49),(21,140),(22,137),(23,138),(24,139),(25,58),(26,59),(27,60),(28,57),(29,120),(30,117),(31,118),(32,119),(33,143),(34,144),(35,141),(36,142),(37,125),(38,126),(39,127),(40,128),(41,79),(42,80),(43,77),(44,78),(45,83),(46,84),(47,81),(48,82),(53,91),(54,92),(55,89),(56,90),(61,99),(62,100),(63,97),(64,98),(65,103),(66,104),(67,101),(68,102),(69,107),(70,108),(71,105),(72,106),(73,111),(74,112),(75,109),(76,110),(85,123),(86,124),(87,121),(88,122),(93,131),(94,132),(95,129),(96,130)], [(1,76),(2,73),(3,74),(4,75),(5,44),(6,41),(7,42),(8,43),(9,110),(10,111),(11,112),(12,109),(13,64),(14,61),(15,62),(16,63),(17,122),(18,123),(19,124),(20,121),(21,68),(22,65),(23,66),(24,67),(25,130),(26,131),(27,132),(28,129),(29,48),(30,45),(31,46),(32,47),(33,71),(34,72),(35,69),(36,70),(37,53),(38,54),(39,55),(40,56),(49,87),(50,88),(51,85),(52,86),(57,95),(58,96),(59,93),(60,94),(77,115),(78,116),(79,113),(80,114),(81,119),(82,120),(83,117),(84,118),(89,127),(90,128),(91,125),(92,126),(97,135),(98,136),(99,133),(100,134),(101,139),(102,140),(103,137),(104,138),(105,143),(106,144),(107,141),(108,142)], [(1,2,3,4),(5,6,7,8),(9,10,11,12),(13,14,15,16),(17,18,19,20),(21,22,23,24),(25,26,27,28),(29,30,31,32),(33,34,35,36),(37,38,39,40),(41,42,43,44),(45,46,47,48),(49,50,51,52),(53,54,55,56),(57,58,59,60),(61,62,63,64),(65,66,67,68),(69,70,71,72),(73,74,75,76),(77,78,79,80),(81,82,83,84),(85,86,87,88),(89,90,91,92),(93,94,95,96),(97,98,99,100),(101,102,103,104),(105,106,107,108),(109,110,111,112),(113,114,115,116),(117,118,119,120),(121,122,123,124),(125,126,127,128),(129,130,131,132),(133,134,135,136),(137,138,139,140),(141,142,143,144)], [(1,16,31),(2,13,32),(3,14,29),(4,15,30),(5,36,57),(6,33,58),(7,34,59),(8,35,60),(9,135,118),(10,136,119),(11,133,120),(12,134,117),(17,137,125),(18,138,126),(19,139,127),(20,140,128),(21,40,49),(22,37,50),(23,38,51),(24,39,52),(25,113,143),(26,114,144),(27,115,141),(28,116,142),(41,71,96),(42,72,93),(43,69,94),(44,70,95),(45,75,62),(46,76,63),(47,73,64),(48,74,61),(53,88,65),(54,85,66),(55,86,67),(56,87,68),(77,107,132),(78,108,129),(79,105,130),(80,106,131),(81,111,98),(82,112,99),(83,109,100),(84,110,97),(89,124,101),(90,121,102),(91,122,103),(92,123,104)], [(1,8,38),(2,5,39),(3,6,40),(4,7,37),(9,115,126),(10,116,127),(11,113,128),(12,114,125),(13,36,52),(14,33,49),(15,34,50),(16,35,51),(17,134,144),(18,135,141),(19,136,142),(20,133,143),(21,29,58),(22,30,59),(23,31,60),(24,32,57),(25,140,120),(26,137,117),(27,138,118),(28,139,119),(41,56,74),(42,53,75),(43,54,76),(44,55,73),(45,93,65),(46,94,66),(47,95,67),(48,96,68),(61,71,87),(62,72,88),(63,69,85),(64,70,86),(77,92,110),(78,89,111),(79,90,112),(80,91,109),(81,129,101),(82,130,102),(83,131,103),(84,132,104),(97,107,123),(98,108,124),(99,105,121),(100,106,122)], [(1,74),(2,75),(3,76),(4,73),(5,53),(6,54),(7,55),(8,56),(9,112),(10,109),(11,110),(12,111),(13,45),(14,46),(15,47),(16,48),(17,129),(18,130),(19,131),(20,132),(21,69),(22,70),(23,71),(24,72),(25,123),(26,124),(27,121),(28,122),(29,63),(30,64),(31,61),(32,62),(33,66),(34,67),(35,68),(36,65),(37,44),(38,41),(39,42),(40,43),(49,94),(50,95),(51,96),(52,93),(57,88),(58,85),(59,86),(60,87),(77,128),(78,125),(79,126),(80,127),(81,134),(82,135),(83,136),(84,133),(89,114),(90,115),(91,116),(92,113),(97,120),(98,117),(99,118),(100,119),(101,144),(102,141),(103,142),(104,143),(105,138),(106,139),(107,140),(108,137)]])
96 conjugacy classes
class 1 2A ··· 2G 2H ··· 2O 3A 3B 3C 3D 4A ··· 4H 4I ··· 4P 6A ··· 6AB 12A ··· 12AF order 1 2 ··· 2 2 ··· 2 3 3 3 3 4 ··· 4 4 ··· 4 6 ··· 6 12 ··· 12 size 1 1 ··· 1 9 ··· 9 2 2 2 2 1 ··· 1 9 ··· 9 2 ··· 2 2 ··· 2
96 irreducible representations
dim 1 1 1 1 1 1 2 2 2 2 type + + + + + + + + image C1 C2 C2 C2 C2 C4 S3 D6 D6 C4×S3 kernel C22×C4×C3⋊S3 C2×C4×C3⋊S3 C22×C3⋊Dic3 C2×C6×C12 C23×C3⋊S3 C22×C3⋊S3 C22×C12 C2×C12 C22×C6 C2×C6 # reps 1 12 1 1 1 16 4 24 4 32
Matrix representation of C22×C4×C3⋊S3 in GL8(𝔽13)
1 0 0 0 0 0 0 0 0 1 0 0 0 0 0 0 0 0 12 0 0 0 0 0 0 0 0 12 0 0 0 0 0 0 0 0 1 0 0 0 0 0 0 0 0 1 0 0 0 0 0 0 0 0 1 0 0 0 0 0 0 0 0 1
,
12 0 0 0 0 0 0 0 0 12 0 0 0 0 0 0 0 0 12 0 0 0 0 0 0 0 0 12 0 0 0 0 0 0 0 0 12 0 0 0 0 0 0 0 0 12 0 0 0 0 0 0 0 0 1 0 0 0 0 0 0 0 0 1
,
5 0 0 0 0 0 0 0 0 5 0 0 0 0 0 0 0 0 12 0 0 0 0 0 0 0 0 12 0 0 0 0 0 0 0 0 12 0 0 0 0 0 0 0 0 12 0 0 0 0 0 0 0 0 1 0 0 0 0 0 0 0 0 1
,
1 0 0 0 0 0 0 0 0 1 0 0 0 0 0 0 0 0 1 0 0 0 0 0 0 0 0 1 0 0 0 0 0 0 0 0 1 0 0 0 0 0 0 0 0 1 0 0 0 0 0 0 0 0 12 1 0 0 0 0 0 0 12 0
,
12 12 0 0 0 0 0 0 1 0 0 0 0 0 0 0 0 0 12 12 0 0 0 0 0 0 1 0 0 0 0 0 0 0 0 0 12 12 0 0 0 0 0 0 1 0 0 0 0 0 0 0 0 0 0 12 0 0 0 0 0 0 1 12
,
1 0 0 0 0 0 0 0 12 12 0 0 0 0 0 0 0 0 12 0 0 0 0 0 0 0 1 1 0 0 0 0 0 0 0 0 12 0 0 0 0 0 0 0 1 1 0 0 0 0 0 0 0 0 0 1 0 0 0 0 0 0 1 0
G:=sub<GL(8,GF(13))| [1,0,0,0,0,0,0,0,0,1,0,0,0,0,0,0,0,0,12,0,0,0,0,0,0,0,0,12,0,0,0,0,0,0,0,0,1,0,0,0,0,0,0,0,0,1,0,0,0,0,0,0,0,0,1,0,0,0,0,0,0,0,0,1],[12,0,0,0,0,0,0,0,0,12,0,0,0,0,0,0,0,0,12,0,0,0,0,0,0,0,0,12,0,0,0,0,0,0,0,0,12,0,0,0,0,0,0,0,0,12,0,0,0,0,0,0,0,0,1,0,0,0,0,0,0,0,0,1],[5,0,0,0,0,0,0,0,0,5,0,0,0,0,0,0,0,0,12,0,0,0,0,0,0,0,0,12,0,0,0,0,0,0,0,0,12,0,0,0,0,0,0,0,0,12,0,0,0,0,0,0,0,0,1,0,0,0,0,0,0,0,0,1],[1,0,0,0,0,0,0,0,0,1,0,0,0,0,0,0,0,0,1,0,0,0,0,0,0,0,0,1,0,0,0,0,0,0,0,0,1,0,0,0,0,0,0,0,0,1,0,0,0,0,0,0,0,0,12,12,0,0,0,0,0,0,1,0],[12,1,0,0,0,0,0,0,12,0,0,0,0,0,0,0,0,0,12,1,0,0,0,0,0,0,12,0,0,0,0,0,0,0,0,0,12,1,0,0,0,0,0,0,12,0,0,0,0,0,0,0,0,0,0,1,0,0,0,0,0,0,12,12],[1,12,0,0,0,0,0,0,0,12,0,0,0,0,0,0,0,0,12,1,0,0,0,0,0,0,0,1,0,0,0,0,0,0,0,0,12,1,0,0,0,0,0,0,0,1,0,0,0,0,0,0,0,0,0,1,0,0,0,0,0,0,1,0] >;
C22×C4×C3⋊S3 in GAP, Magma, Sage, TeX
C_2^2\times C_4\times C_3\rtimes S_3
% in TeX
G:=Group("C2^2xC4xC3:S3");
// GroupNames label
G:=SmallGroup(288,1004);
// by ID
G=gap.SmallGroup(288,1004);
# by ID
G:=PCGroup([7,-2,-2,-2,-2,-2,-3,-3,80,2693,9414]);
// Polycyclic
G:=Group<a,b,c,d,e,f|a^2=b^2=c^4=d^3=e^3=f^2=1,a*b=b*a,a*c=c*a,a*d=d*a,a*e=e*a,a*f=f*a,b*c=c*b,b*d=d*b,b*e=e*b,b*f=f*b,c*d=d*c,c*e=e*c,c*f=f*c,d*e=e*d,f*d*f=d^-1,f*e*f=e^-1>;
// generators/relations
×
𝔽 | 9,921 | 16,989 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.671875 | 3 | CC-MAIN-2024-10 | latest | en | 0.396907 |
https://www.webqc.org/molecularweightcalculated-190210-117.html | 1,582,698,436,000,000,000 | text/html | crawl-data/CC-MAIN-2020-10/segments/1581875146187.93/warc/CC-MAIN-20200226054316-20200226084316-00037.warc.gz | 916,002,700 | 7,865 | #### Chemical Equations Balanced on 02/10/19
Molecular weights calculated on 02/09/19 Molecular weights calculated on 02/11/19
Calculate molecular weight
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45 46 47 48 49 50 51 52 53 54 55 56 57 58 59 60 61 62 63 64 65 66 67 68 69 70 71 72 73 74 75 76 77 78 79 80 81 82 83 84 85 86 87 88 89 90 91 92 93 94 95 96 97 98 99 100 101 102 103 104 105 106 107 108 109 110 111 112 113 114 115 116 117 118 119 120 121 122 123 124 125 126 127 128 129 130 131 132 133 134 135 136 137 138 139 140 141 142 143 144 145 146 147 148 149 150 151 152 153 154 155 156 157 158 159 160 161 162 163 164 165 166 167 168 169 170 171 172 173 174 175 176 177 178 179 180 181 182 183 184 185 186 187 188 189 190 191 192 193 194 195 196 197 198 199 200 201 202 203 204 205 206
Molar mass of [Co(NO3)2]6(H2O) is 1115.67325
Molar mass of [Co(NO3)2](H2O)6 is 291.034675
Molar mass of H3(PO4)3 is 287.937906
Molar mass of (NH4)2S · 2NH3 is 102.20296
Molar mass of C62h108o7 is 965.51672
Molar mass of PNI2 is 298.789402
Molar mass of N2 is 28.0134
Molar mass of [Co(NH3)4](CO3)NO3(H2O) is 267.084355
Molar mass of Mg(NO3)2 is 148,3148
Molar mass of Ba(NO3)2 is 261.3368
Molar mass of Ca(NO3)2 is 164,0878
Molar mass of Al2(SO4)3 is 342,1508772
Molar mass of Na2SO4 is 142.04213856
Molar mass of MgSO4 is 120,3676
Molar mass of Ca(OH)2 is 74,09268
Molar mass of Br2 is 159.808
Molar mass of O2 is 31.9988
Molar mass of Mg2SO4*10H2O is 324.8254
Molar mass of CaBr2 is 199.886
Molar mass of C7H11NO7S is 253.22974
Molar mass of CO is 28.0101
Molar mass of MgSO4*10H2O is 300.5204
Molar mass of NaCl is 58,44276928
Molar mass of (NH4)2SO4 is 132.13952
Molar mass of SO2 is 64.0638
Molar mass of CaCl2 is 110.984
Molar mass of CO2 is 44.0095
Molar mass of CoBr3 is 298.645195
Molar mass of Ba0 is 137,327
Molar mass of Fe(CH3COO)2 is 173.93304
Molar mass of k2(co)3 is 254.996185
Molar mass of h20 is 20.1588
Molar mass of CaSO4 is 136.1406
Molar mass of k2(CO)3 is 162.2269
Molar mass of k2(CO)3 is 162.2269
Molar mass of NaOH is 39.99710928
Molar mass of CaBr2 is 199.886
Molar mass of C4H10 is 58.1222
Molar mass of C6H5Cl is 112.5569
Molar mass of Na2HPO4 is 141,95884056
Molar mass of S4O4 is 192.2576
Molar mass of Cu2SO4 is 223,1546
Molar mass of S2O8 is 192.1252
Molar mass of S2O8 is 192.1252
Molar mass of KOH is 56.10564
Molar mass of C14H9Cl5 is 354.48626
Molar mass of NaBr is 102.89376928
Molar mass of CO2 is 44,0095
Molar mass of C6H8O6 is 176,12412
Molar mass of S4O4 is 192.2576
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Calculate molecular weight
Molecular weights calculated on 02/09/19 Molecular weights calculated on 02/11/19
Molecular masses on 02/03/19
Molecular masses on 01/11/19
Molecular masses on 02/10/18 | 1,717 | 3,497 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.578125 | 3 | CC-MAIN-2020-10 | latest | en | 0.57222 |
https://cmd.inp.nsk.su/old/cmd2/manuals/cernlib/shortwrups/node112.html | 1,669,564,240,000,000,000 | text/html | crawl-data/CC-MAIN-2022-49/segments/1669446710409.16/warc/CC-MAIN-20221127141808-20221127171808-00844.warc.gz | 205,851,315 | 2,319 | Next: E201 Least Squares Up: CERNLIB Previous: E105 Function Interpolation
## E106 Binary Search for Element in Ordered Array
Routine ID: E106
Author(s): F. JamesLibrary: KERNLIB
Submitter: Submitted: 18.10.1974
Language: FortranRevised: 27.11.1984
Integer function subprograms LOCATI and LOCATF perform a binary search in an array of non-decreasing integer or real numbers $a$1≤a2≤...≤an to locate a specified value t.
Structure:
FUNCTION subprograms
User Entry Names: LOCATI, LOCATF
Usage:
In any arithmetic expression, LOCATI(IA,N,IT) or LOCATF(A,N,T)
has the INTEGER value according to the description below.
IA,A
( INTEGER,REAL) One-dimensional array. The numbers IA(j) or A(j) must form a non-decreasing sequence for $j=1,2,...,N$ .
N
( INTEGER) Number n of elements in array IA or A.
IT,T
( INTEGER,REAL) Search value t.
Depending on four possible outcomes of the search, each subprogram returns the following value L $\left(a =IA$ or A, $t =IT$ or T):
$a$j= t for some $j$ with $1 \le j \le N$ $L=j$ $t < a$1 $L = 0$ $a$k< t < ak+1 for some $k$ with $1 \le k \le N-1$ $L=-k$ $a$n< t $L=-N$
If the value t occurs more than once in the array a, the result L may correspond to any of the occurrences.
Method:
Repeated bisection of the subscript range.
Notes:
The number of comparisons performed is approximately proportional to $lnN$ . Therefore, for large N the binary search is considerably faster than a sequential search using a DO loop. However, for N less than about 40 a DO loop is faster.
$•$
Janne Saarela
Mon Apr 3 15:06:23 METDST 1995 | 462 | 1,567 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 18, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.796875 | 3 | CC-MAIN-2022-49 | latest | en | 0.71911 |
https://www.ukessays.com/essays/engineering/manual-operation-motor-load-engineering-essay.php | 1,481,018,983,000,000,000 | text/html | crawl-data/CC-MAIN-2016-50/segments/1480698541896.91/warc/CC-MAIN-20161202170901-00447-ip-10-31-129-80.ec2.internal.warc.gz | 1,043,240,532 | 18,910 | # Manual Operation Motor Load Engineering Essay
Published:
This essay has been submitted by a student. This is not an example of the work written by our professional essay writers.
## CHAPTER 6
The method used to load the motor is torque mode. As many as load points can be applied until the motor is at the rated torque.
Couple the dynamometer to the motor.
The power supply of the dynamometer must be in zero position.
Now turn on the readout unit and the power supply.
With constant field excitation, load the motor in increasing steps until the motor current reaches rated value.
For each step measure and record: Speed, torque, armature voltage and, field voltage and Power.
Manually taken values are thoroughly analysed in MS-EXCEL plotting the graphs for torque-voltage, speed, voltage, and speed and torque.
## 6.2 Analysis For Torque Vs Voltage
The graphs have been plotted for a series of readings.The above graph represents the voltage with respect to torque. From the above graph it can be concluded that there is a linear relationship between torque and voltage which makes a sense that the voltage is at peak of 3.96v at 0 torque and steadily decreased as the torque increased.
Since it is linear the slope can be taken from the eqn
Y=MX+C……………………………………………………………..(1)
Fom the above graph the SLOPE can be caluclated which is given.Considering the two values
Now caluclating for the constant
Where V is the voltage
M is the slope
is the torque.
From the above eqn
Therefor torque can be derived from the eqn.
The above calucalted torque formula is used to caluclate the torque from voltage with the help of LABVIEW.
## SPEED AND ARMATURE VOLTAGE CHARACTERISTICS-MANUAL OPEARTION
The below presented graphs are taken for SPEED vs VOLTAGE manually. The voltage is caluclated by varying the speed with the help of a dc drive. The speed is varied in steps of 100 and the corresponding voltage readings are taken.
## 6.3 ANALYSIS FOR ARMATURE SPEED-VOLTAGE
From the above voltage speed graph we can observe that at low speed the voltage is minimum and the speed of the armature is incresed the voltage increased linearly. The graph almost shows a linear relationship between speed and voltage.
At a speed of 67 the dc voltage is around 0.1 and the speed is increased to 1488 the voltage went high upto 3.4V thus moving linearly with the speed.
Since it is a linear graph slope can be caluclated by
Y=MX+C
C=0.74
## APPLICATION OF LABVIEW FOR MOTOR- LOAD CHARACTERISTICS
From the above graphs and caluclations it can be concluded that it is not possible to manually check for the speed of motor each and every time it is varied. Some motors need to run at different speeds in different conditions as per the requirement. Manually taking the readings may also cause some errors due to human mistakes which may result in loss of time and accuracy in producing the goods. So labview can be used to avoid these errors which continuously keep the track on data. The visual Language is easy to understand with out much difficulty making it easy to understand for man at work. Programs can build in such a way that the conditional loops can be written which act according to the changes in the input.
Thus due to these features such as "Ease of use, , invaluable built-in functions, programming capabilities, flexible user interfaces, unambiguous compilation, comprehensive dataflow tracking features, and debugging tools, and compatibility with many hardware facilities make "LabVIEW a power tool in data acquisition, control and analysis.
## WORKING PROCEDURE - PC MONITORING
A eddy current dynamometer is coupled to the dc shunt wound motor(1500 rpm). An ac tachometer is coupled to the motor which gives out an ac voltage in the range of(0-9v).
The mechanical load is applied to the motor by varying the torque from the eddy current dynamometer and by using a voltage divider voltage is taken which is of range(0-4vdc).The output voltage is pure dc which can be fed to ADC.
The output from ac tachometer is a ac voltage . Ac signal cannot be fed to the adc . The ac signal obtained contains noisy disturbances which when fed directly cannot be digitised. This signal can be converted to dc by using simple precision rectifier circuit. The output from the precision full wave rectifier is a dc voltage. The dc voltage obtained contains nosiy disturbances. Signal conditioning filters the ripple content present in the signal. To eleminate this ripple a RC filter is used.The output from RC filter is a pure dc without ripple which can be fed to ADC card. The output signal from the RC filter can now be fed into ADC card which is of range(0-4v). Thus the two signals ar fed to the ADC card with in the acceptable range of(+/-5v). The signals acquired by the ADC dgitise the signals so that they are understood by the PC. The signals are now analysed with the help of Graphical Programming Language " LABVIEW".A Virual Instrument is built which takes the input signals and process them to give the output voltage.
To measure the motor at a particluar load the dynamometer can set to control in two ways. The first method is by controlling the torque where speed can be measured and vice versa. In both the cases if once the desired point is given, the tester can measure speed, torque, voltage and current depending the way the dynamometer is configured.
## SOFTWARE DESIGN
A LabView program has been designed to show the motor load characteristics which include the monitoring of Speed, Torque and Power. Torque and Speed are calculated from the voltages.
Torque is calculated from the voltage coming from eddy current dynamometer
Speed is calculated from AC Tachometer.
Fig: 7.8 STEP-BY-STEP PROCEDURE FOR LABVIEW PROGRAMMING
## FRONT PANEL DIGRAM FOR THE BLOCK DIAGRAM
The front panel window is the interface to the VI code and is one of the two lab view windows that comprise a virtual instrument. The front panel is of command buttons and status indicators that are used for controlling Vis. Push buttons, Knobs, Controls, Graphs acts as input devices. Controls are meant to simulate instrument input devices and supply data to the block diagram of the VI whereas Indicators simulate instrument output devices and display data the block diagram acquires or generates[25].
## TORQUE ACQUISITION:
The AIN.VI acquires the voltage signal. The output obtained from the voltage divider is fed to adc card. The board number is zero and the channel number has to be specified for it t recognise and acquire the data. The output from the AI is raw information and needs to be converted to voltage units.
The output from the analog input is given to the TOENG.VI which processes the voltage signal received and converts the input signal to appropriate voltage signal.
This Voltage readings can be recorded with the help of a numeric indicator which is connected to the output of the TOENG.VI. The numeric indicator can be connected to the meter which is shown under.
The output from the TO ENG.vi is given to the numeric indicator through which the readings can be recorded . The voltage from the motor is of the range(0-4v) which is compatible with the ADC.
From the manual readings taken for the voltage- torque we observed that a linear relationship exists between them. Thus from the eqn we derived that
To obtain the torque applied to the motor from the voltage signal the program has been designed for the torque equation(1).
A subtract vi is used to take the difference of the voltage and constant which is then divided by the slope Value (M) to get the Torque.
The output from the subtract vi is taken and given as one input to the divide vi and the other input is the M which is the slope. The output from the divide.vi is connected to the numercal indicator and meter indicator which gives the display for the torque value .
The Torque value obtained can be analysed with the help og Graph connected to a wave form chart. The GRAPH XY has two inputs in which the torque is connected to x-axis and the y-axis is connected to the voltage.
The Graph XY can be configured to suit the scale of x-axis and y-axis. The graphs obtained are saved and analysis has been done.
The voltage and torque readings are recorded and analysedwith the help of microsoft excel .
## LABVIEW ANALYSIS FOR TORQUE AND VOLTAGE
The below shown Torque vs Voltage graph has been obtained by running the Lab view program from which the accuracy of the characteristics can be studied.
Fig 7.9.2 torque vs voltage graph
## CONCLUSIONS :
From the graph we can observe that under no load the voltage is maximum and as the load is increased there's a steady drop in the voltage giving a linear relationship.
Another important aspect from the above graph is that the graph clearly shows the increasing steps of the torque. The rated torque of the motor is 4.725 Nm.
As the motor is loaded above the rated torque the voltage kept decreasing and after certain time it couldn't resist the load above the rated torque and hence stalled. When the load is applied above the rated torque the motor starts drawing high currents which can damage the motor condition in long run.
The above graph gives the information of the voltage readings from no load torque to full load torque and the red line represents the voltage calculated from full load torque to no load torque.
The difference of 0.2 nm torque is due to mechanical imbalance of the swing assembly.
## TORQUE VS VOLTAGE -DECREASING LOAD
The motor is started at a full load torque above the rated torque and as the torque is decreased in steps the voltage kept increasing. At full load torque the voltage is minimum and as the torque is decreased the voltage reached the peak to a max of 3.92v.
## SPEED ACQUISITION
The input voltage from the ac tachometer is acquired by the AIN.VI. The output Voltage from AIN.Vi is converted to appropriate voltage with the help of the TOENG.vi. The output voltage from the TOENG.Vi is given to the numerical indicator which continuously transmits the data received and a meter to visualise the data. is connected to the output which shows the voltage readings in the front panel.
Output from gives the voltage which is further converted to speed by applying the Speed formula calculated from the Manual Operation. Voltage obtained is multiplied by 461.73 and added to 0.74 which are calculated using the slope eqn taken from the manual readings.
The graph for Speed Voltage characteristics is plotted by using XY Graph/
C=0.74
The voltage output is then caluclted by the
Thus the output voltage from the numeric indicator is multiples with the Slope M using a numeric multiply vi and then added to constant C . Armature speed is calculated from the above methodology..
The speed of the motor is taken in RPM. The speed is acquired from the ac tacho which gives the ac voltage. This voltage is converted to dc with the help of simple precision rectifier.
TACHO VOLTAGE VS SPEED
## TACHOVOLTAGE VS SPEED IN LABVIEW
The above graph shows the customized view of tacho voltage vs speed graph. The green line represents the minimum voltage and the blue line represents the max speed at no load. As the motor is loaded there is a steady decrease in the speed. As the speed is decreased the voltage decreases. The graph shows that theres no huge variation in the speed as the motor is loaded to rated torque.
From the above graph it proves that there's a linear voltage speed relationship. The speed is maximum at the max voltage. Thus the speed can be controlled by minimizing the voltage.
## 7.4.1 TORQUE VS SPEED CHARACTERISTICS
The above graph shows the variation of motor speed starting from no load torque to full load torque. The speed declines with increasing torque. The graph can be defined by 2 load points. Torque at zero is no load point and 6 is the stall torque. From the graph it is clear that there is no much variation in the speed when loaded even above the rated torque. In shunt motors the speed variation is less when loaded and hence these motors are considered to be constant speed motors.
## TORQUE VS SPEED -LABVIEW
Theoretically the torque speed of a motor has linear relationship. The above graph shows that as the torque is increasing the speed decreases. The graph shows pretty linear relationships. The ups and downs in the graph is due to the Voltage fluctuations and mechanical vibrations in the motor which is causing it to deviate from the original readings.
When the motor is started at full load torque the motor started at very low speeds and as the torque is decreased the speed increased. The graph shows a linear relationship between torque and speed. When loaded above the rated torque the power of the motor is at max drawing heavy currents but as the load is decreased the motor started gaining the speed and reached the maximum at no load torque. The torque speed characteristics are non-linear at higher current levels.
## Torque vs Speed over load
The above graph shows the characteristics of the motor when overloaded. From the graph we can conclude that at no load torque that is when torque is at 0 the motor is at a rated speed of nearly 1500 rpm. As the torque is steadily increased in steps the speed of motor decreased gradually till 800rpm. When the torque is set to 7Nm much higher than the rated torque the motor runs above the rated efficiency. This may lead to rise in temperature, current may exceed above the ampere rating of the motor and the windings are affected. This may lead to electrical shorts in windings and hence damage the motor in long run.
## 7.5 POWER
The POWER of motor is calculated from the SPEED AND TORQUE. The torque is the input from the dynamometer and the output speed from the motor is used to calculate the Power of the motor.
## .
The efficiency of a motor is determined by Power. The speed and torque of a motor gives the efficency of a motor. Depending on the efficiency motors are used for different requirements.
STEP BY STEP PROCEDURE TO ACQUIRE POWER
Torque is acquired from voltage. Speed is acquired from the ac tacho. The product of torque and speed gives the Power. The torque is in NM and needs to be converted to lb-ft to get the standard units for power. Hence the torque is multiplied with 0.74 to get the output in lb-ft.
Speed obtained is in rpm and need not be further changed.
The product of speed and torque is taken by Multiply.VI. The output is divided by 5252 to get the result in HP.
To convert the hp to KW units the obtained results are multiplied by 0.75.
## TORQUE VS POWER CHARACTERISTICS
From the above graph we can observe that the power is minimum at no load and as the torque is increase the power is increased showing a linear graph. The power is 0 at o torque and reached a maximum of 0.9Kw at 6.6 nm. The small deviations in the graph are due to the voltage fluctuations caused by the noisy disturbances and vibrations of the motor. Hence from the above graph it proves heavy loading of the motor causes power to be consumed more
When the decrease in torque is small it cannot offset the RPM and the power still increases. When the decrease in torque is large enough it there is a steady increase in RPM and hence the power start to drop.Â
## SPEED VS POWER USING LABVIEW
From the above graphs we can see that as the speed increases power decreases. At a speed of 1500 RPM the power is at nearly 0KW as no load is applied on the motor. As soon we start loading the motor the POWER kept changing according to the change in speed. At low speed the power is high. This shows that under loaded conditions motor is using its maximum efficiency to turn the motor. If motor is loaded above rated power it starts drawing heavy currents and gets demagnetised before stalling. This Problem can be avoided by designing over load protection
ACCURACY
## TORQUE:
The torque meter is 0.9% inaccurate. This is due to the mechanical imbalance of the meter.
## TACHOMETER
The speed obtained is nearly 0.1% inaccurate.
Thus from the above accuracy we can see that the system is not that inaccurate. It has very minute errors which are almost negligible.
## CONCLUSION:
The design, implementation and investigation of "pc monitoring of dynamometer" measuring device have been studied in detail.
Factors like speed , torque and power have been briefly discussed.
After the whole analysis , it can be clearly observed that the speed of shunt wound motors run nearly at constant speed even though loaded.
Data acquisition terminology is used to log the results which are much more accurate than the manually logged data.
The characteristics of the motor at no load and at high load are thoroughly studied.
Concept of Virtual Instrumentation has been implemented to study the voltage, speed, torque, power characteristics.
With the help of Labview the real time moments of the motor are captured and analysed.
The results obtained practically are not that accurate when compared to the theoretical analysis due to mechanical vibrations and noisy disturbances of motor.
The advantages of Data acquisition through Labview is shown visually with the help of graphs plotted between manually logged results and results obtained in real time using Labview.
The need and importance of virtual instrumentation in industry is shown.
## HURDLES FACED DURING PROJECT
ADC card has been used instead of NI DAQ card which is termed as the standard Data Acqusition Card. With out the DAQ card most of the VI s doesn't work. Only basic VI's work with the ADC card which has been a major constraint in my project. DAQ cards have the feature of signal-conditioning devices with built-in features.
Sampling rate is a major feature which determines how often analog to digital conversion takes place. Due to problem with instrument drivers loaded this feature has an error which was not used to know the sampling rate of the signal.
Labview has built in filtering VI's which reduces the ripple and noisy disturbances associated with the signal. This feature can be only used with NI DAQ card which has been another major constraint.
Some of the VI's built using ADC card is not seen with the systems loaded with NI DAQ cards.
## RECOMMENDATIONS AND FUTURE WORK
Developing a programme which directly controls the motor. Conditional statements can be written to control the entire motor process.
Digital filtering concept can be implemented instead of the electronic circuits which are confusing.
Motor Vibrations can be with the help of hardware designed by National Instruments.
NI Labview FPGA model can be implemented and compact RIO hardware can be used to minimise the vibrations of motors for accurate results. | 3,991 | 18,818 | {"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} | 3.515625 | 4 | CC-MAIN-2016-50 | longest | en | 0.904217 |
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