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# New Experiments¶

1. Run an experiment by selecting [Click for Actions] button beside the dataset that you want to use. Click Predict to begin an experiment.
1. The Experiment Settings form displays and auto-fills with the selected dataset. Optionally enter a custom name for this experiment. If you do not add a name, Driverless AI will create one for you.
2. Optionally specify a validation dataset and/or a test dataset.
• The validation set is used to tune parameters (models, features, etc.). If a validation dataset is not provided, the training data is used (with holdout splits). If a validation dataset is provided, training data is not used for parameter tuning - only for training. A validation dataset can help to improve the generalization performance on shifting data distributions.
• The test dataset is used for the final stage scoring and is the dataset for which model metrics will be computed against. Test set predictions will be available at the end of the experiment. This dataset is not used during training of the modeling pipeline.

Keep in mind that these datasets must have the same number of columns as the training dataset. Also note that if provided, the validation set is not sampled down, so it can lead to large memory usage, even if accuracy=1 (which reduces the train size).

1. Specify the target (response) column. Note that not all explanatory functionality will be available for multiclass classification scenarios (scenarios with more than two outcomes). When the target column is selected, Driverless AI automatically provides the target column type and the number of rows. If this is a classification problem, then the UI shows unique and frequency statistics (Target Freq/Most Freq) for numerical columns. If this is a regression problem, then the UI shows the dataset mean and standard deviation values.

Notes Regarding Frequency:

• For data imported in versions <= 1.0.19, TARGET FREQ and MOST FREQ both represent the count of the least frequent class for numeric target columns and the count of the most frequent class for categorical target columns.
• For data imported in versions 1.0.20-1.0.22, TARGET FREQ and MOST FREQ both represent the frequency of the target class (second class in lexicographic order) for binomial target columns; the count of the most frequent class for categorical multinomial target columns; and the count of the least frequent class for numeric multinomial target columns.
• For data imported in version 1.0.23 (and later), TARGET FREQ is the frequency of the target class for binomial target columns, and MOST FREQ is the most frequent class for multinomial target columns.
1. The next step is to set the parameters and settings for the experiment. (Refer to the Experiment Settings section for more information about these settings.) You can set the parameters individually, or you can let Driverless AI infer the parameters and then override any that you disagree with. Available parameters and settings include the following:
• Dropped Columns: The columns we do not want to use as predictors such as ID columns, columns with data leakage, etc.

• Weight Column: The column that indicates the per row observation weights. If “None” is specified, each row will have an observation weight of 1.

• Fold Column: The column that indicates the fold. If “None” is specified, the folds will be determined by Driverless AI. This is set to “Disabled” if a validation set is used.

• Time Column: The column that provides a time order, if applicable. If “AUTO” is specified, Driverless AI will auto-detect a potential time order. If “OFF” is specified, auto-detection is disabled. This is set to “Disabled” if a validation set is used.

• Specify the Scorer to use for this experiment. The available scorers vary based on whether this is a classification or regression experiment. Scorers include:

• Regression: GINI, MAE, MAPE, MER, MSE, R2, RMSE (default), RMSLE, RMSPE, SMAPE, TOPDECILE
• Classification: ACCURACY, AUC (default), AUCPR, F05, F1, F2, GINI, LOGLOSS, MACROAUC, MCC
• Specify a desired relative Accuracy from 1 to 10

• Specify a desired relative Time from 1 to 10

• Specify a desired relative Interpretability from 1 to 10

Driverless AI will automatically infer the best settings for Accuracy, Time, and Interpretability and provide you with an experiment preview based on those suggestions. If you adjust these knobs, the experiment preview will automatically update based on the new settings.

Expert Settings (optional):

• Optionally specify additional expert settings for the experiment. Refer to the Expert Settings section for more information about these settings. The default values for these options are derived from the environment variables in the config.toml file. Refer to the Setting Environment Variables section for more information.

• Classification or Regression button. Driverless AI automatically determines the problem type based on the response column. Though not recommended, you can override this setting by clicking this button.
• Reproducible: This button allows you to build an experiment with a random seed and get reproducible results. If this is disabled (default), then results will vary between runs.
• Enable GPUs: Specify whether to enable GPUs. (Note that this option is ignored on CPU-only systems.)
1. Click Launch Experiment to start the experiment.

The experiment launches with a randomly generated experiment name. You can change this name at anytime during or after the experiment. Mouse over the name of the experiment to view an edit icon, then type in the desired name.

As the experiment runs, a running status displays in the upper middle portion of the UI. First Driverless AI figures out the backend and determines whether GPUs are running. Then it starts parameter tuning, followed by feature engineering. Finally, Driverless AI builds the scoring pipeline.

## Understanding the Experiment Page¶

In addition to the status, as an experiment is running, the UI also displays the following:

• The iteration data (internal validation) for each cross validation fold along with the specified scorer value. Click on a specific iteration or drag to view a range of iterations. Double click in the graph to reset the view. In this graph, each “column” represents one iteration of the experiment. During the iteration, Driverless AI will train $$n$$ models. (This is called individuals in the experiment preview.) So for any column, you may see the score value for those $$n$$ models for each iteration on the graph.

• The variable importance values. To view variable importance for a specific iteration, just select that iteration in the Iteration Data graph. The Variable Importance list will automatically update to show variable importance information for that iteration. Hover over an entry to view more info. Note: When hovering over an entry, you may notice the term “Internal[…] specification.” This label is used for features that do not need to be translated/explained and ensures that all features are uniquely identified.

The values that display are specific to the variable importance of the model class:

• XGBoost and LightGBM: Gains Variable importance. Gain-based importance is calculated from the gains a specific variable brings to the model. In the case of a decision tree, the gain-based importance will sum up the gains that occurred whenever the data was split by the given variable. The gain-based importance is normalized between 0 and 1. If a variable is never used in the model, the gain-based importance will be 0.

• GLM: The variable importance is the absolute value of the coefficient for each predictor. The variable importance is normalized between 0 and 1. If a variable is never used in the model, the importance will be 0.

• TensorFlow: TensorFlow follows the Gedeon method described in this paper: https://www.ncbi.nlm.nih.gov/pubmed/9327276.

• RuleFit: Sums over a feature’s contribution to each rule. Specifically, Driverless AI:

1. Assigns all features to have zero importance.
2. Scans through all the rules. If a feature is in that rule, Driverless AI adds its contribution (i.e, the absolute values of a rule’s coefficient ) to its overall feature importance.
3. Normalizes the importance.
• CPU/Memory information including Notifications, Logs, and Trace info. (Note that Trace is used for development/debugging and to show what the system is doing at that moment.)

• For classification problems, the lower right section includes a toggle between an ROC curve, Precision-Recall graph, Lift chart, Gains chart, and GPU Usage information (if GPUs are available). For regression problems, the lower right section includes a toggle between a Residuals chart, an Actual vs. Predicted chart, and GPU Usage information (if GPUs are available). (Refer to the Experiment Graphs section for more information.) Upon completion, an Experiment Summary section will populate in the lower right section.

• The bottom portion of the experiment screen will show any warnings that Driverless AI encounters. You can hide this pane by clicking the x icon.

## Finishing/Aborting Experiments¶

You can finish and/or abort experiments that are currently running.

• Finish Click the Finish button to stop a running experiment. Driverless AI will end the experiment and then complete the ensembling and the deployment package.
• Abort: After clicking Finish, you have the option to click Abort, which terminates the experiment. (You will be prompted to confirm the abort.) Aborted experiments will display on the Experiments page as Failed. You can restart aborted experiments by clicking the right side of the experiment, then selecting Restart from Last Checkpoint. This will start a new experiment based on the aborted one. Alternatively, you can started a new experiment based on the aborted one by selecting New Model with Same Params. Refer to Checkpointing, Rerunning, and Retraining for more information.

## Aborting Experiment Report¶

The final step that Driverless AI performs during an experiment is to complete the experiment report. During this step, you can click Abort to skip this report.

## “Pausing” an Experiment¶

A trick for “pausing” an experiment is to:

1. Abort the experiment.
2. On the Experiments page, select Restart from Last Checkpoint for the aborted experiment.
3. On the Expert Settings page, specify 0 for the Ensemble level for final modeling pipeline option in the new experiment’s Expert Settings.


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# Article

Full entry | PDF   (0.2 MB)
Keywords:
partial derivative; high-order approximation; recovery operator
Summary:
Let $\mathcal T_h$ be a triangulation of a bounded polygonal domain $\Omega \subset \Re ^2$, $\mathcal L_h$ the space of the functions from $C(\overline \Omega )$ linear on the triangles from $\mathcal T_h$ and $\Pi _h$ the interpolation operator from $C(\overline \Omega )$ to $\mathcal L_h$. For a unit vector $z$ and an inner vertex $a$ of $\mathcal T_h$, we describe the set of vectors of coefficients such that the related linear combinations of the constant derivatives $\partial \Pi _h(u)/\partial z$ on the triangles surrounding $a$ are equal to $\partial u/\partial z(a)$ for all polynomials $u$ of the total degree less than or equal to two. Then we prove that, generally, the values of the so-called recovery operators approximating the gradient $\nabla u(a)$ cannot be expressed as linear combinations of the constant gradients $\nabla \Pi _h(u)$ on the triangles surrounding $a$.
References:
[1] Ainsworth, M., Craig, A.: A posteriori error estimators in the finite element method. Numer. Math. 60 429-463 (1992). DOI 10.1007/BF01385730 | MR 1142306 | Zbl 0757.65109
[2] Ainsworth, M., Oden, J.: A Posteriori Error Estimation in Finite Element Analysis. Wiley, New York (2000). MR 1885308 | Zbl 1008.65076
[3] Dalík, J.: Local quadratic interpolation in vertices of regular triangulations. East J. Approx. 14 81-102 (2008). MR 2391625 | Zbl 1217.41005
[4] Dalík, J.: Averaging of directional derivatives in vertices of nonobtuse regular triangulations. Submitted for publication to Numer. Math.
[5] Hlaváček, I., Křížek, M., Pištora, V.: How to recover the gradient of linear elements on nonuniform triangulations. Appl. Math. 41 241-267 (1996). MR 1395685
[6] Vacek, J.: Dual variational principles for an elliptic partial differential equation. Appl. Math. 21 5-27 (1976). MR 0412594 | Zbl 0345.35035
[7] Zhang, Z., Naga, A.: A new finite element gradient recovery method: Superconvergence property. SIAM J. Sci. Comp. 26 1192-1213 (2005). DOI 10.1137/S1064827503402837 | MR 2143481 | Zbl 1078.65110
[8] Zienkiewicz, O. C., Cheung, Y. K.: The Finite Element Method in Structural and Continuum Mechanics. McGraw Hill, London (1967). Zbl 0189.24902
[9] Zienkiewicz, O. C., Zhu, J. Z.: The superconvergent patch recovery and a posteriori error estimates. Part 1: The recovery technique. Internat. J. Numer. Methods Engrg. 33 1331-1364 (1992). DOI 10.1002/nme.1620330702 | MR 1161557 | Zbl 0769.73085

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# Triple Integral – Definition, General Forms, and Examples

Through triple integrals, we can now extend our understanding of iterated integrals to expressions and functions with three variables. We can use triple integrals to calculate the volume and mass of different objects – which are important components we need in physics, structural engineering, and other fields. This is why understanding triple integrals are important if we want to work with setups where there are now three variables components.

Triple integrals represent the iterated of functions with three variables, $\boldsymbol{f(x, y, z)}$. Instead of integrating our function over a given area, we’re now integrating it over a three-dimensional figure that has volume.

At this point, it is important that you are familiar with how we set up and evaluate iterated and double integrals. Open the embedded links in a separate tab or window in case you need an immediate refresher.

Through this article, we’ll show you how we can visualize and understand what triple integrals represent. We’ll also provide a thorough explanation of how to evaluate different triple integrals. By the end of the discussion, you should also learn how to solve problems that involve triple integrals.

Let’s begin by visualizing triple integrals in three-dimensional coordinate systems then use these representations to define them.

## What Is a Triple Integral?

A triple integral is an iterated integral with three variables and over a three-dimensional region. We can treat triple integrals’ definition as an extension of that of the double integrals, but this time, we’re integrating over a volume instead of an area.

Similar to double integrals, we can start with a simple triple integral: divide the rectangular space into smaller boxes with the following dimensions: $\Delta x \times \Delta y \times \Delta z$. Adding the volume up and taking the volume’s limit will return the triple integral shown below:

\begin{aligned}\int \int \int_{S} f(x, y, z) \phantom{x} dV\end{aligned}

Keep in mind that for the triple integral to exist for the region $S$, $f(x,y , z)$ must be continuous and integrable on $S$. Let’s say we are integrating over the box with the following bounds:

\begin{aligned}S &= [a, b] \times [c, d] \times, [m, n] \end{aligned}

For this case, $[a, b]$ represents the limits with respect to $x$, $[c, d]$ are the limits for $y$, and $[m, n]$ are the limits for $z$. We can write the triple integral of $f(x, y, z)$ bounded by the box, $S$, as shown below.

\begin{aligned}\int \int \int_{S} f(x, y, z) \phantom{x} dV &= \int_{m}^{n} \int_{c}^{d} \int_{a}^{b} f(x, y, z) \phantom{x} dxdydz \end{aligned}

This is the common form that we’ll encounter when working with triple integrals, but there are six possible forms for triple integrals, in general. Now, let’s see how the expression for the triple integral changes when we’re working with more complex regions such as the one shown below.

Suppose that we’re now integrating the function over a general region, $S$. The image above shows a region, $S$, in space and parallel to the $z$-axis. This region is bounded by the region shown as $T$ and is projection of $S$ onto the $xy$-plane.

\begin{aligned}T &= \{(x, y, z) | (x, y) \in T, z_1(x, y) \leq z \leq z_2(x, y)\}\end{aligned}

This notation highlights the conditions for $S$ and $T$ as well as the limits of integration for $z$ defined by the two other variables: $x$ and $y$. We use these limits to define the triple integral of $f(x, y, z)$ over the space, $S$.

\begin{aligned}\int \int \int_S f(x, y, z) \phantom{x} dV &= \int \int \int_{z_1(x, y)}^{z_2(x, y)} \phantom{x} dA\end{aligned}

As we have learned in the past, we integrate the remaining double integral over $dA$ by integrating with respect to $x$ then $y$. Hence, we have the following:

\begin{aligned}\int \int \int_S f(x, y, z) \phantom{x} dV &= \int_{z_1}^{z_2} \int_{y_1(z)}^{y_2(z)} \int_{z_1(x, y)}^{z_2(x, y)} \phantom{x} dA\end{aligned}

There are two other possible three-dimensional figures that may encounter when working with triple integrals: the space that has a projected region on the $xz$-plane and the space that can be projection on the $yz$-plane.

\begin{aligned}\boldsymbol{xz}\textbf{-plane} &: \{S = {(x, z) \in T, y_1(x, z) \leq y_2(x, z)}\}\\\boldsymbol{yz}\textbf{-plane} &: \{S = {(y, z) \in T, x_1(y, z) \leq x_2(y, z)}\}\end{aligned}

We’re now working with three variables where we project over a two-dimensional figure each time. This means that we’re expecting six possible iterated integrals in Cartesian form and we can find all six combinations by switching the places of $dx$, $dy$, and $dz$.

 Order of Integration Triple Integral Order of Integration Triple Integral $x \rightarrow y \rightarrow z$ $\int_{z_1}^{z_2} \int_{y_1(z)}^{y_2(z)}\int_{x_1(y, z)}^{x_2(y, z)} f(x, y, z)\phantom{x}dx dy dz$ $y \rightarrow z \rightarrow x$ $\int_{x_1}^{x_2} \int_{z_1(x)}^{z_2(x)}\int_{y_1(x, y)}^{y_2(x, y)} f(x, y, z)\phantom{x}dx dy dz$ $x \rightarrow z \rightarrow y$ $\int_{y_1}^{y_2} \int_{z_1(y)}^{z_2(y)}\int_{x_1(y, z)}^{x_2(y, z)} f(x, y, z)\phantom{x}dx dz dy$ $z \rightarrow x \rightarrow y$ $\int_{z_1}^{z_2} \int_{x_1(z)}^{x_2(z)}\int_{y_1(x, z)}^{y_2(x, z)} f(x, y, z)\phantom{x}dx dz dy$ $y \rightarrow x \rightarrow z$ $\int_{z_1}^{z_2} \int_{x_1(z)}^{x_2(z)}\int_{y_1(x, z)}^{y_2(x, z)} f(x, y, z)\phantom{x}dy dx dz$ $z \rightarrow y \rightarrow x$ $\int_{x_1}^{x_2} \int_{y_1(x)}^{y_2(x)}\int_{z_1(x, z)}^{z_2(x, z)} f(x, y, z)\phantom{x}dy dx dz$

Let’s break down the first triple integral shown in this table to help you better understand how we define triple integrals.

\begin{aligned}\int_{z_1}^{z_2} \underbrace{\int_{y_1(z)}^{y_2(z)}\overbrace{\int_{x_1(y, z)}^{x_2(y, z)} f(x, y, z)\phantom{x}dx}^{^{\displaystyle{\text{Resulting function is in terms of $y$ and $z$}}}}dy}_{^{\displaystyle{\text{Resulting function is in terms of $z$}}}}dz\end{aligned}

Similar to double integrals, the bounds of the innermost integral are defined by the remaining variables. These bounds are the region we label as $T$. Now that we’ve shown you the different ways we can define triple integrals, it’s time for us to learn how we can set up the limits of integration for triple integrals and eventually learn how to evaluate them.

## How To Do Triple Integrals?

Use the iterated form of the triple integral to evaluate the function with respect to $x$, $y$, and $z$. Here are some general pointers to remember when working with triple integrals:

• Set up the three pairs of limits of integration correctly.
• Integrate the function with respect to the innermost variable – innermost to the outermost.
• When integrating the functions, only work with one variable while treating the remaining variables as constants.
• Evaluate the resulting expression at the given bounds then repeat the process until we have integrated it three times in a row.

The complexity of the process will actually depend on the given values. As expected, integrating the function, $f(x, y, z)$ over a rectangular box will be straightforward. It becomes more challenging when we have to integrate over a more complex region. In this section, we’ll show you examples to help you understand the process of triple integration.

### Finding the Triple Integral Over a Rectangular Box

When working with a triple integral defined by a rectangular box, we simply use the limits based on the box’s bounds with respect to $x$, $y$, and $z$, respectively. Since we’re working with a rectangular box, $S$, we can define its limits as $[a, b] \times [c, d] \times [m, n]$.

\begin{aligned}\int \int \int_{S} f(x, y, z) \phantom{x} dV &= \int_{m}^{n} \int_{c}^{d} \int_{a}^{b} f(x, y, z) \phantom{x} dxdydz \end{aligned}

Let’s say we want to evaluate $\int \int \int_{S} 2xyz \phantom{x} dV$ over the region, $S = [1, 2] \times [0, 4] \times [1, 3]$. Set up the triple integral by using $[1, 2]$, $[0,4]$, and $[1, 3]$ as the limits of integrations for $x$, $y$, and $z$, respectively.

\begin{aligned}\int \int \int_{S} 2xyz \phantom{x} dV &= \int_{1}^{3} \int_{0}^{4} \int_{1}^{2} 2xyz \phantom{x} dxdydz \end{aligned}

In fact, we can switch the orders for $dx$, $dy$, $dz$, and the value of the triple integral should remain the same. Just make sure that we use the correct limits of integration for each variable! Now, let’s evaluate our triple integral, $S = \int_{1}^{3} \int_{0}^{4} \int_{1}^{2} 2xyz \phantom{x} dxdydz$, by integrating the function with respect to $x$ first since it’s the innermost function. Keep in mind that we’ll be treating $y$ and $z$ as constants.

\begin{aligned}\int_{1}^{3} \int_{0}^{4} \int_{1}^{2} 2xyz \phantom{x} dxdydz &= \int_{1}^{3} \int_{0}^{4} \left[\int_{1}^{2} 2xyz \phantom{x} dx \right ]dydz\\&= \int_{1}^{3} \int_{0}^{4} 2yz\left[\int_{1}^{2} x \phantom{x} dx \right ]dydz\\&= \int_{1}^{3} \int_{0}^{4} 2yz\left[\dfrac{x^2}{2}\right]_{1}^{2}dydz  \\&= \int_{1}^{3} \int_{0}^{4} 2yz\left(\dfrac{2^2}{2} – \dfrac{1^2}{2}\right)dydz\\&= \int_{1}^{3} \int_{0}^{4} 3yz \phantom{x}dydz  \end{aligned}

As we have expected, the resulting function is defined by $y$ and $z$ only. We continue integrating our double integral, now with respect to $y$ this time.

\begin{aligned}\int_{1}^{3} \int_{0}^{4} 3yz \phantom{x}dydz &= \int_{1}^{3} \left[\int_{0}^{4} 3yz \phantom{x}dy \right ]dz\\&= \int_{1}^{3} 3z\left[\int_{0}^{4} y \phantom{x}dy \right ]dz\\&= \int_{1}^{3} 3z\left[\dfrac{y^2}{2}\right ]_{0}^{4}dz\\&= \int_{1}^{3} 3z\left[\dfrac{4^2}{2} – \dfrac{0^2}{2} \right ] dz \\&= \int_{1}^{3} 24z \phantom{x}dz\end{aligned}

We’re now left with a single integral, so let’s continue to evaluate and simplify our expression.

\begin{aligned}\int_{1}^{3} 24z \phantom{x}dz &= 24\left[\dfrac{z^2}{2} \right ]_{1}^{3}\\&= 24 \left[\dfrac{3^2}{2} – \dfrac{1^2}{2}\right ]\\&= 96 \end{aligned}

Hence, we’ve shown you how to evaluate $\int_{1}^{3} \int_{0}^{4} \int_{1}^{2} 2xyz \phantom{x} dxdydz$ and in fact, it is equal to $96$. Let’s now explore a more complex example – evaluating triple integrals over a more complex region.

### Finding the Triple Integral Over a General Region

When working with triple integrals that involve more complex regions, it is important that we account for the projection the region forms with respect to the $xy$, $xz$, or $yz$ planes- depending on what works for our problem.

Let us work on $\int \int \int_{S} z \phantom{x}dV$, where $S$ is a solid figure bounded by the following planes: $x = 0$, $y = 0$, $z = 0$, and $x + y + z = 4$. It is most helpful to sketch two graphs: 1) the actual solid region of $S$ and 2) the projection of $S$ on the plane (for our case, we’ll use the $xy$-plane).

Now, the left graph represents the tetrahedron defining the volume that we’re integrating $z$ over. From the graph, we can see that with respect to $x$, our bounds are from $z = 0$ to $z = 4- x –y$. Now, the right graph shows the projection of $S$ over the $xy$-plane. This occurs when $z = 0$, hence, we have plane bounded by $y =0$ to $y = 4 – x$ as well as $x = 0$ to $x = 4$. We can summarize the limits of integration using the notation shown below.

\begin{aligned}S &= \{(x, y, z) | 0 \leq x \leq 4,  0 \leq y \leq 4 – x, 0 \leq z \leq 4 – x –y \} \end{aligned}

Use these limits of integration to set up the complete expression for $\int \int \int_{S} z \phantom{x}dV$ – arrange the expressions so that we use the correct limits of integration for $dx$, $dy$, and $dz$.

\begin{aligned}\int \int \int_{S} z \phantom{x}dV &= \int_{0}^{4} \int_{0}^{4 – x} \int_{0}^{4 – x – y} z \phantom{x}dzdydx\end{aligned}

Once we’ve set up the triple integral, it’s time for us to evaluate the triple integral by integrating $z$ with respect to $z$ first.

\begin{aligned} \int_{0}^{4} \int_{0}^{4 – x} \int_{0}^{4 – x – y} z \phantom{x}dzdydx &= \int_{0}^{4} \int_{0}^{4 – x} \left[\int_{0}^{4 – x – y} z \phantom{x}dz \right ]dydx\\ &= \int_{0}^{4} \int_{0}^{4 – x} \left[\dfrac{z^2}{2}\right ]_{0}^{4 – x – y} dydx\\&= \dfrac{1}{2}\int_{0}^{4} \int_{0}^{4 – x} (4 – x -y)^2dydx\end{aligned}

Integrate the resulting expression with respect to $y$ then do the same with respect to $x$. We’ve simplified the calculations for you but you can work them out by applying the u-substitution method twice in a row.

\begin{aligned}\dfrac{1}{2}\int_{0}^{4} \int_{0}^{4 – x} (4 – x -y)^2dydx &= \dfrac{1}{2}\int_{0}^{4} \left[\int_{0}^{4 – x} (4 – x -y)^2dy \right ]dx\\&= \dfrac{1}{2} \int_{0}^{4} \left[- \dfrac{(4 – x – y)^3}{3}\right ]_{0}^{4 – x} dx\\&= -\dfrac{1}{6} \int_{0}^{4} [(4 – x – (4 – x))^3 – (4 – x – 0)^3]dx\\&= \dfrac{1}{6} \int_{0}^{4} (4 – x)^3 \phantom{x}dx\\&= \dfrac{1}{6}\left[-\dfrac{(4 -x)^4}{4} \right ]_{0}^{4}\\&= -\dfrac{1}{24}[(4 – 4)^4 – (4 – 0)^4]\\&= \dfrac{32}{3}\end{aligned}

From this, we can see that $\int_{0}^{4} \int_{0}^{4 – x} \int_{0}^{4 – x – y} z \phantom{x}dzdydx$ is equal to $\dfrac{32}{3}$ or approximately equal to $10.67$.

Are you getting the hang of it? Don’t worry, we’ve prepared more examples for you – from triple integrals involving rectangular boxes to integrals involving complex regions. When you’re ready to work on more problems, head over to the next section to master the process of evaluating triple integrals!

Example 1

Calculate the triple integral, $\int_{0}^{1} \int_{0}^{2} \int_{0}^{4} (xz + y) \phantom{x}dxdydz$.

Solution

The triple integral has constants as limits of integrations, so we know that we’re integrating it over a rectangular box. There is no need for us to sketch the region, so we’ll go ahead and integrate the expression three times in a row with respect to $x \rightarrow y \rightarrow z$- in that particular direction.

\begin{aligned}\int_{0}^{1} \int_{0}^{2} \int_{0}^{4} (xz + y) \phantom{x}dxdydz &= \int_{0}^{1} \int_{0}^{2} \left[\int_{0}^{4} (xz + y) \phantom{x}dx \right ]dydz\\&=\int_{0}^{1} \int_{0}^{2} \left[\dfrac{zx^2}{2} + yx\right ]_{0}^{4}dydz\\&=\int_{0}^{1} \int_{0}^{2} \left[\dfrac{z(4^2)}{2} + y(4)\right ] dydz \\&= \int_{0}^{1} \int_{0}^{2} (8z + 4y) \phantom{x}dydz\\&= \int_{0}^{1} \left[8zy + \dfrac{4y^2}{2}\right]_{0}^{2} dz\\&=\int_{0}^{1} (16z + 8) \phantom{x}dz\\&= \left[\dfrac{16z^2}{2} + 8z \right ]_{0}^{1}\\&= 8 + 8 – 0 \\&= 16 \end{aligned}

From this, we can see that $\int_{0}^{1} \int_{0}^{2} \int_{0}^{4} (xz + y) \phantom{x}dxdydz$ is equal to $16$. Apply a similar process when evaluating triple integrals over a rectangular box of any dimensions.

Example 2

Calculate the triple integral, $\int_{0}^{1} \int_{0}^{z^2} \int_{0}^{2} y\sin(z^5) \phantom{x}dxdydz$.

Solution

We’ll apply a similar process when integrating the expression, $\int_{0}^{1} \int_{0}^{z^2} \int_{0}^{2} y\sin(z^4) \phantom{x}dxdydz$. First, let’s integrate the expression with respect to $x$ first and treating the rest of the variables constant.

\begin{aligned}\int_{0}^{1} \int_{0}^{z^2} \int_{0}^{2} y\sin(z^5) \phantom{x}dxdydz &= \int_{0}^{1} \int_{0}^{z^2} \left[\int_{0}^{2} y\sin(z^5) \phantom{x}dx \right ]dydz\\&= \int_{0}^{1} \int_{0}^{z^2} \left[ y\sin(z^5)\int_{0}^{2} \phantom{x}dx \right ]dydz\\&= \int_{0}^{1} \int_{0}^{z^2} \left[ y\sin(z^5)(2 – 0) \right ]dydz\\&= \int_{0}^{1} \int_{0}^{z^2} \left[ 2y\sin(z^5)\right ]dydz\end{aligned}

Now that we have a double integral, we apply a similar process but this time, we integrate with respect to $y$ then with respect to $z$. Apply the $u$-substitution method to integrate the single integral expression. When you let $u = z^5$, we have $du = 5z^4 \phantom{x}dz$.

\begin{aligned}u &= z^5 \\ du &= 5z^4 \phantom{x}dz\\\\ \int z^4 \sin(z^5) \phantom{x}dz &= \dfrac{1}{5}\int \sin u \phantom{x}du\\&= -\dfrac{1}{5} \cos u\\&= -\dfrac{1}{5}\cos z^5\end{aligned}

Use this expression to simplify our triple integral further and find its actual value.

\begin{aligned}\int_{0}^{1} \int_{0}^{z^2} \int_{0}^{2} y\sin(z^5) \phantom{x}dxdydz &= \int_{0}^{1}z^4 \sin(z^5) \phantom{x}dz\\ &= – \dfrac{1}{5} \left[\cos z^5  \right ]_{0}^{1}\\&= -\dfrac{1}{5}[\cos 1 – 1]\\&\approx 0.0919\end{aligned}

This means that the triple integral’s value is approximately equal to $0.0919$.

Example 3

Evaluate the triple integral, $\int \int \int_{S} \sqrt{x^2 + z^2} \phantom{x}dV$, where $S$ is a solid figure bounded by the following:

\begin{aligned}\textbf{paraboloid}&: y = x^2 + z^2\\\textbf{plane}&: y = 9\end{aligned}

Solution

We’ve discussed projecting solids on the $xy$-plane and if we apply this technique to this problem, we’ll end up with a complex triple integral shown below.

\begin{aligned}\int \int \int_{S} \sqrt{x^2 + z^2} \phantom{x}dV &= \int_{-3}^{3} \int_{9}^{x^2} \int_{\sqrt{y – x^2}}^{-\sqrt{y – x^2}} sqrt{x^2 + z^2} \phantom{x}dzdydx \end{aligned}

This example will highlight why it’s helpful to also consider the other projections. Since we’re given a paraboloid, $y = x^2 + z^2$, it’s much easier to project on the $xz$-plane.

As with our previous example, it helps if you sketch the two graphs: the actual 3D figure (paraboloid) and the projected plane (circle). From the graph, we can see that the projected plane, $x^2 + z^2 = 9$, occurs only when $y = x^2 +z^2 \leq 9$. Now looking at the range of $y$ from the 3D-coordinate system, we can see that our bounds are from the paraboloid, $y = x^2 + z^2$, and the plane, $y = 9$. This means that it’s most helpful when $dy$ is the innermost differential.

\begin{aligned}\int \int \int_{S} \sqrt{x^2 + z^2} \phantom{x}dV &=\int \int_{S_{3}} \left[\int_{ x^2 + z^2}^{9 } \sqrt{x^2 + z^2} \phantom{x}dy\right] dA \end{aligned}

We won’t be expanding the rest of the triple integral, for now, so we’ll go ahead and work on integrating $\sqrt{x^2 + z^2}$ with respect to $y$.

\begin{aligned}\int_{x^2 + z^2}^{9} \sqrt{x^2 + z^2} \phantom{x}dy&=\sqrt{x^2 + z^2} \int_{x^2 + z^2}^{9} \phantom{x}dy \\&= \sqrt{x^2 + z^2} \left[y \right ]_{x^2 + z^2}^{9}\\&= \sqrt{x^2 + z^2} \left[ 9 -(x^2 + z^2)\right ]\\&= \sqrt{x^2 + z^2}(9 – x^2 -z^2) \\\\\int \int_{S_{3}} \left[\int_{9}^{x^2 + z^2} \sqrt{x^2 + z^2} \phantom{x}dy\right] dA&= \int \int_{S_3} (9 – x^2 – z^2)\sqrt{x^2 + z^2} \phantom{x}dA \end{aligned}

The reason why we’re not expanding $dA$ (although it’s possible) is that it’s much easier to evaluate the remaining double integral using polar coordinates. Using the projection on the $xz$-plane, we can rewrite $x$ as $r\cos \theta$ and $z$ as $r\sin \theta$, where $2\pi \leq \theta \leq 0$ and $0 \leq r \leq 3$. Rewrite the function and $dA$ in polar form first.

\begin{aligned}\int \int \int_{S} \sqrt{x^2 + z^2} \phantom{x}dV &= \int \int_{S_3} (9 – x^2 – z^2)\sqrt{x^2 + z^2} \phantom{x}dA \\&= \int_{0}^{2\pi} \int_{0}^{3} (9 – r^2) \sqrt{r^2} \phantom{x}rdrd\theta\\&= \int_{0}^{2\pi} \int_{0}^{3} (9r^2 – r^4) \phantom{x}drd\theta\end{aligned}

Evaluate the resulting double integral in polar coordinates. Integrate the expression with respect to $r$ then with respect to $\theta$. Check out our calculations below in case you need a guide in working out the double integral.

\begin{aligned}\int \int \int_{S} \sqrt{x^2 + z^2} \phantom{x}dV &= \int_{0}^{2\pi} \int_{0}^{3} (9r^2 – r^4) \phantom{x}drd\theta\\&=\int_{0}^{2\pi} \left[ \int_{0}^{3} (9r^2 – r^4) \phantom{x}dr\right ]d\theta \\&= \int_{0}^{2\pi} \left[3r^3 – \dfrac{r^5}{5} \right ]_{0}^{3} \phantom{x}d\theta\\&= \int_{0}^{2\pi} \dfrac{162}{5} \phantom{x}d\theta\\&= \dfrac{162}{5} (2\pi)\\&= \dfrac{324\pi}{5}\end{aligned}

This means that $\int \int \int_{S} \sqrt{x^2 + z^2} \phantom{x}dV$ is equal to $\dfrac{324\pi}{5}$ or approximately equal to $203.58$. You can also try setting up the triple integral in Cartesian form and confirm that they return the same value!

Example 4

One of the important applications of triple integrals is finding a solid object’s mass, moments, and center of mass. We can represent these components using the triple integrals shown below.

 Mass \begin{aligned}m &= \int \int \int_{S} \rho(x, y, z) \phantom{x}dV\end{aligned} Moments \begin{aligned}M_{yz} &= \int \int \int_{S} x\rho(x, y, z) \phantom{x}dV\\M_{xz} &= \int \int \int_{S} y\rho(x, y, z) \phantom{x}dV\\M_{xy} &= \int \int \int_{S} z\rho(x, y, z) \phantom{x}dV\end{aligned} Center of Mass \begin{aligned}\overline{x} &= \dfrac{M_{yz}}{m} \\\overline{y} &= \dfrac{M_{xz}}{m}\\ \overline{z} &= \dfrac{M_{xy}}{m}  \end{aligned}

Determine the center of mass of a solid with a constant density, $\rho(x, y, z)$, bounded by the solid, $x = 4y^2$, and the following planes: $x = 2z$, $z = 0$, and $x = 4$.

Solution

First, let’s visualize how the solid (it’s in fact a parabolic cylinder) looks like and its project on the $xy$-plane.

The graphs above highlight the solid, $S$, and its projection on the $xy$-plane. Using these regions, we can set up the bounds of $S$ as shown below.

\begin{aligned}S&= \left\{(x, y, z) | -1 \leq y \leq 1, 4y^2 \leq x \leq 4, 0 \leq z\leq \dfrac{x}{2}\right\}\end{aligned}

We can set up the triple integral representing the mass of the solid first and let $\rho (x, y,z)$ as the constant, $\rho$.

\begin{aligned}m &= \int \int\int_{S} \rho\phantom{x}dV\\&= \int_{-1}^{1}\int_{4y^2}^{4}\int_{0}^{x/2} \rho \phantom{x}dzdxdy\end{aligned}

Evaluate the expression by integrating the function with respect to the following variables in this specific order: $z \rightarrow x \rightarrow y$.

\begin{aligned}\int_{-1}^{1}\int_{4y^2}^{4}\int_{0}^{x/2} \rho \phantom{x}dzdxdy &= \int_{-1}^{1}\int_{4y^2}^{4}\left[\int_{0}^{x/2} \rho \phantom{x}dz \right ]dxdy\\&= \rho\int_{-1}^{1}\int_{4y^2}^{4}\dfrac{x}{2} \phantom{x} dxdy\\&= \dfrac{\rho}{2}\int_{-1}^{1}\left[\dfrac{x^2}{2}\right]_{4y^2}^{4}\phantom{x}dy\\&=\dfrac{\rho}{4}\int_{-1}^{1} (16 – 16y^4)\phantom{x}dy\\&= 4\rho\int_{-1}^{1} (1- y^4)\phantom{x} dy\\&= \dfrac{32\rho}{5}\end{aligned}

In terms of the constant density, $\rho$, the mass of the solid object is equal to $\dfrac{32 \rho}{5}$. Now that we have the mass of the object, it’s time for us to calculate its moments as shown below.

 \begin{aligned}\boldsymbol{M_{yz} = \int \int \int_{S} x\rho(x, y, z)dV}\end{aligned} \begin{aligned}\int_{-1}^{1}\int_{4y^2}^{4}\int_{0}^{x/2} x\rho \phantom{x}dzdxdy &= \int_{-1}^{1}\int_{4y^2}^{4}\left[\int_{0}^{x/2} x\rho \phantom{x}dz \right ]dxdy  \\&= \rho \int_{-1}^{1}\int_{4y^2}^{4} \dfrac{x^2}{2} \phantom{x}dxdy \\&= \dfrac{\rho}{2} \int_{-1}^{1}\left[\dfrac{x^3}{3}\right]_{4y^2}^{4}  \phantom{x}dy \\&=\dfrac{\rho}{6} \int_{-1}^{1}\left[4^3 – (4y^2)^3\right] \phantom{x}dy \\&= 64\cdot\dfrac{\rho}{6} \int_{-1}^{1} (1 – y^6) \phantom{x}dy\\&= \dfrac{32\rho}{3}\left[y – \dfrac{y^7}{7} \right ]_{-1}^{1}\\&= \dfrac{128\rho}{7}   \end{aligned} \begin{aligned}\boldsymbol{M_{xz} = \int \int \int_{S} y\rho(x, y, z)dV}\end{aligned} \begin{aligned}\int_{-1}^{1}\int_{4y^2}^{4}\int_{0}^{x/2} y\rho \phantom{x}dzdxdy &= \int_{-1}^{1}\int_{4y^2}^{4}\left[\int_{0}^{x/2} y\rho \phantom{x}dz \right ]dxdy  \\&= \rho \int_{-1}^{1}\int_{4y^2}^{4} \dfrac{yx}{2} \phantom{x}dxdy \\&= \dfrac{\rho}{2} \int_{-1}^{1}\left[\dfrac{yx^2}{2}\right]_{4y^2}^{4}  \phantom{x}dy \\&=\dfrac{\rho}{4} \int_{-1}^{1}\left[4^2y – (4y^2)^2y\right] \phantom{x}dy \\&= 4\rho\int_{-1}^{1} (y – y^3) \phantom{x}dy\\&= 4\rho\left[\dfrac{y^2}{2} – \dfrac{y^4}{4} \right ]_{-1}^{1}\\&= 0\end{aligned} \begin{aligned}\boldsymbol{M_{xy} = \int \int \int_{S} z\rho(x, y, z)dV}\end{aligned} \begin{aligned}\int_{-1}^{1}\int_{4y^2}^{4}\int_{0}^{x/2} z\rho \phantom{x}dzdxdy &= \int_{-1}^{1}\int_{4y^2}^{4}\left[\int_{0}^{x/2} z\rho \phantom{x}dz \right ]dxdy  \\&= \rho \int_{-1}^{1}\int_{4y^2}^{4} \dfrac{x^2/4}{2} \phantom{x}dxdy \\&= \dfrac{\rho}{8} \int_{-1}^{1}\left[\dfrac{x^3}{3}\right]_{4y^2}^{4}  \phantom{x}dy \\&=\dfrac{\rho}{24} \int_{-1}^{1}\left[4^3 – (4y^2)^3\right] \phantom{x}dy \\&= 64\cdot\dfrac{\rho}{24} \int_{-1}^{1} (1 – y^6) \phantom{x}dy\\&= \dfrac{8\rho}{3}\left[y – \dfrac{y^7}{7} \right ]_{-1}^{1}\\&= \dfrac{32\rho}{7}   \end{aligned}

Now that we have the moments as well as the mass of the object, we can now find the coordinates for its center of mass using the equation, $(\overline{x}, \overline{y}, \overline{z}) = \left(\dfrac{M_{yz}}{m}, \dfrac{M_{xz}}{m}, \dfrac{M_{xy}}{m}\right)$.

 \begin{aligned} \boldsymbol{\overline{x}}\end{aligned} \begin{aligned} \boldsymbol{\overline{y}}\end{aligned} \begin{aligned} \boldsymbol{\overline{z}}\end{aligned} \begin{aligned}\overline{x}&= \dfrac{M_{yz}}{m}\\&= \dfrac{\dfrac{128\rho}{7}}{\dfrac{32\rho}{5}}\\&= \dfrac{20}{7}\end{aligned} \begin{aligned}\overline{x}&= \dfrac{M_{xz}}{m}\\&= \dfrac{0}{\dfrac{32\rho}{5}}\\&= 0\end{aligned} \begin{aligned}\overline{z}&= \dfrac{M_{xy}}{m}\\&= \dfrac{\dfrac{32\rho}{7}}{\dfrac{32\rho}{5}}\\&= \dfrac{5}{7}\end{aligned} \begin{aligned}\boldsymbol{(\overline{x}, \overline{y}, \overline{z})} &= \left(\dfrac{20}{7}, 0, \dfrac{5}{7}\right) \end{aligned}

This means that the solid has a center of mass at $\left(\dfrac{20}{7}, 0, \dfrac{5}{7}\right)$. This problem shows just one of the many applications of triple integrals. This is why it’s important that we understand how to set up and evaluate triple integrals. When you’re ready, we’ve prepared more examples for you to try!

### Practice Questions

1. Calculate the triple integral, $\int_{0}^{2} \int_{0}^{4} \int_{0}^{1} (xy -z) \phantom{x}dxdydz$.
2. Calculate the triple integral, $\int_{0}^{2} \int_{0}^{z^2} \int_{0}^{3} y\cos(z^5) \phantom{x}dxdydz$.
3. Evaluate the triple integral, $\int \int \int_{S} z \phantom{x}dV$, where $S$ is a solid figure bounded by the following planes: $x = 0$, $y = 0$, $z = 0$, and $x + y + z = 9$.
4. Evaluate the triple integral, $\int \int \int_{S} \sqrt{x^2 + z^2} \phantom{x}dV$, where $S$ is a solid figure bounded by the following:
\begin{aligned}\textbf{paraboloid}&: y = x^2 + z^2\\\textbf{plane}&: y = 16\end{aligned}

5. Set up a triple integral to determine the volume of the tetrahedron, $S$, bounded by the following planes: $x = 0$, $x = 4y$, $z = 0$, and $x + 4y + z = 4$.
6. Determine the center of mass of a solid with a constant density, $\rho(x, y, z)$, bounded by the solid, $x = y^2$, and the following planes: $x = z$, $z = 0$, and $x = 1$.

1. $\int_{0}^{9} \int_{0}^{9 – x} \int_{0}^{9 – x – y} z \phantom{x}dzdydx = \dfrac{2187}{8}$
2. $\int_{0}^{2} \int_{0}^{z^2} \int_{0}^{3} y\cos(z^5) \phantom{x}dxdydz = \dfrac{3\sin 32}{10} \approx 0.165$
3. $\int_{0}^{2} \int_{0}^{4} \int_{0}^{1} (xy -z) \phantom{x}dxdydz = 0$
4.$\int \int_{S_3} \left[\int_{x^2 + z^2}^{16} \sqrt{x^2 + z^2}\phantom{x} dy\right] dA = \dfrac{4096\pi}{15} \approx 857.86$
5. $\int_{0}^{2} \int_{x/4}^{1- x/4} \int_{0}^{4 – x – 4y} \phantom{x}dzdydx = \dfrac{4}{3}$
6. $(\overline{x}, \overline{y}, \overline{z}) = \left(\dfrac{5}{7}, 0, \dfrac{5}{14}\right)$


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# Linux – desktop is mirroring home folder and deleting folders removes them permanently, how to restore? (ubuntu)

linuxUbuntu

I tried to make a shortcut (like Windows) to the downloads folder on my desktop, and later realized it caused the actual downloads folder to move, causing issues. In trying to fix them, I ended up with my home folder "home/cchilder/…" on my desktop. When I open Desktop, I see the stuff my desktop actually had before the issue. Of course I don't want to delete my entire computer, so I can't drag and remove all of this and do a simple restore.

When I delete something in home (~) or on the desktop, the operations mirror each other, and deleting a file from the shell deletes it from the desktop too.

I want to know

1. How could this possibly happen, and why?

2. How to restore only the real desktop contents (Desktop folder) and untie the desktop I see from the home folder

Thank you

cchilders@cody_pc:~/Desktop$ls -sFl total 100 4 -rwxr-xr-x 1 cchilders cchilders 271 Jul 18 00:32 atom.desktop* 8 -rw-rw-r-- 1 cchilders cchilders 4552 Aug 29 01:25 Attributes.docx 4 drwxrwxr-x 2 cchilders cchilders 4096 Sep 17 01:04 books/ 4 -rw-rw-r-- 1 cchilders cchilders 222 Apr 11 21:05 cook rice 4-11-15~ 4 -rw-rw-r-- 1 cchilders cchilders 195 Aug 28 01:37 DEMO~ 4 -rw-rw-r-- 1 cchilders cchilders 201 Aug 28 04:11 demos~ 4 drwxrwxr-x 4 cchilders cchilders 4096 Jun 4 18:56 django texts/ 4 drwx------ 3 cchilders cchilders 4096 Sep 17 14:13 Downloads/ 4 -rwxr-xr-x 1 cchilders cchilders 791 Apr 4 12:52 gedit.desktop* 4 -rw-rw-r-- 1 cchilders cchilders 1668 Apr 15 20:00 git-workflow-readme.md 4 -rw-rw-r-- 1 cchilders cchilders 1662 Apr 15 19:51 git-workflow-readme.md~ 8 -rwxr-xr-x 1 cchilders cchilders 4484 Apr 26 10:46 gnome-terminal.desktop* 12 -rwxr-xr-x 1 cchilders cchilders 8469 Mar 29 01:08 google-chrome.desktop* 4 drwxrwxr-x 2 cchilders cchilders 4096 Jun 16 18:01 hcc/ 4 drwxrwxr-x 3 cchilders cchilders 4096 Apr 12 18:33 Lynda/ 4 -rwxr-xr-x 1 cchilders cchilders 712 Mar 29 20:21 nautilus.desktop* 4 drwxrwxr-x 2 cchilders cchilders 4096 Jul 27 20:18 pics backup/ 8 -rw-rw-r-- 1 cchilders cchilders 4223 Jun 28 19:03 quiz 8 -rw-rw-r-- 1 cchilders cchilders 4223 Jun 28 16:51 quiz~  #### Best Answer It seems somehow the XDG_DESKTOP_DIR is pointing to your $HOME directory. If that is the case then it should be pointing to $HOME/Desktop To change this navigate to the .config directory. Ctrl+h shows hidden files/directories in the file manager. Look for a file called user-dirs.dirs, this is the file you will be editing. From this file you can set the default directories for various things like Downloads, Documents, Video, etc. look for the line XDG_DESKTOP_DIR and set it to: XDG_DESKTOP_DIR="$HOME/Desktop"

Save, exit and restart the file manager and possibly your machine.


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# Impedance matching with a cable of same impedance?

Recently in the german magazine "Funkamateur", there were two articles where the author claimed that it is possible to get a better standing wave ratio (VSWR) by using a specific lengths of 50Ω cable.

The first article was about a duo-band antenna for the 70cm and 2m ham radio bands [1]. This casually mentioned that the VSWR could be improved by using a specific length of 50Ω cable and provided a link to a software by the author to compute that length (in fact it was claimed that there are different lengths of cable that would achieve that purpose). It was claimed that this length of cable can transform the impedance (the word used was "Leitungstransformation").

In a second article – prompted by several readers who wrote letters to the editor – the author provided measurements that clearly showed that, indeed, the VSWR was improved with a certain length of cable [2].

Now transmission line theory says that a transformation resulting in a different VSWR can only be achieved by a piece of line with a different impedance that the characteristic impedance of the system, see my recent article on the subject. So there must be another reason why the impedance is changed – as clearly shown by the authors measurements. The author of the article claimed here that there is a discrepance between theory and practice – I'll show in the following that there is a reasonable explanation for the findings without dumping the existing electromagnetics theory.

The hint here is that the author did not use a balun or any other means that prevents current on the outer sheath of the feedline. So I had some fun modelling the setup with the antenna simulation program NEC2. The results should be reproduceable with any other antenna simulation program, at least if it contains a NEC core (e.g. EZNEC) which can model ground and not just free space. I've modelled the (simple two-element) antenna from the original article.

The following image shows the standing wave ratio (VSWR) and the impedance split into real and imaginary part directly at the antenna. The first NEC input file can be used to reproduce the results. Note that I'm using a Sommerfeld/Norton ground model with a medium quality ground. The antenna is located 15m above ground in the model because the antenna is meant to be used on a balcony.

The next image shows the same antenna with a length of feedline (the length is the one for which the author claims that it can improve the VSWR). Note that the feedline is not modelled as a piece of wire but as a transmission line (which NEC provides as an abstraction). Again we show real and imaginary part. The electrical length (NEC uses only the electrical length when modeling transmission lines, no velocity factor is necessary) is about 1.81m computed by the formula in [1] with a velocity-factor of 1. It can be clearly seen that the impedance changes but the VSWR stays the same. The corresponding second NEC input file can be used to reproduce these results.

Now to simulate the conditions at the feedpoint when there is no balun I've modeled a wire between the antenna and the feedpoint at the end of the cable in the third NEC input file. The following image shows the model, we have an additional wire which represents the outside of the coax feedline. It is connected to the antenna at one point and to the end of the coax at the second point. The length of the cable is the physical length. Except for a small increase of the electrical length due to the isolation (not with the usual coax velocity factor, we're simulating only the outer braid!) the cable in the model needs to use the real physical length of the coax as computed by the formula in the first article [1]. We're using 1.484m from [1], table 3 for an Aircell5 with the factor n=1 in the table. Now we see that, indeed, as observed in the second article [2], the standing wave ratio VSWR is indeed smaller. The reason is, that due to a missing balun, the antenna itself has changed: The outer braid of the feedline is part of the antenna.

A view with the Linux program xnec2c shows the currents in the antenna and on the outer braid of the feedline. We can clearly see that there is current on the outer braid of the feedline (the feedline in the picture is the long part to the right, in parallel to that the modelled transmission line is shown).

So the electromagnetics theory is saved, there is a physical explanation of the phenomenon. It would still be interesting to model different influences like height above ground or different ground characteristics – and find out if the results above are reproduceable with different parameters since the author claims that the observed behaviour is well known ([2] p. 35).

 [1] (1, 2, 3, 4) Klaus Warsow. Duoband-Fensterantenne für das 2-m- und 70-cm-Band. Funkamateur, (9):712–714, September 2021.
 [2] (1, 2, 3) Klaus Warsow. Impendanzanpassung mithilfe von Koaxialkabeln. Funkamateur, (1):35–37, January 2022.


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## Abstract

In order to simplify the manufacture of multiphase pump and improve the operating flexibility for gas void fraction (GVF) of the multiphase fluids that the pump transported, an eccentric rotary multiphase pump (ERMP) is presented. In this study, the structural characteristics and working principle of the ERMP are presented first. Then, the kinematic and force models are established for the key components- sliding vane and rotor. The velocity, acceleration, and force equations with shaft rotation angle are derived for each component. Based on the established models, simulations are performed for an ERMP prototype. The simulated results show that the areas opposite the sliding vane and apart from the center of the rotor have larger velocities and wear problem. Moreover, the binding force, pressure difference induced force and the normal force exercise a negative effect on the friction at the sliding vane sides and rotor. Lower shaft speed and smaller eccentric distance of the crankshaft are helpful to reduce this effect. The findings confirm that the proposed ERMP is suitable for multiphase transportation and has a higher mechanical efficiency for its advanced structure and working principle.

Key words: multiphase pump, ERMP, sliding vane, rotor, kinematic and force behavior

## 1. Introduction

Multiphase fluid transportation, which enables the transfer of multiphase fluid in a single pipeline, is considered to be an efficient and economical technology in the petrochemical industry. Due to the elimination of conventional transportation facilities such as bulky separators, fluid pumps, gas compressors and extra pipelines, the cost of the multiphase transportation technology is only about 70% of the conventional transportation technology [1].

Moreover, the multiphase transportation technology is especially helpful in the exploration of the petroleum resources at remote areas for this technology provides a simpler and more compact system. More and more petrochemical companies worldwide are attracted to the use of multiphase pumps for boosting multiphase flow and many investigations on multiphase pumps have been carried out [2–8]. According to the boosting principles of developed pumps, the multiphase pumps can be classified as the rotodynamic pump and positive displacement pump. Rotodynamic pump, such as the helicon-axial pump, transform the mechanical input power into kinetic energy of multiphase fluid by displacing it along suitable designed channels within the pumps and then converted into working pressure [9]. The rotodynamic pump can provide a large flow but is not suitable for handling multiphase fluids that gas void fraction (GVF) is more than 50%. Positive displacement pump, such as the twin-screw pump, works on the principle of enclosing a defined volume on the suction side and moving it to the discharge end [10]. The positive displacement pump can handle the multiphase fluids with a large difference in the inlet GVFs. Several multiphase boosting applications demonstrated that positive displacement pump could operate for multiphase duties successfully [11-15]. However, the rotors of the positive displacement pump are geometrically complex and require higher machining precision that results in a higher manufacture cost.

In order to simplify the manufacture of multiphase pump and improve its operating flexibility for GVF, a novel positive displacement multiphase pump names eccentric rotary multiphase pump (ERMP) is presented. The major components of the proposed ERMP are all geometry simple and easy to manufacture. Due to the synchronically movement of the rotor and the cylinder of the pump, the frictional loss between the major components are largely minimized and the mechanical efficiency of the proposed pump is higher than that of the conventional rotary mechanisms.

In this investigation, the structural characteristics and working process of the proposed ERMP are particularly introduced. Its kinematic behaviors as well as the dynamic behaviors are theoretically formulated and discussed. In addition, the effects of the eccentric distance of the crankshaft on the kinematic and dynamic behaviors of key components of the pump are theoretically analyzed.

## 2. Structural characteristics and working principle

Shown as in Figure 1, the proposed ERMP consists of a rotor, a sliding vane, a housing case, and a crankshaft. The rotor is eccentrically assembly in the cylinder that is assembly with the housing case. Due to the outside wall of the rotor and the inside wall of the cylinder are tangent to each other, a crescent-shaped working chamber is formed by the inside wall of the cylinder and the outside wall of the rotor. The upper end the sliding vane is inserted in the groove of the housing case while the lower end is assembly with the rotor by a joint. In such case, the formed crescent-shaped chamber is divided into an intake chamber and a discharge chamber by the sliding vane. The intake port and discharge port of the pump are connected to the intake chamber and discharge chamber, respectively.

 1-housing case, 2-cylinder, 3-rotor, 4-rotor bearing, 5-crankshaft, 6-sliding vane Figure 1. Structure of the proposed ERMP

During the operation of the pump, the rotor is driven by the crankshaft, which is rotate clockwise around its geometrical axis, to swing in the cylinder and change the volume of the crescent-shaped working chamber. At the same time, the sliding vane reciprocate straightly in the vane slot. Due to the compound motion of the vane, the volumes of the divided intake chamber and discharge chamber vary gradually, and thus the intake -discharge process is resulted, shown as in Figure 2. The volume of the intake chamber increase with the crankshaft angle of θ and result in a vacuum in this area during the intake process. Meanwhile, the volume of the discharge chamber decrease with the crankshaft angle of θ during the discharge process. In such case, the pump inhales the multiphase fluid continuously into the intake chamber through the intake port of the pump during the intake process and discharge the multiphase fluid out of the pump through the discharge port during the discharge process.

 Figure 2. Schematic diagram of the working process

## 3. Kinematic model for the sliding vane and rotor

### 3.1 Velocity and acceleration of the sliding vane

The rotor driven by the crankshaft, which rotates at a uniform speed, exhibits a composite planar motion. Simultaneously, the sliding vane driven by the rotor reciprocates along the vane slot. In order to analyze the kinematic characteristics of the sliding vane and the rotor, a fixed coordinate system and a rotational coordinate system have been established. Shown as in Figure 3, the fixed coordinate system ${\displaystyle XOY}$ locates the origin point at the centrum of pump’s cylinder and ${\displaystyle Y}$-axis crossing through the centerline of sliding vane. The rotational coordinate system ${\displaystyle X_{1}O_{1}Y_{1}}$ locates the origin point at the centrum of the rotor and ${\displaystyle Y_{1}}$-axis crossing the center of round end of the sliding vane. The established rotational coordinate system ${\displaystyle X_{1}O_{1}Y_{1}}$ rotates synchronously with the rotor based on the radius of the eccentric distance of the crankshaft.

 Figure 3. Coordinate systems of the analysis model

The extension length of the sliding vane, ls, can be expressed as,

 ${\displaystyle l_{s}=e(1-cos\theta )+{\frac {e^{2}}{4r}}(1-cos2\theta )}$
(1)

where ${\displaystyle e}$ is the eccentric distance of the crankshaft and ${\displaystyle r}$ is the radius of the rotor. Shown as in Figure 2, the minimum extension length of the sliding vane is zero when the crankshaft angle, ${\displaystyle \theta }$, equals to zero or ${\displaystyle 2\pi }$ while its maximum value is ${\displaystyle 2e}$ when the crankshaft angle equals to ${\displaystyle \pi }$. By differentiating the equation (1), the velocity of the sliding vane, ${\displaystyle v_{s}}$, and acceleration of the sliding vane, ${\displaystyle a_{s}}$, can be expressed as,

 ${\displaystyle v_{s}=e\omega \left(sin\theta +{\frac {e}{2r}}sin2\theta \right)}$
(2)
 ${\displaystyle a_{s}=e{\omega }^{2}\left(cos\theta +{\frac {e}{r}}cos2\theta \right)}$
(3)

where ${\displaystyle \omega }$ is the angular speed of the crankshaft which can be obtained as ${\displaystyle \omega =n\pi /30}$, where ${\displaystyle n}$ is the rotation rate of the pump. Shown as in Figure 3, the round end of the sliding vane that hinged into the rotor rotates reversely along with the rotor. Its relative angular velocity, ${\displaystyle \omega _{r}}$, can be expressed as,

 ${\displaystyle {\omega }_{r}={\frac {-e\omega cos(\theta +\beta )}{d}}}$
(4)

where ${\displaystyle d}$ is the distance from ${\displaystyle O_{1}}$ to point ${\displaystyle A}$, and ${\displaystyle \beta }$ is the intersection angle between the center line of the sliding vane and line ${\displaystyle AO_{1}}$. The value of ${\displaystyle \beta }$ is vary with crankshaft angle ${\displaystyle \theta }$ and it can be obtained as,

 ${\displaystyle \beta =arcsin\left({\frac {esin\theta }{d}}\right)}$
(5)

### 3.2 Velocity and acceleration of the rotor

The rotor of the pump is restrained by both the sliding vane and crankshaft. The movements of the pump include the rotation around the center of the cylinder ${\displaystyle O}$ and rotation around its own center ${\displaystyle O_{1}}$. As the sliding vane and the rotor hinged at the round end, the absolute velocity of the rotor at the round end equals to the velocity of the sliding vane. Based on the rotation of the entrained movement of the rotor, the direction of the entrained velocity, ${\displaystyle v_{e}}$, is vertical to ${\displaystyle OO_{1}}$ and can be obtained as,

 ${\displaystyle v_{e}=e\omega }$
(6)

The movement at point ${\displaystyle A}$ is rotating around the rotational coordinate system ${\displaystyle X_{1}O_{1}Y_{1}}$. Then, the relative velocity at point ${\displaystyle A}$ is vertical to ${\displaystyle O_{1}A}$ and its value can be obtained as,

 ${\displaystyle v_{r}=\left(r-r_{0}\right){\omega }_{p}}$
(7)

where ${\displaystyle r_{0}}$ is the radius of the round end of the sliding vane, ${\displaystyle w_{p}}$ is the rotor’s angular velocity around its own center. According to the velocity vector at point ${\displaystyle A}$, the relative velocity can be expressed as,

 ${\displaystyle v_{r}={\frac {-e\omega \left(r-r_{0}\right)cos\theta }{\sqrt {{\left(r-r_{0}\right)}^{2}-{\left(esin\theta \right)}^{2}}}}}$
(8)

Then, the angular velocity ${\displaystyle w_{p}}$ can be expressed as,

 ${\displaystyle {\omega }_{p}={\frac {-e\omega cos\theta }{\sqrt {{\left(r-r_{0}\right)}^{2}-{\left(esin\theta \right)}^{2}}}}}$
(9)

The coordinate value of point on the head surface of the rotor can be expressed under the rotational coordinate system ${\displaystyle X_{1}O_{1}Y_{1}}$ as,

 ${\displaystyle \left\{{\begin{array}{c}x_{1}=l_{B}cos\alpha \\y_{1}=l_{B}sin\alpha \end{array}}\right.}$
(10)

where ${\displaystyle l_{B}}$ is the distance between point on the head surface and the center of the rotor, ${\displaystyle \alpha }$ is the intersection angle between the coordinate axis ${\displaystyle x_{1}}$ and line ${\displaystyle BO_{1}}$.

 Figure 4. Coordinate systems of the analysis model

Based on the geometry relationship between the fixed coordinate system XOY and the rotational coordinate system ${\displaystyle X_{1}O_{1}Y_{1}}$, the coordinate value of point on the head surface of the rotor can be expressed under the fixed coordinate system ${\displaystyle XOY}$ as,

 ${\textstyle \left\{{\begin{array}{l}x=l_{B}cos\left(\alpha +\beta \right)+esin\theta \\y=l_{B}sin\left(\alpha +\beta \right)+ecos\theta \end{array}}\right.}$
(11)

Then, the velocity of the head surface on the rotor can be obtained by differentiating the above equation as,

 ${\textstyle \left\{{\begin{array}{l}x=\omega \left(ecos\theta -l_{B}sin(\alpha +\beta ){\frac {ecos\theta }{\sqrt {d^{2}-e^{2}{sin}^{2}\theta }}}\right)\\y=\omega \left(-esin\theta -l_{B}cos(\alpha +\beta ){\frac {ecos\theta }{\sqrt {d^{2}-e^{2}{sin}^{2}\theta }}}\right)\end{array}}\right.}$
(12)

## 4. Force model for sliding vane and rotor

The force model established in this research include not only the principle forces but also the often-neglected pressure difference induced force between intake chamber and discharge chamber. In these models, we assume the following:

1. The pressure pulsation of the intake and discharge processes is ignored, which means that the intake and discharge pressures are regarded as constants.
2. The effect of oil film thickness change on friction coefficient is neglected.

### 4.1 Force model for sliding vane

The forces acting on the sliding vane are defined to be positive when their directions are the same as the positive direction of the ${\displaystyle X}$-axis or ${\displaystyle Y}$-axis of the established coordinate system ${\displaystyle XOY}$, shown as in Figure 5. The forces acting on the sliding vane include:

• The normal force (${\displaystyle F_{n}}$) and tangential force (${\displaystyle F_{t}}$) at the round end of the sliding vane.
• The binding forces (${\displaystyle F_{R1}}$ and ${\displaystyle F_{R2}}$) caused by the groove that the sliding vane inserted.
• The friction forces (${\displaystyle F_{f1}}$ and ${\displaystyle F_{f2}}$) caused by the movement of the sliding vane.
• The inertia force (${\displaystyle F_{Iv}}$) of the sliding vane.
• The force (${\displaystyle F_{pb}}$) caused by the pressure difference between the intake chamber and the discharge chamber.

 Figure 5. Forces acting on the sliding vane

The friction forces acting on the sliding vane can be obtained as,

 ${\textstyle \left\{{\begin{array}{c}F_{f1}=u_{s}F_{R1}\\F_{f2}=u_{s}F_{R2}\end{array}}\right.}$
(13)

where ${\displaystyle u_{s}}$ is the friction coefficient between the sliding vane and groove. The pressure difference induced force acing on the sliding vane can be obtained as,

 ${\displaystyle F_{pb}=Hl\left(p_{c}-p_{d}\right)}$
(14)

where ${\displaystyle H}$ is the thickness of the sliding vane, ${\displaystyle l}$ is the extension length of the sliding vane, ${\displaystyle p_{c}}$ is the discharge pressure of the pump, ${\displaystyle p_{d}}$ is the intake pressure of the pump.

Based on the rotation of the rotor, the tangential force at the round end of the sliding vane can be obtained as,

 ${\displaystyle F_{t}={\frac {M_{c}-M_{JZ}}{r-r_{0}}}}$
(15)

where ${\displaystyle r}$ is the radius of the rotor, ${\displaystyle r_{0}}$ is the radius of the round end of the sliding vane, ${\displaystyle M_{c}}$ is the friction torque between the rotor and crankshaft, ${\displaystyle M_{JZ}}$ is the inertia moment acting on the rotor.

According to the conditions of force equilibrium and moment equilibrium on the sliding vane with a fixed axis, the equilibrium equations of the sliding vane can be expressed as,

 ${\displaystyle \left\{{\begin{array}{l}F_{pb}-F_{R1}+F_{R2}+F_{n}sin\beta -F_{t}cos\beta =0\\F_{f1}+F_{f2}-F_{n}cos\beta -F_{t}sin\beta -F_{Iv}=0\\(F_{f1}-F_{f2})H-F_{R1}(l_{0}-2l)-F_{R2}l_{0}+F_{pb}(l_{0}-l)+F_{n}l_{0}sin\beta -F_{t}l_{0}cos\beta =0\end{array}}\right.}$
(16)

where ${\displaystyle l_{0}}$ is the length of the sliding vane.

### 4.2 Force model for rotor

The force acting on the rotor include the normal force and tangential force induced by the sliding vane, force (${\displaystyle F_{pg}}$) induced by the pressure difference of the multiphase flow, inertia force (${\displaystyle F_{Ip}}$) of the rotor.

 Figure 6. Forces acting on the rotor

Different pressure exist between the intake chamber and discharge chamber of the pump during its application. The pressure difference induced force would loaded on the rotor during the application of the pump, shown as in Figure 6. The value of the force can be obtained as,

 ${\displaystyle F_{p{\mbox{g}}}=2rH(p_{c}-p_{d})sin{\frac {\theta +\beta }{2}}}$
(17)

The resistance moment (${\displaystyle M_{pg}}$) caused by the force mentioned above can be expressed as,

 ${\displaystyle M_{pg}=2F_{pg}sin{\frac {\theta +\beta }{2}}}$
(18)

The inertia force of the rotor can be obtained as,

 ${\displaystyle F_{Ip}=me{\omega }^{2}}$
(19)

where ${\displaystyle m}$ is the weight of the rotor.

In this research, the friction torque (${\displaystyle M_{c}}$) between the rotor and crankshaft is considered. Based on the lubrication theory, the friction torque between the rotor and crankshaft can be obtained as,

 ${\displaystyle M_{c}={\frac {2\pi \mu \left(\omega -{\omega }_{p}\right)R_{e}^{3}l_{e}}{{\delta }_{e}}}}$
(20)

where ${\displaystyle R_{e}}$ is the radius of the crankshaft, ${\displaystyle \delta _{e}}$ is the interstice between the rotor and crankshaft, ${\displaystyle l_{e}}$ is the length of the crankshaft, ${\displaystyle \mu }$ is the frictional coefficient.

The resistance moment (${\displaystyle M_{n}}$) caused by the normal force at round end of the sliding vane can be expressed as,

 ${\displaystyle M_{n}=eF_{n}sin\left(\beta +\theta \right)}$
(21)

Based on the research mentioned above, the radial force acting on the rotor can be obtained as,

 ${\displaystyle F={\sqrt {F_{x}^{2}+F_{y}^{2}}}}$
(22)

where ${\displaystyle F_{x}}$ and ${\displaystyle F_{y}}$ are the radial components on the ${\displaystyle X}$-axis and ${\displaystyle Y}$-axis, and can be obtained as,

 ${\displaystyle \left\{{\begin{array}{c}F_{x}=F_{t}cos\beta -F_{n}sin\beta +F_{pg}cos{\frac {\beta +\theta }{2}}\\F_{y}=F_{n}cos\beta +F_{t}sin\beta -F_{pg}sin{\frac {\beta +\theta }{2}}\end{array}}\right.}$
(23)

The resistance moment (${\displaystyle M_{z}}$) acting on the rotor can be obtained as,

 ${\displaystyle M_{z}=M_{c}+M_{n}+M_{pg}}$
(24)

Then, the capacity of the multiphase pump can be obtained as,

 ${\displaystyle P={\frac {n}{60}}{\int }_{0}^{2\pi }M_{z}d\theta }$
(25)

## 5. Simulation results and discussions

In this investigation, the ERMP porotype with rated flow of 6${\displaystyle m^{3}/h}$ is designed for multiphase transportation, where its main dimensions are given in Table 1. Based on the established model, kinematic characteristics of rotor’s components and forces acting on them can be simulated.

Table 1. Operating specifications and main dimensions of the ERMP porotype
 Operating specifications Main dimensions Rated flow 6m${\displaystyle m^{3}/h}$ Length of sliding vane 136mm Rated speed 200/min Width of sliding vane 20mm Working fluid Gas-oil multiphase flow Cylinder radius 90mm Intake pressure 0.1${\displaystyle \sim }$0.3MPa Rotor radius 82mm Discharge pressure 0.3${\displaystyle \sim }$3.0MPa Radius of vane joint 7mm

### 5.1 Simulation results for kinematics of sliding vane

In this study, the eccentric distance of the crankshaft is set to be 0.007m, 0.008m and 0.009m to evaluate the kinematic characteristics of sliding vane. Figure 7 shows the simulation results of kinematics for sliding vane. From figure 7, it can be noted that the movement, velocity, acceleration of the sliding vane and the angular velocity of the round end of sliding vane increase with the eccentric distance. Shown as in Figure 7 (a), the sliding vane reaches its maximum values when the crankshaft rotates to the ${\displaystyle \pi }$ position. The angular velocity of the round end of the sliding vane has the same variation of the movement of sliding vane, shown as in Figure 7 (d). The velocity of the sliding vane vary with degree of crankshaft in sine form, shown as in Figure 7 (b). On the country, the acceleration of the sliding vane vary with degree of crankshaft in cosine form, shown as in Figure 7 (c).

 Figure 7. Kinematic analysis for sliding vane

### 5.2 Simulation results for kinematics of rotor

Based on the established fixed coordinate system (${\displaystyle XOY}$) and rotational coordinate system (${\displaystyle X_{1}O_{1}Y_{1}}$), movements of points on the end surface of rotor can be obtained. Shown as in Figure 8, the motion path of the points are ellipses in different size, except the motion path of the rotor’s center which represents as a circle.

 Figure 8. Movements of points on end surface of rotor

In this study, four points on the contour of rotor (${\displaystyle C_{2}}$, ${\displaystyle C_{3}}$, ${\displaystyle C_{4}}$ and ${\displaystyle C_{5}}$) and four points on the middle position of end surface of rotor (${\displaystyle B_{2}}$, ${\displaystyle B_{3}}$, ${\displaystyle B_{4}}$ and ${\displaystyle B_{5}}$) are selected to simulate velocity of rotor on the end surface, shown as in Figure 8. Figure 9 shows the velocity variations of each selected points on the end surface of rotor. The velocities of each selected points vary with the rotation of the crankshaft in a same trend. The points located at ${\displaystyle 3\pi /2}$ position of the rotor, such as points ${\displaystyle C_{5}}$ and ${\displaystyle B_{5}}$, have larger velocities compared with the points in other area. Besides, the more the points approach to the center of the rotor, the smaller velocities they have. Then, the area located at opposite of the sliding vane (${\displaystyle 3\pi /2}$ position of the rotor) and located at contour part of the rotor, such as area of point ${\displaystyle C_{5}}$, have wear problem during application of the pump and attention need to be paid.

 Figure 9. Velocity variations of rotor under different position

Besides, different value of eccentric distance are set to evaluate the effects of eccentric distance on the kinematic behavior of the rotor, shown as in Figure 10. It can noted in Figure 10, the rotor rotates at a faster velocity, a larger angular velocity and angular acceleration when selecting a larger value of the eccentric distance.

 Figure 10. Variation of kinematic behavior of rotor</

### 5.3 Simulation results for forces on sliding vane

Figure 11 shows the forces acting on the sliding vane. It can be noted that the binding force between the groove (${\displaystyle F_{R1}}$) neutralized most of the pressure difference induced force (${\displaystyle F_{pg}}$). The tangential force (${\displaystyle F_{t}}$) acting on the sliding vane is much less than the pressure difference induced force for the reason that it mainly rest on the rotational inertia of the rotor. Besides, the normal force (${\displaystyle F_{n}}$), which mainly affected by the friction forces (${\displaystyle F_{f1}}$ and ${\displaystyle F_{f2}}$) caused by the movement of the sliding vane, increase with rotation of the crankshaft at the first period and decrease at the last period. Moreover, compared with other forces acting on the sliding vane, the binding force (${\displaystyle F_{R1}}$), pressure difference induced force (${\displaystyle F_{pg}}$), the normal force (${\displaystyle F_{n}}$) and friction forces (${\displaystyle F_{f1}}$) are much larger and need to be carefully analyzed.

 Figure 11. Forces acting on the sliding vane

Figure 12 shows the frictional force and radial force acting on the rotor. Compared with the frictional force acting on the sliding vane, the frictional force acting on the rotor is much smaller, shown as in Figure 12 (a). The radial force acting on the rotor is also much larger than the frictional force acting on the rotor, shown as in Figure 12 (b). The radial force increase with rotation of the crankshaft until reach its maximum value at π position.

 Figure 12. Frictional force and radial force acting on the rotor

Figure 13 shows the resistance moment acting on the rotor. It can be noted that the total resistance moment (${\displaystyle M_{z}}$) is mainly composed by the resistance moment (${\displaystyle M_{pg}}$) caused by the pressure difference induced force, for the reason that the values of these two resistances are almost the same, shown as in Figure 13 (a). Meanwhile, the values of the resistance moment caused by the normal force at round end of the sliding vane (${\displaystyle M_{n}}$) and the friction moment between the rotor and crankshaft (${\displaystyle M_{c}}$) are much smaller than that of the resistance moment (${\displaystyle M_{pg}}$) caused by the pressure difference induced force, shown as in Figure 13 (b). Then, the resistance moment ${\displaystyle M_{n}}$ and ${\displaystyle M_{c}}$ acting on the rotor can be neglected during the analysis of the pump for the resistance moment is mainly contributed by the resistance moment of ${\displaystyle M_{pg}}$ which need carefully analyzed.

 Figure 13. Resistance moment acting on the rotor

## 6. Conclusion

The structural characteristics and working process of the proposed ERMP are introduced. The mathematic models of the kinematic and dynamic behaviors of key components of the pump are established. Based on the work done in this research, the following results and conclusions were obtained:

1. The velocity and acceleration of the sliding vane vary with degree of crankshaft in sine and cosine form, respectively. The sliding vane reaches its maximum value when the crankshaft rotates to the ${\displaystyle \pi }$ position.
2. The areas on the surface of rotor located opposite the sliding vane and apart from the center have larger velocities and wear problem during application of the pump.
3. The binding force, pressure difference induced force, the normal force and friction forces are much larger than other forces and need to be carefully analyzed.
4. The resistance moment is mainly contributed by the resistance moment of ${\displaystyle M_{pg}}$ induced by the pressure difference of the multiphase flow.

## Acknowledgements

The authors acknowledge the support of grant No. 2017QHZ010 from the scientific research-starting project of SWPU China.

## References

[1] Porto F., Larson A. Multiphase pump field trials demonstrate practical applications for the technology. SPE Prod. Facil., 12(3):159–164, 1996.

[2] Chynoweth E. Twin screw pump is developed for multiphase applications. Process Eng. London, 67(12):27, 1986.

[3] Neumann W. Efficient multiphase pump station for onshore application and prospects for offshore application. Proceedings of the Eighth International Pump Users Symposium, Texas A&M Univ., College Station, TX, 43–48, 1991.

[4] Beran W.T. On the threshold: subsea multiphase pumping. Proceedings of the 1994 SPE Annual Technical Conference and Exhibition, SPE, New Orleans, 569–581, 1994.

[5] Salis D.J., Cordner M., Birnov M. Multiphase pumping comes of age. World Pumps, 384:53–54, 1998.

[6] Leporcher E., Delaytermoz A., Renault J.F., Gerbier A., Burger O. Deployment of multiphase pumps on a north sea field. Proceedings of the 2001 SPE Annual Technical Conference and Exhibition, SPE, New Orleans, 1755–1766, 2001.

[7] Dogru A.H., Hamoud A.A., Barlow S.G. Multiphase pump recovers more oil in a mature carbonate reservoir. J. Petrol. Technol., 56(2):64–67, 2004.

[8] Bornemann P. Multiphase pumps – the path to success. World Pumps, 511:18–20, 2009.

[9] Cao S., Peng G., Yu Z. Hydrodynamic design of rotodynamic pump impeller for multiphase pumping by combined approach of inverse design and cfd analysis. ASME J. Fluids Eng., 127(2):330–338, 2005.

[10] Cao F., Peng X., Xing Z., Shu P. Thermodynamic performance simulation of a twin-screw multiphase pump. Proc. Inst. Mech. Eng., Part E, 215(2):157–163, 2001.

[11] Chynoweth E. Twin screw pump is developed for multiphase applications. Process Eng. London, 67(12):27–31, 1986.

[12] Dal P.D., Larson L.A. Multiphase pump field trials demonstrate practical applications for the technology. SPE Prod. Facilities. 12(3):159–164, 1996.

[13] Ryazantsev V.M., Plyasov V.V., Shumakov N.D. Twin-screw pumps for transferring petroleum-water-gas multiphase mixture or diesel fuel. Chem. Petrol. Eng., 40(8):407–410, 2004.

[14] Nakashima C. Y., Oliveira S. Heat transfer in a twin-screw multiphase pump: thermal modeling and one application in the petroleum industry. Energy, 31(15):3415–3425, 2006.

[15] Muller L.D., Rohlfing G., Spelter H. Twin screw pumps help improving oil recovery in mature fields and transfer heavy crude oil over long distances. Oil Gas Eur. Mag., 35(3):127–131, 2009.

### Document information

Published on 27/03/20
Accepted on 24/03/20
Submitted on 10/03/20

Volume 36, Issue 1, 2020
DOI: 10.23967/j.rimni.2020.03.004

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# How To Remove Duplicates From A List Of Lists In Python

To remove duplicates from a list of lists in Python, you can use two ways: using the sorted() function in combination with the set() function and using the map() function in combination with the set() function. Read the following article for more information.

## Remove duplicates from a list of lists in Python

### Use the sorted() function

Syntax:

sorted(object, key=None, reverse=False)

Parameters:

• object: a string, an interpolator, or an iterator object.
• key: sorted() function with key value, then sort. The default is none.
• reverse: default value is False. If reverse = True, the list will sort reverse.

The sorted() function in Python sorts the elements of an object in a particular order (ascending or descending) and returns the object as a sorted list.

Example:

• Initialize a list of arrays where some of the arrays are the same.
• Use the sorted() function to sort the list.
• Convert the list of tuples to a set to eliminate duplicates.
• Note: Remember to convert the nested list to a tuple before converting it to a set because lists are mutable and non-hashable and cannot be a member of a set.
• Finally, use the list() function to return the results to a list.
# The list consists of sublists
myList = [[0, 0, 1], [1, 1, 0], [1, 1, 0], [1, 2, 3], [4, 5, 6]]

# Use the sorted() and set() function
result = list(set(tuple(sorted(i)) for i in myList))

print("List after removing the same sublists: ", result)

Output:

List after removing the same sublists: [(1, 2, 3), (0, 0, 1), (4, 5, 6), (0, 1, 1)]

### Use the map() and set() function

Syntax:

map(func,iterable)

Parameters:

• func: given function to iterate over the elements of iterable.
• iterable: the object you want to browse.

Example:

• Initialize a list of arrays where some of the arrays are the same.
• The map function converts the list to a tuple object that can be hashed (so that the list is a set member).
• Use the set() function to remove duplicates of the same.
• Finally, use the list() function to return the results to a list.
# The list consists of sublists
myList = [[0, 0, 1], [1, 1, 0], [1, 1, 0], [1, 2, 3], [4, 5, 6]]

# Use the map() and set() functions to remove duplicates from a list of lists
result=list(set(map(tuple,myList)))

print("List after removing the same sublists: ", result)

Output:

List after removing the same sublists: [(1, 2, 3), (0, 0, 1), (1, 1, 0), (4, 5, 6)]

## Summary

Thank you for being so interested in my article. Hope you got an idea for the topic of how to remove duplicates from a list of lists in Python. You can use both ways for this theme. Let us know if the article is flawed or if you have other ideas.

Maybe you are interested:


text:
## ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 18 Mensuration Chapter Test

These Solutions are part of ML Aggarwal Class 10 Solutions for ICSE Maths. Here we have given ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 18 Mensuration Chapter Test.

Question 1.
A cylindrical container is to be made of tin sheet. The height of the container is 1 m and its diameter is 70 cm. If the container is open at the top and the tin sheet costs Rs 300 per m2, find the cost of the tin for making the container.
Solution:
Height of container opened at the top (h) = 1 m = 100 cm
and diameter = 70 cm
∴Radius (r) = $$\\ \frac { 70 }{ 2 }$$ = 35 cm
∴Total surface area = 2πrh + πr2

Question 2.
A cylinder of maximum volume is cut out from a wooden cuboid of length 30 cm and cross-section of square of side 14 cm. Find the volume of the cylinder and the volume of wood wasted.
Solution:
Dimensions of the wooden cuboid = 30 cm × 14 cm × 14 cm
Volume = 30 × 14 × 14 = 5880 cm3

Question 3.
Find the volume and the total surface area of a cone having slant height 17 cm and base diameter 30 cm. Take π = 3.14.
Solution:
Slant height of a cone (l) = 17 cm
Diameter of base = 30 cm
Radius (r) = $$\\ \frac { 30 }{ 2 }$$ = 15 cm

Question 4.
Find the volume of a cone given that its height is 8 cm and the area of base 156 cm2.
Solution:
Height of a cone = 8 cm
Area of base = 156 cm
.’. Volume = $$\\ \frac { 1 }{ 3 }$$ × area of base × height

Question 5.
The circumference of the edge of a hemispherical bowl is 132 cm. Find the capacity of the bowl.
Solution:
Circumference of the edge of bowl = 132 cm

Question 6.
The volume of a hemisphere is $$2425 \frac { 1 }{ 2 }$$ cm2. Find the curved surface area.
Solution:
Volume of a hemisphere = $$2425 \frac { 1 }{ 2 }$$ cm3
= $$\\ \frac { 4851 }{ 2 }$$ cm3

Question 7.
A solid wooden toy is in the shape of a right circular cone mounted on a hemisphere. If the radius of the hemisphere is 4.2 cm and the total height of the toy is 10.2 cm, find the volume of the toy
Solution:
A wooden solid toy is of a shape of a right circular cone
mounted on a hemisphere.
Radius of hemisphere (r) = 4.2 cm
Total height = 10.2 cm

Question 8.
A medicine capsule is in the shape of a cylinder of diameter 0.5 cm with two hemispheres stuck to each of its ends. The length of the entire capsule is 2 cm. Find the capacity of the capsule.
Solution:
Diameter of cylindrical part = 0.5 cm
Total length of the capsule = 2 cm

Question 9.
A solid is in the form of a cylinder with hemispherical ends. The total height of the solid is 19 cm and the diameter of the cylinder is 7 cm. Find the volume and the total surface area of the solid.
Solution:
Radius of cylinder = $$\\ \frac { 7 }{ 2 }$$cm
and height of cylinder = 19 – 2 × $$\\ \frac { 7 }{ 2 }$$ cm
= 19 – 7 = 12 cm
and radius of hemisphere = $$\\ \frac { 7 }{ 2 }$$ cm

Question 10.
The radius and height of a right circular cone are in the ratio 5 : 12. If its volume is 2512 cm , find its slant height. (Take π = 3.14).
Solution:
Let radius of cone (r) = 5x
then height (h) = 12x

Question 11.
A cone and a cylinder are of the same height. If diameters of their bases are in the ratio 3 : 2, find the ratio of their volumes.
Solution:
Let height of cone and cylinder = h
Diameter of the base of cone = 3x
Diameter of base of cylinder = 2x

Question 12.
A solid cone of base radius 9 cm and height 10 cm is lowered into a cylindrical jar of radius 10 cm, which contains water sufficient to submerge the cone completely. Find the rise in water level in the jar.
Solution:
Radius of the cone (r) = 9 cm
Height of the cone (h) = 10 cm
Volume of water filled in cone

Question 13.
An iron pillar has some part in the form of a right circular cylinder and the remaining in the form of a right circular cone. The radius of the base of each of cone and cylinder is 8 cm. The cylindrical part is 240 cm high and the conical part is 36 cm high. Find the weight of the pillar if one cu. cm of iron weighs 7.8 grams.
Solution:
Radius of the base of cone = 8 cm

Question 14.
A circus tent is made of canvas and is in the form of right circular cylinder and a right circular cone above it. The diameter and height of the cylindrical part of the tent are 126 m and 5 m respectively. The total height of the tent is 21 m. Find the total cost of the tent if the canvas used costs Rs 36 per square metre.
Solution:

Question 15.
The entire surface of a solid cone of base radius 3 cm and height 4 cm is equal to the entire surface of a solid right circular cylinder of diameter 4 cm. Find the ratio of their
(i) curved surfaces
(ii) volumes.
Solution:
Radius of the base of a cone (r) = 3 cm
Height (h) = 4 cm

Question 16.
A cone is 8.4 cm high and the radius of its base is 2.1 cm. It is melted and recast into a sphere. Find the radius of the sphere.
Solution:
Radius of base of a cone (r) = 2. 1 cm
and height (h) = 8.4 cm

Question 17.
How many lead shots each of diameter 4.2 cm can be obtained from a solid rectangular lead piece with dimensions 66 cm, 42 cm and 21 cm.
Solution:
Dimensions of a solid rectangular lead piece
= 66 cm × 42 cm × 21 cm
.’. Volume = 66 × 42 × 21 cm3
Diameter of a lead shot = 4.2 cm

Question 18.
Find the least number of coins of diameter 2.5 cm and height 3 mm which are to be melted to form a solid cylinder of radius 3 cm and height 5 cm.
Solution:
Radius of a cylinder (r) = 3 cm
Height (h) = 5 cm

Question 19.
A hemisphere of lead of radius 8 cm is cast into a right circular cone of base radius 6 cm. Determine the height of the cone correct to 2 places of decimal.
Solution:
Radius of hemisphere = 8 cm
Volume = $$\frac { 2 }{ 3 } \pi { r }^{ 3 }{ cm }^{ 3 }$$

Question 20.
A vessel in the form of a hemispherical bowl is full of water. The contents are emptied into a cylinder. The internal radii of the bowl and cylinder are respectively 6 cm and 4 cm. Find the height of the water in the cylinder.
Solution:
Radius of hemispherical bowl = 6 cm
.’. Volume of the water in the bowl

Question 21.
The diameter of a metallic sphere is 42 cm. It is metled and drawn into a cylindrical wire of 28 cm diameter. Find the length of the wire.
Solution:
Diameter of sphere = 42 cm
Radius of sphere =$$\\ \frac { 42 }{ 2 }$$ = 21 cm

Question 22.
A sphere of diameter 6 cm is dropped into a right circular cylindrical vessel partly filled with water. The diameter of the cylindrical vessel is 12 cm. If the sphere is completely submerged in water, by how much will the level of water rise in the cylindrical vessel?
Solution:
Radius of sphere = $$\\ \frac { 6 }{ 2 }$$ = 3 cm

Question 23.
A solid sphere of radius 6 cm is metled into a hollow cylinder of uniform thickness. If the external radius of the base of the cylinder is 5 cm and its height is 32 cm, find the uniform thickness of the cylinder.
Solution:
Radius of solid sphere = 6 cm
Volume of solid sphere = $$\frac { 4 }{ 3 } \pi { r }^{ 3 }$$

Question 24.
A solid is in the form of a right circular cone mounted on a hemisphere. The radius of the hemisphere is 3.5 cm and the height of the cone is 4 cm. The solid is placed in a cylindrical vessel, full of water, in such a Way that the whole solid is submerged in water. If the radius of the cylindrical vessel is 5 cm and its height is 10.5 cm, find the volume of water left in the cylindrical vessel.
Solution:
Radius of hemisphere (r) = 3.5 cm
Height of cone (h1) = 4 cm

We hope the ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 18 Mensuration Chapter Test help you. If you have any query regarding ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 18 Mensuration Chapter Test, drop a comment below and we will get back to you at the earliest.


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# 314

A diagram which shows the derivation of Pi, or 3.14, or "314", in Greek isopsephy value, on a scale set equal to units of the diameter of a circle.[1]

In numbers, 314 (LH:14), in hieroglyphics: [2], is the first three digits of ${\displaystyle \pi }$ (3.141596 ...); the isopsephy value of the word "Bible" (Greek βῐiβλος) (NE:314); or the day of Mar 14th (3/14), aka Pi Day[3], which curiously falls on Einstein's birthday.

## Overview

In c.3000BC, Egyptians, as can be historically deduced, determined pi as follows:

${\displaystyle \pi ={\frac {314}{100}}}$

or in Egyptian fractional addition form:

${\displaystyle \pi =3+{\frac {14}{100}}}$

This equals 3.14 in modern "decimal" notation (Stevin, 1585). This can be found by taking a rope that is 314 units in length (as circumference), making it into a circle on the ground, then take another string of 100 units in length (as diameter), and from these measure Pi by unrolling the circle, as shown adjacent, and finding Pi to at least two digits or 3 plus 14/100 units, which is written in Egyptian numbers as follows:

${\displaystyle \pi =}$ ${\displaystyle =3+{\frac {14}{100}}}$

where is 3, means plus, and means 1/100.

### Khufu pyramid | Pi

In 2600BC, the Egyptians, during the construction of the Khufu pyramid, knew the value of pi (${\displaystyle \pi }$) to at least two or three digits, based on the following formula:

${\displaystyle \pi ={\frac {2b}{h}}}$

where h (146.6 m) is the height and b (230.3 meters) is the based of the pyramid, which yields:

${\displaystyle \pi =3.1419}$

This measure is based on the measurements of the height and based of the Khufu.[4]

### 318 themed | Pi

In c.2000BC, Egyptians were using the following for pi (${\displaystyle \pi }$):

${\displaystyle \pi ={\frac {1000}{318}}}$

This equals:

${\displaystyle \pi =3.145}$

Written in Egyptian number hieroglyphs, using the modern "divided by" notation, instead of the mouth symbol , which the Egyptians used for divide, this would read as follows (in shorthand), albeit written as an expanded sum of fractions (in Egyptian):

The lotus symbol: , the Egyptian symbol for 1000, represents or is symbolic of Horus, as the morning sun, born out of the waters of the Nun, the void like waters defined as the "monad" (e.g. by Pythagoras). The number "318" was translated by the Greeks into the words "theta", symbol: Θ, the symbol of the Egyptian sun god and the Ennead sun god family, and the word "Helios", the Greek sun god. The value of pi in this Greek-Egyptian translated sense, would equal:

Said another way, theta, Θ, i.e. 318 (or the sun), is equal to the lotus, i.e. 1000 (or the monad), divided by pi (or 3.14):

In c.1200BC, Egyptians knew pi (${\displaystyle \pi }$) to about four digits, according to the following formula:

${\displaystyle \pi ={\frac {1000}{318+{\frac {318}{1000}}}}}$

In decimal form:

${\displaystyle \pi ={\frac {1000}{318.318}}=3.1415}$

In 750BC, Hesiod wrote his Theogony, was already using Greek words "theta" (Greek letter 9) and "Helios" (Greek sun god) with the isopsephy-coded as sacred number "318". In isopsephy formula:

${\displaystyle \pi ={\frac {1000}{{\text{Theta}}+{\frac {\text{Helios}}{1000}}}}}$

Therefore, if Hesiod was already telling a full story using this double theta name cipher (Helios = Theta = 318), then the "318 cipher" would have already had to have been in place one to four centuries prior, when the Greek alphabet (1200BC) first took form, which is corroborated by the fact that numerical ordering of the Greek alphabet follows the order of steps of the Hermopolis recension (2400BC) of the Heliopolis creation myth (2800BC).

### Other | Pi values

In 1650BC, Rhind Papyrus, copied from a document dated to 1850BC, says the following:

“Cut off 1/9 of a diameter and construct a square upon the remainder; this has the same area as the circle.”
— Ahmes (1650BC), Rhind Papyrus[5]

This yields a formula for the area of a circle that treats ${\displaystyle \pi }$ as 3.16, one of the earliest approximations of ${\displaystyle \pi }$ on papyrus record.

${\displaystyle \left({\frac {16}{9}}\right)^{2}}$ ≈ 3.16.

In 250BC, Archimedes proved, using a 96-side regular polygon, that π falls within the following range: 3.1408 < π < 3.1429.

In 150AD, Ptolemy, in his Almagest, gave a value for π of 3.1416.

### Geometry | Mathematics

The formula that relates the circumference C and the diameter D of a circle is as follows:

${\displaystyle \pi ={\frac {C}{D}}}$

where ${\displaystyle \pi }$ is 3.1415...

### 318

A circle with a diameter of 318.318 units, according (Ѻ) to the formula C = 2πr , has a circumference of 1000, which the Greeks equated with the monad or perfect unity.[6]

Related to ${\displaystyle \pi }$, is the number "318", which firstly is the ratio of 1/π which is:

${\displaystyle {\frac {1}{\pi }}=0.318}$

Theologically, the number "318" is isopsephy-coded into the Greek words: theta, the 9th letter of the Greek alphabet, symbol "Θ", which is code for the Egyptian sun god based creation model, centered around Atum-Ra (or Ra), and the Greek sun god family Ennead, and Helios, the Greek sun god.

Mathematically, the number 318 is found to have several relations in respect to a circle with a circumference of 1,000 units; which the Greeks equated with the "monad" or mind of god.

In Greek religio-mythology, the value of 314 (π) is coded, via mathematical theology, such that a circle with a circumference of 1000, which equates to the "monad" in Pythagoreanism, and a diameter of 318.318, which equates to two thetas (Θ,Θ), the isopsephy value of theta being 318, which is also the isopsephy value of Helios, the Greek sun god, yields an accurate calculation of "${\displaystyle \pi }$" to four digits:

${\displaystyle {\frac {1000}{318.318}}=3.14151}$

which is the best approximation of pi that mathematicians knew at this time. As for what a monad divided by two thetas (or one theta and one Helios; or two Helioses) actually means, in the original Greco-Egyptian conception of things, is a bit puzzling? The Monad translates as the "nous" or mind of god, which derives from the Egyptian Nun. Helios is the Greek sun god, or Greek rescript of either Ra or Horus. Theta, however, is an admixture of about five different sun god themes, in layered cyphers, mostly meant to embody the gist of Heliopolis creation myth. In the original Egyptian conception, ${\displaystyle \pi }$ could render as Nun (1000) divided by the sum of Ra (318), the father "sun god" / noon sun god, and Horus (318), the son of the "sun god" / morning sun god?

### Bible

The name Bible, accordingly, thus renders, in words, via decoded isopsephy, as follows:

${\displaystyle \pi ={\frac {\text{Monad}}{\text{Theta-Helios}}}}$

The word "Bible", accordingly, renders as the book of "book of Pi", or the "book of the father", as pi (Π) is the initial of English name father (pater).

### Isopsephy / Gematria

The following are religio-mythology isopsephy and or gematria of related note:

• Bible [Book of Gods] (βῐ́βλος) = 314 [Π] (isopsephy value)
• 1000 = 1 or Monad [Nous / Nun] (in Pythagorean mathematics)

### Amicable pairs

• Chosen one [Bechir] (בָּחִיר) = 220 [higher amicable pair] (gematria value)
• Christ / Lamb [Kar] (כר) = 220 [lower amicable pair] (gematria value)

### Dates

The following are some notable dates:

• 14 Mar 1988 | Larry Show, at San Francisco Exploratorium, celebrates Pi Day [3]

## Quotes

The following are quotes:

“By the way, today is Einstein’s birthday. So we are at a very fitting place. And it just occurred to me, 3/14 [or] 3.14, is the first approximation of Pi (Π), which refers to ‘infinity’ [god]. So as long as your quantum physics helps me understand infinity, I would be honored [to have you teach me].”
— Deepak Chopra (2010), “Debate: Does God Have a Future”, Audience Q&A [with Leonard Mlodinow[8]] (1:04:43-5:16) [7]
42 is 3 14s [3 x 14 (314) = 42 / 3.141592 ... / Pi / Bible]. It’s meant to get you to explore the number. It encodes pi, among other things. An atheist arguing against a historical Bible is like a vegetarian yelling at someone wearing a synthetic leather jacket. Your base of argumentation is off, unfortunately.”
— The Sozo Guy (2020), Comment to Thims “42 Generations of Christ: Abraham (Ra) to Jesus (Horus)” video[9]

## End matter

### References

1. Reed, Thomas. (2011). “History of Pi” (YT), Thomas Reed, May 30.
2. Egyptian numerals – Dcode.fr.
3. Pi Day – Wikipedia.
4. For whatever reason, many will argue that this Khufu pi is just a coincidence, and that the Egyptians didn't know the value of pi until after the Rhind Papyrus (1650BC). This type of argument, however, goes against the entire 4,000-year cosmology of the Egyptians, which is circle based.
5. Wilson, David. (2000). “The History of Pi”, Rutgers, Spring.
6. Fideler, David. (1993). Jesus Christ, Sun of God: Ancient Cosmology and Early Christian Symbolism (318, 6+ pgs). Quest Books.
7. Shermer, Michael; Harris, Sam; Chopra, Deepak; Houston, Jean. (2010). “Does God Have a Future” (YT) (Ѻ) (woo woo, 14:50-15:24; god = infinity [Chopra and Leonard Mlodinow], 1:03:56-1:05:16; Pi and Einstein's birthday, 1:04:43-1:05:05), Mar 14; aired on Nightline Face-Off, Mar 23.
8. Leonard Mlodinow – Wikipedia.
9. Sozo Guy. (2020). Comment to “42 Generations of Christ: Abraham (Ra) to Jesus (Horus)” (YT), Atheism Reviews, 2017.


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157 Pages

## ProblemEdit

Here are two facts about a substance:

• The entropy is the following function of $T$ and $V$:
$S = R{{V_0 } \over V}\left( {{T \over {T_0 }}} \right)^{\alpha - 1}$
• At constant temperature $T_0$ the work the substance does on its surroundings as it expands from $V_0$ to $V$ is
$\left. W \right|_{T_0 } = RT_0 \ln {V \over {V_0 }}$

(a)Find the Helmholtz free energy F, assuming that it is zero at the state values specified by the subscript 0.

(b)Find the equation of state of this stuff. Is there a point in parameter space where it is ideal?

(c)For this part we’ll simplify the algebra by assuming that $\ln {V \over {V_0 }} = 1$ and also $\alpha = 1$. At what temperature $T'$ would the system do twice as much work in going from $V_0$ to $V$ as it does at $T = T_0$?


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# Thread: Universal Quantifier in Predicate Logic

1. ## Universal Quantifier in Predicate Logic

Hi,

I have the following:
B(x) stands for "x is a baker",
F(x) stands for "x is a farmer",
K(x,y) stands for "x knows y".

Ax: for all x
Ex: there exists an x

(a) Express "Everybody knows a farmer" in terms of predicate logic.

For this I have:

Ax(Ey (F(y) -> K(x,y))

Which I believe translates to "For all X, there exists a farmer Y, such that X knows a Y". Which in turn implies that everybody knows a farmer.

Now, the second question is:

(b) "Everybody knows someone who knows a farmer"

This one confuses me, as the same relation is involved twice.

So far, I have:

Ax(Ey K(x,y) -> (K(y,z) && Ez(F(z))

But this does not seem right. How do I approach this question?

2. ## Re: Universal Quantifier in Predicate Logic

Originally Posted by aprilrocks92
I have the following:
B(x) stands for "x is a baker",
F(x) stands for "x is a farmer",
K(x,y) stands for "x knows y".

(a) Express "Everybody knows a farmer" in terms of predicate logic.
Here is that one: $\left( {\forall x} \right)\left( {\exists y} \right)\left[ {F(y) \wedge K(x,y)} \right]$

Now you post your effort on the other parts.

3. ## Re: Universal Quantifier in Predicate Logic

For all x, there exists a y,z such that if z is a farmer, and y knows z, then x knows y.

(Ax)(Ey,z) [F(z) & K(z, y)]->K(x,y).

As stated above, I'm rather confused as to how this is put.

4. ## Re: Universal Quantifier in Predicate Logic

Originally Posted by aprilrocks92
For all x, there exists a y,z such that if z is a farmer, and y knows z, then x knows y.
(Ax)(Ey,z) [F(z) & K(z, y)]->K(x,y).
What the devil are you on about?

Look at the statement ""Everybody knows someone who knows a farmer""
Where do you see an implication (an if...then) there?

$\left( {\forall x} \right)\left( {\exists y} \right)\left( {\exists z} \right)\left[ {K(x,y) \wedge F(z) \wedge K(y,z)} \right]$


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# Question #6a930

Feb 2, 2017

$x + 2 y = 8 \iff y = - \frac{1}{2} x + 4$

#### Explanation:

$x + 2 y = 8$
$\iff$ subtract both sides by $x$
$2 y = 8 - x$
$\iff$ divide both sides by 2
$y = 4 - \frac{1}{2} x$
The solution is this line