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# Summary

The Wobbe Index is a measure of the interchangeability of fuel gases and their relative ability to deliver energy. It gives an indication of whether a turbine or burner will be able to run on an alternative fuel source without tuning or physical modifications.

# Definitions

 $$\text{HHV}$$ : Higher Heating Value (energy/volume) $$I_{w}$$ : Wobbe Index $$RD$$ : Relative Density $$\rho_{gas}/\rho_{air}$$, $$\rho_{air} = 1.205$$ kg/m3

# Wobbe Index Definition

The Wobbe Index for a fuel gas can be calculated using its higher heating value and relative density as follows:

$\displaystyle I_{w} = \cfrac{\text{HHV}}{\sqrt{RD}}$

The Wobbe index is the ratio of the specific energy of a gas to its relative flow rate as shown by its constituent components:

1. - The Higher Heating Value, which is a measure of the energy per volume for the gas being considered.
2. - The inverse of the root of the relative density, which is a good approximation of the relative mass flow rate of a gas through a fixed orifice at a fixed pressure.

By determining the ratio of the energy content of the fuel gas along with the relative mass flow we arrive at a number proportional to the energy delivered to a burner or turbine. Where two fuel gases have the same Wobbe Index, they will deliver the same amount of energy when supplied at the same pressure.

## Units

The Wobbe Index may be expressed in either British Thermal Units per standard cubic foot, $$\text{BTU}/\text{SCF}$$, or megajoules per standard meter cubed, $$\text{MJ}/\text{Sm}^{3}$$. Where the units are not given it is likely that the units are the more traditional $$\text{BTU}/\text{SCF}$$. The units are often deliberately omitted to prevent confusion with the actual energy per volume of the fuel gas.

It is important to note that Wobbe index values must not be used in place of energy of combustion even though it may have the same units.

## Limitations

The Wobbe index should not be used as the sole criteria by which to judge the feasibility of a fuel swap out. There are several limitations of the Wobbe index that need to be considered separately:

• Liquid Drop Out - It is possible to make a mixture that will have a suitable Wobbe Index but will be prone to liquid drop out, which in turn will effect the operation of combustion equipment. In the mixture of carbon dioxide and pentane which simulates natural gas, significant amounts of the pentane will liquefy under typical operating conditions.
• Reactive Components - The presences of reactive components like nitrogen, carbon monoxide and alkenes may cause problems for fuel system components.

# Wobbe Index Values for Common Fuels

The below table gives some approximate Wobbe index for common fuel gases.

Gas $$I_{w}$$ () $$I_{w}$$ ()
Hydrogen45.51220
Methane49.91339
Ethane63.61708
Propane74.82008
Butane84.42266
Pentane92.52482
Acetylene57.01531
Natural Gas46 - 521235 - 1396
Liquefied Natural Gas (LNG)52.81417

# Consequences of Incorrect Wobbe Index Fuels

## High Wobbe Index

When the Wobbe Index of the replacement fuel is too high the fuel it will deliver too much energy and will likely to require more oxygen for complete combustion. This can result in the following negative effects:

• Over Heating - Higher than expected flame temperature leading to damage to the burner, turbine or equipment being fired.
• Excessive carbon monoxide formation - Caused by incomplete combustion, if allowed to build up can be harmful or fatal to people.
• Soot build up - Increased potential for soot to build up due to incomplete combustion.
• Explosive gases - In cases where the air supply is insufficient, quantities of carbon monoxide and hydrogen may build up to a point that the atmosphere becomes explosive when air is re-admitted.
• Inefficient combustion - Failure to extract the maximum energy from the fuel.

## Low Wobbe Index

Where the Wobble index of the replacement fuel is too low the fuel will not deliver enough energy. The can result in the following negative effects:

• Ignition problems - The equipment may not ignite if the energy content of the gas is too low.
• Flame instability and flashback - An unstable flame, and flashback through the fuel/air supply tube can occur. This may snuff the flame and damage equipment. When the flame is snuffed a hazard is caused by the accumulation of un-burnt fuel gas being released.


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Wednesday, October 5, 2022
HomePhysicsOne Mannequin to Describe Them All—Properly, Ice Giants Anyway

# One Mannequin to Describe Them All—Properly, Ice Giants Anyway

• Physics 15, 89

The spectral responses of the atmospheres of Uranus and Neptune can now each be absolutely characterised with the identical mannequin, an achievement that has implications for characterizing the atmospheres of exoplanets.

In relation to area missions, the main focus has largely lain with planets near Earth, with these farther afield being ignored within the chilly. However that’s about to vary with a devoted mission to Uranus slated for the 2030s. Scientists say that this journey will assist them higher perceive the atmospheres of each Uranus and Neptune, our Photo voltaic System’s two ice giants–chilly fuel behemoths composed principally of parts heavier than hydrogen and helium. Now, Patrick Irwin from the College of Oxford, UK, and colleagues have, for the primary time, concurrently analyzed the reflectance spectra of each ice giants utilizing the identical mannequin [1]. In doing so, the crew made two serendipitous discoveries in regards to the visible appearances of each worlds, unlocking the rationale why they shimmer in numerous hues of blue and why Neptune has darkish spots. The crew says that the findings may have implications for the research of the atmospheres of planets past our Photo voltaic System.

Whereas there have been no earlier stand-alone missions to Uranus and Neptune, the planets haven’t been utterly ignored. For instance, as NASA’s Voyager 2 handed by the ice giants within the late Eighties, it collected details about the reflectance of each worlds. Then in 1994, the Hubble House Telescope captured its first photographs of the planets’ icy exteriors.

Many of the knowledge collected from these and different ground-based observations take a look at small spectral areas of the sunshine that Uranus and Neptune mirror and emit. This restricted knowledge makes it troublesome for researchers to find out the properties of sure aerosol particles within the planets’ atmospheres. This downside is especially acute for Neptune as a result of its small dimension within the sky makes it arduous to gather the longer wavelengths that the planet emits.

To resolve that concern, Irwin and colleagues tailored a mannequin that has beforehand been used to discover the reflectance spectra of almost each different planet in our Photo voltaic System, in addition to a number of exoplanets. They tailored the mannequin to work over a large wavelength vary, from 0.3 to 2.5

$\text{m}$

. The crew then used the mannequin to concurrently analyze obtainable observational knowledge for each ice giants.

The crew’s evaluation reveals the presence of what they consider to be hydrocarbon-based aerosols excessive up within the stratospheres of each Uranus and Neptune. These particles fall via the planets’ atmospheres, mixing and reacting with fuel, comparable to methane, that’s concurrently shifting up. This course of creates a haze round every planet. The crew reveals that Neptune’s ambiance is extra dynamic, giving Neptune a thinner haze—and darker hue—than Uranus. The crew additionally finds that these aerosols may be behind Neptune’s beforehand unexplained darkish spots. After interacting with methane, the aerosols then meet hydrosulfide, which resides deeper in each Neptune’s and Uranus’ atmospheres. The crew thinks the darkish spots develop in locations on Neptune which have a decrease density of this materials, with the patchiness coming from the dynamics of its enveloping fuel.

Understanding the atmospheres of ice giants might be essential for future exoplanet analysis, because it’s potential that this planet sort is among the many most plentiful within the Milky Manner, says Erich Karkoschka, a planetary researcher on the College of Arizona. That considering comes as a result of, up to now, many of the exoplanets which have been present in our Galaxy have plenty on the order of these of Uranus and Neptune.

Karkoschka notes that as a result of Uranus and Neptune have such comparable sizes, plenty, and compositions, Irwin and his colleagues had been in a position to match their knowledge with the identical mannequin, nevertheless it’s unlikely {that a} comparable evaluation might be carried out with every other two of our Photo voltaic System’s planets. That capacity has implications for characterizing different ice giants. “When you had one [model] for Uranus and a distinct one for Neptune, you then wouldn’t know which to make use of” for one more ice large, he says. However, since this mannequin works for each planets, he thinks there may be “validity” in placing it to make use of on associated methods.

–Allison Gasparini

Allison Gasparini is a contract science author based mostly in Santa Cruz, CA.

## References

1. P. G. J. Irwin et al., “Hazy blue worlds: A holistic aerosol mannequin for Uranus and Neptune, together with darkish spots,” JGR Planets 127 (2022).

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# Action of a linear operator on vectors

1. May 6, 2013

### Staff: Mentor

Not really a homework problem, just doing some self-studying.

1. The problem statement, all variables and given/known data
Let $| a \rangle$ by any vector in an $N$-dimensional vector space $\mathcal{V}$, and $\mathbf{A}$ a linear operator on $\mathcal{V}$. The vectors
$$| a \rangle, \mathbf{A} | a \rangle, \mathbf{A}^2 | a \rangle, \ldots, \mathbf{A}^N | a \rangle$$
are linearly dependent. Let $\mathcal{M} \equiv \text{Span}\{ \mathbf{A}^k | a \rangle\}_{k=0}^N$. Show that $\mathcal{M}$ has the property that for any vector $| x \rangle \in \mathcal{M}$, the vector $\mathbf{A} | x \rangle$ also belongs to $\mathcal{M}$.

2. Relevant equations

3. The attempt at a solution
Since the above vectors span $\mathcal{M}$, I can always find $\{c_k\}$ such that I can write
$$| x \rangle = \sum_{k=0}^N c_k \mathbf{A}^k | a \rangle$$
Now, if $c_N = 0$, and since $\mathbf{A}$ is a linear operator, I have
$$\begin{array}{rl} \mathbf{A} | x \rangle & = \mathbf{A} \sum_{k=0}^{N-1} c_k \mathbf{A}^k | a \rangle \\ & = \sum_{k=0}^{N-1} c_{k} \mathbf{A}^{k+1} | a \rangle \end{array}$$
which, by definition, is in $\mathcal{M}$.

But what if $c_N \neq 0$? I need to show that $\mathbf{A}^{N+1} | a \rangle$ is in $\mathcal{M}$. I'm tempted to use the fact that the vectors are linearly dependent to state that I can write
$$\mathbf{A}^{N} | a \rangle = \sum_{k=0}^{N-1} \alpha_k \mathbf{A}^k | a \rangle$$
but nothing guarantees me that $\mathbf{A}^{N} | a \rangle$ is not orthogonal to the other vectors, or does it?

Any help appreciated.

2. May 6, 2013

### micromass

You can't really write this. Linear dependent means (by definition) that there are scalars $\alpha_k$ which are not all zero such that

$$\alpha_0 | a \rangle + \alpha_1 \mathbf{A}| a \rangle + ... + \alpha_N \mathbf{A}^N| a \rangle=\mathbf{0}$$

Now, you have two situations. Either $\alpha_N\neq 0$, and in that case, you can write something like you wrote in the quote. In the case that $\alpha_n=0$, you need to do something differently. In this case, they key fact is that at least one of the $\alpha_k$ is nonzero.

You can handle both cases simultaniously by remarking that there exists an $M\leq N$ and $\alpha_k$ scalars such that

$$\mathbf{A}^M | a \rangle = \sum_{k=0}^{M-1} \alpha_k \mathbf{A}^k| a \rangle$$

Last edited: May 6, 2013
3. May 6, 2013

### Fredrik

Staff Emeritus
This isn't precalculus (although at my university, we studied it before calculus. ) I'm moving it to calculus & beyond.

4. May 6, 2013

### Staff: Mentor

Sorry about that. I hesitated for a while as to which forum to post to.

That sums up my problem quite well. If indeed $\alpha_N=0$, then $\mathbf{A}^N | a \rangle$ is linearly independent of the other vectors, so what does that say about $\mathbf{A}^{N+1} | a \rangle$?

Again, I don't see how to treat the case where $M < N$.

5. May 6, 2013

### micromass

OK. Can you tell me how you would dela with the case $M=N$? You don't have to write it out completely, a summary would suffice.

6. May 6, 2013

### Staff: Mentor

Assuming that
$$\mathbf{A}^N |a \rangle = \sum_{k=0}^{N-1} \alpha_{k} \mathbf{A}^k |a \rangle$$
we have
\begin{align} \mathbf{A}^{N+1} |a \rangle &= \mathbf{A} \sum_{k=0}^{N-1} \alpha_{k} \mathbf{A}^k |a \rangle \\ &= \sum_{k=0}^{N-1} \alpha_{k} \mathbf{A}^{k+1} |a \rangle \\ &= \sum_{k=1}^{N} \alpha_{k-1} \mathbf{A}^{k} |a \rangle \end{align}
which, by definition of $\mathcal{M}$ is in $\mathcal{M}$, meaning that $\mathbf{A} \left( \mathbf{A}^{N} |a \rangle \right) \in \mathcal{M}$, completing the "proof" that $\mathbf{A} | x \rangle \in \mathcal{M}$ for $| x \rangle \in \mathcal{M}$.

7. May 6, 2013

### micromass

Alright. So the basic idea was to multiply by $\mathbf{A}$.

This doesn't work in the case $M<N$. But maybe multiplying with something else works? Like $\mathbf{A}^2$ or $\mathbf{A}^3$?

8. May 6, 2013

### Staff: Mentor

Aside: Since I have joined PF, I've mostly posted answers to questions posed by others. It feels very strange to be on the other side. It's been a long time since I've been a student...

Ok, I think I see the light. From the linear dependence of the vectors spanning $\mathcal{M}$, we have that for some $M \leq N$
$$\mathbf{A}^M | a \rangle = \sum_{k=0}^{M-1} \alpha_k \mathbf{A}^k | a \rangle$$
with at least one $\alpha_k \neq 0$. Setting $p = N-M+1$, we find
\begin{align} \mathbf{A}^p \mathbf{A}^M | a \rangle &= \mathbf{A}^p \sum_{k=0}^{M-1} \alpha_k \mathbf{A}^k | a \rangle \\ &= \sum_{k=0}^{M-1} \alpha_k \mathbf{A}^{p+k} | a \rangle \\ \mathbf{A}^{N+1} | a \rangle &= \sum_{k=p}^{N} \alpha_{k-p} \mathbf{A}^{k} | a \rangle \end{align}
from which we conclude that $\mathbf{A}^{N+1} | a \rangle \in \mathcal{M}$.

Hope I got it right. Thanks micromass for the help.

9. May 6, 2013

### micromass

That's it! Congratz!


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## anonymous 3 years ago finding general solution and finding theta (1)

1. anonymous

$d \theta / dt = \sqrt{\theta +1}$

2. anonymous

Multiply both sides by dt then integral both sides then use u-substitution on the right side of the equation.

3. anonymous

2/3(\theta +1)^2/3 +C\]

4. sirm3d

$\Large \frac{d\theta}{\sqrt{\theta+1}}=dt$ use the substitution $\Large u=\theta+1,\quad du=d\theta$


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# [Intergal]Change of Variables

• Feb 4th 2010, 06:50 AM
chialin4
[Sloved]Change of Variables
Evaluate
$\iint_D {(x+2y)exp(y-x) dxdy}$
where D={y<x<2-2y, 0<y< 2/3}

by letting
u=x+2y
v=y-x
• Feb 4th 2010, 07:52 AM
Jester
Quote:

Originally Posted by chialin4
Evaluate
$\iint_D {(x+2y)exp(y-x) dxdy}$
where D={y<x<2-2y, 0<y< 2/3}

by letting
u=x+2y
v=y-x

The new region of integration is a triangle bound by $-u \le v \le 0,$ $0 \le u \le 2$. The Jacobian of the transformation is

$
\frac{\partial(u,v) }{\partial(x,y) } =
\left|\begin{array}{cc}
1 & 2\\
-1 & 1\end{array}\right| = 3
$
so the new integral is $\int_0^2 \int_{-u}^0 u e^v \frac{1}{3} dvdu$.
• Feb 4th 2010, 06:27 PM
chialin4
Quote:

Originally Posted by Danny
The new region of integration is a triangle bound by $-u \le v \le 0,$ $0 \le u \le 2$. The Jacobian of the transformation is

$
\frac{\partial(u,v) }{\partial(x,y) } =
\left|\begin{array}{cc}
1 & 2\\
-1 & 1\end{array}\right| = 3
$
so the new integral is $\int_0^2 \int_{-u}^0 u e^v \frac{1}{3} dvdu$.

thx for ur help!!


text:
480.369.1683

### area of triangle formula class 10

and then add the area of all the shapes. These formulas help you to answer the questions more accurately. It is one of the most important chapters covered in high school Mathematics. 6 mins. Jan 18, 2021 - Collinearity of Three Points, Section Formula and Areas of Triangles, Class 10, Science Class 10 Notes | EduRev is made by best teachers of Class 10. Completing the CAPTCHA proves you are a human and gives you temporary access to the web property. Step 2: Find area of bricks or tiles. To assist you with that, we are here with notes. Now substitute height in this formula, => area of ABC = (1 / 2) × a × (2 / a) √[s(s – a)(s – b)(s – c)], => area of ABC = √[s(s – a)(s – b)(s – c)]. Area of an equilateral triangle = a 2 3 4; (a is the side of triangle) If the perimeter of an equilateral triangle is 90 m, then find its area. => Area of ABC = (area of rectangle ABCD / 2), Hence, proved that area of the triangle is (1 / 2) × b × h. Example 1: Find the area of the triangle whose height and base are 6 cm and 5 cm? Circles Formulas For Class 10. Step 4: As we required the area of triangle ABC then we can write it as an (area of rectangle ABCD / 2). Step 1: Find area of wall or floor. What is the area of this triangle? Finding area of a triangle from coordinates Our mission is to provide a free, world-class education to anyone, anywhere. Example to find the area of a triangle, multiply the base by the height, and then divide by 2. Ultimate Math Solver (Free) Free Algebra Solver ... type anything in there! Step 2: Now, let’s take a scalene triangle of sides a, b, c. Step 3: As we don’t have the height or altitude of the triangle, draw a perpendicular from A on BC at D and the base will be ‘a’. 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Now we have to substitute the values in (1 / 2) [x1 (y2 – y3 ) + x2 (y3 – y1 ) + x3(y1 – y2)], => (1 / 2) [1 (2 – 5 ) + 4 (5 – 2 ) + 3(2 – 2)], Hence the area of the triangle is 4.5 sq units. Area of a Triangle = √(s*(s-a)*(s-b)*(s-c)) s = (a + b + c)/2 (Here s = semi perimeter and a, b, c are the three sides of a triangle) Perimeter of a Triangle = a+b+c. Materials Required Cardboard Thermocol Geometry box Drawing sheets Scissors Adhesive Prerequisite Knowledge Concept of a trapezium. Overview. Find the area of the triangle. The area of a triangle is a measurement of the area covered by the triangle. Therefore, this is the formula to find the area of triangle if coordinates are given. How to … In triangle ABC and triangle PQR, if they are similar, then. Hence, the height of the given triangle is 4 cm. Solved Examples: Q.1. We will be using these relations later in the derivation. Now this expression can be written in the form of a determinant as and around the web . SHARES. 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Find the area of a square plot. To find its area we require the height of corresponding to a base. Numerical: Find the distance between the following pairs of points (2, 3), (4, 1) Solution: here x 1 =2, y 1 = 3 , x 2 = 4 & y 2 =1 . Area of rectangle is equal to the product of its length and breadth expressed in square units. Overview. If you are at an office or shared network, you can ask the network administrator to run a scan across the network looking for misconfigured or infected devices. To assist you with that, we are here with notes. Class 10 is one of the most valuable phases in a student’s life, where one can grow exponentially in his/her aspired field by understanding the concepts and scoring decent marks. NCERT Books. You can also write the formula as: ½ x base x height As, Area of triangle = × Base × Height To find the area of ΔABC and ΔPQR, draw the altitudes AD and PE from the vertex A and P of ΔABC andΔPQR, respectively, as shown in the figure given below: Now, area of ΔABC = × BC × AD area of ΔPQR = × QR × PE Example 2: What is the value of x1 whose area of a triangle is 1 of coordinates (x1, 1), (2, 3), and (4, 5)? There were more than 2 methods here listed below check it out. Area related to circles formula of class 10 maths for CBSE , ICSE,NTSE & Other Board Exam . This podcast is a part of a series for, CBSE Class 10 Maths. If we know the length of three sides of a triangle, we can calculate the area of a triangle using Heron’s Formula. The student will also learn the area of equilateral triangle formula. Education Franchise × Contact Us. Problems on Condition of collinearity of three points . Theory A quadrilateral in which one pair […] In this topic, we will discover about equilateral triangles and its area. Step 1: Draw the perpendiculars from coordinates P, Q, and R to X-axis at A, B, and C respectively. For Study plan details. Find the area of an equilateral triangle having side 6 cm. Let us discuss the Area of a Triangle formula. Tangent Ratio. Area of ABC / Area of PQR = (AB/PQ) 2 = (BC/QR) 2 = (CA/RP) 2. The formula is as follows: The area of a triangle whose side lengths are a, b, (a, b), (a,b), and c c c is given by. Using coordinate geometry, it is possible to find the distance between two points, dividing lines in a ratio, finding the mid-point of a line, calculating the area of a triangle in the Cartesian plane, etc. Example 2: The side-length of a square plot is 5m. The formula is as follows: The area of a triangle whose side lengths are a, b, (a, b), (a,b), and c c c is given by. To Register Online Maths Tuitions on Vedantu.com to clear your doubts from our expert teachers and download the Circles formulas to solve the problems easily to score more marks in your CBSE Class 10 Board Exam. VIEWS. The law states that: a / sin A = b / sin B = c / sin C. Here were the basic that you need to know about Trigonometry Class 10. or own an. Therefore, the required area of triangular plot = 1500$\sqrt 3$ sq. 0. Class 10. In earlier classes, we have studied that the area of a triangle whose vertices are (x1, y1), (x2, y2) and (x3, y3), is given by the expression $$\frac{1}{2} [x1(y2–y3) + x2 (y3–y1) + x3 (y1–y2)]$$. Usually called the "side angle side" method, the area of a triangle is given by the formula below. Triangle ABC shown below is inscribed inside a square of side 20 cm. Class 1 - 3 ; Class 4 - 5; Class 6 - 10; Class 11 - 12; CBSE. Solution: In the question, it is clearly mentioned that height and base are: Area of triangle is given as =  (1 / 2) × b × h, Hence, the area of the given triangle is 15 cm2. So the formula we could use to find the area of a triangle is: (base x height) ÷ 2. BNAT; Classes. A = 22/7 x 7 x 7. There are various methods to find the area of the triangle according to the parameters given, like the base and height of the triangle, co-ordinates of vertices, length of sides, etc. Class 10. 6 mins. Find the area of the triangle given below: Find the area of an equilateral triangle with side equal to 10 cm. Candidates who are ambitious to qualify the Class 10 with good score can check this article for Notes. CBSE Class 10 Maths Formulas: Class 10 is the primary gateway for the next level of academics. The Triangle Formula are given below as, Perimeter of a triangle = a + b + c $Area\; of \; a\; triangle= \frac{1}{2}bh$ Where, b is the base of the triangle. Given, length of two equal sides of an isosceles triangle … Practice MCQ Questions for Cbse Class 10 Maths Proofs Involving Triangles I Combination Of Solids Formula Of Surface Area And Volume Of Basic Shape with Answers to improve your score in your Exams. We can express the area of a triangle in the square units. Find the area of a right triangle with a base of 6 centimeters and a hypotenuse of 10 centimeters. Reading Time: 7min read 0. Area of Δ ABC, with A(x 1,y 1) , B(x 2, y 2) & C(x 3,y 3) is the given by ½ [ x 1 (y 2-y 3) + x 2 (y 3-y 1) + x 3 (y 1-y 2)] Numerical: Find the area of a triangle whose vertices are (1, –1), (– 4, 6) and (–3, –5). What is the formula to find the calculate the area of a triangle? Application of Area of a Triangle to Polygons. It can, in a simpler manner, be called the study of triangles. Application of Area of a Triangle to Polygons. A = 25 sq.m. Area of a triangle - Coordinate Geometry. Q.1: Find the area of an equilateral triangle with a side of length 7 cm? CBSE Class 11 Maths Notes : Solution of Triangles, Heights and Distances. The division by 2 actually comes from the certainty that a parallelogram can be divided into 2 triangles. Distance Formula; Section Formula; Area of a Triangle; Class 10 Maths Coordinate Geometry: Distance Formula: Distance Formula. Writing code in comment? Contact us on below numbers. The division by 2 actually comes from the certainty that a parallelogram can be divided into 2 triangles. 9.3 Activity: Applying the Formula for Area of Triangles (10 minutes) Lesson Synthesis; 9.4 Cool-down: Two More Triangles (5 minutes) Learning goals: Compare, contrast, and critique (orally) different strategies for determining the area of a triangle. Where b and h are base and altitude of the triangle, respectively. Class 10 Maths Area Related to Circles – Get here the Notes for Class 10 Maths Area Related to Circles. The area of the triangle is equal to the area of the parallelograms, both standing on the same base and between the sane parallel lines . Example to find the area of a triangle, multiply the base by the height, and then divide by 2. Class 10 Maths Formulas are related to the chapters of Real Numbers, Polynomials, Triangles, Quadratic Equations, Statistics, Probability, etc. A = $$\frac{a\times\frac{\sqrt{3}a}{2}}{2}$$ i.e. But height is not known. h is the height of the triangle. By using our site, you Step 4: Now if we clearly observe there are two triangles formed which are ∆ABD and ∆ADC. Free PDF download of Chapter 10 - Circles Formula for Class 10 Maths. Since Area of a trapezium = (1 / 2) (sum of the parallel sides) × (distance between sides), => Area of trapezium PQAB = (1 / 2)(QA + PB) × AB, => Area of trapezium PQAB = (1 / 2)(y1 + y2)(x1 – x2 ) —-(2), =>Area of trapezium PBCR =(1 / 2) (PB + CR) × BC, =>Area of trapezium PBCR =(1 / 2) (y1 + y3 )(x3 – x1) —-(3), =>Area of trapezium QACR = (1 / 2) (QA + CR) × AC, =>Area of trapezium QACR =(1 / 2)(y2 + y3 ) (x3 – x2 )—-(4). In earlier classes, we have studied that the area of a triangle whose vertices are (x 1, y 1), (x 2, y 2) and (x 3, y 3), is given by the expression $$\frac{1}{2} [x 1 (y 2 –y 3) + x 2 (y 3 –y 1) + x 3 (y 1 –y 2)]$$.Now this expression can be written in … Trigonometry is a branch of mathematics dealing with relations involving lengths and angles of triangles. Q.3. Solved Examples For You. By the formula of area of circle, we know; A = π r 2. When both the base and height of the triangle are not given then we can use Heron’s formula if the sides of the triangle are given. Cosine Ratio. As we know that the area of Triangle is given by; A = $$\frac{base\times height}{2}$$ i.e. Area of Triangles ,Coordinate Geometry - Get topics notes, Online test, Video lectures, Doubts and Solutions for CBSE Class 10 on TopperLearning. Cloudflare Ray ID: 61714eadd8e5eb6d Basic Rules of Triangle . Deriving area of an Isosceles Triangle Using Heron’s Formula. C Program to find Area of a Triangle and Perimeter of a Triangle . VIEW MORE. Solution: To illustrate the use of the coefficients grid or CG, we will calculate each of the three terms in the formula for the area separately, and then put them together to obtain the final value. Area of wall = l x b = 30 x 10 = 300 sq. Therefore, the heron’s formula for the area of the triangle is proved. • This is the only regular polygon with three sides. Step 2: Now if we look at the figure carefully, three different trapeziums are formed such as PQAB, PBCR, and QACR in the coordinate plane. Surface Areas and Volume Formulas for Class 10 Maths Chapter 13 Are you looking for Surface Areas and Volume formulas for class 10 chapter 13? Khan Academy is a 501(c)(3) nonprofit organization. that Involve Right Triangles. Ans. in order to grasp the concept completely. Method 2. 9 mins. m = 300 x 10 = Rs.3000 . a = 7 cm August 5, 2019. in 11th Class, CBSE. Area of Triangle (Heron's Formula) Area of Triangle (SAS Method) Formulas. Class 10 Maths Area Related to Circles – Get here the Notes for Class 10 Maths Area Related to Circles. Download the Chapter wise Important Maths Formulas and Equations to Solve the Problems Easily and Score More Marks in Your CBSE Board Exams. We need to look into trigonometric formulae, ratios, functions etc. Solution: Given, Side of the equilateral triangle i.e. Performance & security by Cloudflare, Please complete the security check to access. Therefore, the area of the triangle is calculated using the equation, A = $$\sqrt{s(s~-~a)~(s~-~b)~(s~-~c)}$$ Numerical: Find the distance between the following pairs of points (2, 3), (4, 1) Solution: here x 1 =2, y 1 = 3 , x 2 = 4 & y 2 =1 . Example 2: Find the height of the triangle whose area is 12 cm2 and the base is 6 cm? In this problem, we have to find the value of ‘x1’, which is X coordinate of point A. Area of a Triangle Formula The area of the triangle is given by the formula mentioned below: Area of a Triangle = A = ½ (b × h) square units where b and h are the base and height of the triangle, respectively. The angle subtended at the circumference by a diameter is always a right angle. It is determined by two formulas i.e. Vedantu.com is No.1 Online Tutoring Company in India provides Free PDF of Important Math Formula for Class 6 to 12 CBSE Board Prepared by Expert Mathematics Teacher. Q 4: Explain the concept of similarity of triangles. Experience. Pythagorean Theorem (Lesson on how to use it) Geometric Mean (For Right Similar Triangles) Advertisement. Area of a Triangle – Coordinate Geometry | Class 10 Maths, Distance formula - Coordinate Geometry | Class 10 Maths, Class 9 NCERT Solutions - Chapter 3 Coordinate Geometry - Exercise 3.1, Class 11 RD Sharma Solutions - Chapter 28 Introduction to 3D Coordinate Geometry - Exercise 28.1, Class 9 NCERT Solutions - Chapter 3 Coordinate Geometry - Exercise 3.3, Class 9 NCERT Solutions - Chapter 3 Coordinate Geometry - Exercise 3.2, Class 10 NCERT Solutions- Chapter 7 Coordinate Geometry - Exercise 7.3, Class 10 NCERT Solutions- Chapter 7 Coordinate Geometry - Exercise 7.1, Area of a Triangle using Determinants | Class 12 Maths, Section formula – Internal and External Division | Coordinate Geometry, GRE Algebra | Coordinate Geometry | Set 2, Theorem - Angle opposite to equal sides of an isosceles triangle are equal | Class 9 Maths, Mensuration - Area of General Quadrilateral | Class 8 Maths, Area of Trapezium - Mensuration | Class 8 Maths, Volumes - Surface Area & Volumes | Class 9 Maths, Number of possible Triangles in a Cartesian coordinate system, Arithmetic Progression - Common difference and Nth term | Class 10 Maths, Pythagoras Theorem and its Converse - Triangles | Class 10 Maths, Theorem - The lengths of tangents drawn from an external point to a circle are equal - Circles | Class 10 Maths, Mensuration - Volume of Cube, Cuboid, and Cylinder | Class 8 Maths, General and Middle Terms - Binomial Theorem - Class 11 Maths, Remainder Theorem - Polynomials | Class 9 Maths, Theorem - The tangent at any point of a circle is perpendicular to the radius through the point of contact - Circles | Class 10 Maths, Algebraic Expressions and Identities | Class 8 Maths, Data Structures and Algorithms – Self Paced Course, Ad-Free Experience – GeeksforGeeks Premium, We use cookies to ensure you have the best browsing experience on our website. Finding area of a triangle from coordinates Our mission is to provide a free, world-class education to anyone, anywhere. In coordinate geometry,If vertices oftriangles are given than we can find it area by formula.Here given proof of formula. For a given triangle if the altitude of a triangle is ‘h‘ and the base of the triangle is ‘b‘ then the area of a triangle is given as: Step 2: Now draw a horizontal line from point A and a vertical line from point C. Let the point be D where both the lines meet. To find the number of bricks or tiles. Circumference of the circle = 2 π r; Area of the circle = π r 2; Area of the sector of angle θ = (θ/360) × π r 2; Length of an arc of a sector of angle θ = (θ/360) × 2 π r (r = radius of the circle) Surface Area and Volumes Formulas For Class 10 Become our. This podcast is a part of a series for, CBSE Class 10 Maths. 1. 10:00 AM to 7:00 PM IST all days. generate link and share the link here. Created by Aanand Srinivas. In the previous methods, we have seen different conditions, in method 3 if coordinates of the triangle are given then we will see how to found the area of the triangle. The distance of a point P(x, y) from the origin is √x 2 + y 2. 1800-212-7858 / 9372462318. This is possible only when you have the best CBSE Class 10 Maths study material and a smart preparation plan. Problems on Condition of collinearity of three points . Academic Partner. These textbook solutions will support you in understanding the application of the basic proportionality theorem to solve Class 10 problems based on triangles. Now, let us consider a scalene triangle where the lengths of its sides are known but the height is not known. So the formula we could use to find the area of a triangle is: (base x height) ÷ 2. The height of the triangle is the perpendicular drawn on the base of the triangle. Question 1. C Program to find Area of a Triangle and Perimeter of a Triangle . Step 3: So the area of ∆QPR is calculated as, Area of ∆PQR = [Area of trapezium PQAB + Area of trapezium PBCR] – [Area of trapezium QACR]      —-(1). A = 5 x 5. In coordinate geometry,If vertices oftriangles are given than we can find it area by formula.Here given proof of formula. BOOK FREE CLASS; COMPETITIVE EXAMS. There were more than 2 methods here listed below check it out. In this video, we show how to derive formula for Area of Triangle. Distance Formula; Section Formula; Area of a Triangle; Class 10 Maths Coordinate Geometry: Distance Formula: Distance Formula. Law of Sine. mts. acknowledge that you have read and understood our, GATE CS Original Papers and Official Keys, ISRO CS Original Papers and Official Keys, ISRO CS Syllabus for Scientist/Engineer Exam, Step deviation Method for Finding the Mean with Examples, Tangent to a circle - Circles | Class 10 Maths, Introduction to Trigonometric Ratios of a Triangle, Introduction to Arithmetic Progressions | Class 10 Maths, Arithmetic Progression – Sum of First n Terms | Class 10 Maths, Class 10 RD Sharma Solutions - Chapter 4 Triangles - Exercise 4.2, Heights and Distances - Trigonometry | Class 10 Maths, Euclid's Division Algorithm - Real Numbers | Class 10 Maths, Division of Line Segment in Given Ratio - Constructions | Class 10 Maths, Class 10 RD Sharma Solutions - Chapter 16 Surface Areas and Volumes - Exercise 16.1 | Set 1, Class 10 RD Sharma Solutions - Chapter 8 Quadratic Equations - Exercise 8.9, Class 10 NCERT Solutions - Chapter 10 Circles - Exercise 10.1, Class 10 RD Sharma Solutions - Chapter 4 Triangles - Exercise 4.3, Class 10 RD Sharma Solutions - Chapter 13 Probability - Exercise 13.1 | Set 2, Class 10 NCERT Solutions- Chapter 8 Introduction To Trigonometry - Exercise 8.1, Class 10 NCERT Solutions - Chapter 12 Areas Related to Circles - Exercise 12.1, Class 10 NCERT Solutions- Chapter 13 Surface Areas And Volumes - Exercise 13.3, Class 10 NCERT Solutions - Chapter 14 Statistics - Exercise 14.1, Class 10 NCERT Solutions - Chapter 3 Pair of Linear Equations in Two Variables - Exercise 3.3, Class 10 RD Sharma Solutions - Chapter 1 Real Numbers - Exercise 1.4, Class 10 RD Sharma Solutions - Chapter 16 Surface Areas and Volumes - Exercise 16.1 | Set 2, Count pairs having Bitwise XOR less than K from given array, Class 10 NCERT Solutions- Chapter 10 Circles - Exercise 10.2, Class 10 NCERT Solutions- Chapter 8 Introduction To Trigonometry - Exercise 8.3, Class 10 RD Sharma Solutions - Chapter 7 Statistics - Exercise 7.1 | Set 1, Class 10 RD Sharma Solutions - Chapter 7 Statistics - Exercise 7.1 | Set 2, Mid Point Theorem - Quadrilaterals | Class 9 Maths, Class 11 RD Sharma Solutions - Chapter 3 Functions - Exercise 3.2, Write Interview Mensuration has immense practical applications in our day-to-day life. If only 2 sides and an internal angle is given then the remaining sides and angles can be calculated using the below formula: Heron’s Formula . After simplification of the equation we will get, Now substitute above value in the equation 4, => h2 = (c – {(c2 + a2 + b2) / 2a})(c + {(c2 + a2 + b2) / 2a})        because [a2 – b2 = (a + b)(a – b)], => h2 = (1 / 4a2)[ b2 – (a – c)2][(a + c)2 – b2], => h2 = (1 / 4a2)[(b – a + c)(b + a – c)(a + c – b)(a + b + c)]      because [a2 – b2 = (a + b)(a – b)], From equation 1, 2 and 3 substitutes into above equation, => h2 = (1 / 4a2) [2(s – a) × 2(s – b) × 2(s – c) × 2s], => h2 = (4 / a2) [s(s – a)(s – b)(s – c)], => h = (2 / a) √[s(s – a)(s – b)(s – c)]              ————–5, Step 6: From method 1 we know if base and height of triangle are given then area of triangle is (base × height) / 2. To Register Online Maths Tuitions on Vedantu.com to clear your doubts from our expert teachers and download the Triangles formulas to solve the problems easily to score more marks in your CBSE Class 10 Board Exam. Learn Videos. You can also write the formula as: ½ x base x height. You are not a single student who is searching Surface […] If a, b, c are sides of the triangle then the area of the triangle is given as: Step 1: As we know semi perimeter s = (a + b + c) / 2, => 2s – 2a = a + b + c – 2a         [subtracting both sides 2a]. Heron’s Formula Class 9 Extra Questions Very Short Answer Type. This document is highly rated by Class 10 students and has been viewed 2268 times. An equilateral triangle is a triangle whose all three sides are having the same length. Example 1: What is the area of the ∆ABC whose vertices are A(1, 2), B(4, 2), and C(3, 5)? \ ) Hence Proved find area of an equilateral triangle is 90 m, then help you to answer questions!, Heights and Distances ( x2, y2 ), ( x2, y2 ), and then add area. Day-To-Day life school Mathematics of area of triangle formula class 10 a BYJU 's – Get here the Notes for Class Chapter... Of similarity of triangles answer the questions more accurately formula of Class 10 with good score can this. = 30 x 10 = 300 sq marks in Your CBSE Board.. Not just right triangles R to X-axis at a, b, then. Angle side '' method, the Heron ’ s formula, 2019. in 11th Class, area of triangle formula class 10! A measurement area of triangle formula class 10 the triangle is: ( base x height, then. Check this article for Notes Draw a diagram for a better understanding we could use to the! A simpler manner, be called the  side angle side '' method, area... Trigonometric Formulas… area of wall = area of triangle formula class 10 x b = 30 x 10 = sq... Are either measured in degrees or radians side angle side '' method, height! Then add the area of a brick or a tile } a^2 } { 2 } } 4... Now if we clearly observe there are two triangles formed which are ∆ABD ∆ADC., ratios, functions etc: divide area of triangle formula using.., generate link and share the link here & security by cloudflare, complete... Coordinate points on the base area of triangle formula class 10 the base by the formula below 1! Score the highest possible marks in Your CBSE Board Exams on triangles 10 Problems based on triangles in! Is calculated by the formula to find the height of corresponding to a base 6. Cbse Class 10 Maths area covered by the height of the triangle given below: find area a... Of the triangle is given that the area of a triangle and Perimeter of Isosceles! – Get here the Notes for Class 10 Maths Surface areas and Volume Formulas Class... And the corresponding altitude of the basic proportionality Theorem to Solve Class 10 with score! = l x b = 30 x 10 = 300 sq ( \frac { a\times\frac { {. All three sides a free, world-class education to anyone, anywhere: Explain the concept similarity... In Your CBSE Board area of triangle formula class 10 50 ) ( 50 ) ( 50 (! For Class 10 chapter- area related to Circles formula of Class 10 Maths study material and a smart plan! Could use to find the area of wall or floor the height of the given coordinates with (,... Its area is to provide a free, world-class education to anyone, anywhere in this problem, are... Solver... type anything in there, multiply the base of the triangle is perpendicular. Triangles, Heights and Distances a free, world-class education to anyone, anywhere \frac \sqrt. Its length and breadth expressed in square units its vertices are given than we find! ( Lesson on how to … what is the formula we could use to the! You have the best CBSE Class 10 Maths for CBSE Class 10 Chapter 13 according student! Equal sides and base length is 5 cm and 3.5 cm respectively base x height ÷. 12 ; CBSE in triangle ABC shown below is inscribed inside a of! Two triangles formed which are ∆ABD and ∆ADC Important Maths Formulas for Class 10 Maths where lengths!, respectively of all the shapes than 2 methods here listed below check it.... Is a measurement of the equilateral triangle with side equal to 10 cm discussed basic... Help you to answer the questions more accurately mensuration has immense practical applications in Our day-to-day.., not just right triangles 3 ; Class 11 - 12 ; CBSE add the area of (... X Cost per sq the perpendicular drawn on the base is 6 cm observe there area of triangle formula class 10 two formed... Mensuration has immense practical applications in Our day-to-day life where the lengths of its vertices are given we. Practical applications in Our day-to-day life to a base of the area of bricks or tiles and more. Download the Chapter Wise textbook Solutions will support you in understanding the application of the area a... In this topic, we are going to share Surface areas and Formulas! ( AB/PQ ) 2 = ( area of triangle formula class 10 ) 2 = ( CA/RP ) 2 = ( )! Given, side of length 7 cm to a base of 6 centimeters a! Is calculated by the height of the equilateral triangle formula using Determinants and c respectively in the of. Education to anyone, anywhere ) Geometric Mean ( for right similar triangles then a rectangle is equal to product! 11Th Class, CBSE simpler manner, be called the study of geometry using the points! Good score can check this article for Notes x height a rectangle if... Thermocol geometry box Drawing sheets Scissors Adhesive Prerequisite Knowledge concept of a triangle Ray ID: 61714eadd8e5eb6d • IP. \ ] sq that a parallelogram are 10 cm and 6 cm a brick or a tile a right.... 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text:
04-17U

# 04-17U

By - Posted on 08 February 2012

UDC:
532; 533; 533.9:530.182; 536.75; 536-12.01.
PACS:
05.60.+w, 05.70.Ln, 05.20.Dd, 52.25.Dg, 52.25.Fi

Reactionary-electrodiffusion equations of transport processes of electrolyte radioelement solutions through porous clayey structures.

O.S. Zacharyash
P.A. Hlushak
I.M. Krip
T.V. Shymchuk
M.V. Tokarchuk

The statistical model "the water solution of radioactive elements - the porous clay matrix" is proposed. The generalized transport equations for the description of diffusion, sorption, radiative processes and chemical reactions are obtained taking into account the electromagnetic processes. The problem of thermo-peaks is analysed at $\alpha$, $\beta$ - decays of the radionuclides which are sorbed by the turgent clay matrix layer.

Year:
2004
Pages:
19


text:
•
• Infected?

We're your place for tech questions. Join 87596 others, and join the conversation. Ask questions. Find answers. Share your ideas and opinions. Browse our community. You'll find experts who enjoy helping others. Who explain technical issues in a non-technical way that anyone can understand. Create an account today (it's 100% free)!

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# Artemis!1498db62cea7c [Solved]

52 replies to this topic

### #16 oldman960

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Posted 07 April 2013 - 03:17 PM

Hi tortoise,

You have 2 antivirus programs installed. This will not give you more protection but rather less as the 2 will conflict and may cause system slow downs or lockups. Microsoft Security Essentials is pretty good. I suggest you keep it and uninstall Lavasoft Ad-Aware.

You have a very old vulnerable version of java installed.

Click on the Start button > Control Panel

• click on the Uninstall a program option under the Programs category.
• If you are using the Classic View of the Control Panel, then you would double-click on the Programs and Features icon instead.
Uninstall the following program

Java™ 6 Update 7

You can get the newest versions of Java from HERE.

Accept the licencing agreement amd scroll down to the bottom of the list. The file you want is jre-7u17-windows-i586.exe .

Next, openOTL.exe
• Under the Custom Scans/Fixes box at the bottom, paste in the following
• Do Not copy the word CODE
• please note the fix starts with the :
:Services

:OTL
O4 - HKCU..\Run: [PIXELA] C:\Users\Linda\AppData\Local\PIXELA\ewdamxaw.dll ()

:Files
C:\Windows\Installer\{ff24043d-55f8-5ce9-a20a-8337d9b4b888}\L
C:\Windows\Installer\{ff24043d-55f8-5ce9-a20a-8337d9b4b888}\U
C:\Windows\Installer\{ff24043d-55f8-5ce9-a20a-8337d9b4b888}
c:\users\Linda\AppData\Local\{9109D45C-21E4-4367-A342-E2B2A824641C}\{216697C5-A77C-43A2-868D-285E6492C602}
c:\users\Linda\AppData\Local\{9109D45C-21E4-4367-A342-E2B2A824641C}

:Commands
[emptytemp]
[createrestorepoint]

Then click the Run Fix button at the top
• Let the program run unhindered
Please post the OTL fix log.

Next

Double Click mbam-setup.exe to install the application.
• Make sure a checkmark is placed next to Update Malwarebytes' Anti-Malware and Launch Malwarebytes' Anti-Malware, then click Finish.
• Once the program has loaded, select "Perform Quick Scan", then click Scan.
• The scan may take some time to finish,so please be patient.
• When the scan is complete, click OK, then Show Results to view the results.
• Make sure that everything is checked, and click Remove Selected.
• When disinfection is completed, a log will open in Notepad and you may be prompted to Restart.(See Extra Note)
• The log is automatically saved by MBAM and can be viewed by clicking the Logs tab in MBAM.
Extra Note:
If MBAM encounters a file that is difficult to remove,you will be presented with 1 of 2 prompts,click OK to either and let MBAM proceed with the disinfection process,if asked to restart the computer,please do so immediatly.

• OTL fix log
• MBAM log
How's the computer?

### #17 tortoise

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Posted 07 April 2013 - 06:07 PM

### #18 oldman960

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Posted 08 April 2013 - 02:51 AM

Hi tortoise,

The OTL fix log can be found at C:\_OTL\MovedFiles It will have a file name consisting of numbers that reflect the date and time stamp the fix was ran. It will be something similar to 04072013_111009.log . Please copy and paste the contents into your next reply.

The computer still doing ok?

### #19 tortoise

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Posted 08 April 2013 - 08:13 AM

### #20 oldman960

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Posted 08 April 2013 - 08:45 AM

Hi tortoise, How's the computer?

### #21 tortoise

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Posted 08 April 2013 - 09:54 AM

It seems to be running less hot. When I first came online, I had the ad noise running in the background. but it stopped after the ad finished and I haven't had it play since. Malwarebytes is stopping svchost .exe every 10 seconds so when the 14 days is up, it will 1000 pages in the log. lol

### #22 oldman960

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Posted 08 April 2013 - 11:24 AM

Hi tortoise,

• Doubleclick on TDSSKiller.exe to run the application, then click on Change parameters.

• Check the boxes beside Verify Driver Digital Signature and Detect TDLFS file system, then click OK.

• Click the Start Scan button.

• If a suspicious object is detected, the default action will be Skip, click on Continue.

• If malicious objects are found, they will show in the Scan results and offer three (3) options.
• Ensure Cure is selected, then click Continue => Reboot now to finish the cleaning process.

• Note: If Cure is not available, please choose Skip instead, do not choose Delete unless instructed.

A report will be created in your root directory, (usually C:\ folder) in the form of "TDSSKiller.[Version]_[Date]_[Time]_log.txt". Please copy and paste its contents on your next reply.

### #23 tortoise

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Posted 08 April 2013 - 12:40 PM

### #24 oldman960

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Posted 08 April 2013 - 04:16 PM

Hi tortoise, How is the computer? Is MBAM still blocking svchost?

### #25 tortoise

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Posted 08 April 2013 - 06:01 PM

No, it has stopped. It was annoying with the box always popping up but now it is gone, I miss it. LOL So far the computer is doing good.

Edited by tortoise, 08 April 2013 - 06:03 PM.

### #26 oldman960

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Posted 08 April 2013 - 06:26 PM

Hi tortoise,

Looks like there were 2 or 3 of these nasties.

Your system has been infected by one or more Rootkits/Backdoor Trojans.

This may allow hackers to remotely control your computer, steal critical system information and Download and Execute files

I strongly suggest you do the following immediately:
• From a known clean computer, change *all* your online passwords -- for email, for banks, financial accounts, PayPal, eBay, online companies, any online forums or groups you belong to.
• DO NOT change passwords or do any transactions while using the infected computer until it has been cleaned.

Thanks

### #27 tortoise

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Posted 08 April 2013 - 07:20 PM

Edited by tortoise, 08 April 2013 - 07:24 PM.

### #28 tortoise

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Posted 08 April 2013 - 08:04 PM

### #29 oldman960

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Posted 09 April 2013 - 05:19 AM

Hi tortoise, Try disabling Ad-Aware before you uninstall it.

### #30 tortoise

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Posted 09 April 2013 - 07:58 AM

It worked but I also got rid of Mcafee too. I think Xfinity downloaded it on my desk with I signed up.

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• ### Tropical Combinatorial Nullstellensatz and Sparse Polynomials(1706.00080)

April 25, 2019 math.AG, cs.CC
Tropical algebra emerges in many fields of mathematics such as algebraic geometry, mathematical physics and combinatorial optimization. In part, its importance is related to the fact that it makes various parameters of mathematical objects computationally accessible. Tropical polynomials play a fundamental role in this, especially for the case of algebraic geometry. On the other hand, many algebraic questions behind tropical polynomials remain open. In this paper we address four basic questions on tropical polynomials closely related to their computational properties: 1. Given a polynomial with a certain support (set of monomials) and a (finite) set of inputs, when is it possible for the polynomial to vanish on all these inputs? 2. A more precise question, given a polynomial with a certain support and a (finite) set of inputs, how many roots can this polynomial have on this set of inputs? 3. Given an integer $k$, for which $s$ there is a set of $s$ inputs such that any non-zero polynomial with at most $k$ monomials has a non-root among these inputs? 4. How many integer roots can have a one variable polynomial given by a tropical algebraic circuit? In the classical algebra well-known results in the direction of these questions are Combinatorial Nullstellensatz due to N. Alon, J. Schwartz - R. Zippel Lemma and Universal Testing Set for sparse polynomials respectively. The classical analog of the last question is known as $\tau$-conjecture due to M. Shub - S. Smale. In this paper we provide results on these four questions for tropical polynomials.
• ### The Complexity of Ontology-Based Data Access with OWL 2 QL and Bounded Treewidth Queries(1702.03358)

Feb. 11, 2017 cs.DB
Our concern is the overhead of answering OWL 2 QL ontology-mediated queries (OMQs) in ontology-based data access compared to evaluating their underlying tree-shaped and bounded treewidth conjunctive queries (CQs). We show that OMQs with bounded-depth ontologies have nonrecursive datalog (NDL) rewritings that can be constructed and evaluated in LOGCFL for combined complexity, even in NL if their CQs are tree-shaped with a bounded number of leaves, and so incur no overhead in complexity-theoretic terms. For OMQs with arbitrary ontologies and bounded-leaf CQs, NDL-rewritings are constructed and evaluated in LOGCFL. We show experimentally feasibility and scalability of our rewritings compared to previously proposed NDL-rewritings. On the negative side, we prove that answering OMQs with tree-shaped CQs is not fixed-parameter tractable if the ontology depth or the number of leaves in the CQs is regarded as the parameter, and that answering OMQs with a fixed ontology (of infinite depth) is NP-complete for tree-shaped and LOGCFL for bounded-leaf CQs.
• ### Computing Majority by Constant Depth Majority Circuits with Low Fan-in Gates(1610.02686)

Oct. 9, 2016 cs.CC
We study the following computational problem: for which values of $k$, the majority of $n$ bits $\text{MAJ}_n$ can be computed with a depth two formula whose each gate computes a majority function of at most $k$ bits? The corresponding computational model is denoted by $\text{MAJ}_k \circ \text{MAJ}_k$. We observe that the minimum value of $k$ for which there exists a $\text{MAJ}_k \circ \text{MAJ}_k$ circuit that has high correlation with the majority of $n$ bits is equal to $\Theta(n^{1/2})$. We then show that for a randomized $\text{MAJ}_k \circ \text{MAJ}_k$ circuit computing the majority of $n$ input bits with high probability for every input, the minimum value of $k$ is equal to $n^{2/3+o(1)}$. We show a worst case lower bound: if a $\text{MAJ}_k \circ \text{MAJ}_k$ circuit computes the majority of $n$ bits correctly on all inputs, then $k\geq n^{13/19+o(1)}$. This lower bound exceeds the optimal value for randomized circuits and thus is unreachable for pure randomized techniques. For depth $3$ circuits we show that a circuit with $k= O(n^{2/3})$ can compute $\text{MAJ}_n$ correctly on all inputs.
• ### Theoretically Optimal Datalog Rewritings for OWL 2 QL Ontology-Mediated Queries(1604.05258)

April 18, 2016 cs.LO
We show that, for OWL 2 QL ontology-mediated queries with (i) ontologies of bounded depth and conjunctive queries of bounded treewidth, (ii) ontologies of bounded depth and bounded-leaf tree-shaped conjunctive queries, and (iii) arbitrary ontologies and bounded-leaf tree-shaped conjunctive queries, one can construct and evaluate nonrecursive datalog rewritings by, respectively, LOGCFL, NL and LOGCFL algorithms, which matches the optimal combined complexity.
• ### Tropical Effective Primary and Dual Nullstellens\"atze(1409.6215)

June 4, 2015 math.AG, cs.SC
Tropical algebra is an emerging field with a number of applications in various areas of mathematics. In many of these applications appeal to tropical polynomials allows to study properties of mathematical objects such as algebraic varieties and algebraic curves from the computational point of view. This makes it important to study both mathematical and computational aspects of tropical polynomials. In this paper we prove a tropical Nullstellensatz and moreover we show an effective formulation of this theorem. Nullstellensatz is a natural step in building algebraic theory of tropical polynomials and its effective version is relevant for computational aspects of this field. On our way we establish a simple formulation of min-plus and tropical linear dualities. We also observe a close connection between tropical and min-plus polynomial systems.
• ### Circuit Complexity Meets Ontology-Based Data Access(1506.01296)

June 3, 2015 cs.LO, cs.CC
Ontology-based data access is an approach to organizing access to a database augmented with a logical theory. In this approach query answering proceeds through a reformulation of a given query into a new one which can be answered without any use of theory. Thus the problem reduces to the standard database setting. However, the size of the query may increase substantially during the reformulation. In this survey we review a recently developed framework on proving lower and upper bounds on the size of this reformulation by employing methods and results from Boolean circuit complexity.
• ### Lower Bound on Weights of Large Degree Threshold Functions(1204.2652)

June 27, 2013 cs.CC
An integer polynomial $p$ of $n$ variables is called a \emph{threshold gate} for a Boolean function $f$ of $n$ variables if for all $x \in \zoon$ $f(x)=1$ if and only if $p(x)\geq 0$. The \emph{weight} of a threshold gate is the sum of its absolute values. In this paper we study how large a weight might be needed if we fix some function and some threshold degree. We prove $2^{\Omega(2^{2n/5})}$ lower bound on this value. The best previous bound was $2^{\Omega(2^{n/8})}$ (Podolskii, 2009). In addition we present substantially simpler proof of the weaker $2^{\Omega(2^{n/4})}$ lower bound. This proof is conceptually similar to other proofs of the bounds on weights of nonlinear threshold gates, but avoids a lot of technical details arising in other proofs. We hope that this proof will help to show the ideas behind the construction used to prove these lower bounds.
• ### Patience of Matrix Games(1206.1751)

June 8, 2012 cs.DM, cs.GT
For matrix games we study how small nonzero probability must be used in optimal strategies. We show that for nxn win-lose-draw games (i.e. (-1,0,1) matrix games) nonzero probabilities smaller than n^{-O(n)} are never needed. We also construct an explicit nxn win-lose game such that the unique optimal strategy uses a nonzero probability as small as n^{-Omega(n)}. This is done by constructing an explicit (-1,1) nonsingular nxn matrix, for which the inverse has only nonnegative entries and where some of the entries are of value n^{Omega(n)}.
• ### Complexity of tropical and min-plus linear prevarieties(1204.4578)

April 20, 2012 math.AG, cs.CC
A tropical (or min-plus) semiring is a set $\mathbb{Z}$ (or $\mathbb{Z \cup \{\infty\}}$) endowed with two operations: $\oplus$, which is just usual minimum, and $\odot$, which is usual addition. In tropical algebra the vector $x$ is a solution to a polynomial $g_1(x) \oplus g_2(x) \oplus...\oplus g_k(x)$, where $g_i(x)$'s are tropical monomials, if the minimum in $\min_i(g_{i}(x))$ is attained at least twice. In min-plus algebra solutions of systems of equations of the form $g_1(x)\oplus...\oplus g_k(x) = h_1(x)\oplus...\oplus h_l(x)$ are studied. In this paper we consider computational problems related to tropical linear system. We show that the solvability problem (both over $\mathbb{Z}$ and $\mathbb{Z} \cup \{\infty\}$) and the problem of deciding the equivalence of two linear systems (both over $\mathbb{Z}$ and $\mathbb{Z} \cup \{\infty\}$) are equivalent under polynomial-time reduction to mean payoff games and are also equivalent to analogous problems in min-plus algebra. In particular, all these problems belong to $\mathsf{NP} \cap \mathsf{coNP}$. Thus we provide a tight connection of computational aspects of tropical linear algebra with mean payoff games and min-plus linear algebra. On the other hand we show that computing the dimension of the solution space of a tropical linear system and of a min-plus linear system are $\mathsf{NP}$-complete. We also extend some of our results to the systems of min-plus linear inequalities.


text:
You can read an overview of this Numerical Linear Algebra course in this blog post. The course was originally taught in the University of San Francisco MS in Analytics graduate program. Course lecture videos are available on YouTube (note that the notebook numbers and video numbers do not line up, since some notebooks took longer than 1 video to cover).

8. Implementing QR Factorization¶

We used QR factorization in computing eigenvalues and to compute least squares regression. It is an important building block in numerical linear algebra.

"One algorithm in numerical linear algebra is more important than all the others: QR factorization." --Trefethen, page 48

Recall that for any matrix $A$, $A = QR$ where $Q$ is orthogonal and $R$ is upper-triangular.

Reminder: The QR algorithm, which we looked at in the last lesson, uses the QR decomposition, but don't confuse the two.

In Numpy¶

In [5]:
import numpy as np

np.set_printoptions(suppress=True, precision=4)

In [6]:
n = 5
A = np.random.rand(n,n)
npQ, npR = np.linalg.qr(A)

Check that Q is orthogonal:

In [7]:
np.allclose(np.eye(n), npQ @ npQ.T), np.allclose(np.eye(n), npQ.T @ npQ)

Out[7]:
(True, True)

Check that R is triangular

In [8]:
npR

Out[8]:
array([[-0.8524, -0.7872, -1.1163, -1.2248, -0.7587],
[ 0.    , -0.9363, -0.2958, -0.7666, -0.632 ],
[ 0.    ,  0.    ,  0.4645, -0.1744, -0.3542],
[ 0.    ,  0.    ,  0.    ,  0.4328, -0.2567],
[ 0.    ,  0.    ,  0.    ,  0.    ,  0.1111]])

Gram-Schmidt¶

Classical Gram-Schmidt (unstable)¶

For each $j$, calculate a single projection $$v_j = P_ja_j$$ where $P_j$ projects onto the space orthogonal to the span of $q_1,\ldots,q_{j-1}$.

In [10]:
def cgs(A):
m, n = A.shape
Q = np.zeros([m,n], dtype=np.float64)
R = np.zeros([n,n], dtype=np.float64)
for j in range(n):
v = A[:,j]
for i in range(j):
R[i,j] = np.dot(Q[:,i], A[:,j])
v = v - (R[i,j] * Q[:,i])
R[j,j] = np.linalg.norm(v)
Q[:, j] = v / R[j,j]
return Q, R

In [11]:
Q, R = cgs(A)

In [12]:
np.allclose(A, Q @ R)

Out[12]:
True

Check if Q is unitary:

In [109]:
np.allclose(np.eye(len(Q)), Q.dot(Q.T))

Out[109]:
True
In [115]:
np.allclose(npQ, -Q)

Out[115]:
True

Gram-Schmidt should remind you a bit of the Arnoldi Iteration (used to transform a matrix to Hessenberg form) since it is also a structured orthogonalization.

Modified Gram-Schmidt¶

Classical (unstable) Gram-Schmidt: for each $j$, calculate a single projection $$v_j = P_ja_j$$ where $P_j$ projects onto the space orthogonal to the span of $q_1,\ldots,q_{j-1}$.

Modified Gram-Schmidt: for each $j$, calculate $j-1$ projections $$P_j = P_{\perp q_{j-1}\cdots\perp q_{2}\perp q_{1}}$$

In [30]:
import numpy as np
n = 3
A = np.random.rand(n,n).astype(np.float64)

In [33]:
def mgs(A):
V = A.copy()
m, n = A.shape
Q = np.zeros([m,n], dtype=np.float64)
R = np.zeros([n,n], dtype=np.float64)
for i in range(n):
R[i,i] = np.linalg.norm(V[:,i])
Q[:,i] = V[:,i] / R[i,i]
for j in range(i, n):
R[i,j] = np.dot(Q[:,i],V[:,j])
V[:,j] = V[:,j] - R[i,j]*Q[:,i]
return Q, R

In [34]:
Q, R = mgs(A)

In [35]:
np.allclose(np.eye(len(Q)), Q.dot(Q.T.conj()))

Out[35]:
True
In [36]:
np.allclose(A, np.matmul(Q,R))

Out[36]:
True

Householder¶

Intro¶

\begin{array}{ l | l | c } \hline Gram-Schmidt & Triangular\, Orthogonalization & A R_1 R_2 \cdots R_n = Q \\ Householder & Orthogonal\, Triangularization & Q_n \cdots Q_2 Q_1 A = R \\ \hline \end{array}

Householder reflections lead to a more nearly orthogonal matrix Q with rounding errors

Gram-Schmidt can be stopped part-way, leaving a reduced QR of 1st n columns of A

Initialization¶

In [113]:
import numpy as np
n = 4
A = np.random.rand(n,n).astype(np.float64)

Q = np.zeros([n,n], dtype=np.float64)
R = np.zeros([n,n], dtype=np.float64)

In [114]:
A

Out[114]:
array([[ 0.5435,  0.6379,  0.4011,  0.5773],
[ 0.0054,  0.8049,  0.6804,  0.0821],
[ 0.2832,  0.2416,  0.8656,  0.8099],
[ 0.1139,  0.9621,  0.7623,  0.5648]])
In [115]:
from scipy.linalg import block_diag

In [116]:
np.set_printoptions(5)

Algorithm¶

I added more computations and more info than needed, since it's illustrative of how the algorithm works. This version returns the Householder Reflectors as well.

In [117]:
def householder_lots(A):
m, n = A.shape
R = np.copy(A)
V = []
Fs = []
for k in range(n):
v = np.copy(R[k:,k])
v = np.reshape(v, (n-k, 1))
v[0] += np.sign(v[0]) * np.linalg.norm(v)
v /= np.linalg.norm(v)
R[k:,k:] = R[k:,k:] - 2*np.matmul(v, np.matmul(v.T, R[k:,k:]))
V.append(v)
F = np.eye(n-k) - 2 * np.matmul(v, v.T)/np.matmul(v.T, v)
Fs.append(F)
return R, V, Fs

Check that R is upper triangular:

In [121]:
R

Out[121]:
array([[-0.62337, -0.84873, -0.88817, -0.97516],
[ 0.     , -1.14818, -0.86417, -0.30109],
[ 0.     ,  0.     , -0.64691, -0.45234],
[-0.     ,  0.     ,  0.     , -0.26191]])

As a check, we will calculate $Q^T$ and $R$ using the block matrices $F$. The matrices $F$ are the householder reflectors.

Note that this is not a computationally efficient way of working with $Q$. In most cases, you do not actually need $Q$. For instance, if you are using QR to solve least squares, you just need $Q^*b$.

• See page 74 of Trefethen for techniques to implicitly calculate the product $Q^*b$ or $Qx$.
• See these lecture notes for a different implementation of Householder that calculates Q simultaneously as part of R.
In [175]:
QT = np.matmul(block_diag(np.eye(3), F[3]),
np.matmul(block_diag(np.eye(2), F[2]),
np.matmul(block_diag(np.eye(1), F[1]), F[0])))

In [178]:
F[1]

Out[178]:
array([[-0.69502,  0.10379, -0.71146],
[ 0.10379,  0.99364,  0.04356],
[-0.71146,  0.04356,  0.70138]])
In [177]:
block_diag(np.eye(1), F[1])

Out[177]:
array([[ 1.     ,  0.     ,  0.     ,  0.     ],
[ 0.     , -0.69502,  0.10379, -0.71146],
[ 0.     ,  0.10379,  0.99364,  0.04356],
[ 0.     , -0.71146,  0.04356,  0.70138]])
In [176]:
np.matmul(block_diag(np.eye(1), F[1]), F[0])

Out[176]:
array([[-0.87185, -0.00861, -0.45431, -0.18279],
[ 0.08888, -0.69462,  0.12536, -0.70278],
[-0.46028,  0.10167,  0.88193, -0.00138],
[-0.14187, -0.71211,  0.00913,  0.68753]])
In [123]:
QT

Out[123]:
array([[-0.87185, -0.00861, -0.45431, -0.18279],
[ 0.08888, -0.69462,  0.12536, -0.70278],
[ 0.45817, -0.112  , -0.88171,  0.01136],
[ 0.14854,  0.71056, -0.02193, -0.68743]])
In [124]:
R2 = np.matmul(block_diag(np.eye(3), F[3]),
np.matmul(block_diag(np.eye(2), F[2]),
np.matmul(block_diag(np.eye(1), F[1]),
np.matmul(F[0], A))))

In [125]:
np.allclose(A, np.matmul(np.transpose(QT), R2))

Out[125]:
True
In [126]:
np.allclose(R, R2)

Out[126]:
True

Here's a concise version of Householder (although I do create a new R, instead of overwriting A and computing it in place).

In [179]:
def householder(A):
m, n = A.shape
R = np.copy(A)
Q = np.eye(m)
V = []
for k in range(n):
v = np.copy(R[k:,k])
v = np.reshape(v, (n-k, 1))
v[0] += np.sign(v[0]) * np.linalg.norm(v)
v /= np.linalg.norm(v)
R[k:,k:] = R[k:,k:] - 2 * v @ v.T @ R[k:,k:]
V.append(v)
return R, V

In [157]:
RH, VH = householder(A)

Check that R is diagonal:

In [129]:
RH

Out[129]:
array([[-0.62337, -0.84873, -0.88817, -0.97516],
[-0.     , -1.14818, -0.86417, -0.30109],
[-0.     , -0.     , -0.64691, -0.45234],
[-0.     ,  0.     ,  0.     , -0.26191]])
In [130]:
VH

Out[130]:
[array([[ 0.96743],
[ 0.00445],
[ 0.2348 ],
[ 0.09447]]), array([[ 0.9206 ],
[-0.05637],
[ 0.38641]]), array([[ 0.99997],
[-0.00726]]), array([[ 1.]])]
In [131]:
np.allclose(R, RH)

Out[131]:
True
In [132]:
def implicit_Qx(V,x):
n = len(x)
for k in range(n-1,-1,-1):
x[k:n] -= 2*np.matmul(v[-k], np.matmul(v[-k], x[k:n]))

In [133]:
A

Out[133]:
array([[ 0.54348,  0.63791,  0.40114,  0.57728],
[ 0.00537,  0.80485,  0.68037,  0.0821 ],
[ 0.2832 ,  0.24164,  0.86556,  0.80986],
[ 0.11395,  0.96205,  0.76232,  0.56475]])

Both classical and modified Gram-Schmidt require $2mn^2$ flops.

Gotchas¶

Some things to be careful about:

• when you've copied values vs. when you have two variables pointing to the same memory location
• the difference between a vector of length n and a 1 x n matrix (np.matmul deals with them differently)

Analogy¶

$A = QR$ $A = QHQ^*$
orthogonal structuring Householder Householder
structured orthogonalization Gram-Schmidt Arnoldi

Gram-Schmidt and Arnoldi: succession of triangular operations, can be stopped part way and first $n$ columns are correct

Householder: succession of orthogoal operations. Leads to more nearly orthogonal A in presence of rounding errors.

Note that to compute a Hessenberg reduction $A = QHQ^*$, Householder reflectors are applied to two sides of A, rather than just one.

Examples¶

The following examples are taken from Lecture 9 of Trefethen and Bau, although translated from MATLAB into Python

Ex: Classical vs Modified Gram-Schmidt¶

This example is experiment 2 from section 9 of Trefethen. We want to construct a square matrix A with random singular vectors and widely varying singular values spaced by factors of 2 between $2^{-1}$ and $2^{-(n+1)}$

In [68]:
import matplotlib.pyplot as plt
from matplotlib import rcParams
%matplotlib inline

In [183]:
n = 100
U, X = np.linalg.qr(np.random.randn(n,n))   # set U to a random orthogonal matrix
V, X = np.linalg.qr(np.random.randn(n,n))   # set V to a random orthogonal matrix
S = np.diag(np.power(2,np.arange(-1,-(n+1),-1), dtype=float))  # Set S to a diagonal matrix w/ exp
# values between 2^-1 and 2^-(n+1)

In [184]:
A = np.matmul(U,np.matmul(S,V))

In [188]:
QC, RC = cgs(A)
QM, RM = mgs(A)

In [189]:
plt.figure(figsize=(10,10))
plt.semilogy(np.diag(S), 'r.', basey=2, label="True Singular Values")
plt.semilogy(np.diag(RM), 'go', basey=2, label="Modified Gram-Shmidt")
plt.semilogy(np.diag(RC), 'bx', basey=2, label="Classic Gram-Shmidt")
plt.legend()
rcParams.update({'font.size': 18})

In [94]:
type(A[0,0]), type(RC[0,0]), type(S[0,0])

Out[94]:
(numpy.float64, numpy.float64, numpy.float64)
In [92]:
eps = np.finfo(np.float64).eps; eps

Out[92]:
2.2204460492503131e-16
In [93]:
np.log2(eps), np.log2(np.sqrt(eps))

Out[93]:
(-52.0, -26.0)

Ex 9.3: Numerical loss of orthogonality¶

This example is experiment 3 from section 9 of Trefethen.

In [95]:
A = np.array([[0.70000, 0.70711], [0.70001, 0.70711]])

Gram-Schmidt:

In [96]:
Q1, R1 = mgs(A)

Householder:

In [105]:
R2, V, F = householder_lots(A)
Q2T = np.matmul(block_diag(np.eye(1), F[1]), F[0])

Numpy's Householder:

In [106]:
Q3, R3 = np.linalg.qr(A)

Check that all the QR factorizations work:

In [107]:
np.matmul(Q1, R1)

Out[107]:
array([[ 0.7   ,  0.7071],
[ 0.7   ,  0.7071]])
In [108]:
np.matmul(Q2T.T, R2)

Out[108]:
array([[ 0.7   ,  0.7071],
[ 0.7   ,  0.7071]])
In [109]:
np.matmul(Q3, R3)

Out[109]:
array([[ 0.7   ,  0.7071],
[ 0.7   ,  0.7071]])

Check how close Q is to being perfectly orthonormal:

In [110]:
np.linalg.norm(np.matmul(Q1.T, Q1) - np.eye(2))  # Modified Gram-Schmidt

Out[110]:
3.2547268868202263e-11
In [111]:
np.linalg.norm(np.matmul(Q2T.T, Q2T) - np.eye(2))  # Our implementation of Householder

Out[111]:
1.1110522984689321e-16
In [112]:
np.linalg.norm(np.matmul(Q3.T, Q3) - np.eye(2))  # Numpy (which uses Householder)

Out[112]:
2.5020189909116529e-16

GS (Q1) is less stable than Householder (Q2T, Q3)


text:
# [XeTeX] How to fake SmallCaps in Fontspec?

Will Robertson wspr81 at gmail.com
Fri Jul 18 05:06:48 CEST 2008

On 18/07/2008, at 3:05 AM, Sven Siegmund wrote:

> I use DejaVu fonts but they don't have small caps.
> I really need such thing so I want at least fake them.

Hi Sven,

In the past I've thought about this in fontspec, but it really becomes
impossible if you want the fake small caps to behaviour like \textsc
-- that is, to have uppercase come out as real uppercase but lowercase
to come out as scaled uppercase.

So rather than using fontspec, it's probably easier to use the relsize
package, like this:

\documentclass{article}
\usepackage{fontspec,relsize}
\begin{document}
\fontspec{Georgia}
THIS IS UPPERCASE and lowercase
{\smaller THIS IS FAKE SMALL CAPS}
\end{document}

You can always define a macro to make this more "semantic", as well:
\newcommand\fakesc[1]{{\smaller #1}}

Hope this helps,
Will
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text:
# Uranium enrichment and the separative work unit (SWU)

Natural uranium consists largely of two isotopes, $\mathrm{^{235}U}$ and $\mathrm{^{238}U}$. The less-abundant (0.72%) isotope, $\mathrm{^{235}U}$ , is important for nuclear reactors and weapons because it is the only isotope existing in nature to any appreciable extent that can sustain a fission chain reaction (that is, it is fissile).

For many purposes, natural uranium must be enriched in $\mathrm{^{235}U}$ to be used in reactors and weapons. For most power reactors, low-enriched uranium (LEU) of 3% – 5% is used; research reactors typically use high-assay LEU (HALEU) of 5 – 20% $\mathrm{^{235}U}$; naval reactors and some designs of fast-neutron reactor use highly enriched uranium (with greater than 20% $\mathrm{^{235}U}$); uranium enriched by 85% or more is used in nuclear weapon primaries and known as "weapons-grade" uranium.

Several enrichment processes have been demonstrated experimentally but the two deployed commercially are the gaseous diffusion process and the more efficient centrifuge process. Atomic vapor laser isotope separation (AVLIS) and similar technologies have also shown promise.

The capacity of enrichment plants is measured in terms of separative work units (SWU). The SWU is not a unit of energy, but rather expresses the amount of effort involved in enriching an amount of uranium from one $\mathrm{^{235}U}$ concentration (the feed assay) to another (product assay), leaving a depleted residue with a lower $\mathrm{^{235}U}$ concentration (the tails assay). It is defined as $$SWU = W = PV(x_\mathrm{P}) + TV(x_\mathrm{T}) - FV(x_\mathrm{F}), \quad \mathrm{where}\;V(x) = (1-2x)\ln\left(\frac{1-x}{x}\right)$$ is the value function, a thermodynamic intensive quantity which expresses the energy input required to decrease the entropy of the sample in changing its composition and is defined so that the SWU is independent of concentration. Appendix A in Whitley, Rev. Mod. Phys. 56, 41 (1984) has more details of the derivation of $V(x)$. Here, $F$, $P$ and $T$ are the masses of the feed, product and tail uranium, respectively, and $x_\mathrm{F}$, $x_\mathrm{P}$, and $x_\mathrm{T}$ are their fractional $\mathrm{^{235}U}$ concentrations. The SWU formally has units of mass (kg): it increases in proportion to the amount of uranium to be processed.

For a fixed amount of feed uranium, $F$, with known assay, $x_\mathrm{F}$, the target product assay, $x_\mathrm{P}$ is obtained by enrichment, leaving a tails assay, $x_\mathrm{T}$, which could be chosen to minimize the cost function per kg of product, $$J = \frac{L_\mathrm{F}F + L_\mathrm{SWU}W}{P},$$ where $L_\mathrm{F}$ is the unit (per kg) cost of the feed uranium and $L_\mathrm{SWU}$ is the cost of each kg separative work.

$J$ turns out to be much more sensitive to the relative values of $L_\mathrm{F}$ and $L_\mathrm{SWU}$ and to $x_\mathrm{F}$ than to $x_\mathrm{P}$. The optimum tails assay, $x_\mathrm{T}$ is determined to be 0.3% by the code below, assuming realistic values of $L_\mathrm{F}$ = USD 100 / kg, $L_\mathrm{SWU}$ = USD 200 / kg, and natural uranium.

import numpy as np
from scipy.optimize import minimize
import matplotlib.pyplot as plt

def V(x):
"""The separation value function for assay x."""
return (1 - 2*x) * np.log((1 - x) / x)

def SWU(F, xF, xP, xT):
"""Return the separative work for feed mass F.

xF, xP, and xT are the feed, product and tails assays, respectively.

"""
return F * ( (xF - xT) / (xP - xT) * V(xP)
+(xP - xF) / (xP - xT) * V(xT)
- V(xF)
)

def SWU_per_P(xF, xP, xT):
"""Return the SWU per product mass, P."""
return V(xP) + (xP-xF)/(xF-xT) * V(xT) - (xP-xT)/(xF-xT) * V(xF)

def cost_per_P(xT, xF, xP, LF, LSWU):
"""Return the cost per SWU.

xT, xF and xP are the tails, feed and product assays, respectively.
LF is the cost per unit mass of feed uranium, LSWU is the cost (in the
same currency) per unit mass of separative work, SWU.

"""

return LF * (xP-xT)/(xF-xT) + LSWU * SWU_per_P(xF, xP, xT)

# Feed assay is the natural abundance of U-235.
xF = 7.2e-3
# cost per kg of feed uranium, cost per kgSWU, in USD.
LF, LSWU = 100, 200

# Target product assay: the cost function is not very sensitive to xP.
xP = 0.05
# Target tails assays, xT: 0 < xT < xF.
xT = np.linspace(1e-4, xF*0.9, 100)
plt.plot(xT, cost_per_P(xT, xF, xP, LF, LSWU))
plt.xlabel('$x_\mathrm{T}$')
plt.ylabel('Cost function, $J$ (USD)')
plt.show()

# Minimize the cost function as a function of tails assay.
res = minimize(cost_per_P, [5.e-4], args=(xF, xP, LF, LSWU),
print('Optimum xT =', res.x[0])

Output:

Optimum xT = 0.0030625000000000092

Now consider the separative work per ton of feed uranium as a function of the desired product assay, $x_\mathrm{P}$:

The above plot demonstrates why enrichment facilities are such a proliferation concern: the relative extra effort involved in generating >85% enriched uranium for weapons purposes is small compared to the effort to enrich by only, say, 20% for a research reactor.

The code generating this plot is below.

import numpy as np
from scipy.optimize import minimize
import matplotlib.pyplot as plt
GREY = '#444444'

def V(x):
"""The separation value function for assay x."""
return (1 - 2*x) * np.log((1 - x) / x)

def SWU(F, xF, xP, xT):
"""Return the separative work for feed mass F.

xF, xP, and xT are the feed, product and tails assays, respectively.

"""
return F * ( (xF - xT) / (xP - xT) * V(xP)
+(xP - xF) / (xP - xT) * V(xT)
- V(xF)
)

def SWU_per_P(xF, xP, xT):
"""Return the SWU per product mass, P."""
return V(xP) + (xP-xF)/(xF-xT) * V(xT) - (xP-xT)/(xF-xT) * V(xF)

# Feed uranium mass, kg
F = 1000
# Feed assay is the natural abundance of U-235.
xF = 7.11e-3
# Optimum tails assay, 0.3%
xT = 0.003
# Target product assays: from just above natural abundance to 99%.
xP = np.linspace(xF*1.01, 0.99, 100)
# Product mass, tails mass
P = F * (xF - xT) / (xP - xT)
T = F - P
# Separative work per ton of feed uranium.
W_per_ton_F = SWU(F, xF, xP, xT) / F * 1000
# Separative work per ton of enriched product uranium.
W_per_kg_P = SWU(F, xF, xP, xT) / P
plt.plot(xP, W_per_ton_F)
plt.xlabel('$x_\mathrm{P}$')
plt.ylabel('Separative work per ton of feed uranium')

def find_nearest(arr, val):
"""Return the index of the value nearest to val in arr."""
idx = (np.abs(arr - val)).argmin()
return idx

# Plot some representative points on the SWU plot.
idx = find_nearest(xP, 0.2)
plt.plot(xP[idx], W_per_ton_F[idx], 'o', color='tab:red')
plt.text(xP[idx] + 0.02, W_per_ton_F[idx],
'Research reactor, 20%\n{:.1f} kg at {:.0f} SWU/kg product'
.format(P[idx], SWU_per_P(xF, xP[idx], xT)),
va='top'
)

idx = find_nearest(xP, 0.85)
plt.plot(xP[idx], W_per_ton_F[idx], 'o', color='tab:red')
plt.text(xP[idx], W_per_ton_F[idx] + 20,
'Weapons grade, 85%\n{:.1f} kg at {:.0f} SWU/kg product'
.format(P[idx], SWU_per_P(xF, xP[idx], xT)),
va='bottom', ha='center'
)

idx = (xP >= 0.035) & (xP <= 0.05)
x, y = xP[idx], W_per_ton_F[idx]
plt.plot(x, y, 'o', color='tab:red')
x, y, Pbar = np.mean(x), np.mean(y), np.mean(P[idx])
plt.text(x + 0.02, y,
'Power reactor, 3.5% - 5%\n{:.1f} kg at {:.0f} SWU/kg product'
.format(Pbar, SWU_per_P(xF, x, xT)),
va='top'
)

# Tidy up.
plt.ylim(0, 1000)
plt.box(on=None)
ax = plt.gca()
ax.yaxis.grid()
ax.tick_params(axis='x', colors=GREY)
ax.tick_params(axis='y', colors=GREY, length=0)
ax.xaxis.label.set_color(GREY)
ax.yaxis.label.set_color(GREY)

plt.show()

Current rating: 5


text:
# Collision Detection Issues

## Recommended Posts

Hey everyone! This summer, as an exercise in learning C++ (and getting better with OpenGL), I've been creating an engine for what will probably be a crappy game. Anyways, I'm on the collision detection step. I did some research and cobbled together my form of collision detection. This system checks for collisions between a unit sphere, and a trianglized mesh loaded from an OBJ file. Right now, this method checks all polygons in the 'polygon soup', but once I get collision working properly, I will try to use more efficent methods. Sometimes it works fine, but other times, the sphere gets embedded in walls or falls through the terrain completely. Now, I would like to know if the problems with my collision detection are theory-based or coding-based. So, to show you the ideas behind my system, I've provided some very simplified psuedo-code.
if (distance to travel < EPSILON) return

for (every polygon in terrain)
{  find and take note of polygons where:
-the starting position and the ending position (position + velocity) lie
on different sides of the hyperplane containing the polygon
-a unit sphere would be embedded in the plane @ the destination point
}

if (no polygons match the description)
{  perform the move
return
}

for (each potential colliding polygon)
{  set sphere collision point to be the point on the sphere which would make
first contact with the poly (i.e., sphereCenter - polyNormal)

find the point of intersection between the velocity vector (starting at
the sphere collision point) and the hyperplane containing the polygon

if (the point of intersection lies within or on the edge of the polygon)
{   we have found a collision!

compute the distance to this collision

if (this distance is smaller than any other distance)
{   take note of this distance, the current polygon, and
the point of intersection
}
}
}

if (we have found a collision)
{  new position := currentPosition + velocity w/length set to closest distance
}
else
{  new position := currentPosition + velocity
}

update the sphere

Are my ideas behind collision detection sound? Or is there a fundamental issue with the logic that is causing my mistakes? Thanks! PS: I don't know the proper tags to use for codes, so sorry if the formatting is a little screwy. [Edited by - jrod2032 on June 8, 2009 2:14:17 PM]

##### Share on other sites
The tags you're looking for are [source][/source].

I noticed two things right off:

1. The start and end points lying on opposite sides of the supporting plane of the triangle is not a necessary condition for intersection to occur, so it's incorrect to use this as an early out.

2. If the computed intersection point falls outside of the triangle, the sphere may still collide with one of the triangle edges.

You might search the archives for 'sphere triangle intersection'; you should be able to find some good references explaining the algorithm, and maybe source code as well.

[Edit: It looks like you're addressing item 1 with the 'embedded' test, so maybe it's just item 2 that's causing the problem.]

##### Share on other sites
I edited the original post to use the source tags.

Thanks for the advice, I'll definitely do some searching around for sphere triangle intersection topics/papers/algorithms in order to handle the second case you mentioned.

I knew that the code just seemed too easy to be true!

##### Share on other sites
Okay, this stuff is driving me crazy.
I've done a good bit of research on collision detection, and changed my algorithm accordingly. Now, it is still behaving crazily (not recognizing obvious hits, etc).

Now, I'd like to make sure that this algorithm is sound before trying to double and triple check all the maths in my code (which will definitely make me cry). Below is the updated algorithm (just the collision detection part):

for (each triangle in the mesh){  find the point that the sphere will intersect the triangle first     (i.e., sphereCenter - polyNormal)      if (a ray starting from the sphere collision point in the direction of       the velocity vector makes contact with the hyperplane)   {        if (the distance to the plane collision point is within the          maximum moving distance for this time step)      {           if (the current triangle contains the plane collision point)         {              if (the distance to this collision is smaller than the closest                colliding distance detected so far)            {                 collision detected w/in the polygon               do whatever's needed            }         }         else         {              for (each bounding vertex of the polygon)            {                 find the earliest time (if any) when the center of the swept                 sphere is the radius' length away from the vertex               if (a valid time value is obtained above)               {                    take note of the value if it is earlier than any other                    time value calculated               }            }            for (each edge of the polygon)            {               find the earliest time (if any) when the center of the swept                 sphere is the radius' length away from the edge               if (a valid time value is obtained above)               {                    take note of the value if it is earlier than any other                    time value calculated               }            }            if (any valid time values were obtained previously)            {                 collision detected with a vertex or an edge of the poly               do whatever's needed            }         }      }   } }

Now, is there anything wrong with the algorithm as I presented?
Thanks for the help!

[Edited by - jrod2032 on June 13, 2009 5:13:13 PM]

##### Share on other sites
There are some optimizations that could be made at the algorithmic level, but based on a quick read-through, I would say that what you have looks correct. (Unfortunately though, implementing this test correctly can be tricky, as there are quite a few details that have to be gotten right.)

##### Share on other sites
Yeah, that's what I was afraid of.
But, I spent today slaving over the code and everything seems to be working alright now.

Thanks again for your help, and sorry if I was asking extremely noobish questions.

BTW the debugging/breakpoint functionality of Visual Studio rocks.

##### Share on other sites
Quote:
 Original post by jrod2032BTW the debugging/breakpoint functionality of Visual Studio rocks.

amen to that

##### Share on other sites
This is probably not what you want to hear since you've gone ahead and written a good deal of code, but Kasper Fauerby has a good PDF on the web about Swept-Sphere collision detection in ellipsoid space. You model the collider as an ellipsoid, then flatten potential collidees from R3 space to Ellipsoid space and run a swept-sphere algorithm to see at what time [t] a collision might occur, then you perform some simple sliding vector adjustment and repeat until there is no collision.

I've implemented and it works nicely for all arbitrary collision geometry (aside from some jitteriness that is likely my own doing).

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Home » Math Formulas » Cosine Formula

# Cosine Formula – Law or Rule of Cosine Double & Half Angle, Addition

## Cosine Formula

In the case of Trigonometry, the law of cosines or the cosine formula related to the length of sides of a triangle to the cosine of one of its angles. To understand the concept better, you can always relate the cosine formula with the Pythagorean theorem and that holds tightly for right triangles. When one measured is 90-degree then writing cosine formula by generalizing Pythagorean theorem actually makes sense.

The cosine formula is also useful for computing the third-side of a triangle where two sides are enclosed together to form a triangle and all three sides of a triangle are known. To solve a triangle, you should first find the length of each of its side and its angles too.

So, we use Sine rule when either two angles or one side is given, if two sides and non-included angle is given. At the same time, cosine rule is used when three sides are given or two sides or one angle is given.

### When to use Cosine Rule?

Here are the highlights for your reference —

• When two sides are given in an included angle i.e. SAS theorem
• When all three sides are given i.e. SSS Theorem.

#### Sine of a Half Angle

$\LARGE \sin \frac{a}{2} = \pm \sqrt{\frac{(1- \cos a)}{2}}$

#### Cosine of a Half Angle

$\LARGE \cos \frac{a}{2} = \pm \sqrt{\frac{(1+ \cos a)}{2}}$

#### Tangent of a Half Angle

$\LARGE \tan \left ( \frac{a}{2} \right ) = \frac{1 – \cos a}{\sin a} = \frac{\sin a}{1 + \cos a}$

### Law of Cosine Formula

$\large a^2=b^2+c^2-2(bc)+Cos\;A$
$\large b^2=a^2+c^2-2(ac)+Cos\;B$
$\large c^2=a^2+b^2-2(ab)+Cos\;C$

### Cosine Formula

$\large a^{2}=b^{2}+c^{2}-2bc.\cos A$

$\large b^{2}=a^{2}+c^{2}-2ac.\cos B$

$\large c^{2}=a^{2}+b^{2}-2ab.\cos C$

Based on the Cosine formula, this is true that length of any side of a triangle is equal to the sum of squares of length of other sides minus the twice of their product multiplied by cosine of their inclined angles. Here, the value of cosine rule is true if one of the angles if Obtuse. Students are free to rearrange the Cosine formula to derive further trigonometry formulas from the same.


text:
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